ReliaWiki - User contributions [en]

File:6.15.png

2012-11-26T16:14:05Z

Dingzhou Cao: uploaded a new version of "File:6.15.png"

Preventive Maintenance

2012-09-14T18:45:15Z

Dingzhou Cao: /* Optimum Age Replacement Policy */

<noinclude>{{Banner BlockSim Articles}}{{Navigation box}}
''This article, which discusses preventive maintenance in BlockSim 8, also appears in the [[Introduction_to_Repairable_Systems#Preventive_Maintenance_2|System Analysis Reference]] book.''
</noinclude>
Preventive maintenance (PM) is a schedule of planned maintenance actions aimed at the prevention of breakdowns and failures. The primary goal of preventive maintenance is to prevent the failure of equipment before it actually occurs. It is designed to preserve and enhance equipment reliability by replacing worn components before they actually fail. Preventive maintenance activities include equipment checks, partial or complete overhauls at specified periods, oil changes, lubrication and so on. In addition, workers can record equipment deterioration so they know to replace or repair worn parts before they cause system failure. Recent technological advances in tools for inspection and diagnosis have enabled even more accurate and effective equipment maintenance. The ideal preventive maintenance program would prevent all equipment failure before it occurs.

===Value of Preventive Maintenance===

There are multiple misconceptions about preventive maintenance. One such misconception is that PM is unduly costly. This logic dictates that it would cost more for regularly scheduled downtime and maintenance than it would normally cost to operate equipment until repair is absolutely necessary. This may be true for some components; however, one should compare not only the costs but the long-term benefits and savings associated with preventive maintenance. Without preventive maintenance, for example, costs for lost production time from unscheduled equipment breakdown will be incurred. Also, preventive maintenance will result in savings due to an increase of effective system service life.

Long-term benefits of preventive maintenance include:

:• Improved system reliability. 
:• Decreased cost of replacement. 
:• Decreased system downtime. 
:• Better spares inventory management. 

Long-term effects and cost comparisons usually favor preventive maintenance over performing maintenance actions only when the system fails.

====When Does Preventive Maintenance Make Sense?====

Preventive maintenance is a logical choice if, and only if, the following two conditions are met:

*Condition #1: The component in question has an increasing failure rate. In other words, the failure rate of the component increases with time, implying wear-out. Preventive maintenance of a component that is assumed to have an exponential distribution (which implies a constant failure rate) does not make sense! 
*Condition #2: The overall cost of the preventive maintenance action must be less than the overall cost of a corrective action. 

If both of these conditions are met, then preventive maintenance makes sense. Additionally, based on the costs ratios, an optimum time for such action can be easily computed for a single component. This is detailed in later sections.

====The Fallacy of "Constant Failure Rate" and "Preventive Replacement"====

Even though we alluded to the fact in the last section, it is important to make it explicitly clear that if a component has a constant failure rate (i.e., defined by an exponential distribution), then preventive maintenance of the component will have no effect on the component's failure occurrences. To illustrate this, consider a component with an <math>MTTF</math> = <math>100</math> hours, or <math>\lambda =0.01</math>, and with preventive replacement every 50 hours. The reliability vs. time graph for this case is illustrated in the following figure, where the component is replaced every 50 hours, thereby resetting the component's reliability to one. At first glance, it may seem that the preventive maintenance action is actually maintaining the component at a higher reliability.

[[Image:7i2.png|center|400px|Reliability vs. time for a single component with an <math>MTTF =100</math> hours, or <math>\lambda =0.01</math> , and with preventive replacement every 50 hours.]]

However, consider the following cases for a single component:

'''Case 1''': The component's reliability from 0 to 60 hours:

*With preventive maintenance, the component was replaced with a new one at 50 hours so the overall reliability is based on the reliability of the new component for 10 hours, <math>R(t=10)=90.48%\,\!</math>, times the reliability of the previous component, <math>R(t=50)=60.65%\,\!</math>. The result is <math>R(t=60)=54.88%.\,\!</math> 
*Without preventive maintenance, the reliability would be the reliability of the same component operating to 60 hours, or <math>R(t=60)=54.88%\,\!</math>. 
 
'''Case 2''': The component's reliability from 50 to 60 hours:
*With preventive maintenance, the component was replaced at 50 hours, so this is solely based on the reliability of the new component for a mission of 10 hours, or <math>R(t=10)=90.48%\,\!</math>. 
*Without preventive maintenance, the reliability would be the conditional reliability of the same component operating to 60 hours, having already survived to 50 hours, or <math>{{R}_{C}}(T=60|50)=R(60)/R(50)=90.48%\,\!</math> .

As can be seen, both cases — with and without preventive maintenance — yield the same results.

===Determining Preventive Replacement Time===

As mentioned earlier, if the component has an increasing failure rate, then a carefully designed preventive maintenance program is beneficial to system availability. Otherwise, the costs of preventive maintenance might actually outweigh the benefits. The objective of a good preventive maintenance program is to either minimize the overall costs (or downtime, etc.) or meet a reliability objective. In order to achieve this, an appropriate interval (time) for scheduled maintenance must be determined. One way to do that is to use the optimum age replacement model, as presented next. The model adheres to the conditions discussed previously:
 
:• The component is exhibiting behavior associated with a wear-out mode. That is, the failure rate of the component is increasing with time. 
:• The cost for planned replacements is significantly less than the cost for unplanned replacements. 

The following figure shows the Cost Per Unit Time vs. Time plot and it can be seen that the corrective replacement costs increase as the replacement interval increases. In other words, the less often you perform a PM action, the higher your corrective costs will be. Obviously, as we let a component operate for longer times, its failure rate increases to a point that it is more likely to fail, thus requiring more corrective actions. The opposite is true for the preventive replacement costs. The longer you wait to perform a PM, the less the costs; if you do PM too often, the costs increase. If we combine both costs, we can see that there is an optimum point that minimizes the costs. In other words, one must strike a balance between the risk (costs) associated with a failure while maximizing the time between PM actions.

[[Image:costpertime.png|center|450px|Cost curves for preventive and corrective replacement.]]

===Optimum Age Replacement Policy===

To determine the optimum time for such a preventive maintenance action (replacement), we need to mathematically formulate a model that describes the associated costs and risks. In developing the model, it is assumed that if the unit fails before time <math>t\,\!</math> , a corrective action will occur and if it does not fail by time <math>t</math> , a preventive action will occur. In other words, the unit is replaced upon failure or after a time of operation, <math>t\,\!</math> , whichever occurs first.
Thus, the optimum replacement time can be found by minimizing the cost per unit time, <math>CPUT\left( t \right).\,\!</math> <math>CPUT\left( t \right)\,\!</math> is given by:

::<math>\begin{align}
CPUT\left( t \right)= & \frac{\text{Total Expected Replacement Cost per Cycle}}{\text{Expected Cycle Length}} \\
= & \frac{{{C}_{P}}\cdot R\left( t \right)+{{C}_{U}}\cdot \left[ 1-R\left( t \right) \right]}{\mathop{}_{0}^{t}R\left( s \right)ds}
\end{align}</math>

where:

:* <math>R(t)\,\!</math> = reliability at time <math>t\,\!</math> . 
:* <math>{{C}_{P}}\,\!</math> = cost of planned replacement. 
:* <math>{{C}_{U}}\,\!</math> = cost of unplanned replacement. 

The optimum replacement time interval, <math>t\,\!</math> , is the time that minimizes <math>CPUT\left( t \right).\,\!</math> This can be found by solving for <math>t\,\!</math> such that:

::<math>\frac{\partial \left[ CPUT(t) \right]}{\partial t}=0</math>

Or by solving for a <math>t\,\!</math> that satisfies the following equation:

::<math>\frac{\partial \left[ \tfrac{{{C}_{P}}\cdot R\left( t \right)+{{C}_{U}}\cdot \left[ 1-R\left( t \right) \right]}{\mathop{}_{0}^{t}R\left( s \right)ds} \right]}{\partial t}=0</math>

Interested readers can refer to Barlow and Hunter [[Appendix_B:_References |[2]]] for more details on this model.

In BlockSim 8, you can use the Optimum Replacement window to determine the optimum replacement time either for an individual block or for multiple blocks in a diagram simultaneously. When working with multiple blocks, the calculations can be for individual blocks or for one or more groups of blocks. For each item that is included in the optimization calculations, you will need to specify the cost for a planned replacement and the cost for an unplanned replacement. This is done by calculating the costs for replacement based on the item settings using equations or simulation and then, if desired, manually entering any additional costs for either type of replacement in the corresponding columns of the table.

The equations used to calculate the costs of planned and unplanned tasks for each item based on its associated URD are as follows:

*For the cost of planned tasks, here denoted as PM cost:

::<math>\begin{align}
\text{PM Cost}= \left(\text{PM Down Time Rate}+ \text{Block Level Down Time Rate} \right)\\ \cdot \left( \text{MTTPM}+\text{Pool Delay} +\text{Crew Delay} \right) \\
+ \text{Crew Labor Rate} \cdot \text{MTTPM} + \text{Cost per PM} + \text{Cost per Pool} +\text{Cost per Crew}
\end{align}</math>

:Only PM tasks based on item age or system age (fixed or dynamic intervals) are considered. If there is more than one PM task based on item age, only the first one is considered.
*For the cost of the unplanned task, here denoted as CM cost:
::<math>\begin{align}
\text{CM Cost}= \left(\text{CM Down Time Rate}+ \text{Block Level Down Time Rate} \right)\\ \cdot \left( \text{MTTR}+\text{Pool Delay} +\text{Crew Delay} \right) \\
+ \text{Crew Labor Rate} \cdot \text{MTTR} + \text{Cost per CM} + \text{Cost per Pool} +\text{Cost per Crew} \\
+\text{Block Level Cost per Failure}
\end{align}</math>

When using simulation, for costs associated with planned replacements, all preventive tasks based on item age or system age (fixed or dynamic intervals) are considered. Because each item is simulated as a system (i.e., in isolation from any other item), tasks triggered in other ways are not considered.

===Example: Optimum Replacement Time===
{{:Optimum_Replacement_Time_Example}}

Preventive Maintenance

2012-08-23T01:39:54Z

Dingzhou Cao: /* Optimum Age Replacement Policy */

<noinclude>{{Banner BlockSim Articles}}{{Navigation box}}
''This article, which discusses preventive maintenance in BlockSim 8, also appears in the [[Introduction_to_Repairable_Systems#Preventive_Maintenance_2|System Analysis Reference]] book.''
</noinclude>
Preventive maintenance (PM) is a schedule of planned maintenance actions aimed at the prevention of breakdowns and failures. The primary goal of preventive maintenance is to prevent the failure of equipment before it actually occurs. It is designed to preserve and enhance equipment reliability by replacing worn components before they actually fail. Preventive maintenance activities include equipment checks, partial or complete overhauls at specified periods, oil changes, lubrication and so on. In addition, workers can record equipment deterioration so they know to replace or repair worn parts before they cause system failure. Recent technological advances in tools for inspection and diagnosis have enabled even more accurate and effective equipment maintenance. The ideal preventive maintenance program would prevent all equipment failure before it occurs.

===Value of Preventive Maintenance===

There are multiple misconceptions about preventive maintenance. One such misconception is that PM is unduly costly. This logic dictates that it would cost more for regularly scheduled downtime and maintenance than it would normally cost to operate equipment until repair is absolutely necessary. This may be true for some components; however, one should compare not only the costs but the long-term benefits and savings associated with preventive maintenance. Without preventive maintenance, for example, costs for lost production time from unscheduled equipment breakdown will be incurred. Also, preventive maintenance will result in savings due to an increase of effective system service life.

Long-term benefits of preventive maintenance include:

:• Improved system reliability. 
:• Decreased cost of replacement. 
:• Decreased system downtime. 
:• Better spares inventory management. 

Long-term effects and cost comparisons usually favor preventive maintenance over performing maintenance actions only when the system fails.

====When Does Preventive Maintenance Make Sense?====

Preventive maintenance is a logical choice if, and only if, the following two conditions are met:

*Condition #1: The component in question has an increasing failure rate. In other words, the failure rate of the component increases with time, implying wear-out. Preventive maintenance of a component that is assumed to have an exponential distribution (which implies a constant failure rate) does not make sense! 
*Condition #2: The overall cost of the preventive maintenance action must be less than the overall cost of a corrective action. 

If both of these conditions are met, then preventive maintenance makes sense. Additionally, based on the costs ratios, an optimum time for such action can be easily computed for a single component. This is detailed in later sections.

====The Fallacy of "Constant Failure Rate" and "Preventive Replacement"====

Even though we alluded to the fact in the last section, it is important to make it explicitly clear that if a component has a constant failure rate (i.e., defined by an exponential distribution), then preventive maintenance of the component will have no effect on the component's failure occurrences. To illustrate this, consider a component with an <math>MTTF</math> = <math>100</math> hours, or <math>\lambda =0.01</math>, and with preventive replacement every 50 hours. The reliability vs. time graph for this case is illustrated in the following figure, where the component is replaced every 50 hours, thereby resetting the component's reliability to one. At first glance, it may seem that the preventive maintenance action is actually maintaining the component at a higher reliability.

[[Image:7i2.png|center|400px|Reliability vs. time for a single component with an <math>MTTF =100</math> hours, or <math>\lambda =0.01</math> , and with preventive replacement every 50 hours.]]

However, consider the following cases for a single component:

'''Case 1''': The component's reliability from 0 to 60 hours:

*With preventive maintenance, the component was replaced with a new one at 50 hours so the overall reliability is based on the reliability of the new component for 10 hours, <math>R(t=10)=90.48%\,\!</math>, times the reliability of the previous component, <math>R(t=50)=60.65%\,\!</math>. The result is <math>R(t=60)=54.88%.\,\!</math> 
*Without preventive maintenance, the reliability would be the reliability of the same component operating to 60 hours, or <math>R(t=60)=54.88%\,\!</math>. 
 
'''Case 2''': The component's reliability from 50 to 60 hours:
*With preventive maintenance, the component was replaced at 50 hours, so this is solely based on the reliability of the new component for a mission of 10 hours, or <math>R(t=10)=90.48%\,\!</math>. 
*Without preventive maintenance, the reliability would be the conditional reliability of the same component operating to 60 hours, having already survived to 50 hours, or <math>{{R}_{C}}(T=60|50)=R(60)/R(50)=90.48%\,\!</math> .

As can be seen, both cases — with and without preventive maintenance — yield the same results.

===Determining Preventive Replacement Time===

As mentioned earlier, if the component has an increasing failure rate, then a carefully designed preventive maintenance program is beneficial to system availability. Otherwise, the costs of preventive maintenance might actually outweigh the benefits. The objective of a good preventive maintenance program is to either minimize the overall costs (or downtime, etc.) or meet a reliability objective. In order to achieve this, an appropriate interval (time) for scheduled maintenance must be determined. One way to do that is to use the optimum age replacement model, as presented next. The model adheres to the conditions discussed previously:
 
:• The component is exhibiting behavior associated with a wear-out mode. That is, the failure rate of the component is increasing with time. 
:• The cost for planned replacements is significantly less than the cost for unplanned replacements. 

The following figure shows the Cost Per Unit Time vs. Time plot and it can be seen that the corrective replacement costs increase as the replacement interval increases. In other words, the less often you perform a PM action, the higher your corrective costs will be. Obviously, as we let a component operate for longer times, its failure rate increases to a point that it is more likely to fail, thus requiring more corrective actions. The opposite is true for the preventive replacement costs. The longer you wait to perform a PM, the less the costs; if you do PM too often, the costs increase. If we combine both costs, we can see that there is an optimum point that minimizes the costs. In other words, one must strike a balance between the risk (costs) associated with a failure while maximizing the time between PM actions.

[[Image:costpertime.png|center|450px|Cost curves for preventive and corrective replacement.]]

===Optimum Age Replacement Policy===

To determine the optimum time for such a preventive maintenance action (replacement), we need to mathematically formulate a model that describes the associated costs and risks. In developing the model, it is assumed that if the unit fails before time <math>t\,\!</math> , a corrective action will occur and if it does not fail by time <math>t</math> , a preventive action will occur. In other words, the unit is replaced upon failure or after a time of operation, <math>t\,\!</math> , whichever occurs first.
Thus, the optimum replacement time can be found by minimizing the cost per unit time, <math>CPUT\left( t \right).\,\!</math> <math>CPUT\left( t \right)\,\!</math> is given by:

::<math>\begin{align}
CPUT\left( t \right)= & \frac{\text{Total Expected Replacement Cost per Cycle}}{\text{Expected Cycle Length}} \\
= & \frac{{{C}_{P}}\cdot R\left( t \right)+{{C}_{U}}\cdot \left[ 1-R\left( t \right) \right]}{\mathop{}_{0}^{t}R\left( s \right)ds}
\end{align}</math>

where:

:* <math>R(t)\,\!</math> = reliability at time <math>t\,\!</math> . 
:* <math>{{C}_{P}}\,\!</math> = cost of planned replacement. 
:* <math>{{C}_{U}}\,\!</math> = cost of unplanned replacement. 

The optimum replacement time interval, <math>t\,\!</math> , is the time that minimizes <math>CPUT\left( t \right).\,\!</math> This can be found by solving for <math>t\,\!</math> such that:

::<math>\frac{\partial \left[ CPUT(t) \right]}{\partial t}=0</math>

Or by solving for a <math>t\,\!</math> that satisfies the following equation:

::<math>\frac{\partial \left[ \tfrac{{{C}_{P}}\cdot R\left( t \right)+{{C}_{U}}\cdot \left[ 1-R\left( t \right) \right]}{\mathop{}_{0}^{t}R\left( s \right)ds} \right]}{\partial t}=0</math>

Interested readers can refer to [[Appendix_B:_References |Barlow and Hunter [2]]] for more details on this model.

In BlockSim 8, you can use the Optimum Replacement window to determine the optimum replacement time either for an individual block or for multiple blocks in a diagram simultaneously. When working with multiple blocks, the calculations can be for individual blocks or for one or more groups of blocks. For each item that is included in the optimization calculations, you will need to specify the cost for a planned replacement and the cost for an unplanned replacement. This is done by calculating the costs for replacement based on the item settings using equations or simulation and then, if desired, manually entering any additional costs for either type of replacement in the corresponding columns of the table.

The equations used to calculate the costs of planned and unplanned tasks for each item based on its associated URD are as follows:

*For the cost of planned tasks, here denoted as PM cost:

::<math>\begin{align}
\text{PM Cost}= \left(\text{PM Down Time Rate}+ \text{Block Level Down Time Rate} \right)\\ \cdot \left( \text{MTTPM}+\text{Pool Delay} +\text{Crew Delay} \right) \\
+ \text{Crew Labor Rate} \cdot \text{MTTPM} + \text{Cost per PM} + \text{Cost per Pool} +\text{Cost per Crew}
\end{align}</math>

:Only PM tasks based on item age are considered. If there is more than one PM task based on item age, only the first one is considered.
*For the cost of the unplanned task, here denoted as CM cost:
::<math>\begin{align}
\text{CM Cost}= \left(\text{CM Down Time Rate}+ \text{Block Level Down Time Rate} \right)\\ \cdot \left( \text{MTTR}+\text{Pool Delay} +\text{Crew Delay} \right) \\
+ \text{Crew Labor Rate} \cdot \text{MTTR} + \text{Cost per CM} + \text{Cost per Pool} +\text{Cost per Crew} \\
+\text{Block Level Cost per Failure}
\end{align}</math>

When using simulation, for costs associated with planned replacements, all preventive tasks based on item age or system age are considered. Because each item is simulated as a system (i.e., in isolation from any other item), tasks triggered in other ways are not considered.

===Example: Optimum Replacement Time===
{{:Optimum_Replacement_Time_Example}}

Time-Dependent System Reliability (Analytical)

2012-06-21T15:57:12Z

Dingzhou Cao: /* Solution to Example 5 */

{{Template:bsbook|5}}

In the previous chapter, different system configuration types were examined, as well as different methods for obtaining the system's reliability function analytically. Because the reliabilities in the problems presented were treated as probabilities (e.g. <math>P(A)</math> , <math>{{R}_{i}}</math> ), the reliability values and equations presented were referred to as static (not time-dependent). Thus, in the prior chapter, the life distributions of the components were not incorporated in the process of calculating the system reliability. In this chapter, time dependency in the reliability function will be introduced. We will develop the models necessary to observe the reliability over the life of the system, instead of at just one point in time. In addition, performance measures such as failure rate, MTTF and warranty time will be estimated for the entire system. The methods of obtaining the reliability function analytically remain identical to the ones presented in the previous chapter, with the exception that the reliabilities will be functions of time. In other words, instead of dealing with <math>{{R}_{i}}</math> , we will use <math>{{R}_{i}}(t)</math> . All examples in this chapter assume that no repairs are performed on the components.

=Analytical Life Predictions=
The analytical approach presented in the prior chapter involved the determination of a mathematical expression that describes the reliability of the system, expressed in terms of the reliabilities of its components. So far we have estimated only static system reliability (at a fixed time). For example, in the case of a system with three components in series, the system's reliability equation was given by:
 

::<math>{{R}_{s}}={{R}_{1}}\cdot {{R}_{2}}\cdot {{R}_{3}} \ </math>
 

The values of <math>{{R}_{1}}</math> , <math>{{R}_{2}}</math> and <math>{{R}_{3}}</math> were given for a common time and the reliability of the system was estimated for that time. However, since the component failure characteristics can be described by distributions, the system reliability is actually time-dependent. In this case, the equation above can be rewritten as:
 

::<math>{{R}_{s}}(t)={{R}_{1}}(t)\cdot {{R}_{2}}(t)\cdot {{R}_{3}}(t)</math>

 

The reliability of the system for any mission time can now be estimated. Assuming a Weibull life distribution for each component, the first equation above can now be expressed in terms of each component's reliability function, or:

 

[[Category:Completed Theoretical Review]]

::<math>{{R}_{s}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}}\cdot {{e}^{-{{\left( \tfrac{t}{{{\eta }_{2}}} \right)}^{{{\beta }_{2}}}}}}\cdot {{e}^{-{{\left( \tfrac{t}{{{\eta }_{3}}} \right)}^{{{\beta }_{3}}}}}}</math>

 

In the same manner, any life distribution can be substituted into the system reliability equation. Suppose that the times-to-failure of the first component are described with a Weibull distribution, the times-to-failure of the second component with an exponential distribution and the times-to-failure of the third component with a normal distribution. Then the first equation above can be written as:
 

::<math>{{R}_{s}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}}\cdot {{e}^{-{{\lambda }_{2}}t}}\cdot \left[ 1-\Phi \left( \frac{t-{{\mu }_{3}}}{{{\sigma }_{3}}} \right) \right]</math>

 

It can be seen that the biggest challenge is in obtaining the system's reliability function in terms of component reliabilities, which has already been discussed in depth. Once this has been achieved, calculating the reliability of the system for any mission duration is just a matter of substituting the corresponding component reliability functions into the system reliability equation.
===Advantages of the Analytical Method===
The primary advantage of the analytical solution is that it produces a mathematical expression that describes the reliability of the system. Once the system's reliability function has been determined, other calculations can then be performed to obtain metrics of interest for the system. Such calculations include:
 
:• Determination of the system's <math>pdf.</math> 
:• Determination of warranty periods. 
:• Determination of the system's failure rate. 
:• Determination of the system's MTTF. 
 
In addition, optimization and reliability allocation techniques can be used to aid engineers in their design improvement efforts. Another advantage in using analytical techniques is the ability to perform static calculations and analyze systems with a mixture of static and time-dependent components. Finally, the reliability importance of components over time can be calculated with this methodology.
===Disadvantages of the Analytical Method===
The biggest disadvantage of the analytical method is that formulations can become very complicated. The more complicated a system is, the larger and more difficult it will be to analytically formulate an expression for the system's reliability. For particularly detailed systems this process can be quite time-consuming, even with the use of computers. Furthermore, when the maintainability of the system or some of its components must be taken into consideration, analytical solutions become intractable. In these situations, the use of simulation methods may be more advantageous than attempting to develop a solution analytically. Simulation methods are presented in later chapters.
===Looking at a Simple ''Complex'' System Analytically===

The complexity involved in an analytical solution can be best illustrated by looking at the simple ''complex'' system with 15 components, as shown in Figure " An RBD of a complex system".
 
[[Image:5-1.png|center|400px|<div align="center"> An RBD of a complex system.</div>]]

 
The system reliability for this system (computed using BlockSim) is shown next. The first solution is provided using BlockSim's symbolic solution. In symbolic mode, BlockSim breaks the equation into segments, identified by tokens, that need to be substituted into the final system equation for a complete solution. This creates algebraic solutions that are more compact than if the substitutions were made.
 

::<math>\begin{align}
{{R}_{System}}= & D2\cdot D3\cdot {{R}_{L}} \\
D3= & +{{R}_{K}}\cdot IK \\
IK= & +{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}} \\
D2 = & +{{R}_{A}}\cdot {{R}_{E}}\cdot IE \\
IE = & -D1\cdot {{R}_{M}}\cdot {{R}_{N}}+{{R}_{M}}\cdot {{R}_{N}}+D1 \\
D1 = & +{{R}_{D}}\cdot ID \\
ID = & -{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}}
\end{align}</math>
 
Substituting the terms yields:
 

::<math>\begin{align}
{{R}_{System}}= & {{R}_{A}}\cdot {{R}_{E}}\cdot {{R}_{L}}\cdot {{R}_{K}} \\
& \cdot \{({{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})\cdot {{R}_{M}}\cdot {{R}_{N}} \\
& +{{R}_{M}}\cdot {{R}_{N}}-{{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}}\} \\
& \cdot \{{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}}\}
\end{align}</math>

 
BlockSim's automatic algebraic simplification would yield the following format for the above solution:
 

::<math>\begin{align}
{{R}_{System}}= & (({{R}_{A}}\cdot {{R}_{E}}(-({{R}_{D}}(-{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})){{R}_{M}}\cdot {{R}_{N}} \\
& +{{R}_{M}}\cdot {{R}_{N}} \\
& +({{R}_{D}}(-{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})))) \\
& ({{R}_{K}}({{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& +RI\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}})){{R}_{L}}) \
\end{align}</math>

 
In this equation, each <math>{{R}_{i}}</math> represents the reliability function of a block. For example, if <math>{{R}_{A}}</math> has a Weibull distribution, then each <math>{{R}_{A}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{A}}} \right)}^{{{\beta }_{A}}}}}}</math> and so forth. Substitution of each component's reliability function in the last <math>{{R}_{System}}</math> equation above will result in an analytical expression for the system reliability as a function of time, or <math>{{R}_{s}}(t)</math> , which is the same as <math>(1-cd{{f}_{System}}).</math>

===Obtaining Other Functions of Interest===
Once the system reliability equation (or the cumulative density function, <math>cdf</math> ) has been determined, other functions and metrics of interest can be derived.
 
Consider the following simple system:
 
[[Image:5-2.png|center|150px|<div align="center"> Simple two-component system. </div>]]

 
Furthermore, assume that component 1 follows an exponential distribution with a mean of 10,000 (<math>\mu =10,000,</math> <math>\lambda =1/10,000)</math> and component 2 follows a Weibull distribution with <math>\beta =6</math> and <math>\eta =10,000</math> . The reliability equation of this system is:

 
::<math>\begin{align}
{{R}_{S}}(t)= & {{R}_{1}}(t)\cdot {{R}_{2}}(t) \\
= & {{e}^{-\lambda t}}\cdot {{e}^{-{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \\
= & {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \
\end{align}</math>

 
The system <math>cdf</math> is:

 
::<math>\begin{align}
{{F}_{S}}(t)= & 1-({{R}_{1}}(t)\cdot {{R}_{2}}(t)) \\
= & 1-\left( {{e}^{-\lambda t}}\cdot {{e}^{-{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \right) \\
= & 1-\left( {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)
\end{align}</math>

 
{{analytical system pdf}}

====Conditional Reliability====
Conditional reliability is the probability of a system successfully completing another mission following the successful completion of a previous mission. The time of the previous mission and the time for the mission to be undertaken must be taken into account for conditional reliability calculations. The system's conditional reliability function is given by:

::<math>R(T,t)=\frac{R(T+t)}{R(T)}</math>

Equation above gives the reliability for a new mission of duration <math>t</math> having already accumulated <math>T</math> hours of operation up to the start of this new mission. The system is evaluated to assure that it will start the next mission successfully.
 
For the system in Figure "Simple two-component system.", the reliability for mission of <math>t=1,000</math> hours, having an age of <math>T=500</math> hours, is:
 

::<math>\begin{align}
{{R}_{S}}(T=500,t=1000)= & \frac{R(T+t)}{R(T)} \\
= & \frac{R(1500)}{R(500)} \\
= & \frac{{{e}^{-\tfrac{1500}{10,000}}}\cdot {{e}^{-{{\left( \tfrac{1500}{10,000} \right)}^{6}}}}}{{{e}^{-\tfrac{500}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{500}{10,000} \right)}^{6}}}}} \\
= & 0.9048=90.48%
\end{align}</math>

 

====Conditional Reliability for Components====

Now in this formulation, it was assumed that the accumulated age was equivalent for both units. That is, both started life at zero and aged to 500. It is possible to consider an individual component that has already accumulated some age (used component) in the same formulation. To illustrate this, assume that component 2 started life with an age of T=100. Then the reliability equation of the system, as given in <math>{{R}_{S}}(t)= {{R}_{1}}(t)\cdot {{R}_{2}}(t)</math>, would need to be modified to include a conditional term for 2, or:
 

::<math>{{R}_{S}}(t)={{R}_{1}}(t)\cdot \frac{{{R}_{2}}({{T}_{2}}+t)}{{{R}_{2}}({{T}_{2}})} \ </math>

 

In BlockSim, the start age input box may be used to specify a starting age greater than zero.
{{system failure rate analytical}}

====System Mean Life (Mean Time To Failure)====
The mean life (or mean time to failure, MTTF) can be obtained by integrating the system reliability function from zero to infinity:
 

::<math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math>

 

The mean time is a performance index and does not provide any information about the behavior of the failure distribution of the system.
 
For the system in Figure "Simple two-component system":
 

::<math>\begin{align}
MTTF= & \int_{0}^{\infty }\left( {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)dt \\
= & 5978.9
\end{align}</math>

 

====Warranty Period and BX Life====
Sometimes it is desirable to know the time value associated with a certain reliability. Warranty periods are often calculated by determining what percentage of the failure population can be covered financially and estimating the time at which this portion of the population will fail. Similarly, engineering specifications may call for a certain BX life, which also represents a time period during which a certain proportion of the population will fail. For example, the B10 life is the time in which 10% of the population will fail.
This is obtained by setting <math>{{R}_{S}}(t)</math> to the desired value and solving for <math>t.</math>
 
For the system in Figure "Simple two-component system":
 

::<math>{{R}_{s}}\left( t \right)={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>
 

To compute the time by which reliability would be equal to 90%, equation above is recast as follows and solved for <math>t.</math>

 
::<math>0.90={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>

 

In this case, <math>t=1053.59</math> . Equivalently, the B10 life for this system is also <math>1053.59</math> .
Except for some trivial cases, a closed form solution for <math>t</math> cannot be obtained. Thus, it is necessary to solve for <math>t</math> using numerical methods. BlockSim uses numerical methods.

===Example 1===
Consider a system consisting of three exponential units in series with the following failure rates (in failures per hour): <math>{{\lambda }_{1}}</math> = 0.0002, <math>{{\lambda }_{2}}</math> = 0.0005 and <math>{{\lambda }_{3}}</math> = 0.0001.
 
:• Obtain the reliability equation for the system. 
:• What is the reliability of the system after 150 hours of operation? 
:• Obtain the system's <math>pdf.</math> 
:• Obtain the system's failure rate equation. 
:• What is the MTTF for the system? 
:• What should the warranty period be for a 90% reliability? 
====Solution to Example 1====
 
:• The analytical expression for the reliability of the system is given by:

::<math>\begin{align}
{{R}_{s}}(t)= & {{R}_{1}}(t)\cdot {{R}_{2}}(t)\cdot {{R}_{3}}(t) \\
= & {{e}^{-{{\lambda }_{1}}t}}\cdot {{e}^{-{{\lambda }_{2}}t}}\cdot {{e}^{-{{\lambda }_{1}}t}} \\
= & {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}
\end{align}</math>

 

At 150 hours of operation, the reliability of the system is:
 

::<math>\begin{align}
{{R}_{s}}(t)= & {{e}^{-(0.0002+0.0005+0.0001)150}} \\
= & 0.8869\text{ or }88.69%
\end{align}</math>

 

:• In order to obtain the system's <math>pdf</math> , the derivative of the reliability equation given in the first equation above is taken with respect to time, or:
 
::<math>\begin{align}
{{f}_{s}}(t)= & -\frac{d[{{R}_{s}}(t)]}{dt} \\
= & -\frac{d\left[ {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}} \right]}{dt} \\
= & ({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})\cdot {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}
\end{align}</math>

:• The system's failure rate can now be obtained simply by dividing the system's <math>pdf</math> given in the equation above by the system's reliability function given in the first equation above, and:

::<math>\begin{align}
{{\lambda }_{s}}\left( t \right)= & \frac{{{f}_{s}}\left( t \right)}{{{R}_{s}}\left( t \right)} \\
= & \frac{({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})\cdot {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}}{{{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}} \\
= & ({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}) \\
= & 0.0008\text{ }fr/hr
\end{align}</math>

 
:• Combining <math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math> and the first equation above, the system's MTTF can be obtained:

 
::<math>\begin{align}
MTTF= & \int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \\
= & \int_{0}^{\infty }{{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}dt \\
= & \frac{1}{({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})} \\
= & 1250\text{ }hr
\end{align}</math>
 

:• Solving the first equation above with respect to time will yield the corresponding warranty period for a 90% reliability. In this case, the system reliability equation is simple and a closed form solution exists. The warranty time can now be found by solving:

 
::<math>\begin{align}
t= & -\frac{\ln (R)}{{{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}} \\
= & -\frac{\ln (0.9)}{0.0008} \\
= & 131.7\text{ }hr
\end{align}</math>
 

Thus, the warranty period should be 132 hours.

===Example 2===
Consider the system shown in Figure "Complex bridge system in Example 2".
 
[[Image:BS5.5.png|center|400px|<div align="center"> Complex bridge system in Example 2. </div>]]
 

Components <math>A</math> through <math>E</math> are Weibull distributed with <math>\beta =1.2</math> and <math>\eta =1230</math> hours. The starting and ending blocks cannot fail. 
Determine the following:
 
:• The reliability equation for the system and its corresponding plot. 
:• The system's <math>pdf</math> and its corresponding plot. 
:• The system's failure rate equation and the corresponding plot. 
:• The <math>MTTF</math> . 
:• The warranty time for a 90% reliability. 
:• The reliability for a 200-hour mission, if it is known that the system has already successfully operated for 200 hours. 
====Solution====
 
The first step is to obtain the reliability function for the system. The methods described in the previous chapter can be employed, such as the event space or path-tracing methods. Using BlockSim, the following reliability equation is obtained:
 

::<math>\begin{align}
{{R}_{s}}(t)= & ({{R}_{Start}}\cdot {{R}_{End}}(2{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}))
\end{align}</math>
 

Note that since the starting and ending blocks cannot fail, <math>{{R}_{Start}}=1</math> and <math>{{R}_{End}}=1,</math> equation above can be reduced to:

 

::<math>\begin{align}
{{R}_{s}}(t)= & 2\cdot {{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}
\end{align}</math>

 

Where <math>{{R}_{A}}</math> is the reliability equation for Component A, or:

 

::<math>{{R}_{A}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{A}}} \right)}^{{{\beta }_{A}}}}}}</math>

 

::<math>{{R}_{B}}</math> is the reliability equation for Component <math>B</math> , etc.
 

Since the components in this example are identical, the system reliability equation can be further reduced to:

 

::<math>{{R}_{s}}(t)=2R{{(t)}^{2}}+2R{{(t)}^{3}}-5R{{(t)}^{4}}+2R{{(t)}^{5}}</math>

 

Or, in terms of the failure distribution:

 

::<math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>

 

The corresponding plot is given in Figure "Reliability plot for the system".
 
[[Image:BS5.6.png|center|300px|<div align="center"> Reliability plot for the system. </div>]]
 
 

In order to obtain the system's <math>pdf</math> , the derivative of the reliability equation given above is taken with respect to time, resulting in:
 

::<math>\begin{align}
{{f}_{s}}(t)= & 4\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+6\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \\
& -20\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+10\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}
\end{align}</math>

 

The <math>pdf</math> can now be plotted for different time values, <math>t</math> , as shown in Figure "''pdf'' plot for the system".
 

The system's failure rate can now be obtained by dividing the system's <math>pdf</math> given in equation above by the system's reliability function given in <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>, or:

 

::<math>\begin{align}
{{\lambda }_{s}}(t)= & \frac{4\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+6\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}{2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}} \\
& +\frac{-20\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+10\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}{2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}
\end{align}</math>

 

The corresponding plot is given in Figure "Failure rate for the system".
 
[[Image:BS5.7.png|center|300px|<div align="center"> ''pdf'' plot for the system.</div>]]
 

The <math>MTTF</math> of the system is obtained by integrating the system's reliability function given by <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math> from time zero to infinity, as given by <math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math>. Using BlockSim's Analytical QCP, an <math>MTTF</math> of 1007.8 hours is calculated, as shown in Figure "MTTF of the system".
 

The warranty time can be obtained by solving <math>{{R}_{s}}(t)</math> with respect to time for a system reliability <math>{{R}_{s}}=0.9</math> . Using the Analytical QCP and selecting the 
Warranty Time option, a time of 372.72 hours is obtained, as shown in Figure "Time at which ''R''=0.9 or 90% for the system".
 
[[Image:BS5.8.png|center|300px|<div align="center"> Failure rate for the system.</div>]]
 
[[Image:Fig 5.9.PNG|thumb|center|400px|<div align="center"> MTTF of the system. </div>]]
 
[[Image:Fig 5.10.PNG|thumb|center|400px|<div align="center"> Time at which ''R''=0.9 or 90% for the system.</div>]]
 
[[Image:Fig 5.11.PNG|thumb|center|400px|<div align="center"> Conditional reliability calculation for the system.</div>]]
 

Lastly, the conditional reliability can be obtained using <math>R(T,t)=\frac{R(T+t)}{R(T)}</math> and <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>, or:
 

::<math>\begin{align}
R(200,200)= & \frac{R(400)}{R(200)} \\
= & \frac{0.883825}{0.975321} \\
= & 0.906189
\end{align}</math>

 

This can be calculated using BlockSim's Analytical QCP, as shown in Figure "Conditional reliability calculation for the system".
 

=Approximating the System CDF=

In many cases, it is valuable to fit a distribution that represents the system's times-to-failure. This can be useful when the system is part of a larger assembly and may be used for repeated calculations or in calculations for other systems. In cases such as this, it can be useful to characterize the system's behavior by fitting a distribution to the overall system and calculating parameters for this distribution. This is equivalent to fitting a single distribution to describe <math>{{R}_{S}}(t</math> ). In essence, it is like reducing the entire system to a component in order to simplify calculations.
 
For the system in Figure below:
[[Image:5-2.png|center|150px|<div align="center">Simple two-component system. </div>]]
 

::<math>{{R}_{S}}(t)={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>

 
To compute an approximate reliability function for this system, <math>{{R}_{A}}(t)\simeq {{R}_{S}}(t)</math> , one would compute <math>n</math> pairs of reliability and time values and then fit a single distribution to the data, or:

 

::<math>\begin{align}
{{R}_{S}}(t= & 10,396.7)=10% \\
{{R}_{S}}(t= & 9,361.9)=20% \\
& ... \\
{{R}_{S}}(t= & 1,053.6)=90%
\end{align}</math>

 

A single distribution, <math>{{R}_{A}}(t)</math> , that approximates <math>{{R}_{S}}(t)</math> can now be computed from these pairs using life data analysis methods. If using the Weibull++ software, one would enter the values as free form data.

===Example 3===
Compute a single Weibull distribution approximation for the system in Example 2.
====Solution to Example 3====
The system in the previous example, shown in Figure "Complex bridge system in Example 2", can be approximated by use of a 2-parameter Weibull distribution with <math>\beta =2.02109</math> and <math>\eta =1123.51</math> . In BlockSim, this is accomplished by representing the entire system as one distribution by going to the Distribution Fit area, as shown next.

 
[[Image:Fig 5.13.PNG|center|200px|<div align="center"> Representing a system with a distribution.</div>]]

 
[[Image:Fig 5.12.PNG|thumb|center|350px|<div align="center"> Distribution Estimator.</div>]]
 
Clicking inside the Distribution Fit area opens the Distribution Estimator window (Figure "Distribution Fitting window"). In this window you can select a distribution to represent the data. BlockSim will then generate a number of system failure times based on the system's reliability function. The system's reliability function can be used to solve for a time value associated with that unreliability value. The distribution of the generated time values can then be fitted to a probability distribution function.
 

Consider a value of <math>F(t)=0.11</math> . Using the system's reliability equation and solving for time, the corresponding time-to-failure for a 0.11 unreliability can be calculated. 

For the system of Example 2, the time for a 0.11 unreliability is 389.786 hours. 

When enough points have been generated, the selected distribution will be fitted to this data set and the distribution's parameters will be returned. In addition, if ReliaSoft's Weibull++ is installed, the generated data can be viewed/analyzed using a Weibull++ instance, as shown in Figure "Using Weibull++ to calculate distribution parameters".
 
[[Image:Fig 5.14.PNG|thumb|center|450px|<div align="center"> Using Weibull++ to calculate distribution parameters.</div>]]
 
It is recommended that the analyst examine the fit to ascertain the applicability of the approximation.

=Duty Cycle=

Components of a system may not operate continuously during a system's mission, or may be subjected to loads greater or lesser than the rated loads during system operation. To model this, a factor called the Duty Cycle ( <math>{{d}_{c}}</math> ) is used. The duty cycle may also be used to account for changes in environmental stress, such as temperature changes, that may effect the operation of a component. The duty cycle is a positive value, with a default value of 1 representing continuous operation at rated load, and any values other than 1 representing other load values with respect to the rated load value (or total operating time). A duty cycle value higher than 1 indicates a load in excess of the rated value. A duty cycle value lower than 1 indicates that the component is operating at a load lower than the rated load or not operating continuously during the system's mission. For instance, a duty cycle of 0.5 may be used for a component that operates only half of the time during the system's mission.
 
The reliability metrics for a component with a duty cycle are calculated as follows. Let <math>{{d}_{c}}</math> represent the duty cycle during a particular mission of the component, <math>t</math> represent the mission time and <math>{t}'</math> represent the accumulated age. Then:
 

::<math>{t}'={{d}_{c}}\times t</math>

 

The reliability equation for the component is:
 

::<math>R({t}')=R({{d}_{c}}\times t)</math>
 

The component ''pdf'' is:
 

::<math>f({t}')=-\frac{d(R({t}'))}{dt}=-\frac{d(R({{d}_{c}}\times t))}{dt}={{d}_{c}}f({{d}_{c}}\times t)</math>
 

The failure rate of the component is:
 

::<math>\lambda ({t}')=\frac{f({t}')}{R({t}')}=\frac{{{d}_{c}}f({{d}_{c}}\times t)}{R({{d}_{c}}\times t)}={{d}_{c}}\lambda ({{d}_{c}}\times t)</math>

===Example 4===
Consider a computer system with three components: a processor, a hard drive and a CD drive in series as shown next. Assume that all three components follow a Weibull failure distribution with the parameters <math>{{\beta }_{1}}=1.5</math> and <math>{{\eta }_{1}}=5000</math> for the processor, <math>{{\beta }_{2}}=2.5</math> and <math>{{\eta }_{2}}=3000</math> for the hard drive, and <math>{{\beta }_{3}}=2</math> and <math>{{\eta }_{3}}=4000</math> for the CD drive. Determine the reliability of the computer system after one year (365 days) of operation, assuming that the CD drive is used only 30% of the time.
 
[[Image:BS5ex4.png|center|250px]]

====Solution to Example 4====
The reliability of the processor after 365 days of operation is given by:

::<math>\begin{align}
{{R}_{processor}}(365)= & {{e}^{-{{\left( \tfrac{365}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}} \\
= & {{e}^{-{{\left( \tfrac{365}{5000} \right)}^{1.5}}}} \\
= & 0.9805\text{ or }98.05%
\end{align}</math>

The reliability of the hard drive after 365 days of operation is given by:

::<math>\begin{align}
{{R}_{harddrive}}(365)= & {{e}^{-{{\left( \tfrac{365}{{{\eta }_{2}}} \right)}^{{{\beta }_{2}}}}}} \\
= & {{e}^{-{{\left( \tfrac{365}{3000} \right)}^{2.5}}}} \\
= & 0.9948\text{ or }99.48%
\end{align}</math>

The reliability of the CD drive after 365 days of operation (taking into account the 30% operation using a duty cycle of 0.3) is given by:

::<math>\begin{align}
{{R}_{CDdrive}}(365)= & {{e}^{-{{\left( \tfrac{{{d}_{c}}\times 365}{{{\eta }_{3}}} \right)}^{{{\beta }_{3}}}}}} \\
= & {{e}^{-{{\left( \tfrac{0.3\times 365}{4000} \right)}^{2}}}} \\
= & 0.9993\text{ or }99.93%
\end{align}</math>
 
[[Image:Fig 5.15.PNG|center|thumb|400px|<div align="center"> Result for the computer system reliability.</div>]]
 

Thus the reliability of the computer system after 365 days of operation is:

::<math>\begin{align}
{{R}_{s}}(365)= & {{R}_{processor}}(365)\cdot {{R}_{harddrive}}(365)\cdot {{R}_{CDdrive}}(365) \\
= & 0.9805\cdot 0.9948\cdot 0.9993 \\
= & 0.9747\text{ or }97.47%
\end{align}</math>
 
This result can be obtained in BlockSim as shown in Figure "Result for the computer system reliability".
 

=Load Sharing=
As presented in earlier chapters, a reliability block diagram (RBD) allows you to graphically represent how the components within a system are reliability-wise connected. In most cases, independence is assumed across the components within the system. For example, the failure of component A does not affect the failure of component B. However, if a system consists of components that are sharing a load, then the assumption of independence no longer holds true.
 

If one component fails, then the component(s) that are still operating will have to assume the failed unit's portion of the load. Therefore, the reliabilities of the surviving unit(s) will change. Calculating the system reliability is no longer an easy proposition. In the case of load sharing components, the change of the failure distributions of the surviving components must be known in order to determine the system's reliability.
 

To illustrate this, consider the a system of two units connected reliability-wise in parallel (Figure "Two units connected reliability-wise in parallel").
 

[[Image:5-16.png|center|200px|<div align="center"> Two units connected reliability-wise in parallel.</div>]]
 

Assume that the units must supply an output of 8 volts and that if both units are operational, each unit is to supply 50% of the total output. If one of the units fails, then the surviving unit supplies 100%. Furthermore, assume that having to supply the entire load has a negative impact on the reliability characteristics of the surviving unit. Since the reliability characteristics of the unit change based on whether both or only one is operating, a life distribution along with a life-stress relationship (as discussed in [http://reliawiki.com/index.php/Statistical_Background#A_Brief_Introduction_to_Life-Stress_Relationships|A Brief Introduction to Life-Stress Relationships]) will be needed to model each component.
 

To illustrate the steps needed, we will create the model starting from raw data. Assume that a total of 20 units were tested to failure at 7, 10 and 15 volts. The test data set is presented in the next table.
 

[[Image:5-17.png|center|450px|]]
 

For this example, Units 1 and 2 are the same component. Therefore, only one set of data was collected. However, it is possible that the load sharing components in a system may not be the same. If that were the case, data would need to be collected for each component.
 

The data set in Table 1 was analyzed using ReliaSoft's ALTA software (as shown in Figure "Using ALTA to calculate component parameters") with the Inverse Power Law as the underlying life-stress relationship and Weibull as the life distribution.
 
 

The estimated model parameters, <math>\beta </math> , <math>K</math> and <math>n</math> , are shown next.
 

::<math>\begin{align}
\beta = & 1.9239 \\
K= & 3.2387\times {{10}^{-7}} \\
n= & 3.4226
\end{align}</math>
 

:Or:
 
::<math>\begin{align}
{{R}_{1}}(t,{{S}_{1}})= & {{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} \\
= & {{e}^{-{{\left( 3.2387\times {{10}^{-7}}S_{1}^{3.4226}t \right)}^{1.9239}}}} \
\end{align}</math>.

 
::<math>{{f}_{1}}(t,{{S}_{1}})=\beta KS_{1}^{n}{{\left( KS_{1}^{n}t \right)}^{\beta -1}}{{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} \ </math>
 

And for this case:

::<math>\begin{align}
{{R}_{1}}(t,{{S}_{1}})= & {{R}_{2}}(t,{{S}_{2}}) \\
{{f}_{1}}(t,{{S}_{1}})= & {{f}_{2}}(t,{{S}_{2}})
\end{align}</math>
 
 
[[Image:Fig 5.17.PNG|thumb|center|450px|<div align="center"> Using ALTA to calculate component parameters.</div>]]
 

Figure "Reliability curves for different voltage output conditions" shows a plot of <math>{{R}_{1}}(t,{{S}_{1}})={{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} = {{e}^{-{{\left( 3.2387\times {{10}^{-7}}S_{1}^{3.4226}t \right)}^{1.9239}}}}\ </math>.
 

Now that the failure properties have been determined using the test data, the reliability of the system at some time, <math>t</math> , can be calculated using the following equation:

 
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>

 
:Where:
 

::<math>\begin{align}
{{S}_{1}}= & {{P}_{1}}S \\
{{S}_{2}}= & {{P}_{2}}S
\end{align}</math>

 
[[Image:5-18.png|center|300px|<div align="center"> Reliability curves for different voltage output conditions. </div>]]
 
[[Image:BS5.19.png|center|300px| Illustrating <math>{{t}_{e}}</math>]]
 

And:
 

:• <math>S</math> is the total load (or required output). 
:• <math>{{P}_{1}}</math> and <math>{{P}_{2}}</math> are the portion of the total load that each unit supports when both units are operational. In this case, <math>{{P}_{1}}={{P}_{2}}=0.5=50%.</math> 
:• <math>{{S}_{1}}</math> and <math>{{S}_{2}}</math> represent the portions of the load that Unit 1 and Unit 2 must support when both units are operational. 
:• <math>{{t}_{{{1}_{e}}}}</math> is the equivalent operating time for Unit 1 if it had been operating at <math>S</math> instead of <math>{{S}_{1}}</math> . A graphical representation of the equivalent time is shown in Figure 5.19, where the curve marked by L represents the low stress (load) and the curve marked by H represents the high stress (load). 
 

::<math>{{t}_{1e}}</math> can be calculated by:
 
::<math>\begin{align}
{{R}_{1}}(t)= & {{R}_{1}}({{t}_{1e}}) \\
{{e}^{-{{(tKS_{1}^{n})}^{\beta }}}}= & {{e}^{-{{({{t}_{1e}}K{{S}^{n}})}^{\beta }}}} \\
tS_{1}^{n}= & {{t}_{1e}}{{S}^{n}} \\
{{t}_{1e}}= & t{{\left( \frac{{{S}_{1}}}{S} \right)}^{n}},\text{ }{{S}_{1}}={{P}_{1}}S \\
\therefore & {{t}_{1e}}=tP_{1}^{n}
\end{align}</math>

 
::<math>{{t}_{2e}}</math> can be calculated the same way, or:

 
::<math>\begin{align}
{{R}_{2}}(t)= & {{R}_{2}}({{t}_{2e}}) \\
\therefore & {{t}_{2e}}=tP_{2}^{n}
\end{align}</math>
 

In this example, the reliability equations for Unit 1 and Unit 2 are the same since they are the same type of component and demonstrate the same failure properties. In addition, the total output is divided equally between the two units (when both units are operating), so <math>{{t}_{1e}}</math> and <math>{{t}_{2e}}</math> will also be the same.
 
The next step is to determine the reliability of the system after 8,760 hours, <math>R(t=8,760)</math> .
Using
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>
 

the system reliability is found to be:
 

::<math>\begin{align}
R(t=8760)= & 0.8567 \\
= & 85.67%
\end{align}</math>
 

===Load Sharing in BlockSim===
BlockSim uses this formulation when computing reliabilities of units in a load sharing configuration. When using the System Reliability Equation window, BlockSim returns a single token for the reliability of units in a load sharing configuration (as well as in the case of standby redundancy, discussed in the next section). As an example, consider the following RBD with Unit 1 in series with a container that includes two load sharing units.
 
[[Image:BS5.19.2.png|center|150px|]]
 
BlockSim will return the system equation as:
 

::<math>{{R}_{System}}=+{{R}_{LS}}\cdot {{R}_{1}}</math>
 

Where <math>{{R}_{LS}}</math> implies a form similar to
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>
 .
BlockSim allows for <math>k</math> -out-of- <math>n</math> units in a load sharing configuration.

===Example 5===
A component has five possible failure modes, <math>A</math> , <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> , <math>{{B}_{C}}</math> and <math>C</math> , and the <math>B</math> modes are interdependent. The system will fail if mode <math>A</math> occurs, mode <math>C</math> occurs or two out of the three <math>B</math> modes occur.
 

Modes <math>A</math> and <math>C</math> each have a Weibull distribution, with a <math>\beta =2</math> and <math>\eta =10,000</math> and 15,000 respectively. Events <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> and <math>{{B}_{C}}</math> each have an exponential distribution with a mean of 10,000 hours.
 

If any <math>B</math> event occurs (i.e. <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> or <math>{{B}_{C}}</math> ), the remaining <math>B</math> events are more likely to occur. Specifically, the mean times of the remaining <math>B</math> events are halved. Determine the reliability at 1000 hours for this component.
====Solution to Example 5====
The first step is to create the RBD. Modes <math>A</math> , <math>C</math> and a load sharing container with the <math>{{B}_{i}}</math> modes must be drawn in series, as illustrated next.
 
[[Image:BS5.19.3.png|center|200px|]]
 

The next step is to define the properties for each block, including those for the container. Based on the problem statement, the <math>B</math> modes are in a 2-out-of-3 load sharing redundancy. When all three are working (i.e. when no <math>B</math> mode has occurred), each block has an exponential distribution with <math>\mu=10,000</math>.
 

The next step is to set the weight proportionality factor. This factor defines the portion of the load that the particular item carries while operating, as well as the load that shifts to the remaining units upon failure of the item. To illustrate, assume three units (1, 2 and 3) are in a load sharing container with weight proportionality factors of 1, 2 and 3 respectively (and a 1-out-of-3 requirement).
 

:• Unit 1 carries <math>\left( \tfrac{1}{1+2+3} \right)=0.166</math> or 16.6% of the total load.
:• Unit 2 carries <math>\left( \tfrac{2}{1+2+3} \right)=0.333</math> or 33.3% of the total load.
:• Unit 3 carries <math>\left( \tfrac{3}{1+2+3} \right)=0.50</math> or 50% of the total load.
 

The actual load on each unit then becomes the product of the entire load defined for the container times the portion carried by that unit. For example, if the container load is 100 lbs, then the portion assigned to Unit 1 will be <math>100\cdot 0.166=16.6</math> lbs.
 

In the current example, all units share the same load and thus have equal weight proportionality factors. Because these factors are relative, if the same number is used for all three items then the results will be the same. Thus, weight proportional factor is set equal to 1 for each item.
 

 
[[Image:Fig 5.23_2.PNG|thumb|center|450px|<div align="center"> Defining Weight Proportional Factor. </div>]]
 

The last properties that need to be defined are the total load and the 2-out-of-3 redundancy. The total load is dependent on how the parameters were computed. In this case, total load was assumed to be 3 when the parameters were computed (i.e. the load per item was 1 when all worked and 1.5 when two worked). This is defined at the container level, as shown in Figure "Defining Weight Proportional Factor".
When all of the parameters have been specified in BlockSim, the reliability at 1,000 hours can be determined. From the Analytical QCP, this is found to be 98.1048%.

=Standby Components=

In the previous section, the case of a system with load sharing components was presented. This is a form of redundancy with dependent components. That is, the failure of one component affects the failure of the other(s). This section presents another form of redundancy: standby redundancy. In standby redundancy the redundant components are set to be under a lighter load condition (or no load) while not needed and under the operating load when they are activated.
 

In standby redundancy the components are set to have two states: an active state and a standby state. Components in standby redundancy have two failure distributions, one for each state. When in the standby state, they have a quiescent (or dormant) failure distribution and when operating, they have an active failure distribution.
 

In the case that both quiescent and active failure distributions are the same, the units are in a simple parallel configuration (also called a hot standby configuration). When the rate of failure of the standby component is lower in quiescent mode than in active mode, that is called a warm standby configuration. When the rate of failure of the standby component is zero in quiescent mode (i.e. the component cannot fail when in standby), that is called a cold standby configuration.

===Simple Standby Configuration===

Consider two components in a standby configuration. Component 1 is the active component with a Weibull failure distribution with parameters <math>\beta </math> = 1.5 and <math>\eta </math> = 1,000. Component 2 is the standby component. When Component 2 is operating, it also has a Weibull failure distribution with <math>\beta </math> = 1.5 and <math>\eta </math> = 1,000. Furthermore, assume the following cases for the quiescent distribution.
 
:• Case 1: The quiescent distribution is the same as the active distribution (hot standby). 
:• Case 2: The quiescent distribution is a Weibull distribution with <math>\beta </math> = 1.5 and <math>\eta </math> = 2000 (warm standby). 
:• Case 3: The component cannot fail in quiescent mode (cold standby). 
 
In this case, the reliability of the system at some time, <math>t</math> , can be obtained using the following equation:
 

::<math>R(t)={{R}_{1}}(t)+\underset{0}{\overset{t}{\mathop \int }}\,{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x)\cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}dx \ </math>
 

Where:
 
:• <math>{{R}_{1}}</math> is the reliability of the active component. 
:• <math>{{f}_{1}}</math> is the <math>pdf</math> of the active component. 
:• <math>{{R}_{2;SB}}</math> is the reliability of the standby component when in standby mode (quiescent reliability). 
:• <math>{{R}_{2;A}}</math> is the reliability of the standby component when in active mode. 
:• <math>{{t}_{e}}</math> is the equivalent operating time for the standby unit if it had been operating at an active mode, such that: 

 
<math>{{R}_{2;SB}}(x)={{R}_{2;A}}({{t}_{e}})</math> 
 
The second equation above can be solved for <math>{{t}_{e}}</math> and substituted into the first equation above.
Figure "Standby container" illustrates the example as entered in BlockSim using a standby container.
 
 
[[Image:5.24.png|center|150px|<div align="center"> Standby container.</div>]]

 

The active and standby blocks are within a container, which is used to specify standby redundancy. Since the standby component has two distributions (active and quiescent), the Block Properties window of the standby block has two pages for specifying each one. Figures "Defining the active failure distribution" and "Defining the quiescent failure distribution" illustrate these pages.

 

The system reliability results for 1000 hours are given in the following table:
 
[[Image:5-24.png|center|150px|]]
 

Note that even though the <math>\beta </math> value for the quiescent distribution is the same as in the active distribution, it is possible that the two can be different. That is, the failure modes present during the quiescent mode could be different from the modes present during the active mode. In that sense, the two distribution types can be different as well (e.g. lognormal when quiescent and Weibull when active).
 

In many cases when considering standby systems, a switching device may also be present that switches from the failed active component to the standby component. The reliability of the switch can also be incorporated into
<math>R(t)={{R}_{1}}(t)+\underset{0}{\overset{t}{\mathop \int }}\,{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x)\cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}dx \ </math>

as presented in the next section.
 

BlockSim's System Reliability Equation window returns a single token for the reliability of units in a standby configuration. This is the same as the load sharing case presented in the previous section.
 

[[Image: Fig 5.25.PNG|thumb|center|400px|<div align="center"> Defining the active failure distribution </div>]]
 
 

[[Image:Fig 5.26.PNG|thumb|center|400px|<div align="center"> Defining the quiescent failure distribution </div>]]
 

===Reliability of Standby Systems with a Switching Device===

In many cases when dealing with standby systems, a switching device is present that will switch to the standby component when the active component fails. Therefore, the failure properties of the switch must also be included in the analysis.
 

[[Image:BS5.26.2.png|center|150px|]]
 

In most cases when the reliability of a switch is to be included in the analysis, two probabilities can be considered. The first and most common one is the probability of the switch performing the action (i.e. switching) when requested to do so. This is called Switch Probability per Request in BlockSim and is expressed as a static probability (e.g. 90%). The second probability is the quiescent reliability of the switch. This is the reliability of the switch as it ages (e.g. the switch might wear out with age due to corrosion, material degradation, etc.). Thus it is possible for the switch to fail before the active component fails. However, a switch failure does not cause the system to fail, but rather causes the system to fail only if the switch is needed and the switch has failed. For example, if the active component does not fail until the mission end time and the switch fails, then the system does not fail. However, if the active component fails and the switch has also failed, then the system cannot be switched to the standby component and it therefore fails.
 

In analyzing standby components with a switching device, either or both failure probabilities (during the switching or while waiting to switch) can be considered for the switch, since each probability can represent different failure modes. For example, the switch probability per request may represent software-related issues or the probability of detecting the failure of an active component, and the quiescent probability may represent wear-out type failures of the switch.
 

To illustrate the formulation, consider the previous example that assumes perfect switching. To examine the effects of including an imperfect switch, assume that when the active component fails there is a 90% probability that the switch will switch from the active component to the standby component. In addition, assume that the switch can also fail due to a wear-out failure mode described by a Weibull distribution with <math>\beta </math> = 1.7 and <math>\eta </math> = 5000.
 

Therefore, the reliability of the system at some time, <math>t</math> , is given by the following equation.

::<math>\begin{align}
R(t)= & {{R}_{1}}(t) \\
& +\underset{0}{\overset{t}{\mathop \int }}\,\{{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x) \\
& \cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}\cdot {{R}_{SW;Q}}(x)\cdot {{R}_{SW;REQ}}(x)\}dx
\end{align}</math>

Where:
 
:• <math>{{R}_{1}}</math> is the reliability of the active component. 
:• <math>{{f}_{1}}</math> is the <math>pdf</math> of the active component. 
:• <math>{{R}_{2;SB}}</math> is the reliability of the standby component when in standby mode (quiescent reliability). 
:• <math>{{R}_{2;A}}</math> is the reliability of the standby component when in active mode. 
:• <math>{{R}_{SW;Q}}</math> is the quiescent reliability of the switch. 
:• <math>{{R}_{SW;REQ}}</math> is the switch probability per request. 
:• <math>{{t}_{e}}</math> is the equivalent operating time for the standby unit if it had been operating at an active mode. 
This problem can be solved in BlockSim by including these probabilities in the container's properties, as shown in Figures "Standby container (switch) failure probabilities while attempting to switch" and "Standby container (switch) failure distribution while waiting to switch". In BlockSim, the standby container is acting as the switch.

 

[[Image:Fig 5.28_2.PNG|thumb|center|400px|<div align="center"> Standby container (switch) failure probabilities while attempting to switch </div> ]]
 

 

[[Image:Fig 5.27.PNG|thumb|center|500px|<div align="center"> Standby container (switch) failure distribution while waiting to switch </div>]]
 

 

Note that there are additional properties that can be specified in BlockSim for a switch, such as Switch Restart Probability, No. of Restarts and Switch Delay Time. In many applications, the switch is re-tested (or re-cycled) if it fails to switch the first time. In these cases, it might be possible that it switches in the second or third, or <math>{{n}^{th}}</math> attempt.
 
The Switch Restart Probability specifies each additional attempt's probability of successfully switching and the Finite Restarts specifies the total number of attempts. Note that the Switch Restart Probability specifies the probability of success of each trial (or attempt). The probability of success of <math>n</math> consecutive trials is calculated by BlockSim using the binomial distribution and this probability is then incorporated into equation above. The Switch Delay Time property is related to repairable systems and is considered in BlockSim only when using simulation. When using the analytical solution (i.e. for a non-repairable system), this property is ignored.
 

Solving the analytical solution (as given by Equation above), the following results are obtained.
 

[[Image:5-30.png|center|350px|]]
 

From the table above, it can be seen that the presence of a switching device has a significant effect on the reliability of a standby system. It is therefore important when modeling standby redundancy to incorporate the switching device reliability properties. It should be noted that this methodology is not the same as treating the switching device as another series component with the standby subsystem. This would be valid only if the failure of the switch resulted in the failure of system (e.g. switch failing open). In equation above, the Switch Probability per Request and quiescent probability are present only in the second term of the equation. Treating these two failure modes as a series configuration with the standby subsystem would imply that they are also present when the active component is functioning (i.e. first term of equation above). This is invalid and would result in the underestimation of the reliability of the system. In other words, these two failure modes become significant only when the active component fails.
 

As an example, and if we consider the warm standby case, the reliability of the system without the switch is 70.57% at 1000 hours. If the system was modeled so that the switching device was in series with the warm standby subsystem, the result would have been:

 

::<math>\begin{align}
{{R}_{S}}(1000)= & {{R}_{Standby}}(1000)\cdot {{R}_{sw,Q(1000)}}\cdot {{R}_{sw,req}} \\
= & 0.7057\cdot 0.9372\cdot 0.9 \\
= & 0.5952
\end{align}</math>

 

In the case where a switch failure mode causes the standby subsystem to fail, then this mode can be modeled as an individual block in series with the standby subsystem.

===Example 6===
Consider a car with four new tires and a full-size spare. Assume the following failure characteristics:
 
:• The tires follow a Weibull distribution with a <math>\beta=4</math> and an <math>\eta =</math> 40,000 miles while on the car due to wear. 
:• The tires also have a probability of failing due to puncture or other causes. For this, assume a constant rate for this occurrence with a probability of 1 every 50,000 miles. 
:• When not on the car (i.e. is a spare), a tire's probability of failing also has a Weibull distribution with a <math>\beta =</math> 2 and <math>\eta =</math> 120,000 miles. 
 
Assume a mission of 1,000 miles. If a tire fails during this trip, it will be replaced with the spare. However, the spare will not be repaired during the trip. In other words, the trip will continue with the spare on the car and if the spare fails the system will fail. Determine the probability of system failure.
====Solution to Example 6====
Active failure distribution for tires:
 
:• Due to wear-out, Weibull <math>\beta =4</math> and <math>\eta =40,000</math> miles. 
:• Due to random puncture, exponential <math>\mu =50,000.</math> 
:• The quiescent failure distribution is a Weibull distribution with <math>\beta =2</math> and <math>\eta =120,000</math> miles. 
 
The block diagram for each tire has two blocks in series, one block representing the wear-out mode and the other the random puncture mode, as shown next:
 
 
[[Image:small5.png|center|250px|]]
 

There are five tires, four active and one standby (represented in the diagram by a standby container with a 4-out-of-5 requirement), as shown next:
 
[[Image:BStirecontainer.png|center|250px|]]

For the standby Wear block, set the active failure and the quiescent distributions, but for the Puncture block, set only the active puncture distribution (because the tire cannot fail due to puncture while stored). Using BlockSim, the probability of system failure is found to be 0.003 or 0.3%.
 

=Note Regarding Numerical Integration Solutions=

Load sharing and standby solutions in BlockSim are performed using numerical integration routines. As with any numerical analysis routine, the solution error depends on the number of iterations performed, the step size chosen and related factors, plus the behavior of the underlying function. By default, BlockSim uses a certain set of preset factors. In general, these defaults are sufficient for most problems. If a higher precision or verification of the precision for a specific problem is required, BlockSim's preset options can be modified and/or the integration error can be assessed using the Integration Parameters... option for each container. For more details, you can refer to the documentation on the Algorithm Setup window in the BlockSim 8 User's help files.

Time-Dependent System Reliability (Analytical)

2012-06-21T15:56:07Z

Dingzhou Cao: /* Solution to Example 5 */

{{Template:bsbook|5}}

In the previous chapter, different system configuration types were examined, as well as different methods for obtaining the system's reliability function analytically. Because the reliabilities in the problems presented were treated as probabilities (e.g. <math>P(A)</math> , <math>{{R}_{i}}</math> ), the reliability values and equations presented were referred to as static (not time-dependent). Thus, in the prior chapter, the life distributions of the components were not incorporated in the process of calculating the system reliability. In this chapter, time dependency in the reliability function will be introduced. We will develop the models necessary to observe the reliability over the life of the system, instead of at just one point in time. In addition, performance measures such as failure rate, MTTF and warranty time will be estimated for the entire system. The methods of obtaining the reliability function analytically remain identical to the ones presented in the previous chapter, with the exception that the reliabilities will be functions of time. In other words, instead of dealing with <math>{{R}_{i}}</math> , we will use <math>{{R}_{i}}(t)</math> . All examples in this chapter assume that no repairs are performed on the components.

=Analytical Life Predictions=
The analytical approach presented in the prior chapter involved the determination of a mathematical expression that describes the reliability of the system, expressed in terms of the reliabilities of its components. So far we have estimated only static system reliability (at a fixed time). For example, in the case of a system with three components in series, the system's reliability equation was given by:
 

::<math>{{R}_{s}}={{R}_{1}}\cdot {{R}_{2}}\cdot {{R}_{3}} \ </math>
 

The values of <math>{{R}_{1}}</math> , <math>{{R}_{2}}</math> and <math>{{R}_{3}}</math> were given for a common time and the reliability of the system was estimated for that time. However, since the component failure characteristics can be described by distributions, the system reliability is actually time-dependent. In this case, the equation above can be rewritten as:
 

::<math>{{R}_{s}}(t)={{R}_{1}}(t)\cdot {{R}_{2}}(t)\cdot {{R}_{3}}(t)</math>

 

The reliability of the system for any mission time can now be estimated. Assuming a Weibull life distribution for each component, the first equation above can now be expressed in terms of each component's reliability function, or:

 

[[Category:Completed Theoretical Review]]

::<math>{{R}_{s}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}}\cdot {{e}^{-{{\left( \tfrac{t}{{{\eta }_{2}}} \right)}^{{{\beta }_{2}}}}}}\cdot {{e}^{-{{\left( \tfrac{t}{{{\eta }_{3}}} \right)}^{{{\beta }_{3}}}}}}</math>

 

In the same manner, any life distribution can be substituted into the system reliability equation. Suppose that the times-to-failure of the first component are described with a Weibull distribution, the times-to-failure of the second component with an exponential distribution and the times-to-failure of the third component with a normal distribution. Then the first equation above can be written as:
 

::<math>{{R}_{s}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}}\cdot {{e}^{-{{\lambda }_{2}}t}}\cdot \left[ 1-\Phi \left( \frac{t-{{\mu }_{3}}}{{{\sigma }_{3}}} \right) \right]</math>

 

It can be seen that the biggest challenge is in obtaining the system's reliability function in terms of component reliabilities, which has already been discussed in depth. Once this has been achieved, calculating the reliability of the system for any mission duration is just a matter of substituting the corresponding component reliability functions into the system reliability equation.
===Advantages of the Analytical Method===
The primary advantage of the analytical solution is that it produces a mathematical expression that describes the reliability of the system. Once the system's reliability function has been determined, other calculations can then be performed to obtain metrics of interest for the system. Such calculations include:
 
:• Determination of the system's <math>pdf.</math> 
:• Determination of warranty periods. 
:• Determination of the system's failure rate. 
:• Determination of the system's MTTF. 
 
In addition, optimization and reliability allocation techniques can be used to aid engineers in their design improvement efforts. Another advantage in using analytical techniques is the ability to perform static calculations and analyze systems with a mixture of static and time-dependent components. Finally, the reliability importance of components over time can be calculated with this methodology.
===Disadvantages of the Analytical Method===
The biggest disadvantage of the analytical method is that formulations can become very complicated. The more complicated a system is, the larger and more difficult it will be to analytically formulate an expression for the system's reliability. For particularly detailed systems this process can be quite time-consuming, even with the use of computers. Furthermore, when the maintainability of the system or some of its components must be taken into consideration, analytical solutions become intractable. In these situations, the use of simulation methods may be more advantageous than attempting to develop a solution analytically. Simulation methods are presented in later chapters.
===Looking at a Simple ''Complex'' System Analytically===

The complexity involved in an analytical solution can be best illustrated by looking at the simple ''complex'' system with 15 components, as shown in Figure " An RBD of a complex system".
 
[[Image:5-1.png|center|400px|<div align="center"> An RBD of a complex system.</div>]]

 
The system reliability for this system (computed using BlockSim) is shown next. The first solution is provided using BlockSim's symbolic solution. In symbolic mode, BlockSim breaks the equation into segments, identified by tokens, that need to be substituted into the final system equation for a complete solution. This creates algebraic solutions that are more compact than if the substitutions were made.
 

::<math>\begin{align}
{{R}_{System}}= & D2\cdot D3\cdot {{R}_{L}} \\
D3= & +{{R}_{K}}\cdot IK \\
IK= & +{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}} \\
D2 = & +{{R}_{A}}\cdot {{R}_{E}}\cdot IE \\
IE = & -D1\cdot {{R}_{M}}\cdot {{R}_{N}}+{{R}_{M}}\cdot {{R}_{N}}+D1 \\
D1 = & +{{R}_{D}}\cdot ID \\
ID = & -{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}}
\end{align}</math>
 
Substituting the terms yields:
 

::<math>\begin{align}
{{R}_{System}}= & {{R}_{A}}\cdot {{R}_{E}}\cdot {{R}_{L}}\cdot {{R}_{K}} \\
& \cdot \{({{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})\cdot {{R}_{M}}\cdot {{R}_{N}} \\
& +{{R}_{M}}\cdot {{R}_{N}}-{{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}}\} \\
& \cdot \{{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}}\}
\end{align}</math>

 
BlockSim's automatic algebraic simplification would yield the following format for the above solution:
 

::<math>\begin{align}
{{R}_{System}}= & (({{R}_{A}}\cdot {{R}_{E}}(-({{R}_{D}}(-{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})){{R}_{M}}\cdot {{R}_{N}} \\
& +{{R}_{M}}\cdot {{R}_{N}} \\
& +({{R}_{D}}(-{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})))) \\
& ({{R}_{K}}({{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& +RI\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}})){{R}_{L}}) \
\end{align}</math>

 
In this equation, each <math>{{R}_{i}}</math> represents the reliability function of a block. For example, if <math>{{R}_{A}}</math> has a Weibull distribution, then each <math>{{R}_{A}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{A}}} \right)}^{{{\beta }_{A}}}}}}</math> and so forth. Substitution of each component's reliability function in the last <math>{{R}_{System}}</math> equation above will result in an analytical expression for the system reliability as a function of time, or <math>{{R}_{s}}(t)</math> , which is the same as <math>(1-cd{{f}_{System}}).</math>

===Obtaining Other Functions of Interest===
Once the system reliability equation (or the cumulative density function, <math>cdf</math> ) has been determined, other functions and metrics of interest can be derived.
 
Consider the following simple system:
 
[[Image:5-2.png|center|150px|<div align="center"> Simple two-component system. </div>]]

 
Furthermore, assume that component 1 follows an exponential distribution with a mean of 10,000 (<math>\mu =10,000,</math> <math>\lambda =1/10,000)</math> and component 2 follows a Weibull distribution with <math>\beta =6</math> and <math>\eta =10,000</math> . The reliability equation of this system is:

 
::<math>\begin{align}
{{R}_{S}}(t)= & {{R}_{1}}(t)\cdot {{R}_{2}}(t) \\
= & {{e}^{-\lambda t}}\cdot {{e}^{-{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \\
= & {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \
\end{align}</math>

 
The system <math>cdf</math> is:

 
::<math>\begin{align}
{{F}_{S}}(t)= & 1-({{R}_{1}}(t)\cdot {{R}_{2}}(t)) \\
= & 1-\left( {{e}^{-\lambda t}}\cdot {{e}^{-{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \right) \\
= & 1-\left( {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)
\end{align}</math>

 
{{analytical system pdf}}

====Conditional Reliability====
Conditional reliability is the probability of a system successfully completing another mission following the successful completion of a previous mission. The time of the previous mission and the time for the mission to be undertaken must be taken into account for conditional reliability calculations. The system's conditional reliability function is given by:

::<math>R(T,t)=\frac{R(T+t)}{R(T)}</math>

Equation above gives the reliability for a new mission of duration <math>t</math> having already accumulated <math>T</math> hours of operation up to the start of this new mission. The system is evaluated to assure that it will start the next mission successfully.
 
For the system in Figure "Simple two-component system.", the reliability for mission of <math>t=1,000</math> hours, having an age of <math>T=500</math> hours, is:
 

::<math>\begin{align}
{{R}_{S}}(T=500,t=1000)= & \frac{R(T+t)}{R(T)} \\
= & \frac{R(1500)}{R(500)} \\
= & \frac{{{e}^{-\tfrac{1500}{10,000}}}\cdot {{e}^{-{{\left( \tfrac{1500}{10,000} \right)}^{6}}}}}{{{e}^{-\tfrac{500}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{500}{10,000} \right)}^{6}}}}} \\
= & 0.9048=90.48%
\end{align}</math>

 

====Conditional Reliability for Components====

Now in this formulation, it was assumed that the accumulated age was equivalent for both units. That is, both started life at zero and aged to 500. It is possible to consider an individual component that has already accumulated some age (used component) in the same formulation. To illustrate this, assume that component 2 started life with an age of T=100. Then the reliability equation of the system, as given in <math>{{R}_{S}}(t)= {{R}_{1}}(t)\cdot {{R}_{2}}(t)</math>, would need to be modified to include a conditional term for 2, or:
 

::<math>{{R}_{S}}(t)={{R}_{1}}(t)\cdot \frac{{{R}_{2}}({{T}_{2}}+t)}{{{R}_{2}}({{T}_{2}})} \ </math>

 

In BlockSim, the start age input box may be used to specify a starting age greater than zero.
{{system failure rate analytical}}

====System Mean Life (Mean Time To Failure)====
The mean life (or mean time to failure, MTTF) can be obtained by integrating the system reliability function from zero to infinity:
 

::<math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math>

 

The mean time is a performance index and does not provide any information about the behavior of the failure distribution of the system.
 
For the system in Figure "Simple two-component system":
 

::<math>\begin{align}
MTTF= & \int_{0}^{\infty }\left( {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)dt \\
= & 5978.9
\end{align}</math>

 

====Warranty Period and BX Life====
Sometimes it is desirable to know the time value associated with a certain reliability. Warranty periods are often calculated by determining what percentage of the failure population can be covered financially and estimating the time at which this portion of the population will fail. Similarly, engineering specifications may call for a certain BX life, which also represents a time period during which a certain proportion of the population will fail. For example, the B10 life is the time in which 10% of the population will fail.
This is obtained by setting <math>{{R}_{S}}(t)</math> to the desired value and solving for <math>t.</math>
 
For the system in Figure "Simple two-component system":
 

::<math>{{R}_{s}}\left( t \right)={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>
 

To compute the time by which reliability would be equal to 90%, equation above is recast as follows and solved for <math>t.</math>

 
::<math>0.90={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>

 

In this case, <math>t=1053.59</math> . Equivalently, the B10 life for this system is also <math>1053.59</math> .
Except for some trivial cases, a closed form solution for <math>t</math> cannot be obtained. Thus, it is necessary to solve for <math>t</math> using numerical methods. BlockSim uses numerical methods.

===Example 1===
Consider a system consisting of three exponential units in series with the following failure rates (in failures per hour): <math>{{\lambda }_{1}}</math> = 0.0002, <math>{{\lambda }_{2}}</math> = 0.0005 and <math>{{\lambda }_{3}}</math> = 0.0001.
 
:• Obtain the reliability equation for the system. 
:• What is the reliability of the system after 150 hours of operation? 
:• Obtain the system's <math>pdf.</math> 
:• Obtain the system's failure rate equation. 
:• What is the MTTF for the system? 
:• What should the warranty period be for a 90% reliability? 
====Solution to Example 1====
 
:• The analytical expression for the reliability of the system is given by:

::<math>\begin{align}
{{R}_{s}}(t)= & {{R}_{1}}(t)\cdot {{R}_{2}}(t)\cdot {{R}_{3}}(t) \\
= & {{e}^{-{{\lambda }_{1}}t}}\cdot {{e}^{-{{\lambda }_{2}}t}}\cdot {{e}^{-{{\lambda }_{1}}t}} \\
= & {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}
\end{align}</math>

 

At 150 hours of operation, the reliability of the system is:
 

::<math>\begin{align}
{{R}_{s}}(t)= & {{e}^{-(0.0002+0.0005+0.0001)150}} \\
= & 0.8869\text{ or }88.69%
\end{align}</math>

 

:• In order to obtain the system's <math>pdf</math> , the derivative of the reliability equation given in the first equation above is taken with respect to time, or:
 
::<math>\begin{align}
{{f}_{s}}(t)= & -\frac{d[{{R}_{s}}(t)]}{dt} \\
= & -\frac{d\left[ {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}} \right]}{dt} \\
= & ({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})\cdot {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}
\end{align}</math>

:• The system's failure rate can now be obtained simply by dividing the system's <math>pdf</math> given in the equation above by the system's reliability function given in the first equation above, and:

::<math>\begin{align}
{{\lambda }_{s}}\left( t \right)= & \frac{{{f}_{s}}\left( t \right)}{{{R}_{s}}\left( t \right)} \\
= & \frac{({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})\cdot {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}}{{{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}} \\
= & ({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}) \\
= & 0.0008\text{ }fr/hr
\end{align}</math>

 
:• Combining <math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math> and the first equation above, the system's MTTF can be obtained:

 
::<math>\begin{align}
MTTF= & \int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \\
= & \int_{0}^{\infty }{{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}dt \\
= & \frac{1}{({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})} \\
= & 1250\text{ }hr
\end{align}</math>
 

:• Solving the first equation above with respect to time will yield the corresponding warranty period for a 90% reliability. In this case, the system reliability equation is simple and a closed form solution exists. The warranty time can now be found by solving:

 
::<math>\begin{align}
t= & -\frac{\ln (R)}{{{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}} \\
= & -\frac{\ln (0.9)}{0.0008} \\
= & 131.7\text{ }hr
\end{align}</math>
 

Thus, the warranty period should be 132 hours.

===Example 2===
Consider the system shown in Figure "Complex bridge system in Example 2".
 
[[Image:BS5.5.png|center|400px|<div align="center"> Complex bridge system in Example 2. </div>]]
 

Components <math>A</math> through <math>E</math> are Weibull distributed with <math>\beta =1.2</math> and <math>\eta =1230</math> hours. The starting and ending blocks cannot fail. 
Determine the following:
 
:• The reliability equation for the system and its corresponding plot. 
:• The system's <math>pdf</math> and its corresponding plot. 
:• The system's failure rate equation and the corresponding plot. 
:• The <math>MTTF</math> . 
:• The warranty time for a 90% reliability. 
:• The reliability for a 200-hour mission, if it is known that the system has already successfully operated for 200 hours. 
====Solution====
 
The first step is to obtain the reliability function for the system. The methods described in the previous chapter can be employed, such as the event space or path-tracing methods. Using BlockSim, the following reliability equation is obtained:
 

::<math>\begin{align}
{{R}_{s}}(t)= & ({{R}_{Start}}\cdot {{R}_{End}}(2{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}))
\end{align}</math>
 

Note that since the starting and ending blocks cannot fail, <math>{{R}_{Start}}=1</math> and <math>{{R}_{End}}=1,</math> equation above can be reduced to:

 

::<math>\begin{align}
{{R}_{s}}(t)= & 2\cdot {{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}
\end{align}</math>

 

Where <math>{{R}_{A}}</math> is the reliability equation for Component A, or:

 

::<math>{{R}_{A}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{A}}} \right)}^{{{\beta }_{A}}}}}}</math>

 

::<math>{{R}_{B}}</math> is the reliability equation for Component <math>B</math> , etc.
 

Since the components in this example are identical, the system reliability equation can be further reduced to:

 

::<math>{{R}_{s}}(t)=2R{{(t)}^{2}}+2R{{(t)}^{3}}-5R{{(t)}^{4}}+2R{{(t)}^{5}}</math>

 

Or, in terms of the failure distribution:

 

::<math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>

 

The corresponding plot is given in Figure "Reliability plot for the system".
 
[[Image:BS5.6.png|center|300px|<div align="center"> Reliability plot for the system. </div>]]
 
 

In order to obtain the system's <math>pdf</math> , the derivative of the reliability equation given above is taken with respect to time, resulting in:
 

::<math>\begin{align}
{{f}_{s}}(t)= & 4\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+6\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \\
& -20\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+10\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}
\end{align}</math>

 

The <math>pdf</math> can now be plotted for different time values, <math>t</math> , as shown in Figure "''pdf'' plot for the system".
 

The system's failure rate can now be obtained by dividing the system's <math>pdf</math> given in equation above by the system's reliability function given in <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>, or:

 

::<math>\begin{align}
{{\lambda }_{s}}(t)= & \frac{4\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+6\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}{2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}} \\
& +\frac{-20\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+10\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}{2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}
\end{align}</math>

 

The corresponding plot is given in Figure "Failure rate for the system".
 
[[Image:BS5.7.png|center|300px|<div align="center"> ''pdf'' plot for the system.</div>]]
 

The <math>MTTF</math> of the system is obtained by integrating the system's reliability function given by <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math> from time zero to infinity, as given by <math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math>. Using BlockSim's Analytical QCP, an <math>MTTF</math> of 1007.8 hours is calculated, as shown in Figure "MTTF of the system".
 

The warranty time can be obtained by solving <math>{{R}_{s}}(t)</math> with respect to time for a system reliability <math>{{R}_{s}}=0.9</math> . Using the Analytical QCP and selecting the 
Warranty Time option, a time of 372.72 hours is obtained, as shown in Figure "Time at which ''R''=0.9 or 90% for the system".
 
[[Image:BS5.8.png|center|300px|<div align="center"> Failure rate for the system.</div>]]
 
[[Image:Fig 5.9.PNG|thumb|center|400px|<div align="center"> MTTF of the system. </div>]]
 
[[Image:Fig 5.10.PNG|thumb|center|400px|<div align="center"> Time at which ''R''=0.9 or 90% for the system.</div>]]
 
[[Image:Fig 5.11.PNG|thumb|center|400px|<div align="center"> Conditional reliability calculation for the system.</div>]]
 

Lastly, the conditional reliability can be obtained using <math>R(T,t)=\frac{R(T+t)}{R(T)}</math> and <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>, or:
 

::<math>\begin{align}
R(200,200)= & \frac{R(400)}{R(200)} \\
= & \frac{0.883825}{0.975321} \\
= & 0.906189
\end{align}</math>

 

This can be calculated using BlockSim's Analytical QCP, as shown in Figure "Conditional reliability calculation for the system".
 

=Approximating the System CDF=

In many cases, it is valuable to fit a distribution that represents the system's times-to-failure. This can be useful when the system is part of a larger assembly and may be used for repeated calculations or in calculations for other systems. In cases such as this, it can be useful to characterize the system's behavior by fitting a distribution to the overall system and calculating parameters for this distribution. This is equivalent to fitting a single distribution to describe <math>{{R}_{S}}(t</math> ). In essence, it is like reducing the entire system to a component in order to simplify calculations.
 
For the system in Figure below:
[[Image:5-2.png|center|150px|<div align="center">Simple two-component system. </div>]]
 

::<math>{{R}_{S}}(t)={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>

 
To compute an approximate reliability function for this system, <math>{{R}_{A}}(t)\simeq {{R}_{S}}(t)</math> , one would compute <math>n</math> pairs of reliability and time values and then fit a single distribution to the data, or:

 

::<math>\begin{align}
{{R}_{S}}(t= & 10,396.7)=10% \\
{{R}_{S}}(t= & 9,361.9)=20% \\
& ... \\
{{R}_{S}}(t= & 1,053.6)=90%
\end{align}</math>

 

A single distribution, <math>{{R}_{A}}(t)</math> , that approximates <math>{{R}_{S}}(t)</math> can now be computed from these pairs using life data analysis methods. If using the Weibull++ software, one would enter the values as free form data.

===Example 3===
Compute a single Weibull distribution approximation for the system in Example 2.
====Solution to Example 3====
The system in the previous example, shown in Figure "Complex bridge system in Example 2", can be approximated by use of a 2-parameter Weibull distribution with <math>\beta =2.02109</math> and <math>\eta =1123.51</math> . In BlockSim, this is accomplished by representing the entire system as one distribution by going to the Distribution Fit area, as shown next.

 
[[Image:Fig 5.13.PNG|center|200px|<div align="center"> Representing a system with a distribution.</div>]]

 
[[Image:Fig 5.12.PNG|thumb|center|350px|<div align="center"> Distribution Estimator.</div>]]
 
Clicking inside the Distribution Fit area opens the Distribution Estimator window (Figure "Distribution Fitting window"). In this window you can select a distribution to represent the data. BlockSim will then generate a number of system failure times based on the system's reliability function. The system's reliability function can be used to solve for a time value associated with that unreliability value. The distribution of the generated time values can then be fitted to a probability distribution function.
 

Consider a value of <math>F(t)=0.11</math> . Using the system's reliability equation and solving for time, the corresponding time-to-failure for a 0.11 unreliability can be calculated. 

For the system of Example 2, the time for a 0.11 unreliability is 389.786 hours. 

When enough points have been generated, the selected distribution will be fitted to this data set and the distribution's parameters will be returned. In addition, if ReliaSoft's Weibull++ is installed, the generated data can be viewed/analyzed using a Weibull++ instance, as shown in Figure "Using Weibull++ to calculate distribution parameters".
 
[[Image:Fig 5.14.PNG|thumb|center|450px|<div align="center"> Using Weibull++ to calculate distribution parameters.</div>]]
 
It is recommended that the analyst examine the fit to ascertain the applicability of the approximation.

=Duty Cycle=

Components of a system may not operate continuously during a system's mission, or may be subjected to loads greater or lesser than the rated loads during system operation. To model this, a factor called the Duty Cycle ( <math>{{d}_{c}}</math> ) is used. The duty cycle may also be used to account for changes in environmental stress, such as temperature changes, that may effect the operation of a component. The duty cycle is a positive value, with a default value of 1 representing continuous operation at rated load, and any values other than 1 representing other load values with respect to the rated load value (or total operating time). A duty cycle value higher than 1 indicates a load in excess of the rated value. A duty cycle value lower than 1 indicates that the component is operating at a load lower than the rated load or not operating continuously during the system's mission. For instance, a duty cycle of 0.5 may be used for a component that operates only half of the time during the system's mission.
 
The reliability metrics for a component with a duty cycle are calculated as follows. Let <math>{{d}_{c}}</math> represent the duty cycle during a particular mission of the component, <math>t</math> represent the mission time and <math>{t}'</math> represent the accumulated age. Then:
 

::<math>{t}'={{d}_{c}}\times t</math>

 

The reliability equation for the component is:
 

::<math>R({t}')=R({{d}_{c}}\times t)</math>
 

The component ''pdf'' is:
 

::<math>f({t}')=-\frac{d(R({t}'))}{dt}=-\frac{d(R({{d}_{c}}\times t))}{dt}={{d}_{c}}f({{d}_{c}}\times t)</math>
 

The failure rate of the component is:
 

::<math>\lambda ({t}')=\frac{f({t}')}{R({t}')}=\frac{{{d}_{c}}f({{d}_{c}}\times t)}{R({{d}_{c}}\times t)}={{d}_{c}}\lambda ({{d}_{c}}\times t)</math>

===Example 4===
Consider a computer system with three components: a processor, a hard drive and a CD drive in series as shown next. Assume that all three components follow a Weibull failure distribution with the parameters <math>{{\beta }_{1}}=1.5</math> and <math>{{\eta }_{1}}=5000</math> for the processor, <math>{{\beta }_{2}}=2.5</math> and <math>{{\eta }_{2}}=3000</math> for the hard drive, and <math>{{\beta }_{3}}=2</math> and <math>{{\eta }_{3}}=4000</math> for the CD drive. Determine the reliability of the computer system after one year (365 days) of operation, assuming that the CD drive is used only 30% of the time.
 
[[Image:BS5ex4.png|center|250px]]

====Solution to Example 4====
The reliability of the processor after 365 days of operation is given by:

::<math>\begin{align}
{{R}_{processor}}(365)= & {{e}^{-{{\left( \tfrac{365}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}} \\
= & {{e}^{-{{\left( \tfrac{365}{5000} \right)}^{1.5}}}} \\
= & 0.9805\text{ or }98.05%
\end{align}</math>

The reliability of the hard drive after 365 days of operation is given by:

::<math>\begin{align}
{{R}_{harddrive}}(365)= & {{e}^{-{{\left( \tfrac{365}{{{\eta }_{2}}} \right)}^{{{\beta }_{2}}}}}} \\
= & {{e}^{-{{\left( \tfrac{365}{3000} \right)}^{2.5}}}} \\
= & 0.9948\text{ or }99.48%
\end{align}</math>

The reliability of the CD drive after 365 days of operation (taking into account the 30% operation using a duty cycle of 0.3) is given by:

::<math>\begin{align}
{{R}_{CDdrive}}(365)= & {{e}^{-{{\left( \tfrac{{{d}_{c}}\times 365}{{{\eta }_{3}}} \right)}^{{{\beta }_{3}}}}}} \\
= & {{e}^{-{{\left( \tfrac{0.3\times 365}{4000} \right)}^{2}}}} \\
= & 0.9993\text{ or }99.93%
\end{align}</math>
 
[[Image:Fig 5.15.PNG|center|thumb|400px|<div align="center"> Result for the computer system reliability.</div>]]
 

Thus the reliability of the computer system after 365 days of operation is:

::<math>\begin{align}
{{R}_{s}}(365)= & {{R}_{processor}}(365)\cdot {{R}_{harddrive}}(365)\cdot {{R}_{CDdrive}}(365) \\
= & 0.9805\cdot 0.9948\cdot 0.9993 \\
= & 0.9747\text{ or }97.47%
\end{align}</math>
 
This result can be obtained in BlockSim as shown in Figure "Result for the computer system reliability".
 

=Load Sharing=
As presented in earlier chapters, a reliability block diagram (RBD) allows you to graphically represent how the components within a system are reliability-wise connected. In most cases, independence is assumed across the components within the system. For example, the failure of component A does not affect the failure of component B. However, if a system consists of components that are sharing a load, then the assumption of independence no longer holds true.
 

If one component fails, then the component(s) that are still operating will have to assume the failed unit's portion of the load. Therefore, the reliabilities of the surviving unit(s) will change. Calculating the system reliability is no longer an easy proposition. In the case of load sharing components, the change of the failure distributions of the surviving components must be known in order to determine the system's reliability.
 

To illustrate this, consider the a system of two units connected reliability-wise in parallel (Figure "Two units connected reliability-wise in parallel").
 

[[Image:5-16.png|center|200px|<div align="center"> Two units connected reliability-wise in parallel.</div>]]
 

Assume that the units must supply an output of 8 volts and that if both units are operational, each unit is to supply 50% of the total output. If one of the units fails, then the surviving unit supplies 100%. Furthermore, assume that having to supply the entire load has a negative impact on the reliability characteristics of the surviving unit. Since the reliability characteristics of the unit change based on whether both or only one is operating, a life distribution along with a life-stress relationship (as discussed in [http://reliawiki.com/index.php/Statistical_Background#A_Brief_Introduction_to_Life-Stress_Relationships|A Brief Introduction to Life-Stress Relationships]) will be needed to model each component.
 

To illustrate the steps needed, we will create the model starting from raw data. Assume that a total of 20 units were tested to failure at 7, 10 and 15 volts. The test data set is presented in the next table.
 

[[Image:5-17.png|center|450px|]]
 

For this example, Units 1 and 2 are the same component. Therefore, only one set of data was collected. However, it is possible that the load sharing components in a system may not be the same. If that were the case, data would need to be collected for each component.
 

The data set in Table 1 was analyzed using ReliaSoft's ALTA software (as shown in Figure "Using ALTA to calculate component parameters") with the Inverse Power Law as the underlying life-stress relationship and Weibull as the life distribution.
 
 

The estimated model parameters, <math>\beta </math> , <math>K</math> and <math>n</math> , are shown next.
 

::<math>\begin{align}
\beta = & 1.9239 \\
K= & 3.2387\times {{10}^{-7}} \\
n= & 3.4226
\end{align}</math>
 

:Or:
 
::<math>\begin{align}
{{R}_{1}}(t,{{S}_{1}})= & {{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} \\
= & {{e}^{-{{\left( 3.2387\times {{10}^{-7}}S_{1}^{3.4226}t \right)}^{1.9239}}}} \
\end{align}</math>.

 
::<math>{{f}_{1}}(t,{{S}_{1}})=\beta KS_{1}^{n}{{\left( KS_{1}^{n}t \right)}^{\beta -1}}{{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} \ </math>
 

And for this case:

::<math>\begin{align}
{{R}_{1}}(t,{{S}_{1}})= & {{R}_{2}}(t,{{S}_{2}}) \\
{{f}_{1}}(t,{{S}_{1}})= & {{f}_{2}}(t,{{S}_{2}})
\end{align}</math>
 
 
[[Image:Fig 5.17.PNG|thumb|center|450px|<div align="center"> Using ALTA to calculate component parameters.</div>]]
 

Figure "Reliability curves for different voltage output conditions" shows a plot of <math>{{R}_{1}}(t,{{S}_{1}})={{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} = {{e}^{-{{\left( 3.2387\times {{10}^{-7}}S_{1}^{3.4226}t \right)}^{1.9239}}}}\ </math>.
 

Now that the failure properties have been determined using the test data, the reliability of the system at some time, <math>t</math> , can be calculated using the following equation:

 
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>

 
:Where:
 

::<math>\begin{align}
{{S}_{1}}= & {{P}_{1}}S \\
{{S}_{2}}= & {{P}_{2}}S
\end{align}</math>

 
[[Image:5-18.png|center|300px|<div align="center"> Reliability curves for different voltage output conditions. </div>]]
 
[[Image:BS5.19.png|center|300px| Illustrating <math>{{t}_{e}}</math>]]
 

And:
 

:• <math>S</math> is the total load (or required output). 
:• <math>{{P}_{1}}</math> and <math>{{P}_{2}}</math> are the portion of the total load that each unit supports when both units are operational. In this case, <math>{{P}_{1}}={{P}_{2}}=0.5=50%.</math> 
:• <math>{{S}_{1}}</math> and <math>{{S}_{2}}</math> represent the portions of the load that Unit 1 and Unit 2 must support when both units are operational. 
:• <math>{{t}_{{{1}_{e}}}}</math> is the equivalent operating time for Unit 1 if it had been operating at <math>S</math> instead of <math>{{S}_{1}}</math> . A graphical representation of the equivalent time is shown in Figure 5.19, where the curve marked by L represents the low stress (load) and the curve marked by H represents the high stress (load). 
 

::<math>{{t}_{1e}}</math> can be calculated by:
 
::<math>\begin{align}
{{R}_{1}}(t)= & {{R}_{1}}({{t}_{1e}}) \\
{{e}^{-{{(tKS_{1}^{n})}^{\beta }}}}= & {{e}^{-{{({{t}_{1e}}K{{S}^{n}})}^{\beta }}}} \\
tS_{1}^{n}= & {{t}_{1e}}{{S}^{n}} \\
{{t}_{1e}}= & t{{\left( \frac{{{S}_{1}}}{S} \right)}^{n}},\text{ }{{S}_{1}}={{P}_{1}}S \\
\therefore & {{t}_{1e}}=tP_{1}^{n}
\end{align}</math>

 
::<math>{{t}_{2e}}</math> can be calculated the same way, or:

 
::<math>\begin{align}
{{R}_{2}}(t)= & {{R}_{2}}({{t}_{2e}}) \\
\therefore & {{t}_{2e}}=tP_{2}^{n}
\end{align}</math>
 

In this example, the reliability equations for Unit 1 and Unit 2 are the same since they are the same type of component and demonstrate the same failure properties. In addition, the total output is divided equally between the two units (when both units are operating), so <math>{{t}_{1e}}</math> and <math>{{t}_{2e}}</math> will also be the same.
 
The next step is to determine the reliability of the system after 8,760 hours, <math>R(t=8,760)</math> .
Using
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>
 

the system reliability is found to be:
 

::<math>\begin{align}
R(t=8760)= & 0.8567 \\
= & 85.67%
\end{align}</math>
 

===Load Sharing in BlockSim===
BlockSim uses this formulation when computing reliabilities of units in a load sharing configuration. When using the System Reliability Equation window, BlockSim returns a single token for the reliability of units in a load sharing configuration (as well as in the case of standby redundancy, discussed in the next section). As an example, consider the following RBD with Unit 1 in series with a container that includes two load sharing units.
 
[[Image:BS5.19.2.png|center|150px|]]
 
BlockSim will return the system equation as:
 

::<math>{{R}_{System}}=+{{R}_{LS}}\cdot {{R}_{1}}</math>
 

Where <math>{{R}_{LS}}</math> implies a form similar to
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>
 .
BlockSim allows for <math>k</math> -out-of- <math>n</math> units in a load sharing configuration.

===Example 5===
A component has five possible failure modes, <math>A</math> , <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> , <math>{{B}_{C}}</math> and <math>C</math> , and the <math>B</math> modes are interdependent. The system will fail if mode <math>A</math> occurs, mode <math>C</math> occurs or two out of the three <math>B</math> modes occur.
 

Modes <math>A</math> and <math>C</math> each have a Weibull distribution, with a <math>\beta =2</math> and <math>\eta =10,000</math> and 15,000 respectively. Events <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> and <math>{{B}_{C}}</math> each have an exponential distribution with a mean of 10,000 hours.
 

If any <math>B</math> event occurs (i.e. <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> or <math>{{B}_{C}}</math> ), the remaining <math>B</math> events are more likely to occur. Specifically, the mean times of the remaining <math>B</math> events are halved. Determine the reliability at 1000 hours for this component.
====Solution to Example 5====
The first step is to create the RBD. Modes <math>A</math> , <math>C</math> and a load sharing container with the <math>{{B}_{i}}</math> modes must be drawn in series, as illustrated next.
 
[[Image:BS5.19.3.png|center|200px|]]
 

The next step is to define the properties for each block, including those for the container. Setting the failure distributions for modes <math>A</math> and <math>C</math> is simple. The more difficult part is setting the properties for the container and the contained blocks. Based on the problem statement, the <math>B</math> modes are in a 2-out-of-3 load sharing redundancy. When all three are working (i.e. when no <math>B</math> mode has occurred), each block has an exponential distribution with <math>\mu=10,000</math>.
 

The next step is to set the weight proportionality factor. This factor defines the portion of the load that the particular item carries while operating, as well as the load that shifts to the remaining units upon failure of the item. To illustrate, assume three units (1, 2 and 3) are in a load sharing container with weight proportionality factors of 1, 2 and 3 respectively (and a 1-out-of-3 requirement).
 

:• Unit 1 carries <math>\left( \tfrac{1}{1+2+3} \right)=0.166</math> or 16.6% of the total load.
:• Unit 2 carries <math>\left( \tfrac{2}{1+2+3} \right)=0.333</math> or 33.3% of the total load.
:• Unit 3 carries <math>\left( \tfrac{3}{1+2+3} \right)=0.50</math> or 50% of the total load.
 

The actual load on each unit then becomes the product of the entire load defined for the container times the portion carried by that unit. For example, if the container load is 100 lbs, then the portion assigned to Unit 1 will be <math>100\cdot 0.166=16.6</math> lbs.
 

In the current example, all units share the same load and thus have equal weight proportionality factors. Because these factors are relative, if the same number is used for all three items then the results will be the same. Thus, weight proportional factor is set equal to 1 for each item.
 

 
[[Image:Fig 5.23_2.PNG|thumb|center|450px|<div align="center"> Defining Weight Proportional Factor. </div>]]
 

The last properties that need to be defined are the total load and the 2-out-of-3 redundancy. The total load is dependent on how the parameters were computed. In this case, total load was assumed to be 3 when the parameters were computed (i.e. the load per item was 1 when all worked and 1.5 when two worked). This is defined at the container level, as shown in Figure "Defining Weight Proportional Factor".
When all of the parameters have been specified in BlockSim, the reliability at 1,000 hours can be determined. From the Analytical QCP, this is found to be 98.1048%.

=Standby Components=

In the previous section, the case of a system with load sharing components was presented. This is a form of redundancy with dependent components. That is, the failure of one component affects the failure of the other(s). This section presents another form of redundancy: standby redundancy. In standby redundancy the redundant components are set to be under a lighter load condition (or no load) while not needed and under the operating load when they are activated.
 

In standby redundancy the components are set to have two states: an active state and a standby state. Components in standby redundancy have two failure distributions, one for each state. When in the standby state, they have a quiescent (or dormant) failure distribution and when operating, they have an active failure distribution.
 

In the case that both quiescent and active failure distributions are the same, the units are in a simple parallel configuration (also called a hot standby configuration). When the rate of failure of the standby component is lower in quiescent mode than in active mode, that is called a warm standby configuration. When the rate of failure of the standby component is zero in quiescent mode (i.e. the component cannot fail when in standby), that is called a cold standby configuration.

===Simple Standby Configuration===

Consider two components in a standby configuration. Component 1 is the active component with a Weibull failure distribution with parameters <math>\beta </math> = 1.5 and <math>\eta </math> = 1,000. Component 2 is the standby component. When Component 2 is operating, it also has a Weibull failure distribution with <math>\beta </math> = 1.5 and <math>\eta </math> = 1,000. Furthermore, assume the following cases for the quiescent distribution.
 
:• Case 1: The quiescent distribution is the same as the active distribution (hot standby). 
:• Case 2: The quiescent distribution is a Weibull distribution with <math>\beta </math> = 1.5 and <math>\eta </math> = 2000 (warm standby). 
:• Case 3: The component cannot fail in quiescent mode (cold standby). 
 
In this case, the reliability of the system at some time, <math>t</math> , can be obtained using the following equation:
 

::<math>R(t)={{R}_{1}}(t)+\underset{0}{\overset{t}{\mathop \int }}\,{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x)\cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}dx \ </math>
 

Where:
 
:• <math>{{R}_{1}}</math> is the reliability of the active component. 
:• <math>{{f}_{1}}</math> is the <math>pdf</math> of the active component. 
:• <math>{{R}_{2;SB}}</math> is the reliability of the standby component when in standby mode (quiescent reliability). 
:• <math>{{R}_{2;A}}</math> is the reliability of the standby component when in active mode. 
:• <math>{{t}_{e}}</math> is the equivalent operating time for the standby unit if it had been operating at an active mode, such that: 

 
<math>{{R}_{2;SB}}(x)={{R}_{2;A}}({{t}_{e}})</math> 
 
The second equation above can be solved for <math>{{t}_{e}}</math> and substituted into the first equation above.
Figure "Standby container" illustrates the example as entered in BlockSim using a standby container.
 
 
[[Image:5.24.png|center|150px|<div align="center"> Standby container.</div>]]

 

The active and standby blocks are within a container, which is used to specify standby redundancy. Since the standby component has two distributions (active and quiescent), the Block Properties window of the standby block has two pages for specifying each one. Figures "Defining the active failure distribution" and "Defining the quiescent failure distribution" illustrate these pages.

 

The system reliability results for 1000 hours are given in the following table:
 
[[Image:5-24.png|center|150px|]]
 

Note that even though the <math>\beta </math> value for the quiescent distribution is the same as in the active distribution, it is possible that the two can be different. That is, the failure modes present during the quiescent mode could be different from the modes present during the active mode. In that sense, the two distribution types can be different as well (e.g. lognormal when quiescent and Weibull when active).
 

In many cases when considering standby systems, a switching device may also be present that switches from the failed active component to the standby component. The reliability of the switch can also be incorporated into
<math>R(t)={{R}_{1}}(t)+\underset{0}{\overset{t}{\mathop \int }}\,{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x)\cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}dx \ </math>

as presented in the next section.
 

BlockSim's System Reliability Equation window returns a single token for the reliability of units in a standby configuration. This is the same as the load sharing case presented in the previous section.
 

[[Image: Fig 5.25.PNG|thumb|center|400px|<div align="center"> Defining the active failure distribution </div>]]
 
 

[[Image:Fig 5.26.PNG|thumb|center|400px|<div align="center"> Defining the quiescent failure distribution </div>]]
 

===Reliability of Standby Systems with a Switching Device===

In many cases when dealing with standby systems, a switching device is present that will switch to the standby component when the active component fails. Therefore, the failure properties of the switch must also be included in the analysis.
 

[[Image:BS5.26.2.png|center|150px|]]
 

In most cases when the reliability of a switch is to be included in the analysis, two probabilities can be considered. The first and most common one is the probability of the switch performing the action (i.e. switching) when requested to do so. This is called Switch Probability per Request in BlockSim and is expressed as a static probability (e.g. 90%). The second probability is the quiescent reliability of the switch. This is the reliability of the switch as it ages (e.g. the switch might wear out with age due to corrosion, material degradation, etc.). Thus it is possible for the switch to fail before the active component fails. However, a switch failure does not cause the system to fail, but rather causes the system to fail only if the switch is needed and the switch has failed. For example, if the active component does not fail until the mission end time and the switch fails, then the system does not fail. However, if the active component fails and the switch has also failed, then the system cannot be switched to the standby component and it therefore fails.
 

In analyzing standby components with a switching device, either or both failure probabilities (during the switching or while waiting to switch) can be considered for the switch, since each probability can represent different failure modes. For example, the switch probability per request may represent software-related issues or the probability of detecting the failure of an active component, and the quiescent probability may represent wear-out type failures of the switch.
 

To illustrate the formulation, consider the previous example that assumes perfect switching. To examine the effects of including an imperfect switch, assume that when the active component fails there is a 90% probability that the switch will switch from the active component to the standby component. In addition, assume that the switch can also fail due to a wear-out failure mode described by a Weibull distribution with <math>\beta </math> = 1.7 and <math>\eta </math> = 5000.
 

Therefore, the reliability of the system at some time, <math>t</math> , is given by the following equation.

::<math>\begin{align}
R(t)= & {{R}_{1}}(t) \\
& +\underset{0}{\overset{t}{\mathop \int }}\,\{{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x) \\
& \cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}\cdot {{R}_{SW;Q}}(x)\cdot {{R}_{SW;REQ}}(x)\}dx
\end{align}</math>

Where:
 
:• <math>{{R}_{1}}</math> is the reliability of the active component. 
:• <math>{{f}_{1}}</math> is the <math>pdf</math> of the active component. 
:• <math>{{R}_{2;SB}}</math> is the reliability of the standby component when in standby mode (quiescent reliability). 
:• <math>{{R}_{2;A}}</math> is the reliability of the standby component when in active mode. 
:• <math>{{R}_{SW;Q}}</math> is the quiescent reliability of the switch. 
:• <math>{{R}_{SW;REQ}}</math> is the switch probability per request. 
:• <math>{{t}_{e}}</math> is the equivalent operating time for the standby unit if it had been operating at an active mode. 
This problem can be solved in BlockSim by including these probabilities in the container's properties, as shown in Figures "Standby container (switch) failure probabilities while attempting to switch" and "Standby container (switch) failure distribution while waiting to switch". In BlockSim, the standby container is acting as the switch.

 

[[Image:Fig 5.28_2.PNG|thumb|center|400px|<div align="center"> Standby container (switch) failure probabilities while attempting to switch </div> ]]
 

 

[[Image:Fig 5.27.PNG|thumb|center|500px|<div align="center"> Standby container (switch) failure distribution while waiting to switch </div>]]
 

 

Note that there are additional properties that can be specified in BlockSim for a switch, such as Switch Restart Probability, No. of Restarts and Switch Delay Time. In many applications, the switch is re-tested (or re-cycled) if it fails to switch the first time. In these cases, it might be possible that it switches in the second or third, or <math>{{n}^{th}}</math> attempt.
 
The Switch Restart Probability specifies each additional attempt's probability of successfully switching and the Finite Restarts specifies the total number of attempts. Note that the Switch Restart Probability specifies the probability of success of each trial (or attempt). The probability of success of <math>n</math> consecutive trials is calculated by BlockSim using the binomial distribution and this probability is then incorporated into equation above. The Switch Delay Time property is related to repairable systems and is considered in BlockSim only when using simulation. When using the analytical solution (i.e. for a non-repairable system), this property is ignored.
 

Solving the analytical solution (as given by Equation above), the following results are obtained.
 

[[Image:5-30.png|center|350px|]]
 

From the table above, it can be seen that the presence of a switching device has a significant effect on the reliability of a standby system. It is therefore important when modeling standby redundancy to incorporate the switching device reliability properties. It should be noted that this methodology is not the same as treating the switching device as another series component with the standby subsystem. This would be valid only if the failure of the switch resulted in the failure of system (e.g. switch failing open). In equation above, the Switch Probability per Request and quiescent probability are present only in the second term of the equation. Treating these two failure modes as a series configuration with the standby subsystem would imply that they are also present when the active component is functioning (i.e. first term of equation above). This is invalid and would result in the underestimation of the reliability of the system. In other words, these two failure modes become significant only when the active component fails.
 

As an example, and if we consider the warm standby case, the reliability of the system without the switch is 70.57% at 1000 hours. If the system was modeled so that the switching device was in series with the warm standby subsystem, the result would have been:

 

::<math>\begin{align}
{{R}_{S}}(1000)= & {{R}_{Standby}}(1000)\cdot {{R}_{sw,Q(1000)}}\cdot {{R}_{sw,req}} \\
= & 0.7057\cdot 0.9372\cdot 0.9 \\
= & 0.5952
\end{align}</math>

 

In the case where a switch failure mode causes the standby subsystem to fail, then this mode can be modeled as an individual block in series with the standby subsystem.

===Example 6===
Consider a car with four new tires and a full-size spare. Assume the following failure characteristics:
 
:• The tires follow a Weibull distribution with a <math>\beta=4</math> and an <math>\eta =</math> 40,000 miles while on the car due to wear. 
:• The tires also have a probability of failing due to puncture or other causes. For this, assume a constant rate for this occurrence with a probability of 1 every 50,000 miles. 
:• When not on the car (i.e. is a spare), a tire's probability of failing also has a Weibull distribution with a <math>\beta =</math> 2 and <math>\eta =</math> 120,000 miles. 
 
Assume a mission of 1,000 miles. If a tire fails during this trip, it will be replaced with the spare. However, the spare will not be repaired during the trip. In other words, the trip will continue with the spare on the car and if the spare fails the system will fail. Determine the probability of system failure.
====Solution to Example 6====
Active failure distribution for tires:
 
:• Due to wear-out, Weibull <math>\beta =4</math> and <math>\eta =40,000</math> miles. 
:• Due to random puncture, exponential <math>\mu =50,000.</math> 
:• The quiescent failure distribution is a Weibull distribution with <math>\beta =2</math> and <math>\eta =120,000</math> miles. 
 
The block diagram for each tire has two blocks in series, one block representing the wear-out mode and the other the random puncture mode, as shown next:
 
 
[[Image:small5.png|center|250px|]]
 

There are five tires, four active and one standby (represented in the diagram by a standby container with a 4-out-of-5 requirement), as shown next:
 
[[Image:BStirecontainer.png|center|250px|]]

For the standby Wear block, set the active failure and the quiescent distributions, but for the Puncture block, set only the active puncture distribution (because the tire cannot fail due to puncture while stored). Using BlockSim, the probability of system failure is found to be 0.003 or 0.3%.
 

=Note Regarding Numerical Integration Solutions=

Load sharing and standby solutions in BlockSim are performed using numerical integration routines. As with any numerical analysis routine, the solution error depends on the number of iterations performed, the step size chosen and related factors, plus the behavior of the underlying function. By default, BlockSim uses a certain set of preset factors. In general, these defaults are sufficient for most problems. If a higher precision or verification of the precision for a specific problem is required, BlockSim's preset options can be modified and/or the integration error can be assessed using the Integration Parameters... option for each container. For more details, you can refer to the documentation on the Algorithm Setup window in the BlockSim 8 User's help files.

Time-Dependent System Reliability (Analytical)

2012-06-21T15:54:07Z

Dingzhou Cao: /* Solution to Example 5 */

{{Template:bsbook|5}}

In the previous chapter, different system configuration types were examined, as well as different methods for obtaining the system's reliability function analytically. Because the reliabilities in the problems presented were treated as probabilities (e.g. <math>P(A)</math> , <math>{{R}_{i}}</math> ), the reliability values and equations presented were referred to as static (not time-dependent). Thus, in the prior chapter, the life distributions of the components were not incorporated in the process of calculating the system reliability. In this chapter, time dependency in the reliability function will be introduced. We will develop the models necessary to observe the reliability over the life of the system, instead of at just one point in time. In addition, performance measures such as failure rate, MTTF and warranty time will be estimated for the entire system. The methods of obtaining the reliability function analytically remain identical to the ones presented in the previous chapter, with the exception that the reliabilities will be functions of time. In other words, instead of dealing with <math>{{R}_{i}}</math> , we will use <math>{{R}_{i}}(t)</math> . All examples in this chapter assume that no repairs are performed on the components.

=Analytical Life Predictions=
The analytical approach presented in the prior chapter involved the determination of a mathematical expression that describes the reliability of the system, expressed in terms of the reliabilities of its components. So far we have estimated only static system reliability (at a fixed time). For example, in the case of a system with three components in series, the system's reliability equation was given by:
 

::<math>{{R}_{s}}={{R}_{1}}\cdot {{R}_{2}}\cdot {{R}_{3}} \ </math>
 

The values of <math>{{R}_{1}}</math> , <math>{{R}_{2}}</math> and <math>{{R}_{3}}</math> were given for a common time and the reliability of the system was estimated for that time. However, since the component failure characteristics can be described by distributions, the system reliability is actually time-dependent. In this case, the equation above can be rewritten as:
 

::<math>{{R}_{s}}(t)={{R}_{1}}(t)\cdot {{R}_{2}}(t)\cdot {{R}_{3}}(t)</math>

 

The reliability of the system for any mission time can now be estimated. Assuming a Weibull life distribution for each component, the first equation above can now be expressed in terms of each component's reliability function, or:

 

[[Category:Completed Theoretical Review]]

::<math>{{R}_{s}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}}\cdot {{e}^{-{{\left( \tfrac{t}{{{\eta }_{2}}} \right)}^{{{\beta }_{2}}}}}}\cdot {{e}^{-{{\left( \tfrac{t}{{{\eta }_{3}}} \right)}^{{{\beta }_{3}}}}}}</math>

 

In the same manner, any life distribution can be substituted into the system reliability equation. Suppose that the times-to-failure of the first component are described with a Weibull distribution, the times-to-failure of the second component with an exponential distribution and the times-to-failure of the third component with a normal distribution. Then the first equation above can be written as:
 

::<math>{{R}_{s}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}}\cdot {{e}^{-{{\lambda }_{2}}t}}\cdot \left[ 1-\Phi \left( \frac{t-{{\mu }_{3}}}{{{\sigma }_{3}}} \right) \right]</math>

 

It can be seen that the biggest challenge is in obtaining the system's reliability function in terms of component reliabilities, which has already been discussed in depth. Once this has been achieved, calculating the reliability of the system for any mission duration is just a matter of substituting the corresponding component reliability functions into the system reliability equation.
===Advantages of the Analytical Method===
The primary advantage of the analytical solution is that it produces a mathematical expression that describes the reliability of the system. Once the system's reliability function has been determined, other calculations can then be performed to obtain metrics of interest for the system. Such calculations include:
 
:• Determination of the system's <math>pdf.</math> 
:• Determination of warranty periods. 
:• Determination of the system's failure rate. 
:• Determination of the system's MTTF. 
 
In addition, optimization and reliability allocation techniques can be used to aid engineers in their design improvement efforts. Another advantage in using analytical techniques is the ability to perform static calculations and analyze systems with a mixture of static and time-dependent components. Finally, the reliability importance of components over time can be calculated with this methodology.
===Disadvantages of the Analytical Method===
The biggest disadvantage of the analytical method is that formulations can become very complicated. The more complicated a system is, the larger and more difficult it will be to analytically formulate an expression for the system's reliability. For particularly detailed systems this process can be quite time-consuming, even with the use of computers. Furthermore, when the maintainability of the system or some of its components must be taken into consideration, analytical solutions become intractable. In these situations, the use of simulation methods may be more advantageous than attempting to develop a solution analytically. Simulation methods are presented in later chapters.
===Looking at a Simple ''Complex'' System Analytically===

The complexity involved in an analytical solution can be best illustrated by looking at the simple ''complex'' system with 15 components, as shown in Figure " An RBD of a complex system".
 
[[Image:5-1.png|center|400px|<div align="center"> An RBD of a complex system.</div>]]

 
The system reliability for this system (computed using BlockSim) is shown next. The first solution is provided using BlockSim's symbolic solution. In symbolic mode, BlockSim breaks the equation into segments, identified by tokens, that need to be substituted into the final system equation for a complete solution. This creates algebraic solutions that are more compact than if the substitutions were made.
 

::<math>\begin{align}
{{R}_{System}}= & D2\cdot D3\cdot {{R}_{L}} \\
D3= & +{{R}_{K}}\cdot IK \\
IK= & +{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}} \\
D2 = & +{{R}_{A}}\cdot {{R}_{E}}\cdot IE \\
IE = & -D1\cdot {{R}_{M}}\cdot {{R}_{N}}+{{R}_{M}}\cdot {{R}_{N}}+D1 \\
D1 = & +{{R}_{D}}\cdot ID \\
ID = & -{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}}
\end{align}</math>
 
Substituting the terms yields:
 

::<math>\begin{align}
{{R}_{System}}= & {{R}_{A}}\cdot {{R}_{E}}\cdot {{R}_{L}}\cdot {{R}_{K}} \\
& \cdot \{({{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})\cdot {{R}_{M}}\cdot {{R}_{N}} \\
& +{{R}_{M}}\cdot {{R}_{N}}-{{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}}\} \\
& \cdot \{{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}}\}
\end{align}</math>

 
BlockSim's automatic algebraic simplification would yield the following format for the above solution:
 

::<math>\begin{align}
{{R}_{System}}= & (({{R}_{A}}\cdot {{R}_{E}}(-({{R}_{D}}(-{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})){{R}_{M}}\cdot {{R}_{N}} \\
& +{{R}_{M}}\cdot {{R}_{N}} \\
& +({{R}_{D}}(-{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})))) \\
& ({{R}_{K}}({{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& +RI\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}})){{R}_{L}}) \
\end{align}</math>

 
In this equation, each <math>{{R}_{i}}</math> represents the reliability function of a block. For example, if <math>{{R}_{A}}</math> has a Weibull distribution, then each <math>{{R}_{A}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{A}}} \right)}^{{{\beta }_{A}}}}}}</math> and so forth. Substitution of each component's reliability function in the last <math>{{R}_{System}}</math> equation above will result in an analytical expression for the system reliability as a function of time, or <math>{{R}_{s}}(t)</math> , which is the same as <math>(1-cd{{f}_{System}}).</math>

===Obtaining Other Functions of Interest===
Once the system reliability equation (or the cumulative density function, <math>cdf</math> ) has been determined, other functions and metrics of interest can be derived.
 
Consider the following simple system:
 
[[Image:5-2.png|center|150px|<div align="center"> Simple two-component system. </div>]]

 
Furthermore, assume that component 1 follows an exponential distribution with a mean of 10,000 (<math>\mu =10,000,</math> <math>\lambda =1/10,000)</math> and component 2 follows a Weibull distribution with <math>\beta =6</math> and <math>\eta =10,000</math> . The reliability equation of this system is:

 
::<math>\begin{align}
{{R}_{S}}(t)= & {{R}_{1}}(t)\cdot {{R}_{2}}(t) \\
= & {{e}^{-\lambda t}}\cdot {{e}^{-{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \\
= & {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \
\end{align}</math>

 
The system <math>cdf</math> is:

 
::<math>\begin{align}
{{F}_{S}}(t)= & 1-({{R}_{1}}(t)\cdot {{R}_{2}}(t)) \\
= & 1-\left( {{e}^{-\lambda t}}\cdot {{e}^{-{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \right) \\
= & 1-\left( {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)
\end{align}</math>

 
{{analytical system pdf}}

====Conditional Reliability====
Conditional reliability is the probability of a system successfully completing another mission following the successful completion of a previous mission. The time of the previous mission and the time for the mission to be undertaken must be taken into account for conditional reliability calculations. The system's conditional reliability function is given by:

::<math>R(T,t)=\frac{R(T+t)}{R(T)}</math>

Equation above gives the reliability for a new mission of duration <math>t</math> having already accumulated <math>T</math> hours of operation up to the start of this new mission. The system is evaluated to assure that it will start the next mission successfully.
 
For the system in Figure "Simple two-component system.", the reliability for mission of <math>t=1,000</math> hours, having an age of <math>T=500</math> hours, is:
 

::<math>\begin{align}
{{R}_{S}}(T=500,t=1000)= & \frac{R(T+t)}{R(T)} \\
= & \frac{R(1500)}{R(500)} \\
= & \frac{{{e}^{-\tfrac{1500}{10,000}}}\cdot {{e}^{-{{\left( \tfrac{1500}{10,000} \right)}^{6}}}}}{{{e}^{-\tfrac{500}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{500}{10,000} \right)}^{6}}}}} \\
= & 0.9048=90.48%
\end{align}</math>

 

====Conditional Reliability for Components====

Now in this formulation, it was assumed that the accumulated age was equivalent for both units. That is, both started life at zero and aged to 500. It is possible to consider an individual component that has already accumulated some age (used component) in the same formulation. To illustrate this, assume that component 2 started life with an age of T=100. Then the reliability equation of the system, as given in <math>{{R}_{S}}(t)= {{R}_{1}}(t)\cdot {{R}_{2}}(t)</math>, would need to be modified to include a conditional term for 2, or:
 

::<math>{{R}_{S}}(t)={{R}_{1}}(t)\cdot \frac{{{R}_{2}}({{T}_{2}}+t)}{{{R}_{2}}({{T}_{2}})} \ </math>

 

In BlockSim, the start age input box may be used to specify a starting age greater than zero.
{{system failure rate analytical}}

====System Mean Life (Mean Time To Failure)====
The mean life (or mean time to failure, MTTF) can be obtained by integrating the system reliability function from zero to infinity:
 

::<math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math>

 

The mean time is a performance index and does not provide any information about the behavior of the failure distribution of the system.
 
For the system in Figure "Simple two-component system":
 

::<math>\begin{align}
MTTF= & \int_{0}^{\infty }\left( {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)dt \\
= & 5978.9
\end{align}</math>

 

====Warranty Period and BX Life====
Sometimes it is desirable to know the time value associated with a certain reliability. Warranty periods are often calculated by determining what percentage of the failure population can be covered financially and estimating the time at which this portion of the population will fail. Similarly, engineering specifications may call for a certain BX life, which also represents a time period during which a certain proportion of the population will fail. For example, the B10 life is the time in which 10% of the population will fail.
This is obtained by setting <math>{{R}_{S}}(t)</math> to the desired value and solving for <math>t.</math>
 
For the system in Figure "Simple two-component system":
 

::<math>{{R}_{s}}\left( t \right)={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>
 

To compute the time by which reliability would be equal to 90%, equation above is recast as follows and solved for <math>t.</math>

 
::<math>0.90={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>

 

In this case, <math>t=1053.59</math> . Equivalently, the B10 life for this system is also <math>1053.59</math> .
Except for some trivial cases, a closed form solution for <math>t</math> cannot be obtained. Thus, it is necessary to solve for <math>t</math> using numerical methods. BlockSim uses numerical methods.

===Example 1===
Consider a system consisting of three exponential units in series with the following failure rates (in failures per hour): <math>{{\lambda }_{1}}</math> = 0.0002, <math>{{\lambda }_{2}}</math> = 0.0005 and <math>{{\lambda }_{3}}</math> = 0.0001.
 
:• Obtain the reliability equation for the system. 
:• What is the reliability of the system after 150 hours of operation? 
:• Obtain the system's <math>pdf.</math> 
:• Obtain the system's failure rate equation. 
:• What is the MTTF for the system? 
:• What should the warranty period be for a 90% reliability? 
====Solution to Example 1====
 
:• The analytical expression for the reliability of the system is given by:

::<math>\begin{align}
{{R}_{s}}(t)= & {{R}_{1}}(t)\cdot {{R}_{2}}(t)\cdot {{R}_{3}}(t) \\
= & {{e}^{-{{\lambda }_{1}}t}}\cdot {{e}^{-{{\lambda }_{2}}t}}\cdot {{e}^{-{{\lambda }_{1}}t}} \\
= & {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}
\end{align}</math>

 

At 150 hours of operation, the reliability of the system is:
 

::<math>\begin{align}
{{R}_{s}}(t)= & {{e}^{-(0.0002+0.0005+0.0001)150}} \\
= & 0.8869\text{ or }88.69%
\end{align}</math>

 

:• In order to obtain the system's <math>pdf</math> , the derivative of the reliability equation given in the first equation above is taken with respect to time, or:
 
::<math>\begin{align}
{{f}_{s}}(t)= & -\frac{d[{{R}_{s}}(t)]}{dt} \\
= & -\frac{d\left[ {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}} \right]}{dt} \\
= & ({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})\cdot {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}
\end{align}</math>

:• The system's failure rate can now be obtained simply by dividing the system's <math>pdf</math> given in the equation above by the system's reliability function given in the first equation above, and:

::<math>\begin{align}
{{\lambda }_{s}}\left( t \right)= & \frac{{{f}_{s}}\left( t \right)}{{{R}_{s}}\left( t \right)} \\
= & \frac{({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})\cdot {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}}{{{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}} \\
= & ({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}) \\
= & 0.0008\text{ }fr/hr
\end{align}</math>

 
:• Combining <math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math> and the first equation above, the system's MTTF can be obtained:

 
::<math>\begin{align}
MTTF= & \int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \\
= & \int_{0}^{\infty }{{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}dt \\
= & \frac{1}{({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})} \\
= & 1250\text{ }hr
\end{align}</math>
 

:• Solving the first equation above with respect to time will yield the corresponding warranty period for a 90% reliability. In this case, the system reliability equation is simple and a closed form solution exists. The warranty time can now be found by solving:

 
::<math>\begin{align}
t= & -\frac{\ln (R)}{{{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}} \\
= & -\frac{\ln (0.9)}{0.0008} \\
= & 131.7\text{ }hr
\end{align}</math>
 

Thus, the warranty period should be 132 hours.

===Example 2===
Consider the system shown in Figure "Complex bridge system in Example 2".
 
[[Image:BS5.5.png|center|400px|<div align="center"> Complex bridge system in Example 2. </div>]]
 

Components <math>A</math> through <math>E</math> are Weibull distributed with <math>\beta =1.2</math> and <math>\eta =1230</math> hours. The starting and ending blocks cannot fail. 
Determine the following:
 
:• The reliability equation for the system and its corresponding plot. 
:• The system's <math>pdf</math> and its corresponding plot. 
:• The system's failure rate equation and the corresponding plot. 
:• The <math>MTTF</math> . 
:• The warranty time for a 90% reliability. 
:• The reliability for a 200-hour mission, if it is known that the system has already successfully operated for 200 hours. 
====Solution====
 
The first step is to obtain the reliability function for the system. The methods described in the previous chapter can be employed, such as the event space or path-tracing methods. Using BlockSim, the following reliability equation is obtained:
 

::<math>\begin{align}
{{R}_{s}}(t)= & ({{R}_{Start}}\cdot {{R}_{End}}(2{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}))
\end{align}</math>
 

Note that since the starting and ending blocks cannot fail, <math>{{R}_{Start}}=1</math> and <math>{{R}_{End}}=1,</math> equation above can be reduced to:

 

::<math>\begin{align}
{{R}_{s}}(t)= & 2\cdot {{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}
\end{align}</math>

 

Where <math>{{R}_{A}}</math> is the reliability equation for Component A, or:

 

::<math>{{R}_{A}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{A}}} \right)}^{{{\beta }_{A}}}}}}</math>

 

::<math>{{R}_{B}}</math> is the reliability equation for Component <math>B</math> , etc.
 

Since the components in this example are identical, the system reliability equation can be further reduced to:

 

::<math>{{R}_{s}}(t)=2R{{(t)}^{2}}+2R{{(t)}^{3}}-5R{{(t)}^{4}}+2R{{(t)}^{5}}</math>

 

Or, in terms of the failure distribution:

 

::<math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>

 

The corresponding plot is given in Figure "Reliability plot for the system".
 
[[Image:BS5.6.png|center|300px|<div align="center"> Reliability plot for the system. </div>]]
 
 

In order to obtain the system's <math>pdf</math> , the derivative of the reliability equation given above is taken with respect to time, resulting in:
 

::<math>\begin{align}
{{f}_{s}}(t)= & 4\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+6\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \\
& -20\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+10\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}
\end{align}</math>

 

The <math>pdf</math> can now be plotted for different time values, <math>t</math> , as shown in Figure "''pdf'' plot for the system".
 

The system's failure rate can now be obtained by dividing the system's <math>pdf</math> given in equation above by the system's reliability function given in <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>, or:

 

::<math>\begin{align}
{{\lambda }_{s}}(t)= & \frac{4\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+6\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}{2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}} \\
& +\frac{-20\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+10\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}{2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}
\end{align}</math>

 

The corresponding plot is given in Figure "Failure rate for the system".
 
[[Image:BS5.7.png|center|300px|<div align="center"> ''pdf'' plot for the system.</div>]]
 

The <math>MTTF</math> of the system is obtained by integrating the system's reliability function given by <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math> from time zero to infinity, as given by <math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math>. Using BlockSim's Analytical QCP, an <math>MTTF</math> of 1007.8 hours is calculated, as shown in Figure "MTTF of the system".
 

The warranty time can be obtained by solving <math>{{R}_{s}}(t)</math> with respect to time for a system reliability <math>{{R}_{s}}=0.9</math> . Using the Analytical QCP and selecting the 
Warranty Time option, a time of 372.72 hours is obtained, as shown in Figure "Time at which ''R''=0.9 or 90% for the system".
 
[[Image:BS5.8.png|center|300px|<div align="center"> Failure rate for the system.</div>]]
 
[[Image:Fig 5.9.PNG|thumb|center|400px|<div align="center"> MTTF of the system. </div>]]
 
[[Image:Fig 5.10.PNG|thumb|center|400px|<div align="center"> Time at which ''R''=0.9 or 90% for the system.</div>]]
 
[[Image:Fig 5.11.PNG|thumb|center|400px|<div align="center"> Conditional reliability calculation for the system.</div>]]
 

Lastly, the conditional reliability can be obtained using <math>R(T,t)=\frac{R(T+t)}{R(T)}</math> and <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>, or:
 

::<math>\begin{align}
R(200,200)= & \frac{R(400)}{R(200)} \\
= & \frac{0.883825}{0.975321} \\
= & 0.906189
\end{align}</math>

 

This can be calculated using BlockSim's Analytical QCP, as shown in Figure "Conditional reliability calculation for the system".
 

=Approximating the System CDF=

In many cases, it is valuable to fit a distribution that represents the system's times-to-failure. This can be useful when the system is part of a larger assembly and may be used for repeated calculations or in calculations for other systems. In cases such as this, it can be useful to characterize the system's behavior by fitting a distribution to the overall system and calculating parameters for this distribution. This is equivalent to fitting a single distribution to describe <math>{{R}_{S}}(t</math> ). In essence, it is like reducing the entire system to a component in order to simplify calculations.
 
For the system in Figure below:
[[Image:5-2.png|center|150px|<div align="center">Simple two-component system. </div>]]
 

::<math>{{R}_{S}}(t)={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>

 
To compute an approximate reliability function for this system, <math>{{R}_{A}}(t)\simeq {{R}_{S}}(t)</math> , one would compute <math>n</math> pairs of reliability and time values and then fit a single distribution to the data, or:

 

::<math>\begin{align}
{{R}_{S}}(t= & 10,396.7)=10% \\
{{R}_{S}}(t= & 9,361.9)=20% \\
& ... \\
{{R}_{S}}(t= & 1,053.6)=90%
\end{align}</math>

 

A single distribution, <math>{{R}_{A}}(t)</math> , that approximates <math>{{R}_{S}}(t)</math> can now be computed from these pairs using life data analysis methods. If using the Weibull++ software, one would enter the values as free form data.

===Example 3===
Compute a single Weibull distribution approximation for the system in Example 2.
====Solution to Example 3====
The system in the previous example, shown in Figure "Complex bridge system in Example 2", can be approximated by use of a 2-parameter Weibull distribution with <math>\beta =2.02109</math> and <math>\eta =1123.51</math> . In BlockSim, this is accomplished by representing the entire system as one distribution by going to the Distribution Fit area, as shown next.

 
[[Image:Fig 5.13.PNG|center|200px|<div align="center"> Representing a system with a distribution.</div>]]

 
[[Image:Fig 5.12.PNG|thumb|center|350px|<div align="center"> Distribution Estimator.</div>]]
 
Clicking inside the Distribution Fit area opens the Distribution Estimator window (Figure "Distribution Fitting window"). In this window you can select a distribution to represent the data. BlockSim will then generate a number of system failure times based on the system's reliability function. The system's reliability function can be used to solve for a time value associated with that unreliability value. The distribution of the generated time values can then be fitted to a probability distribution function.
 

Consider a value of <math>F(t)=0.11</math> . Using the system's reliability equation and solving for time, the corresponding time-to-failure for a 0.11 unreliability can be calculated. 

For the system of Example 2, the time for a 0.11 unreliability is 389.786 hours. 

When enough points have been generated, the selected distribution will be fitted to this data set and the distribution's parameters will be returned. In addition, if ReliaSoft's Weibull++ is installed, the generated data can be viewed/analyzed using a Weibull++ instance, as shown in Figure "Using Weibull++ to calculate distribution parameters".
 
[[Image:Fig 5.14.PNG|thumb|center|450px|<div align="center"> Using Weibull++ to calculate distribution parameters.</div>]]
 
It is recommended that the analyst examine the fit to ascertain the applicability of the approximation.

=Duty Cycle=

Components of a system may not operate continuously during a system's mission, or may be subjected to loads greater or lesser than the rated loads during system operation. To model this, a factor called the Duty Cycle ( <math>{{d}_{c}}</math> ) is used. The duty cycle may also be used to account for changes in environmental stress, such as temperature changes, that may effect the operation of a component. The duty cycle is a positive value, with a default value of 1 representing continuous operation at rated load, and any values other than 1 representing other load values with respect to the rated load value (or total operating time). A duty cycle value higher than 1 indicates a load in excess of the rated value. A duty cycle value lower than 1 indicates that the component is operating at a load lower than the rated load or not operating continuously during the system's mission. For instance, a duty cycle of 0.5 may be used for a component that operates only half of the time during the system's mission.
 
The reliability metrics for a component with a duty cycle are calculated as follows. Let <math>{{d}_{c}}</math> represent the duty cycle during a particular mission of the component, <math>t</math> represent the mission time and <math>{t}'</math> represent the accumulated age. Then:
 

::<math>{t}'={{d}_{c}}\times t</math>

 

The reliability equation for the component is:
 

::<math>R({t}')=R({{d}_{c}}\times t)</math>
 

The component ''pdf'' is:
 

::<math>f({t}')=-\frac{d(R({t}'))}{dt}=-\frac{d(R({{d}_{c}}\times t))}{dt}={{d}_{c}}f({{d}_{c}}\times t)</math>
 

The failure rate of the component is:
 

::<math>\lambda ({t}')=\frac{f({t}')}{R({t}')}=\frac{{{d}_{c}}f({{d}_{c}}\times t)}{R({{d}_{c}}\times t)}={{d}_{c}}\lambda ({{d}_{c}}\times t)</math>

===Example 4===
Consider a computer system with three components: a processor, a hard drive and a CD drive in series as shown next. Assume that all three components follow a Weibull failure distribution with the parameters <math>{{\beta }_{1}}=1.5</math> and <math>{{\eta }_{1}}=5000</math> for the processor, <math>{{\beta }_{2}}=2.5</math> and <math>{{\eta }_{2}}=3000</math> for the hard drive, and <math>{{\beta }_{3}}=2</math> and <math>{{\eta }_{3}}=4000</math> for the CD drive. Determine the reliability of the computer system after one year (365 days) of operation, assuming that the CD drive is used only 30% of the time.
 
[[Image:BS5ex4.png|center|250px]]

====Solution to Example 4====
The reliability of the processor after 365 days of operation is given by:

::<math>\begin{align}
{{R}_{processor}}(365)= & {{e}^{-{{\left( \tfrac{365}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}} \\
= & {{e}^{-{{\left( \tfrac{365}{5000} \right)}^{1.5}}}} \\
= & 0.9805\text{ or }98.05%
\end{align}</math>

The reliability of the hard drive after 365 days of operation is given by:

::<math>\begin{align}
{{R}_{harddrive}}(365)= & {{e}^{-{{\left( \tfrac{365}{{{\eta }_{2}}} \right)}^{{{\beta }_{2}}}}}} \\
= & {{e}^{-{{\left( \tfrac{365}{3000} \right)}^{2.5}}}} \\
= & 0.9948\text{ or }99.48%
\end{align}</math>

The reliability of the CD drive after 365 days of operation (taking into account the 30% operation using a duty cycle of 0.3) is given by:

::<math>\begin{align}
{{R}_{CDdrive}}(365)= & {{e}^{-{{\left( \tfrac{{{d}_{c}}\times 365}{{{\eta }_{3}}} \right)}^{{{\beta }_{3}}}}}} \\
= & {{e}^{-{{\left( \tfrac{0.3\times 365}{4000} \right)}^{2}}}} \\
= & 0.9993\text{ or }99.93%
\end{align}</math>
 
[[Image:Fig 5.15.PNG|center|thumb|400px|<div align="center"> Result for the computer system reliability.</div>]]
 

Thus the reliability of the computer system after 365 days of operation is:

::<math>\begin{align}
{{R}_{s}}(365)= & {{R}_{processor}}(365)\cdot {{R}_{harddrive}}(365)\cdot {{R}_{CDdrive}}(365) \\
= & 0.9805\cdot 0.9948\cdot 0.9993 \\
= & 0.9747\text{ or }97.47%
\end{align}</math>
 
This result can be obtained in BlockSim as shown in Figure "Result for the computer system reliability".
 

=Load Sharing=
As presented in earlier chapters, a reliability block diagram (RBD) allows you to graphically represent how the components within a system are reliability-wise connected. In most cases, independence is assumed across the components within the system. For example, the failure of component A does not affect the failure of component B. However, if a system consists of components that are sharing a load, then the assumption of independence no longer holds true.
 

If one component fails, then the component(s) that are still operating will have to assume the failed unit's portion of the load. Therefore, the reliabilities of the surviving unit(s) will change. Calculating the system reliability is no longer an easy proposition. In the case of load sharing components, the change of the failure distributions of the surviving components must be known in order to determine the system's reliability.
 

To illustrate this, consider the a system of two units connected reliability-wise in parallel (Figure "Two units connected reliability-wise in parallel").
 

[[Image:5-16.png|center|200px|<div align="center"> Two units connected reliability-wise in parallel.</div>]]
 

Assume that the units must supply an output of 8 volts and that if both units are operational, each unit is to supply 50% of the total output. If one of the units fails, then the surviving unit supplies 100%. Furthermore, assume that having to supply the entire load has a negative impact on the reliability characteristics of the surviving unit. Since the reliability characteristics of the unit change based on whether both or only one is operating, a life distribution along with a life-stress relationship (as discussed in [http://reliawiki.com/index.php/Statistical_Background#A_Brief_Introduction_to_Life-Stress_Relationships|A Brief Introduction to Life-Stress Relationships]) will be needed to model each component.
 

To illustrate the steps needed, we will create the model starting from raw data. Assume that a total of 20 units were tested to failure at 7, 10 and 15 volts. The test data set is presented in the next table.
 

[[Image:5-17.png|center|450px|]]
 

For this example, Units 1 and 2 are the same component. Therefore, only one set of data was collected. However, it is possible that the load sharing components in a system may not be the same. If that were the case, data would need to be collected for each component.
 

The data set in Table 1 was analyzed using ReliaSoft's ALTA software (as shown in Figure "Using ALTA to calculate component parameters") with the Inverse Power Law as the underlying life-stress relationship and Weibull as the life distribution.
 
 

The estimated model parameters, <math>\beta </math> , <math>K</math> and <math>n</math> , are shown next.
 

::<math>\begin{align}
\beta = & 1.9239 \\
K= & 3.2387\times {{10}^{-7}} \\
n= & 3.4226
\end{align}</math>
 

:Or:
 
::<math>\begin{align}
{{R}_{1}}(t,{{S}_{1}})= & {{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} \\
= & {{e}^{-{{\left( 3.2387\times {{10}^{-7}}S_{1}^{3.4226}t \right)}^{1.9239}}}} \
\end{align}</math>.

 
::<math>{{f}_{1}}(t,{{S}_{1}})=\beta KS_{1}^{n}{{\left( KS_{1}^{n}t \right)}^{\beta -1}}{{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} \ </math>
 

And for this case:

::<math>\begin{align}
{{R}_{1}}(t,{{S}_{1}})= & {{R}_{2}}(t,{{S}_{2}}) \\
{{f}_{1}}(t,{{S}_{1}})= & {{f}_{2}}(t,{{S}_{2}})
\end{align}</math>
 
 
[[Image:Fig 5.17.PNG|thumb|center|450px|<div align="center"> Using ALTA to calculate component parameters.</div>]]
 

Figure "Reliability curves for different voltage output conditions" shows a plot of <math>{{R}_{1}}(t,{{S}_{1}})={{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} = {{e}^{-{{\left( 3.2387\times {{10}^{-7}}S_{1}^{3.4226}t \right)}^{1.9239}}}}\ </math>.
 

Now that the failure properties have been determined using the test data, the reliability of the system at some time, <math>t</math> , can be calculated using the following equation:

 
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>

 
:Where:
 

::<math>\begin{align}
{{S}_{1}}= & {{P}_{1}}S \\
{{S}_{2}}= & {{P}_{2}}S
\end{align}</math>

 
[[Image:5-18.png|center|300px|<div align="center"> Reliability curves for different voltage output conditions. </div>]]
 
[[Image:BS5.19.png|center|300px| Illustrating <math>{{t}_{e}}</math>]]
 

And:
 

:• <math>S</math> is the total load (or required output). 
:• <math>{{P}_{1}}</math> and <math>{{P}_{2}}</math> are the portion of the total load that each unit supports when both units are operational. In this case, <math>{{P}_{1}}={{P}_{2}}=0.5=50%.</math> 
:• <math>{{S}_{1}}</math> and <math>{{S}_{2}}</math> represent the portions of the load that Unit 1 and Unit 2 must support when both units are operational. 
:• <math>{{t}_{{{1}_{e}}}}</math> is the equivalent operating time for Unit 1 if it had been operating at <math>S</math> instead of <math>{{S}_{1}}</math> . A graphical representation of the equivalent time is shown in Figure 5.19, where the curve marked by L represents the low stress (load) and the curve marked by H represents the high stress (load). 
 

::<math>{{t}_{1e}}</math> can be calculated by:
 
::<math>\begin{align}
{{R}_{1}}(t)= & {{R}_{1}}({{t}_{1e}}) \\
{{e}^{-{{(tKS_{1}^{n})}^{\beta }}}}= & {{e}^{-{{({{t}_{1e}}K{{S}^{n}})}^{\beta }}}} \\
tS_{1}^{n}= & {{t}_{1e}}{{S}^{n}} \\
{{t}_{1e}}= & t{{\left( \frac{{{S}_{1}}}{S} \right)}^{n}},\text{ }{{S}_{1}}={{P}_{1}}S \\
\therefore & {{t}_{1e}}=tP_{1}^{n}
\end{align}</math>

 
::<math>{{t}_{2e}}</math> can be calculated the same way, or:

 
::<math>\begin{align}
{{R}_{2}}(t)= & {{R}_{2}}({{t}_{2e}}) \\
\therefore & {{t}_{2e}}=tP_{2}^{n}
\end{align}</math>
 

In this example, the reliability equations for Unit 1 and Unit 2 are the same since they are the same type of component and demonstrate the same failure properties. In addition, the total output is divided equally between the two units (when both units are operating), so <math>{{t}_{1e}}</math> and <math>{{t}_{2e}}</math> will also be the same.
 
The next step is to determine the reliability of the system after 8,760 hours, <math>R(t=8,760)</math> .
Using
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>
 

the system reliability is found to be:
 

::<math>\begin{align}
R(t=8760)= & 0.8567 \\
= & 85.67%
\end{align}</math>
 

===Load Sharing in BlockSim===
BlockSim uses this formulation when computing reliabilities of units in a load sharing configuration. When using the System Reliability Equation window, BlockSim returns a single token for the reliability of units in a load sharing configuration (as well as in the case of standby redundancy, discussed in the next section). As an example, consider the following RBD with Unit 1 in series with a container that includes two load sharing units.
 
[[Image:BS5.19.2.png|center|150px|]]
 
BlockSim will return the system equation as:
 

::<math>{{R}_{System}}=+{{R}_{LS}}\cdot {{R}_{1}}</math>
 

Where <math>{{R}_{LS}}</math> implies a form similar to
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>
 .
BlockSim allows for <math>k</math> -out-of- <math>n</math> units in a load sharing configuration.

===Example 5===
A component has five possible failure modes, <math>A</math> , <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> , <math>{{B}_{C}}</math> and <math>C</math> , and the <math>B</math> modes are interdependent. The system will fail if mode <math>A</math> occurs, mode <math>C</math> occurs or two out of the three <math>B</math> modes occur.
 

Modes <math>A</math> and <math>C</math> each have a Weibull distribution, with a <math>\beta =2</math> and <math>\eta =10,000</math> and 15,000 respectively. Events <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> and <math>{{B}_{C}}</math> each have an exponential distribution with a mean of 10,000 hours.
 

If any <math>B</math> event occurs (i.e. <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> or <math>{{B}_{C}}</math> ), the remaining <math>B</math> events are more likely to occur. Specifically, the mean times of the remaining <math>B</math> events are halved. Determine the reliability at 1000 hours for this component.
====Solution to Example 5====
The first step is to create the RBD. Modes <math>A</math> , <math>C</math> and a load sharing container with the <math>{{B}_{i}}</math> modes must be drawn in series, as illustrated next.
 
[[Image:BS5.19.3.png|center|200px|]]
 

The next step is to define the properties for each block, including those for the container. Setting the failure distributions for modes <math>A</math> and <math>C</math> is simple. The more difficult part is setting the properties for the container and the contained blocks. Based on the problem statement, the <math>B</math> modes are in a 2-out-of-3 load sharing redundancy. When all three are working (i.e. when no <math>B</math> mode has occurred), each block has an exponential distribution with <math>\mu=10,000</math>.
 

The next step is to set the weight proportionality factor. This factor defines the portion of the load that the particular item carries while operating, as well as the load that shifts to the remaining units upon failure of the item. To illustrate, assume three units (1, 2 and 3) are in a load sharing container with weight proportionality factors of 1, 2 and 3 respectively (and a 1-out-of-3 requirement).
 

:• Unit 1 carries <math>\left( \tfrac{1}{1+2+3} \right)=0.166</math> or 16.6% of the total load.
:• Unit 2 carries <math>\left( \tfrac{2}{1+2+3} \right)=0.333</math> or 33.3% of the total load.
:• Unit 3 carries <math>\left( \tfrac{3}{1+2+3} \right)=0.50</math> or 50% of the total load.
 

The actual load on each unit then becomes the product of the entire load defined for the container times the portion carried by that unit. For example, if the container load is 100 lbs, then the portion assigned to Unit 1 will be <math>100\cdot 0.166=16.6</math> lbs.
 

In the current example, all units share the same load and thus have equal weight proportionality factors. Because these factors are relative, if the same number is used for all three items then the results will be the same. Thus, weight proportional factor is set equal to 1 for each item.
 

 
[[Image:Fig 5.23_2.PNG|thumb|center|450px|<div align="center"> Defining Weight Proportional Factor. </div>]]
 

The last properties that need to be defined are the total load and the 2-out-of-3 redundancy. The total load is dependent on how the parameters were computed. In this case, total load was assumed to be 3 when the parameters were computed (i.e. the load per item was 1 when all worked and 1.5 when two worked). This is defined at the container level, as shown in Figure "Defining Weight Proportional Factor".
When all of the parameters have been specified in BlockSim, the reliability at 1,000 hours can be determined. From the Analytical QCP, this is found to be 93.87%.

=Standby Components=

In the previous section, the case of a system with load sharing components was presented. This is a form of redundancy with dependent components. That is, the failure of one component affects the failure of the other(s). This section presents another form of redundancy: standby redundancy. In standby redundancy the redundant components are set to be under a lighter load condition (or no load) while not needed and under the operating load when they are activated.
 

In standby redundancy the components are set to have two states: an active state and a standby state. Components in standby redundancy have two failure distributions, one for each state. When in the standby state, they have a quiescent (or dormant) failure distribution and when operating, they have an active failure distribution.
 

In the case that both quiescent and active failure distributions are the same, the units are in a simple parallel configuration (also called a hot standby configuration). When the rate of failure of the standby component is lower in quiescent mode than in active mode, that is called a warm standby configuration. When the rate of failure of the standby component is zero in quiescent mode (i.e. the component cannot fail when in standby), that is called a cold standby configuration.

===Simple Standby Configuration===

Consider two components in a standby configuration. Component 1 is the active component with a Weibull failure distribution with parameters <math>\beta </math> = 1.5 and <math>\eta </math> = 1,000. Component 2 is the standby component. When Component 2 is operating, it also has a Weibull failure distribution with <math>\beta </math> = 1.5 and <math>\eta </math> = 1,000. Furthermore, assume the following cases for the quiescent distribution.
 
:• Case 1: The quiescent distribution is the same as the active distribution (hot standby). 
:• Case 2: The quiescent distribution is a Weibull distribution with <math>\beta </math> = 1.5 and <math>\eta </math> = 2000 (warm standby). 
:• Case 3: The component cannot fail in quiescent mode (cold standby). 
 
In this case, the reliability of the system at some time, <math>t</math> , can be obtained using the following equation:
 

::<math>R(t)={{R}_{1}}(t)+\underset{0}{\overset{t}{\mathop \int }}\,{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x)\cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}dx \ </math>
 

Where:
 
:• <math>{{R}_{1}}</math> is the reliability of the active component. 
:• <math>{{f}_{1}}</math> is the <math>pdf</math> of the active component. 
:• <math>{{R}_{2;SB}}</math> is the reliability of the standby component when in standby mode (quiescent reliability). 
:• <math>{{R}_{2;A}}</math> is the reliability of the standby component when in active mode. 
:• <math>{{t}_{e}}</math> is the equivalent operating time for the standby unit if it had been operating at an active mode, such that: 

 
<math>{{R}_{2;SB}}(x)={{R}_{2;A}}({{t}_{e}})</math> 
 
The second equation above can be solved for <math>{{t}_{e}}</math> and substituted into the first equation above.
Figure "Standby container" illustrates the example as entered in BlockSim using a standby container.
 
 
[[Image:5.24.png|center|150px|<div align="center"> Standby container.</div>]]

 

The active and standby blocks are within a container, which is used to specify standby redundancy. Since the standby component has two distributions (active and quiescent), the Block Properties window of the standby block has two pages for specifying each one. Figures "Defining the active failure distribution" and "Defining the quiescent failure distribution" illustrate these pages.

 

The system reliability results for 1000 hours are given in the following table:
 
[[Image:5-24.png|center|150px|]]
 

Note that even though the <math>\beta </math> value for the quiescent distribution is the same as in the active distribution, it is possible that the two can be different. That is, the failure modes present during the quiescent mode could be different from the modes present during the active mode. In that sense, the two distribution types can be different as well (e.g. lognormal when quiescent and Weibull when active).
 

In many cases when considering standby systems, a switching device may also be present that switches from the failed active component to the standby component. The reliability of the switch can also be incorporated into
<math>R(t)={{R}_{1}}(t)+\underset{0}{\overset{t}{\mathop \int }}\,{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x)\cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}dx \ </math>

as presented in the next section.
 

BlockSim's System Reliability Equation window returns a single token for the reliability of units in a standby configuration. This is the same as the load sharing case presented in the previous section.
 

[[Image: Fig 5.25.PNG|thumb|center|400px|<div align="center"> Defining the active failure distribution </div>]]
 
 

[[Image:Fig 5.26.PNG|thumb|center|400px|<div align="center"> Defining the quiescent failure distribution </div>]]
 

===Reliability of Standby Systems with a Switching Device===

In many cases when dealing with standby systems, a switching device is present that will switch to the standby component when the active component fails. Therefore, the failure properties of the switch must also be included in the analysis.
 

[[Image:BS5.26.2.png|center|150px|]]
 

In most cases when the reliability of a switch is to be included in the analysis, two probabilities can be considered. The first and most common one is the probability of the switch performing the action (i.e. switching) when requested to do so. This is called Switch Probability per Request in BlockSim and is expressed as a static probability (e.g. 90%). The second probability is the quiescent reliability of the switch. This is the reliability of the switch as it ages (e.g. the switch might wear out with age due to corrosion, material degradation, etc.). Thus it is possible for the switch to fail before the active component fails. However, a switch failure does not cause the system to fail, but rather causes the system to fail only if the switch is needed and the switch has failed. For example, if the active component does not fail until the mission end time and the switch fails, then the system does not fail. However, if the active component fails and the switch has also failed, then the system cannot be switched to the standby component and it therefore fails.
 

In analyzing standby components with a switching device, either or both failure probabilities (during the switching or while waiting to switch) can be considered for the switch, since each probability can represent different failure modes. For example, the switch probability per request may represent software-related issues or the probability of detecting the failure of an active component, and the quiescent probability may represent wear-out type failures of the switch.
 

To illustrate the formulation, consider the previous example that assumes perfect switching. To examine the effects of including an imperfect switch, assume that when the active component fails there is a 90% probability that the switch will switch from the active component to the standby component. In addition, assume that the switch can also fail due to a wear-out failure mode described by a Weibull distribution with <math>\beta </math> = 1.7 and <math>\eta </math> = 5000.
 

Therefore, the reliability of the system at some time, <math>t</math> , is given by the following equation.

::<math>\begin{align}
R(t)= & {{R}_{1}}(t) \\
& +\underset{0}{\overset{t}{\mathop \int }}\,\{{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x) \\
& \cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}\cdot {{R}_{SW;Q}}(x)\cdot {{R}_{SW;REQ}}(x)\}dx
\end{align}</math>

Where:
 
:• <math>{{R}_{1}}</math> is the reliability of the active component. 
:• <math>{{f}_{1}}</math> is the <math>pdf</math> of the active component. 
:• <math>{{R}_{2;SB}}</math> is the reliability of the standby component when in standby mode (quiescent reliability). 
:• <math>{{R}_{2;A}}</math> is the reliability of the standby component when in active mode. 
:• <math>{{R}_{SW;Q}}</math> is the quiescent reliability of the switch. 
:• <math>{{R}_{SW;REQ}}</math> is the switch probability per request. 
:• <math>{{t}_{e}}</math> is the equivalent operating time for the standby unit if it had been operating at an active mode. 
This problem can be solved in BlockSim by including these probabilities in the container's properties, as shown in Figures "Standby container (switch) failure probabilities while attempting to switch" and "Standby container (switch) failure distribution while waiting to switch". In BlockSim, the standby container is acting as the switch.

 

[[Image:Fig 5.28_2.PNG|thumb|center|400px|<div align="center"> Standby container (switch) failure probabilities while attempting to switch </div> ]]
 

 

[[Image:Fig 5.27.PNG|thumb|center|500px|<div align="center"> Standby container (switch) failure distribution while waiting to switch </div>]]
 

 

Note that there are additional properties that can be specified in BlockSim for a switch, such as Switch Restart Probability, No. of Restarts and Switch Delay Time. In many applications, the switch is re-tested (or re-cycled) if it fails to switch the first time. In these cases, it might be possible that it switches in the second or third, or <math>{{n}^{th}}</math> attempt.
 
The Switch Restart Probability specifies each additional attempt's probability of successfully switching and the Finite Restarts specifies the total number of attempts. Note that the Switch Restart Probability specifies the probability of success of each trial (or attempt). The probability of success of <math>n</math> consecutive trials is calculated by BlockSim using the binomial distribution and this probability is then incorporated into equation above. The Switch Delay Time property is related to repairable systems and is considered in BlockSim only when using simulation. When using the analytical solution (i.e. for a non-repairable system), this property is ignored.
 

Solving the analytical solution (as given by Equation above), the following results are obtained.
 

[[Image:5-30.png|center|350px|]]
 

From the table above, it can be seen that the presence of a switching device has a significant effect on the reliability of a standby system. It is therefore important when modeling standby redundancy to incorporate the switching device reliability properties. It should be noted that this methodology is not the same as treating the switching device as another series component with the standby subsystem. This would be valid only if the failure of the switch resulted in the failure of system (e.g. switch failing open). In equation above, the Switch Probability per Request and quiescent probability are present only in the second term of the equation. Treating these two failure modes as a series configuration with the standby subsystem would imply that they are also present when the active component is functioning (i.e. first term of equation above). This is invalid and would result in the underestimation of the reliability of the system. In other words, these two failure modes become significant only when the active component fails.
 

As an example, and if we consider the warm standby case, the reliability of the system without the switch is 70.57% at 1000 hours. If the system was modeled so that the switching device was in series with the warm standby subsystem, the result would have been:

 

::<math>\begin{align}
{{R}_{S}}(1000)= & {{R}_{Standby}}(1000)\cdot {{R}_{sw,Q(1000)}}\cdot {{R}_{sw,req}} \\
= & 0.7057\cdot 0.9372\cdot 0.9 \\
= & 0.5952
\end{align}</math>

 

In the case where a switch failure mode causes the standby subsystem to fail, then this mode can be modeled as an individual block in series with the standby subsystem.

===Example 6===
Consider a car with four new tires and a full-size spare. Assume the following failure characteristics:
 
:• The tires follow a Weibull distribution with a <math>\beta=4</math> and an <math>\eta =</math> 40,000 miles while on the car due to wear. 
:• The tires also have a probability of failing due to puncture or other causes. For this, assume a constant rate for this occurrence with a probability of 1 every 50,000 miles. 
:• When not on the car (i.e. is a spare), a tire's probability of failing also has a Weibull distribution with a <math>\beta =</math> 2 and <math>\eta =</math> 120,000 miles. 
 
Assume a mission of 1,000 miles. If a tire fails during this trip, it will be replaced with the spare. However, the spare will not be repaired during the trip. In other words, the trip will continue with the spare on the car and if the spare fails the system will fail. Determine the probability of system failure.
====Solution to Example 6====
Active failure distribution for tires:
 
:• Due to wear-out, Weibull <math>\beta =4</math> and <math>\eta =40,000</math> miles. 
:• Due to random puncture, exponential <math>\mu =50,000.</math> 
:• The quiescent failure distribution is a Weibull distribution with <math>\beta =2</math> and <math>\eta =120,000</math> miles. 
 
The block diagram for each tire has two blocks in series, one block representing the wear-out mode and the other the random puncture mode, as shown next:
 
 
[[Image:small5.png|center|250px|]]
 

There are five tires, four active and one standby (represented in the diagram by a standby container with a 4-out-of-5 requirement), as shown next:
 
[[Image:BStirecontainer.png|center|250px|]]

For the standby Wear block, set the active failure and the quiescent distributions, but for the Puncture block, set only the active puncture distribution (because the tire cannot fail due to puncture while stored). Using BlockSim, the probability of system failure is found to be 0.003 or 0.3%.
 

=Note Regarding Numerical Integration Solutions=

Load sharing and standby solutions in BlockSim are performed using numerical integration routines. As with any numerical analysis routine, the solution error depends on the number of iterations performed, the step size chosen and related factors, plus the behavior of the underlying function. By default, BlockSim uses a certain set of preset factors. In general, these defaults are sufficient for most problems. If a higher precision or verification of the precision for a specific problem is required, BlockSim's preset options can be modified and/or the integration error can be assessed using the Integration Parameters... option for each container. For more details, you can refer to the documentation on the Algorithm Setup window in the BlockSim 8 User's help files.

Time-Dependent System Reliability (Analytical)

2012-06-21T15:49:00Z

Dingzhou Cao: /* Solution to Example 5 */

{{Template:bsbook|5}}

In the previous chapter, different system configuration types were examined, as well as different methods for obtaining the system's reliability function analytically. Because the reliabilities in the problems presented were treated as probabilities (e.g. <math>P(A)</math> , <math>{{R}_{i}}</math> ), the reliability values and equations presented were referred to as static (not time-dependent). Thus, in the prior chapter, the life distributions of the components were not incorporated in the process of calculating the system reliability. In this chapter, time dependency in the reliability function will be introduced. We will develop the models necessary to observe the reliability over the life of the system, instead of at just one point in time. In addition, performance measures such as failure rate, MTTF and warranty time will be estimated for the entire system. The methods of obtaining the reliability function analytically remain identical to the ones presented in the previous chapter, with the exception that the reliabilities will be functions of time. In other words, instead of dealing with <math>{{R}_{i}}</math> , we will use <math>{{R}_{i}}(t)</math> . All examples in this chapter assume that no repairs are performed on the components.

=Analytical Life Predictions=
The analytical approach presented in the prior chapter involved the determination of a mathematical expression that describes the reliability of the system, expressed in terms of the reliabilities of its components. So far we have estimated only static system reliability (at a fixed time). For example, in the case of a system with three components in series, the system's reliability equation was given by:
 

::<math>{{R}_{s}}={{R}_{1}}\cdot {{R}_{2}}\cdot {{R}_{3}} \ </math>
 

The values of <math>{{R}_{1}}</math> , <math>{{R}_{2}}</math> and <math>{{R}_{3}}</math> were given for a common time and the reliability of the system was estimated for that time. However, since the component failure characteristics can be described by distributions, the system reliability is actually time-dependent. In this case, the equation above can be rewritten as:
 

::<math>{{R}_{s}}(t)={{R}_{1}}(t)\cdot {{R}_{2}}(t)\cdot {{R}_{3}}(t)</math>

 

The reliability of the system for any mission time can now be estimated. Assuming a Weibull life distribution for each component, the first equation above can now be expressed in terms of each component's reliability function, or:

 

[[Category:Completed Theoretical Review]]

::<math>{{R}_{s}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}}\cdot {{e}^{-{{\left( \tfrac{t}{{{\eta }_{2}}} \right)}^{{{\beta }_{2}}}}}}\cdot {{e}^{-{{\left( \tfrac{t}{{{\eta }_{3}}} \right)}^{{{\beta }_{3}}}}}}</math>

 

In the same manner, any life distribution can be substituted into the system reliability equation. Suppose that the times-to-failure of the first component are described with a Weibull distribution, the times-to-failure of the second component with an exponential distribution and the times-to-failure of the third component with a normal distribution. Then the first equation above can be written as:
 

::<math>{{R}_{s}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}}\cdot {{e}^{-{{\lambda }_{2}}t}}\cdot \left[ 1-\Phi \left( \frac{t-{{\mu }_{3}}}{{{\sigma }_{3}}} \right) \right]</math>

 

It can be seen that the biggest challenge is in obtaining the system's reliability function in terms of component reliabilities, which has already been discussed in depth. Once this has been achieved, calculating the reliability of the system for any mission duration is just a matter of substituting the corresponding component reliability functions into the system reliability equation.
===Advantages of the Analytical Method===
The primary advantage of the analytical solution is that it produces a mathematical expression that describes the reliability of the system. Once the system's reliability function has been determined, other calculations can then be performed to obtain metrics of interest for the system. Such calculations include:
 
:• Determination of the system's <math>pdf.</math> 
:• Determination of warranty periods. 
:• Determination of the system's failure rate. 
:• Determination of the system's MTTF. 
 
In addition, optimization and reliability allocation techniques can be used to aid engineers in their design improvement efforts. Another advantage in using analytical techniques is the ability to perform static calculations and analyze systems with a mixture of static and time-dependent components. Finally, the reliability importance of components over time can be calculated with this methodology.
===Disadvantages of the Analytical Method===
The biggest disadvantage of the analytical method is that formulations can become very complicated. The more complicated a system is, the larger and more difficult it will be to analytically formulate an expression for the system's reliability. For particularly detailed systems this process can be quite time-consuming, even with the use of computers. Furthermore, when the maintainability of the system or some of its components must be taken into consideration, analytical solutions become intractable. In these situations, the use of simulation methods may be more advantageous than attempting to develop a solution analytically. Simulation methods are presented in later chapters.
===Looking at a Simple ''Complex'' System Analytically===

The complexity involved in an analytical solution can be best illustrated by looking at the simple ''complex'' system with 15 components, as shown in Figure " An RBD of a complex system".
 
[[Image:5-1.png|center|400px|<div align="center"> An RBD of a complex system.</div>]]

 
The system reliability for this system (computed using BlockSim) is shown next. The first solution is provided using BlockSim's symbolic solution. In symbolic mode, BlockSim breaks the equation into segments, identified by tokens, that need to be substituted into the final system equation for a complete solution. This creates algebraic solutions that are more compact than if the substitutions were made.
 

::<math>\begin{align}
{{R}_{System}}= & D2\cdot D3\cdot {{R}_{L}} \\
D3= & +{{R}_{K}}\cdot IK \\
IK= & +{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}} \\
D2 = & +{{R}_{A}}\cdot {{R}_{E}}\cdot IE \\
IE = & -D1\cdot {{R}_{M}}\cdot {{R}_{N}}+{{R}_{M}}\cdot {{R}_{N}}+D1 \\
D1 = & +{{R}_{D}}\cdot ID \\
ID = & -{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}}
\end{align}</math>
 
Substituting the terms yields:
 

::<math>\begin{align}
{{R}_{System}}= & {{R}_{A}}\cdot {{R}_{E}}\cdot {{R}_{L}}\cdot {{R}_{K}} \\
& \cdot \{({{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})\cdot {{R}_{M}}\cdot {{R}_{N}} \\
& +{{R}_{M}}\cdot {{R}_{N}}-{{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}}\} \\
& \cdot \{{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}}\}
\end{align}</math>

 
BlockSim's automatic algebraic simplification would yield the following format for the above solution:
 

::<math>\begin{align}
{{R}_{System}}= & (({{R}_{A}}\cdot {{R}_{E}}(-({{R}_{D}}(-{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})){{R}_{M}}\cdot {{R}_{N}} \\
& +{{R}_{M}}\cdot {{R}_{N}} \\
& +({{R}_{D}}(-{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})))) \\
& ({{R}_{K}}({{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& +RI\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}})){{R}_{L}}) \
\end{align}</math>

 
In this equation, each <math>{{R}_{i}}</math> represents the reliability function of a block. For example, if <math>{{R}_{A}}</math> has a Weibull distribution, then each <math>{{R}_{A}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{A}}} \right)}^{{{\beta }_{A}}}}}}</math> and so forth. Substitution of each component's reliability function in the last <math>{{R}_{System}}</math> equation above will result in an analytical expression for the system reliability as a function of time, or <math>{{R}_{s}}(t)</math> , which is the same as <math>(1-cd{{f}_{System}}).</math>

===Obtaining Other Functions of Interest===
Once the system reliability equation (or the cumulative density function, <math>cdf</math> ) has been determined, other functions and metrics of interest can be derived.
 
Consider the following simple system:
 
[[Image:5-2.png|center|150px|<div align="center"> Simple two-component system. </div>]]

 
Furthermore, assume that component 1 follows an exponential distribution with a mean of 10,000 (<math>\mu =10,000,</math> <math>\lambda =1/10,000)</math> and component 2 follows a Weibull distribution with <math>\beta =6</math> and <math>\eta =10,000</math> . The reliability equation of this system is:

 
::<math>\begin{align}
{{R}_{S}}(t)= & {{R}_{1}}(t)\cdot {{R}_{2}}(t) \\
= & {{e}^{-\lambda t}}\cdot {{e}^{-{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \\
= & {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \
\end{align}</math>

 
The system <math>cdf</math> is:

 
::<math>\begin{align}
{{F}_{S}}(t)= & 1-({{R}_{1}}(t)\cdot {{R}_{2}}(t)) \\
= & 1-\left( {{e}^{-\lambda t}}\cdot {{e}^{-{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \right) \\
= & 1-\left( {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)
\end{align}</math>

 
{{analytical system pdf}}

====Conditional Reliability====
Conditional reliability is the probability of a system successfully completing another mission following the successful completion of a previous mission. The time of the previous mission and the time for the mission to be undertaken must be taken into account for conditional reliability calculations. The system's conditional reliability function is given by:

::<math>R(T,t)=\frac{R(T+t)}{R(T)}</math>

Equation above gives the reliability for a new mission of duration <math>t</math> having already accumulated <math>T</math> hours of operation up to the start of this new mission. The system is evaluated to assure that it will start the next mission successfully.
 
For the system in Figure "Simple two-component system.", the reliability for mission of <math>t=1,000</math> hours, having an age of <math>T=500</math> hours, is:
 

::<math>\begin{align}
{{R}_{S}}(T=500,t=1000)= & \frac{R(T+t)}{R(T)} \\
= & \frac{R(1500)}{R(500)} \\
= & \frac{{{e}^{-\tfrac{1500}{10,000}}}\cdot {{e}^{-{{\left( \tfrac{1500}{10,000} \right)}^{6}}}}}{{{e}^{-\tfrac{500}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{500}{10,000} \right)}^{6}}}}} \\
= & 0.9048=90.48%
\end{align}</math>

 

====Conditional Reliability for Components====

Now in this formulation, it was assumed that the accumulated age was equivalent for both units. That is, both started life at zero and aged to 500. It is possible to consider an individual component that has already accumulated some age (used component) in the same formulation. To illustrate this, assume that component 2 started life with an age of T=100. Then the reliability equation of the system, as given in <math>{{R}_{S}}(t)= {{R}_{1}}(t)\cdot {{R}_{2}}(t)</math>, would need to be modified to include a conditional term for 2, or:
 

::<math>{{R}_{S}}(t)={{R}_{1}}(t)\cdot \frac{{{R}_{2}}({{T}_{2}}+t)}{{{R}_{2}}({{T}_{2}})} \ </math>

 

In BlockSim, the start age input box may be used to specify a starting age greater than zero.
{{system failure rate analytical}}

====System Mean Life (Mean Time To Failure)====
The mean life (or mean time to failure, MTTF) can be obtained by integrating the system reliability function from zero to infinity:
 

::<math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math>

 

The mean time is a performance index and does not provide any information about the behavior of the failure distribution of the system.
 
For the system in Figure "Simple two-component system":
 

::<math>\begin{align}
MTTF= & \int_{0}^{\infty }\left( {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)dt \\
= & 5978.9
\end{align}</math>

 

====Warranty Period and BX Life====
Sometimes it is desirable to know the time value associated with a certain reliability. Warranty periods are often calculated by determining what percentage of the failure population can be covered financially and estimating the time at which this portion of the population will fail. Similarly, engineering specifications may call for a certain BX life, which also represents a time period during which a certain proportion of the population will fail. For example, the B10 life is the time in which 10% of the population will fail.
This is obtained by setting <math>{{R}_{S}}(t)</math> to the desired value and solving for <math>t.</math>
 
For the system in Figure "Simple two-component system":
 

::<math>{{R}_{s}}\left( t \right)={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>
 

To compute the time by which reliability would be equal to 90%, equation above is recast as follows and solved for <math>t.</math>

 
::<math>0.90={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>

 

In this case, <math>t=1053.59</math> . Equivalently, the B10 life for this system is also <math>1053.59</math> .
Except for some trivial cases, a closed form solution for <math>t</math> cannot be obtained. Thus, it is necessary to solve for <math>t</math> using numerical methods. BlockSim uses numerical methods.

===Example 1===
Consider a system consisting of three exponential units in series with the following failure rates (in failures per hour): <math>{{\lambda }_{1}}</math> = 0.0002, <math>{{\lambda }_{2}}</math> = 0.0005 and <math>{{\lambda }_{3}}</math> = 0.0001.
 
:• Obtain the reliability equation for the system. 
:• What is the reliability of the system after 150 hours of operation? 
:• Obtain the system's <math>pdf.</math> 
:• Obtain the system's failure rate equation. 
:• What is the MTTF for the system? 
:• What should the warranty period be for a 90% reliability? 
====Solution to Example 1====
 
:• The analytical expression for the reliability of the system is given by:

::<math>\begin{align}
{{R}_{s}}(t)= & {{R}_{1}}(t)\cdot {{R}_{2}}(t)\cdot {{R}_{3}}(t) \\
= & {{e}^{-{{\lambda }_{1}}t}}\cdot {{e}^{-{{\lambda }_{2}}t}}\cdot {{e}^{-{{\lambda }_{1}}t}} \\
= & {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}
\end{align}</math>

 

At 150 hours of operation, the reliability of the system is:
 

::<math>\begin{align}
{{R}_{s}}(t)= & {{e}^{-(0.0002+0.0005+0.0001)150}} \\
= & 0.8869\text{ or }88.69%
\end{align}</math>

 

:• In order to obtain the system's <math>pdf</math> , the derivative of the reliability equation given in the first equation above is taken with respect to time, or:
 
::<math>\begin{align}
{{f}_{s}}(t)= & -\frac{d[{{R}_{s}}(t)]}{dt} \\
= & -\frac{d\left[ {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}} \right]}{dt} \\
= & ({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})\cdot {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}
\end{align}</math>

:• The system's failure rate can now be obtained simply by dividing the system's <math>pdf</math> given in the equation above by the system's reliability function given in the first equation above, and:

::<math>\begin{align}
{{\lambda }_{s}}\left( t \right)= & \frac{{{f}_{s}}\left( t \right)}{{{R}_{s}}\left( t \right)} \\
= & \frac{({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})\cdot {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}}{{{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}} \\
= & ({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}) \\
= & 0.0008\text{ }fr/hr
\end{align}</math>

 
:• Combining <math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math> and the first equation above, the system's MTTF can be obtained:

 
::<math>\begin{align}
MTTF= & \int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \\
= & \int_{0}^{\infty }{{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}dt \\
= & \frac{1}{({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})} \\
= & 1250\text{ }hr
\end{align}</math>
 

:• Solving the first equation above with respect to time will yield the corresponding warranty period for a 90% reliability. In this case, the system reliability equation is simple and a closed form solution exists. The warranty time can now be found by solving:

 
::<math>\begin{align}
t= & -\frac{\ln (R)}{{{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}} \\
= & -\frac{\ln (0.9)}{0.0008} \\
= & 131.7\text{ }hr
\end{align}</math>
 

Thus, the warranty period should be 132 hours.

===Example 2===
Consider the system shown in Figure "Complex bridge system in Example 2".
 
[[Image:BS5.5.png|center|400px|<div align="center"> Complex bridge system in Example 2. </div>]]
 

Components <math>A</math> through <math>E</math> are Weibull distributed with <math>\beta =1.2</math> and <math>\eta =1230</math> hours. The starting and ending blocks cannot fail. 
Determine the following:
 
:• The reliability equation for the system and its corresponding plot. 
:• The system's <math>pdf</math> and its corresponding plot. 
:• The system's failure rate equation and the corresponding plot. 
:• The <math>MTTF</math> . 
:• The warranty time for a 90% reliability. 
:• The reliability for a 200-hour mission, if it is known that the system has already successfully operated for 200 hours. 
====Solution====
 
The first step is to obtain the reliability function for the system. The methods described in the previous chapter can be employed, such as the event space or path-tracing methods. Using BlockSim, the following reliability equation is obtained:
 

::<math>\begin{align}
{{R}_{s}}(t)= & ({{R}_{Start}}\cdot {{R}_{End}}(2{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}))
\end{align}</math>
 

Note that since the starting and ending blocks cannot fail, <math>{{R}_{Start}}=1</math> and <math>{{R}_{End}}=1,</math> equation above can be reduced to:

 

::<math>\begin{align}
{{R}_{s}}(t)= & 2\cdot {{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}
\end{align}</math>

 

Where <math>{{R}_{A}}</math> is the reliability equation for Component A, or:

 

::<math>{{R}_{A}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{A}}} \right)}^{{{\beta }_{A}}}}}}</math>

 

::<math>{{R}_{B}}</math> is the reliability equation for Component <math>B</math> , etc.
 

Since the components in this example are identical, the system reliability equation can be further reduced to:

 

::<math>{{R}_{s}}(t)=2R{{(t)}^{2}}+2R{{(t)}^{3}}-5R{{(t)}^{4}}+2R{{(t)}^{5}}</math>

 

Or, in terms of the failure distribution:

 

::<math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>

 

The corresponding plot is given in Figure "Reliability plot for the system".
 
[[Image:BS5.6.png|center|300px|<div align="center"> Reliability plot for the system. </div>]]
 
 

In order to obtain the system's <math>pdf</math> , the derivative of the reliability equation given above is taken with respect to time, resulting in:
 

::<math>\begin{align}
{{f}_{s}}(t)= & 4\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+6\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \\
& -20\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+10\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}
\end{align}</math>

 

The <math>pdf</math> can now be plotted for different time values, <math>t</math> , as shown in Figure "''pdf'' plot for the system".
 

The system's failure rate can now be obtained by dividing the system's <math>pdf</math> given in equation above by the system's reliability function given in <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>, or:

 

::<math>\begin{align}
{{\lambda }_{s}}(t)= & \frac{4\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+6\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}{2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}} \\
& +\frac{-20\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+10\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}{2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}
\end{align}</math>

 

The corresponding plot is given in Figure "Failure rate for the system".
 
[[Image:BS5.7.png|center|300px|<div align="center"> ''pdf'' plot for the system.</div>]]
 

The <math>MTTF</math> of the system is obtained by integrating the system's reliability function given by <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math> from time zero to infinity, as given by <math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math>. Using BlockSim's Analytical QCP, an <math>MTTF</math> of 1007.8 hours is calculated, as shown in Figure "MTTF of the system".
 

The warranty time can be obtained by solving <math>{{R}_{s}}(t)</math> with respect to time for a system reliability <math>{{R}_{s}}=0.9</math> . Using the Analytical QCP and selecting the 
Warranty Time option, a time of 372.72 hours is obtained, as shown in Figure "Time at which ''R''=0.9 or 90% for the system".
 
[[Image:BS5.8.png|center|300px|<div align="center"> Failure rate for the system.</div>]]
 
[[Image:Fig 5.9.PNG|thumb|center|400px|<div align="center"> MTTF of the system. </div>]]
 
[[Image:Fig 5.10.PNG|thumb|center|400px|<div align="center"> Time at which ''R''=0.9 or 90% for the system.</div>]]
 
[[Image:Fig 5.11.PNG|thumb|center|400px|<div align="center"> Conditional reliability calculation for the system.</div>]]
 

Lastly, the conditional reliability can be obtained using <math>R(T,t)=\frac{R(T+t)}{R(T)}</math> and <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>, or:
 

::<math>\begin{align}
R(200,200)= & \frac{R(400)}{R(200)} \\
= & \frac{0.883825}{0.975321} \\
= & 0.906189
\end{align}</math>

 

This can be calculated using BlockSim's Analytical QCP, as shown in Figure "Conditional reliability calculation for the system".
 

=Approximating the System CDF=

In many cases, it is valuable to fit a distribution that represents the system's times-to-failure. This can be useful when the system is part of a larger assembly and may be used for repeated calculations or in calculations for other systems. In cases such as this, it can be useful to characterize the system's behavior by fitting a distribution to the overall system and calculating parameters for this distribution. This is equivalent to fitting a single distribution to describe <math>{{R}_{S}}(t</math> ). In essence, it is like reducing the entire system to a component in order to simplify calculations.
 
For the system in Figure below:
[[Image:5-2.png|center|150px|<div align="center">Simple two-component system. </div>]]
 

::<math>{{R}_{S}}(t)={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>

 
To compute an approximate reliability function for this system, <math>{{R}_{A}}(t)\simeq {{R}_{S}}(t)</math> , one would compute <math>n</math> pairs of reliability and time values and then fit a single distribution to the data, or:

 

::<math>\begin{align}
{{R}_{S}}(t= & 10,396.7)=10% \\
{{R}_{S}}(t= & 9,361.9)=20% \\
& ... \\
{{R}_{S}}(t= & 1,053.6)=90%
\end{align}</math>

 

A single distribution, <math>{{R}_{A}}(t)</math> , that approximates <math>{{R}_{S}}(t)</math> can now be computed from these pairs using life data analysis methods. If using the Weibull++ software, one would enter the values as free form data.

===Example 3===
Compute a single Weibull distribution approximation for the system in Example 2.
====Solution to Example 3====
The system in the previous example, shown in Figure "Complex bridge system in Example 2", can be approximated by use of a 2-parameter Weibull distribution with <math>\beta =2.02109</math> and <math>\eta =1123.51</math> . In BlockSim, this is accomplished by representing the entire system as one distribution by going to the Distribution Fit area, as shown next.

 
[[Image:Fig 5.13.PNG|center|200px|<div align="center"> Representing a system with a distribution.</div>]]

 
[[Image:Fig 5.12.PNG|thumb|center|350px|<div align="center"> Distribution Estimator.</div>]]
 
Clicking inside the Distribution Fit area opens the Distribution Estimator window (Figure "Distribution Fitting window"). In this window you can select a distribution to represent the data. BlockSim will then generate a number of system failure times based on the system's reliability function. The system's reliability function can be used to solve for a time value associated with that unreliability value. The distribution of the generated time values can then be fitted to a probability distribution function.
 

Consider a value of <math>F(t)=0.11</math> . Using the system's reliability equation and solving for time, the corresponding time-to-failure for a 0.11 unreliability can be calculated. 

For the system of Example 2, the time for a 0.11 unreliability is 389.786 hours. 

When enough points have been generated, the selected distribution will be fitted to this data set and the distribution's parameters will be returned. In addition, if ReliaSoft's Weibull++ is installed, the generated data can be viewed/analyzed using a Weibull++ instance, as shown in Figure "Using Weibull++ to calculate distribution parameters".
 
[[Image:Fig 5.14.PNG|thumb|center|450px|<div align="center"> Using Weibull++ to calculate distribution parameters.</div>]]
 
It is recommended that the analyst examine the fit to ascertain the applicability of the approximation.

=Duty Cycle=

Components of a system may not operate continuously during a system's mission, or may be subjected to loads greater or lesser than the rated loads during system operation. To model this, a factor called the Duty Cycle ( <math>{{d}_{c}}</math> ) is used. The duty cycle may also be used to account for changes in environmental stress, such as temperature changes, that may effect the operation of a component. The duty cycle is a positive value, with a default value of 1 representing continuous operation at rated load, and any values other than 1 representing other load values with respect to the rated load value (or total operating time). A duty cycle value higher than 1 indicates a load in excess of the rated value. A duty cycle value lower than 1 indicates that the component is operating at a load lower than the rated load or not operating continuously during the system's mission. For instance, a duty cycle of 0.5 may be used for a component that operates only half of the time during the system's mission.
 
The reliability metrics for a component with a duty cycle are calculated as follows. Let <math>{{d}_{c}}</math> represent the duty cycle during a particular mission of the component, <math>t</math> represent the mission time and <math>{t}'</math> represent the accumulated age. Then:
 

::<math>{t}'={{d}_{c}}\times t</math>

 

The reliability equation for the component is:
 

::<math>R({t}')=R({{d}_{c}}\times t)</math>
 

The component ''pdf'' is:
 

::<math>f({t}')=-\frac{d(R({t}'))}{dt}=-\frac{d(R({{d}_{c}}\times t))}{dt}={{d}_{c}}f({{d}_{c}}\times t)</math>
 

The failure rate of the component is:
 

::<math>\lambda ({t}')=\frac{f({t}')}{R({t}')}=\frac{{{d}_{c}}f({{d}_{c}}\times t)}{R({{d}_{c}}\times t)}={{d}_{c}}\lambda ({{d}_{c}}\times t)</math>

===Example 4===
Consider a computer system with three components: a processor, a hard drive and a CD drive in series as shown next. Assume that all three components follow a Weibull failure distribution with the parameters <math>{{\beta }_{1}}=1.5</math> and <math>{{\eta }_{1}}=5000</math> for the processor, <math>{{\beta }_{2}}=2.5</math> and <math>{{\eta }_{2}}=3000</math> for the hard drive, and <math>{{\beta }_{3}}=2</math> and <math>{{\eta }_{3}}=4000</math> for the CD drive. Determine the reliability of the computer system after one year (365 days) of operation, assuming that the CD drive is used only 30% of the time.
 
[[Image:BS5ex4.png|center|250px]]

====Solution to Example 4====
The reliability of the processor after 365 days of operation is given by:

::<math>\begin{align}
{{R}_{processor}}(365)= & {{e}^{-{{\left( \tfrac{365}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}} \\
= & {{e}^{-{{\left( \tfrac{365}{5000} \right)}^{1.5}}}} \\
= & 0.9805\text{ or }98.05%
\end{align}</math>

The reliability of the hard drive after 365 days of operation is given by:

::<math>\begin{align}
{{R}_{harddrive}}(365)= & {{e}^{-{{\left( \tfrac{365}{{{\eta }_{2}}} \right)}^{{{\beta }_{2}}}}}} \\
= & {{e}^{-{{\left( \tfrac{365}{3000} \right)}^{2.5}}}} \\
= & 0.9948\text{ or }99.48%
\end{align}</math>

The reliability of the CD drive after 365 days of operation (taking into account the 30% operation using a duty cycle of 0.3) is given by:

::<math>\begin{align}
{{R}_{CDdrive}}(365)= & {{e}^{-{{\left( \tfrac{{{d}_{c}}\times 365}{{{\eta }_{3}}} \right)}^{{{\beta }_{3}}}}}} \\
= & {{e}^{-{{\left( \tfrac{0.3\times 365}{4000} \right)}^{2}}}} \\
= & 0.9993\text{ or }99.93%
\end{align}</math>
 
[[Image:Fig 5.15.PNG|center|thumb|400px|<div align="center"> Result for the computer system reliability.</div>]]
 

Thus the reliability of the computer system after 365 days of operation is:

::<math>\begin{align}
{{R}_{s}}(365)= & {{R}_{processor}}(365)\cdot {{R}_{harddrive}}(365)\cdot {{R}_{CDdrive}}(365) \\
= & 0.9805\cdot 0.9948\cdot 0.9993 \\
= & 0.9747\text{ or }97.47%
\end{align}</math>
 
This result can be obtained in BlockSim as shown in Figure "Result for the computer system reliability".
 

=Load Sharing=
As presented in earlier chapters, a reliability block diagram (RBD) allows you to graphically represent how the components within a system are reliability-wise connected. In most cases, independence is assumed across the components within the system. For example, the failure of component A does not affect the failure of component B. However, if a system consists of components that are sharing a load, then the assumption of independence no longer holds true.
 

If one component fails, then the component(s) that are still operating will have to assume the failed unit's portion of the load. Therefore, the reliabilities of the surviving unit(s) will change. Calculating the system reliability is no longer an easy proposition. In the case of load sharing components, the change of the failure distributions of the surviving components must be known in order to determine the system's reliability.
 

To illustrate this, consider the a system of two units connected reliability-wise in parallel (Figure "Two units connected reliability-wise in parallel").
 

[[Image:5-16.png|center|200px|<div align="center"> Two units connected reliability-wise in parallel.</div>]]
 

Assume that the units must supply an output of 8 volts and that if both units are operational, each unit is to supply 50% of the total output. If one of the units fails, then the surviving unit supplies 100%. Furthermore, assume that having to supply the entire load has a negative impact on the reliability characteristics of the surviving unit. Since the reliability characteristics of the unit change based on whether both or only one is operating, a life distribution along with a life-stress relationship (as discussed in [http://reliawiki.com/index.php/Statistical_Background#A_Brief_Introduction_to_Life-Stress_Relationships|A Brief Introduction to Life-Stress Relationships]) will be needed to model each component.
 

To illustrate the steps needed, we will create the model starting from raw data. Assume that a total of 20 units were tested to failure at 7, 10 and 15 volts. The test data set is presented in the next table.
 

[[Image:5-17.png|center|450px|]]
 

For this example, Units 1 and 2 are the same component. Therefore, only one set of data was collected. However, it is possible that the load sharing components in a system may not be the same. If that were the case, data would need to be collected for each component.
 

The data set in Table 1 was analyzed using ReliaSoft's ALTA software (as shown in Figure "Using ALTA to calculate component parameters") with the Inverse Power Law as the underlying life-stress relationship and Weibull as the life distribution.
 
 

The estimated model parameters, <math>\beta </math> , <math>K</math> and <math>n</math> , are shown next.
 

::<math>\begin{align}
\beta = & 1.9239 \\
K= & 3.2387\times {{10}^{-7}} \\
n= & 3.4226
\end{align}</math>
 

:Or:
 
::<math>\begin{align}
{{R}_{1}}(t,{{S}_{1}})= & {{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} \\
= & {{e}^{-{{\left( 3.2387\times {{10}^{-7}}S_{1}^{3.4226}t \right)}^{1.9239}}}} \
\end{align}</math>.

 
::<math>{{f}_{1}}(t,{{S}_{1}})=\beta KS_{1}^{n}{{\left( KS_{1}^{n}t \right)}^{\beta -1}}{{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} \ </math>
 

And for this case:

::<math>\begin{align}
{{R}_{1}}(t,{{S}_{1}})= & {{R}_{2}}(t,{{S}_{2}}) \\
{{f}_{1}}(t,{{S}_{1}})= & {{f}_{2}}(t,{{S}_{2}})
\end{align}</math>
 
 
[[Image:Fig 5.17.PNG|thumb|center|450px|<div align="center"> Using ALTA to calculate component parameters.</div>]]
 

Figure "Reliability curves for different voltage output conditions" shows a plot of <math>{{R}_{1}}(t,{{S}_{1}})={{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} = {{e}^{-{{\left( 3.2387\times {{10}^{-7}}S_{1}^{3.4226}t \right)}^{1.9239}}}}\ </math>.
 

Now that the failure properties have been determined using the test data, the reliability of the system at some time, <math>t</math> , can be calculated using the following equation:

 
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>

 
:Where:
 

::<math>\begin{align}
{{S}_{1}}= & {{P}_{1}}S \\
{{S}_{2}}= & {{P}_{2}}S
\end{align}</math>

 
[[Image:5-18.png|center|300px|<div align="center"> Reliability curves for different voltage output conditions. </div>]]
 
[[Image:BS5.19.png|center|300px| Illustrating <math>{{t}_{e}}</math>]]
 

And:
 

:• <math>S</math> is the total load (or required output). 
:• <math>{{P}_{1}}</math> and <math>{{P}_{2}}</math> are the portion of the total load that each unit supports when both units are operational. In this case, <math>{{P}_{1}}={{P}_{2}}=0.5=50%.</math> 
:• <math>{{S}_{1}}</math> and <math>{{S}_{2}}</math> represent the portions of the load that Unit 1 and Unit 2 must support when both units are operational. 
:• <math>{{t}_{{{1}_{e}}}}</math> is the equivalent operating time for Unit 1 if it had been operating at <math>S</math> instead of <math>{{S}_{1}}</math> . A graphical representation of the equivalent time is shown in Figure 5.19, where the curve marked by L represents the low stress (load) and the curve marked by H represents the high stress (load). 
 

::<math>{{t}_{1e}}</math> can be calculated by:
 
::<math>\begin{align}
{{R}_{1}}(t)= & {{R}_{1}}({{t}_{1e}}) \\
{{e}^{-{{(tKS_{1}^{n})}^{\beta }}}}= & {{e}^{-{{({{t}_{1e}}K{{S}^{n}})}^{\beta }}}} \\
tS_{1}^{n}= & {{t}_{1e}}{{S}^{n}} \\
{{t}_{1e}}= & t{{\left( \frac{{{S}_{1}}}{S} \right)}^{n}},\text{ }{{S}_{1}}={{P}_{1}}S \\
\therefore & {{t}_{1e}}=tP_{1}^{n}
\end{align}</math>

 
::<math>{{t}_{2e}}</math> can be calculated the same way, or:

 
::<math>\begin{align}
{{R}_{2}}(t)= & {{R}_{2}}({{t}_{2e}}) \\
\therefore & {{t}_{2e}}=tP_{2}^{n}
\end{align}</math>
 

In this example, the reliability equations for Unit 1 and Unit 2 are the same since they are the same type of component and demonstrate the same failure properties. In addition, the total output is divided equally between the two units (when both units are operating), so <math>{{t}_{1e}}</math> and <math>{{t}_{2e}}</math> will also be the same.
 
The next step is to determine the reliability of the system after 8,760 hours, <math>R(t=8,760)</math> .
Using
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>
 

the system reliability is found to be:
 

::<math>\begin{align}
R(t=8760)= & 0.8567 \\
= & 85.67%
\end{align}</math>
 

===Load Sharing in BlockSim===
BlockSim uses this formulation when computing reliabilities of units in a load sharing configuration. When using the System Reliability Equation window, BlockSim returns a single token for the reliability of units in a load sharing configuration (as well as in the case of standby redundancy, discussed in the next section). As an example, consider the following RBD with Unit 1 in series with a container that includes two load sharing units.
 
[[Image:BS5.19.2.png|center|150px|]]
 
BlockSim will return the system equation as:
 

::<math>{{R}_{System}}=+{{R}_{LS}}\cdot {{R}_{1}}</math>
 

Where <math>{{R}_{LS}}</math> implies a form similar to
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>
 .
BlockSim allows for <math>k</math> -out-of- <math>n</math> units in a load sharing configuration.

===Example 5===
A component has five possible failure modes, <math>A</math> , <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> , <math>{{B}_{C}}</math> and <math>C</math> , and the <math>B</math> modes are interdependent. The system will fail if mode <math>A</math> occurs, mode <math>C</math> occurs or two out of the three <math>B</math> modes occur.
 

Modes <math>A</math> and <math>C</math> each have a Weibull distribution, with a <math>\beta =2</math> and <math>\eta =10,000</math> and 15,000 respectively. Events <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> and <math>{{B}_{C}}</math> each have an exponential distribution with a mean of 10,000 hours.
 

If any <math>B</math> event occurs (i.e. <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> or <math>{{B}_{C}}</math> ), the remaining <math>B</math> events are more likely to occur. Specifically, the mean times of the remaining <math>B</math> events are halved. Determine the reliability at 1000 hours for this component.
====Solution to Example 5====
The first step is to create the RBD. Modes <math>A</math> , <math>C</math> and a load sharing container with the <math>{{B}_{i}}</math> modes must be drawn in series, as illustrated next.
 
[[Image:BS5.19.3.png|center|200px|]]
 

The next step is to define the properties for each block, including those for the container. Setting the failure distributions for modes <math>A</math> and <math>C</math> is simple. The more difficult part is setting the properties for the container and the contained blocks. Based on the problem statement, the <math>B</math> modes are in a 2-out-of-3 load sharing redundancy. When all three are working (i.e. when no <math>B</math> mode has occurred), each block has an exponential distribution with <math>\mu=10,000</math>.
 

Assume a Power Life-Stress relationship for the components. Then:
 

::<math>\begin{align}
{{\mu }_{1}}= & \frac{1}{KV_{1}^{n}} \ \\
{{\mu }_{2}}= & \frac{1}{KV_{2}^{n}} \
\end{align}</math>
 

Substituting <math>{{\mu }_{1}}=10,000</math> and <math>{{V}_{1}}=1</math> in <math>{{\mu }_{1}}= \frac{1}{KV_{1}^{n}}</math> and casting it in terms of <math>K</math> yields:
 

::<math>\begin{align}
10,000= & \frac{1}{K} \ \\
K = & \frac{1}{10,000}=0.0001
\end{align}</math>

 
Substituting <math>{{\mu }_{2}}=5,000</math> , <math>{{V}_{2}}=1.5</math> (because if one fails, then each survivor takes on an additional 0.5 units of load) and <math>10,000= \frac{1}{K}</math> for <math>K</math> in <math>{{\mu }_{2}}= \frac{1}{KV_{2}^{n}}</math> yields:

 
::<math>\begin{align}
5,000= & \frac{1}{0.0001\cdot {{(1.5)}^{n}}} \\
0.5= & {{(1.5)}^{-n}} \\
\ln (0.5)= & -n\ln (1.5) \\
n = & 1.7095
\end{align}</math>
 

This also could have been computed in ALTA, as shown in Figure "Calculation performed in ALTA".

The next step is to set the weight proportionality factor. This factor defines the portion of the load that the particular item carries while operating, as well as the load that shifts to the remaining units upon failure of the item. To illustrate, assume three units (1, 2 and 3) are in a load sharing container with weight proportionality factors of 1, 2 and 3 respectively (and a 1-out-of-3 requirement).
 

:• Unit 1 carries <math>\left( \tfrac{1}{1+2+3} \right)=0.166</math> or 16.6% of the total load.
:• Unit 2 carries <math>\left( \tfrac{2}{1+2+3} \right)=0.333</math> or 33.3% of the total load.
:• Unit 3 carries <math>\left( \tfrac{3}{1+2+3} \right)=0.50</math> or 50% of the total load.
 

The actual load on each unit then becomes the product of the entire load defined for the container times the portion carried by that unit. For example, if the container load is 100 lbs, then the portion assigned to Unit 1 will be <math>100\cdot 0.166=16.6</math> lbs.
 

In the current example, all units share the same load and thus have equal weight proportionality factors. Because these factors are relative, if the same number is used for all three items then the results will be the same. Thus, weight proportional factor is set equal to 1 for each item.
 

[[Image:5.20.PNG|thumb|center|500px|<div align="center"> Calculation performed in ALTA.</div>]]
 

 
[[Image:Fig 5.23_2.PNG|thumb|center|450px|<div align="center"> Defining Weight Proportional Factor. </div>]]
 

The last properties that need to be defined are the total load and the 2-out-of-3 redundancy. The total load is dependent on how the parameters were computed. In this case, total load was assumed to be 3 when the parameters were computed (i.e. the load per item was 1 when all worked and 1.5 when two worked). This is defined at the container level, as shown in Figure "Defining Weight Proportional Factor".
When all of the parameters have been specified in BlockSim, the reliability at 1,000 hours can be determined. From the Analytical QCP, this is found to be 93.87%.

=Standby Components=

In the previous section, the case of a system with load sharing components was presented. This is a form of redundancy with dependent components. That is, the failure of one component affects the failure of the other(s). This section presents another form of redundancy: standby redundancy. In standby redundancy the redundant components are set to be under a lighter load condition (or no load) while not needed and under the operating load when they are activated.
 

In standby redundancy the components are set to have two states: an active state and a standby state. Components in standby redundancy have two failure distributions, one for each state. When in the standby state, they have a quiescent (or dormant) failure distribution and when operating, they have an active failure distribution.
 

In the case that both quiescent and active failure distributions are the same, the units are in a simple parallel configuration (also called a hot standby configuration). When the rate of failure of the standby component is lower in quiescent mode than in active mode, that is called a warm standby configuration. When the rate of failure of the standby component is zero in quiescent mode (i.e. the component cannot fail when in standby), that is called a cold standby configuration.

===Simple Standby Configuration===

Consider two components in a standby configuration. Component 1 is the active component with a Weibull failure distribution with parameters <math>\beta </math> = 1.5 and <math>\eta </math> = 1,000. Component 2 is the standby component. When Component 2 is operating, it also has a Weibull failure distribution with <math>\beta </math> = 1.5 and <math>\eta </math> = 1,000. Furthermore, assume the following cases for the quiescent distribution.
 
:• Case 1: The quiescent distribution is the same as the active distribution (hot standby). 
:• Case 2: The quiescent distribution is a Weibull distribution with <math>\beta </math> = 1.5 and <math>\eta </math> = 2000 (warm standby). 
:• Case 3: The component cannot fail in quiescent mode (cold standby). 
 
In this case, the reliability of the system at some time, <math>t</math> , can be obtained using the following equation:
 

::<math>R(t)={{R}_{1}}(t)+\underset{0}{\overset{t}{\mathop \int }}\,{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x)\cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}dx \ </math>
 

Where:
 
:• <math>{{R}_{1}}</math> is the reliability of the active component. 
:• <math>{{f}_{1}}</math> is the <math>pdf</math> of the active component. 
:• <math>{{R}_{2;SB}}</math> is the reliability of the standby component when in standby mode (quiescent reliability). 
:• <math>{{R}_{2;A}}</math> is the reliability of the standby component when in active mode. 
:• <math>{{t}_{e}}</math> is the equivalent operating time for the standby unit if it had been operating at an active mode, such that: 

 
<math>{{R}_{2;SB}}(x)={{R}_{2;A}}({{t}_{e}})</math> 
 
The second equation above can be solved for <math>{{t}_{e}}</math> and substituted into the first equation above.
Figure "Standby container" illustrates the example as entered in BlockSim using a standby container.
 
 
[[Image:5.24.png|center|150px|<div align="center"> Standby container.</div>]]

 

The active and standby blocks are within a container, which is used to specify standby redundancy. Since the standby component has two distributions (active and quiescent), the Block Properties window of the standby block has two pages for specifying each one. Figures "Defining the active failure distribution" and "Defining the quiescent failure distribution" illustrate these pages.

 

The system reliability results for 1000 hours are given in the following table:
 
[[Image:5-24.png|center|150px|]]
 

Note that even though the <math>\beta </math> value for the quiescent distribution is the same as in the active distribution, it is possible that the two can be different. That is, the failure modes present during the quiescent mode could be different from the modes present during the active mode. In that sense, the two distribution types can be different as well (e.g. lognormal when quiescent and Weibull when active).
 

In many cases when considering standby systems, a switching device may also be present that switches from the failed active component to the standby component. The reliability of the switch can also be incorporated into
<math>R(t)={{R}_{1}}(t)+\underset{0}{\overset{t}{\mathop \int }}\,{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x)\cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}dx \ </math>

as presented in the next section.
 

BlockSim's System Reliability Equation window returns a single token for the reliability of units in a standby configuration. This is the same as the load sharing case presented in the previous section.
 

[[Image: Fig 5.25.PNG|thumb|center|400px|<div align="center"> Defining the active failure distribution </div>]]
 
 

[[Image:Fig 5.26.PNG|thumb|center|400px|<div align="center"> Defining the quiescent failure distribution </div>]]
 

===Reliability of Standby Systems with a Switching Device===

In many cases when dealing with standby systems, a switching device is present that will switch to the standby component when the active component fails. Therefore, the failure properties of the switch must also be included in the analysis.
 

[[Image:BS5.26.2.png|center|150px|]]
 

In most cases when the reliability of a switch is to be included in the analysis, two probabilities can be considered. The first and most common one is the probability of the switch performing the action (i.e. switching) when requested to do so. This is called Switch Probability per Request in BlockSim and is expressed as a static probability (e.g. 90%). The second probability is the quiescent reliability of the switch. This is the reliability of the switch as it ages (e.g. the switch might wear out with age due to corrosion, material degradation, etc.). Thus it is possible for the switch to fail before the active component fails. However, a switch failure does not cause the system to fail, but rather causes the system to fail only if the switch is needed and the switch has failed. For example, if the active component does not fail until the mission end time and the switch fails, then the system does not fail. However, if the active component fails and the switch has also failed, then the system cannot be switched to the standby component and it therefore fails.
 

In analyzing standby components with a switching device, either or both failure probabilities (during the switching or while waiting to switch) can be considered for the switch, since each probability can represent different failure modes. For example, the switch probability per request may represent software-related issues or the probability of detecting the failure of an active component, and the quiescent probability may represent wear-out type failures of the switch.
 

To illustrate the formulation, consider the previous example that assumes perfect switching. To examine the effects of including an imperfect switch, assume that when the active component fails there is a 90% probability that the switch will switch from the active component to the standby component. In addition, assume that the switch can also fail due to a wear-out failure mode described by a Weibull distribution with <math>\beta </math> = 1.7 and <math>\eta </math> = 5000.
 

Therefore, the reliability of the system at some time, <math>t</math> , is given by the following equation.

::<math>\begin{align}
R(t)= & {{R}_{1}}(t) \\
& +\underset{0}{\overset{t}{\mathop \int }}\,\{{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x) \\
& \cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}\cdot {{R}_{SW;Q}}(x)\cdot {{R}_{SW;REQ}}(x)\}dx
\end{align}</math>

Where:
 
:• <math>{{R}_{1}}</math> is the reliability of the active component. 
:• <math>{{f}_{1}}</math> is the <math>pdf</math> of the active component. 
:• <math>{{R}_{2;SB}}</math> is the reliability of the standby component when in standby mode (quiescent reliability). 
:• <math>{{R}_{2;A}}</math> is the reliability of the standby component when in active mode. 
:• <math>{{R}_{SW;Q}}</math> is the quiescent reliability of the switch. 
:• <math>{{R}_{SW;REQ}}</math> is the switch probability per request. 
:• <math>{{t}_{e}}</math> is the equivalent operating time for the standby unit if it had been operating at an active mode. 
This problem can be solved in BlockSim by including these probabilities in the container's properties, as shown in Figures "Standby container (switch) failure probabilities while attempting to switch" and "Standby container (switch) failure distribution while waiting to switch". In BlockSim, the standby container is acting as the switch.

 

[[Image:Fig 5.28_2.PNG|thumb|center|400px|<div align="center"> Standby container (switch) failure probabilities while attempting to switch </div> ]]
 

 

[[Image:Fig 5.27.PNG|thumb|center|500px|<div align="center"> Standby container (switch) failure distribution while waiting to switch </div>]]
 

 

Note that there are additional properties that can be specified in BlockSim for a switch, such as Switch Restart Probability, No. of Restarts and Switch Delay Time. In many applications, the switch is re-tested (or re-cycled) if it fails to switch the first time. In these cases, it might be possible that it switches in the second or third, or <math>{{n}^{th}}</math> attempt.
 
The Switch Restart Probability specifies each additional attempt's probability of successfully switching and the Finite Restarts specifies the total number of attempts. Note that the Switch Restart Probability specifies the probability of success of each trial (or attempt). The probability of success of <math>n</math> consecutive trials is calculated by BlockSim using the binomial distribution and this probability is then incorporated into equation above. The Switch Delay Time property is related to repairable systems and is considered in BlockSim only when using simulation. When using the analytical solution (i.e. for a non-repairable system), this property is ignored.
 

Solving the analytical solution (as given by Equation above), the following results are obtained.
 

[[Image:5-30.png|center|350px|]]
 

From the table above, it can be seen that the presence of a switching device has a significant effect on the reliability of a standby system. It is therefore important when modeling standby redundancy to incorporate the switching device reliability properties. It should be noted that this methodology is not the same as treating the switching device as another series component with the standby subsystem. This would be valid only if the failure of the switch resulted in the failure of system (e.g. switch failing open). In equation above, the Switch Probability per Request and quiescent probability are present only in the second term of the equation. Treating these two failure modes as a series configuration with the standby subsystem would imply that they are also present when the active component is functioning (i.e. first term of equation above). This is invalid and would result in the underestimation of the reliability of the system. In other words, these two failure modes become significant only when the active component fails.
 

As an example, and if we consider the warm standby case, the reliability of the system without the switch is 70.57% at 1000 hours. If the system was modeled so that the switching device was in series with the warm standby subsystem, the result would have been:

 

::<math>\begin{align}
{{R}_{S}}(1000)= & {{R}_{Standby}}(1000)\cdot {{R}_{sw,Q(1000)}}\cdot {{R}_{sw,req}} \\
= & 0.7057\cdot 0.9372\cdot 0.9 \\
= & 0.5952
\end{align}</math>

 

In the case where a switch failure mode causes the standby subsystem to fail, then this mode can be modeled as an individual block in series with the standby subsystem.

===Example 6===
Consider a car with four new tires and a full-size spare. Assume the following failure characteristics:
 
:• The tires follow a Weibull distribution with a <math>\beta=4</math> and an <math>\eta =</math> 40,000 miles while on the car due to wear. 
:• The tires also have a probability of failing due to puncture or other causes. For this, assume a constant rate for this occurrence with a probability of 1 every 50,000 miles. 
:• When not on the car (i.e. is a spare), a tire's probability of failing also has a Weibull distribution with a <math>\beta =</math> 2 and <math>\eta =</math> 120,000 miles. 
 
Assume a mission of 1,000 miles. If a tire fails during this trip, it will be replaced with the spare. However, the spare will not be repaired during the trip. In other words, the trip will continue with the spare on the car and if the spare fails the system will fail. Determine the probability of system failure.
====Solution to Example 6====
Active failure distribution for tires:
 
:• Due to wear-out, Weibull <math>\beta =4</math> and <math>\eta =40,000</math> miles. 
:• Due to random puncture, exponential <math>\mu =50,000.</math> 
:• The quiescent failure distribution is a Weibull distribution with <math>\beta =2</math> and <math>\eta =120,000</math> miles. 
 
The block diagram for each tire has two blocks in series, one block representing the wear-out mode and the other the random puncture mode, as shown next:
 
 
[[Image:small5.png|center|250px|]]
 

There are five tires, four active and one standby (represented in the diagram by a standby container with a 4-out-of-5 requirement), as shown next:
 
[[Image:BStirecontainer.png|center|250px|]]

For the standby Wear block, set the active failure and the quiescent distributions, but for the Puncture block, set only the active puncture distribution (because the tire cannot fail due to puncture while stored). Using BlockSim, the probability of system failure is found to be 0.003 or 0.3%.
 

=Note Regarding Numerical Integration Solutions=

Load sharing and standby solutions in BlockSim are performed using numerical integration routines. As with any numerical analysis routine, the solution error depends on the number of iterations performed, the step size chosen and related factors, plus the behavior of the underlying function. By default, BlockSim uses a certain set of preset factors. In general, these defaults are sufficient for most problems. If a higher precision or verification of the precision for a specific problem is required, BlockSim's preset options can be modified and/or the integration error can be assessed using the Integration Parameters... option for each container. For more details, you can refer to the documentation on the Algorithm Setup window in the BlockSim 8 User's help files.

Time-Dependent System Reliability (Analytical)

2012-06-21T15:47:33Z

Dingzhou Cao: /* Solution to Example 5 */

{{Template:bsbook|5}}

In the previous chapter, different system configuration types were examined, as well as different methods for obtaining the system's reliability function analytically. Because the reliabilities in the problems presented were treated as probabilities (e.g. <math>P(A)</math> , <math>{{R}_{i}}</math> ), the reliability values and equations presented were referred to as static (not time-dependent). Thus, in the prior chapter, the life distributions of the components were not incorporated in the process of calculating the system reliability. In this chapter, time dependency in the reliability function will be introduced. We will develop the models necessary to observe the reliability over the life of the system, instead of at just one point in time. In addition, performance measures such as failure rate, MTTF and warranty time will be estimated for the entire system. The methods of obtaining the reliability function analytically remain identical to the ones presented in the previous chapter, with the exception that the reliabilities will be functions of time. In other words, instead of dealing with <math>{{R}_{i}}</math> , we will use <math>{{R}_{i}}(t)</math> . All examples in this chapter assume that no repairs are performed on the components.

=Analytical Life Predictions=
The analytical approach presented in the prior chapter involved the determination of a mathematical expression that describes the reliability of the system, expressed in terms of the reliabilities of its components. So far we have estimated only static system reliability (at a fixed time). For example, in the case of a system with three components in series, the system's reliability equation was given by:
 

::<math>{{R}_{s}}={{R}_{1}}\cdot {{R}_{2}}\cdot {{R}_{3}} \ </math>
 

The values of <math>{{R}_{1}}</math> , <math>{{R}_{2}}</math> and <math>{{R}_{3}}</math> were given for a common time and the reliability of the system was estimated for that time. However, since the component failure characteristics can be described by distributions, the system reliability is actually time-dependent. In this case, the equation above can be rewritten as:
 

::<math>{{R}_{s}}(t)={{R}_{1}}(t)\cdot {{R}_{2}}(t)\cdot {{R}_{3}}(t)</math>

 

The reliability of the system for any mission time can now be estimated. Assuming a Weibull life distribution for each component, the first equation above can now be expressed in terms of each component's reliability function, or:

 

[[Category:Completed Theoretical Review]]

::<math>{{R}_{s}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}}\cdot {{e}^{-{{\left( \tfrac{t}{{{\eta }_{2}}} \right)}^{{{\beta }_{2}}}}}}\cdot {{e}^{-{{\left( \tfrac{t}{{{\eta }_{3}}} \right)}^{{{\beta }_{3}}}}}}</math>

 

In the same manner, any life distribution can be substituted into the system reliability equation. Suppose that the times-to-failure of the first component are described with a Weibull distribution, the times-to-failure of the second component with an exponential distribution and the times-to-failure of the third component with a normal distribution. Then the first equation above can be written as:
 

::<math>{{R}_{s}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}}\cdot {{e}^{-{{\lambda }_{2}}t}}\cdot \left[ 1-\Phi \left( \frac{t-{{\mu }_{3}}}{{{\sigma }_{3}}} \right) \right]</math>

 

It can be seen that the biggest challenge is in obtaining the system's reliability function in terms of component reliabilities, which has already been discussed in depth. Once this has been achieved, calculating the reliability of the system for any mission duration is just a matter of substituting the corresponding component reliability functions into the system reliability equation.
===Advantages of the Analytical Method===
The primary advantage of the analytical solution is that it produces a mathematical expression that describes the reliability of the system. Once the system's reliability function has been determined, other calculations can then be performed to obtain metrics of interest for the system. Such calculations include:
 
:• Determination of the system's <math>pdf.</math> 
:• Determination of warranty periods. 
:• Determination of the system's failure rate. 
:• Determination of the system's MTTF. 
 
In addition, optimization and reliability allocation techniques can be used to aid engineers in their design improvement efforts. Another advantage in using analytical techniques is the ability to perform static calculations and analyze systems with a mixture of static and time-dependent components. Finally, the reliability importance of components over time can be calculated with this methodology.
===Disadvantages of the Analytical Method===
The biggest disadvantage of the analytical method is that formulations can become very complicated. The more complicated a system is, the larger and more difficult it will be to analytically formulate an expression for the system's reliability. For particularly detailed systems this process can be quite time-consuming, even with the use of computers. Furthermore, when the maintainability of the system or some of its components must be taken into consideration, analytical solutions become intractable. In these situations, the use of simulation methods may be more advantageous than attempting to develop a solution analytically. Simulation methods are presented in later chapters.
===Looking at a Simple ''Complex'' System Analytically===

The complexity involved in an analytical solution can be best illustrated by looking at the simple ''complex'' system with 15 components, as shown in Figure " An RBD of a complex system".
 
[[Image:5-1.png|center|400px|<div align="center"> An RBD of a complex system.</div>]]

 
The system reliability for this system (computed using BlockSim) is shown next. The first solution is provided using BlockSim's symbolic solution. In symbolic mode, BlockSim breaks the equation into segments, identified by tokens, that need to be substituted into the final system equation for a complete solution. This creates algebraic solutions that are more compact than if the substitutions were made.
 

::<math>\begin{align}
{{R}_{System}}= & D2\cdot D3\cdot {{R}_{L}} \\
D3= & +{{R}_{K}}\cdot IK \\
IK= & +{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}} \\
D2 = & +{{R}_{A}}\cdot {{R}_{E}}\cdot IE \\
IE = & -D1\cdot {{R}_{M}}\cdot {{R}_{N}}+{{R}_{M}}\cdot {{R}_{N}}+D1 \\
D1 = & +{{R}_{D}}\cdot ID \\
ID = & -{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}}
\end{align}</math>
 
Substituting the terms yields:
 

::<math>\begin{align}
{{R}_{System}}= & {{R}_{A}}\cdot {{R}_{E}}\cdot {{R}_{L}}\cdot {{R}_{K}} \\
& \cdot \{({{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})\cdot {{R}_{M}}\cdot {{R}_{N}} \\
& +{{R}_{M}}\cdot {{R}_{N}}-{{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}}\} \\
& \cdot \{{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}}\}
\end{align}</math>

 
BlockSim's automatic algebraic simplification would yield the following format for the above solution:
 

::<math>\begin{align}
{{R}_{System}}= & (({{R}_{A}}\cdot {{R}_{E}}(-({{R}_{D}}(-{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})){{R}_{M}}\cdot {{R}_{N}} \\
& +{{R}_{M}}\cdot {{R}_{N}} \\
& +({{R}_{D}}(-{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})))) \\
& ({{R}_{K}}({{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& +RI\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}})){{R}_{L}}) \
\end{align}</math>

 
In this equation, each <math>{{R}_{i}}</math> represents the reliability function of a block. For example, if <math>{{R}_{A}}</math> has a Weibull distribution, then each <math>{{R}_{A}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{A}}} \right)}^{{{\beta }_{A}}}}}}</math> and so forth. Substitution of each component's reliability function in the last <math>{{R}_{System}}</math> equation above will result in an analytical expression for the system reliability as a function of time, or <math>{{R}_{s}}(t)</math> , which is the same as <math>(1-cd{{f}_{System}}).</math>

===Obtaining Other Functions of Interest===
Once the system reliability equation (or the cumulative density function, <math>cdf</math> ) has been determined, other functions and metrics of interest can be derived.
 
Consider the following simple system:
 
[[Image:5-2.png|center|150px|<div align="center"> Simple two-component system. </div>]]

 
Furthermore, assume that component 1 follows an exponential distribution with a mean of 10,000 (<math>\mu =10,000,</math> <math>\lambda =1/10,000)</math> and component 2 follows a Weibull distribution with <math>\beta =6</math> and <math>\eta =10,000</math> . The reliability equation of this system is:

 
::<math>\begin{align}
{{R}_{S}}(t)= & {{R}_{1}}(t)\cdot {{R}_{2}}(t) \\
= & {{e}^{-\lambda t}}\cdot {{e}^{-{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \\
= & {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \
\end{align}</math>

 
The system <math>cdf</math> is:

 
::<math>\begin{align}
{{F}_{S}}(t)= & 1-({{R}_{1}}(t)\cdot {{R}_{2}}(t)) \\
= & 1-\left( {{e}^{-\lambda t}}\cdot {{e}^{-{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \right) \\
= & 1-\left( {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)
\end{align}</math>

 
{{analytical system pdf}}

====Conditional Reliability====
Conditional reliability is the probability of a system successfully completing another mission following the successful completion of a previous mission. The time of the previous mission and the time for the mission to be undertaken must be taken into account for conditional reliability calculations. The system's conditional reliability function is given by:

::<math>R(T,t)=\frac{R(T+t)}{R(T)}</math>

Equation above gives the reliability for a new mission of duration <math>t</math> having already accumulated <math>T</math> hours of operation up to the start of this new mission. The system is evaluated to assure that it will start the next mission successfully.
 
For the system in Figure "Simple two-component system.", the reliability for mission of <math>t=1,000</math> hours, having an age of <math>T=500</math> hours, is:
 

::<math>\begin{align}
{{R}_{S}}(T=500,t=1000)= & \frac{R(T+t)}{R(T)} \\
= & \frac{R(1500)}{R(500)} \\
= & \frac{{{e}^{-\tfrac{1500}{10,000}}}\cdot {{e}^{-{{\left( \tfrac{1500}{10,000} \right)}^{6}}}}}{{{e}^{-\tfrac{500}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{500}{10,000} \right)}^{6}}}}} \\
= & 0.9048=90.48%
\end{align}</math>

 

====Conditional Reliability for Components====

Now in this formulation, it was assumed that the accumulated age was equivalent for both units. That is, both started life at zero and aged to 500. It is possible to consider an individual component that has already accumulated some age (used component) in the same formulation. To illustrate this, assume that component 2 started life with an age of T=100. Then the reliability equation of the system, as given in <math>{{R}_{S}}(t)= {{R}_{1}}(t)\cdot {{R}_{2}}(t)</math>, would need to be modified to include a conditional term for 2, or:
 

::<math>{{R}_{S}}(t)={{R}_{1}}(t)\cdot \frac{{{R}_{2}}({{T}_{2}}+t)}{{{R}_{2}}({{T}_{2}})} \ </math>

 

In BlockSim, the start age input box may be used to specify a starting age greater than zero.
{{system failure rate analytical}}

====System Mean Life (Mean Time To Failure)====
The mean life (or mean time to failure, MTTF) can be obtained by integrating the system reliability function from zero to infinity:
 

::<math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math>

 

The mean time is a performance index and does not provide any information about the behavior of the failure distribution of the system.
 
For the system in Figure "Simple two-component system":
 

::<math>\begin{align}
MTTF= & \int_{0}^{\infty }\left( {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)dt \\
= & 5978.9
\end{align}</math>

 

====Warranty Period and BX Life====
Sometimes it is desirable to know the time value associated with a certain reliability. Warranty periods are often calculated by determining what percentage of the failure population can be covered financially and estimating the time at which this portion of the population will fail. Similarly, engineering specifications may call for a certain BX life, which also represents a time period during which a certain proportion of the population will fail. For example, the B10 life is the time in which 10% of the population will fail.
This is obtained by setting <math>{{R}_{S}}(t)</math> to the desired value and solving for <math>t.</math>
 
For the system in Figure "Simple two-component system":
 

::<math>{{R}_{s}}\left( t \right)={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>
 

To compute the time by which reliability would be equal to 90%, equation above is recast as follows and solved for <math>t.</math>

 
::<math>0.90={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>

 

In this case, <math>t=1053.59</math> . Equivalently, the B10 life for this system is also <math>1053.59</math> .
Except for some trivial cases, a closed form solution for <math>t</math> cannot be obtained. Thus, it is necessary to solve for <math>t</math> using numerical methods. BlockSim uses numerical methods.

===Example 1===
Consider a system consisting of three exponential units in series with the following failure rates (in failures per hour): <math>{{\lambda }_{1}}</math> = 0.0002, <math>{{\lambda }_{2}}</math> = 0.0005 and <math>{{\lambda }_{3}}</math> = 0.0001.
 
:• Obtain the reliability equation for the system. 
:• What is the reliability of the system after 150 hours of operation? 
:• Obtain the system's <math>pdf.</math> 
:• Obtain the system's failure rate equation. 
:• What is the MTTF for the system? 
:• What should the warranty period be for a 90% reliability? 
====Solution to Example 1====
 
:• The analytical expression for the reliability of the system is given by:

::<math>\begin{align}
{{R}_{s}}(t)= & {{R}_{1}}(t)\cdot {{R}_{2}}(t)\cdot {{R}_{3}}(t) \\
= & {{e}^{-{{\lambda }_{1}}t}}\cdot {{e}^{-{{\lambda }_{2}}t}}\cdot {{e}^{-{{\lambda }_{1}}t}} \\
= & {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}
\end{align}</math>

 

At 150 hours of operation, the reliability of the system is:
 

::<math>\begin{align}
{{R}_{s}}(t)= & {{e}^{-(0.0002+0.0005+0.0001)150}} \\
= & 0.8869\text{ or }88.69%
\end{align}</math>

 

:• In order to obtain the system's <math>pdf</math> , the derivative of the reliability equation given in the first equation above is taken with respect to time, or:
 
::<math>\begin{align}
{{f}_{s}}(t)= & -\frac{d[{{R}_{s}}(t)]}{dt} \\
= & -\frac{d\left[ {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}} \right]}{dt} \\
= & ({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})\cdot {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}
\end{align}</math>

:• The system's failure rate can now be obtained simply by dividing the system's <math>pdf</math> given in the equation above by the system's reliability function given in the first equation above, and:

::<math>\begin{align}
{{\lambda }_{s}}\left( t \right)= & \frac{{{f}_{s}}\left( t \right)}{{{R}_{s}}\left( t \right)} \\
= & \frac{({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})\cdot {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}}{{{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}} \\
= & ({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}) \\
= & 0.0008\text{ }fr/hr
\end{align}</math>

 
:• Combining <math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math> and the first equation above, the system's MTTF can be obtained:

 
::<math>\begin{align}
MTTF= & \int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \\
= & \int_{0}^{\infty }{{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}dt \\
= & \frac{1}{({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})} \\
= & 1250\text{ }hr
\end{align}</math>
 

:• Solving the first equation above with respect to time will yield the corresponding warranty period for a 90% reliability. In this case, the system reliability equation is simple and a closed form solution exists. The warranty time can now be found by solving:

 
::<math>\begin{align}
t= & -\frac{\ln (R)}{{{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}} \\
= & -\frac{\ln (0.9)}{0.0008} \\
= & 131.7\text{ }hr
\end{align}</math>
 

Thus, the warranty period should be 132 hours.

===Example 2===
Consider the system shown in Figure "Complex bridge system in Example 2".
 
[[Image:BS5.5.png|center|400px|<div align="center"> Complex bridge system in Example 2. </div>]]
 

Components <math>A</math> through <math>E</math> are Weibull distributed with <math>\beta =1.2</math> and <math>\eta =1230</math> hours. The starting and ending blocks cannot fail. 
Determine the following:
 
:• The reliability equation for the system and its corresponding plot. 
:• The system's <math>pdf</math> and its corresponding plot. 
:• The system's failure rate equation and the corresponding plot. 
:• The <math>MTTF</math> . 
:• The warranty time for a 90% reliability. 
:• The reliability for a 200-hour mission, if it is known that the system has already successfully operated for 200 hours. 
====Solution====
 
The first step is to obtain the reliability function for the system. The methods described in the previous chapter can be employed, such as the event space or path-tracing methods. Using BlockSim, the following reliability equation is obtained:
 

::<math>\begin{align}
{{R}_{s}}(t)= & ({{R}_{Start}}\cdot {{R}_{End}}(2{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}))
\end{align}</math>
 

Note that since the starting and ending blocks cannot fail, <math>{{R}_{Start}}=1</math> and <math>{{R}_{End}}=1,</math> equation above can be reduced to:

 

::<math>\begin{align}
{{R}_{s}}(t)= & 2\cdot {{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}
\end{align}</math>

 

Where <math>{{R}_{A}}</math> is the reliability equation for Component A, or:

 

::<math>{{R}_{A}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{A}}} \right)}^{{{\beta }_{A}}}}}}</math>

 

::<math>{{R}_{B}}</math> is the reliability equation for Component <math>B</math> , etc.
 

Since the components in this example are identical, the system reliability equation can be further reduced to:

 

::<math>{{R}_{s}}(t)=2R{{(t)}^{2}}+2R{{(t)}^{3}}-5R{{(t)}^{4}}+2R{{(t)}^{5}}</math>

 

Or, in terms of the failure distribution:

 

::<math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>

 

The corresponding plot is given in Figure "Reliability plot for the system".
 
[[Image:BS5.6.png|center|300px|<div align="center"> Reliability plot for the system. </div>]]
 
 

In order to obtain the system's <math>pdf</math> , the derivative of the reliability equation given above is taken with respect to time, resulting in:
 

::<math>\begin{align}
{{f}_{s}}(t)= & 4\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+6\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \\
& -20\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+10\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}
\end{align}</math>

 

The <math>pdf</math> can now be plotted for different time values, <math>t</math> , as shown in Figure "''pdf'' plot for the system".
 

The system's failure rate can now be obtained by dividing the system's <math>pdf</math> given in equation above by the system's reliability function given in <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>, or:

 

::<math>\begin{align}
{{\lambda }_{s}}(t)= & \frac{4\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+6\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}{2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}} \\
& +\frac{-20\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+10\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}{2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}
\end{align}</math>

 

The corresponding plot is given in Figure "Failure rate for the system".
 
[[Image:BS5.7.png|center|300px|<div align="center"> ''pdf'' plot for the system.</div>]]
 

The <math>MTTF</math> of the system is obtained by integrating the system's reliability function given by <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math> from time zero to infinity, as given by <math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math>. Using BlockSim's Analytical QCP, an <math>MTTF</math> of 1007.8 hours is calculated, as shown in Figure "MTTF of the system".
 

The warranty time can be obtained by solving <math>{{R}_{s}}(t)</math> with respect to time for a system reliability <math>{{R}_{s}}=0.9</math> . Using the Analytical QCP and selecting the 
Warranty Time option, a time of 372.72 hours is obtained, as shown in Figure "Time at which ''R''=0.9 or 90% for the system".
 
[[Image:BS5.8.png|center|300px|<div align="center"> Failure rate for the system.</div>]]
 
[[Image:Fig 5.9.PNG|thumb|center|400px|<div align="center"> MTTF of the system. </div>]]
 
[[Image:Fig 5.10.PNG|thumb|center|400px|<div align="center"> Time at which ''R''=0.9 or 90% for the system.</div>]]
 
[[Image:Fig 5.11.PNG|thumb|center|400px|<div align="center"> Conditional reliability calculation for the system.</div>]]
 

Lastly, the conditional reliability can be obtained using <math>R(T,t)=\frac{R(T+t)}{R(T)}</math> and <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>, or:
 

::<math>\begin{align}
R(200,200)= & \frac{R(400)}{R(200)} \\
= & \frac{0.883825}{0.975321} \\
= & 0.906189
\end{align}</math>

 

This can be calculated using BlockSim's Analytical QCP, as shown in Figure "Conditional reliability calculation for the system".
 

=Approximating the System CDF=

In many cases, it is valuable to fit a distribution that represents the system's times-to-failure. This can be useful when the system is part of a larger assembly and may be used for repeated calculations or in calculations for other systems. In cases such as this, it can be useful to characterize the system's behavior by fitting a distribution to the overall system and calculating parameters for this distribution. This is equivalent to fitting a single distribution to describe <math>{{R}_{S}}(t</math> ). In essence, it is like reducing the entire system to a component in order to simplify calculations.
 
For the system in Figure below:
[[Image:5-2.png|center|150px|<div align="center">Simple two-component system. </div>]]
 

::<math>{{R}_{S}}(t)={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>

 
To compute an approximate reliability function for this system, <math>{{R}_{A}}(t)\simeq {{R}_{S}}(t)</math> , one would compute <math>n</math> pairs of reliability and time values and then fit a single distribution to the data, or:

 

::<math>\begin{align}
{{R}_{S}}(t= & 10,396.7)=10% \\
{{R}_{S}}(t= & 9,361.9)=20% \\
& ... \\
{{R}_{S}}(t= & 1,053.6)=90%
\end{align}</math>

 

A single distribution, <math>{{R}_{A}}(t)</math> , that approximates <math>{{R}_{S}}(t)</math> can now be computed from these pairs using life data analysis methods. If using the Weibull++ software, one would enter the values as free form data.

===Example 3===
Compute a single Weibull distribution approximation for the system in Example 2.
====Solution to Example 3====
The system in the previous example, shown in Figure "Complex bridge system in Example 2", can be approximated by use of a 2-parameter Weibull distribution with <math>\beta =2.02109</math> and <math>\eta =1123.51</math> . In BlockSim, this is accomplished by representing the entire system as one distribution by going to the Distribution Fit area, as shown next.

 
[[Image:Fig 5.13.PNG|center|200px|<div align="center"> Representing a system with a distribution.</div>]]

 
[[Image:Fig 5.12.PNG|thumb|center|350px|<div align="center"> Distribution Estimator.</div>]]
 
Clicking inside the Distribution Fit area opens the Distribution Estimator window (Figure "Distribution Fitting window"). In this window you can select a distribution to represent the data. BlockSim will then generate a number of system failure times based on the system's reliability function. The system's reliability function can be used to solve for a time value associated with that unreliability value. The distribution of the generated time values can then be fitted to a probability distribution function.
 

Consider a value of <math>F(t)=0.11</math> . Using the system's reliability equation and solving for time, the corresponding time-to-failure for a 0.11 unreliability can be calculated. 

For the system of Example 2, the time for a 0.11 unreliability is 389.786 hours. 

When enough points have been generated, the selected distribution will be fitted to this data set and the distribution's parameters will be returned. In addition, if ReliaSoft's Weibull++ is installed, the generated data can be viewed/analyzed using a Weibull++ instance, as shown in Figure "Using Weibull++ to calculate distribution parameters".
 
[[Image:Fig 5.14.PNG|thumb|center|450px|<div align="center"> Using Weibull++ to calculate distribution parameters.</div>]]
 
It is recommended that the analyst examine the fit to ascertain the applicability of the approximation.

=Duty Cycle=

Components of a system may not operate continuously during a system's mission, or may be subjected to loads greater or lesser than the rated loads during system operation. To model this, a factor called the Duty Cycle ( <math>{{d}_{c}}</math> ) is used. The duty cycle may also be used to account for changes in environmental stress, such as temperature changes, that may effect the operation of a component. The duty cycle is a positive value, with a default value of 1 representing continuous operation at rated load, and any values other than 1 representing other load values with respect to the rated load value (or total operating time). A duty cycle value higher than 1 indicates a load in excess of the rated value. A duty cycle value lower than 1 indicates that the component is operating at a load lower than the rated load or not operating continuously during the system's mission. For instance, a duty cycle of 0.5 may be used for a component that operates only half of the time during the system's mission.
 
The reliability metrics for a component with a duty cycle are calculated as follows. Let <math>{{d}_{c}}</math> represent the duty cycle during a particular mission of the component, <math>t</math> represent the mission time and <math>{t}'</math> represent the accumulated age. Then:
 

::<math>{t}'={{d}_{c}}\times t</math>

 

The reliability equation for the component is:
 

::<math>R({t}')=R({{d}_{c}}\times t)</math>
 

The component ''pdf'' is:
 

::<math>f({t}')=-\frac{d(R({t}'))}{dt}=-\frac{d(R({{d}_{c}}\times t))}{dt}={{d}_{c}}f({{d}_{c}}\times t)</math>
 

The failure rate of the component is:
 

::<math>\lambda ({t}')=\frac{f({t}')}{R({t}')}=\frac{{{d}_{c}}f({{d}_{c}}\times t)}{R({{d}_{c}}\times t)}={{d}_{c}}\lambda ({{d}_{c}}\times t)</math>

===Example 4===
Consider a computer system with three components: a processor, a hard drive and a CD drive in series as shown next. Assume that all three components follow a Weibull failure distribution with the parameters <math>{{\beta }_{1}}=1.5</math> and <math>{{\eta }_{1}}=5000</math> for the processor, <math>{{\beta }_{2}}=2.5</math> and <math>{{\eta }_{2}}=3000</math> for the hard drive, and <math>{{\beta }_{3}}=2</math> and <math>{{\eta }_{3}}=4000</math> for the CD drive. Determine the reliability of the computer system after one year (365 days) of operation, assuming that the CD drive is used only 30% of the time.
 
[[Image:BS5ex4.png|center|250px]]

====Solution to Example 4====
The reliability of the processor after 365 days of operation is given by:

::<math>\begin{align}
{{R}_{processor}}(365)= & {{e}^{-{{\left( \tfrac{365}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}} \\
= & {{e}^{-{{\left( \tfrac{365}{5000} \right)}^{1.5}}}} \\
= & 0.9805\text{ or }98.05%
\end{align}</math>

The reliability of the hard drive after 365 days of operation is given by:

::<math>\begin{align}
{{R}_{harddrive}}(365)= & {{e}^{-{{\left( \tfrac{365}{{{\eta }_{2}}} \right)}^{{{\beta }_{2}}}}}} \\
= & {{e}^{-{{\left( \tfrac{365}{3000} \right)}^{2.5}}}} \\
= & 0.9948\text{ or }99.48%
\end{align}</math>

The reliability of the CD drive after 365 days of operation (taking into account the 30% operation using a duty cycle of 0.3) is given by:

::<math>\begin{align}
{{R}_{CDdrive}}(365)= & {{e}^{-{{\left( \tfrac{{{d}_{c}}\times 365}{{{\eta }_{3}}} \right)}^{{{\beta }_{3}}}}}} \\
= & {{e}^{-{{\left( \tfrac{0.3\times 365}{4000} \right)}^{2}}}} \\
= & 0.9993\text{ or }99.93%
\end{align}</math>
 
[[Image:Fig 5.15.PNG|center|thumb|400px|<div align="center"> Result for the computer system reliability.</div>]]
 

Thus the reliability of the computer system after 365 days of operation is:

::<math>\begin{align}
{{R}_{s}}(365)= & {{R}_{processor}}(365)\cdot {{R}_{harddrive}}(365)\cdot {{R}_{CDdrive}}(365) \\
= & 0.9805\cdot 0.9948\cdot 0.9993 \\
= & 0.9747\text{ or }97.47%
\end{align}</math>
 
This result can be obtained in BlockSim as shown in Figure "Result for the computer system reliability".
 

=Load Sharing=
As presented in earlier chapters, a reliability block diagram (RBD) allows you to graphically represent how the components within a system are reliability-wise connected. In most cases, independence is assumed across the components within the system. For example, the failure of component A does not affect the failure of component B. However, if a system consists of components that are sharing a load, then the assumption of independence no longer holds true.
 

If one component fails, then the component(s) that are still operating will have to assume the failed unit's portion of the load. Therefore, the reliabilities of the surviving unit(s) will change. Calculating the system reliability is no longer an easy proposition. In the case of load sharing components, the change of the failure distributions of the surviving components must be known in order to determine the system's reliability.
 

To illustrate this, consider the a system of two units connected reliability-wise in parallel (Figure "Two units connected reliability-wise in parallel").
 

[[Image:5-16.png|center|200px|<div align="center"> Two units connected reliability-wise in parallel.</div>]]
 

Assume that the units must supply an output of 8 volts and that if both units are operational, each unit is to supply 50% of the total output. If one of the units fails, then the surviving unit supplies 100%. Furthermore, assume that having to supply the entire load has a negative impact on the reliability characteristics of the surviving unit. Since the reliability characteristics of the unit change based on whether both or only one is operating, a life distribution along with a life-stress relationship (as discussed in [http://reliawiki.com/index.php/Statistical_Background#A_Brief_Introduction_to_Life-Stress_Relationships|A Brief Introduction to Life-Stress Relationships]) will be needed to model each component.
 

To illustrate the steps needed, we will create the model starting from raw data. Assume that a total of 20 units were tested to failure at 7, 10 and 15 volts. The test data set is presented in the next table.
 

[[Image:5-17.png|center|450px|]]
 

For this example, Units 1 and 2 are the same component. Therefore, only one set of data was collected. However, it is possible that the load sharing components in a system may not be the same. If that were the case, data would need to be collected for each component.
 

The data set in Table 1 was analyzed using ReliaSoft's ALTA software (as shown in Figure "Using ALTA to calculate component parameters") with the Inverse Power Law as the underlying life-stress relationship and Weibull as the life distribution.
 
 

The estimated model parameters, <math>\beta </math> , <math>K</math> and <math>n</math> , are shown next.
 

::<math>\begin{align}
\beta = & 1.9239 \\
K= & 3.2387\times {{10}^{-7}} \\
n= & 3.4226
\end{align}</math>
 

:Or:
 
::<math>\begin{align}
{{R}_{1}}(t,{{S}_{1}})= & {{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} \\
= & {{e}^{-{{\left( 3.2387\times {{10}^{-7}}S_{1}^{3.4226}t \right)}^{1.9239}}}} \
\end{align}</math>.

 
::<math>{{f}_{1}}(t,{{S}_{1}})=\beta KS_{1}^{n}{{\left( KS_{1}^{n}t \right)}^{\beta -1}}{{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} \ </math>
 

And for this case:

::<math>\begin{align}
{{R}_{1}}(t,{{S}_{1}})= & {{R}_{2}}(t,{{S}_{2}}) \\
{{f}_{1}}(t,{{S}_{1}})= & {{f}_{2}}(t,{{S}_{2}})
\end{align}</math>
 
 
[[Image:Fig 5.17.PNG|thumb|center|450px|<div align="center"> Using ALTA to calculate component parameters.</div>]]
 

Figure "Reliability curves for different voltage output conditions" shows a plot of <math>{{R}_{1}}(t,{{S}_{1}})={{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} = {{e}^{-{{\left( 3.2387\times {{10}^{-7}}S_{1}^{3.4226}t \right)}^{1.9239}}}}\ </math>.
 

Now that the failure properties have been determined using the test data, the reliability of the system at some time, <math>t</math> , can be calculated using the following equation:

 
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>

 
:Where:
 

::<math>\begin{align}
{{S}_{1}}= & {{P}_{1}}S \\
{{S}_{2}}= & {{P}_{2}}S
\end{align}</math>

 
[[Image:5-18.png|center|300px|<div align="center"> Reliability curves for different voltage output conditions. </div>]]
 
[[Image:BS5.19.png|center|300px| Illustrating <math>{{t}_{e}}</math>]]
 

And:
 

:• <math>S</math> is the total load (or required output). 
:• <math>{{P}_{1}}</math> and <math>{{P}_{2}}</math> are the portion of the total load that each unit supports when both units are operational. In this case, <math>{{P}_{1}}={{P}_{2}}=0.5=50%.</math> 
:• <math>{{S}_{1}}</math> and <math>{{S}_{2}}</math> represent the portions of the load that Unit 1 and Unit 2 must support when both units are operational. 
:• <math>{{t}_{{{1}_{e}}}}</math> is the equivalent operating time for Unit 1 if it had been operating at <math>S</math> instead of <math>{{S}_{1}}</math> . A graphical representation of the equivalent time is shown in Figure 5.19, where the curve marked by L represents the low stress (load) and the curve marked by H represents the high stress (load). 
 

::<math>{{t}_{1e}}</math> can be calculated by:
 
::<math>\begin{align}
{{R}_{1}}(t)= & {{R}_{1}}({{t}_{1e}}) \\
{{e}^{-{{(tKS_{1}^{n})}^{\beta }}}}= & {{e}^{-{{({{t}_{1e}}K{{S}^{n}})}^{\beta }}}} \\
tS_{1}^{n}= & {{t}_{1e}}{{S}^{n}} \\
{{t}_{1e}}= & t{{\left( \frac{{{S}_{1}}}{S} \right)}^{n}},\text{ }{{S}_{1}}={{P}_{1}}S \\
\therefore & {{t}_{1e}}=tP_{1}^{n}
\end{align}</math>

 
::<math>{{t}_{2e}}</math> can be calculated the same way, or:

 
::<math>\begin{align}
{{R}_{2}}(t)= & {{R}_{2}}({{t}_{2e}}) \\
\therefore & {{t}_{2e}}=tP_{2}^{n}
\end{align}</math>
 

In this example, the reliability equations for Unit 1 and Unit 2 are the same since they are the same type of component and demonstrate the same failure properties. In addition, the total output is divided equally between the two units (when both units are operating), so <math>{{t}_{1e}}</math> and <math>{{t}_{2e}}</math> will also be the same.
 
The next step is to determine the reliability of the system after 8,760 hours, <math>R(t=8,760)</math> .
Using
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>
 

the system reliability is found to be:
 

::<math>\begin{align}
R(t=8760)= & 0.8567 \\
= & 85.67%
\end{align}</math>
 

===Load Sharing in BlockSim===
BlockSim uses this formulation when computing reliabilities of units in a load sharing configuration. When using the System Reliability Equation window, BlockSim returns a single token for the reliability of units in a load sharing configuration (as well as in the case of standby redundancy, discussed in the next section). As an example, consider the following RBD with Unit 1 in series with a container that includes two load sharing units.
 
[[Image:BS5.19.2.png|center|150px|]]
 
BlockSim will return the system equation as:
 

::<math>{{R}_{System}}=+{{R}_{LS}}\cdot {{R}_{1}}</math>
 

Where <math>{{R}_{LS}}</math> implies a form similar to
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>
 .
BlockSim allows for <math>k</math> -out-of- <math>n</math> units in a load sharing configuration.

===Example 5===
A component has five possible failure modes, <math>A</math> , <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> , <math>{{B}_{C}}</math> and <math>C</math> , and the <math>B</math> modes are interdependent. The system will fail if mode <math>A</math> occurs, mode <math>C</math> occurs or two out of the three <math>B</math> modes occur.
 

Modes <math>A</math> and <math>C</math> each have a Weibull distribution, with a <math>\beta =2</math> and <math>\eta =10,000</math> and 15,000 respectively. Events <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> and <math>{{B}_{C}}</math> each have an exponential distribution with a mean of 10,000 hours.
 

If any <math>B</math> event occurs (i.e. <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> or <math>{{B}_{C}}</math> ), the remaining <math>B</math> events are more likely to occur. Specifically, the mean times of the remaining <math>B</math> events are halved. Determine the reliability at 1000 hours for this component.
====Solution to Example 5====
The first step is to create the RBD. Modes <math>A</math> , <math>C</math> and a load sharing container with the <math>{{B}_{i}}</math> modes must be drawn in series, as illustrated next.
 
[[Image:BS5.19.3.png|center|200px|]]
 

The next step is to define the properties for each block, including those for the container. Setting the failure distributions for modes <math>A</math> and <math>C</math> is simple. The more difficult part is setting the properties for the container and the contained blocks. Based on the problem statement, the <math>B</math> modes are in a 2-out-of-3 load sharing redundancy. When all three are working (i.e. when no <math>B</math> mode has occurred), each block has an exponential distribution with <math>\mu=10,000</math>. The current load for each block is 1 now (total load is 3). If one <math>B</math> mode occurs, then the two surviving units would take all the load and each with load 1.5 (3/2=1.5). </math>
 

Assume a Power Life-Stress relationship for the components. Then:
 

::<math>\begin{align}
{{\mu }_{1}}= & \frac{1}{KV_{1}^{n}} \ \\
{{\mu }_{2}}= & \frac{1}{KV_{2}^{n}} \
\end{align}</math>
 

Substituting <math>{{\mu }_{1}}=10,000</math> and <math>{{V}_{1}}=1</math> in <math>{{\mu }_{1}}= \frac{1}{KV_{1}^{n}}</math> and casting it in terms of <math>K</math> yields:
 

::<math>\begin{align}
10,000= & \frac{1}{K} \ \\
K = & \frac{1}{10,000}=0.0001
\end{align}</math>

 
Substituting <math>{{\mu }_{2}}=5,000</math> , <math>{{V}_{2}}=1.5</math> (because if one fails, then each survivor takes on an additional 0.5 units of load) and <math>10,000= \frac{1}{K}</math> for <math>K</math> in <math>{{\mu }_{2}}= \frac{1}{KV_{2}^{n}}</math> yields:

 
::<math>\begin{align}
5,000= & \frac{1}{0.0001\cdot {{(1.5)}^{n}}} \\
0.5= & {{(1.5)}^{-n}} \\
\ln (0.5)= & -n\ln (1.5) \\
n = & 1.7095
\end{align}</math>
 

This also could have been computed in ALTA, as shown in Figure "Calculation performed in ALTA".

The next step is to set the weight proportionality factor. This factor defines the portion of the load that the particular item carries while operating, as well as the load that shifts to the remaining units upon failure of the item. To illustrate, assume three units (1, 2 and 3) are in a load sharing container with weight proportionality factors of 1, 2 and 3 respectively (and a 1-out-of-3 requirement).
 

:• Unit 1 carries <math>\left( \tfrac{1}{1+2+3} \right)=0.166</math> or 16.6% of the total load.
:• Unit 2 carries <math>\left( \tfrac{2}{1+2+3} \right)=0.333</math> or 33.3% of the total load.
:• Unit 3 carries <math>\left( \tfrac{3}{1+2+3} \right)=0.50</math> or 50% of the total load.
 

The actual load on each unit then becomes the product of the entire load defined for the container times the portion carried by that unit. For example, if the container load is 100 lbs, then the portion assigned to Unit 1 will be <math>100\cdot 0.166=16.6</math> lbs.
 

In the current example, all units share the same load and thus have equal weight proportionality factors. Because these factors are relative, if the same number is used for all three items then the results will be the same. Thus, weight proportional factor is set equal to 1 for each item.
 

[[Image:5.20.PNG|thumb|center|500px|<div align="center"> Calculation performed in ALTA.</div>]]
 

 
[[Image:Fig 5.23_2.PNG|thumb|center|450px|<div align="center"> Defining Weight Proportional Factor. </div>]]
 

The last properties that need to be defined are the total load and the 2-out-of-3 redundancy. The total load is dependent on how the parameters were computed. In this case, total load was assumed to be 3 when the parameters were computed (i.e. the load per item was 1 when all worked and 1.5 when two worked). This is defined at the container level, as shown in Figure "Defining Weight Proportional Factor".
When all of the parameters have been specified in BlockSim, the reliability at 1,000 hours can be determined. From the Analytical QCP, this is found to be 93.87%.

=Standby Components=

In the previous section, the case of a system with load sharing components was presented. This is a form of redundancy with dependent components. That is, the failure of one component affects the failure of the other(s). This section presents another form of redundancy: standby redundancy. In standby redundancy the redundant components are set to be under a lighter load condition (or no load) while not needed and under the operating load when they are activated.
 

In standby redundancy the components are set to have two states: an active state and a standby state. Components in standby redundancy have two failure distributions, one for each state. When in the standby state, they have a quiescent (or dormant) failure distribution and when operating, they have an active failure distribution.
 

In the case that both quiescent and active failure distributions are the same, the units are in a simple parallel configuration (also called a hot standby configuration). When the rate of failure of the standby component is lower in quiescent mode than in active mode, that is called a warm standby configuration. When the rate of failure of the standby component is zero in quiescent mode (i.e. the component cannot fail when in standby), that is called a cold standby configuration.

===Simple Standby Configuration===

Consider two components in a standby configuration. Component 1 is the active component with a Weibull failure distribution with parameters <math>\beta </math> = 1.5 and <math>\eta </math> = 1,000. Component 2 is the standby component. When Component 2 is operating, it also has a Weibull failure distribution with <math>\beta </math> = 1.5 and <math>\eta </math> = 1,000. Furthermore, assume the following cases for the quiescent distribution.
 
:• Case 1: The quiescent distribution is the same as the active distribution (hot standby). 
:• Case 2: The quiescent distribution is a Weibull distribution with <math>\beta </math> = 1.5 and <math>\eta </math> = 2000 (warm standby). 
:• Case 3: The component cannot fail in quiescent mode (cold standby). 
 
In this case, the reliability of the system at some time, <math>t</math> , can be obtained using the following equation:
 

::<math>R(t)={{R}_{1}}(t)+\underset{0}{\overset{t}{\mathop \int }}\,{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x)\cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}dx \ </math>
 

Where:
 
:• <math>{{R}_{1}}</math> is the reliability of the active component. 
:• <math>{{f}_{1}}</math> is the <math>pdf</math> of the active component. 
:• <math>{{R}_{2;SB}}</math> is the reliability of the standby component when in standby mode (quiescent reliability). 
:• <math>{{R}_{2;A}}</math> is the reliability of the standby component when in active mode. 
:• <math>{{t}_{e}}</math> is the equivalent operating time for the standby unit if it had been operating at an active mode, such that: 

 
<math>{{R}_{2;SB}}(x)={{R}_{2;A}}({{t}_{e}})</math> 
 
The second equation above can be solved for <math>{{t}_{e}}</math> and substituted into the first equation above.
Figure "Standby container" illustrates the example as entered in BlockSim using a standby container.
 
 
[[Image:5.24.png|center|150px|<div align="center"> Standby container.</div>]]

 

The active and standby blocks are within a container, which is used to specify standby redundancy. Since the standby component has two distributions (active and quiescent), the Block Properties window of the standby block has two pages for specifying each one. Figures "Defining the active failure distribution" and "Defining the quiescent failure distribution" illustrate these pages.

 

The system reliability results for 1000 hours are given in the following table:
 
[[Image:5-24.png|center|150px|]]
 

Note that even though the <math>\beta </math> value for the quiescent distribution is the same as in the active distribution, it is possible that the two can be different. That is, the failure modes present during the quiescent mode could be different from the modes present during the active mode. In that sense, the two distribution types can be different as well (e.g. lognormal when quiescent and Weibull when active).
 

In many cases when considering standby systems, a switching device may also be present that switches from the failed active component to the standby component. The reliability of the switch can also be incorporated into
<math>R(t)={{R}_{1}}(t)+\underset{0}{\overset{t}{\mathop \int }}\,{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x)\cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}dx \ </math>

as presented in the next section.
 

BlockSim's System Reliability Equation window returns a single token for the reliability of units in a standby configuration. This is the same as the load sharing case presented in the previous section.
 

[[Image: Fig 5.25.PNG|thumb|center|400px|<div align="center"> Defining the active failure distribution </div>]]
 
 

[[Image:Fig 5.26.PNG|thumb|center|400px|<div align="center"> Defining the quiescent failure distribution </div>]]
 

===Reliability of Standby Systems with a Switching Device===

In many cases when dealing with standby systems, a switching device is present that will switch to the standby component when the active component fails. Therefore, the failure properties of the switch must also be included in the analysis.
 

[[Image:BS5.26.2.png|center|150px|]]
 

In most cases when the reliability of a switch is to be included in the analysis, two probabilities can be considered. The first and most common one is the probability of the switch performing the action (i.e. switching) when requested to do so. This is called Switch Probability per Request in BlockSim and is expressed as a static probability (e.g. 90%). The second probability is the quiescent reliability of the switch. This is the reliability of the switch as it ages (e.g. the switch might wear out with age due to corrosion, material degradation, etc.). Thus it is possible for the switch to fail before the active component fails. However, a switch failure does not cause the system to fail, but rather causes the system to fail only if the switch is needed and the switch has failed. For example, if the active component does not fail until the mission end time and the switch fails, then the system does not fail. However, if the active component fails and the switch has also failed, then the system cannot be switched to the standby component and it therefore fails.
 

In analyzing standby components with a switching device, either or both failure probabilities (during the switching or while waiting to switch) can be considered for the switch, since each probability can represent different failure modes. For example, the switch probability per request may represent software-related issues or the probability of detecting the failure of an active component, and the quiescent probability may represent wear-out type failures of the switch.
 

To illustrate the formulation, consider the previous example that assumes perfect switching. To examine the effects of including an imperfect switch, assume that when the active component fails there is a 90% probability that the switch will switch from the active component to the standby component. In addition, assume that the switch can also fail due to a wear-out failure mode described by a Weibull distribution with <math>\beta </math> = 1.7 and <math>\eta </math> = 5000.
 

Therefore, the reliability of the system at some time, <math>t</math> , is given by the following equation.

::<math>\begin{align}
R(t)= & {{R}_{1}}(t) \\
& +\underset{0}{\overset{t}{\mathop \int }}\,\{{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x) \\
& \cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}\cdot {{R}_{SW;Q}}(x)\cdot {{R}_{SW;REQ}}(x)\}dx
\end{align}</math>

Where:
 
:• <math>{{R}_{1}}</math> is the reliability of the active component. 
:• <math>{{f}_{1}}</math> is the <math>pdf</math> of the active component. 
:• <math>{{R}_{2;SB}}</math> is the reliability of the standby component when in standby mode (quiescent reliability). 
:• <math>{{R}_{2;A}}</math> is the reliability of the standby component when in active mode. 
:• <math>{{R}_{SW;Q}}</math> is the quiescent reliability of the switch. 
:• <math>{{R}_{SW;REQ}}</math> is the switch probability per request. 
:• <math>{{t}_{e}}</math> is the equivalent operating time for the standby unit if it had been operating at an active mode. 
This problem can be solved in BlockSim by including these probabilities in the container's properties, as shown in Figures "Standby container (switch) failure probabilities while attempting to switch" and "Standby container (switch) failure distribution while waiting to switch". In BlockSim, the standby container is acting as the switch.

 

[[Image:Fig 5.28_2.PNG|thumb|center|400px|<div align="center"> Standby container (switch) failure probabilities while attempting to switch </div> ]]
 

 

[[Image:Fig 5.27.PNG|thumb|center|500px|<div align="center"> Standby container (switch) failure distribution while waiting to switch </div>]]
 

 

Note that there are additional properties that can be specified in BlockSim for a switch, such as Switch Restart Probability, No. of Restarts and Switch Delay Time. In many applications, the switch is re-tested (or re-cycled) if it fails to switch the first time. In these cases, it might be possible that it switches in the second or third, or <math>{{n}^{th}}</math> attempt.
 
The Switch Restart Probability specifies each additional attempt's probability of successfully switching and the Finite Restarts specifies the total number of attempts. Note that the Switch Restart Probability specifies the probability of success of each trial (or attempt). The probability of success of <math>n</math> consecutive trials is calculated by BlockSim using the binomial distribution and this probability is then incorporated into equation above. The Switch Delay Time property is related to repairable systems and is considered in BlockSim only when using simulation. When using the analytical solution (i.e. for a non-repairable system), this property is ignored.
 

Solving the analytical solution (as given by Equation above), the following results are obtained.
 

[[Image:5-30.png|center|350px|]]
 

From the table above, it can be seen that the presence of a switching device has a significant effect on the reliability of a standby system. It is therefore important when modeling standby redundancy to incorporate the switching device reliability properties. It should be noted that this methodology is not the same as treating the switching device as another series component with the standby subsystem. This would be valid only if the failure of the switch resulted in the failure of system (e.g. switch failing open). In equation above, the Switch Probability per Request and quiescent probability are present only in the second term of the equation. Treating these two failure modes as a series configuration with the standby subsystem would imply that they are also present when the active component is functioning (i.e. first term of equation above). This is invalid and would result in the underestimation of the reliability of the system. In other words, these two failure modes become significant only when the active component fails.
 

As an example, and if we consider the warm standby case, the reliability of the system without the switch is 70.57% at 1000 hours. If the system was modeled so that the switching device was in series with the warm standby subsystem, the result would have been:

 

::<math>\begin{align}
{{R}_{S}}(1000)= & {{R}_{Standby}}(1000)\cdot {{R}_{sw,Q(1000)}}\cdot {{R}_{sw,req}} \\
= & 0.7057\cdot 0.9372\cdot 0.9 \\
= & 0.5952
\end{align}</math>

 

In the case where a switch failure mode causes the standby subsystem to fail, then this mode can be modeled as an individual block in series with the standby subsystem.

===Example 6===
Consider a car with four new tires and a full-size spare. Assume the following failure characteristics:
 
:• The tires follow a Weibull distribution with a <math>\beta=4</math> and an <math>\eta =</math> 40,000 miles while on the car due to wear. 
:• The tires also have a probability of failing due to puncture or other causes. For this, assume a constant rate for this occurrence with a probability of 1 every 50,000 miles. 
:• When not on the car (i.e. is a spare), a tire's probability of failing also has a Weibull distribution with a <math>\beta =</math> 2 and <math>\eta =</math> 120,000 miles. 
 
Assume a mission of 1,000 miles. If a tire fails during this trip, it will be replaced with the spare. However, the spare will not be repaired during the trip. In other words, the trip will continue with the spare on the car and if the spare fails the system will fail. Determine the probability of system failure.
====Solution to Example 6====
Active failure distribution for tires:
 
:• Due to wear-out, Weibull <math>\beta =4</math> and <math>\eta =40,000</math> miles. 
:• Due to random puncture, exponential <math>\mu =50,000.</math> 
:• The quiescent failure distribution is a Weibull distribution with <math>\beta =2</math> and <math>\eta =120,000</math> miles. 
 
The block diagram for each tire has two blocks in series, one block representing the wear-out mode and the other the random puncture mode, as shown next:
 
 
[[Image:small5.png|center|250px|]]
 

There are five tires, four active and one standby (represented in the diagram by a standby container with a 4-out-of-5 requirement), as shown next:
 
[[Image:BStirecontainer.png|center|250px|]]

For the standby Wear block, set the active failure and the quiescent distributions, but for the Puncture block, set only the active puncture distribution (because the tire cannot fail due to puncture while stored). Using BlockSim, the probability of system failure is found to be 0.003 or 0.3%.
 

=Note Regarding Numerical Integration Solutions=

Load sharing and standby solutions in BlockSim are performed using numerical integration routines. As with any numerical analysis routine, the solution error depends on the number of iterations performed, the step size chosen and related factors, plus the behavior of the underlying function. By default, BlockSim uses a certain set of preset factors. In general, these defaults are sufficient for most problems. If a higher precision or verification of the precision for a specific problem is required, BlockSim's preset options can be modified and/or the integration error can be assessed using the Integration Parameters... option for each container. For more details, you can refer to the documentation on the Algorithm Setup window in the BlockSim 8 User's help files.

Using FMRA to Estimate Baseline Reliability

2012-04-26T00:06:25Z

Dingzhou Cao: /* Illustrating the FMRA Process */

==Introduction==
As you may have seen when exploring the different standards and templates for FMEA, the analysis method can be modified to meet different objectives. Regardless of what the objective is, at the end of the day the FMEA process will produce a wealth of information from a cross-functional team that should be leveraged for subsequent Design for Reliability (DFR) activities.

Most of us are familiar with the traditional approaches of using information from the design FMEA (DFMEA) as an input to design verification plans (DVP&Rs), process FMEAs and process control plans. Some practitioners also have experience with using FMEA data to generate fault trees for advanced risk analysis. What we have not done to date is use the DFMEA as the starting point in our reliability analysis, and as an integral part of our DFR process. Along these lines, then, we would like to introduce you to a new type of analysis based on the DFMEA called '''Failure Modes and Reliability Analysis''' (FMRA).

<center>[[File:fmra_relationships.png]]</center>

==FMEA from a DFR and Reliability Perspective==
<div style="width: 320px; float: right;">[[File:fmra_build_configuration.png]]</div>On its own, the DFMEA activity accomplishes its objectives of identifying potential failures, assessing risk and initiating corrective actions to improve the design. In doing so, the analysis also produces a wealth of information that can be effectively leveraged by other activities. As an example, and early on in the design process, the DFMEA can be used to generate a baseline estimate of the design’s reliability, which is sought as part of the overall DFR program.

As a starting point when other reliability information is not yet available, the ''quantitative'' probability of occurrence for each failure cause can be obtained from the ''qualitative'' occurrence rating that has been assigned by the FMEA team as part of the traditional risk priority number (RPN) calculation. For the purposes of this analysis, the traditional FMEA occurrence scale can be expanded to include a quantitative value for each rating in the scale. For example, if the FMEA team assumed that the occurrence rating labeled “Rare” implies 1:100,000 then ''Probability of Occurrence = 0.0010%''.

This can be treated as a fixed probability (Q) that is the same regardless of how long the product operates. Alternatively, and for better reliability modeling, a life distribution could be used to describe this probability. For example, if the FMEA team assumed that the probability of occurrence '''by 1,000 hours of operation''' is “Rare” (1:100,000), then an exponential distribution could be easily substituted by computing a lambda for a time=1,000 where the unreliability (i.e., probability of occurrence of this failure cause) is 0.0010%.

Then, for each item within the existing FMEA, a fault tree or reliability block diagram (RBD) can be easily constructed relating the probability of occurrence of each cause to the probability of failure of the item. For example, if the team assumes that the item will fail if any one of the failure causes occurs, this could be modeled with a series configuration RBD or an OR gate in a fault tree. The model can then be expanded to the entire system using combinations of RBDs and fault trees, rolling up from the FMEA causes to the system level.

Note that for series reliability-wise configurations, the Xfmea/RCM++ software can automatically construct the RBDs (in the background) based on the system configuration and failure causes defined in the FMEAs. For more complex configurations, the software allows users to view and modify the configurations in a synchronized view of the FMRA in BlockSim.

<center>[[File:fmra_roll_up.png]]</center>

With this approach, the reliability modeling subject matter expert leverages the work done by the FMEA team to automatically create a baseline reliability model. Obviously, and at this stage, the results obtained at the system level are solely based on the probabilities of occurrence defined for each cause in the FMEA, which may or may not be correct. During this modeling activity, an overall assessment of the validity of these values can be performed and communicated back to the FMEA team for reassessment, modification or further information gathering actions (e.g., reliability testing). (See later discussion in the “FMRA Vetting Process” section.)

==Illustrating the FMRA Process==
To illustrate the FMRA process, consider a simple example based on the assembly/component DFMEAs for a single light pendant chandelier. The following picture shows the rating scale that was used by the team when they assigned an occurrence rating to each failure cause identified in the FMEA. In addition to the qualitative ratings and criteria (e.g., 1 = 1 in 1 Million), the scale also has a quantitative value associated with each rating (e.g., 1 in 1 Million = 0.000001).

<center>[[File:fmra_occurrence_scale_large.png]]</center>

Based on this probability of occurrence and its corresponding probabilistic definition, one can easily build a one-parameter exponential reliability model for each failure cause as follows:
 
<center><math>\text{Probability }\!\!~\!\!\text{ of }\!\!~\!\!\text{ Failure }\!\!~\!\!\text{ at }\!\!~\!\!\text{ Time}~t=Q\left( t \right)</math>

</center>

Assuming an exponential distribution, its single parameter λ can be estimated as follows:
 
<center><math>1-Q\left( T \right)=R\left( t \right)={{e}^{-\lambda t}}</math>
</center>
 
<center><math>\frac{-\text{ln}(1-Q(t))}{t}=\lambda </math></center>

Note that the exponential distribution is the default choice because a single probability value/time is the only information available. When better information is obtained, other more appropriate models should be utilized to describe the reliability. (See later discussion in the “FMRA Vetting Process” section.)

Now, assuming that any one of these causes could cause the component to fail (reliability-wise in series), an initial reliability estimate at any given time could easily be obtained by combining the causes to get the reliability of the component and then combining components and assemblies until we reach the system level. We will call this the “first draft” FMRA.
Note that more complex configurations may be appropriate to describe the reliability-wise relationships of the failure causes and/or components. These can be implemented in the analysis after the first draft is completed. (See later discussion in the “Next Steps” section.)

<center>[[File:fmra_table_3.png]]</center>

It is extremely important to note at this point that, even though we just computed a system reliability value, this first draft FMRA value may be nowhere close to the true reliability. What we need to do now is go back through the first draft and review each entry and result. We will call this subsequent step the “FMRA Vetting” process.

==The FMRA Vetting Process==
The first draft of the FMRA is just that, a draft. It needs to be thoroughly reviewed and vetted before proceeding. The list that follows outlines items that need to be considered in the vetting process.

<div style="border: 1px; background: #f7f7f7; padding-left: 10px; padding-right: 10px; padding-top: 10px; padding-bottom: 10px; border-left-style: Solid; border-left-width: 1px; border-right-style: Solid; border-right-width: 1px; border-top-style: Solid; border-top-width: 1px; border-bottom-style: Solid; border-bottom-width: 1px; border-left-color: #DADADA; border-right-color: #DADADA; border-top-color: #DADADA; border-bottom-color: #DADADA;">'''Tip:''' Within the Xfmea/RCM++ software, you can create a baseline (i.e., an exact replica of the project at a specific point in time) before any major change to the analysis. You can restore the baselines whenever it may be needed, which allow you to view the project as it was at the previous point in time.</div>

When you are ready to begin modifying the first draft of the FMRA in the Xfmea/RCM++ software, one option is to create a copy of the project so the original DFMEA can remain unchanged and you can modify the FMRA in a separate project. The drawback of this approach is that you will now dissociate the FMRA from the DFMEA and changes made to either will not be reflected in both. As an alternative, the list below includes tips for ways that you can adjust the FMRA for the purposes of a more accurate reliability calculation while still maintaining synchronization with the original DFMEA.

===Clean Up===

'''1) Discount/eliminate issues that have no impact on reliability.''' Depending on how the FMEA was done, there may be multiple items, functions, failures or causes in the DFMEA that have no impact on reliability. For example:

:a) A failure in the DFMEA could be "Design fails to aesthetically please the customer" with multiple causes such as "Red color is disliked by X% of the population," etc. While these are important considerations during design, the color of the chandelier is irrelevant from a reliability perspective.

:b) Other failures could be process issues that will be addressed in manufacturing, or items that will not impact reliability if the appropriate controls are put in place.

In short, we need to remove failures that are not reliability-related from our FMRA. If you wish to maintain synchronization with the original DFMEA, instead of deleting records from the original FMEA, you can set their reliability to 100% (i.e., cannot fail). Doing so excludes the issues from the reliability analysis but keeps the FMRA and DFMEA synchronized in the same project and tightly integrated.

'''2) Include other contributing items not considered in the FMEA that have an impact on reliability.''' There may be other failures including interfaces that were not included in the DFMEA. These will need to be added into the FMRA.

'''3) Account for common causes that may appear multiple times in the FMEA.''' Depending on how the DFMEA was done, a single cause may appear more than once. From a reliability perspective, the item only fails once, and a single failure should not be counted multiple times. If you wish to maintain synchronization with the original DFMEA, you can use the mirroring functionality to ensure that common cause failure modes are handled appropriately in the analysis.

===Review and Validate Inputs===
'''4) Review each occurrence rating assigned during the DFMEA and its derived reliability equivalent.''' The resulting values are only as good as the inputs provided (“garbage in, garbage out”). In this case, the inputs are the qualitative occurrence ratings from the FMEA team. The team could be wrong on some or all of them.

:a) Compare/cross-reference the occurrence rating value with other FMEAs done by other teams on similar items and similar environments. In Xfmea/RCM++, it is easy to search through all FMEAs stored in the same Synthesis repository.

:b) Review any available historical data, published data and warranty data, as well as all related analysis and models that have been performed to describe the reliability of the item at the expected use conditions.

::*Look for similar models in the Synthesis repository (i.e., Weibull++, ALTA or other analyses on similar items).
::*Look for reference data in published standards (e.g., standards based reliability prediction).
::*Cross-reference with data from the failure reporting, analysis and corrective action system (FRACAS).

:c) Get expert opinion and use the Quick Parameter Estimator within Xfmea/RCM++ to translate these opinions into usable models.
:d) Use physics of failure, computer simulation, finite element analysis and other tools and methods.
:e) In cases where no reasonable assessment can be made, testing may need to be performed. Make this testing a part of the reliability plan and set aside a budget for it.
:f) Do a common-sense reality check on the values given! As an example, in the chandelier FMRA example shown above, the reliability of the bulb was calculated as 93% after 5,000 hours of operation. If this is an incandescent bulb, this value may be overly optimistic.

'''5) Question the dangerous exponential assumption.''' Even though we used an exponential model for the initial transition to a reliability model, remember that this distribution assumes a constant failure rate. For the majority of items, this assumption is invalid. If wearout is present or suspected, you may need to replace these initial models with distributions that have non-constant failure rates (e.g., Weibull or lognormal). In the absence of data, you can use the Quick Parameter Estimator within Xfmea/RCM++ to translate these into different models. For example, if beta is known for a specific failure mode, use a 1-parameter Weibull distribution coupled with the stated probability.

'''6) Comparatively review and rank all values to further identify inconsistencies.''' Re-compute the FMRA based on the modifications performed so far. Use the color-coding feature to look at causes/failure modes that are high unreliability contributors. Assess if this is valid. For each item in question, repeat the above steps.

'''7) Review all items in question with the original FMEA team, revise the DFMEA as appropriate and regenerate the FMRA.''' At this point, the DFMEA will likely be modified to take the new information into account, and then the FMRA can be regenerated. Depending on the number of changes and the extent of your participation, you may have to go through the vetting process again.

<center>[[File:fmra_colors.png]]</center>

==Next Steps==
Once vetted, the baseline reliability value is now your initial design reliability. Compare this with the target reliability, keeping in mind that this first baseline estimate is usually optimistic. Furthermore, the product needs to go through manufacturing, and the manufacturing process isn’t going to increase the reliability, thus you want your initial baseline reliability value to exceed your reliability target value.

If the initial baseline is not sufficient for the target, you may expand the analysis to include RBDs in BlockSim and continue with different types of analysis including reliability importance and reliability allocation. Reliability importance analysis provides more advanced methods to identify the issues that have the biggest contribution on the overall reliability. Reliability allocation analysis provides reliability requirements for each item (all the way down to the cause level) so that the target is met. For causes that have higher reliability requirements (which translate back to lower probability of occurrence), a review of the new requirements with the FMEA team is advised so that additional corrective actions can be assigned if necessary to assure that the new requirements are met. This may again require a revision of the FMEA and the FMRA.

From a knowledge base perspective, the FMEA and related FMRA should continuously be updated as new information becomes available, including the addition of new failure modes uncovered during testing and the revision of the underlying reliability models based on data obtained from testing. After the product’s release, this process should continue with field information.

Time-Dependent System Reliability (Analytical)

2012-03-21T21:27:47Z

Dingzhou Cao: /* Solution to Example 4 */

{{Template:bsbook|5}}

In the previous chapter, different system configuration types were examined, as well as different methods for obtaining the system's reliability function analytically. Because the reliabilities in the problems presented were treated as probabilities (e.g. <math>P(A)</math> , <math>{{R}_{i}}</math> ), the reliability values and equations presented were referred to as static (not time-dependent). Thus, in the prior chapter, the life distributions of the components were not incorporated in the process of calculating the system reliability. In this chapter, time dependency in the reliability function will be introduced. We will develop the models necessary to observe the reliability over the life of the system, instead of at just one point in time. In addition, performance measures such as failure rate, MTTF and warranty time will be estimated for the entire system. The methods of obtaining the reliability function analytically remain identical to the ones presented in the previous chapter, with the exception that the reliabilities will be functions of time. In other words, instead of dealing with <math>{{R}_{i}}</math> , we will use <math>{{R}_{i}}(t)</math> . All examples in this chapter assume that no repairs are performed on the components.

=Analytical Life Predictions=
The analytical approach presented in the prior chapter involved the determination of a mathematical expression that describes the reliability of the system, expressed in terms of the reliabilities of its components. So far we have estimated only static system reliability (at a fixed time). For example, in the case of a system with three components in series, the system's reliability equation was given by:
 

::<math>{{R}_{s}}={{R}_{1}}\cdot {{R}_{2}}\cdot {{R}_{3}} \ </math>
 

The values of <math>{{R}_{1}}</math> , <math>{{R}_{2}}</math> and <math>{{R}_{3}}</math> were given for a common time and the reliability of the system was estimated for that time. However, since the component failure characteristics can be described by distributions, the system reliability is actually time-dependent. In this case, the equation above can be rewritten as:
 

::<math>{{R}_{s}}(t)={{R}_{1}}(t)\cdot {{R}_{2}}(t)\cdot {{R}_{3}}(t)</math>

 

The reliability of the system for any mission time can now be estimated. Assuming a Weibull life distribution for each component, the first equation above can now be expressed in terms of each component's reliability function, or:

 

[[Category:Completed Theoretical Review]]

::<math>{{R}_{s}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}}\cdot {{e}^{-{{\left( \tfrac{t}{{{\eta }_{2}}} \right)}^{{{\beta }_{2}}}}}}\cdot {{e}^{-{{\left( \tfrac{t}{{{\eta }_{3}}} \right)}^{{{\beta }_{3}}}}}}</math>

 

In the same manner, any life distribution can be substituted into the system reliability equation. Suppose that the times-to-failure of the first component are described with a Weibull distribution, the times-to-failure of the second component with an exponential distribution and the times-to-failure of the third component with a normal distribution. Then the first equation above can be written as:
 

::<math>{{R}_{s}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}}\cdot {{e}^{-{{\lambda }_{2}}t}}\cdot \left[ 1-\Phi \left( \frac{t-{{\mu }_{3}}}{{{\sigma }_{3}}} \right) \right]</math>

 

It can be seen that the biggest challenge is in obtaining the system's reliability function in terms of component reliabilities, which has already been discussed in depth. Once this has been achieved, calculating the reliability of the system for any mission duration is just a matter of substituting the corresponding component reliability functions into the system reliability equation.
===Advantages of the Analytical Method===
The primary advantage of the analytical solution is that it produces a mathematical expression that describes the reliability of the system. Once the system's reliability function has been determined, other calculations can then be performed to obtain metrics of interest for the system. Such calculations include:
 
:• Determination of the system's <math>pdf.</math> 
:• Determination of warranty periods. 
:• Determination of the system's failure rate. 
:• Determination of the system's MTTF. 
 
In addition, optimization and reliability allocation techniques can be used to aid engineers in their design improvement efforts. Another advantage in using analytical techniques is the ability to perform static calculations and analyze systems with a mixture of static and time-dependent components. Finally, the reliability importance of components over time can be calculated with this methodology.
===Disadvantages of the Analytical Method===
The biggest disadvantage of the analytical method is that formulations can become very complicated. The more complicated a system is, the larger and more difficult it will be to analytically formulate an expression for the system's reliability. For particularly detailed systems this process can be quite time-consuming, even with the use of computers. Furthermore, when the maintainability of the system or some of its components must be taken into consideration, analytical solutions become intractable. In these situations, the use of simulation methods may be more advantageous than attempting to develop a solution analytically. Simulation methods are presented in later chapters.
===Looking at a Simple ''Complex'' System Analytically===

The complexity involved in an analytical solution can be best illustrated by looking at the simple ''complex'' system with 15 components, as shown in Figure " An RBD of a complex system".
 
[[Image:5-1.png|thumb|center|400px|<div align="center"> An RBD of a complex system.</div>]]

 
The system reliability for this system (computed using BlockSim) is shown next. The first solution is provided using BlockSim's symbolic solution. In symbolic mode, BlockSim breaks the equation into segments, identified by tokens, that need to be substituted into the final system equation for a complete solution. This creates algebraic solutions that are more compact than if the substitutions were made.
 

::<math>\begin{align}
{{R}_{System}}= & D2\cdot D3\cdot {{R}_{L}} \\
D3= & +{{R}_{K}}\cdot IK \\
IK= & +{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}} \\
D2 = & +{{R}_{A}}\cdot {{R}_{E}}\cdot IE \\
IE = & -D1\cdot {{R}_{M}}\cdot {{R}_{N}}+{{R}_{M}}\cdot {{R}_{N}}+D1 \\
D1 = & +{{R}_{D}}\cdot ID \\
ID = & -{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}}
\end{align}</math>
 
Substituting the terms yields:
 

::<math>\begin{align}
{{R}_{System}}= & {{R}_{A}}\cdot {{R}_{E}}\cdot {{R}_{L}}\cdot {{R}_{K}} \\
& \cdot \{({{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})\cdot {{R}_{M}}\cdot {{R}_{N}} \\
& +{{R}_{M}}\cdot {{R}_{N}}-{{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}}\} \\
& \cdot \{{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}}\}
\end{align}</math>

 
BlockSim's automatic algebraic simplification would yield the following format for the above solution:
 

::<math>\begin{align}
{{R}_{System}}= & (({{R}_{A}}\cdot {{R}_{E}}(-({{R}_{D}}(-{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})){{R}_{M}}\cdot {{R}_{N}} \\
& +{{R}_{M}}\cdot {{R}_{N}} \\
& +({{R}_{D}}(-{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})))) \\
& ({{R}_{K}}({{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& +RI\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}})){{R}_{L}}) \
\end{align}</math>

 
In this equation, each <math>{{R}_{i}}</math> represents the reliability function of a block. For example, if <math>{{R}_{A}}</math> has a Weibull distribution, then each <math>{{R}_{A}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{A}}} \right)}^{{{\beta }_{A}}}}}}</math> and so forth. Substitution of each component's reliability function in the last <math>{{R}_{System}}</math> equation above will result in an analytical expression for the system reliability as a function of time, or <math>{{R}_{s}}(t)</math> , which is the same as <math>(1-cd{{f}_{System}}).</math>

===Obtaining Other Functions of Interest===
Once the system reliability equation (or the cumulative density function, <math>cdf</math> ) has been determined, other functions and metrics of interest can be derived.
 
Consider the following simple system:
 
[[Image:5-2.png|thumb|center|300px|<div align="center"> Simple two-component system. </div>]]

 
Furthermore, assume that component 1 follows an exponential distribution with a mean of 10,000 (<math>\mu =10,000,</math> <math>\lambda =1/10,000)</math> and component 2 follows a Weibull distribution with <math>\beta =6</math> and <math>\eta =10,000</math> . The reliability equation of this system is:

 
::<math>\begin{align}
{{R}_{S}}(t)= & {{R}_{1}}(t)\cdot {{R}_{2}}(t) \\
= & {{e}^{-\lambda t}}\cdot {{e}^{-{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \\
= & {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \
\end{align}</math>

 
The system <math>cdf</math> is:

 
::<math>\begin{align}
{{F}_{S}}(t)= & 1-({{R}_{1}}(t)\cdot {{R}_{2}}(t)) \\
= & 1-\left( {{e}^{-\lambda t}}\cdot {{e}^{-{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \right) \\
= & 1-\left( {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)
\end{align}</math>

 
{{analytical system pdf}}

====Conditional Reliability====
Conditional reliability is the probability of a system successfully completing another mission following the successful completion of a previous mission. The time of the previous mission and the time for the mission to be undertaken must be taken into account for conditional reliability calculations. The system's conditional reliability function is given by:

::<math>R(T,t)=\frac{R(T+t)}{R(T)}</math>

Equation above gives the reliability for a new mission of duration <math>t</math> having already accumulated <math>T</math> hours of operation up to the start of this new mission. The system is evaluated to assure that it will start the next mission successfully.
 
For the system in Figure "Simple two-component system.", the reliability for mission of <math>t=1,000</math> hours, having an age of <math>T=500</math> hours, is:
 

::<math>\begin{align}
{{R}_{S}}(T=500,t=1000)= & \frac{R(T+t)}{R(T)} \\
= & \frac{R(1500)}{R(500)} \\
= & \frac{{{e}^{-\tfrac{1500}{10,000}}}\cdot {{e}^{-{{\left( \tfrac{1500}{10,000} \right)}^{6}}}}}{{{e}^{-\tfrac{500}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{500}{10,000} \right)}^{6}}}}} \\
= & 0.9048=90.48%
\end{align}</math>

 

====Conditional Reliability for Components====

Now in this formulation, it was assumed that the accumulated age was equivalent for both units. That is, both started life at zero and aged to 500. It is possible to consider an individual component that has already accumulated some age (used component) in the same formulation. To illustrate this, assume that component 2 started life with an age of T=100. Then the reliability equation of the system, as given in <math>{{R}_{S}}(t)= {{R}_{1}}(t)\cdot {{R}_{2}}(t)</math>, would need to be modified to include a conditional term for 2, or:
 

::<math>{{R}_{S}}(t)={{R}_{1}}(t)\cdot \frac{{{R}_{2}}({{T}_{2}}+t)}{{{R}_{2}}({{T}_{2}})} \ </math>

 

In BlockSim, the start age input box may be used to specify a starting age greater than zero.
{{system failure rate analytical}}

====System Mean Life (Mean Time To Failure)====
The mean life (or mean time to failure, MTTF) can be obtained by integrating the system reliability function from zero to infinity:
 

::<math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math>

 

The mean time is a performance index and does not provide any information about the behavior of the failure distribution of the system.
 
For the system in Figure "Simple two-component system":
 

::<math>\begin{align}
MTTF= & \int_{0}^{\infty }\left( {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)dt \\
= & 5978.9
\end{align}</math>

 

====Warranty Period and BX Life====
Sometimes it is desirable to know the time value associated with a certain reliability. Warranty periods are often calculated by determining what percentage of the failure population can be covered financially and estimating the time at which this portion of the population will fail. Similarly, engineering specifications may call for a certain BX life, which also represents a time period during which a certain proportion of the population will fail. For example, the B10 life is the time in which 10% of the population will fail.
This is obtained by setting <math>{{R}_{S}}(t)</math> to the desired value and solving for <math>t.</math>
 
For the system in Figure "Simple two-component system":
 

::<math>{{R}_{s}}\left( t \right)={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>
 

To compute the time by which reliability would be equal to 90%, equation above is recast as follows and solved for <math>t.</math>

 
::<math>0.90={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>

 

In this case, <math>t=1053.59</math> . Equivalently, the B10 life for this system is also <math>1053.59</math> .
Except for some trivial cases, a closed form solution for <math>t</math> cannot be obtained. Thus, it is necessary to solve for <math>t</math> using numerical methods. BlockSim uses numerical methods.

===Example 1===
Consider a system consisting of three exponential units in series with the following failure rates (in failures per hour): <math>{{\lambda }_{1}}</math> = 0.0002, <math>{{\lambda }_{2}}</math> = 0.0005 and <math>{{\lambda }_{3}}</math> = 0.0001.
 
:• Obtain the reliability equation for the system. 
:• What is the reliability of the system after 150 hours of operation? 
:• Obtain the system's <math>pdf.</math> 
:• Obtain the system's failure rate equation. 
:• What is the MTTF for the system? 
:• What should the warranty period be for a 90% reliability? 
====Solution to Example 1====
 
:• The analytical expression for the reliability of the system is given by:

::<math>\begin{align}
{{R}_{s}}(t)= & {{R}_{1}}(t)\cdot {{R}_{2}}(t)\cdot {{R}_{3}}(t) \\
= & {{e}^{-{{\lambda }_{1}}t}}\cdot {{e}^{-{{\lambda }_{2}}t}}\cdot {{e}^{-{{\lambda }_{1}}t}} \\
= & {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}
\end{align}</math>

 

At 150 hours of operation, the reliability of the system is:
 

::<math>\begin{align}
{{R}_{s}}(t)= & {{e}^{-(0.0002+0.0005+0.0001)150}} \\
= & 0.8869\text{ or }88.69%
\end{align}</math>

 

:• In order to obtain the system's <math>pdf</math> , the derivative of the reliability equation given in the first equation above is taken with respect to time, or:
 
::<math>\begin{align}
{{f}_{s}}(t)= & -\frac{d[{{R}_{s}}(t)]}{dt} \\
= & -\frac{d\left[ {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}} \right]}{dt} \\
= & ({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})\cdot {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}
\end{align}</math>

:• The system's failure rate can now be obtained simply by dividing the system's <math>pdf</math> given in the equation above by the system's reliability function given in the first equation above, and:

::<math>\begin{align}
{{\lambda }_{s}}\left( t \right)= & \frac{{{f}_{s}}\left( t \right)}{{{R}_{s}}\left( t \right)} \\
= & \frac{({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})\cdot {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}}{{{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}} \\
= & ({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}) \\
= & 0.0008\text{ }fr/hr
\end{align}</math>

 
:• Combining <math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math> and the first equation above, the system's MTTF can be obtained:

 
::<math>\begin{align}
MTTF= & \int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \\
= & \int_{0}^{\infty }{{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}dt \\
= & \frac{1}{({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})} \\
= & 1250\text{ }hr
\end{align}</math>
 

:• Solving the first equation above with respect to time will yield the corresponding warranty period for a 90% reliability. In this case, the system reliability equation is simple and a closed form solution exists. The warranty time can now be found by solving:

 
::<math>\begin{align}
t= & -\frac{\ln (R)}{{{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}} \\
= & -\frac{\ln (0.9)}{0.0008} \\
= & 131.7\text{ }hr
\end{align}</math>
 

Thus, the warranty period should be 132 hours.

===Example 2===
Consider the system shown in Figure "Complex bridge system in Example 2".
 
[[Image:BS5.5.png|thumb|center|300px|<div align="center"> Complex bridge system in Example 2. </div>]]
 

Components <math>A</math> through <math>E</math> are Weibull distributed with <math>\beta =1.2</math> and <math>\eta =1230</math> hours. The starting and ending blocks cannot fail. 
Determine the following:
 
:• The reliability equation for the system and its corresponding plot. 
:• The system's <math>pdf</math> and its corresponding plot. 
:• The system's failure rate equation and the corresponding plot. 
:• The <math>MTTF</math> . 
:• The warranty time for a 90% reliability. 
:• The reliability for a 200-hour mission, if it is known that the system has already successfully operated for 200 hours. 
====Solution====
 
The first step is to obtain the reliability function for the system. The methods described in the previous chapter can be employed, such as the event space or path-tracing methods. Using BlockSim, the following reliability equation is obtained:
 

::<math>\begin{align}
{{R}_{s}}(t)= & ({{R}_{Start}}\cdot {{R}_{End}}(2{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}))
\end{align}</math>
 

Note that since the starting and ending blocks cannot fail, <math>{{R}_{Start}}=1</math> and <math>{{R}_{End}}=1,</math> equation above can be reduced to:

 

::<math>\begin{align}
{{R}_{s}}(t)= & 2\cdot {{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}
\end{align}</math>

 

Where <math>{{R}_{A}}</math> is the reliability equation for Component A, or:

 

::<math>{{R}_{A}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{A}}} \right)}^{{{\beta }_{A}}}}}}</math>

 

::<math>{{R}_{B}}</math> is the reliability equation for Component <math>B</math> , etc.
 

Since the components in this example are identical, the system reliability equation can be further reduced to:

 

::<math>{{R}_{s}}(t)=2R{{(t)}^{2}}+2R{{(t)}^{3}}-5R{{(t)}^{4}}+2R{{(t)}^{5}}</math>

 

Or, in terms of the failure distribution:

 

::<math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>

 

The corresponding plot is given in Figure "Reliability plot for the system".
 
[[Image:BS5.6.png|thumb|center|300px|<div align="center"> Reliability plot for the system. </div>]]
 
 

In order to obtain the system's <math>pdf</math> , the derivative of the reliability equation given above is taken with respect to time, resulting in:
 

::<math>\begin{align}
{{f}_{s}}(t)= & 4\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+6\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \\
& -20\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+10\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}
\end{align}</math>

 

The <math>pdf</math> can now be plotted for different time values, <math>t</math> , as shown in Figure "''pdf'' plot for the system".
 

The system's failure rate can now be obtained by dividing the system's <math>pdf</math> given in equation above by the system's reliability function given in <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>, or:

 

::<math>\begin{align}
{{\lambda }_{s}}(t)= & \frac{4\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+6\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}{2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}} \\
& +\frac{-20\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+10\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}{2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}
\end{align}</math>

 

The corresponding plot is given in Figure "Failure rate for the system".
 
[[Image:BS5.7.gif|thumb|center|300px|<div align="center"> ''pdf'' plot for the system.</div>]]
 

The <math>MTTF</math> of the system is obtained by integrating the system's reliability function given by <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math> from time zero to infinity, as given by <math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math>. Using BlockSim's Analytical QCP, an <math>MTTF</math> of 1007.8 hours is calculated, as shown in Figure "MTTF of the system".
 

The warranty time can be obtained by solving <math>{{R}_{s}}(t)</math> with respect to time for a system reliability <math>{{R}_{s}}=0.9</math> . Using the Analytical QCP and selecting the 
Warranty Time option, a time of 372.72 hours is obtained, as shown in Figure "Time at which ''R''=0.9 or 90% for the system".
 
[[Image:BS5.8.gif|thumb|center|400px|<div align="center"> Failure rate for the system.</div>]]
 
[[Image:Fig 5.9.PNG|thumb|center|400px|<div align="center"> MTTF of the system. </div>]]
 
[[Image:Fig 5.10.PNG|thumb|center|400px|<div align="center"> Time at which ''R''=0.9 or 90% for the system.</div>]]
 
[[Image:Fig 5.11.PNG|thumb|center|400px|<div align="center"> Conditional reliability calculation for the system.</div>]]
 

Lastly, the conditional reliability can be obtained using <math>R(T,t)=\frac{R(T+t)}{R(T)}</math> and <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>, or:
 

::<math>\begin{align}
R(200,200)= & \frac{R(400)}{R(200)} \\
= & \frac{0.883825}{0.975321} \\
= & 0.906189
\end{align}</math>

 

This can be calculated using BlockSim's Analytical QCP, as shown in Figure "Conditional reliability calculation for the system".
 

=Approximating the System CDF=

In many cases, it is valuable to fit a distribution that represents the system's times-to-failure. This can be useful when the system is part of a larger assembly and may be used for repeated calculations or in calculations for other systems. In cases such as this, it can be useful to characterize the system's behavior by fitting a distribution to the overall system and calculating parameters for this distribution. This is equivalent to fitting a single distribution to describe <math>{{R}_{S}}(t</math> ). In essence, it is like reducing the entire system to a component in order to simplify calculations.
 
For the system in Figure below:
[[Image:5-2.png|thumb|center|300px|<div align="center">Simple two-component system. </div>]]
 

::<math>{{R}_{S}}(t)={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>

 
To compute an approximate reliability function for this system, <math>{{R}_{A}}(t)\simeq {{R}_{S}}(t)</math> , one would compute <math>n</math> pairs of reliability and time values and then fit a single distribution to the data, or:

 

::<math>\begin{align}
{{R}_{S}}(t= & 10,396.7)=10% \\
{{R}_{S}}(t= & 9,361.9)=20% \\
& ... \\
{{R}_{S}}(t= & 1,053.6)=90%
\end{align}</math>

 

A single distribution, <math>{{R}_{A}}(t)</math> , that approximates <math>{{R}_{S}}(t)</math> can now be computed from these pairs using life data analysis methods. If using the Weibull++ software, one would enter the values as free form data.

===Example 3===
Compute a single Weibull distribution approximation for the system in Example 2.
====Solution to Example 3====
The system in the previous example, shown in Figure "Complex bridge system in Example 2", can be approximated by use of a 2-parameter Weibull distribution with <math>\beta =2.02109</math> and <math>\eta =1123.51</math> . In BlockSim, this is accomplished by representing the entire system as one distribution by going to the Distribution Fit window Figure " Representing a system with a distribution".

 
[[Image:Fig 5.13.PNG|thumb|center|400px|<div align="center"> Representing a system with a distribution.</div>]]
 
[[Image:Fig 5.12.PNG|thumb|center|300px|<div align="center"> Distribution Fitting window.</div>]]
 
by clicking the Distribution Fit Window, the Distribution Estimator window will appear (Figure "Distribution Fitting window").
 
In this window you can select a distribution to represent the data. BlockSim will then generate a number of system failure times based on the system's reliability function. The system's reliability function can be used to solve for a time value associated with that unreliability value. The distribution of the generated time values can then be fitted to a probability distribution function.
 

Consider a value of <math>F(t)=0.11</math> . Using the system's reliability equation and solving for time, the corresponding time-to-failure for a 0.11 unreliability can be calculated. 
For the system of Example 2, the time for a 0.11 unreliability is 389.786 hours.
 
When enough points have been generated, the selected distribution will be fitted to this data set and the distribution's parameters will be returned. In addition, if ReliaSoft's Weibull++ is installed, the generated data can be viewed/analyzed using a Weibull++ instance, as shown in Figure "Using Weibull++ to calculate distribution parameters".
 
[[Image:Fig 5.14.PNG|thumb|center|400px|<div align="center"> Using Weibull++ to calculate distribution parameters.</div>]]
 
It is recommended that the analyst examine the fit to ascertain the applicability of the approximation.

=Duty Cycle=

Components of a system may not operate continuously during a system's mission, or may be subjected to loads greater or lesser than the rated loads during system operation. To model this, a factor called the Duty Cycle ( <math>{{d}_{c}}</math> ) is used. The duty cycle may also be used to account for changes in environmental stress, such as temperature changes, that may effect the operation of a component. The duty cycle is a positive value, with a default value of 1 representing continuous operation at rated load, and any values other than 1 representing other load values with respect to the rated load value (or total operating time). A duty cycle value higher than 1 indicates a load in excess of the rated value. A duty cycle value lower than 1 indicates that the component is operating at a load lower than the rated load or not operating continuously during the system's mission. For instance, a duty cycle of 0.5 may be used for a component that operates only half of the time during the system's mission.
 
The reliability metrics for a component with a duty cycle are calculated as follows. Let <math>{{d}_{c}}</math> represent the duty cycle during a particular mission of the component, <math>t</math> represent the mission time and <math>{t}'</math> represent the accumulated age. Then:
 

::<math>{t}'={{d}_{c}}\times t</math>

 

The reliability equation for the component is:
 

::<math>R({t}')=R({{d}_{c}}\times t)</math>
 

The component ''pdf'' is:
 

::<math>f({t}')=-\frac{d(R({t}'))}{dt}=-\frac{d(R({{d}_{c}}\times t))}{dt}={{d}_{c}}f({{d}_{c}}\times t)</math>
 

The failure rate of the component is:
 

::<math>\lambda ({t}')=\frac{f({t}')}{R({t}')}=\frac{{{d}_{c}}f({{d}_{c}}\times t)}{R({{d}_{c}}\times t)}={{d}_{c}}\lambda ({{d}_{c}}\times t)</math>

===Example 4===
Consider a computer system with three components: a processor, a hard drive and a CD drive in series as shown next. Assume that all three components follow a Weibull failure distribution with the parameters <math>{{\beta }_{1}}=1.5</math> and <math>{{\eta }_{1}}=5000</math> for the processor, <math>{{\beta }_{2}}=2.5</math> and <math>{{\eta }_{2}}=3000</math> for the hard drive, and <math>{{\beta }_{3}}=2</math> and <math>{{\eta }_{3}}=4000</math> for the CD drive. Determine the reliability of the computer system after one year (365 days) of operation, assuming that the CD drive is used only 30% of the time.
 
[[Image:BS5ex4.png|thumb|center|300px|]]

====Solution to Example 4====
The reliability of the processor after 365 days of operation is given by:

::<math>\begin{align}
{{R}_{processor}}(365)= & {{e}^{-{{\left( \tfrac{365}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}} \\
= & {{e}^{-{{\left( \tfrac{365}{5000} \right)}^{1.5}}}} \\
= & 0.9805\text{ or }98.05%
\end{align}</math>

The reliability of the hard drive after 365 days of operation is given by:

::<math>\begin{align}
{{R}_{harddrive}}(365)= & {{e}^{-{{\left( \tfrac{365}{{{\eta }_{2}}} \right)}^{{{\beta }_{2}}}}}} \\
= & {{e}^{-{{\left( \tfrac{365}{3000} \right)}^{2.5}}}} \\
= & 0.9948\text{ or }99.48%
\end{align}</math>

The reliability of the CD drive after 365 days of operation (taking into account the 30% operation using a duty cycle of 0.3) is given by:

::<math>\begin{align}
{{R}_{CDdrive}}(365)= & {{e}^{-{{\left( \tfrac{{{d}_{c}}\times 365}{{{\eta }_{3}}} \right)}^{{{\beta }_{3}}}}}} \\
= & {{e}^{-{{\left( \tfrac{0.3\times 365}{4000} \right)}^{2}}}} \\
= & 0.9993\text{ or }99.93%
\end{align}</math>
 
[[Image:Fig 5.15.PNG|thumb|center|400px|<div align="center"> Result for the computer system reliability.</div>]]
 

Thus the reliability of the computer system after 365 days of operation is:

::<math>\begin{align}
{{R}_{s}}(365)= & {{R}_{processor}}(365)\cdot {{R}_{harddrive}}(365)\cdot {{R}_{CDdrive}}(365) \\
= & 0.9805\cdot 0.9948\cdot 0.9993 \\
= & 0.9747\text{ or }97.47%
\end{align}</math>
 
This result can be obtained in BlockSim as shown in Figure "Result for the computer system reliability".
 

=Load Sharing=
As presented in earlier chapters, a reliability block diagram (RBD) allows you to graphically represent how the components within a system are reliability-wise connected. In most cases, independence is assumed across the components within the system. For example, the failure of component A does not affect the failure of component B. However, if a system consists of components that are sharing a load, then the assumption of independence no longer holds true.
 

If one component fails, then the component(s) that are still operating will have to assume the failed unit's portion of the load. Therefore, the reliabilities of the surviving unit(s) will change. Calculating the system reliability is no longer an easy proposition. In the case of load sharing components, the change of the failure distributions of the surviving components must be known in order to determine the system's reliability.
 

To illustrate this, consider the a system of two units connected reliability-wise in parallel (Figure "Two units connected reliability-wise in parallel").
 

[[Image:5-16.png|thumb|center|300px|<div align="center"> Two units connected reliability-wise in parallel.</div>]]
 

Assume that the units must supply an output of 8 volts and that if both units are operational, each unit is to supply 50% of the total output. If one of the units fails, then the surviving unit supplies 100%. Furthermore, assume that having to supply the entire load has a negative impact on the reliability characteristics of the surviving unit. Since the reliability characteristics of the unit change based on whether both or only one is operating, a life distribution along with a life-stress relationship (as discussed in [http://reliawiki.com/index.php/Statistical_Background#A_Brief_Introduction_to_Life-Stress_Relationships|A Brief Introduction to Life-Stress Relationships]) will be needed to model each component.
 

To illustrate the steps needed, we will create the model starting from raw data. Assume that a total of 20 units were tested to failure at 7, 10 and 15 volts. The test data set is presented in the next table.
 

[[Image:5-17.png|thumb|center|300px|]]
 

For this example, Units 1 and 2 are the same component. Therefore, only one set of data was collected. However, it is possible that the load sharing components in a system may not be the same. If that were the case, data would need to be collected for each component.
 

The data set in Table 1 was analyzed using ReliaSoft's ALTA software (as shown in Figure "Using ALTA to calculate component parameters") with the Inverse Power Law as the underlying life-stress relationship and Weibull as the life distribution.
 
 

The estimated model parameters, <math>\beta </math> , <math>K</math> and <math>n</math> , are shown next.
 

::<math>\begin{align}
\beta = & 1.9239 \\
K= & 3.2387\times {{10}^{-7}} \\
n= & 3.4226
\end{align}</math>
 

:Or:
 
::<math>\begin{align}
{{R}_{1}}(t,{{S}_{1}})= & {{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} \\
= & {{e}^{-{{\left( 3.2387\times {{10}^{-7}}S_{1}^{3.4226}t \right)}^{1.9239}}}} \
\end{align}</math>.

 
::<math>{{f}_{1}}(t,{{S}_{1}})=\beta KS_{1}^{n}{{\left( KS_{1}^{n}t \right)}^{\beta -1}}{{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} \ </math>
 

And for this case:

::<math>\begin{align}
{{R}_{1}}(t,{{S}_{1}})= & {{R}_{2}}(t,{{S}_{2}}) \\
{{f}_{1}}(t,{{S}_{1}})= & {{f}_{2}}(t,{{S}_{2}})
\end{align}</math>
 
 
[[Image:Fig 5.17.PNG|thumb|center|500px|<div align="center"> Using ALTA to calculate component parameters.</div>]]
 

Figure "Reliability curves for different voltage output conditions" shows a plot of <math>{{R}_{1}}(t,{{S}_{1}})={{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} = {{e}^{-{{\left( 3.2387\times {{10}^{-7}}S_{1}^{3.4226}t \right)}^{1.9239}}}}\ </math>.
 

Now that the failure properties have been determined using the test data, the reliability of the system at some time, <math>t</math> , can be calculated using the following equation:

 
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>

 
:Where:
 

::<math>\begin{align}
{{S}_{1}}= & {{P}_{1}}S \\
{{S}_{2}}= & {{P}_{2}}S
\end{align}</math>

 
[[Image:5-18.png|thumb|center|300px|<div align="center"> Reliability curves for different voltage output conditions. </div>]]
 
[[Image:BS5.19.png|thumb|center|395px| Illustrating <math>{{t}_{e}}</math>]]
 

And:
 

:• <math>S</math> is the total load (or required output). 
:• <math>{{P}_{1}}</math> and <math>{{P}_{2}}</math> are the portion of the total load that each unit supports when both units are operational. In this case, <math>{{P}_{1}}={{P}_{2}}=0.5=50%.</math> 
:• <math>{{S}_{1}}</math> and <math>{{S}_{2}}</math> represent the portions of the load that Unit 1 and Unit 2 must support when both units are operational. 
:• <math>{{t}_{{{1}_{e}}}}</math> is the equivalent operating time for Unit 1 if it had been operating at <math>S</math> instead of <math>{{S}_{1}}</math> . A graphical representation of the equivalent time is shown in Figure 5.19, where the curve marked by L represents the low stress (load) and the curve marked by H represents the high stress (load). 
 

::<math>{{t}_{1e}}</math> can be calculated by:
 
::<math>\begin{align}
{{R}_{1}}(t)= & {{R}_{1}}({{t}_{1e}}) \\
{{e}^{-{{(tKS_{1}^{n})}^{\beta }}}}= & {{e}^{-{{({{t}_{1e}}K{{S}^{n}})}^{\beta }}}} \\
tS_{1}^{n}= & {{t}_{1e}}{{S}^{n}} \\
{{t}_{1e}}= & t{{\left( \frac{{{S}_{1}}}{S} \right)}^{n}},\text{ }{{S}_{1}}={{P}_{1}}S \\
\therefore & {{t}_{1e}}=tP_{1}^{n}
\end{align}</math>

 
::<math>{{t}_{2e}}</math> can be calculated the same way, or:

 
::<math>\begin{align}
{{R}_{2}}(t)= & {{R}_{2}}({{t}_{2e}}) \\
\therefore & {{t}_{2e}}=tP_{2}^{n}
\end{align}</math>
 

In this example, the reliability equations for Unit 1 and Unit 2 are the same since they are the same type of component and demonstrate the same failure properties. In addition, the total output is divided equally between the two units (when both units are operating), so <math>{{t}_{1e}}</math> and <math>{{t}_{2e}}</math> will also be the same.
 
The next step is to determine the reliability of the system after 8,760 hours, <math>R(t=8,760)</math> .
Using
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>
 

the system reliability is found to be:
 

::<math>\begin{align}
R(t=8760)= & 0.8567 \\
= & 85.67%
\end{align}</math>
 

===Load Sharing in BlockSim===
BlockSim uses this formulation when computing reliabilities of units in a load sharing configuration. When using the System Reliability Equation window, BlockSim returns a single token for the reliability of units in a load sharing configuration (as well as in the case of standby redundancy, discussed in the next section). As an example, consider the following RBD with Unit 1 in series with a container that includes two load sharing units.
 
[[Image:BS5.19.2.png|thumb|center|200px|]]
 
BlockSim will return the system equation as:
 

::<math>{{R}_{System}}=+{{R}_{LS}}\cdot {{R}_{1}}</math>
 

Where <math>{{R}_{LS}}</math> implies a form similar to
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>
 .
BlockSim allows for <math>k</math> -out-of- <math>n</math> units in a load sharing configuration.

===Example 5===
A component has five possible failure modes, <math>A</math> , <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> , <math>{{B}_{C}}</math> and <math>C</math> , and the <math>B</math> modes are interdependent. The system will fail if mode <math>A</math> occurs, mode <math>C</math> occurs or two out of the three <math>B</math> modes occur.
 

Modes <math>A</math> and <math>C</math> each have a Weibull distribution, with a <math>\beta =2</math> and <math>\eta =10,000</math> and 15,000 respectively. Events <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> and <math>{{B}_{C}}</math> each have an exponential distribution with a mean of 10,000 hours.
 

If any <math>B</math> event occurs (i.e. <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> or <math>{{B}_{C}}</math> ), the remaining <math>B</math> events are more likely to occur. Specifically, the mean times of the remaining <math>B</math> events are halved. Determine the reliability at 1000 hours for this component.
====Solution to Example 5====
The first step is to create the RBD. Modes <math>A</math> , <math>C</math> and a load sharing container with the <math>{{B}_{i}}</math> modes must be drawn in series, as illustrated next.
 
[[Image:BS5.19.3.png|thumb|center|200px|]]
 

The next step is to define the properties for each block, including those for the container. Setting the failure distributions for modes <math>A</math> and <math>C</math> is simple. The more difficult part is setting the properties for the container and the contained blocks. Based on the problem statement, the <math>B</math> modes are in a 2-out-of-3 load sharing redundancy. When all three are working (i.e. when no <math>B</math> mode has occurred), each block has an exponential distribution with <math>\mu=10,000</math>. If one <math>B</math> mode occurs, then the two surviving units have an exponential distribution with <math>\mu =5,000.</math>
 

Assume a Power Life-Stress relationship for the components. Then:
 

::<math>\begin{align}
{{\mu }_{1}}= & \frac{1}{KV_{1}^{n}} \ \\
{{\mu }_{2}}= & \frac{1}{KV_{2}^{n}} \
\end{align}</math>
 

Substituting <math>{{\mu }_{1}}=10,000</math> and <math>{{V}_{1}}=1</math> in <math>{{\mu }_{1}}= \frac{1}{KV_{1}^{n}}</math> and casting it in terms of <math>K</math> yields:
 

::<math>\begin{align}
10,000= & \frac{1}{K} \ \\
K = & \frac{1}{10,000}=0.0001
\end{align}</math>

 
Substituting <math>{{\mu }_{2}}=5,000</math> , <math>{{V}_{2}}=1.5</math> (because if one fails, then each survivor takes on an additional 0.5 units of load) and <math>10,000= \frac{1}{K}</math> for <math>K</math> in <math>{{\mu }_{2}}= \frac{1}{KV_{2}^{n}}</math> yields:

 
::<math>\begin{align}
5,000= & \frac{1}{0.0001\cdot {{(1.5)}^{n}}} \\
0.5= & {{(1.5)}^{-n}} \\
\ln (0.5)= & -n\ln (1.5) \\
n = & 1.7095
\end{align}</math>
 

This also could have been computed in ALTA, as shown in Figure "Calculation performed in ALTA".

The next step is to set the weight proportionality factor. This factor defines the portion of the load that the particular item carries while operating, as well as the load that shifts to the remaining units upon failure of the item. To illustrate, assume three units (1, 2 and 3) are in a load sharing container with weight proportionality factors of 1, 2 and 3 respectively (and a 1-out-of-3 requirement).
 

:• Unit 1 carries <math>\left( \tfrac{1}{1+2+3} \right)=0.166</math> or 16.6% of the total load.
:• Unit 2 carries <math>\left( \tfrac{2}{1+2+3} \right)=0.333</math> or 33.3% of the total load.
:• Unit 3 carries <math>\left( \tfrac{3}{1+2+3} \right)=0.50</math> or 50% of the total load.
 

The actual load on each unit then becomes the product of the entire load defined for the container times the portion carried by that unit. For example, if the container load is 100 lbs, then the portion assigned to Unit 1 will be <math>100\cdot 0.166=16.6</math> lbs.
 

In the current example, all units share the same load and thus have equal weight proportionality factors. Because these factors are relative, if the same number is used for all three items then the results will be the same. Thus, weight proportional factor is set equal to 1 for each item.
 

[[Image:5.20.PNG|thumb|center|500px|<div align="center"> Calculation performed in ALTA.</div>]]
 

 
[[Image:Fig 5.23_2.PNG|thumb|center|400px|<div align="center"> Defining Weight Proportional Factor. </div>]]
 

The last properties that need to be defined are the total load and the 2-out-of-3 redundancy. The total load is dependent on how the parameters were computed. In this case, total load was assumed to be 3 when the parameters were computed (i.e. the load per item was 1 when all worked and 1.5 when two worked). This is defined at the container level, as shown in Figure "Defining Weight Proportional Factor".
When all of the parameters have been specified in BlockSim, the reliability at 1,000 hours can be determined. From the Analytical QCP, this is found to be 93.87%.

=Standby Components=

In the previous section, the case of a system with load sharing components was presented. This is a form of redundancy with dependent components. That is, the failure of one component affects the failure of the other(s). This section presents another form of redundancy: standby redundancy. In standby redundancy the redundant components are set to be under a lighter load condition (or no load) while not needed and under the operating load when they are activated.
 

In standby redundancy the components are set to have two states: an active state and a standby state. Components in standby redundancy have two failure distributions, one for each state. When in the standby state, they have a quiescent (or dormant) failure distribution and when operating, they have an active failure distribution.
 

In the case that both quiescent and active failure distributions are the same, the units are in a simple parallel configuration (also called a hot standby configuration). When the rate of failure of the standby component is lower in quiescent mode than in active mode, that is called a warm standby configuration. When the rate of failure of the standby component is zero in quiescent mode (i.e. the component cannot fail when in standby), that is called a cold standby configuration.

===Simple Standby Configuration===

Consider two components in a standby configuration. Component 1 is the active component with a Weibull failure distribution with parameters <math>\beta </math> = 1.5 and <math>\eta </math> = 1,000. Component 2 is the standby component. When Component 2 is operating, it also has a Weibull failure distribution with <math>\beta </math> = 1.5 and <math>\eta </math> = 1,000. Furthermore, assume the following cases for the quiescent distribution.
 
:• Case 1: The quiescent distribution is the same as the active distribution (hot standby). 
:• Case 2: The quiescent distribution is a Weibull distribution with <math>\beta </math> = 1.5 and <math>\eta </math> = 2000 (warm standby). 
:• Case 3: The component cannot fail in quiescent mode (cold standby). 
 
In this case, the reliability of the system at some time, <math>t</math> , can be obtained using the following equation:
 

::<math>R(t)={{R}_{1}}(t)+\underset{0}{\overset{t}{\mathop \int }}\,{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x)\cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}dx \ </math>
 

Where:
 
:• <math>{{R}_{1}}</math> is the reliability of the active component. 
:• <math>{{f}_{1}}</math> is the <math>pdf</math> of the active component. 
:• <math>{{R}_{2;SB}}</math> is the reliability of the standby component when in standby mode (quiescent reliability). 
:• <math>{{R}_{2;A}}</math> is the reliability of the standby component when in active mode. 
:• <math>{{t}_{e}}</math> is the equivalent operating time for the standby unit if it had been operating at an active mode, such that: 

 
<math>{{R}_{2;SB}}(x)={{R}_{2;A}}({{t}_{e}})</math> 
 
The second equation above can be solved for <math>{{t}_{e}}</math> and substituted into the first equation above.
Figure "Standby container" illustrates the example as entered in BlockSim using a standby container.
 
 
[[Image:5.24.gif|thumb|center|400px|<div align="center"> Standby container.</div>]]

 

The active and standby blocks are within a container, which is used to specify standby redundancy. Since the standby component has two distributions (active and quiescent), the Block Properties window of the standby block has two pages for specifying each one. Figures "Defining the active failure distribution" and "Defining the quiescent failure distribution" illustrate these pages.

 

The system reliability results for 1000 hours are given in the following table:
 
[[Image:5-24.png|thumb|center|400px|]]
 

Note that even though the <math>\beta </math> value for the quiescent distribution is the same as in the active distribution, it is possible that the two can be different. That is, the failure modes present during the quiescent mode could be different from the modes present during the active mode. In that sense, the two distribution types can be different as well (e.g. lognormal when quiescent and Weibull when active).
 

In many cases when considering standby systems, a switching device may also be present that switches from the failed active component to the standby component. The reliability of the switch can also be incorporated into
<math>R(t)={{R}_{1}}(t)+\underset{0}{\overset{t}{\mathop \int }}\,{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x)\cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}dx \ </math>

as presented in the next section.
 

BlockSim's System Reliability Equation window returns a single token for the reliability of units in a standby configuration. This is the same as the load sharing case presented in the previous section.
 

[[Image: Fig 5.25.PNG|thumb|center|400px|<div align="center"> Defining the active failure distribution </div>]]
 
 

[[Image:Fig 5.26.PNG|thumb|center|400px|<div align="center"> Defining the quiescent failure distribution </div>]]
 

===Reliability of Standby Systems with a Switching Device===

In many cases when dealing with standby systems, a switching device is present that will switch to the standby component when the active component fails. Therefore, the failure properties of the switch must also be included in the analysis.
 

[[Image:BS5.26.2.png|thumb|center|300px|]]
 

In most cases when the reliability of a switch is to be included in the analysis, two probabilities can be considered. The first and most common one is the probability of the switch performing the action (i.e. switching) when requested to do so. This is called Switch Probability per Request in BlockSim and is expressed as a static probability (e.g. 90%). The second probability is the quiescent reliability of the switch. This is the reliability of the switch as it ages (e.g. the switch might wear out with age due to corrosion, material degradation, etc.). Thus it is possible for the switch to fail before the active component fails. However, a switch failure does not cause the system to fail, but rather causes the system to fail only if the switch is needed and the switch has failed. For example, if the active component does not fail until the mission end time and the switch fails, then the system does not fail. However, if the active component fails and the switch has also failed, then the system cannot be switched to the standby component and it therefore fails.
 

In analyzing standby components with a switching device, either or both failure probabilities (during the switching or while waiting to switch) can be considered for the switch, since each probability can represent different failure modes. For example, the switch probability per request may represent software-related issues or the probability of detecting the failure of an active component, and the quiescent probability may represent wear-out type failures of the switch.
 

To illustrate the formulation, consider the previous example that assumes perfect switching. To examine the effects of including an imperfect switch, assume that when the active component fails there is a 90% probability that the switch will switch from the active component to the standby component. In addition, assume that the switch can also fail due to a wear-out failure mode described by a Weibull distribution with <math>\beta </math> = 1.7 and <math>\eta </math> = 5000.
 

Therefore, the reliability of the system at some time, <math>t</math> , is given by the following equation.

::<math>\begin{align}
R(t)= & {{R}_{1}}(t) \\
& +\underset{0}{\overset{t}{\mathop \int }}\,\{{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x) \\
& \cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}\cdot {{R}_{SW;Q}}(x)\cdot {{R}_{SW;REQ}}(x)\}dx
\end{align}</math>

Where:
 
:• <math>{{R}_{1}}</math> is the reliability of the active component. 
:• <math>{{f}_{1}}</math> is the <math>pdf</math> of the active component. 
:• <math>{{R}_{2;SB}}</math> is the reliability of the standby component when in standby mode (quiescent reliability). 
:• <math>{{R}_{2;A}}</math> is the reliability of the standby component when in active mode. 
:• <math>{{R}_{SW;Q}}</math> is the quiescent reliability of the switch. 
:• <math>{{R}_{SW;REQ}}</math> is the switch probability per request. 
:• <math>{{t}_{e}}</math> is the equivalent operating time for the standby unit if it had been operating at an active mode. 
This problem can be solved in BlockSim by including these probabilities in the container's properties, as shown in Figures "Standby container (switch) failure probabilities while attempting to switch" and "Standby container (switch) failure distribution while waiting to switch". In BlockSim, the standby container is acting as the switch.

 

[[Image:Fig 5.28_2.PNG|thumb|center|400px|<div align="center"> Standby container (switch) failure probabilities while attempting to switch </div> ]]
 

 

[[Image:Fig 5.27.PNG|thumb|center|400px|<div align="center"> Standby container (switch) failure distribution while waiting to switch </div>]]
 

 

Note that there are additional properties that can be specified in BlockSim for a switch, such as Switch Restart Probability, No. of Restarts and Switch Delay Time. In many applications, the switch is re-tested (or re-cycled) if it fails to switch the first time. In these cases, it might be possible that it switches in the second or third, or <math>{{n}^{th}}</math> attempt.
 
The Switch Restart Probability specifies each additional attempt's probability of successfully switching and the Finite Restarts specifies the total number of attempts. Note that the Switch Restart Probability specifies the probability of success of each trial (or attempt). The probability of success of <math>n</math> consecutive trials is calculated by BlockSim using the binomial distribution and this probability is then incorporated into equation above. The Switch Delay Time property is related to repairable systems and is considered in BlockSim only when using simulation. When using the analytical solution (i.e. for a non-repairable system), this property is ignored.
 

Solving the analytical solution (as given by Equation above), the following results are obtained.
 

[[Image:5-30.png|thumb|center|400px|]]
 

From the table above, it can be seen that the presence of a switching device has a significant effect on the reliability of a standby system. It is therefore important when modeling standby redundancy to incorporate the switching device reliability properties. It should be noted that this methodology is not the same as treating the switching device as another series component with the standby subsystem. This would be valid only if the failure of the switch resulted in the failure of system (e.g. switch failing open). In equation above, the Switch Probability per Request and quiescent probability are present only in the second term of the equation. Treating these two failure modes as a series configuration with the standby subsystem would imply that they are also present when the active component is functioning (i.e. first term of equation above). This is invalid and would result in the underestimation of the reliability of the system. In other words, these two failure modes become significant only when the active component fails.
 

As an example, and if we consider the warm standby case, the reliability of the system without the switch is 70.57% at 1000 hours. If the system was modeled so that the switching device was in series with the warm standby subsystem, the result would have been:

 

::<math>\begin{align}
{{R}_{S}}(1000)= & {{R}_{Standby}}(1000)\cdot {{R}_{sw,Q(1000)}}\cdot {{R}_{sw,req}} \\
= & 0.7057\cdot 0.9372\cdot 0.9 \\
= & 0.5952
\end{align}</math>

 

In the case where a switch failure mode causes the standby subsystem to fail, then this mode can be modeled as an individual block in series with the standby subsystem.

===Example 6===
Consider a car with four new tires and a full-size spare. Assume the following failure characteristics:
 
:• The tires follow a Weibull distribution with a <math>\beta=4</math> and an <math>\eta =</math> 40,000 miles while on the car due to wear. 
:• The tires also have a probability of failing due to puncture or other causes. For this, assume a constant rate for this occurrence with a probability of 1 every 50,000 miles. 
:• When not on the car (i.e. is a spare), a tire's probability of failing also has a Weibull distribution with a <math>\beta =</math> 2 and <math>\eta =</math> 120,000 miles. 
 
Assume a mission of 1,000 miles. If a tire fails during this trip, it will be replaced with the spare. However, the spare will not be repaired during the trip. In other words, the trip will continue with the spare on the car and if the spare fails the system will fail. Determine the probability of system failure.
====Solution to Example 6====
Active failure distribution for tires:
 
:• Due to wear-out, Weibull <math>\beta =4</math> and <math>\eta =40,000</math> miles. 
:• Due to random puncture, exponential <math>\mu =50,000.</math> 
:• The quiescent failure distribution is a Weibull distribution with <math>\beta =2</math> and <math>\eta =120,000</math> miles. 
 
The block diagram for each tire has two blocks in series, one block representing the wear-out mode and the other the random puncture mode, as shown next:
 
 
[[Image:small5.gif|thumb|center|400px|]]
 

There are five tires, four active and one standby (represented in the diagram by a standby container with a 4-out-of-5 requirement), as shown next:
 
[[Image:BStirecontainer.png|thumb|center|400px|]]

For the standby Wear block, set the active failure and the quiescent distributions, but for the Puncture block, set only the active puncture distribution (because the tire cannot fail due to puncture while stored). Using BlockSim, the probability of system failure is found to be 0.003 or 0.3%.
 

=Note Regarding Numerical Integration Solutions=

Load sharing and standby solutions in BlockSim are performed using numerical integration routines. As with any numerical analysis routine, the solution error depends on the number of iterations performed, the step size chosen and related factors, plus the behavior of the underlying function. By default, BlockSim uses a certain set of preset factors. In general, these defaults are sufficient for most problems. If a higher precision or verification of the precision for a specific problem is required, BlockSim's preset options can be modified and/or the integration error can be assessed using the Integration Parameters... option for each container. For more details, you can refer to the documentation on the Algorithm Setup window in the BlockSim 8 User's help files.

Time-Dependent System Reliability (Analytical)

2012-03-21T21:27:32Z

Dingzhou Cao: /* Solution to Example 3 */

{{Template:bsbook|5}}

In the previous chapter, different system configuration types were examined, as well as different methods for obtaining the system's reliability function analytically. Because the reliabilities in the problems presented were treated as probabilities (e.g. <math>P(A)</math> , <math>{{R}_{i}}</math> ), the reliability values and equations presented were referred to as static (not time-dependent). Thus, in the prior chapter, the life distributions of the components were not incorporated in the process of calculating the system reliability. In this chapter, time dependency in the reliability function will be introduced. We will develop the models necessary to observe the reliability over the life of the system, instead of at just one point in time. In addition, performance measures such as failure rate, MTTF and warranty time will be estimated for the entire system. The methods of obtaining the reliability function analytically remain identical to the ones presented in the previous chapter, with the exception that the reliabilities will be functions of time. In other words, instead of dealing with <math>{{R}_{i}}</math> , we will use <math>{{R}_{i}}(t)</math> . All examples in this chapter assume that no repairs are performed on the components.

=Analytical Life Predictions=
The analytical approach presented in the prior chapter involved the determination of a mathematical expression that describes the reliability of the system, expressed in terms of the reliabilities of its components. So far we have estimated only static system reliability (at a fixed time). For example, in the case of a system with three components in series, the system's reliability equation was given by:
 

::<math>{{R}_{s}}={{R}_{1}}\cdot {{R}_{2}}\cdot {{R}_{3}} \ </math>
 

The values of <math>{{R}_{1}}</math> , <math>{{R}_{2}}</math> and <math>{{R}_{3}}</math> were given for a common time and the reliability of the system was estimated for that time. However, since the component failure characteristics can be described by distributions, the system reliability is actually time-dependent. In this case, the equation above can be rewritten as:
 

::<math>{{R}_{s}}(t)={{R}_{1}}(t)\cdot {{R}_{2}}(t)\cdot {{R}_{3}}(t)</math>

 

The reliability of the system for any mission time can now be estimated. Assuming a Weibull life distribution for each component, the first equation above can now be expressed in terms of each component's reliability function, or:

 

[[Category:Completed Theoretical Review]]

::<math>{{R}_{s}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}}\cdot {{e}^{-{{\left( \tfrac{t}{{{\eta }_{2}}} \right)}^{{{\beta }_{2}}}}}}\cdot {{e}^{-{{\left( \tfrac{t}{{{\eta }_{3}}} \right)}^{{{\beta }_{3}}}}}}</math>

 

In the same manner, any life distribution can be substituted into the system reliability equation. Suppose that the times-to-failure of the first component are described with a Weibull distribution, the times-to-failure of the second component with an exponential distribution and the times-to-failure of the third component with a normal distribution. Then the first equation above can be written as:
 

::<math>{{R}_{s}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}}\cdot {{e}^{-{{\lambda }_{2}}t}}\cdot \left[ 1-\Phi \left( \frac{t-{{\mu }_{3}}}{{{\sigma }_{3}}} \right) \right]</math>

 

It can be seen that the biggest challenge is in obtaining the system's reliability function in terms of component reliabilities, which has already been discussed in depth. Once this has been achieved, calculating the reliability of the system for any mission duration is just a matter of substituting the corresponding component reliability functions into the system reliability equation.
===Advantages of the Analytical Method===
The primary advantage of the analytical solution is that it produces a mathematical expression that describes the reliability of the system. Once the system's reliability function has been determined, other calculations can then be performed to obtain metrics of interest for the system. Such calculations include:
 
:• Determination of the system's <math>pdf.</math> 
:• Determination of warranty periods. 
:• Determination of the system's failure rate. 
:• Determination of the system's MTTF. 
 
In addition, optimization and reliability allocation techniques can be used to aid engineers in their design improvement efforts. Another advantage in using analytical techniques is the ability to perform static calculations and analyze systems with a mixture of static and time-dependent components. Finally, the reliability importance of components over time can be calculated with this methodology.
===Disadvantages of the Analytical Method===
The biggest disadvantage of the analytical method is that formulations can become very complicated. The more complicated a system is, the larger and more difficult it will be to analytically formulate an expression for the system's reliability. For particularly detailed systems this process can be quite time-consuming, even with the use of computers. Furthermore, when the maintainability of the system or some of its components must be taken into consideration, analytical solutions become intractable. In these situations, the use of simulation methods may be more advantageous than attempting to develop a solution analytically. Simulation methods are presented in later chapters.
===Looking at a Simple ''Complex'' System Analytically===

The complexity involved in an analytical solution can be best illustrated by looking at the simple ''complex'' system with 15 components, as shown in Figure " An RBD of a complex system".
 
[[Image:5-1.png|thumb|center|400px|<div align="center"> An RBD of a complex system.</div>]]

 
The system reliability for this system (computed using BlockSim) is shown next. The first solution is provided using BlockSim's symbolic solution. In symbolic mode, BlockSim breaks the equation into segments, identified by tokens, that need to be substituted into the final system equation for a complete solution. This creates algebraic solutions that are more compact than if the substitutions were made.
 

::<math>\begin{align}
{{R}_{System}}= & D2\cdot D3\cdot {{R}_{L}} \\
D3= & +{{R}_{K}}\cdot IK \\
IK= & +{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}} \\
D2 = & +{{R}_{A}}\cdot {{R}_{E}}\cdot IE \\
IE = & -D1\cdot {{R}_{M}}\cdot {{R}_{N}}+{{R}_{M}}\cdot {{R}_{N}}+D1 \\
D1 = & +{{R}_{D}}\cdot ID \\
ID = & -{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}}
\end{align}</math>
 
Substituting the terms yields:
 

::<math>\begin{align}
{{R}_{System}}= & {{R}_{A}}\cdot {{R}_{E}}\cdot {{R}_{L}}\cdot {{R}_{K}} \\
& \cdot \{({{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})\cdot {{R}_{M}}\cdot {{R}_{N}} \\
& +{{R}_{M}}\cdot {{R}_{N}}-{{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}}\} \\
& \cdot \{{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}}\}
\end{align}</math>

 
BlockSim's automatic algebraic simplification would yield the following format for the above solution:
 

::<math>\begin{align}
{{R}_{System}}= & (({{R}_{A}}\cdot {{R}_{E}}(-({{R}_{D}}(-{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})){{R}_{M}}\cdot {{R}_{N}} \\
& +{{R}_{M}}\cdot {{R}_{N}} \\
& +({{R}_{D}}(-{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})))) \\
& ({{R}_{K}}({{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& +RI\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}})){{R}_{L}}) \
\end{align}</math>

 
In this equation, each <math>{{R}_{i}}</math> represents the reliability function of a block. For example, if <math>{{R}_{A}}</math> has a Weibull distribution, then each <math>{{R}_{A}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{A}}} \right)}^{{{\beta }_{A}}}}}}</math> and so forth. Substitution of each component's reliability function in the last <math>{{R}_{System}}</math> equation above will result in an analytical expression for the system reliability as a function of time, or <math>{{R}_{s}}(t)</math> , which is the same as <math>(1-cd{{f}_{System}}).</math>

===Obtaining Other Functions of Interest===
Once the system reliability equation (or the cumulative density function, <math>cdf</math> ) has been determined, other functions and metrics of interest can be derived.
 
Consider the following simple system:
 
[[Image:5-2.png|thumb|center|300px|<div align="center"> Simple two-component system. </div>]]

 
Furthermore, assume that component 1 follows an exponential distribution with a mean of 10,000 (<math>\mu =10,000,</math> <math>\lambda =1/10,000)</math> and component 2 follows a Weibull distribution with <math>\beta =6</math> and <math>\eta =10,000</math> . The reliability equation of this system is:

 
::<math>\begin{align}
{{R}_{S}}(t)= & {{R}_{1}}(t)\cdot {{R}_{2}}(t) \\
= & {{e}^{-\lambda t}}\cdot {{e}^{-{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \\
= & {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \
\end{align}</math>

 
The system <math>cdf</math> is:

 
::<math>\begin{align}
{{F}_{S}}(t)= & 1-({{R}_{1}}(t)\cdot {{R}_{2}}(t)) \\
= & 1-\left( {{e}^{-\lambda t}}\cdot {{e}^{-{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \right) \\
= & 1-\left( {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)
\end{align}</math>

 
{{analytical system pdf}}

====Conditional Reliability====
Conditional reliability is the probability of a system successfully completing another mission following the successful completion of a previous mission. The time of the previous mission and the time for the mission to be undertaken must be taken into account for conditional reliability calculations. The system's conditional reliability function is given by:

::<math>R(T,t)=\frac{R(T+t)}{R(T)}</math>

Equation above gives the reliability for a new mission of duration <math>t</math> having already accumulated <math>T</math> hours of operation up to the start of this new mission. The system is evaluated to assure that it will start the next mission successfully.
 
For the system in Figure "Simple two-component system.", the reliability for mission of <math>t=1,000</math> hours, having an age of <math>T=500</math> hours, is:
 

::<math>\begin{align}
{{R}_{S}}(T=500,t=1000)= & \frac{R(T+t)}{R(T)} \\
= & \frac{R(1500)}{R(500)} \\
= & \frac{{{e}^{-\tfrac{1500}{10,000}}}\cdot {{e}^{-{{\left( \tfrac{1500}{10,000} \right)}^{6}}}}}{{{e}^{-\tfrac{500}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{500}{10,000} \right)}^{6}}}}} \\
= & 0.9048=90.48%
\end{align}</math>

 

====Conditional Reliability for Components====

Now in this formulation, it was assumed that the accumulated age was equivalent for both units. That is, both started life at zero and aged to 500. It is possible to consider an individual component that has already accumulated some age (used component) in the same formulation. To illustrate this, assume that component 2 started life with an age of T=100. Then the reliability equation of the system, as given in <math>{{R}_{S}}(t)= {{R}_{1}}(t)\cdot {{R}_{2}}(t)</math>, would need to be modified to include a conditional term for 2, or:
 

::<math>{{R}_{S}}(t)={{R}_{1}}(t)\cdot \frac{{{R}_{2}}({{T}_{2}}+t)}{{{R}_{2}}({{T}_{2}})} \ </math>

 

In BlockSim, the start age input box may be used to specify a starting age greater than zero.
{{system failure rate analytical}}

====System Mean Life (Mean Time To Failure)====
The mean life (or mean time to failure, MTTF) can be obtained by integrating the system reliability function from zero to infinity:
 

::<math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math>

 

The mean time is a performance index and does not provide any information about the behavior of the failure distribution of the system.
 
For the system in Figure "Simple two-component system":
 

::<math>\begin{align}
MTTF= & \int_{0}^{\infty }\left( {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)dt \\
= & 5978.9
\end{align}</math>

 

====Warranty Period and BX Life====
Sometimes it is desirable to know the time value associated with a certain reliability. Warranty periods are often calculated by determining what percentage of the failure population can be covered financially and estimating the time at which this portion of the population will fail. Similarly, engineering specifications may call for a certain BX life, which also represents a time period during which a certain proportion of the population will fail. For example, the B10 life is the time in which 10% of the population will fail.
This is obtained by setting <math>{{R}_{S}}(t)</math> to the desired value and solving for <math>t.</math>
 
For the system in Figure "Simple two-component system":
 

::<math>{{R}_{s}}\left( t \right)={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>
 

To compute the time by which reliability would be equal to 90%, equation above is recast as follows and solved for <math>t.</math>

 
::<math>0.90={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>

 

In this case, <math>t=1053.59</math> . Equivalently, the B10 life for this system is also <math>1053.59</math> .
Except for some trivial cases, a closed form solution for <math>t</math> cannot be obtained. Thus, it is necessary to solve for <math>t</math> using numerical methods. BlockSim uses numerical methods.

===Example 1===
Consider a system consisting of three exponential units in series with the following failure rates (in failures per hour): <math>{{\lambda }_{1}}</math> = 0.0002, <math>{{\lambda }_{2}}</math> = 0.0005 and <math>{{\lambda }_{3}}</math> = 0.0001.
 
:• Obtain the reliability equation for the system. 
:• What is the reliability of the system after 150 hours of operation? 
:• Obtain the system's <math>pdf.</math> 
:• Obtain the system's failure rate equation. 
:• What is the MTTF for the system? 
:• What should the warranty period be for a 90% reliability? 
====Solution to Example 1====
 
:• The analytical expression for the reliability of the system is given by:

::<math>\begin{align}
{{R}_{s}}(t)= & {{R}_{1}}(t)\cdot {{R}_{2}}(t)\cdot {{R}_{3}}(t) \\
= & {{e}^{-{{\lambda }_{1}}t}}\cdot {{e}^{-{{\lambda }_{2}}t}}\cdot {{e}^{-{{\lambda }_{1}}t}} \\
= & {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}
\end{align}</math>

 

At 150 hours of operation, the reliability of the system is:
 

::<math>\begin{align}
{{R}_{s}}(t)= & {{e}^{-(0.0002+0.0005+0.0001)150}} \\
= & 0.8869\text{ or }88.69%
\end{align}</math>

 

:• In order to obtain the system's <math>pdf</math> , the derivative of the reliability equation given in the first equation above is taken with respect to time, or:
 
::<math>\begin{align}
{{f}_{s}}(t)= & -\frac{d[{{R}_{s}}(t)]}{dt} \\
= & -\frac{d\left[ {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}} \right]}{dt} \\
= & ({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})\cdot {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}
\end{align}</math>

:• The system's failure rate can now be obtained simply by dividing the system's <math>pdf</math> given in the equation above by the system's reliability function given in the first equation above, and:

::<math>\begin{align}
{{\lambda }_{s}}\left( t \right)= & \frac{{{f}_{s}}\left( t \right)}{{{R}_{s}}\left( t \right)} \\
= & \frac{({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})\cdot {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}}{{{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}} \\
= & ({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}) \\
= & 0.0008\text{ }fr/hr
\end{align}</math>

 
:• Combining <math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math> and the first equation above, the system's MTTF can be obtained:

 
::<math>\begin{align}
MTTF= & \int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \\
= & \int_{0}^{\infty }{{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}dt \\
= & \frac{1}{({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})} \\
= & 1250\text{ }hr
\end{align}</math>
 

:• Solving the first equation above with respect to time will yield the corresponding warranty period for a 90% reliability. In this case, the system reliability equation is simple and a closed form solution exists. The warranty time can now be found by solving:

 
::<math>\begin{align}
t= & -\frac{\ln (R)}{{{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}} \\
= & -\frac{\ln (0.9)}{0.0008} \\
= & 131.7\text{ }hr
\end{align}</math>
 

Thus, the warranty period should be 132 hours.

===Example 2===
Consider the system shown in Figure "Complex bridge system in Example 2".
 
[[Image:BS5.5.png|thumb|center|300px|<div align="center"> Complex bridge system in Example 2. </div>]]
 

Components <math>A</math> through <math>E</math> are Weibull distributed with <math>\beta =1.2</math> and <math>\eta =1230</math> hours. The starting and ending blocks cannot fail. 
Determine the following:
 
:• The reliability equation for the system and its corresponding plot. 
:• The system's <math>pdf</math> and its corresponding plot. 
:• The system's failure rate equation and the corresponding plot. 
:• The <math>MTTF</math> . 
:• The warranty time for a 90% reliability. 
:• The reliability for a 200-hour mission, if it is known that the system has already successfully operated for 200 hours. 
====Solution====
 
The first step is to obtain the reliability function for the system. The methods described in the previous chapter can be employed, such as the event space or path-tracing methods. Using BlockSim, the following reliability equation is obtained:
 

::<math>\begin{align}
{{R}_{s}}(t)= & ({{R}_{Start}}\cdot {{R}_{End}}(2{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}))
\end{align}</math>
 

Note that since the starting and ending blocks cannot fail, <math>{{R}_{Start}}=1</math> and <math>{{R}_{End}}=1,</math> equation above can be reduced to:

 

::<math>\begin{align}
{{R}_{s}}(t)= & 2\cdot {{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}
\end{align}</math>

 

Where <math>{{R}_{A}}</math> is the reliability equation for Component A, or:

 

::<math>{{R}_{A}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{A}}} \right)}^{{{\beta }_{A}}}}}}</math>

 

::<math>{{R}_{B}}</math> is the reliability equation for Component <math>B</math> , etc.
 

Since the components in this example are identical, the system reliability equation can be further reduced to:

 

::<math>{{R}_{s}}(t)=2R{{(t)}^{2}}+2R{{(t)}^{3}}-5R{{(t)}^{4}}+2R{{(t)}^{5}}</math>

 

Or, in terms of the failure distribution:

 

::<math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>

 

The corresponding plot is given in Figure "Reliability plot for the system".
 
[[Image:BS5.6.png|thumb|center|300px|<div align="center"> Reliability plot for the system. </div>]]
 
 

In order to obtain the system's <math>pdf</math> , the derivative of the reliability equation given above is taken with respect to time, resulting in:
 

::<math>\begin{align}
{{f}_{s}}(t)= & 4\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+6\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \\
& -20\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+10\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}
\end{align}</math>

 

The <math>pdf</math> can now be plotted for different time values, <math>t</math> , as shown in Figure "''pdf'' plot for the system".
 

The system's failure rate can now be obtained by dividing the system's <math>pdf</math> given in equation above by the system's reliability function given in <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>, or:

 

::<math>\begin{align}
{{\lambda }_{s}}(t)= & \frac{4\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+6\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}{2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}} \\
& +\frac{-20\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+10\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}{2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}
\end{align}</math>

 

The corresponding plot is given in Figure "Failure rate for the system".
 
[[Image:BS5.7.gif|thumb|center|300px|<div align="center"> ''pdf'' plot for the system.</div>]]
 

The <math>MTTF</math> of the system is obtained by integrating the system's reliability function given by <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math> from time zero to infinity, as given by <math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math>. Using BlockSim's Analytical QCP, an <math>MTTF</math> of 1007.8 hours is calculated, as shown in Figure "MTTF of the system".
 

The warranty time can be obtained by solving <math>{{R}_{s}}(t)</math> with respect to time for a system reliability <math>{{R}_{s}}=0.9</math> . Using the Analytical QCP and selecting the 
Warranty Time option, a time of 372.72 hours is obtained, as shown in Figure "Time at which ''R''=0.9 or 90% for the system".
 
[[Image:BS5.8.gif|thumb|center|400px|<div align="center"> Failure rate for the system.</div>]]
 
[[Image:Fig 5.9.PNG|thumb|center|400px|<div align="center"> MTTF of the system. </div>]]
 
[[Image:Fig 5.10.PNG|thumb|center|400px|<div align="center"> Time at which ''R''=0.9 or 90% for the system.</div>]]
 
[[Image:Fig 5.11.PNG|thumb|center|400px|<div align="center"> Conditional reliability calculation for the system.</div>]]
 

Lastly, the conditional reliability can be obtained using <math>R(T,t)=\frac{R(T+t)}{R(T)}</math> and <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>, or:
 

::<math>\begin{align}
R(200,200)= & \frac{R(400)}{R(200)} \\
= & \frac{0.883825}{0.975321} \\
= & 0.906189
\end{align}</math>

 

This can be calculated using BlockSim's Analytical QCP, as shown in Figure "Conditional reliability calculation for the system".
 

=Approximating the System CDF=

In many cases, it is valuable to fit a distribution that represents the system's times-to-failure. This can be useful when the system is part of a larger assembly and may be used for repeated calculations or in calculations for other systems. In cases such as this, it can be useful to characterize the system's behavior by fitting a distribution to the overall system and calculating parameters for this distribution. This is equivalent to fitting a single distribution to describe <math>{{R}_{S}}(t</math> ). In essence, it is like reducing the entire system to a component in order to simplify calculations.
 
For the system in Figure below:
[[Image:5-2.png|thumb|center|300px|<div align="center">Simple two-component system. </div>]]
 

::<math>{{R}_{S}}(t)={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>

 
To compute an approximate reliability function for this system, <math>{{R}_{A}}(t)\simeq {{R}_{S}}(t)</math> , one would compute <math>n</math> pairs of reliability and time values and then fit a single distribution to the data, or:

 

::<math>\begin{align}
{{R}_{S}}(t= & 10,396.7)=10% \\
{{R}_{S}}(t= & 9,361.9)=20% \\
& ... \\
{{R}_{S}}(t= & 1,053.6)=90%
\end{align}</math>

 

A single distribution, <math>{{R}_{A}}(t)</math> , that approximates <math>{{R}_{S}}(t)</math> can now be computed from these pairs using life data analysis methods. If using the Weibull++ software, one would enter the values as free form data.

===Example 3===
Compute a single Weibull distribution approximation for the system in Example 2.
====Solution to Example 3====
The system in the previous example, shown in Figure "Complex bridge system in Example 2", can be approximated by use of a 2-parameter Weibull distribution with <math>\beta =2.02109</math> and <math>\eta =1123.51</math> . In BlockSim, this is accomplished by representing the entire system as one distribution by going to the Distribution Fit window Figure " Representing a system with a distribution".

 
[[Image:Fig 5.13.PNG|thumb|center|400px|<div align="center"> Representing a system with a distribution.</div>]]
 
[[Image:Fig 5.12.PNG|thumb|center|300px|<div align="center"> Distribution Fitting window.</div>]]
 
by clicking the Distribution Fit Window, the Distribution Estimator window will appear (Figure "Distribution Fitting window").
 
In this window you can select a distribution to represent the data. BlockSim will then generate a number of system failure times based on the system's reliability function. The system's reliability function can be used to solve for a time value associated with that unreliability value. The distribution of the generated time values can then be fitted to a probability distribution function.
 

Consider a value of <math>F(t)=0.11</math> . Using the system's reliability equation and solving for time, the corresponding time-to-failure for a 0.11 unreliability can be calculated. 
For the system of Example 2, the time for a 0.11 unreliability is 389.786 hours.
 
When enough points have been generated, the selected distribution will be fitted to this data set and the distribution's parameters will be returned. In addition, if ReliaSoft's Weibull++ is installed, the generated data can be viewed/analyzed using a Weibull++ instance, as shown in Figure "Using Weibull++ to calculate distribution parameters".
 
[[Image:Fig 5.14.PNG|thumb|center|400px|<div align="center"> Using Weibull++ to calculate distribution parameters.</div>]]
 
It is recommended that the analyst examine the fit to ascertain the applicability of the approximation.

=Duty Cycle=

Components of a system may not operate continuously during a system's mission, or may be subjected to loads greater or lesser than the rated loads during system operation. To model this, a factor called the Duty Cycle ( <math>{{d}_{c}}</math> ) is used. The duty cycle may also be used to account for changes in environmental stress, such as temperature changes, that may effect the operation of a component. The duty cycle is a positive value, with a default value of 1 representing continuous operation at rated load, and any values other than 1 representing other load values with respect to the rated load value (or total operating time). A duty cycle value higher than 1 indicates a load in excess of the rated value. A duty cycle value lower than 1 indicates that the component is operating at a load lower than the rated load or not operating continuously during the system's mission. For instance, a duty cycle of 0.5 may be used for a component that operates only half of the time during the system's mission.
 
The reliability metrics for a component with a duty cycle are calculated as follows. Let <math>{{d}_{c}}</math> represent the duty cycle during a particular mission of the component, <math>t</math> represent the mission time and <math>{t}'</math> represent the accumulated age. Then:
 

::<math>{t}'={{d}_{c}}\times t</math>

 

The reliability equation for the component is:
 

::<math>R({t}')=R({{d}_{c}}\times t)</math>
 

The component ''pdf'' is:
 

::<math>f({t}')=-\frac{d(R({t}'))}{dt}=-\frac{d(R({{d}_{c}}\times t))}{dt}={{d}_{c}}f({{d}_{c}}\times t)</math>
 

The failure rate of the component is:
 

::<math>\lambda ({t}')=\frac{f({t}')}{R({t}')}=\frac{{{d}_{c}}f({{d}_{c}}\times t)}{R({{d}_{c}}\times t)}={{d}_{c}}\lambda ({{d}_{c}}\times t)</math>

===Example 4===
Consider a computer system with three components: a processor, a hard drive and a CD drive in series as shown next. Assume that all three components follow a Weibull failure distribution with the parameters <math>{{\beta }_{1}}=1.5</math> and <math>{{\eta }_{1}}=5000</math> for the processor, <math>{{\beta }_{2}}=2.5</math> and <math>{{\eta }_{2}}=3000</math> for the hard drive, and <math>{{\beta }_{3}}=2</math> and <math>{{\eta }_{3}}=4000</math> for the CD drive. Determine the reliability of the computer system after one year (365 days) of operation, assuming that the CD drive is used only 30% of the time.
 
[[Image:BS5ex4.png|thumb|center|300px|]]

====Solution to Example 4====
The reliability of the processor after 365 days of operation is given by:

::<math>\begin{align}
{{R}_{processor}}(365)= & {{e}^{-{{\left( \tfrac{365}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}} \\
= & {{e}^{-{{\left( \tfrac{365}{5000} \right)}^{1.5}}}} \\
= & 0.9805\text{ or }98.05%
\end{align}</math>

The reliability of the hard drive after 365 days of operation is given by:

::<math>\begin{align}
{{R}_{harddrive}}(365)= & {{e}^{-{{\left( \tfrac{365}{{{\eta }_{2}}} \right)}^{{{\beta }_{2}}}}}} \\
= & {{e}^{-{{\left( \tfrac{365}{3000} \right)}^{2.5}}}} \\
= & 0.9948\text{ or }99.48%
\end{align}</math>

The reliability of the CD drive after 365 days of operation (taking into account the 30% operation using a duty cycle of 0.3) is given by:

::<math>\begin{align}
{{R}_{CDdrive}}(365)= & {{e}^{-{{\left( \tfrac{{{d}_{c}}\times 365}{{{\eta }_{3}}} \right)}^{{{\beta }_{3}}}}}} \\
= & {{e}^{-{{\left( \tfrac{0.3\times 365}{4000} \right)}^{2}}}} \\
= & 0.9993\text{ or }99.93%
\end{align}</math>
 
[[Image:Fig 5.15.PNG|thumb|center|400px|<div align="center"> Fig 5.15 Result for the computer system reliability.</div>]]
 

Thus the reliability of the computer system after 365 days of operation is:

::<math>\begin{align}
{{R}_{s}}(365)= & {{R}_{processor}}(365)\cdot {{R}_{harddrive}}(365)\cdot {{R}_{CDdrive}}(365) \\
= & 0.9805\cdot 0.9948\cdot 0.9993 \\
= & 0.9747\text{ or }97.47%
\end{align}</math>
 
This result can be obtained in BlockSim as shown in Figure "Result for the computer system reliability".
 

=Load Sharing=
As presented in earlier chapters, a reliability block diagram (RBD) allows you to graphically represent how the components within a system are reliability-wise connected. In most cases, independence is assumed across the components within the system. For example, the failure of component A does not affect the failure of component B. However, if a system consists of components that are sharing a load, then the assumption of independence no longer holds true.
 

If one component fails, then the component(s) that are still operating will have to assume the failed unit's portion of the load. Therefore, the reliabilities of the surviving unit(s) will change. Calculating the system reliability is no longer an easy proposition. In the case of load sharing components, the change of the failure distributions of the surviving components must be known in order to determine the system's reliability.
 

To illustrate this, consider the a system of two units connected reliability-wise in parallel (Figure "Two units connected reliability-wise in parallel").
 

[[Image:5-16.png|thumb|center|300px|<div align="center"> Two units connected reliability-wise in parallel.</div>]]
 

Assume that the units must supply an output of 8 volts and that if both units are operational, each unit is to supply 50% of the total output. If one of the units fails, then the surviving unit supplies 100%. Furthermore, assume that having to supply the entire load has a negative impact on the reliability characteristics of the surviving unit. Since the reliability characteristics of the unit change based on whether both or only one is operating, a life distribution along with a life-stress relationship (as discussed in [http://reliawiki.com/index.php/Statistical_Background#A_Brief_Introduction_to_Life-Stress_Relationships|A Brief Introduction to Life-Stress Relationships]) will be needed to model each component.
 

To illustrate the steps needed, we will create the model starting from raw data. Assume that a total of 20 units were tested to failure at 7, 10 and 15 volts. The test data set is presented in the next table.
 

[[Image:5-17.png|thumb|center|300px|]]
 

For this example, Units 1 and 2 are the same component. Therefore, only one set of data was collected. However, it is possible that the load sharing components in a system may not be the same. If that were the case, data would need to be collected for each component.
 

The data set in Table 1 was analyzed using ReliaSoft's ALTA software (as shown in Figure "Using ALTA to calculate component parameters") with the Inverse Power Law as the underlying life-stress relationship and Weibull as the life distribution.
 
 

The estimated model parameters, <math>\beta </math> , <math>K</math> and <math>n</math> , are shown next.
 

::<math>\begin{align}
\beta = & 1.9239 \\
K= & 3.2387\times {{10}^{-7}} \\
n= & 3.4226
\end{align}</math>
 

:Or:
 
::<math>\begin{align}
{{R}_{1}}(t,{{S}_{1}})= & {{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} \\
= & {{e}^{-{{\left( 3.2387\times {{10}^{-7}}S_{1}^{3.4226}t \right)}^{1.9239}}}} \
\end{align}</math>.

 
::<math>{{f}_{1}}(t,{{S}_{1}})=\beta KS_{1}^{n}{{\left( KS_{1}^{n}t \right)}^{\beta -1}}{{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} \ </math>
 

And for this case:

::<math>\begin{align}
{{R}_{1}}(t,{{S}_{1}})= & {{R}_{2}}(t,{{S}_{2}}) \\
{{f}_{1}}(t,{{S}_{1}})= & {{f}_{2}}(t,{{S}_{2}})
\end{align}</math>
 
 
[[Image:Fig 5.17.PNG|thumb|center|500px|<div align="center"> Using ALTA to calculate component parameters.</div>]]
 

Figure "Reliability curves for different voltage output conditions" shows a plot of <math>{{R}_{1}}(t,{{S}_{1}})={{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} = {{e}^{-{{\left( 3.2387\times {{10}^{-7}}S_{1}^{3.4226}t \right)}^{1.9239}}}}\ </math>.
 

Now that the failure properties have been determined using the test data, the reliability of the system at some time, <math>t</math> , can be calculated using the following equation:

 
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>

 
:Where:
 

::<math>\begin{align}
{{S}_{1}}= & {{P}_{1}}S \\
{{S}_{2}}= & {{P}_{2}}S
\end{align}</math>

 
[[Image:5-18.png|thumb|center|300px|<div align="center"> Reliability curves for different voltage output conditions. </div>]]
 
[[Image:BS5.19.png|thumb|center|395px| Illustrating <math>{{t}_{e}}</math>]]
 

And:
 

:• <math>S</math> is the total load (or required output). 
:• <math>{{P}_{1}}</math> and <math>{{P}_{2}}</math> are the portion of the total load that each unit supports when both units are operational. In this case, <math>{{P}_{1}}={{P}_{2}}=0.5=50%.</math> 
:• <math>{{S}_{1}}</math> and <math>{{S}_{2}}</math> represent the portions of the load that Unit 1 and Unit 2 must support when both units are operational. 
:• <math>{{t}_{{{1}_{e}}}}</math> is the equivalent operating time for Unit 1 if it had been operating at <math>S</math> instead of <math>{{S}_{1}}</math> . A graphical representation of the equivalent time is shown in Figure 5.19, where the curve marked by L represents the low stress (load) and the curve marked by H represents the high stress (load). 
 

::<math>{{t}_{1e}}</math> can be calculated by:
 
::<math>\begin{align}
{{R}_{1}}(t)= & {{R}_{1}}({{t}_{1e}}) \\
{{e}^{-{{(tKS_{1}^{n})}^{\beta }}}}= & {{e}^{-{{({{t}_{1e}}K{{S}^{n}})}^{\beta }}}} \\
tS_{1}^{n}= & {{t}_{1e}}{{S}^{n}} \\
{{t}_{1e}}= & t{{\left( \frac{{{S}_{1}}}{S} \right)}^{n}},\text{ }{{S}_{1}}={{P}_{1}}S \\
\therefore & {{t}_{1e}}=tP_{1}^{n}
\end{align}</math>

 
::<math>{{t}_{2e}}</math> can be calculated the same way, or:

 
::<math>\begin{align}
{{R}_{2}}(t)= & {{R}_{2}}({{t}_{2e}}) \\
\therefore & {{t}_{2e}}=tP_{2}^{n}
\end{align}</math>
 

In this example, the reliability equations for Unit 1 and Unit 2 are the same since they are the same type of component and demonstrate the same failure properties. In addition, the total output is divided equally between the two units (when both units are operating), so <math>{{t}_{1e}}</math> and <math>{{t}_{2e}}</math> will also be the same.
 
The next step is to determine the reliability of the system after 8,760 hours, <math>R(t=8,760)</math> .
Using
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>
 

the system reliability is found to be:
 

::<math>\begin{align}
R(t=8760)= & 0.8567 \\
= & 85.67%
\end{align}</math>
 

===Load Sharing in BlockSim===
BlockSim uses this formulation when computing reliabilities of units in a load sharing configuration. When using the System Reliability Equation window, BlockSim returns a single token for the reliability of units in a load sharing configuration (as well as in the case of standby redundancy, discussed in the next section). As an example, consider the following RBD with Unit 1 in series with a container that includes two load sharing units.
 
[[Image:BS5.19.2.png|thumb|center|200px|]]
 
BlockSim will return the system equation as:
 

::<math>{{R}_{System}}=+{{R}_{LS}}\cdot {{R}_{1}}</math>
 

Where <math>{{R}_{LS}}</math> implies a form similar to
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>
 .
BlockSim allows for <math>k</math> -out-of- <math>n</math> units in a load sharing configuration.

===Example 5===
A component has five possible failure modes, <math>A</math> , <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> , <math>{{B}_{C}}</math> and <math>C</math> , and the <math>B</math> modes are interdependent. The system will fail if mode <math>A</math> occurs, mode <math>C</math> occurs or two out of the three <math>B</math> modes occur.
 

Modes <math>A</math> and <math>C</math> each have a Weibull distribution, with a <math>\beta =2</math> and <math>\eta =10,000</math> and 15,000 respectively. Events <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> and <math>{{B}_{C}}</math> each have an exponential distribution with a mean of 10,000 hours.
 

If any <math>B</math> event occurs (i.e. <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> or <math>{{B}_{C}}</math> ), the remaining <math>B</math> events are more likely to occur. Specifically, the mean times of the remaining <math>B</math> events are halved. Determine the reliability at 1000 hours for this component.
====Solution to Example 5====
The first step is to create the RBD. Modes <math>A</math> , <math>C</math> and a load sharing container with the <math>{{B}_{i}}</math> modes must be drawn in series, as illustrated next.
 
[[Image:BS5.19.3.png|thumb|center|200px|]]
 

The next step is to define the properties for each block, including those for the container. Setting the failure distributions for modes <math>A</math> and <math>C</math> is simple. The more difficult part is setting the properties for the container and the contained blocks. Based on the problem statement, the <math>B</math> modes are in a 2-out-of-3 load sharing redundancy. When all three are working (i.e. when no <math>B</math> mode has occurred), each block has an exponential distribution with <math>\mu=10,000</math>. If one <math>B</math> mode occurs, then the two surviving units have an exponential distribution with <math>\mu =5,000.</math>
 

Assume a Power Life-Stress relationship for the components. Then:
 

::<math>\begin{align}
{{\mu }_{1}}= & \frac{1}{KV_{1}^{n}} \ \\
{{\mu }_{2}}= & \frac{1}{KV_{2}^{n}} \
\end{align}</math>
 

Substituting <math>{{\mu }_{1}}=10,000</math> and <math>{{V}_{1}}=1</math> in <math>{{\mu }_{1}}= \frac{1}{KV_{1}^{n}}</math> and casting it in terms of <math>K</math> yields:
 

::<math>\begin{align}
10,000= & \frac{1}{K} \ \\
K = & \frac{1}{10,000}=0.0001
\end{align}</math>

 
Substituting <math>{{\mu }_{2}}=5,000</math> , <math>{{V}_{2}}=1.5</math> (because if one fails, then each survivor takes on an additional 0.5 units of load) and <math>10,000= \frac{1}{K}</math> for <math>K</math> in <math>{{\mu }_{2}}= \frac{1}{KV_{2}^{n}}</math> yields:

 
::<math>\begin{align}
5,000= & \frac{1}{0.0001\cdot {{(1.5)}^{n}}} \\
0.5= & {{(1.5)}^{-n}} \\
\ln (0.5)= & -n\ln (1.5) \\
n = & 1.7095
\end{align}</math>
 

This also could have been computed in ALTA, as shown in Figure "Calculation performed in ALTA".

The next step is to set the weight proportionality factor. This factor defines the portion of the load that the particular item carries while operating, as well as the load that shifts to the remaining units upon failure of the item. To illustrate, assume three units (1, 2 and 3) are in a load sharing container with weight proportionality factors of 1, 2 and 3 respectively (and a 1-out-of-3 requirement).
 

:• Unit 1 carries <math>\left( \tfrac{1}{1+2+3} \right)=0.166</math> or 16.6% of the total load.
:• Unit 2 carries <math>\left( \tfrac{2}{1+2+3} \right)=0.333</math> or 33.3% of the total load.
:• Unit 3 carries <math>\left( \tfrac{3}{1+2+3} \right)=0.50</math> or 50% of the total load.
 

The actual load on each unit then becomes the product of the entire load defined for the container times the portion carried by that unit. For example, if the container load is 100 lbs, then the portion assigned to Unit 1 will be <math>100\cdot 0.166=16.6</math> lbs.
 

In the current example, all units share the same load and thus have equal weight proportionality factors. Because these factors are relative, if the same number is used for all three items then the results will be the same. Thus, weight proportional factor is set equal to 1 for each item.
 

[[Image:5.20.PNG|thumb|center|500px|<div align="center"> Calculation performed in ALTA.</div>]]
 

 
[[Image:Fig 5.23_2.PNG|thumb|center|400px|<div align="center"> Defining Weight Proportional Factor. </div>]]
 

The last properties that need to be defined are the total load and the 2-out-of-3 redundancy. The total load is dependent on how the parameters were computed. In this case, total load was assumed to be 3 when the parameters were computed (i.e. the load per item was 1 when all worked and 1.5 when two worked). This is defined at the container level, as shown in Figure "Defining Weight Proportional Factor".
When all of the parameters have been specified in BlockSim, the reliability at 1,000 hours can be determined. From the Analytical QCP, this is found to be 93.87%.

=Standby Components=

In the previous section, the case of a system with load sharing components was presented. This is a form of redundancy with dependent components. That is, the failure of one component affects the failure of the other(s). This section presents another form of redundancy: standby redundancy. In standby redundancy the redundant components are set to be under a lighter load condition (or no load) while not needed and under the operating load when they are activated.
 

In standby redundancy the components are set to have two states: an active state and a standby state. Components in standby redundancy have two failure distributions, one for each state. When in the standby state, they have a quiescent (or dormant) failure distribution and when operating, they have an active failure distribution.
 

In the case that both quiescent and active failure distributions are the same, the units are in a simple parallel configuration (also called a hot standby configuration). When the rate of failure of the standby component is lower in quiescent mode than in active mode, that is called a warm standby configuration. When the rate of failure of the standby component is zero in quiescent mode (i.e. the component cannot fail when in standby), that is called a cold standby configuration.

===Simple Standby Configuration===

Consider two components in a standby configuration. Component 1 is the active component with a Weibull failure distribution with parameters <math>\beta </math> = 1.5 and <math>\eta </math> = 1,000. Component 2 is the standby component. When Component 2 is operating, it also has a Weibull failure distribution with <math>\beta </math> = 1.5 and <math>\eta </math> = 1,000. Furthermore, assume the following cases for the quiescent distribution.
 
:• Case 1: The quiescent distribution is the same as the active distribution (hot standby). 
:• Case 2: The quiescent distribution is a Weibull distribution with <math>\beta </math> = 1.5 and <math>\eta </math> = 2000 (warm standby). 
:• Case 3: The component cannot fail in quiescent mode (cold standby). 
 
In this case, the reliability of the system at some time, <math>t</math> , can be obtained using the following equation:
 

::<math>R(t)={{R}_{1}}(t)+\underset{0}{\overset{t}{\mathop \int }}\,{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x)\cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}dx \ </math>
 

Where:
 
:• <math>{{R}_{1}}</math> is the reliability of the active component. 
:• <math>{{f}_{1}}</math> is the <math>pdf</math> of the active component. 
:• <math>{{R}_{2;SB}}</math> is the reliability of the standby component when in standby mode (quiescent reliability). 
:• <math>{{R}_{2;A}}</math> is the reliability of the standby component when in active mode. 
:• <math>{{t}_{e}}</math> is the equivalent operating time for the standby unit if it had been operating at an active mode, such that: 

 
<math>{{R}_{2;SB}}(x)={{R}_{2;A}}({{t}_{e}})</math> 
 
The second equation above can be solved for <math>{{t}_{e}}</math> and substituted into the first equation above.
Figure "Standby container" illustrates the example as entered in BlockSim using a standby container.
 
 
[[Image:5.24.gif|thumb|center|400px|<div align="center"> Standby container.</div>]]

 

The active and standby blocks are within a container, which is used to specify standby redundancy. Since the standby component has two distributions (active and quiescent), the Block Properties window of the standby block has two pages for specifying each one. Figures "Defining the active failure distribution" and "Defining the quiescent failure distribution" illustrate these pages.

 

The system reliability results for 1000 hours are given in the following table:
 
[[Image:5-24.png|thumb|center|400px|]]
 

Note that even though the <math>\beta </math> value for the quiescent distribution is the same as in the active distribution, it is possible that the two can be different. That is, the failure modes present during the quiescent mode could be different from the modes present during the active mode. In that sense, the two distribution types can be different as well (e.g. lognormal when quiescent and Weibull when active).
 

In many cases when considering standby systems, a switching device may also be present that switches from the failed active component to the standby component. The reliability of the switch can also be incorporated into
<math>R(t)={{R}_{1}}(t)+\underset{0}{\overset{t}{\mathop \int }}\,{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x)\cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}dx \ </math>

as presented in the next section.
 

BlockSim's System Reliability Equation window returns a single token for the reliability of units in a standby configuration. This is the same as the load sharing case presented in the previous section.
 

[[Image: Fig 5.25.PNG|thumb|center|400px|<div align="center"> Defining the active failure distribution </div>]]
 
 

[[Image:Fig 5.26.PNG|thumb|center|400px|<div align="center"> Defining the quiescent failure distribution </div>]]
 

===Reliability of Standby Systems with a Switching Device===

In many cases when dealing with standby systems, a switching device is present that will switch to the standby component when the active component fails. Therefore, the failure properties of the switch must also be included in the analysis.
 

[[Image:BS5.26.2.png|thumb|center|300px|]]
 

In most cases when the reliability of a switch is to be included in the analysis, two probabilities can be considered. The first and most common one is the probability of the switch performing the action (i.e. switching) when requested to do so. This is called Switch Probability per Request in BlockSim and is expressed as a static probability (e.g. 90%). The second probability is the quiescent reliability of the switch. This is the reliability of the switch as it ages (e.g. the switch might wear out with age due to corrosion, material degradation, etc.). Thus it is possible for the switch to fail before the active component fails. However, a switch failure does not cause the system to fail, but rather causes the system to fail only if the switch is needed and the switch has failed. For example, if the active component does not fail until the mission end time and the switch fails, then the system does not fail. However, if the active component fails and the switch has also failed, then the system cannot be switched to the standby component and it therefore fails.
 

In analyzing standby components with a switching device, either or both failure probabilities (during the switching or while waiting to switch) can be considered for the switch, since each probability can represent different failure modes. For example, the switch probability per request may represent software-related issues or the probability of detecting the failure of an active component, and the quiescent probability may represent wear-out type failures of the switch.
 

To illustrate the formulation, consider the previous example that assumes perfect switching. To examine the effects of including an imperfect switch, assume that when the active component fails there is a 90% probability that the switch will switch from the active component to the standby component. In addition, assume that the switch can also fail due to a wear-out failure mode described by a Weibull distribution with <math>\beta </math> = 1.7 and <math>\eta </math> = 5000.
 

Therefore, the reliability of the system at some time, <math>t</math> , is given by the following equation.

::<math>\begin{align}
R(t)= & {{R}_{1}}(t) \\
& +\underset{0}{\overset{t}{\mathop \int }}\,\{{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x) \\
& \cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}\cdot {{R}_{SW;Q}}(x)\cdot {{R}_{SW;REQ}}(x)\}dx
\end{align}</math>

Where:
 
:• <math>{{R}_{1}}</math> is the reliability of the active component. 
:• <math>{{f}_{1}}</math> is the <math>pdf</math> of the active component. 
:• <math>{{R}_{2;SB}}</math> is the reliability of the standby component when in standby mode (quiescent reliability). 
:• <math>{{R}_{2;A}}</math> is the reliability of the standby component when in active mode. 
:• <math>{{R}_{SW;Q}}</math> is the quiescent reliability of the switch. 
:• <math>{{R}_{SW;REQ}}</math> is the switch probability per request. 
:• <math>{{t}_{e}}</math> is the equivalent operating time for the standby unit if it had been operating at an active mode. 
This problem can be solved in BlockSim by including these probabilities in the container's properties, as shown in Figures "Standby container (switch) failure probabilities while attempting to switch" and "Standby container (switch) failure distribution while waiting to switch". In BlockSim, the standby container is acting as the switch.

 

[[Image:Fig 5.28_2.PNG|thumb|center|400px|<div align="center"> Standby container (switch) failure probabilities while attempting to switch </div> ]]
 

 

[[Image:Fig 5.27.PNG|thumb|center|400px|<div align="center"> Standby container (switch) failure distribution while waiting to switch </div>]]
 

 

Note that there are additional properties that can be specified in BlockSim for a switch, such as Switch Restart Probability, No. of Restarts and Switch Delay Time. In many applications, the switch is re-tested (or re-cycled) if it fails to switch the first time. In these cases, it might be possible that it switches in the second or third, or <math>{{n}^{th}}</math> attempt.
 
The Switch Restart Probability specifies each additional attempt's probability of successfully switching and the Finite Restarts specifies the total number of attempts. Note that the Switch Restart Probability specifies the probability of success of each trial (or attempt). The probability of success of <math>n</math> consecutive trials is calculated by BlockSim using the binomial distribution and this probability is then incorporated into equation above. The Switch Delay Time property is related to repairable systems and is considered in BlockSim only when using simulation. When using the analytical solution (i.e. for a non-repairable system), this property is ignored.
 

Solving the analytical solution (as given by Equation above), the following results are obtained.
 

[[Image:5-30.png|thumb|center|400px|]]
 

From the table above, it can be seen that the presence of a switching device has a significant effect on the reliability of a standby system. It is therefore important when modeling standby redundancy to incorporate the switching device reliability properties. It should be noted that this methodology is not the same as treating the switching device as another series component with the standby subsystem. This would be valid only if the failure of the switch resulted in the failure of system (e.g. switch failing open). In equation above, the Switch Probability per Request and quiescent probability are present only in the second term of the equation. Treating these two failure modes as a series configuration with the standby subsystem would imply that they are also present when the active component is functioning (i.e. first term of equation above). This is invalid and would result in the underestimation of the reliability of the system. In other words, these two failure modes become significant only when the active component fails.
 

As an example, and if we consider the warm standby case, the reliability of the system without the switch is 70.57% at 1000 hours. If the system was modeled so that the switching device was in series with the warm standby subsystem, the result would have been:

 

::<math>\begin{align}
{{R}_{S}}(1000)= & {{R}_{Standby}}(1000)\cdot {{R}_{sw,Q(1000)}}\cdot {{R}_{sw,req}} \\
= & 0.7057\cdot 0.9372\cdot 0.9 \\
= & 0.5952
\end{align}</math>

 

In the case where a switch failure mode causes the standby subsystem to fail, then this mode can be modeled as an individual block in series with the standby subsystem.

===Example 6===
Consider a car with four new tires and a full-size spare. Assume the following failure characteristics:
 
:• The tires follow a Weibull distribution with a <math>\beta=4</math> and an <math>\eta =</math> 40,000 miles while on the car due to wear. 
:• The tires also have a probability of failing due to puncture or other causes. For this, assume a constant rate for this occurrence with a probability of 1 every 50,000 miles. 
:• When not on the car (i.e. is a spare), a tire's probability of failing also has a Weibull distribution with a <math>\beta =</math> 2 and <math>\eta =</math> 120,000 miles. 
 
Assume a mission of 1,000 miles. If a tire fails during this trip, it will be replaced with the spare. However, the spare will not be repaired during the trip. In other words, the trip will continue with the spare on the car and if the spare fails the system will fail. Determine the probability of system failure.
====Solution to Example 6====
Active failure distribution for tires:
 
:• Due to wear-out, Weibull <math>\beta =4</math> and <math>\eta =40,000</math> miles. 
:• Due to random puncture, exponential <math>\mu =50,000.</math> 
:• The quiescent failure distribution is a Weibull distribution with <math>\beta =2</math> and <math>\eta =120,000</math> miles. 
 
The block diagram for each tire has two blocks in series, one block representing the wear-out mode and the other the random puncture mode, as shown next:
 
 
[[Image:small5.gif|thumb|center|400px|]]
 

There are five tires, four active and one standby (represented in the diagram by a standby container with a 4-out-of-5 requirement), as shown next:
 
[[Image:BStirecontainer.png|thumb|center|400px|]]

For the standby Wear block, set the active failure and the quiescent distributions, but for the Puncture block, set only the active puncture distribution (because the tire cannot fail due to puncture while stored). Using BlockSim, the probability of system failure is found to be 0.003 or 0.3%.
 

=Note Regarding Numerical Integration Solutions=

Load sharing and standby solutions in BlockSim are performed using numerical integration routines. As with any numerical analysis routine, the solution error depends on the number of iterations performed, the step size chosen and related factors, plus the behavior of the underlying function. By default, BlockSim uses a certain set of preset factors. In general, these defaults are sufficient for most problems. If a higher precision or verification of the precision for a specific problem is required, BlockSim's preset options can be modified and/or the integration error can be assessed using the Integration Parameters... option for each container. For more details, you can refer to the documentation on the Algorithm Setup window in the BlockSim 8 User's help files.

Time-Dependent System Reliability (Analytical)

2012-03-21T21:27:18Z

Dingzhou Cao: /* Solution to Example 3 */

{{Template:bsbook|5}}

In the previous chapter, different system configuration types were examined, as well as different methods for obtaining the system's reliability function analytically. Because the reliabilities in the problems presented were treated as probabilities (e.g. <math>P(A)</math> , <math>{{R}_{i}}</math> ), the reliability values and equations presented were referred to as static (not time-dependent). Thus, in the prior chapter, the life distributions of the components were not incorporated in the process of calculating the system reliability. In this chapter, time dependency in the reliability function will be introduced. We will develop the models necessary to observe the reliability over the life of the system, instead of at just one point in time. In addition, performance measures such as failure rate, MTTF and warranty time will be estimated for the entire system. The methods of obtaining the reliability function analytically remain identical to the ones presented in the previous chapter, with the exception that the reliabilities will be functions of time. In other words, instead of dealing with <math>{{R}_{i}}</math> , we will use <math>{{R}_{i}}(t)</math> . All examples in this chapter assume that no repairs are performed on the components.

=Analytical Life Predictions=
The analytical approach presented in the prior chapter involved the determination of a mathematical expression that describes the reliability of the system, expressed in terms of the reliabilities of its components. So far we have estimated only static system reliability (at a fixed time). For example, in the case of a system with three components in series, the system's reliability equation was given by:
 

::<math>{{R}_{s}}={{R}_{1}}\cdot {{R}_{2}}\cdot {{R}_{3}} \ </math>
 

The values of <math>{{R}_{1}}</math> , <math>{{R}_{2}}</math> and <math>{{R}_{3}}</math> were given for a common time and the reliability of the system was estimated for that time. However, since the component failure characteristics can be described by distributions, the system reliability is actually time-dependent. In this case, the equation above can be rewritten as:
 

::<math>{{R}_{s}}(t)={{R}_{1}}(t)\cdot {{R}_{2}}(t)\cdot {{R}_{3}}(t)</math>

 

The reliability of the system for any mission time can now be estimated. Assuming a Weibull life distribution for each component, the first equation above can now be expressed in terms of each component's reliability function, or:

 

[[Category:Completed Theoretical Review]]

::<math>{{R}_{s}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}}\cdot {{e}^{-{{\left( \tfrac{t}{{{\eta }_{2}}} \right)}^{{{\beta }_{2}}}}}}\cdot {{e}^{-{{\left( \tfrac{t}{{{\eta }_{3}}} \right)}^{{{\beta }_{3}}}}}}</math>

 

In the same manner, any life distribution can be substituted into the system reliability equation. Suppose that the times-to-failure of the first component are described with a Weibull distribution, the times-to-failure of the second component with an exponential distribution and the times-to-failure of the third component with a normal distribution. Then the first equation above can be written as:
 

::<math>{{R}_{s}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}}\cdot {{e}^{-{{\lambda }_{2}}t}}\cdot \left[ 1-\Phi \left( \frac{t-{{\mu }_{3}}}{{{\sigma }_{3}}} \right) \right]</math>

 

It can be seen that the biggest challenge is in obtaining the system's reliability function in terms of component reliabilities, which has already been discussed in depth. Once this has been achieved, calculating the reliability of the system for any mission duration is just a matter of substituting the corresponding component reliability functions into the system reliability equation.
===Advantages of the Analytical Method===
The primary advantage of the analytical solution is that it produces a mathematical expression that describes the reliability of the system. Once the system's reliability function has been determined, other calculations can then be performed to obtain metrics of interest for the system. Such calculations include:
 
:• Determination of the system's <math>pdf.</math> 
:• Determination of warranty periods. 
:• Determination of the system's failure rate. 
:• Determination of the system's MTTF. 
 
In addition, optimization and reliability allocation techniques can be used to aid engineers in their design improvement efforts. Another advantage in using analytical techniques is the ability to perform static calculations and analyze systems with a mixture of static and time-dependent components. Finally, the reliability importance of components over time can be calculated with this methodology.
===Disadvantages of the Analytical Method===
The biggest disadvantage of the analytical method is that formulations can become very complicated. The more complicated a system is, the larger and more difficult it will be to analytically formulate an expression for the system's reliability. For particularly detailed systems this process can be quite time-consuming, even with the use of computers. Furthermore, when the maintainability of the system or some of its components must be taken into consideration, analytical solutions become intractable. In these situations, the use of simulation methods may be more advantageous than attempting to develop a solution analytically. Simulation methods are presented in later chapters.
===Looking at a Simple ''Complex'' System Analytically===

The complexity involved in an analytical solution can be best illustrated by looking at the simple ''complex'' system with 15 components, as shown in Figure " An RBD of a complex system".
 
[[Image:5-1.png|thumb|center|400px|<div align="center"> An RBD of a complex system.</div>]]

 
The system reliability for this system (computed using BlockSim) is shown next. The first solution is provided using BlockSim's symbolic solution. In symbolic mode, BlockSim breaks the equation into segments, identified by tokens, that need to be substituted into the final system equation for a complete solution. This creates algebraic solutions that are more compact than if the substitutions were made.
 

::<math>\begin{align}
{{R}_{System}}= & D2\cdot D3\cdot {{R}_{L}} \\
D3= & +{{R}_{K}}\cdot IK \\
IK= & +{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}} \\
D2 = & +{{R}_{A}}\cdot {{R}_{E}}\cdot IE \\
IE = & -D1\cdot {{R}_{M}}\cdot {{R}_{N}}+{{R}_{M}}\cdot {{R}_{N}}+D1 \\
D1 = & +{{R}_{D}}\cdot ID \\
ID = & -{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}}
\end{align}</math>
 
Substituting the terms yields:
 

::<math>\begin{align}
{{R}_{System}}= & {{R}_{A}}\cdot {{R}_{E}}\cdot {{R}_{L}}\cdot {{R}_{K}} \\
& \cdot \{({{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})\cdot {{R}_{M}}\cdot {{R}_{N}} \\
& +{{R}_{M}}\cdot {{R}_{N}}-{{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}}\} \\
& \cdot \{{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}}\}
\end{align}</math>

 
BlockSim's automatic algebraic simplification would yield the following format for the above solution:
 

::<math>\begin{align}
{{R}_{System}}= & (({{R}_{A}}\cdot {{R}_{E}}(-({{R}_{D}}(-{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})){{R}_{M}}\cdot {{R}_{N}} \\
& +{{R}_{M}}\cdot {{R}_{N}} \\
& +({{R}_{D}}(-{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})))) \\
& ({{R}_{K}}({{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& +RI\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}})){{R}_{L}}) \
\end{align}</math>

 
In this equation, each <math>{{R}_{i}}</math> represents the reliability function of a block. For example, if <math>{{R}_{A}}</math> has a Weibull distribution, then each <math>{{R}_{A}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{A}}} \right)}^{{{\beta }_{A}}}}}}</math> and so forth. Substitution of each component's reliability function in the last <math>{{R}_{System}}</math> equation above will result in an analytical expression for the system reliability as a function of time, or <math>{{R}_{s}}(t)</math> , which is the same as <math>(1-cd{{f}_{System}}).</math>

===Obtaining Other Functions of Interest===
Once the system reliability equation (or the cumulative density function, <math>cdf</math> ) has been determined, other functions and metrics of interest can be derived.
 
Consider the following simple system:
 
[[Image:5-2.png|thumb|center|300px|<div align="center"> Simple two-component system. </div>]]

 
Furthermore, assume that component 1 follows an exponential distribution with a mean of 10,000 (<math>\mu =10,000,</math> <math>\lambda =1/10,000)</math> and component 2 follows a Weibull distribution with <math>\beta =6</math> and <math>\eta =10,000</math> . The reliability equation of this system is:

 
::<math>\begin{align}
{{R}_{S}}(t)= & {{R}_{1}}(t)\cdot {{R}_{2}}(t) \\
= & {{e}^{-\lambda t}}\cdot {{e}^{-{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \\
= & {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \
\end{align}</math>

 
The system <math>cdf</math> is:

 
::<math>\begin{align}
{{F}_{S}}(t)= & 1-({{R}_{1}}(t)\cdot {{R}_{2}}(t)) \\
= & 1-\left( {{e}^{-\lambda t}}\cdot {{e}^{-{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \right) \\
= & 1-\left( {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)
\end{align}</math>

 
{{analytical system pdf}}

====Conditional Reliability====
Conditional reliability is the probability of a system successfully completing another mission following the successful completion of a previous mission. The time of the previous mission and the time for the mission to be undertaken must be taken into account for conditional reliability calculations. The system's conditional reliability function is given by:

::<math>R(T,t)=\frac{R(T+t)}{R(T)}</math>

Equation above gives the reliability for a new mission of duration <math>t</math> having already accumulated <math>T</math> hours of operation up to the start of this new mission. The system is evaluated to assure that it will start the next mission successfully.
 
For the system in Figure "Simple two-component system.", the reliability for mission of <math>t=1,000</math> hours, having an age of <math>T=500</math> hours, is:
 

::<math>\begin{align}
{{R}_{S}}(T=500,t=1000)= & \frac{R(T+t)}{R(T)} \\
= & \frac{R(1500)}{R(500)} \\
= & \frac{{{e}^{-\tfrac{1500}{10,000}}}\cdot {{e}^{-{{\left( \tfrac{1500}{10,000} \right)}^{6}}}}}{{{e}^{-\tfrac{500}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{500}{10,000} \right)}^{6}}}}} \\
= & 0.9048=90.48%
\end{align}</math>

 

====Conditional Reliability for Components====

Now in this formulation, it was assumed that the accumulated age was equivalent for both units. That is, both started life at zero and aged to 500. It is possible to consider an individual component that has already accumulated some age (used component) in the same formulation. To illustrate this, assume that component 2 started life with an age of T=100. Then the reliability equation of the system, as given in <math>{{R}_{S}}(t)= {{R}_{1}}(t)\cdot {{R}_{2}}(t)</math>, would need to be modified to include a conditional term for 2, or:
 

::<math>{{R}_{S}}(t)={{R}_{1}}(t)\cdot \frac{{{R}_{2}}({{T}_{2}}+t)}{{{R}_{2}}({{T}_{2}})} \ </math>

 

In BlockSim, the start age input box may be used to specify a starting age greater than zero.
{{system failure rate analytical}}

====System Mean Life (Mean Time To Failure)====
The mean life (or mean time to failure, MTTF) can be obtained by integrating the system reliability function from zero to infinity:
 

::<math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math>

 

The mean time is a performance index and does not provide any information about the behavior of the failure distribution of the system.
 
For the system in Figure "Simple two-component system":
 

::<math>\begin{align}
MTTF= & \int_{0}^{\infty }\left( {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)dt \\
= & 5978.9
\end{align}</math>

 

====Warranty Period and BX Life====
Sometimes it is desirable to know the time value associated with a certain reliability. Warranty periods are often calculated by determining what percentage of the failure population can be covered financially and estimating the time at which this portion of the population will fail. Similarly, engineering specifications may call for a certain BX life, which also represents a time period during which a certain proportion of the population will fail. For example, the B10 life is the time in which 10% of the population will fail.
This is obtained by setting <math>{{R}_{S}}(t)</math> to the desired value and solving for <math>t.</math>
 
For the system in Figure "Simple two-component system":
 

::<math>{{R}_{s}}\left( t \right)={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>
 

To compute the time by which reliability would be equal to 90%, equation above is recast as follows and solved for <math>t.</math>

 
::<math>0.90={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>

 

In this case, <math>t=1053.59</math> . Equivalently, the B10 life for this system is also <math>1053.59</math> .
Except for some trivial cases, a closed form solution for <math>t</math> cannot be obtained. Thus, it is necessary to solve for <math>t</math> using numerical methods. BlockSim uses numerical methods.

===Example 1===
Consider a system consisting of three exponential units in series with the following failure rates (in failures per hour): <math>{{\lambda }_{1}}</math> = 0.0002, <math>{{\lambda }_{2}}</math> = 0.0005 and <math>{{\lambda }_{3}}</math> = 0.0001.
 
:• Obtain the reliability equation for the system. 
:• What is the reliability of the system after 150 hours of operation? 
:• Obtain the system's <math>pdf.</math> 
:• Obtain the system's failure rate equation. 
:• What is the MTTF for the system? 
:• What should the warranty period be for a 90% reliability? 
====Solution to Example 1====
 
:• The analytical expression for the reliability of the system is given by:

::<math>\begin{align}
{{R}_{s}}(t)= & {{R}_{1}}(t)\cdot {{R}_{2}}(t)\cdot {{R}_{3}}(t) \\
= & {{e}^{-{{\lambda }_{1}}t}}\cdot {{e}^{-{{\lambda }_{2}}t}}\cdot {{e}^{-{{\lambda }_{1}}t}} \\
= & {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}
\end{align}</math>

 

At 150 hours of operation, the reliability of the system is:
 

::<math>\begin{align}
{{R}_{s}}(t)= & {{e}^{-(0.0002+0.0005+0.0001)150}} \\
= & 0.8869\text{ or }88.69%
\end{align}</math>

 

:• In order to obtain the system's <math>pdf</math> , the derivative of the reliability equation given in the first equation above is taken with respect to time, or:
 
::<math>\begin{align}
{{f}_{s}}(t)= & -\frac{d[{{R}_{s}}(t)]}{dt} \\
= & -\frac{d\left[ {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}} \right]}{dt} \\
= & ({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})\cdot {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}
\end{align}</math>

:• The system's failure rate can now be obtained simply by dividing the system's <math>pdf</math> given in the equation above by the system's reliability function given in the first equation above, and:

::<math>\begin{align}
{{\lambda }_{s}}\left( t \right)= & \frac{{{f}_{s}}\left( t \right)}{{{R}_{s}}\left( t \right)} \\
= & \frac{({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})\cdot {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}}{{{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}} \\
= & ({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}) \\
= & 0.0008\text{ }fr/hr
\end{align}</math>

 
:• Combining <math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math> and the first equation above, the system's MTTF can be obtained:

 
::<math>\begin{align}
MTTF= & \int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \\
= & \int_{0}^{\infty }{{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}dt \\
= & \frac{1}{({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})} \\
= & 1250\text{ }hr
\end{align}</math>
 

:• Solving the first equation above with respect to time will yield the corresponding warranty period for a 90% reliability. In this case, the system reliability equation is simple and a closed form solution exists. The warranty time can now be found by solving:

 
::<math>\begin{align}
t= & -\frac{\ln (R)}{{{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}} \\
= & -\frac{\ln (0.9)}{0.0008} \\
= & 131.7\text{ }hr
\end{align}</math>
 

Thus, the warranty period should be 132 hours.

===Example 2===
Consider the system shown in Figure "Complex bridge system in Example 2".
 
[[Image:BS5.5.png|thumb|center|300px|<div align="center"> Complex bridge system in Example 2. </div>]]
 

Components <math>A</math> through <math>E</math> are Weibull distributed with <math>\beta =1.2</math> and <math>\eta =1230</math> hours. The starting and ending blocks cannot fail. 
Determine the following:
 
:• The reliability equation for the system and its corresponding plot. 
:• The system's <math>pdf</math> and its corresponding plot. 
:• The system's failure rate equation and the corresponding plot. 
:• The <math>MTTF</math> . 
:• The warranty time for a 90% reliability. 
:• The reliability for a 200-hour mission, if it is known that the system has already successfully operated for 200 hours. 
====Solution====
 
The first step is to obtain the reliability function for the system. The methods described in the previous chapter can be employed, such as the event space or path-tracing methods. Using BlockSim, the following reliability equation is obtained:
 

::<math>\begin{align}
{{R}_{s}}(t)= & ({{R}_{Start}}\cdot {{R}_{End}}(2{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}))
\end{align}</math>
 

Note that since the starting and ending blocks cannot fail, <math>{{R}_{Start}}=1</math> and <math>{{R}_{End}}=1,</math> equation above can be reduced to:

 

::<math>\begin{align}
{{R}_{s}}(t)= & 2\cdot {{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}
\end{align}</math>

 

Where <math>{{R}_{A}}</math> is the reliability equation for Component A, or:

 

::<math>{{R}_{A}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{A}}} \right)}^{{{\beta }_{A}}}}}}</math>

 

::<math>{{R}_{B}}</math> is the reliability equation for Component <math>B</math> , etc.
 

Since the components in this example are identical, the system reliability equation can be further reduced to:

 

::<math>{{R}_{s}}(t)=2R{{(t)}^{2}}+2R{{(t)}^{3}}-5R{{(t)}^{4}}+2R{{(t)}^{5}}</math>

 

Or, in terms of the failure distribution:

 

::<math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>

 

The corresponding plot is given in Figure "Reliability plot for the system".
 
[[Image:BS5.6.png|thumb|center|300px|<div align="center"> Reliability plot for the system. </div>]]
 
 

In order to obtain the system's <math>pdf</math> , the derivative of the reliability equation given above is taken with respect to time, resulting in:
 

::<math>\begin{align}
{{f}_{s}}(t)= & 4\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+6\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \\
& -20\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+10\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}
\end{align}</math>

 

The <math>pdf</math> can now be plotted for different time values, <math>t</math> , as shown in Figure "''pdf'' plot for the system".
 

The system's failure rate can now be obtained by dividing the system's <math>pdf</math> given in equation above by the system's reliability function given in <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>, or:

 

::<math>\begin{align}
{{\lambda }_{s}}(t)= & \frac{4\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+6\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}{2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}} \\
& +\frac{-20\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+10\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}{2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}
\end{align}</math>

 

The corresponding plot is given in Figure "Failure rate for the system".
 
[[Image:BS5.7.gif|thumb|center|300px|<div align="center"> ''pdf'' plot for the system.</div>]]
 

The <math>MTTF</math> of the system is obtained by integrating the system's reliability function given by <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math> from time zero to infinity, as given by <math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math>. Using BlockSim's Analytical QCP, an <math>MTTF</math> of 1007.8 hours is calculated, as shown in Figure "MTTF of the system".
 

The warranty time can be obtained by solving <math>{{R}_{s}}(t)</math> with respect to time for a system reliability <math>{{R}_{s}}=0.9</math> . Using the Analytical QCP and selecting the 
Warranty Time option, a time of 372.72 hours is obtained, as shown in Figure "Time at which ''R''=0.9 or 90% for the system".
 
[[Image:BS5.8.gif|thumb|center|400px|<div align="center"> Failure rate for the system.</div>]]
 
[[Image:Fig 5.9.PNG|thumb|center|400px|<div align="center"> MTTF of the system. </div>]]
 
[[Image:Fig 5.10.PNG|thumb|center|400px|<div align="center"> Time at which ''R''=0.9 or 90% for the system.</div>]]
 
[[Image:Fig 5.11.PNG|thumb|center|400px|<div align="center"> Conditional reliability calculation for the system.</div>]]
 

Lastly, the conditional reliability can be obtained using <math>R(T,t)=\frac{R(T+t)}{R(T)}</math> and <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>, or:
 

::<math>\begin{align}
R(200,200)= & \frac{R(400)}{R(200)} \\
= & \frac{0.883825}{0.975321} \\
= & 0.906189
\end{align}</math>

 

This can be calculated using BlockSim's Analytical QCP, as shown in Figure "Conditional reliability calculation for the system".
 

=Approximating the System CDF=

In many cases, it is valuable to fit a distribution that represents the system's times-to-failure. This can be useful when the system is part of a larger assembly and may be used for repeated calculations or in calculations for other systems. In cases such as this, it can be useful to characterize the system's behavior by fitting a distribution to the overall system and calculating parameters for this distribution. This is equivalent to fitting a single distribution to describe <math>{{R}_{S}}(t</math> ). In essence, it is like reducing the entire system to a component in order to simplify calculations.
 
For the system in Figure below:
[[Image:5-2.png|thumb|center|300px|<div align="center">Simple two-component system. </div>]]
 

::<math>{{R}_{S}}(t)={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>

 
To compute an approximate reliability function for this system, <math>{{R}_{A}}(t)\simeq {{R}_{S}}(t)</math> , one would compute <math>n</math> pairs of reliability and time values and then fit a single distribution to the data, or:

 

::<math>\begin{align}
{{R}_{S}}(t= & 10,396.7)=10% \\
{{R}_{S}}(t= & 9,361.9)=20% \\
& ... \\
{{R}_{S}}(t= & 1,053.6)=90%
\end{align}</math>

 

A single distribution, <math>{{R}_{A}}(t)</math> , that approximates <math>{{R}_{S}}(t)</math> can now be computed from these pairs using life data analysis methods. If using the Weibull++ software, one would enter the values as free form data.

===Example 3===
Compute a single Weibull distribution approximation for the system in Example 2.
====Solution to Example 3====
The system in the previous example, shown in Figure "Complex bridge system in Example 2", can be approximated by use of a 2-parameter Weibull distribution with <math>\beta =2.02109</math> and <math>\eta =1123.51</math> . In BlockSim, this is accomplished by representing the entire system as one distribution by going to the Distribution Fit window Figure " Representing a system with a distribution".

 
[[Image:Fig 5.13.PNG|thumb|center|400px|<div align="center"> Representing a system with a distribution.</div>]]
 
[[Image:Fig 5.12.PNG|thumb|center|300px|<div align="center"> Distribution Fitting window.</div>]]
 
by clicking the Distribution Fit Window, the Distribution Estimator window will appear (Figure "Distribution Fitting window").
 
In this window you can select a distribution to represent the data. BlockSim will then generate a number of system failure times based on the system's reliability function. The system's reliability function can be used to solve for a time value associated with that unreliability value. The distribution of the generated time values can then be fitted to a probability distribution function.
 

Consider a value of <math>F(t)=0.11</math> . Using the system's reliability equation and solving for time, the corresponding time-to-failure for a 0.11 unreliability can be calculated. 
For the system of Example 2, the time for a 0.11 unreliability is 389.786 hours.
 
When enough points have been generated, the selected distribution will be fitted to this data set and the distribution's parameters will be returned. In addition, if ReliaSoft's Weibull++ is installed, the generated data can be viewed/analyzed using a Weibull++ instance, as shown in Figure "Using Weibull++ to calculate distribution parameters".
 
[[Image:Fig 5.14.PNG|thumb|center|400px|<div align="center"> Fig 5.14 Using Weibull++ to calculate distribution parameters.</div>]]
 
It is recommended that the analyst examine the fit to ascertain the applicability of the approximation.

=Duty Cycle=

Components of a system may not operate continuously during a system's mission, or may be subjected to loads greater or lesser than the rated loads during system operation. To model this, a factor called the Duty Cycle ( <math>{{d}_{c}}</math> ) is used. The duty cycle may also be used to account for changes in environmental stress, such as temperature changes, that may effect the operation of a component. The duty cycle is a positive value, with a default value of 1 representing continuous operation at rated load, and any values other than 1 representing other load values with respect to the rated load value (or total operating time). A duty cycle value higher than 1 indicates a load in excess of the rated value. A duty cycle value lower than 1 indicates that the component is operating at a load lower than the rated load or not operating continuously during the system's mission. For instance, a duty cycle of 0.5 may be used for a component that operates only half of the time during the system's mission.
 
The reliability metrics for a component with a duty cycle are calculated as follows. Let <math>{{d}_{c}}</math> represent the duty cycle during a particular mission of the component, <math>t</math> represent the mission time and <math>{t}'</math> represent the accumulated age. Then:
 

::<math>{t}'={{d}_{c}}\times t</math>

 

The reliability equation for the component is:
 

::<math>R({t}')=R({{d}_{c}}\times t)</math>
 

The component ''pdf'' is:
 

::<math>f({t}')=-\frac{d(R({t}'))}{dt}=-\frac{d(R({{d}_{c}}\times t))}{dt}={{d}_{c}}f({{d}_{c}}\times t)</math>
 

The failure rate of the component is:
 

::<math>\lambda ({t}')=\frac{f({t}')}{R({t}')}=\frac{{{d}_{c}}f({{d}_{c}}\times t)}{R({{d}_{c}}\times t)}={{d}_{c}}\lambda ({{d}_{c}}\times t)</math>

===Example 4===
Consider a computer system with three components: a processor, a hard drive and a CD drive in series as shown next. Assume that all three components follow a Weibull failure distribution with the parameters <math>{{\beta }_{1}}=1.5</math> and <math>{{\eta }_{1}}=5000</math> for the processor, <math>{{\beta }_{2}}=2.5</math> and <math>{{\eta }_{2}}=3000</math> for the hard drive, and <math>{{\beta }_{3}}=2</math> and <math>{{\eta }_{3}}=4000</math> for the CD drive. Determine the reliability of the computer system after one year (365 days) of operation, assuming that the CD drive is used only 30% of the time.
 
[[Image:BS5ex4.png|thumb|center|300px|]]

====Solution to Example 4====
The reliability of the processor after 365 days of operation is given by:

::<math>\begin{align}
{{R}_{processor}}(365)= & {{e}^{-{{\left( \tfrac{365}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}} \\
= & {{e}^{-{{\left( \tfrac{365}{5000} \right)}^{1.5}}}} \\
= & 0.9805\text{ or }98.05%
\end{align}</math>

The reliability of the hard drive after 365 days of operation is given by:

::<math>\begin{align}
{{R}_{harddrive}}(365)= & {{e}^{-{{\left( \tfrac{365}{{{\eta }_{2}}} \right)}^{{{\beta }_{2}}}}}} \\
= & {{e}^{-{{\left( \tfrac{365}{3000} \right)}^{2.5}}}} \\
= & 0.9948\text{ or }99.48%
\end{align}</math>

The reliability of the CD drive after 365 days of operation (taking into account the 30% operation using a duty cycle of 0.3) is given by:

::<math>\begin{align}
{{R}_{CDdrive}}(365)= & {{e}^{-{{\left( \tfrac{{{d}_{c}}\times 365}{{{\eta }_{3}}} \right)}^{{{\beta }_{3}}}}}} \\
= & {{e}^{-{{\left( \tfrac{0.3\times 365}{4000} \right)}^{2}}}} \\
= & 0.9993\text{ or }99.93%
\end{align}</math>
 
[[Image:Fig 5.15.PNG|thumb|center|400px|<div align="center"> Fig 5.15 Result for the computer system reliability.</div>]]
 

Thus the reliability of the computer system after 365 days of operation is:

::<math>\begin{align}
{{R}_{s}}(365)= & {{R}_{processor}}(365)\cdot {{R}_{harddrive}}(365)\cdot {{R}_{CDdrive}}(365) \\
= & 0.9805\cdot 0.9948\cdot 0.9993 \\
= & 0.9747\text{ or }97.47%
\end{align}</math>
 
This result can be obtained in BlockSim as shown in Figure "Result for the computer system reliability".
 

=Load Sharing=
As presented in earlier chapters, a reliability block diagram (RBD) allows you to graphically represent how the components within a system are reliability-wise connected. In most cases, independence is assumed across the components within the system. For example, the failure of component A does not affect the failure of component B. However, if a system consists of components that are sharing a load, then the assumption of independence no longer holds true.
 

If one component fails, then the component(s) that are still operating will have to assume the failed unit's portion of the load. Therefore, the reliabilities of the surviving unit(s) will change. Calculating the system reliability is no longer an easy proposition. In the case of load sharing components, the change of the failure distributions of the surviving components must be known in order to determine the system's reliability.
 

To illustrate this, consider the a system of two units connected reliability-wise in parallel (Figure "Two units connected reliability-wise in parallel").
 

[[Image:5-16.png|thumb|center|300px|<div align="center"> Two units connected reliability-wise in parallel.</div>]]
 

Assume that the units must supply an output of 8 volts and that if both units are operational, each unit is to supply 50% of the total output. If one of the units fails, then the surviving unit supplies 100%. Furthermore, assume that having to supply the entire load has a negative impact on the reliability characteristics of the surviving unit. Since the reliability characteristics of the unit change based on whether both or only one is operating, a life distribution along with a life-stress relationship (as discussed in [http://reliawiki.com/index.php/Statistical_Background#A_Brief_Introduction_to_Life-Stress_Relationships|A Brief Introduction to Life-Stress Relationships]) will be needed to model each component.
 

To illustrate the steps needed, we will create the model starting from raw data. Assume that a total of 20 units were tested to failure at 7, 10 and 15 volts. The test data set is presented in the next table.
 

[[Image:5-17.png|thumb|center|300px|]]
 

For this example, Units 1 and 2 are the same component. Therefore, only one set of data was collected. However, it is possible that the load sharing components in a system may not be the same. If that were the case, data would need to be collected for each component.
 

The data set in Table 1 was analyzed using ReliaSoft's ALTA software (as shown in Figure "Using ALTA to calculate component parameters") with the Inverse Power Law as the underlying life-stress relationship and Weibull as the life distribution.
 
 

The estimated model parameters, <math>\beta </math> , <math>K</math> and <math>n</math> , are shown next.
 

::<math>\begin{align}
\beta = & 1.9239 \\
K= & 3.2387\times {{10}^{-7}} \\
n= & 3.4226
\end{align}</math>
 

:Or:
 
::<math>\begin{align}
{{R}_{1}}(t,{{S}_{1}})= & {{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} \\
= & {{e}^{-{{\left( 3.2387\times {{10}^{-7}}S_{1}^{3.4226}t \right)}^{1.9239}}}} \
\end{align}</math>.

 
::<math>{{f}_{1}}(t,{{S}_{1}})=\beta KS_{1}^{n}{{\left( KS_{1}^{n}t \right)}^{\beta -1}}{{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} \ </math>
 

And for this case:

::<math>\begin{align}
{{R}_{1}}(t,{{S}_{1}})= & {{R}_{2}}(t,{{S}_{2}}) \\
{{f}_{1}}(t,{{S}_{1}})= & {{f}_{2}}(t,{{S}_{2}})
\end{align}</math>
 
 
[[Image:Fig 5.17.PNG|thumb|center|500px|<div align="center"> Using ALTA to calculate component parameters.</div>]]
 

Figure "Reliability curves for different voltage output conditions" shows a plot of <math>{{R}_{1}}(t,{{S}_{1}})={{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} = {{e}^{-{{\left( 3.2387\times {{10}^{-7}}S_{1}^{3.4226}t \right)}^{1.9239}}}}\ </math>.
 

Now that the failure properties have been determined using the test data, the reliability of the system at some time, <math>t</math> , can be calculated using the following equation:

 
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>

 
:Where:
 

::<math>\begin{align}
{{S}_{1}}= & {{P}_{1}}S \\
{{S}_{2}}= & {{P}_{2}}S
\end{align}</math>

 
[[Image:5-18.png|thumb|center|300px|<div align="center"> Reliability curves for different voltage output conditions. </div>]]
 
[[Image:BS5.19.png|thumb|center|395px| Illustrating <math>{{t}_{e}}</math>]]
 

And:
 

:• <math>S</math> is the total load (or required output). 
:• <math>{{P}_{1}}</math> and <math>{{P}_{2}}</math> are the portion of the total load that each unit supports when both units are operational. In this case, <math>{{P}_{1}}={{P}_{2}}=0.5=50%.</math> 
:• <math>{{S}_{1}}</math> and <math>{{S}_{2}}</math> represent the portions of the load that Unit 1 and Unit 2 must support when both units are operational. 
:• <math>{{t}_{{{1}_{e}}}}</math> is the equivalent operating time for Unit 1 if it had been operating at <math>S</math> instead of <math>{{S}_{1}}</math> . A graphical representation of the equivalent time is shown in Figure 5.19, where the curve marked by L represents the low stress (load) and the curve marked by H represents the high stress (load). 
 

::<math>{{t}_{1e}}</math> can be calculated by:
 
::<math>\begin{align}
{{R}_{1}}(t)= & {{R}_{1}}({{t}_{1e}}) \\
{{e}^{-{{(tKS_{1}^{n})}^{\beta }}}}= & {{e}^{-{{({{t}_{1e}}K{{S}^{n}})}^{\beta }}}} \\
tS_{1}^{n}= & {{t}_{1e}}{{S}^{n}} \\
{{t}_{1e}}= & t{{\left( \frac{{{S}_{1}}}{S} \right)}^{n}},\text{ }{{S}_{1}}={{P}_{1}}S \\
\therefore & {{t}_{1e}}=tP_{1}^{n}
\end{align}</math>

 
::<math>{{t}_{2e}}</math> can be calculated the same way, or:

 
::<math>\begin{align}
{{R}_{2}}(t)= & {{R}_{2}}({{t}_{2e}}) \\
\therefore & {{t}_{2e}}=tP_{2}^{n}
\end{align}</math>
 

In this example, the reliability equations for Unit 1 and Unit 2 are the same since they are the same type of component and demonstrate the same failure properties. In addition, the total output is divided equally between the two units (when both units are operating), so <math>{{t}_{1e}}</math> and <math>{{t}_{2e}}</math> will also be the same.
 
The next step is to determine the reliability of the system after 8,760 hours, <math>R(t=8,760)</math> .
Using
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>
 

the system reliability is found to be:
 

::<math>\begin{align}
R(t=8760)= & 0.8567 \\
= & 85.67%
\end{align}</math>
 

===Load Sharing in BlockSim===
BlockSim uses this formulation when computing reliabilities of units in a load sharing configuration. When using the System Reliability Equation window, BlockSim returns a single token for the reliability of units in a load sharing configuration (as well as in the case of standby redundancy, discussed in the next section). As an example, consider the following RBD with Unit 1 in series with a container that includes two load sharing units.
 
[[Image:BS5.19.2.png|thumb|center|200px|]]
 
BlockSim will return the system equation as:
 

::<math>{{R}_{System}}=+{{R}_{LS}}\cdot {{R}_{1}}</math>
 

Where <math>{{R}_{LS}}</math> implies a form similar to
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>
 .
BlockSim allows for <math>k</math> -out-of- <math>n</math> units in a load sharing configuration.

===Example 5===
A component has five possible failure modes, <math>A</math> , <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> , <math>{{B}_{C}}</math> and <math>C</math> , and the <math>B</math> modes are interdependent. The system will fail if mode <math>A</math> occurs, mode <math>C</math> occurs or two out of the three <math>B</math> modes occur.
 

Modes <math>A</math> and <math>C</math> each have a Weibull distribution, with a <math>\beta =2</math> and <math>\eta =10,000</math> and 15,000 respectively. Events <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> and <math>{{B}_{C}}</math> each have an exponential distribution with a mean of 10,000 hours.
 

If any <math>B</math> event occurs (i.e. <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> or <math>{{B}_{C}}</math> ), the remaining <math>B</math> events are more likely to occur. Specifically, the mean times of the remaining <math>B</math> events are halved. Determine the reliability at 1000 hours for this component.
====Solution to Example 5====
The first step is to create the RBD. Modes <math>A</math> , <math>C</math> and a load sharing container with the <math>{{B}_{i}}</math> modes must be drawn in series, as illustrated next.
 
[[Image:BS5.19.3.png|thumb|center|200px|]]
 

The next step is to define the properties for each block, including those for the container. Setting the failure distributions for modes <math>A</math> and <math>C</math> is simple. The more difficult part is setting the properties for the container and the contained blocks. Based on the problem statement, the <math>B</math> modes are in a 2-out-of-3 load sharing redundancy. When all three are working (i.e. when no <math>B</math> mode has occurred), each block has an exponential distribution with <math>\mu=10,000</math>. If one <math>B</math> mode occurs, then the two surviving units have an exponential distribution with <math>\mu =5,000.</math>
 

Assume a Power Life-Stress relationship for the components. Then:
 

::<math>\begin{align}
{{\mu }_{1}}= & \frac{1}{KV_{1}^{n}} \ \\
{{\mu }_{2}}= & \frac{1}{KV_{2}^{n}} \
\end{align}</math>
 

Substituting <math>{{\mu }_{1}}=10,000</math> and <math>{{V}_{1}}=1</math> in <math>{{\mu }_{1}}= \frac{1}{KV_{1}^{n}}</math> and casting it in terms of <math>K</math> yields:
 

::<math>\begin{align}
10,000= & \frac{1}{K} \ \\
K = & \frac{1}{10,000}=0.0001
\end{align}</math>

 
Substituting <math>{{\mu }_{2}}=5,000</math> , <math>{{V}_{2}}=1.5</math> (because if one fails, then each survivor takes on an additional 0.5 units of load) and <math>10,000= \frac{1}{K}</math> for <math>K</math> in <math>{{\mu }_{2}}= \frac{1}{KV_{2}^{n}}</math> yields:

 
::<math>\begin{align}
5,000= & \frac{1}{0.0001\cdot {{(1.5)}^{n}}} \\
0.5= & {{(1.5)}^{-n}} \\
\ln (0.5)= & -n\ln (1.5) \\
n = & 1.7095
\end{align}</math>
 

This also could have been computed in ALTA, as shown in Figure "Calculation performed in ALTA".

The next step is to set the weight proportionality factor. This factor defines the portion of the load that the particular item carries while operating, as well as the load that shifts to the remaining units upon failure of the item. To illustrate, assume three units (1, 2 and 3) are in a load sharing container with weight proportionality factors of 1, 2 and 3 respectively (and a 1-out-of-3 requirement).
 

:• Unit 1 carries <math>\left( \tfrac{1}{1+2+3} \right)=0.166</math> or 16.6% of the total load.
:• Unit 2 carries <math>\left( \tfrac{2}{1+2+3} \right)=0.333</math> or 33.3% of the total load.
:• Unit 3 carries <math>\left( \tfrac{3}{1+2+3} \right)=0.50</math> or 50% of the total load.
 

The actual load on each unit then becomes the product of the entire load defined for the container times the portion carried by that unit. For example, if the container load is 100 lbs, then the portion assigned to Unit 1 will be <math>100\cdot 0.166=16.6</math> lbs.
 

In the current example, all units share the same load and thus have equal weight proportionality factors. Because these factors are relative, if the same number is used for all three items then the results will be the same. Thus, weight proportional factor is set equal to 1 for each item.
 

[[Image:5.20.PNG|thumb|center|500px|<div align="center"> Calculation performed in ALTA.</div>]]
 

 
[[Image:Fig 5.23_2.PNG|thumb|center|400px|<div align="center"> Defining Weight Proportional Factor. </div>]]
 

The last properties that need to be defined are the total load and the 2-out-of-3 redundancy. The total load is dependent on how the parameters were computed. In this case, total load was assumed to be 3 when the parameters were computed (i.e. the load per item was 1 when all worked and 1.5 when two worked). This is defined at the container level, as shown in Figure "Defining Weight Proportional Factor".
When all of the parameters have been specified in BlockSim, the reliability at 1,000 hours can be determined. From the Analytical QCP, this is found to be 93.87%.

=Standby Components=

In the previous section, the case of a system with load sharing components was presented. This is a form of redundancy with dependent components. That is, the failure of one component affects the failure of the other(s). This section presents another form of redundancy: standby redundancy. In standby redundancy the redundant components are set to be under a lighter load condition (or no load) while not needed and under the operating load when they are activated.
 

In standby redundancy the components are set to have two states: an active state and a standby state. Components in standby redundancy have two failure distributions, one for each state. When in the standby state, they have a quiescent (or dormant) failure distribution and when operating, they have an active failure distribution.
 

In the case that both quiescent and active failure distributions are the same, the units are in a simple parallel configuration (also called a hot standby configuration). When the rate of failure of the standby component is lower in quiescent mode than in active mode, that is called a warm standby configuration. When the rate of failure of the standby component is zero in quiescent mode (i.e. the component cannot fail when in standby), that is called a cold standby configuration.

===Simple Standby Configuration===

Consider two components in a standby configuration. Component 1 is the active component with a Weibull failure distribution with parameters <math>\beta </math> = 1.5 and <math>\eta </math> = 1,000. Component 2 is the standby component. When Component 2 is operating, it also has a Weibull failure distribution with <math>\beta </math> = 1.5 and <math>\eta </math> = 1,000. Furthermore, assume the following cases for the quiescent distribution.
 
:• Case 1: The quiescent distribution is the same as the active distribution (hot standby). 
:• Case 2: The quiescent distribution is a Weibull distribution with <math>\beta </math> = 1.5 and <math>\eta </math> = 2000 (warm standby). 
:• Case 3: The component cannot fail in quiescent mode (cold standby). 
 
In this case, the reliability of the system at some time, <math>t</math> , can be obtained using the following equation:
 

::<math>R(t)={{R}_{1}}(t)+\underset{0}{\overset{t}{\mathop \int }}\,{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x)\cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}dx \ </math>
 

Where:
 
:• <math>{{R}_{1}}</math> is the reliability of the active component. 
:• <math>{{f}_{1}}</math> is the <math>pdf</math> of the active component. 
:• <math>{{R}_{2;SB}}</math> is the reliability of the standby component when in standby mode (quiescent reliability). 
:• <math>{{R}_{2;A}}</math> is the reliability of the standby component when in active mode. 
:• <math>{{t}_{e}}</math> is the equivalent operating time for the standby unit if it had been operating at an active mode, such that: 

 
<math>{{R}_{2;SB}}(x)={{R}_{2;A}}({{t}_{e}})</math> 
 
The second equation above can be solved for <math>{{t}_{e}}</math> and substituted into the first equation above.
Figure "Standby container" illustrates the example as entered in BlockSim using a standby container.
 
 
[[Image:5.24.gif|thumb|center|400px|<div align="center"> Standby container.</div>]]

 

The active and standby blocks are within a container, which is used to specify standby redundancy. Since the standby component has two distributions (active and quiescent), the Block Properties window of the standby block has two pages for specifying each one. Figures "Defining the active failure distribution" and "Defining the quiescent failure distribution" illustrate these pages.

 

The system reliability results for 1000 hours are given in the following table:
 
[[Image:5-24.png|thumb|center|400px|]]
 

Note that even though the <math>\beta </math> value for the quiescent distribution is the same as in the active distribution, it is possible that the two can be different. That is, the failure modes present during the quiescent mode could be different from the modes present during the active mode. In that sense, the two distribution types can be different as well (e.g. lognormal when quiescent and Weibull when active).
 

In many cases when considering standby systems, a switching device may also be present that switches from the failed active component to the standby component. The reliability of the switch can also be incorporated into
<math>R(t)={{R}_{1}}(t)+\underset{0}{\overset{t}{\mathop \int }}\,{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x)\cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}dx \ </math>

as presented in the next section.
 

BlockSim's System Reliability Equation window returns a single token for the reliability of units in a standby configuration. This is the same as the load sharing case presented in the previous section.
 

[[Image: Fig 5.25.PNG|thumb|center|400px|<div align="center"> Defining the active failure distribution </div>]]
 
 

[[Image:Fig 5.26.PNG|thumb|center|400px|<div align="center"> Defining the quiescent failure distribution </div>]]
 

===Reliability of Standby Systems with a Switching Device===

In many cases when dealing with standby systems, a switching device is present that will switch to the standby component when the active component fails. Therefore, the failure properties of the switch must also be included in the analysis.
 

[[Image:BS5.26.2.png|thumb|center|300px|]]
 

In most cases when the reliability of a switch is to be included in the analysis, two probabilities can be considered. The first and most common one is the probability of the switch performing the action (i.e. switching) when requested to do so. This is called Switch Probability per Request in BlockSim and is expressed as a static probability (e.g. 90%). The second probability is the quiescent reliability of the switch. This is the reliability of the switch as it ages (e.g. the switch might wear out with age due to corrosion, material degradation, etc.). Thus it is possible for the switch to fail before the active component fails. However, a switch failure does not cause the system to fail, but rather causes the system to fail only if the switch is needed and the switch has failed. For example, if the active component does not fail until the mission end time and the switch fails, then the system does not fail. However, if the active component fails and the switch has also failed, then the system cannot be switched to the standby component and it therefore fails.
 

In analyzing standby components with a switching device, either or both failure probabilities (during the switching or while waiting to switch) can be considered for the switch, since each probability can represent different failure modes. For example, the switch probability per request may represent software-related issues or the probability of detecting the failure of an active component, and the quiescent probability may represent wear-out type failures of the switch.
 

To illustrate the formulation, consider the previous example that assumes perfect switching. To examine the effects of including an imperfect switch, assume that when the active component fails there is a 90% probability that the switch will switch from the active component to the standby component. In addition, assume that the switch can also fail due to a wear-out failure mode described by a Weibull distribution with <math>\beta </math> = 1.7 and <math>\eta </math> = 5000.
 

Therefore, the reliability of the system at some time, <math>t</math> , is given by the following equation.

::<math>\begin{align}
R(t)= & {{R}_{1}}(t) \\
& +\underset{0}{\overset{t}{\mathop \int }}\,\{{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x) \\
& \cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}\cdot {{R}_{SW;Q}}(x)\cdot {{R}_{SW;REQ}}(x)\}dx
\end{align}</math>

Where:
 
:• <math>{{R}_{1}}</math> is the reliability of the active component. 
:• <math>{{f}_{1}}</math> is the <math>pdf</math> of the active component. 
:• <math>{{R}_{2;SB}}</math> is the reliability of the standby component when in standby mode (quiescent reliability). 
:• <math>{{R}_{2;A}}</math> is the reliability of the standby component when in active mode. 
:• <math>{{R}_{SW;Q}}</math> is the quiescent reliability of the switch. 
:• <math>{{R}_{SW;REQ}}</math> is the switch probability per request. 
:• <math>{{t}_{e}}</math> is the equivalent operating time for the standby unit if it had been operating at an active mode. 
This problem can be solved in BlockSim by including these probabilities in the container's properties, as shown in Figures "Standby container (switch) failure probabilities while attempting to switch" and "Standby container (switch) failure distribution while waiting to switch". In BlockSim, the standby container is acting as the switch.

 

[[Image:Fig 5.28_2.PNG|thumb|center|400px|<div align="center"> Standby container (switch) failure probabilities while attempting to switch </div> ]]
 

 

[[Image:Fig 5.27.PNG|thumb|center|400px|<div align="center"> Standby container (switch) failure distribution while waiting to switch </div>]]
 

 

Note that there are additional properties that can be specified in BlockSim for a switch, such as Switch Restart Probability, No. of Restarts and Switch Delay Time. In many applications, the switch is re-tested (or re-cycled) if it fails to switch the first time. In these cases, it might be possible that it switches in the second or third, or <math>{{n}^{th}}</math> attempt.
 
The Switch Restart Probability specifies each additional attempt's probability of successfully switching and the Finite Restarts specifies the total number of attempts. Note that the Switch Restart Probability specifies the probability of success of each trial (or attempt). The probability of success of <math>n</math> consecutive trials is calculated by BlockSim using the binomial distribution and this probability is then incorporated into equation above. The Switch Delay Time property is related to repairable systems and is considered in BlockSim only when using simulation. When using the analytical solution (i.e. for a non-repairable system), this property is ignored.
 

Solving the analytical solution (as given by Equation above), the following results are obtained.
 

[[Image:5-30.png|thumb|center|400px|]]
 

From the table above, it can be seen that the presence of a switching device has a significant effect on the reliability of a standby system. It is therefore important when modeling standby redundancy to incorporate the switching device reliability properties. It should be noted that this methodology is not the same as treating the switching device as another series component with the standby subsystem. This would be valid only if the failure of the switch resulted in the failure of system (e.g. switch failing open). In equation above, the Switch Probability per Request and quiescent probability are present only in the second term of the equation. Treating these two failure modes as a series configuration with the standby subsystem would imply that they are also present when the active component is functioning (i.e. first term of equation above). This is invalid and would result in the underestimation of the reliability of the system. In other words, these two failure modes become significant only when the active component fails.
 

As an example, and if we consider the warm standby case, the reliability of the system without the switch is 70.57% at 1000 hours. If the system was modeled so that the switching device was in series with the warm standby subsystem, the result would have been:

 

::<math>\begin{align}
{{R}_{S}}(1000)= & {{R}_{Standby}}(1000)\cdot {{R}_{sw,Q(1000)}}\cdot {{R}_{sw,req}} \\
= & 0.7057\cdot 0.9372\cdot 0.9 \\
= & 0.5952
\end{align}</math>

 

In the case where a switch failure mode causes the standby subsystem to fail, then this mode can be modeled as an individual block in series with the standby subsystem.

===Example 6===
Consider a car with four new tires and a full-size spare. Assume the following failure characteristics:
 
:• The tires follow a Weibull distribution with a <math>\beta=4</math> and an <math>\eta =</math> 40,000 miles while on the car due to wear. 
:• The tires also have a probability of failing due to puncture or other causes. For this, assume a constant rate for this occurrence with a probability of 1 every 50,000 miles. 
:• When not on the car (i.e. is a spare), a tire's probability of failing also has a Weibull distribution with a <math>\beta =</math> 2 and <math>\eta =</math> 120,000 miles. 
 
Assume a mission of 1,000 miles. If a tire fails during this trip, it will be replaced with the spare. However, the spare will not be repaired during the trip. In other words, the trip will continue with the spare on the car and if the spare fails the system will fail. Determine the probability of system failure.
====Solution to Example 6====
Active failure distribution for tires:
 
:• Due to wear-out, Weibull <math>\beta =4</math> and <math>\eta =40,000</math> miles. 
:• Due to random puncture, exponential <math>\mu =50,000.</math> 
:• The quiescent failure distribution is a Weibull distribution with <math>\beta =2</math> and <math>\eta =120,000</math> miles. 
 
The block diagram for each tire has two blocks in series, one block representing the wear-out mode and the other the random puncture mode, as shown next:
 
 
[[Image:small5.gif|thumb|center|400px|]]
 

There are five tires, four active and one standby (represented in the diagram by a standby container with a 4-out-of-5 requirement), as shown next:
 
[[Image:BStirecontainer.png|thumb|center|400px|]]

For the standby Wear block, set the active failure and the quiescent distributions, but for the Puncture block, set only the active puncture distribution (because the tire cannot fail due to puncture while stored). Using BlockSim, the probability of system failure is found to be 0.003 or 0.3%.
 

=Note Regarding Numerical Integration Solutions=

Load sharing and standby solutions in BlockSim are performed using numerical integration routines. As with any numerical analysis routine, the solution error depends on the number of iterations performed, the step size chosen and related factors, plus the behavior of the underlying function. By default, BlockSim uses a certain set of preset factors. In general, these defaults are sufficient for most problems. If a higher precision or verification of the precision for a specific problem is required, BlockSim's preset options can be modified and/or the integration error can be assessed using the Integration Parameters... option for each container. For more details, you can refer to the documentation on the Algorithm Setup window in the BlockSim 8 User's help files.

Time-Dependent System Reliability (Analytical)

2012-03-21T21:26:59Z

Dingzhou Cao: /* Example 2 */

{{Template:bsbook|5}}

In the previous chapter, different system configuration types were examined, as well as different methods for obtaining the system's reliability function analytically. Because the reliabilities in the problems presented were treated as probabilities (e.g. <math>P(A)</math> , <math>{{R}_{i}}</math> ), the reliability values and equations presented were referred to as static (not time-dependent). Thus, in the prior chapter, the life distributions of the components were not incorporated in the process of calculating the system reliability. In this chapter, time dependency in the reliability function will be introduced. We will develop the models necessary to observe the reliability over the life of the system, instead of at just one point in time. In addition, performance measures such as failure rate, MTTF and warranty time will be estimated for the entire system. The methods of obtaining the reliability function analytically remain identical to the ones presented in the previous chapter, with the exception that the reliabilities will be functions of time. In other words, instead of dealing with <math>{{R}_{i}}</math> , we will use <math>{{R}_{i}}(t)</math> . All examples in this chapter assume that no repairs are performed on the components.

=Analytical Life Predictions=
The analytical approach presented in the prior chapter involved the determination of a mathematical expression that describes the reliability of the system, expressed in terms of the reliabilities of its components. So far we have estimated only static system reliability (at a fixed time). For example, in the case of a system with three components in series, the system's reliability equation was given by:
 

::<math>{{R}_{s}}={{R}_{1}}\cdot {{R}_{2}}\cdot {{R}_{3}} \ </math>
 

The values of <math>{{R}_{1}}</math> , <math>{{R}_{2}}</math> and <math>{{R}_{3}}</math> were given for a common time and the reliability of the system was estimated for that time. However, since the component failure characteristics can be described by distributions, the system reliability is actually time-dependent. In this case, the equation above can be rewritten as:
 

::<math>{{R}_{s}}(t)={{R}_{1}}(t)\cdot {{R}_{2}}(t)\cdot {{R}_{3}}(t)</math>

 

The reliability of the system for any mission time can now be estimated. Assuming a Weibull life distribution for each component, the first equation above can now be expressed in terms of each component's reliability function, or:

 

[[Category:Completed Theoretical Review]]

::<math>{{R}_{s}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}}\cdot {{e}^{-{{\left( \tfrac{t}{{{\eta }_{2}}} \right)}^{{{\beta }_{2}}}}}}\cdot {{e}^{-{{\left( \tfrac{t}{{{\eta }_{3}}} \right)}^{{{\beta }_{3}}}}}}</math>

 

In the same manner, any life distribution can be substituted into the system reliability equation. Suppose that the times-to-failure of the first component are described with a Weibull distribution, the times-to-failure of the second component with an exponential distribution and the times-to-failure of the third component with a normal distribution. Then the first equation above can be written as:
 

::<math>{{R}_{s}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}}\cdot {{e}^{-{{\lambda }_{2}}t}}\cdot \left[ 1-\Phi \left( \frac{t-{{\mu }_{3}}}{{{\sigma }_{3}}} \right) \right]</math>

 

It can be seen that the biggest challenge is in obtaining the system's reliability function in terms of component reliabilities, which has already been discussed in depth. Once this has been achieved, calculating the reliability of the system for any mission duration is just a matter of substituting the corresponding component reliability functions into the system reliability equation.
===Advantages of the Analytical Method===
The primary advantage of the analytical solution is that it produces a mathematical expression that describes the reliability of the system. Once the system's reliability function has been determined, other calculations can then be performed to obtain metrics of interest for the system. Such calculations include:
 
:• Determination of the system's <math>pdf.</math> 
:• Determination of warranty periods. 
:• Determination of the system's failure rate. 
:• Determination of the system's MTTF. 
 
In addition, optimization and reliability allocation techniques can be used to aid engineers in their design improvement efforts. Another advantage in using analytical techniques is the ability to perform static calculations and analyze systems with a mixture of static and time-dependent components. Finally, the reliability importance of components over time can be calculated with this methodology.
===Disadvantages of the Analytical Method===
The biggest disadvantage of the analytical method is that formulations can become very complicated. The more complicated a system is, the larger and more difficult it will be to analytically formulate an expression for the system's reliability. For particularly detailed systems this process can be quite time-consuming, even with the use of computers. Furthermore, when the maintainability of the system or some of its components must be taken into consideration, analytical solutions become intractable. In these situations, the use of simulation methods may be more advantageous than attempting to develop a solution analytically. Simulation methods are presented in later chapters.
===Looking at a Simple ''Complex'' System Analytically===

The complexity involved in an analytical solution can be best illustrated by looking at the simple ''complex'' system with 15 components, as shown in Figure " An RBD of a complex system".
 
[[Image:5-1.png|thumb|center|400px|<div align="center"> An RBD of a complex system.</div>]]

 
The system reliability for this system (computed using BlockSim) is shown next. The first solution is provided using BlockSim's symbolic solution. In symbolic mode, BlockSim breaks the equation into segments, identified by tokens, that need to be substituted into the final system equation for a complete solution. This creates algebraic solutions that are more compact than if the substitutions were made.
 

::<math>\begin{align}
{{R}_{System}}= & D2\cdot D3\cdot {{R}_{L}} \\
D3= & +{{R}_{K}}\cdot IK \\
IK= & +{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}} \\
D2 = & +{{R}_{A}}\cdot {{R}_{E}}\cdot IE \\
IE = & -D1\cdot {{R}_{M}}\cdot {{R}_{N}}+{{R}_{M}}\cdot {{R}_{N}}+D1 \\
D1 = & +{{R}_{D}}\cdot ID \\
ID = & -{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}}
\end{align}</math>
 
Substituting the terms yields:
 

::<math>\begin{align}
{{R}_{System}}= & {{R}_{A}}\cdot {{R}_{E}}\cdot {{R}_{L}}\cdot {{R}_{K}} \\
& \cdot \{({{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})\cdot {{R}_{M}}\cdot {{R}_{N}} \\
& +{{R}_{M}}\cdot {{R}_{N}}-{{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}}\} \\
& \cdot \{{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}}\}
\end{align}</math>

 
BlockSim's automatic algebraic simplification would yield the following format for the above solution:
 

::<math>\begin{align}
{{R}_{System}}= & (({{R}_{A}}\cdot {{R}_{E}}(-({{R}_{D}}(-{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})){{R}_{M}}\cdot {{R}_{N}} \\
& +{{R}_{M}}\cdot {{R}_{N}} \\
& +({{R}_{D}}(-{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})))) \\
& ({{R}_{K}}({{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& +RI\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}})){{R}_{L}}) \
\end{align}</math>

 
In this equation, each <math>{{R}_{i}}</math> represents the reliability function of a block. For example, if <math>{{R}_{A}}</math> has a Weibull distribution, then each <math>{{R}_{A}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{A}}} \right)}^{{{\beta }_{A}}}}}}</math> and so forth. Substitution of each component's reliability function in the last <math>{{R}_{System}}</math> equation above will result in an analytical expression for the system reliability as a function of time, or <math>{{R}_{s}}(t)</math> , which is the same as <math>(1-cd{{f}_{System}}).</math>

===Obtaining Other Functions of Interest===
Once the system reliability equation (or the cumulative density function, <math>cdf</math> ) has been determined, other functions and metrics of interest can be derived.
 
Consider the following simple system:
 
[[Image:5-2.png|thumb|center|300px|<div align="center"> Simple two-component system. </div>]]

 
Furthermore, assume that component 1 follows an exponential distribution with a mean of 10,000 (<math>\mu =10,000,</math> <math>\lambda =1/10,000)</math> and component 2 follows a Weibull distribution with <math>\beta =6</math> and <math>\eta =10,000</math> . The reliability equation of this system is:

 
::<math>\begin{align}
{{R}_{S}}(t)= & {{R}_{1}}(t)\cdot {{R}_{2}}(t) \\
= & {{e}^{-\lambda t}}\cdot {{e}^{-{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \\
= & {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \
\end{align}</math>

 
The system <math>cdf</math> is:

 
::<math>\begin{align}
{{F}_{S}}(t)= & 1-({{R}_{1}}(t)\cdot {{R}_{2}}(t)) \\
= & 1-\left( {{e}^{-\lambda t}}\cdot {{e}^{-{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \right) \\
= & 1-\left( {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)
\end{align}</math>

 
{{analytical system pdf}}

====Conditional Reliability====
Conditional reliability is the probability of a system successfully completing another mission following the successful completion of a previous mission. The time of the previous mission and the time for the mission to be undertaken must be taken into account for conditional reliability calculations. The system's conditional reliability function is given by:

::<math>R(T,t)=\frac{R(T+t)}{R(T)}</math>

Equation above gives the reliability for a new mission of duration <math>t</math> having already accumulated <math>T</math> hours of operation up to the start of this new mission. The system is evaluated to assure that it will start the next mission successfully.
 
For the system in Figure "Simple two-component system.", the reliability for mission of <math>t=1,000</math> hours, having an age of <math>T=500</math> hours, is:
 

::<math>\begin{align}
{{R}_{S}}(T=500,t=1000)= & \frac{R(T+t)}{R(T)} \\
= & \frac{R(1500)}{R(500)} \\
= & \frac{{{e}^{-\tfrac{1500}{10,000}}}\cdot {{e}^{-{{\left( \tfrac{1500}{10,000} \right)}^{6}}}}}{{{e}^{-\tfrac{500}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{500}{10,000} \right)}^{6}}}}} \\
= & 0.9048=90.48%
\end{align}</math>

 

====Conditional Reliability for Components====

Now in this formulation, it was assumed that the accumulated age was equivalent for both units. That is, both started life at zero and aged to 500. It is possible to consider an individual component that has already accumulated some age (used component) in the same formulation. To illustrate this, assume that component 2 started life with an age of T=100. Then the reliability equation of the system, as given in <math>{{R}_{S}}(t)= {{R}_{1}}(t)\cdot {{R}_{2}}(t)</math>, would need to be modified to include a conditional term for 2, or:
 

::<math>{{R}_{S}}(t)={{R}_{1}}(t)\cdot \frac{{{R}_{2}}({{T}_{2}}+t)}{{{R}_{2}}({{T}_{2}})} \ </math>

 

In BlockSim, the start age input box may be used to specify a starting age greater than zero.
{{system failure rate analytical}}

====System Mean Life (Mean Time To Failure)====
The mean life (or mean time to failure, MTTF) can be obtained by integrating the system reliability function from zero to infinity:
 

::<math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math>

 

The mean time is a performance index and does not provide any information about the behavior of the failure distribution of the system.
 
For the system in Figure "Simple two-component system":
 

::<math>\begin{align}
MTTF= & \int_{0}^{\infty }\left( {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)dt \\
= & 5978.9
\end{align}</math>

 

====Warranty Period and BX Life====
Sometimes it is desirable to know the time value associated with a certain reliability. Warranty periods are often calculated by determining what percentage of the failure population can be covered financially and estimating the time at which this portion of the population will fail. Similarly, engineering specifications may call for a certain BX life, which also represents a time period during which a certain proportion of the population will fail. For example, the B10 life is the time in which 10% of the population will fail.
This is obtained by setting <math>{{R}_{S}}(t)</math> to the desired value and solving for <math>t.</math>
 
For the system in Figure "Simple two-component system":
 

::<math>{{R}_{s}}\left( t \right)={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>
 

To compute the time by which reliability would be equal to 90%, equation above is recast as follows and solved for <math>t.</math>

 
::<math>0.90={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>

 

In this case, <math>t=1053.59</math> . Equivalently, the B10 life for this system is also <math>1053.59</math> .
Except for some trivial cases, a closed form solution for <math>t</math> cannot be obtained. Thus, it is necessary to solve for <math>t</math> using numerical methods. BlockSim uses numerical methods.

===Example 1===
Consider a system consisting of three exponential units in series with the following failure rates (in failures per hour): <math>{{\lambda }_{1}}</math> = 0.0002, <math>{{\lambda }_{2}}</math> = 0.0005 and <math>{{\lambda }_{3}}</math> = 0.0001.
 
:• Obtain the reliability equation for the system. 
:• What is the reliability of the system after 150 hours of operation? 
:• Obtain the system's <math>pdf.</math> 
:• Obtain the system's failure rate equation. 
:• What is the MTTF for the system? 
:• What should the warranty period be for a 90% reliability? 
====Solution to Example 1====
 
:• The analytical expression for the reliability of the system is given by:

::<math>\begin{align}
{{R}_{s}}(t)= & {{R}_{1}}(t)\cdot {{R}_{2}}(t)\cdot {{R}_{3}}(t) \\
= & {{e}^{-{{\lambda }_{1}}t}}\cdot {{e}^{-{{\lambda }_{2}}t}}\cdot {{e}^{-{{\lambda }_{1}}t}} \\
= & {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}
\end{align}</math>

 

At 150 hours of operation, the reliability of the system is:
 

::<math>\begin{align}
{{R}_{s}}(t)= & {{e}^{-(0.0002+0.0005+0.0001)150}} \\
= & 0.8869\text{ or }88.69%
\end{align}</math>

 

:• In order to obtain the system's <math>pdf</math> , the derivative of the reliability equation given in the first equation above is taken with respect to time, or:
 
::<math>\begin{align}
{{f}_{s}}(t)= & -\frac{d[{{R}_{s}}(t)]}{dt} \\
= & -\frac{d\left[ {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}} \right]}{dt} \\
= & ({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})\cdot {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}
\end{align}</math>

:• The system's failure rate can now be obtained simply by dividing the system's <math>pdf</math> given in the equation above by the system's reliability function given in the first equation above, and:

::<math>\begin{align}
{{\lambda }_{s}}\left( t \right)= & \frac{{{f}_{s}}\left( t \right)}{{{R}_{s}}\left( t \right)} \\
= & \frac{({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})\cdot {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}}{{{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}} \\
= & ({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}) \\
= & 0.0008\text{ }fr/hr
\end{align}</math>

 
:• Combining <math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math> and the first equation above, the system's MTTF can be obtained:

 
::<math>\begin{align}
MTTF= & \int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \\
= & \int_{0}^{\infty }{{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}dt \\
= & \frac{1}{({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})} \\
= & 1250\text{ }hr
\end{align}</math>
 

:• Solving the first equation above with respect to time will yield the corresponding warranty period for a 90% reliability. In this case, the system reliability equation is simple and a closed form solution exists. The warranty time can now be found by solving:

 
::<math>\begin{align}
t= & -\frac{\ln (R)}{{{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}} \\
= & -\frac{\ln (0.9)}{0.0008} \\
= & 131.7\text{ }hr
\end{align}</math>
 

Thus, the warranty period should be 132 hours.

===Example 2===
Consider the system shown in Figure "Complex bridge system in Example 2".
 
[[Image:BS5.5.png|thumb|center|300px|<div align="center"> Complex bridge system in Example 2. </div>]]
 

Components <math>A</math> through <math>E</math> are Weibull distributed with <math>\beta =1.2</math> and <math>\eta =1230</math> hours. The starting and ending blocks cannot fail. 
Determine the following:
 
:• The reliability equation for the system and its corresponding plot. 
:• The system's <math>pdf</math> and its corresponding plot. 
:• The system's failure rate equation and the corresponding plot. 
:• The <math>MTTF</math> . 
:• The warranty time for a 90% reliability. 
:• The reliability for a 200-hour mission, if it is known that the system has already successfully operated for 200 hours. 
====Solution====
 
The first step is to obtain the reliability function for the system. The methods described in the previous chapter can be employed, such as the event space or path-tracing methods. Using BlockSim, the following reliability equation is obtained:
 

::<math>\begin{align}
{{R}_{s}}(t)= & ({{R}_{Start}}\cdot {{R}_{End}}(2{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}))
\end{align}</math>
 

Note that since the starting and ending blocks cannot fail, <math>{{R}_{Start}}=1</math> and <math>{{R}_{End}}=1,</math> equation above can be reduced to:

 

::<math>\begin{align}
{{R}_{s}}(t)= & 2\cdot {{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}
\end{align}</math>

 

Where <math>{{R}_{A}}</math> is the reliability equation for Component A, or:

 

::<math>{{R}_{A}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{A}}} \right)}^{{{\beta }_{A}}}}}}</math>

 

::<math>{{R}_{B}}</math> is the reliability equation for Component <math>B</math> , etc.
 

Since the components in this example are identical, the system reliability equation can be further reduced to:

 

::<math>{{R}_{s}}(t)=2R{{(t)}^{2}}+2R{{(t)}^{3}}-5R{{(t)}^{4}}+2R{{(t)}^{5}}</math>

 

Or, in terms of the failure distribution:

 

::<math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>

 

The corresponding plot is given in Figure "Reliability plot for the system".
 
[[Image:BS5.6.png|thumb|center|300px|<div align="center"> Reliability plot for the system. </div>]]
 
 

In order to obtain the system's <math>pdf</math> , the derivative of the reliability equation given above is taken with respect to time, resulting in:
 

::<math>\begin{align}
{{f}_{s}}(t)= & 4\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+6\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \\
& -20\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+10\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}
\end{align}</math>

 

The <math>pdf</math> can now be plotted for different time values, <math>t</math> , as shown in Figure "''pdf'' plot for the system".
 

The system's failure rate can now be obtained by dividing the system's <math>pdf</math> given in equation above by the system's reliability function given in <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>, or:

 

::<math>\begin{align}
{{\lambda }_{s}}(t)= & \frac{4\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+6\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}{2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}} \\
& +\frac{-20\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+10\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}{2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}
\end{align}</math>

 

The corresponding plot is given in Figure "Failure rate for the system".
 
[[Image:BS5.7.gif|thumb|center|300px|<div align="center"> ''pdf'' plot for the system.</div>]]
 

The <math>MTTF</math> of the system is obtained by integrating the system's reliability function given by <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math> from time zero to infinity, as given by <math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math>. Using BlockSim's Analytical QCP, an <math>MTTF</math> of 1007.8 hours is calculated, as shown in Figure "MTTF of the system".
 

The warranty time can be obtained by solving <math>{{R}_{s}}(t)</math> with respect to time for a system reliability <math>{{R}_{s}}=0.9</math> . Using the Analytical QCP and selecting the 
Warranty Time option, a time of 372.72 hours is obtained, as shown in Figure "Time at which ''R''=0.9 or 90% for the system".
 
[[Image:BS5.8.gif|thumb|center|400px|<div align="center"> Failure rate for the system.</div>]]
 
[[Image:Fig 5.9.PNG|thumb|center|400px|<div align="center"> MTTF of the system. </div>]]
 
[[Image:Fig 5.10.PNG|thumb|center|400px|<div align="center"> Time at which ''R''=0.9 or 90% for the system.</div>]]
 
[[Image:Fig 5.11.PNG|thumb|center|400px|<div align="center"> Conditional reliability calculation for the system.</div>]]
 

Lastly, the conditional reliability can be obtained using <math>R(T,t)=\frac{R(T+t)}{R(T)}</math> and <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>, or:
 

::<math>\begin{align}
R(200,200)= & \frac{R(400)}{R(200)} \\
= & \frac{0.883825}{0.975321} \\
= & 0.906189
\end{align}</math>

 

This can be calculated using BlockSim's Analytical QCP, as shown in Figure "Conditional reliability calculation for the system".
 

=Approximating the System CDF=

In many cases, it is valuable to fit a distribution that represents the system's times-to-failure. This can be useful when the system is part of a larger assembly and may be used for repeated calculations or in calculations for other systems. In cases such as this, it can be useful to characterize the system's behavior by fitting a distribution to the overall system and calculating parameters for this distribution. This is equivalent to fitting a single distribution to describe <math>{{R}_{S}}(t</math> ). In essence, it is like reducing the entire system to a component in order to simplify calculations.
 
For the system in Figure below:
[[Image:5-2.png|thumb|center|300px|<div align="center">Simple two-component system. </div>]]
 

::<math>{{R}_{S}}(t)={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>

 
To compute an approximate reliability function for this system, <math>{{R}_{A}}(t)\simeq {{R}_{S}}(t)</math> , one would compute <math>n</math> pairs of reliability and time values and then fit a single distribution to the data, or:

 

::<math>\begin{align}
{{R}_{S}}(t= & 10,396.7)=10% \\
{{R}_{S}}(t= & 9,361.9)=20% \\
& ... \\
{{R}_{S}}(t= & 1,053.6)=90%
\end{align}</math>

 

A single distribution, <math>{{R}_{A}}(t)</math> , that approximates <math>{{R}_{S}}(t)</math> can now be computed from these pairs using life data analysis methods. If using the Weibull++ software, one would enter the values as free form data.

===Example 3===
Compute a single Weibull distribution approximation for the system in Example 2.
====Solution to Example 3====
The system in the previous example, shown in Figure "Complex bridge system in Example 2", can be approximated by use of a 2-parameter Weibull distribution with <math>\beta =2.02109</math> and <math>\eta =1123.51</math> . In BlockSim, this is accomplished by representing the entire system as one distribution by going to the Distribution Fit window Figure " Representing a system with a distribution".

 
[[Image:Fig 5.13.PNG|thumb|center|400px|<div align="center"> Fig 5.12 Representing a system with a distribution.</div>]]
 
[[Image:Fig 5.12.PNG|thumb|center|300px|<div align="center"> Fig 5.13 Distribution Fitting window.</div>]]
 
by clicking the Distribution Fit Window, the Distribution Estimator window will appear (Figure "Distribution Fitting window").
 
In this window you can select a distribution to represent the data. BlockSim will then generate a number of system failure times based on the system's reliability function. The system's reliability function can be used to solve for a time value associated with that unreliability value. The distribution of the generated time values can then be fitted to a probability distribution function.
 

Consider a value of <math>F(t)=0.11</math> . Using the system's reliability equation and solving for time, the corresponding time-to-failure for a 0.11 unreliability can be calculated. 
For the system of Example 2, the time for a 0.11 unreliability is 389.786 hours.
 
When enough points have been generated, the selected distribution will be fitted to this data set and the distribution's parameters will be returned. In addition, if ReliaSoft's Weibull++ is installed, the generated data can be viewed/analyzed using a Weibull++ instance, as shown in Figure "Using Weibull++ to calculate distribution parameters".
 
[[Image:Fig 5.14.PNG|thumb|center|400px|<div align="center"> Fig 5.14 Using Weibull++ to calculate distribution parameters.</div>]]
 
It is recommended that the analyst examine the fit to ascertain the applicability of the approximation.

=Duty Cycle=

Components of a system may not operate continuously during a system's mission, or may be subjected to loads greater or lesser than the rated loads during system operation. To model this, a factor called the Duty Cycle ( <math>{{d}_{c}}</math> ) is used. The duty cycle may also be used to account for changes in environmental stress, such as temperature changes, that may effect the operation of a component. The duty cycle is a positive value, with a default value of 1 representing continuous operation at rated load, and any values other than 1 representing other load values with respect to the rated load value (or total operating time). A duty cycle value higher than 1 indicates a load in excess of the rated value. A duty cycle value lower than 1 indicates that the component is operating at a load lower than the rated load or not operating continuously during the system's mission. For instance, a duty cycle of 0.5 may be used for a component that operates only half of the time during the system's mission.
 
The reliability metrics for a component with a duty cycle are calculated as follows. Let <math>{{d}_{c}}</math> represent the duty cycle during a particular mission of the component, <math>t</math> represent the mission time and <math>{t}'</math> represent the accumulated age. Then:
 

::<math>{t}'={{d}_{c}}\times t</math>

 

The reliability equation for the component is:
 

::<math>R({t}')=R({{d}_{c}}\times t)</math>
 

The component ''pdf'' is:
 

::<math>f({t}')=-\frac{d(R({t}'))}{dt}=-\frac{d(R({{d}_{c}}\times t))}{dt}={{d}_{c}}f({{d}_{c}}\times t)</math>
 

The failure rate of the component is:
 

::<math>\lambda ({t}')=\frac{f({t}')}{R({t}')}=\frac{{{d}_{c}}f({{d}_{c}}\times t)}{R({{d}_{c}}\times t)}={{d}_{c}}\lambda ({{d}_{c}}\times t)</math>

===Example 4===
Consider a computer system with three components: a processor, a hard drive and a CD drive in series as shown next. Assume that all three components follow a Weibull failure distribution with the parameters <math>{{\beta }_{1}}=1.5</math> and <math>{{\eta }_{1}}=5000</math> for the processor, <math>{{\beta }_{2}}=2.5</math> and <math>{{\eta }_{2}}=3000</math> for the hard drive, and <math>{{\beta }_{3}}=2</math> and <math>{{\eta }_{3}}=4000</math> for the CD drive. Determine the reliability of the computer system after one year (365 days) of operation, assuming that the CD drive is used only 30% of the time.
 
[[Image:BS5ex4.png|thumb|center|300px|]]

====Solution to Example 4====
The reliability of the processor after 365 days of operation is given by:

::<math>\begin{align}
{{R}_{processor}}(365)= & {{e}^{-{{\left( \tfrac{365}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}} \\
= & {{e}^{-{{\left( \tfrac{365}{5000} \right)}^{1.5}}}} \\
= & 0.9805\text{ or }98.05%
\end{align}</math>

The reliability of the hard drive after 365 days of operation is given by:

::<math>\begin{align}
{{R}_{harddrive}}(365)= & {{e}^{-{{\left( \tfrac{365}{{{\eta }_{2}}} \right)}^{{{\beta }_{2}}}}}} \\
= & {{e}^{-{{\left( \tfrac{365}{3000} \right)}^{2.5}}}} \\
= & 0.9948\text{ or }99.48%
\end{align}</math>

The reliability of the CD drive after 365 days of operation (taking into account the 30% operation using a duty cycle of 0.3) is given by:

::<math>\begin{align}
{{R}_{CDdrive}}(365)= & {{e}^{-{{\left( \tfrac{{{d}_{c}}\times 365}{{{\eta }_{3}}} \right)}^{{{\beta }_{3}}}}}} \\
= & {{e}^{-{{\left( \tfrac{0.3\times 365}{4000} \right)}^{2}}}} \\
= & 0.9993\text{ or }99.93%
\end{align}</math>
 
[[Image:Fig 5.15.PNG|thumb|center|400px|<div align="center"> Fig 5.15 Result for the computer system reliability.</div>]]
 

Thus the reliability of the computer system after 365 days of operation is:

::<math>\begin{align}
{{R}_{s}}(365)= & {{R}_{processor}}(365)\cdot {{R}_{harddrive}}(365)\cdot {{R}_{CDdrive}}(365) \\
= & 0.9805\cdot 0.9948\cdot 0.9993 \\
= & 0.9747\text{ or }97.47%
\end{align}</math>
 
This result can be obtained in BlockSim as shown in Figure "Result for the computer system reliability".
 

=Load Sharing=
As presented in earlier chapters, a reliability block diagram (RBD) allows you to graphically represent how the components within a system are reliability-wise connected. In most cases, independence is assumed across the components within the system. For example, the failure of component A does not affect the failure of component B. However, if a system consists of components that are sharing a load, then the assumption of independence no longer holds true.
 

If one component fails, then the component(s) that are still operating will have to assume the failed unit's portion of the load. Therefore, the reliabilities of the surviving unit(s) will change. Calculating the system reliability is no longer an easy proposition. In the case of load sharing components, the change of the failure distributions of the surviving components must be known in order to determine the system's reliability.
 

To illustrate this, consider the a system of two units connected reliability-wise in parallel (Figure "Two units connected reliability-wise in parallel").
 

[[Image:5-16.png|thumb|center|300px|<div align="center"> Two units connected reliability-wise in parallel.</div>]]
 

Assume that the units must supply an output of 8 volts and that if both units are operational, each unit is to supply 50% of the total output. If one of the units fails, then the surviving unit supplies 100%. Furthermore, assume that having to supply the entire load has a negative impact on the reliability characteristics of the surviving unit. Since the reliability characteristics of the unit change based on whether both or only one is operating, a life distribution along with a life-stress relationship (as discussed in [http://reliawiki.com/index.php/Statistical_Background#A_Brief_Introduction_to_Life-Stress_Relationships|A Brief Introduction to Life-Stress Relationships]) will be needed to model each component.
 

To illustrate the steps needed, we will create the model starting from raw data. Assume that a total of 20 units were tested to failure at 7, 10 and 15 volts. The test data set is presented in the next table.
 

[[Image:5-17.png|thumb|center|300px|]]
 

For this example, Units 1 and 2 are the same component. Therefore, only one set of data was collected. However, it is possible that the load sharing components in a system may not be the same. If that were the case, data would need to be collected for each component.
 

The data set in Table 1 was analyzed using ReliaSoft's ALTA software (as shown in Figure "Using ALTA to calculate component parameters") with the Inverse Power Law as the underlying life-stress relationship and Weibull as the life distribution.
 
 

The estimated model parameters, <math>\beta </math> , <math>K</math> and <math>n</math> , are shown next.
 

::<math>\begin{align}
\beta = & 1.9239 \\
K= & 3.2387\times {{10}^{-7}} \\
n= & 3.4226
\end{align}</math>
 

:Or:
 
::<math>\begin{align}
{{R}_{1}}(t,{{S}_{1}})= & {{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} \\
= & {{e}^{-{{\left( 3.2387\times {{10}^{-7}}S_{1}^{3.4226}t \right)}^{1.9239}}}} \
\end{align}</math>.

 
::<math>{{f}_{1}}(t,{{S}_{1}})=\beta KS_{1}^{n}{{\left( KS_{1}^{n}t \right)}^{\beta -1}}{{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} \ </math>
 

And for this case:

::<math>\begin{align}
{{R}_{1}}(t,{{S}_{1}})= & {{R}_{2}}(t,{{S}_{2}}) \\
{{f}_{1}}(t,{{S}_{1}})= & {{f}_{2}}(t,{{S}_{2}})
\end{align}</math>
 
 
[[Image:Fig 5.17.PNG|thumb|center|500px|<div align="center"> Using ALTA to calculate component parameters.</div>]]
 

Figure "Reliability curves for different voltage output conditions" shows a plot of <math>{{R}_{1}}(t,{{S}_{1}})={{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} = {{e}^{-{{\left( 3.2387\times {{10}^{-7}}S_{1}^{3.4226}t \right)}^{1.9239}}}}\ </math>.
 

Now that the failure properties have been determined using the test data, the reliability of the system at some time, <math>t</math> , can be calculated using the following equation:

 
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>

 
:Where:
 

::<math>\begin{align}
{{S}_{1}}= & {{P}_{1}}S \\
{{S}_{2}}= & {{P}_{2}}S
\end{align}</math>

 
[[Image:5-18.png|thumb|center|300px|<div align="center"> Reliability curves for different voltage output conditions. </div>]]
 
[[Image:BS5.19.png|thumb|center|395px| Illustrating <math>{{t}_{e}}</math>]]
 

And:
 

:• <math>S</math> is the total load (or required output). 
:• <math>{{P}_{1}}</math> and <math>{{P}_{2}}</math> are the portion of the total load that each unit supports when both units are operational. In this case, <math>{{P}_{1}}={{P}_{2}}=0.5=50%.</math> 
:• <math>{{S}_{1}}</math> and <math>{{S}_{2}}</math> represent the portions of the load that Unit 1 and Unit 2 must support when both units are operational. 
:• <math>{{t}_{{{1}_{e}}}}</math> is the equivalent operating time for Unit 1 if it had been operating at <math>S</math> instead of <math>{{S}_{1}}</math> . A graphical representation of the equivalent time is shown in Figure 5.19, where the curve marked by L represents the low stress (load) and the curve marked by H represents the high stress (load). 
 

::<math>{{t}_{1e}}</math> can be calculated by:
 
::<math>\begin{align}
{{R}_{1}}(t)= & {{R}_{1}}({{t}_{1e}}) \\
{{e}^{-{{(tKS_{1}^{n})}^{\beta }}}}= & {{e}^{-{{({{t}_{1e}}K{{S}^{n}})}^{\beta }}}} \\
tS_{1}^{n}= & {{t}_{1e}}{{S}^{n}} \\
{{t}_{1e}}= & t{{\left( \frac{{{S}_{1}}}{S} \right)}^{n}},\text{ }{{S}_{1}}={{P}_{1}}S \\
\therefore & {{t}_{1e}}=tP_{1}^{n}
\end{align}</math>

 
::<math>{{t}_{2e}}</math> can be calculated the same way, or:

 
::<math>\begin{align}
{{R}_{2}}(t)= & {{R}_{2}}({{t}_{2e}}) \\
\therefore & {{t}_{2e}}=tP_{2}^{n}
\end{align}</math>
 

In this example, the reliability equations for Unit 1 and Unit 2 are the same since they are the same type of component and demonstrate the same failure properties. In addition, the total output is divided equally between the two units (when both units are operating), so <math>{{t}_{1e}}</math> and <math>{{t}_{2e}}</math> will also be the same.
 
The next step is to determine the reliability of the system after 8,760 hours, <math>R(t=8,760)</math> .
Using
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>
 

the system reliability is found to be:
 

::<math>\begin{align}
R(t=8760)= & 0.8567 \\
= & 85.67%
\end{align}</math>
 

===Load Sharing in BlockSim===
BlockSim uses this formulation when computing reliabilities of units in a load sharing configuration. When using the System Reliability Equation window, BlockSim returns a single token for the reliability of units in a load sharing configuration (as well as in the case of standby redundancy, discussed in the next section). As an example, consider the following RBD with Unit 1 in series with a container that includes two load sharing units.
 
[[Image:BS5.19.2.png|thumb|center|200px|]]
 
BlockSim will return the system equation as:
 

::<math>{{R}_{System}}=+{{R}_{LS}}\cdot {{R}_{1}}</math>
 

Where <math>{{R}_{LS}}</math> implies a form similar to
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>
 .
BlockSim allows for <math>k</math> -out-of- <math>n</math> units in a load sharing configuration.

===Example 5===
A component has five possible failure modes, <math>A</math> , <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> , <math>{{B}_{C}}</math> and <math>C</math> , and the <math>B</math> modes are interdependent. The system will fail if mode <math>A</math> occurs, mode <math>C</math> occurs or two out of the three <math>B</math> modes occur.
 

Modes <math>A</math> and <math>C</math> each have a Weibull distribution, with a <math>\beta =2</math> and <math>\eta =10,000</math> and 15,000 respectively. Events <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> and <math>{{B}_{C}}</math> each have an exponential distribution with a mean of 10,000 hours.
 

If any <math>B</math> event occurs (i.e. <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> or <math>{{B}_{C}}</math> ), the remaining <math>B</math> events are more likely to occur. Specifically, the mean times of the remaining <math>B</math> events are halved. Determine the reliability at 1000 hours for this component.
====Solution to Example 5====
The first step is to create the RBD. Modes <math>A</math> , <math>C</math> and a load sharing container with the <math>{{B}_{i}}</math> modes must be drawn in series, as illustrated next.
 
[[Image:BS5.19.3.png|thumb|center|200px|]]
 

The next step is to define the properties for each block, including those for the container. Setting the failure distributions for modes <math>A</math> and <math>C</math> is simple. The more difficult part is setting the properties for the container and the contained blocks. Based on the problem statement, the <math>B</math> modes are in a 2-out-of-3 load sharing redundancy. When all three are working (i.e. when no <math>B</math> mode has occurred), each block has an exponential distribution with <math>\mu=10,000</math>. If one <math>B</math> mode occurs, then the two surviving units have an exponential distribution with <math>\mu =5,000.</math>
 

Assume a Power Life-Stress relationship for the components. Then:
 

::<math>\begin{align}
{{\mu }_{1}}= & \frac{1}{KV_{1}^{n}} \ \\
{{\mu }_{2}}= & \frac{1}{KV_{2}^{n}} \
\end{align}</math>
 

Substituting <math>{{\mu }_{1}}=10,000</math> and <math>{{V}_{1}}=1</math> in <math>{{\mu }_{1}}= \frac{1}{KV_{1}^{n}}</math> and casting it in terms of <math>K</math> yields:
 

::<math>\begin{align}
10,000= & \frac{1}{K} \ \\
K = & \frac{1}{10,000}=0.0001
\end{align}</math>

 
Substituting <math>{{\mu }_{2}}=5,000</math> , <math>{{V}_{2}}=1.5</math> (because if one fails, then each survivor takes on an additional 0.5 units of load) and <math>10,000= \frac{1}{K}</math> for <math>K</math> in <math>{{\mu }_{2}}= \frac{1}{KV_{2}^{n}}</math> yields:

 
::<math>\begin{align}
5,000= & \frac{1}{0.0001\cdot {{(1.5)}^{n}}} \\
0.5= & {{(1.5)}^{-n}} \\
\ln (0.5)= & -n\ln (1.5) \\
n = & 1.7095
\end{align}</math>
 

This also could have been computed in ALTA, as shown in Figure "Calculation performed in ALTA".

The next step is to set the weight proportionality factor. This factor defines the portion of the load that the particular item carries while operating, as well as the load that shifts to the remaining units upon failure of the item. To illustrate, assume three units (1, 2 and 3) are in a load sharing container with weight proportionality factors of 1, 2 and 3 respectively (and a 1-out-of-3 requirement).
 

:• Unit 1 carries <math>\left( \tfrac{1}{1+2+3} \right)=0.166</math> or 16.6% of the total load.
:• Unit 2 carries <math>\left( \tfrac{2}{1+2+3} \right)=0.333</math> or 33.3% of the total load.
:• Unit 3 carries <math>\left( \tfrac{3}{1+2+3} \right)=0.50</math> or 50% of the total load.
 

The actual load on each unit then becomes the product of the entire load defined for the container times the portion carried by that unit. For example, if the container load is 100 lbs, then the portion assigned to Unit 1 will be <math>100\cdot 0.166=16.6</math> lbs.
 

In the current example, all units share the same load and thus have equal weight proportionality factors. Because these factors are relative, if the same number is used for all three items then the results will be the same. Thus, weight proportional factor is set equal to 1 for each item.
 

[[Image:5.20.PNG|thumb|center|500px|<div align="center"> Calculation performed in ALTA.</div>]]
 

 
[[Image:Fig 5.23_2.PNG|thumb|center|400px|<div align="center"> Defining Weight Proportional Factor. </div>]]
 

The last properties that need to be defined are the total load and the 2-out-of-3 redundancy. The total load is dependent on how the parameters were computed. In this case, total load was assumed to be 3 when the parameters were computed (i.e. the load per item was 1 when all worked and 1.5 when two worked). This is defined at the container level, as shown in Figure "Defining Weight Proportional Factor".
When all of the parameters have been specified in BlockSim, the reliability at 1,000 hours can be determined. From the Analytical QCP, this is found to be 93.87%.

=Standby Components=

In the previous section, the case of a system with load sharing components was presented. This is a form of redundancy with dependent components. That is, the failure of one component affects the failure of the other(s). This section presents another form of redundancy: standby redundancy. In standby redundancy the redundant components are set to be under a lighter load condition (or no load) while not needed and under the operating load when they are activated.
 

In standby redundancy the components are set to have two states: an active state and a standby state. Components in standby redundancy have two failure distributions, one for each state. When in the standby state, they have a quiescent (or dormant) failure distribution and when operating, they have an active failure distribution.
 

In the case that both quiescent and active failure distributions are the same, the units are in a simple parallel configuration (also called a hot standby configuration). When the rate of failure of the standby component is lower in quiescent mode than in active mode, that is called a warm standby configuration. When the rate of failure of the standby component is zero in quiescent mode (i.e. the component cannot fail when in standby), that is called a cold standby configuration.

===Simple Standby Configuration===

Consider two components in a standby configuration. Component 1 is the active component with a Weibull failure distribution with parameters <math>\beta </math> = 1.5 and <math>\eta </math> = 1,000. Component 2 is the standby component. When Component 2 is operating, it also has a Weibull failure distribution with <math>\beta </math> = 1.5 and <math>\eta </math> = 1,000. Furthermore, assume the following cases for the quiescent distribution.
 
:• Case 1: The quiescent distribution is the same as the active distribution (hot standby). 
:• Case 2: The quiescent distribution is a Weibull distribution with <math>\beta </math> = 1.5 and <math>\eta </math> = 2000 (warm standby). 
:• Case 3: The component cannot fail in quiescent mode (cold standby). 
 
In this case, the reliability of the system at some time, <math>t</math> , can be obtained using the following equation:
 

::<math>R(t)={{R}_{1}}(t)+\underset{0}{\overset{t}{\mathop \int }}\,{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x)\cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}dx \ </math>
 

Where:
 
:• <math>{{R}_{1}}</math> is the reliability of the active component. 
:• <math>{{f}_{1}}</math> is the <math>pdf</math> of the active component. 
:• <math>{{R}_{2;SB}}</math> is the reliability of the standby component when in standby mode (quiescent reliability). 
:• <math>{{R}_{2;A}}</math> is the reliability of the standby component when in active mode. 
:• <math>{{t}_{e}}</math> is the equivalent operating time for the standby unit if it had been operating at an active mode, such that: 

 
<math>{{R}_{2;SB}}(x)={{R}_{2;A}}({{t}_{e}})</math> 
 
The second equation above can be solved for <math>{{t}_{e}}</math> and substituted into the first equation above.
Figure "Standby container" illustrates the example as entered in BlockSim using a standby container.
 
 
[[Image:5.24.gif|thumb|center|400px|<div align="center"> Standby container.</div>]]

 

The active and standby blocks are within a container, which is used to specify standby redundancy. Since the standby component has two distributions (active and quiescent), the Block Properties window of the standby block has two pages for specifying each one. Figures "Defining the active failure distribution" and "Defining the quiescent failure distribution" illustrate these pages.

 

The system reliability results for 1000 hours are given in the following table:
 
[[Image:5-24.png|thumb|center|400px|]]
 

Note that even though the <math>\beta </math> value for the quiescent distribution is the same as in the active distribution, it is possible that the two can be different. That is, the failure modes present during the quiescent mode could be different from the modes present during the active mode. In that sense, the two distribution types can be different as well (e.g. lognormal when quiescent and Weibull when active).
 

In many cases when considering standby systems, a switching device may also be present that switches from the failed active component to the standby component. The reliability of the switch can also be incorporated into
<math>R(t)={{R}_{1}}(t)+\underset{0}{\overset{t}{\mathop \int }}\,{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x)\cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}dx \ </math>

as presented in the next section.
 

BlockSim's System Reliability Equation window returns a single token for the reliability of units in a standby configuration. This is the same as the load sharing case presented in the previous section.
 

[[Image: Fig 5.25.PNG|thumb|center|400px|<div align="center"> Defining the active failure distribution </div>]]
 
 

[[Image:Fig 5.26.PNG|thumb|center|400px|<div align="center"> Defining the quiescent failure distribution </div>]]
 

===Reliability of Standby Systems with a Switching Device===

In many cases when dealing with standby systems, a switching device is present that will switch to the standby component when the active component fails. Therefore, the failure properties of the switch must also be included in the analysis.
 

[[Image:BS5.26.2.png|thumb|center|300px|]]
 

In most cases when the reliability of a switch is to be included in the analysis, two probabilities can be considered. The first and most common one is the probability of the switch performing the action (i.e. switching) when requested to do so. This is called Switch Probability per Request in BlockSim and is expressed as a static probability (e.g. 90%). The second probability is the quiescent reliability of the switch. This is the reliability of the switch as it ages (e.g. the switch might wear out with age due to corrosion, material degradation, etc.). Thus it is possible for the switch to fail before the active component fails. However, a switch failure does not cause the system to fail, but rather causes the system to fail only if the switch is needed and the switch has failed. For example, if the active component does not fail until the mission end time and the switch fails, then the system does not fail. However, if the active component fails and the switch has also failed, then the system cannot be switched to the standby component and it therefore fails.
 

In analyzing standby components with a switching device, either or both failure probabilities (during the switching or while waiting to switch) can be considered for the switch, since each probability can represent different failure modes. For example, the switch probability per request may represent software-related issues or the probability of detecting the failure of an active component, and the quiescent probability may represent wear-out type failures of the switch.
 

To illustrate the formulation, consider the previous example that assumes perfect switching. To examine the effects of including an imperfect switch, assume that when the active component fails there is a 90% probability that the switch will switch from the active component to the standby component. In addition, assume that the switch can also fail due to a wear-out failure mode described by a Weibull distribution with <math>\beta </math> = 1.7 and <math>\eta </math> = 5000.
 

Therefore, the reliability of the system at some time, <math>t</math> , is given by the following equation.

::<math>\begin{align}
R(t)= & {{R}_{1}}(t) \\
& +\underset{0}{\overset{t}{\mathop \int }}\,\{{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x) \\
& \cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}\cdot {{R}_{SW;Q}}(x)\cdot {{R}_{SW;REQ}}(x)\}dx
\end{align}</math>

Where:
 
:• <math>{{R}_{1}}</math> is the reliability of the active component. 
:• <math>{{f}_{1}}</math> is the <math>pdf</math> of the active component. 
:• <math>{{R}_{2;SB}}</math> is the reliability of the standby component when in standby mode (quiescent reliability). 
:• <math>{{R}_{2;A}}</math> is the reliability of the standby component when in active mode. 
:• <math>{{R}_{SW;Q}}</math> is the quiescent reliability of the switch. 
:• <math>{{R}_{SW;REQ}}</math> is the switch probability per request. 
:• <math>{{t}_{e}}</math> is the equivalent operating time for the standby unit if it had been operating at an active mode. 
This problem can be solved in BlockSim by including these probabilities in the container's properties, as shown in Figures "Standby container (switch) failure probabilities while attempting to switch" and "Standby container (switch) failure distribution while waiting to switch". In BlockSim, the standby container is acting as the switch.

 

[[Image:Fig 5.28_2.PNG|thumb|center|400px|<div align="center"> Standby container (switch) failure probabilities while attempting to switch </div> ]]
 

 

[[Image:Fig 5.27.PNG|thumb|center|400px|<div align="center"> Standby container (switch) failure distribution while waiting to switch </div>]]
 

 

Note that there are additional properties that can be specified in BlockSim for a switch, such as Switch Restart Probability, No. of Restarts and Switch Delay Time. In many applications, the switch is re-tested (or re-cycled) if it fails to switch the first time. In these cases, it might be possible that it switches in the second or third, or <math>{{n}^{th}}</math> attempt.
 
The Switch Restart Probability specifies each additional attempt's probability of successfully switching and the Finite Restarts specifies the total number of attempts. Note that the Switch Restart Probability specifies the probability of success of each trial (or attempt). The probability of success of <math>n</math> consecutive trials is calculated by BlockSim using the binomial distribution and this probability is then incorporated into equation above. The Switch Delay Time property is related to repairable systems and is considered in BlockSim only when using simulation. When using the analytical solution (i.e. for a non-repairable system), this property is ignored.
 

Solving the analytical solution (as given by Equation above), the following results are obtained.
 

[[Image:5-30.png|thumb|center|400px|]]
 

From the table above, it can be seen that the presence of a switching device has a significant effect on the reliability of a standby system. It is therefore important when modeling standby redundancy to incorporate the switching device reliability properties. It should be noted that this methodology is not the same as treating the switching device as another series component with the standby subsystem. This would be valid only if the failure of the switch resulted in the failure of system (e.g. switch failing open). In equation above, the Switch Probability per Request and quiescent probability are present only in the second term of the equation. Treating these two failure modes as a series configuration with the standby subsystem would imply that they are also present when the active component is functioning (i.e. first term of equation above). This is invalid and would result in the underestimation of the reliability of the system. In other words, these two failure modes become significant only when the active component fails.
 

As an example, and if we consider the warm standby case, the reliability of the system without the switch is 70.57% at 1000 hours. If the system was modeled so that the switching device was in series with the warm standby subsystem, the result would have been:

 

::<math>\begin{align}
{{R}_{S}}(1000)= & {{R}_{Standby}}(1000)\cdot {{R}_{sw,Q(1000)}}\cdot {{R}_{sw,req}} \\
= & 0.7057\cdot 0.9372\cdot 0.9 \\
= & 0.5952
\end{align}</math>

 

In the case where a switch failure mode causes the standby subsystem to fail, then this mode can be modeled as an individual block in series with the standby subsystem.

===Example 6===
Consider a car with four new tires and a full-size spare. Assume the following failure characteristics:
 
:• The tires follow a Weibull distribution with a <math>\beta=4</math> and an <math>\eta =</math> 40,000 miles while on the car due to wear. 
:• The tires also have a probability of failing due to puncture or other causes. For this, assume a constant rate for this occurrence with a probability of 1 every 50,000 miles. 
:• When not on the car (i.e. is a spare), a tire's probability of failing also has a Weibull distribution with a <math>\beta =</math> 2 and <math>\eta =</math> 120,000 miles. 
 
Assume a mission of 1,000 miles. If a tire fails during this trip, it will be replaced with the spare. However, the spare will not be repaired during the trip. In other words, the trip will continue with the spare on the car and if the spare fails the system will fail. Determine the probability of system failure.
====Solution to Example 6====
Active failure distribution for tires:
 
:• Due to wear-out, Weibull <math>\beta =4</math> and <math>\eta =40,000</math> miles. 
:• Due to random puncture, exponential <math>\mu =50,000.</math> 
:• The quiescent failure distribution is a Weibull distribution with <math>\beta =2</math> and <math>\eta =120,000</math> miles. 
 
The block diagram for each tire has two blocks in series, one block representing the wear-out mode and the other the random puncture mode, as shown next:
 
 
[[Image:small5.gif|thumb|center|400px|]]
 

There are five tires, four active and one standby (represented in the diagram by a standby container with a 4-out-of-5 requirement), as shown next:
 
[[Image:BStirecontainer.png|thumb|center|400px|]]

For the standby Wear block, set the active failure and the quiescent distributions, but for the Puncture block, set only the active puncture distribution (because the tire cannot fail due to puncture while stored). Using BlockSim, the probability of system failure is found to be 0.003 or 0.3%.
 

=Note Regarding Numerical Integration Solutions=

Load sharing and standby solutions in BlockSim are performed using numerical integration routines. As with any numerical analysis routine, the solution error depends on the number of iterations performed, the step size chosen and related factors, plus the behavior of the underlying function. By default, BlockSim uses a certain set of preset factors. In general, these defaults are sufficient for most problems. If a higher precision or verification of the precision for a specific problem is required, BlockSim's preset options can be modified and/or the integration error can be assessed using the Integration Parameters... option for each container. For more details, you can refer to the documentation on the Algorithm Setup window in the BlockSim 8 User's help files.

Time-Dependent System Reliability (Analytical)

2012-03-21T21:26:43Z

Dingzhou Cao: /* Obtaining Other Functions of Interest */

{{Template:bsbook|5}}

In the previous chapter, different system configuration types were examined, as well as different methods for obtaining the system's reliability function analytically. Because the reliabilities in the problems presented were treated as probabilities (e.g. <math>P(A)</math> , <math>{{R}_{i}}</math> ), the reliability values and equations presented were referred to as static (not time-dependent). Thus, in the prior chapter, the life distributions of the components were not incorporated in the process of calculating the system reliability. In this chapter, time dependency in the reliability function will be introduced. We will develop the models necessary to observe the reliability over the life of the system, instead of at just one point in time. In addition, performance measures such as failure rate, MTTF and warranty time will be estimated for the entire system. The methods of obtaining the reliability function analytically remain identical to the ones presented in the previous chapter, with the exception that the reliabilities will be functions of time. In other words, instead of dealing with <math>{{R}_{i}}</math> , we will use <math>{{R}_{i}}(t)</math> . All examples in this chapter assume that no repairs are performed on the components.

=Analytical Life Predictions=
The analytical approach presented in the prior chapter involved the determination of a mathematical expression that describes the reliability of the system, expressed in terms of the reliabilities of its components. So far we have estimated only static system reliability (at a fixed time). For example, in the case of a system with three components in series, the system's reliability equation was given by:
 

::<math>{{R}_{s}}={{R}_{1}}\cdot {{R}_{2}}\cdot {{R}_{3}} \ </math>
 

The values of <math>{{R}_{1}}</math> , <math>{{R}_{2}}</math> and <math>{{R}_{3}}</math> were given for a common time and the reliability of the system was estimated for that time. However, since the component failure characteristics can be described by distributions, the system reliability is actually time-dependent. In this case, the equation above can be rewritten as:
 

::<math>{{R}_{s}}(t)={{R}_{1}}(t)\cdot {{R}_{2}}(t)\cdot {{R}_{3}}(t)</math>

 

The reliability of the system for any mission time can now be estimated. Assuming a Weibull life distribution for each component, the first equation above can now be expressed in terms of each component's reliability function, or:

 

[[Category:Completed Theoretical Review]]

::<math>{{R}_{s}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}}\cdot {{e}^{-{{\left( \tfrac{t}{{{\eta }_{2}}} \right)}^{{{\beta }_{2}}}}}}\cdot {{e}^{-{{\left( \tfrac{t}{{{\eta }_{3}}} \right)}^{{{\beta }_{3}}}}}}</math>

 

In the same manner, any life distribution can be substituted into the system reliability equation. Suppose that the times-to-failure of the first component are described with a Weibull distribution, the times-to-failure of the second component with an exponential distribution and the times-to-failure of the third component with a normal distribution. Then the first equation above can be written as:
 

::<math>{{R}_{s}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}}\cdot {{e}^{-{{\lambda }_{2}}t}}\cdot \left[ 1-\Phi \left( \frac{t-{{\mu }_{3}}}{{{\sigma }_{3}}} \right) \right]</math>

 

It can be seen that the biggest challenge is in obtaining the system's reliability function in terms of component reliabilities, which has already been discussed in depth. Once this has been achieved, calculating the reliability of the system for any mission duration is just a matter of substituting the corresponding component reliability functions into the system reliability equation.
===Advantages of the Analytical Method===
The primary advantage of the analytical solution is that it produces a mathematical expression that describes the reliability of the system. Once the system's reliability function has been determined, other calculations can then be performed to obtain metrics of interest for the system. Such calculations include:
 
:• Determination of the system's <math>pdf.</math> 
:• Determination of warranty periods. 
:• Determination of the system's failure rate. 
:• Determination of the system's MTTF. 
 
In addition, optimization and reliability allocation techniques can be used to aid engineers in their design improvement efforts. Another advantage in using analytical techniques is the ability to perform static calculations and analyze systems with a mixture of static and time-dependent components. Finally, the reliability importance of components over time can be calculated with this methodology.
===Disadvantages of the Analytical Method===
The biggest disadvantage of the analytical method is that formulations can become very complicated. The more complicated a system is, the larger and more difficult it will be to analytically formulate an expression for the system's reliability. For particularly detailed systems this process can be quite time-consuming, even with the use of computers. Furthermore, when the maintainability of the system or some of its components must be taken into consideration, analytical solutions become intractable. In these situations, the use of simulation methods may be more advantageous than attempting to develop a solution analytically. Simulation methods are presented in later chapters.
===Looking at a Simple ''Complex'' System Analytically===

The complexity involved in an analytical solution can be best illustrated by looking at the simple ''complex'' system with 15 components, as shown in Figure " An RBD of a complex system".
 
[[Image:5-1.png|thumb|center|400px|<div align="center"> An RBD of a complex system.</div>]]

 
The system reliability for this system (computed using BlockSim) is shown next. The first solution is provided using BlockSim's symbolic solution. In symbolic mode, BlockSim breaks the equation into segments, identified by tokens, that need to be substituted into the final system equation for a complete solution. This creates algebraic solutions that are more compact than if the substitutions were made.
 

::<math>\begin{align}
{{R}_{System}}= & D2\cdot D3\cdot {{R}_{L}} \\
D3= & +{{R}_{K}}\cdot IK \\
IK= & +{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}} \\
D2 = & +{{R}_{A}}\cdot {{R}_{E}}\cdot IE \\
IE = & -D1\cdot {{R}_{M}}\cdot {{R}_{N}}+{{R}_{M}}\cdot {{R}_{N}}+D1 \\
D1 = & +{{R}_{D}}\cdot ID \\
ID = & -{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}}
\end{align}</math>
 
Substituting the terms yields:
 

::<math>\begin{align}
{{R}_{System}}= & {{R}_{A}}\cdot {{R}_{E}}\cdot {{R}_{L}}\cdot {{R}_{K}} \\
& \cdot \{({{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})\cdot {{R}_{M}}\cdot {{R}_{N}} \\
& +{{R}_{M}}\cdot {{R}_{N}}-{{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}}\} \\
& \cdot \{{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}}\}
\end{align}</math>

 
BlockSim's automatic algebraic simplification would yield the following format for the above solution:
 

::<math>\begin{align}
{{R}_{System}}= & (({{R}_{A}}\cdot {{R}_{E}}(-({{R}_{D}}(-{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})){{R}_{M}}\cdot {{R}_{N}} \\
& +{{R}_{M}}\cdot {{R}_{N}} \\
& +({{R}_{D}}(-{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})))) \\
& ({{R}_{K}}({{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& +RI\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}})){{R}_{L}}) \
\end{align}</math>

 
In this equation, each <math>{{R}_{i}}</math> represents the reliability function of a block. For example, if <math>{{R}_{A}}</math> has a Weibull distribution, then each <math>{{R}_{A}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{A}}} \right)}^{{{\beta }_{A}}}}}}</math> and so forth. Substitution of each component's reliability function in the last <math>{{R}_{System}}</math> equation above will result in an analytical expression for the system reliability as a function of time, or <math>{{R}_{s}}(t)</math> , which is the same as <math>(1-cd{{f}_{System}}).</math>

===Obtaining Other Functions of Interest===
Once the system reliability equation (or the cumulative density function, <math>cdf</math> ) has been determined, other functions and metrics of interest can be derived.
 
Consider the following simple system:
 
[[Image:5-2.png|thumb|center|300px|<div align="center"> Simple two-component system. </div>]]

 
Furthermore, assume that component 1 follows an exponential distribution with a mean of 10,000 (<math>\mu =10,000,</math> <math>\lambda =1/10,000)</math> and component 2 follows a Weibull distribution with <math>\beta =6</math> and <math>\eta =10,000</math> . The reliability equation of this system is:

 
::<math>\begin{align}
{{R}_{S}}(t)= & {{R}_{1}}(t)\cdot {{R}_{2}}(t) \\
= & {{e}^{-\lambda t}}\cdot {{e}^{-{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \\
= & {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \
\end{align}</math>

 
The system <math>cdf</math> is:

 
::<math>\begin{align}
{{F}_{S}}(t)= & 1-({{R}_{1}}(t)\cdot {{R}_{2}}(t)) \\
= & 1-\left( {{e}^{-\lambda t}}\cdot {{e}^{-{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \right) \\
= & 1-\left( {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)
\end{align}</math>

 
{{analytical system pdf}}

====Conditional Reliability====
Conditional reliability is the probability of a system successfully completing another mission following the successful completion of a previous mission. The time of the previous mission and the time for the mission to be undertaken must be taken into account for conditional reliability calculations. The system's conditional reliability function is given by:

::<math>R(T,t)=\frac{R(T+t)}{R(T)}</math>

Equation above gives the reliability for a new mission of duration <math>t</math> having already accumulated <math>T</math> hours of operation up to the start of this new mission. The system is evaluated to assure that it will start the next mission successfully.
 
For the system in Figure "Simple two-component system.", the reliability for mission of <math>t=1,000</math> hours, having an age of <math>T=500</math> hours, is:
 

::<math>\begin{align}
{{R}_{S}}(T=500,t=1000)= & \frac{R(T+t)}{R(T)} \\
= & \frac{R(1500)}{R(500)} \\
= & \frac{{{e}^{-\tfrac{1500}{10,000}}}\cdot {{e}^{-{{\left( \tfrac{1500}{10,000} \right)}^{6}}}}}{{{e}^{-\tfrac{500}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{500}{10,000} \right)}^{6}}}}} \\
= & 0.9048=90.48%
\end{align}</math>

 

====Conditional Reliability for Components====

Now in this formulation, it was assumed that the accumulated age was equivalent for both units. That is, both started life at zero and aged to 500. It is possible to consider an individual component that has already accumulated some age (used component) in the same formulation. To illustrate this, assume that component 2 started life with an age of T=100. Then the reliability equation of the system, as given in <math>{{R}_{S}}(t)= {{R}_{1}}(t)\cdot {{R}_{2}}(t)</math>, would need to be modified to include a conditional term for 2, or:
 

::<math>{{R}_{S}}(t)={{R}_{1}}(t)\cdot \frac{{{R}_{2}}({{T}_{2}}+t)}{{{R}_{2}}({{T}_{2}})} \ </math>

 

In BlockSim, the start age input box may be used to specify a starting age greater than zero.
{{system failure rate analytical}}

====System Mean Life (Mean Time To Failure)====
The mean life (or mean time to failure, MTTF) can be obtained by integrating the system reliability function from zero to infinity:
 

::<math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math>

 

The mean time is a performance index and does not provide any information about the behavior of the failure distribution of the system.
 
For the system in Figure "Simple two-component system":
 

::<math>\begin{align}
MTTF= & \int_{0}^{\infty }\left( {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)dt \\
= & 5978.9
\end{align}</math>

 

====Warranty Period and BX Life====
Sometimes it is desirable to know the time value associated with a certain reliability. Warranty periods are often calculated by determining what percentage of the failure population can be covered financially and estimating the time at which this portion of the population will fail. Similarly, engineering specifications may call for a certain BX life, which also represents a time period during which a certain proportion of the population will fail. For example, the B10 life is the time in which 10% of the population will fail.
This is obtained by setting <math>{{R}_{S}}(t)</math> to the desired value and solving for <math>t.</math>
 
For the system in Figure "Simple two-component system":
 

::<math>{{R}_{s}}\left( t \right)={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>
 

To compute the time by which reliability would be equal to 90%, equation above is recast as follows and solved for <math>t.</math>

 
::<math>0.90={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>

 

In this case, <math>t=1053.59</math> . Equivalently, the B10 life for this system is also <math>1053.59</math> .
Except for some trivial cases, a closed form solution for <math>t</math> cannot be obtained. Thus, it is necessary to solve for <math>t</math> using numerical methods. BlockSim uses numerical methods.

===Example 1===
Consider a system consisting of three exponential units in series with the following failure rates (in failures per hour): <math>{{\lambda }_{1}}</math> = 0.0002, <math>{{\lambda }_{2}}</math> = 0.0005 and <math>{{\lambda }_{3}}</math> = 0.0001.
 
:• Obtain the reliability equation for the system. 
:• What is the reliability of the system after 150 hours of operation? 
:• Obtain the system's <math>pdf.</math> 
:• Obtain the system's failure rate equation. 
:• What is the MTTF for the system? 
:• What should the warranty period be for a 90% reliability? 
====Solution to Example 1====
 
:• The analytical expression for the reliability of the system is given by:

::<math>\begin{align}
{{R}_{s}}(t)= & {{R}_{1}}(t)\cdot {{R}_{2}}(t)\cdot {{R}_{3}}(t) \\
= & {{e}^{-{{\lambda }_{1}}t}}\cdot {{e}^{-{{\lambda }_{2}}t}}\cdot {{e}^{-{{\lambda }_{1}}t}} \\
= & {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}
\end{align}</math>

 

At 150 hours of operation, the reliability of the system is:
 

::<math>\begin{align}
{{R}_{s}}(t)= & {{e}^{-(0.0002+0.0005+0.0001)150}} \\
= & 0.8869\text{ or }88.69%
\end{align}</math>

 

:• In order to obtain the system's <math>pdf</math> , the derivative of the reliability equation given in the first equation above is taken with respect to time, or:
 
::<math>\begin{align}
{{f}_{s}}(t)= & -\frac{d[{{R}_{s}}(t)]}{dt} \\
= & -\frac{d\left[ {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}} \right]}{dt} \\
= & ({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})\cdot {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}
\end{align}</math>

:• The system's failure rate can now be obtained simply by dividing the system's <math>pdf</math> given in the equation above by the system's reliability function given in the first equation above, and:

::<math>\begin{align}
{{\lambda }_{s}}\left( t \right)= & \frac{{{f}_{s}}\left( t \right)}{{{R}_{s}}\left( t \right)} \\
= & \frac{({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})\cdot {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}}{{{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}} \\
= & ({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}) \\
= & 0.0008\text{ }fr/hr
\end{align}</math>

 
:• Combining <math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math> and the first equation above, the system's MTTF can be obtained:

 
::<math>\begin{align}
MTTF= & \int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \\
= & \int_{0}^{\infty }{{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}dt \\
= & \frac{1}{({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})} \\
= & 1250\text{ }hr
\end{align}</math>
 

:• Solving the first equation above with respect to time will yield the corresponding warranty period for a 90% reliability. In this case, the system reliability equation is simple and a closed form solution exists. The warranty time can now be found by solving:

 
::<math>\begin{align}
t= & -\frac{\ln (R)}{{{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}} \\
= & -\frac{\ln (0.9)}{0.0008} \\
= & 131.7\text{ }hr
\end{align}</math>
 

Thus, the warranty period should be 132 hours.

===Example 2===
Consider the system shown in Figure "Complex bridge system in Example 2".
 
[[Image:BS5.5.png|thumb|center|300px|<div align="center"> Fig 5.5 Complex bridge system in Example 2. </div>]]
 

Components <math>A</math> through <math>E</math> are Weibull distributed with <math>\beta =1.2</math> and <math>\eta =1230</math> hours. The starting and ending blocks cannot fail. 
Determine the following:
 
:• The reliability equation for the system and its corresponding plot. 
:• The system's <math>pdf</math> and its corresponding plot. 
:• The system's failure rate equation and the corresponding plot. 
:• The <math>MTTF</math> . 
:• The warranty time for a 90% reliability. 
:• The reliability for a 200-hour mission, if it is known that the system has already successfully operated for 200 hours. 
====Solution====
 
The first step is to obtain the reliability function for the system. The methods described in the previous chapter can be employed, such as the event space or path-tracing methods. Using BlockSim, the following reliability equation is obtained:
 

::<math>\begin{align}
{{R}_{s}}(t)= & ({{R}_{Start}}\cdot {{R}_{End}}(2{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}))
\end{align}</math>
 

Note that since the starting and ending blocks cannot fail, <math>{{R}_{Start}}=1</math> and <math>{{R}_{End}}=1,</math> equation above can be reduced to:

 

::<math>\begin{align}
{{R}_{s}}(t)= & 2\cdot {{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}
\end{align}</math>

 

Where <math>{{R}_{A}}</math> is the reliability equation for Component A, or:

 

::<math>{{R}_{A}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{A}}} \right)}^{{{\beta }_{A}}}}}}</math>

 

::<math>{{R}_{B}}</math> is the reliability equation for Component <math>B</math> , etc.
 

Since the components in this example are identical, the system reliability equation can be further reduced to:

 

::<math>{{R}_{s}}(t)=2R{{(t)}^{2}}+2R{{(t)}^{3}}-5R{{(t)}^{4}}+2R{{(t)}^{5}}</math>

 

Or, in terms of the failure distribution:

 

::<math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>

 

The corresponding plot is given in Figure "Reliability plot for the system".
 
[[Image:BS5.6.png|thumb|center|300px|<div align="center"> Reliability plot for the system. </div>]]
 
 

In order to obtain the system's <math>pdf</math> , the derivative of the reliability equation given above is taken with respect to time, resulting in:
 

::<math>\begin{align}
{{f}_{s}}(t)= & 4\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+6\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \\
& -20\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+10\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}
\end{align}</math>

 

The <math>pdf</math> can now be plotted for different time values, <math>t</math> , as shown in Figure "''pdf'' plot for the system".
 

The system's failure rate can now be obtained by dividing the system's <math>pdf</math> given in equation above by the system's reliability function given in <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>, or:

 

::<math>\begin{align}
{{\lambda }_{s}}(t)= & \frac{4\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+6\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}{2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}} \\
& +\frac{-20\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+10\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}{2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}
\end{align}</math>

 

The corresponding plot is given in Figure "Failure rate for the system".
 
[[Image:BS5.7.gif|thumb|center|300px|<div align="center"> ''pdf'' plot for the system.</div>]]
 

The <math>MTTF</math> of the system is obtained by integrating the system's reliability function given by <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math> from time zero to infinity, as given by <math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math>. Using BlockSim's Analytical QCP, an <math>MTTF</math> of 1007.8 hours is calculated, as shown in Figure "MTTF of the system".
 

The warranty time can be obtained by solving <math>{{R}_{s}}(t)</math> with respect to time for a system reliability <math>{{R}_{s}}=0.9</math> . Using the Analytical QCP and selecting the 
Warranty Time option, a time of 372.72 hours is obtained, as shown in Figure "Time at which ''R''=0.9 or 90% for the system".
 
[[Image:BS5.8.gif|thumb|center|400px|<div align="center"> Failure rate for the system.</div>]]
 
[[Image:Fig 5.9.PNG|thumb|center|400px|<div align="center"> MTTF of the system. </div>]]
 
[[Image:Fig 5.10.PNG|thumb|center|400px|<div align="center"> Time at which ''R''=0.9 or 90% for the system.</div>]]
 
[[Image:Fig 5.11.PNG|thumb|center|400px|<div align="center"> Conditional reliability calculation for the system.</div>]]
 

Lastly, the conditional reliability can be obtained using <math>R(T,t)=\frac{R(T+t)}{R(T)}</math> and <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>, or:
 

::<math>\begin{align}
R(200,200)= & \frac{R(400)}{R(200)} \\
= & \frac{0.883825}{0.975321} \\
= & 0.906189
\end{align}</math>

 

This can be calculated using BlockSim's Analytical QCP, as shown in Figure "Conditional reliability calculation for the system".
 

=Approximating the System CDF=

In many cases, it is valuable to fit a distribution that represents the system's times-to-failure. This can be useful when the system is part of a larger assembly and may be used for repeated calculations or in calculations for other systems. In cases such as this, it can be useful to characterize the system's behavior by fitting a distribution to the overall system and calculating parameters for this distribution. This is equivalent to fitting a single distribution to describe <math>{{R}_{S}}(t</math> ). In essence, it is like reducing the entire system to a component in order to simplify calculations.
 
For the system in Figure below:
[[Image:5-2.png|thumb|center|300px|<div align="center">Simple two-component system. </div>]]
 

::<math>{{R}_{S}}(t)={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>

 
To compute an approximate reliability function for this system, <math>{{R}_{A}}(t)\simeq {{R}_{S}}(t)</math> , one would compute <math>n</math> pairs of reliability and time values and then fit a single distribution to the data, or:

 

::<math>\begin{align}
{{R}_{S}}(t= & 10,396.7)=10% \\
{{R}_{S}}(t= & 9,361.9)=20% \\
& ... \\
{{R}_{S}}(t= & 1,053.6)=90%
\end{align}</math>

 

A single distribution, <math>{{R}_{A}}(t)</math> , that approximates <math>{{R}_{S}}(t)</math> can now be computed from these pairs using life data analysis methods. If using the Weibull++ software, one would enter the values as free form data.

===Example 3===
Compute a single Weibull distribution approximation for the system in Example 2.
====Solution to Example 3====
The system in the previous example, shown in Figure "Complex bridge system in Example 2", can be approximated by use of a 2-parameter Weibull distribution with <math>\beta =2.02109</math> and <math>\eta =1123.51</math> . In BlockSim, this is accomplished by representing the entire system as one distribution by going to the Distribution Fit window Figure " Representing a system with a distribution".

 
[[Image:Fig 5.13.PNG|thumb|center|400px|<div align="center"> Fig 5.12 Representing a system with a distribution.</div>]]
 
[[Image:Fig 5.12.PNG|thumb|center|300px|<div align="center"> Fig 5.13 Distribution Fitting window.</div>]]
 
by clicking the Distribution Fit Window, the Distribution Estimator window will appear (Figure "Distribution Fitting window").
 
In this window you can select a distribution to represent the data. BlockSim will then generate a number of system failure times based on the system's reliability function. The system's reliability function can be used to solve for a time value associated with that unreliability value. The distribution of the generated time values can then be fitted to a probability distribution function.
 

Consider a value of <math>F(t)=0.11</math> . Using the system's reliability equation and solving for time, the corresponding time-to-failure for a 0.11 unreliability can be calculated. 
For the system of Example 2, the time for a 0.11 unreliability is 389.786 hours.
 
When enough points have been generated, the selected distribution will be fitted to this data set and the distribution's parameters will be returned. In addition, if ReliaSoft's Weibull++ is installed, the generated data can be viewed/analyzed using a Weibull++ instance, as shown in Figure "Using Weibull++ to calculate distribution parameters".
 
[[Image:Fig 5.14.PNG|thumb|center|400px|<div align="center"> Fig 5.14 Using Weibull++ to calculate distribution parameters.</div>]]
 
It is recommended that the analyst examine the fit to ascertain the applicability of the approximation.

=Duty Cycle=

Components of a system may not operate continuously during a system's mission, or may be subjected to loads greater or lesser than the rated loads during system operation. To model this, a factor called the Duty Cycle ( <math>{{d}_{c}}</math> ) is used. The duty cycle may also be used to account for changes in environmental stress, such as temperature changes, that may effect the operation of a component. The duty cycle is a positive value, with a default value of 1 representing continuous operation at rated load, and any values other than 1 representing other load values with respect to the rated load value (or total operating time). A duty cycle value higher than 1 indicates a load in excess of the rated value. A duty cycle value lower than 1 indicates that the component is operating at a load lower than the rated load or not operating continuously during the system's mission. For instance, a duty cycle of 0.5 may be used for a component that operates only half of the time during the system's mission.
 
The reliability metrics for a component with a duty cycle are calculated as follows. Let <math>{{d}_{c}}</math> represent the duty cycle during a particular mission of the component, <math>t</math> represent the mission time and <math>{t}'</math> represent the accumulated age. Then:
 

::<math>{t}'={{d}_{c}}\times t</math>

 

The reliability equation for the component is:
 

::<math>R({t}')=R({{d}_{c}}\times t)</math>
 

The component ''pdf'' is:
 

::<math>f({t}')=-\frac{d(R({t}'))}{dt}=-\frac{d(R({{d}_{c}}\times t))}{dt}={{d}_{c}}f({{d}_{c}}\times t)</math>
 

The failure rate of the component is:
 

::<math>\lambda ({t}')=\frac{f({t}')}{R({t}')}=\frac{{{d}_{c}}f({{d}_{c}}\times t)}{R({{d}_{c}}\times t)}={{d}_{c}}\lambda ({{d}_{c}}\times t)</math>

===Example 4===
Consider a computer system with three components: a processor, a hard drive and a CD drive in series as shown next. Assume that all three components follow a Weibull failure distribution with the parameters <math>{{\beta }_{1}}=1.5</math> and <math>{{\eta }_{1}}=5000</math> for the processor, <math>{{\beta }_{2}}=2.5</math> and <math>{{\eta }_{2}}=3000</math> for the hard drive, and <math>{{\beta }_{3}}=2</math> and <math>{{\eta }_{3}}=4000</math> for the CD drive. Determine the reliability of the computer system after one year (365 days) of operation, assuming that the CD drive is used only 30% of the time.
 
[[Image:BS5ex4.png|thumb|center|300px|]]

====Solution to Example 4====
The reliability of the processor after 365 days of operation is given by:

::<math>\begin{align}
{{R}_{processor}}(365)= & {{e}^{-{{\left( \tfrac{365}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}} \\
= & {{e}^{-{{\left( \tfrac{365}{5000} \right)}^{1.5}}}} \\
= & 0.9805\text{ or }98.05%
\end{align}</math>

The reliability of the hard drive after 365 days of operation is given by:

::<math>\begin{align}
{{R}_{harddrive}}(365)= & {{e}^{-{{\left( \tfrac{365}{{{\eta }_{2}}} \right)}^{{{\beta }_{2}}}}}} \\
= & {{e}^{-{{\left( \tfrac{365}{3000} \right)}^{2.5}}}} \\
= & 0.9948\text{ or }99.48%
\end{align}</math>

The reliability of the CD drive after 365 days of operation (taking into account the 30% operation using a duty cycle of 0.3) is given by:

::<math>\begin{align}
{{R}_{CDdrive}}(365)= & {{e}^{-{{\left( \tfrac{{{d}_{c}}\times 365}{{{\eta }_{3}}} \right)}^{{{\beta }_{3}}}}}} \\
= & {{e}^{-{{\left( \tfrac{0.3\times 365}{4000} \right)}^{2}}}} \\
= & 0.9993\text{ or }99.93%
\end{align}</math>
 
[[Image:Fig 5.15.PNG|thumb|center|400px|<div align="center"> Fig 5.15 Result for the computer system reliability.</div>]]
 

Thus the reliability of the computer system after 365 days of operation is:

::<math>\begin{align}
{{R}_{s}}(365)= & {{R}_{processor}}(365)\cdot {{R}_{harddrive}}(365)\cdot {{R}_{CDdrive}}(365) \\
= & 0.9805\cdot 0.9948\cdot 0.9993 \\
= & 0.9747\text{ or }97.47%
\end{align}</math>
 
This result can be obtained in BlockSim as shown in Figure "Result for the computer system reliability".
 

=Load Sharing=
As presented in earlier chapters, a reliability block diagram (RBD) allows you to graphically represent how the components within a system are reliability-wise connected. In most cases, independence is assumed across the components within the system. For example, the failure of component A does not affect the failure of component B. However, if a system consists of components that are sharing a load, then the assumption of independence no longer holds true.
 

If one component fails, then the component(s) that are still operating will have to assume the failed unit's portion of the load. Therefore, the reliabilities of the surviving unit(s) will change. Calculating the system reliability is no longer an easy proposition. In the case of load sharing components, the change of the failure distributions of the surviving components must be known in order to determine the system's reliability.
 

To illustrate this, consider the a system of two units connected reliability-wise in parallel (Figure "Two units connected reliability-wise in parallel").
 

[[Image:5-16.png|thumb|center|300px|<div align="center"> Two units connected reliability-wise in parallel.</div>]]
 

Assume that the units must supply an output of 8 volts and that if both units are operational, each unit is to supply 50% of the total output. If one of the units fails, then the surviving unit supplies 100%. Furthermore, assume that having to supply the entire load has a negative impact on the reliability characteristics of the surviving unit. Since the reliability characteristics of the unit change based on whether both or only one is operating, a life distribution along with a life-stress relationship (as discussed in [http://reliawiki.com/index.php/Statistical_Background#A_Brief_Introduction_to_Life-Stress_Relationships|A Brief Introduction to Life-Stress Relationships]) will be needed to model each component.
 

To illustrate the steps needed, we will create the model starting from raw data. Assume that a total of 20 units were tested to failure at 7, 10 and 15 volts. The test data set is presented in the next table.
 

[[Image:5-17.png|thumb|center|300px|]]
 

For this example, Units 1 and 2 are the same component. Therefore, only one set of data was collected. However, it is possible that the load sharing components in a system may not be the same. If that were the case, data would need to be collected for each component.
 

The data set in Table 1 was analyzed using ReliaSoft's ALTA software (as shown in Figure "Using ALTA to calculate component parameters") with the Inverse Power Law as the underlying life-stress relationship and Weibull as the life distribution.
 
 

The estimated model parameters, <math>\beta </math> , <math>K</math> and <math>n</math> , are shown next.
 

::<math>\begin{align}
\beta = & 1.9239 \\
K= & 3.2387\times {{10}^{-7}} \\
n= & 3.4226
\end{align}</math>
 

:Or:
 
::<math>\begin{align}
{{R}_{1}}(t,{{S}_{1}})= & {{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} \\
= & {{e}^{-{{\left( 3.2387\times {{10}^{-7}}S_{1}^{3.4226}t \right)}^{1.9239}}}} \
\end{align}</math>.

 
::<math>{{f}_{1}}(t,{{S}_{1}})=\beta KS_{1}^{n}{{\left( KS_{1}^{n}t \right)}^{\beta -1}}{{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} \ </math>
 

And for this case:

::<math>\begin{align}
{{R}_{1}}(t,{{S}_{1}})= & {{R}_{2}}(t,{{S}_{2}}) \\
{{f}_{1}}(t,{{S}_{1}})= & {{f}_{2}}(t,{{S}_{2}})
\end{align}</math>
 
 
[[Image:Fig 5.17.PNG|thumb|center|500px|<div align="center"> Using ALTA to calculate component parameters.</div>]]
 

Figure "Reliability curves for different voltage output conditions" shows a plot of <math>{{R}_{1}}(t,{{S}_{1}})={{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} = {{e}^{-{{\left( 3.2387\times {{10}^{-7}}S_{1}^{3.4226}t \right)}^{1.9239}}}}\ </math>.
 

Now that the failure properties have been determined using the test data, the reliability of the system at some time, <math>t</math> , can be calculated using the following equation:

 
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>

 
:Where:
 

::<math>\begin{align}
{{S}_{1}}= & {{P}_{1}}S \\
{{S}_{2}}= & {{P}_{2}}S
\end{align}</math>

 
[[Image:5-18.png|thumb|center|300px|<div align="center"> Reliability curves for different voltage output conditions. </div>]]
 
[[Image:BS5.19.png|thumb|center|395px| Illustrating <math>{{t}_{e}}</math>]]
 

And:
 

:• <math>S</math> is the total load (or required output). 
:• <math>{{P}_{1}}</math> and <math>{{P}_{2}}</math> are the portion of the total load that each unit supports when both units are operational. In this case, <math>{{P}_{1}}={{P}_{2}}=0.5=50%.</math> 
:• <math>{{S}_{1}}</math> and <math>{{S}_{2}}</math> represent the portions of the load that Unit 1 and Unit 2 must support when both units are operational. 
:• <math>{{t}_{{{1}_{e}}}}</math> is the equivalent operating time for Unit 1 if it had been operating at <math>S</math> instead of <math>{{S}_{1}}</math> . A graphical representation of the equivalent time is shown in Figure 5.19, where the curve marked by L represents the low stress (load) and the curve marked by H represents the high stress (load). 
 

::<math>{{t}_{1e}}</math> can be calculated by:
 
::<math>\begin{align}
{{R}_{1}}(t)= & {{R}_{1}}({{t}_{1e}}) \\
{{e}^{-{{(tKS_{1}^{n})}^{\beta }}}}= & {{e}^{-{{({{t}_{1e}}K{{S}^{n}})}^{\beta }}}} \\
tS_{1}^{n}= & {{t}_{1e}}{{S}^{n}} \\
{{t}_{1e}}= & t{{\left( \frac{{{S}_{1}}}{S} \right)}^{n}},\text{ }{{S}_{1}}={{P}_{1}}S \\
\therefore & {{t}_{1e}}=tP_{1}^{n}
\end{align}</math>

 
::<math>{{t}_{2e}}</math> can be calculated the same way, or:

 
::<math>\begin{align}
{{R}_{2}}(t)= & {{R}_{2}}({{t}_{2e}}) \\
\therefore & {{t}_{2e}}=tP_{2}^{n}
\end{align}</math>
 

In this example, the reliability equations for Unit 1 and Unit 2 are the same since they are the same type of component and demonstrate the same failure properties. In addition, the total output is divided equally between the two units (when both units are operating), so <math>{{t}_{1e}}</math> and <math>{{t}_{2e}}</math> will also be the same.
 
The next step is to determine the reliability of the system after 8,760 hours, <math>R(t=8,760)</math> .
Using
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>
 

the system reliability is found to be:
 

::<math>\begin{align}
R(t=8760)= & 0.8567 \\
= & 85.67%
\end{align}</math>
 

===Load Sharing in BlockSim===
BlockSim uses this formulation when computing reliabilities of units in a load sharing configuration. When using the System Reliability Equation window, BlockSim returns a single token for the reliability of units in a load sharing configuration (as well as in the case of standby redundancy, discussed in the next section). As an example, consider the following RBD with Unit 1 in series with a container that includes two load sharing units.
 
[[Image:BS5.19.2.png|thumb|center|200px|]]
 
BlockSim will return the system equation as:
 

::<math>{{R}_{System}}=+{{R}_{LS}}\cdot {{R}_{1}}</math>
 

Where <math>{{R}_{LS}}</math> implies a form similar to
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>
 .
BlockSim allows for <math>k</math> -out-of- <math>n</math> units in a load sharing configuration.

===Example 5===
A component has five possible failure modes, <math>A</math> , <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> , <math>{{B}_{C}}</math> and <math>C</math> , and the <math>B</math> modes are interdependent. The system will fail if mode <math>A</math> occurs, mode <math>C</math> occurs or two out of the three <math>B</math> modes occur.
 

Modes <math>A</math> and <math>C</math> each have a Weibull distribution, with a <math>\beta =2</math> and <math>\eta =10,000</math> and 15,000 respectively. Events <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> and <math>{{B}_{C}}</math> each have an exponential distribution with a mean of 10,000 hours.
 

If any <math>B</math> event occurs (i.e. <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> or <math>{{B}_{C}}</math> ), the remaining <math>B</math> events are more likely to occur. Specifically, the mean times of the remaining <math>B</math> events are halved. Determine the reliability at 1000 hours for this component.
====Solution to Example 5====
The first step is to create the RBD. Modes <math>A</math> , <math>C</math> and a load sharing container with the <math>{{B}_{i}}</math> modes must be drawn in series, as illustrated next.
 
[[Image:BS5.19.3.png|thumb|center|200px|]]
 

The next step is to define the properties for each block, including those for the container. Setting the failure distributions for modes <math>A</math> and <math>C</math> is simple. The more difficult part is setting the properties for the container and the contained blocks. Based on the problem statement, the <math>B</math> modes are in a 2-out-of-3 load sharing redundancy. When all three are working (i.e. when no <math>B</math> mode has occurred), each block has an exponential distribution with <math>\mu=10,000</math>. If one <math>B</math> mode occurs, then the two surviving units have an exponential distribution with <math>\mu =5,000.</math>
 

Assume a Power Life-Stress relationship for the components. Then:
 

::<math>\begin{align}
{{\mu }_{1}}= & \frac{1}{KV_{1}^{n}} \ \\
{{\mu }_{2}}= & \frac{1}{KV_{2}^{n}} \
\end{align}</math>
 

Substituting <math>{{\mu }_{1}}=10,000</math> and <math>{{V}_{1}}=1</math> in <math>{{\mu }_{1}}= \frac{1}{KV_{1}^{n}}</math> and casting it in terms of <math>K</math> yields:
 

::<math>\begin{align}
10,000= & \frac{1}{K} \ \\
K = & \frac{1}{10,000}=0.0001
\end{align}</math>

 
Substituting <math>{{\mu }_{2}}=5,000</math> , <math>{{V}_{2}}=1.5</math> (because if one fails, then each survivor takes on an additional 0.5 units of load) and <math>10,000= \frac{1}{K}</math> for <math>K</math> in <math>{{\mu }_{2}}= \frac{1}{KV_{2}^{n}}</math> yields:

 
::<math>\begin{align}
5,000= & \frac{1}{0.0001\cdot {{(1.5)}^{n}}} \\
0.5= & {{(1.5)}^{-n}} \\
\ln (0.5)= & -n\ln (1.5) \\
n = & 1.7095
\end{align}</math>
 

This also could have been computed in ALTA, as shown in Figure "Calculation performed in ALTA".

The next step is to set the weight proportionality factor. This factor defines the portion of the load that the particular item carries while operating, as well as the load that shifts to the remaining units upon failure of the item. To illustrate, assume three units (1, 2 and 3) are in a load sharing container with weight proportionality factors of 1, 2 and 3 respectively (and a 1-out-of-3 requirement).
 

:• Unit 1 carries <math>\left( \tfrac{1}{1+2+3} \right)=0.166</math> or 16.6% of the total load.
:• Unit 2 carries <math>\left( \tfrac{2}{1+2+3} \right)=0.333</math> or 33.3% of the total load.
:• Unit 3 carries <math>\left( \tfrac{3}{1+2+3} \right)=0.50</math> or 50% of the total load.
 

The actual load on each unit then becomes the product of the entire load defined for the container times the portion carried by that unit. For example, if the container load is 100 lbs, then the portion assigned to Unit 1 will be <math>100\cdot 0.166=16.6</math> lbs.
 

In the current example, all units share the same load and thus have equal weight proportionality factors. Because these factors are relative, if the same number is used for all three items then the results will be the same. Thus, weight proportional factor is set equal to 1 for each item.
 

[[Image:5.20.PNG|thumb|center|500px|<div align="center"> Calculation performed in ALTA.</div>]]
 

 
[[Image:Fig 5.23_2.PNG|thumb|center|400px|<div align="center"> Defining Weight Proportional Factor. </div>]]
 

The last properties that need to be defined are the total load and the 2-out-of-3 redundancy. The total load is dependent on how the parameters were computed. In this case, total load was assumed to be 3 when the parameters were computed (i.e. the load per item was 1 when all worked and 1.5 when two worked). This is defined at the container level, as shown in Figure "Defining Weight Proportional Factor".
When all of the parameters have been specified in BlockSim, the reliability at 1,000 hours can be determined. From the Analytical QCP, this is found to be 93.87%.

=Standby Components=

In the previous section, the case of a system with load sharing components was presented. This is a form of redundancy with dependent components. That is, the failure of one component affects the failure of the other(s). This section presents another form of redundancy: standby redundancy. In standby redundancy the redundant components are set to be under a lighter load condition (or no load) while not needed and under the operating load when they are activated.
 

In standby redundancy the components are set to have two states: an active state and a standby state. Components in standby redundancy have two failure distributions, one for each state. When in the standby state, they have a quiescent (or dormant) failure distribution and when operating, they have an active failure distribution.
 

In the case that both quiescent and active failure distributions are the same, the units are in a simple parallel configuration (also called a hot standby configuration). When the rate of failure of the standby component is lower in quiescent mode than in active mode, that is called a warm standby configuration. When the rate of failure of the standby component is zero in quiescent mode (i.e. the component cannot fail when in standby), that is called a cold standby configuration.

===Simple Standby Configuration===

Consider two components in a standby configuration. Component 1 is the active component with a Weibull failure distribution with parameters <math>\beta </math> = 1.5 and <math>\eta </math> = 1,000. Component 2 is the standby component. When Component 2 is operating, it also has a Weibull failure distribution with <math>\beta </math> = 1.5 and <math>\eta </math> = 1,000. Furthermore, assume the following cases for the quiescent distribution.
 
:• Case 1: The quiescent distribution is the same as the active distribution (hot standby). 
:• Case 2: The quiescent distribution is a Weibull distribution with <math>\beta </math> = 1.5 and <math>\eta </math> = 2000 (warm standby). 
:• Case 3: The component cannot fail in quiescent mode (cold standby). 
 
In this case, the reliability of the system at some time, <math>t</math> , can be obtained using the following equation:
 

::<math>R(t)={{R}_{1}}(t)+\underset{0}{\overset{t}{\mathop \int }}\,{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x)\cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}dx \ </math>
 

Where:
 
:• <math>{{R}_{1}}</math> is the reliability of the active component. 
:• <math>{{f}_{1}}</math> is the <math>pdf</math> of the active component. 
:• <math>{{R}_{2;SB}}</math> is the reliability of the standby component when in standby mode (quiescent reliability). 
:• <math>{{R}_{2;A}}</math> is the reliability of the standby component when in active mode. 
:• <math>{{t}_{e}}</math> is the equivalent operating time for the standby unit if it had been operating at an active mode, such that: 

 
<math>{{R}_{2;SB}}(x)={{R}_{2;A}}({{t}_{e}})</math> 
 
The second equation above can be solved for <math>{{t}_{e}}</math> and substituted into the first equation above.
Figure "Standby container" illustrates the example as entered in BlockSim using a standby container.
 
 
[[Image:5.24.gif|thumb|center|400px|<div align="center"> Standby container.</div>]]

 

The active and standby blocks are within a container, which is used to specify standby redundancy. Since the standby component has two distributions (active and quiescent), the Block Properties window of the standby block has two pages for specifying each one. Figures "Defining the active failure distribution" and "Defining the quiescent failure distribution" illustrate these pages.

 

The system reliability results for 1000 hours are given in the following table:
 
[[Image:5-24.png|thumb|center|400px|]]
 

Note that even though the <math>\beta </math> value for the quiescent distribution is the same as in the active distribution, it is possible that the two can be different. That is, the failure modes present during the quiescent mode could be different from the modes present during the active mode. In that sense, the two distribution types can be different as well (e.g. lognormal when quiescent and Weibull when active).
 

In many cases when considering standby systems, a switching device may also be present that switches from the failed active component to the standby component. The reliability of the switch can also be incorporated into
<math>R(t)={{R}_{1}}(t)+\underset{0}{\overset{t}{\mathop \int }}\,{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x)\cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}dx \ </math>

as presented in the next section.
 

BlockSim's System Reliability Equation window returns a single token for the reliability of units in a standby configuration. This is the same as the load sharing case presented in the previous section.
 

[[Image: Fig 5.25.PNG|thumb|center|400px|<div align="center"> Defining the active failure distribution </div>]]
 
 

[[Image:Fig 5.26.PNG|thumb|center|400px|<div align="center"> Defining the quiescent failure distribution </div>]]
 

===Reliability of Standby Systems with a Switching Device===

In many cases when dealing with standby systems, a switching device is present that will switch to the standby component when the active component fails. Therefore, the failure properties of the switch must also be included in the analysis.
 

[[Image:BS5.26.2.png|thumb|center|300px|]]
 

In most cases when the reliability of a switch is to be included in the analysis, two probabilities can be considered. The first and most common one is the probability of the switch performing the action (i.e. switching) when requested to do so. This is called Switch Probability per Request in BlockSim and is expressed as a static probability (e.g. 90%). The second probability is the quiescent reliability of the switch. This is the reliability of the switch as it ages (e.g. the switch might wear out with age due to corrosion, material degradation, etc.). Thus it is possible for the switch to fail before the active component fails. However, a switch failure does not cause the system to fail, but rather causes the system to fail only if the switch is needed and the switch has failed. For example, if the active component does not fail until the mission end time and the switch fails, then the system does not fail. However, if the active component fails and the switch has also failed, then the system cannot be switched to the standby component and it therefore fails.
 

In analyzing standby components with a switching device, either or both failure probabilities (during the switching or while waiting to switch) can be considered for the switch, since each probability can represent different failure modes. For example, the switch probability per request may represent software-related issues or the probability of detecting the failure of an active component, and the quiescent probability may represent wear-out type failures of the switch.
 

To illustrate the formulation, consider the previous example that assumes perfect switching. To examine the effects of including an imperfect switch, assume that when the active component fails there is a 90% probability that the switch will switch from the active component to the standby component. In addition, assume that the switch can also fail due to a wear-out failure mode described by a Weibull distribution with <math>\beta </math> = 1.7 and <math>\eta </math> = 5000.
 

Therefore, the reliability of the system at some time, <math>t</math> , is given by the following equation.

::<math>\begin{align}
R(t)= & {{R}_{1}}(t) \\
& +\underset{0}{\overset{t}{\mathop \int }}\,\{{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x) \\
& \cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}\cdot {{R}_{SW;Q}}(x)\cdot {{R}_{SW;REQ}}(x)\}dx
\end{align}</math>

Where:
 
:• <math>{{R}_{1}}</math> is the reliability of the active component. 
:• <math>{{f}_{1}}</math> is the <math>pdf</math> of the active component. 
:• <math>{{R}_{2;SB}}</math> is the reliability of the standby component when in standby mode (quiescent reliability). 
:• <math>{{R}_{2;A}}</math> is the reliability of the standby component when in active mode. 
:• <math>{{R}_{SW;Q}}</math> is the quiescent reliability of the switch. 
:• <math>{{R}_{SW;REQ}}</math> is the switch probability per request. 
:• <math>{{t}_{e}}</math> is the equivalent operating time for the standby unit if it had been operating at an active mode. 
This problem can be solved in BlockSim by including these probabilities in the container's properties, as shown in Figures "Standby container (switch) failure probabilities while attempting to switch" and "Standby container (switch) failure distribution while waiting to switch". In BlockSim, the standby container is acting as the switch.

 

[[Image:Fig 5.28_2.PNG|thumb|center|400px|<div align="center"> Standby container (switch) failure probabilities while attempting to switch </div> ]]
 

 

[[Image:Fig 5.27.PNG|thumb|center|400px|<div align="center"> Standby container (switch) failure distribution while waiting to switch </div>]]
 

 

Note that there are additional properties that can be specified in BlockSim for a switch, such as Switch Restart Probability, No. of Restarts and Switch Delay Time. In many applications, the switch is re-tested (or re-cycled) if it fails to switch the first time. In these cases, it might be possible that it switches in the second or third, or <math>{{n}^{th}}</math> attempt.
 
The Switch Restart Probability specifies each additional attempt's probability of successfully switching and the Finite Restarts specifies the total number of attempts. Note that the Switch Restart Probability specifies the probability of success of each trial (or attempt). The probability of success of <math>n</math> consecutive trials is calculated by BlockSim using the binomial distribution and this probability is then incorporated into equation above. The Switch Delay Time property is related to repairable systems and is considered in BlockSim only when using simulation. When using the analytical solution (i.e. for a non-repairable system), this property is ignored.
 

Solving the analytical solution (as given by Equation above), the following results are obtained.
 

[[Image:5-30.png|thumb|center|400px|]]
 

From the table above, it can be seen that the presence of a switching device has a significant effect on the reliability of a standby system. It is therefore important when modeling standby redundancy to incorporate the switching device reliability properties. It should be noted that this methodology is not the same as treating the switching device as another series component with the standby subsystem. This would be valid only if the failure of the switch resulted in the failure of system (e.g. switch failing open). In equation above, the Switch Probability per Request and quiescent probability are present only in the second term of the equation. Treating these two failure modes as a series configuration with the standby subsystem would imply that they are also present when the active component is functioning (i.e. first term of equation above). This is invalid and would result in the underestimation of the reliability of the system. In other words, these two failure modes become significant only when the active component fails.
 

As an example, and if we consider the warm standby case, the reliability of the system without the switch is 70.57% at 1000 hours. If the system was modeled so that the switching device was in series with the warm standby subsystem, the result would have been:

 

::<math>\begin{align}
{{R}_{S}}(1000)= & {{R}_{Standby}}(1000)\cdot {{R}_{sw,Q(1000)}}\cdot {{R}_{sw,req}} \\
= & 0.7057\cdot 0.9372\cdot 0.9 \\
= & 0.5952
\end{align}</math>

 

In the case where a switch failure mode causes the standby subsystem to fail, then this mode can be modeled as an individual block in series with the standby subsystem.

===Example 6===
Consider a car with four new tires and a full-size spare. Assume the following failure characteristics:
 
:• The tires follow a Weibull distribution with a <math>\beta=4</math> and an <math>\eta =</math> 40,000 miles while on the car due to wear. 
:• The tires also have a probability of failing due to puncture or other causes. For this, assume a constant rate for this occurrence with a probability of 1 every 50,000 miles. 
:• When not on the car (i.e. is a spare), a tire's probability of failing also has a Weibull distribution with a <math>\beta =</math> 2 and <math>\eta =</math> 120,000 miles. 
 
Assume a mission of 1,000 miles. If a tire fails during this trip, it will be replaced with the spare. However, the spare will not be repaired during the trip. In other words, the trip will continue with the spare on the car and if the spare fails the system will fail. Determine the probability of system failure.
====Solution to Example 6====
Active failure distribution for tires:
 
:• Due to wear-out, Weibull <math>\beta =4</math> and <math>\eta =40,000</math> miles. 
:• Due to random puncture, exponential <math>\mu =50,000.</math> 
:• The quiescent failure distribution is a Weibull distribution with <math>\beta =2</math> and <math>\eta =120,000</math> miles. 
 
The block diagram for each tire has two blocks in series, one block representing the wear-out mode and the other the random puncture mode, as shown next:
 
 
[[Image:small5.gif|thumb|center|400px|]]
 

There are five tires, four active and one standby (represented in the diagram by a standby container with a 4-out-of-5 requirement), as shown next:
 
[[Image:BStirecontainer.png|thumb|center|400px|]]

For the standby Wear block, set the active failure and the quiescent distributions, but for the Puncture block, set only the active puncture distribution (because the tire cannot fail due to puncture while stored). Using BlockSim, the probability of system failure is found to be 0.003 or 0.3%.
 

=Note Regarding Numerical Integration Solutions=

Load sharing and standby solutions in BlockSim are performed using numerical integration routines. As with any numerical analysis routine, the solution error depends on the number of iterations performed, the step size chosen and related factors, plus the behavior of the underlying function. By default, BlockSim uses a certain set of preset factors. In general, these defaults are sufficient for most problems. If a higher precision or verification of the precision for a specific problem is required, BlockSim's preset options can be modified and/or the integration error can be assessed using the Integration Parameters... option for each container. For more details, you can refer to the documentation on the Algorithm Setup window in the BlockSim 8 User's help files.

Time-Dependent System Reliability (Analytical)

2012-03-21T21:26:27Z

Dingzhou Cao: /* Looking at a Simple Complex System Analytically */

{{Template:bsbook|5}}

In the previous chapter, different system configuration types were examined, as well as different methods for obtaining the system's reliability function analytically. Because the reliabilities in the problems presented were treated as probabilities (e.g. <math>P(A)</math> , <math>{{R}_{i}}</math> ), the reliability values and equations presented were referred to as static (not time-dependent). Thus, in the prior chapter, the life distributions of the components were not incorporated in the process of calculating the system reliability. In this chapter, time dependency in the reliability function will be introduced. We will develop the models necessary to observe the reliability over the life of the system, instead of at just one point in time. In addition, performance measures such as failure rate, MTTF and warranty time will be estimated for the entire system. The methods of obtaining the reliability function analytically remain identical to the ones presented in the previous chapter, with the exception that the reliabilities will be functions of time. In other words, instead of dealing with <math>{{R}_{i}}</math> , we will use <math>{{R}_{i}}(t)</math> . All examples in this chapter assume that no repairs are performed on the components.

=Analytical Life Predictions=
The analytical approach presented in the prior chapter involved the determination of a mathematical expression that describes the reliability of the system, expressed in terms of the reliabilities of its components. So far we have estimated only static system reliability (at a fixed time). For example, in the case of a system with three components in series, the system's reliability equation was given by:
 

::<math>{{R}_{s}}={{R}_{1}}\cdot {{R}_{2}}\cdot {{R}_{3}} \ </math>
 

The values of <math>{{R}_{1}}</math> , <math>{{R}_{2}}</math> and <math>{{R}_{3}}</math> were given for a common time and the reliability of the system was estimated for that time. However, since the component failure characteristics can be described by distributions, the system reliability is actually time-dependent. In this case, the equation above can be rewritten as:
 

::<math>{{R}_{s}}(t)={{R}_{1}}(t)\cdot {{R}_{2}}(t)\cdot {{R}_{3}}(t)</math>

 

The reliability of the system for any mission time can now be estimated. Assuming a Weibull life distribution for each component, the first equation above can now be expressed in terms of each component's reliability function, or:

 

[[Category:Completed Theoretical Review]]

::<math>{{R}_{s}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}}\cdot {{e}^{-{{\left( \tfrac{t}{{{\eta }_{2}}} \right)}^{{{\beta }_{2}}}}}}\cdot {{e}^{-{{\left( \tfrac{t}{{{\eta }_{3}}} \right)}^{{{\beta }_{3}}}}}}</math>

 

In the same manner, any life distribution can be substituted into the system reliability equation. Suppose that the times-to-failure of the first component are described with a Weibull distribution, the times-to-failure of the second component with an exponential distribution and the times-to-failure of the third component with a normal distribution. Then the first equation above can be written as:
 

::<math>{{R}_{s}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}}\cdot {{e}^{-{{\lambda }_{2}}t}}\cdot \left[ 1-\Phi \left( \frac{t-{{\mu }_{3}}}{{{\sigma }_{3}}} \right) \right]</math>

 

It can be seen that the biggest challenge is in obtaining the system's reliability function in terms of component reliabilities, which has already been discussed in depth. Once this has been achieved, calculating the reliability of the system for any mission duration is just a matter of substituting the corresponding component reliability functions into the system reliability equation.
===Advantages of the Analytical Method===
The primary advantage of the analytical solution is that it produces a mathematical expression that describes the reliability of the system. Once the system's reliability function has been determined, other calculations can then be performed to obtain metrics of interest for the system. Such calculations include:
 
:• Determination of the system's <math>pdf.</math> 
:• Determination of warranty periods. 
:• Determination of the system's failure rate. 
:• Determination of the system's MTTF. 
 
In addition, optimization and reliability allocation techniques can be used to aid engineers in their design improvement efforts. Another advantage in using analytical techniques is the ability to perform static calculations and analyze systems with a mixture of static and time-dependent components. Finally, the reliability importance of components over time can be calculated with this methodology.
===Disadvantages of the Analytical Method===
The biggest disadvantage of the analytical method is that formulations can become very complicated. The more complicated a system is, the larger and more difficult it will be to analytically formulate an expression for the system's reliability. For particularly detailed systems this process can be quite time-consuming, even with the use of computers. Furthermore, when the maintainability of the system or some of its components must be taken into consideration, analytical solutions become intractable. In these situations, the use of simulation methods may be more advantageous than attempting to develop a solution analytically. Simulation methods are presented in later chapters.
===Looking at a Simple ''Complex'' System Analytically===

The complexity involved in an analytical solution can be best illustrated by looking at the simple ''complex'' system with 15 components, as shown in Figure " An RBD of a complex system".
 
[[Image:5-1.png|thumb|center|400px|<div align="center"> An RBD of a complex system.</div>]]

 
The system reliability for this system (computed using BlockSim) is shown next. The first solution is provided using BlockSim's symbolic solution. In symbolic mode, BlockSim breaks the equation into segments, identified by tokens, that need to be substituted into the final system equation for a complete solution. This creates algebraic solutions that are more compact than if the substitutions were made.
 

::<math>\begin{align}
{{R}_{System}}= & D2\cdot D3\cdot {{R}_{L}} \\
D3= & +{{R}_{K}}\cdot IK \\
IK= & +{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}} \\
D2 = & +{{R}_{A}}\cdot {{R}_{E}}\cdot IE \\
IE = & -D1\cdot {{R}_{M}}\cdot {{R}_{N}}+{{R}_{M}}\cdot {{R}_{N}}+D1 \\
D1 = & +{{R}_{D}}\cdot ID \\
ID = & -{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}}
\end{align}</math>
 
Substituting the terms yields:
 

::<math>\begin{align}
{{R}_{System}}= & {{R}_{A}}\cdot {{R}_{E}}\cdot {{R}_{L}}\cdot {{R}_{K}} \\
& \cdot \{({{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})\cdot {{R}_{M}}\cdot {{R}_{N}} \\
& +{{R}_{M}}\cdot {{R}_{N}}-{{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}}\} \\
& \cdot \{{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}}\}
\end{align}</math>

 
BlockSim's automatic algebraic simplification would yield the following format for the above solution:
 

::<math>\begin{align}
{{R}_{System}}= & (({{R}_{A}}\cdot {{R}_{E}}(-({{R}_{D}}(-{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})){{R}_{M}}\cdot {{R}_{N}} \\
& +{{R}_{M}}\cdot {{R}_{N}} \\
& +({{R}_{D}}(-{{R}_{B}}\cdot {{R}_{C}}+{{R}_{B}}+{{R}_{C}})))) \\
& ({{R}_{K}}({{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{O}}\cdot {{R}_{G}}\cdot {{R}_{F}}-{{R}_{I}}\cdot {{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& -{{R}_{I}}\cdot {{R}_{O}}\cdot {{R}_{F}}\cdot {{R}_{H}}-{{R}_{J}}\cdot {{R}_{G}}\cdot {{R}_{F}}\cdot {{R}_{H}} \\
& +RI\cdot {{R}_{O}}\cdot {{R}_{F}} \\
& +{{R}_{I}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{F}}\cdot {{R}_{H}}+{{R}_{J}}\cdot {{R}_{G}})){{R}_{L}}) \
\end{align}</math>

 
In this equation, each <math>{{R}_{i}}</math> represents the reliability function of a block. For example, if <math>{{R}_{A}}</math> has a Weibull distribution, then each <math>{{R}_{A}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{A}}} \right)}^{{{\beta }_{A}}}}}}</math> and so forth. Substitution of each component's reliability function in the last <math>{{R}_{System}}</math> equation above will result in an analytical expression for the system reliability as a function of time, or <math>{{R}_{s}}(t)</math> , which is the same as <math>(1-cd{{f}_{System}}).</math>

===Obtaining Other Functions of Interest===
Once the system reliability equation (or the cumulative density function, <math>cdf</math> ) has been determined, other functions and metrics of interest can be derived.
 
Consider the following simple system:
 
[[Image:5-2.png|thumb|center|300px|<div align="center"> Fig 5.2 Simple two-component system. </div>]]

 
Furthermore, assume that component 1 follows an exponential distribution with a mean of 10,000 (<math>\mu =10,000,</math> <math>\lambda =1/10,000)</math> and component 2 follows a Weibull distribution with <math>\beta =6</math> and <math>\eta =10,000</math> . The reliability equation of this system is:

 
::<math>\begin{align}
{{R}_{S}}(t)= & {{R}_{1}}(t)\cdot {{R}_{2}}(t) \\
= & {{e}^{-\lambda t}}\cdot {{e}^{-{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \\
= & {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \
\end{align}</math>

 
The system <math>cdf</math> is:

 
::<math>\begin{align}
{{F}_{S}}(t)= & 1-({{R}_{1}}(t)\cdot {{R}_{2}}(t)) \\
= & 1-\left( {{e}^{-\lambda t}}\cdot {{e}^{-{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \right) \\
= & 1-\left( {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)
\end{align}</math>

 
{{analytical system pdf}}

====Conditional Reliability====
Conditional reliability is the probability of a system successfully completing another mission following the successful completion of a previous mission. The time of the previous mission and the time for the mission to be undertaken must be taken into account for conditional reliability calculations. The system's conditional reliability function is given by:

::<math>R(T,t)=\frac{R(T+t)}{R(T)}</math>

Equation above gives the reliability for a new mission of duration <math>t</math> having already accumulated <math>T</math> hours of operation up to the start of this new mission. The system is evaluated to assure that it will start the next mission successfully.
 
For the system in Figure "Simple two-component system.", the reliability for mission of <math>t=1,000</math> hours, having an age of <math>T=500</math> hours, is:
 

::<math>\begin{align}
{{R}_{S}}(T=500,t=1000)= & \frac{R(T+t)}{R(T)} \\
= & \frac{R(1500)}{R(500)} \\
= & \frac{{{e}^{-\tfrac{1500}{10,000}}}\cdot {{e}^{-{{\left( \tfrac{1500}{10,000} \right)}^{6}}}}}{{{e}^{-\tfrac{500}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{500}{10,000} \right)}^{6}}}}} \\
= & 0.9048=90.48%
\end{align}</math>

 

====Conditional Reliability for Components====

Now in this formulation, it was assumed that the accumulated age was equivalent for both units. That is, both started life at zero and aged to 500. It is possible to consider an individual component that has already accumulated some age (used component) in the same formulation. To illustrate this, assume that component 2 started life with an age of T=100. Then the reliability equation of the system, as given in <math>{{R}_{S}}(t)= {{R}_{1}}(t)\cdot {{R}_{2}}(t)</math>, would need to be modified to include a conditional term for 2, or:
 

::<math>{{R}_{S}}(t)={{R}_{1}}(t)\cdot \frac{{{R}_{2}}({{T}_{2}}+t)}{{{R}_{2}}({{T}_{2}})} \ </math>

 

In BlockSim, the start age input box may be used to specify a starting age greater than zero.
{{system failure rate analytical}}

====System Mean Life (Mean Time To Failure)====
The mean life (or mean time to failure, MTTF) can be obtained by integrating the system reliability function from zero to infinity:
 

::<math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math>

 

The mean time is a performance index and does not provide any information about the behavior of the failure distribution of the system.
 
For the system in Figure "Simple two-component system":
 

::<math>\begin{align}
MTTF= & \int_{0}^{\infty }\left( {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)dt \\
= & 5978.9
\end{align}</math>

 

====Warranty Period and BX Life====
Sometimes it is desirable to know the time value associated with a certain reliability. Warranty periods are often calculated by determining what percentage of the failure population can be covered financially and estimating the time at which this portion of the population will fail. Similarly, engineering specifications may call for a certain BX life, which also represents a time period during which a certain proportion of the population will fail. For example, the B10 life is the time in which 10% of the population will fail.
This is obtained by setting <math>{{R}_{S}}(t)</math> to the desired value and solving for <math>t.</math>
 
For the system in Figure "Simple two-component system":
 

::<math>{{R}_{s}}\left( t \right)={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>
 

To compute the time by which reliability would be equal to 90%, equation above is recast as follows and solved for <math>t.</math>

 
::<math>0.90={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>

 

In this case, <math>t=1053.59</math> . Equivalently, the B10 life for this system is also <math>1053.59</math> .
Except for some trivial cases, a closed form solution for <math>t</math> cannot be obtained. Thus, it is necessary to solve for <math>t</math> using numerical methods. BlockSim uses numerical methods.

===Example 1===
Consider a system consisting of three exponential units in series with the following failure rates (in failures per hour): <math>{{\lambda }_{1}}</math> = 0.0002, <math>{{\lambda }_{2}}</math> = 0.0005 and <math>{{\lambda }_{3}}</math> = 0.0001.
 
:• Obtain the reliability equation for the system. 
:• What is the reliability of the system after 150 hours of operation? 
:• Obtain the system's <math>pdf.</math> 
:• Obtain the system's failure rate equation. 
:• What is the MTTF for the system? 
:• What should the warranty period be for a 90% reliability? 
====Solution to Example 1====
 
:• The analytical expression for the reliability of the system is given by:

::<math>\begin{align}
{{R}_{s}}(t)= & {{R}_{1}}(t)\cdot {{R}_{2}}(t)\cdot {{R}_{3}}(t) \\
= & {{e}^{-{{\lambda }_{1}}t}}\cdot {{e}^{-{{\lambda }_{2}}t}}\cdot {{e}^{-{{\lambda }_{1}}t}} \\
= & {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}
\end{align}</math>

 

At 150 hours of operation, the reliability of the system is:
 

::<math>\begin{align}
{{R}_{s}}(t)= & {{e}^{-(0.0002+0.0005+0.0001)150}} \\
= & 0.8869\text{ or }88.69%
\end{align}</math>

 

:• In order to obtain the system's <math>pdf</math> , the derivative of the reliability equation given in the first equation above is taken with respect to time, or:
 
::<math>\begin{align}
{{f}_{s}}(t)= & -\frac{d[{{R}_{s}}(t)]}{dt} \\
= & -\frac{d\left[ {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}} \right]}{dt} \\
= & ({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})\cdot {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}
\end{align}</math>

:• The system's failure rate can now be obtained simply by dividing the system's <math>pdf</math> given in the equation above by the system's reliability function given in the first equation above, and:

::<math>\begin{align}
{{\lambda }_{s}}\left( t \right)= & \frac{{{f}_{s}}\left( t \right)}{{{R}_{s}}\left( t \right)} \\
= & \frac{({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})\cdot {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}}{{{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}} \\
= & ({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}) \\
= & 0.0008\text{ }fr/hr
\end{align}</math>

 
:• Combining <math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math> and the first equation above, the system's MTTF can be obtained:

 
::<math>\begin{align}
MTTF= & \int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \\
= & \int_{0}^{\infty }{{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}dt \\
= & \frac{1}{({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})} \\
= & 1250\text{ }hr
\end{align}</math>
 

:• Solving the first equation above with respect to time will yield the corresponding warranty period for a 90% reliability. In this case, the system reliability equation is simple and a closed form solution exists. The warranty time can now be found by solving:

 
::<math>\begin{align}
t= & -\frac{\ln (R)}{{{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}} \\
= & -\frac{\ln (0.9)}{0.0008} \\
= & 131.7\text{ }hr
\end{align}</math>
 

Thus, the warranty period should be 132 hours.

===Example 2===
Consider the system shown in Figure "Complex bridge system in Example 2".
 
[[Image:BS5.5.png|thumb|center|300px|<div align="center"> Fig 5.5 Complex bridge system in Example 2. </div>]]
 

Components <math>A</math> through <math>E</math> are Weibull distributed with <math>\beta =1.2</math> and <math>\eta =1230</math> hours. The starting and ending blocks cannot fail. 
Determine the following:
 
:• The reliability equation for the system and its corresponding plot. 
:• The system's <math>pdf</math> and its corresponding plot. 
:• The system's failure rate equation and the corresponding plot. 
:• The <math>MTTF</math> . 
:• The warranty time for a 90% reliability. 
:• The reliability for a 200-hour mission, if it is known that the system has already successfully operated for 200 hours. 
====Solution====
 
The first step is to obtain the reliability function for the system. The methods described in the previous chapter can be employed, such as the event space or path-tracing methods. Using BlockSim, the following reliability equation is obtained:
 

::<math>\begin{align}
{{R}_{s}}(t)= & ({{R}_{Start}}\cdot {{R}_{End}}(2{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}))
\end{align}</math>
 

Note that since the starting and ending blocks cannot fail, <math>{{R}_{Start}}=1</math> and <math>{{R}_{End}}=1,</math> equation above can be reduced to:

 

::<math>\begin{align}
{{R}_{s}}(t)= & 2\cdot {{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& -{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{B}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}} \\
& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}
\end{align}</math>

 

Where <math>{{R}_{A}}</math> is the reliability equation for Component A, or:

 

::<math>{{R}_{A}}(t)={{e}^{-{{\left( \tfrac{t}{{{\eta }_{A}}} \right)}^{{{\beta }_{A}}}}}}</math>

 

::<math>{{R}_{B}}</math> is the reliability equation for Component <math>B</math> , etc.
 

Since the components in this example are identical, the system reliability equation can be further reduced to:

 

::<math>{{R}_{s}}(t)=2R{{(t)}^{2}}+2R{{(t)}^{3}}-5R{{(t)}^{4}}+2R{{(t)}^{5}}</math>

 

Or, in terms of the failure distribution:

 

::<math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>

 

The corresponding plot is given in Figure "Reliability plot for the system".
 
[[Image:BS5.6.png|thumb|center|300px|<div align="center"> Reliability plot for the system. </div>]]
 
 

In order to obtain the system's <math>pdf</math> , the derivative of the reliability equation given above is taken with respect to time, resulting in:
 

::<math>\begin{align}
{{f}_{s}}(t)= & 4\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+6\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \\
& -20\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+10\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}
\end{align}</math>

 

The <math>pdf</math> can now be plotted for different time values, <math>t</math> , as shown in Figure "''pdf'' plot for the system".
 

The system's failure rate can now be obtained by dividing the system's <math>pdf</math> given in equation above by the system's reliability function given in <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>, or:

 

::<math>\begin{align}
{{\lambda }_{s}}(t)= & \frac{4\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+6\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}{2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}} \\
& +\frac{-20\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+10\cdot \tfrac{\beta }{\eta }{{\left( \tfrac{t}{\eta } \right)}^{\beta -1}}{{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}{2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}}
\end{align}</math>

 

The corresponding plot is given in Figure "Failure rate for the system".
 
[[Image:BS5.7.gif|thumb|center|300px|<div align="center"> ''pdf'' plot for the system.</div>]]
 

The <math>MTTF</math> of the system is obtained by integrating the system's reliability function given by <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math> from time zero to infinity, as given by <math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ </math>. Using BlockSim's Analytical QCP, an <math>MTTF</math> of 1007.8 hours is calculated, as shown in Figure "MTTF of the system".
 

The warranty time can be obtained by solving <math>{{R}_{s}}(t)</math> with respect to time for a system reliability <math>{{R}_{s}}=0.9</math> . Using the Analytical QCP and selecting the 
Warranty Time option, a time of 372.72 hours is obtained, as shown in Figure "Time at which ''R''=0.9 or 90% for the system".
 
[[Image:BS5.8.gif|thumb|center|400px|<div align="center"> Failure rate for the system.</div>]]
 
[[Image:Fig 5.9.PNG|thumb|center|400px|<div align="center"> MTTF of the system. </div>]]
 
[[Image:Fig 5.10.PNG|thumb|center|400px|<div align="center"> Time at which ''R''=0.9 or 90% for the system.</div>]]
 
[[Image:Fig 5.11.PNG|thumb|center|400px|<div align="center"> Conditional reliability calculation for the system.</div>]]
 

Lastly, the conditional reliability can be obtained using <math>R(T,t)=\frac{R(T+t)}{R(T)}</math> and <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}</math>, or:
 

::<math>\begin{align}
R(200,200)= & \frac{R(400)}{R(200)} \\
= & \frac{0.883825}{0.975321} \\
= & 0.906189
\end{align}</math>

 

This can be calculated using BlockSim's Analytical QCP, as shown in Figure "Conditional reliability calculation for the system".
 

=Approximating the System CDF=

In many cases, it is valuable to fit a distribution that represents the system's times-to-failure. This can be useful when the system is part of a larger assembly and may be used for repeated calculations or in calculations for other systems. In cases such as this, it can be useful to characterize the system's behavior by fitting a distribution to the overall system and calculating parameters for this distribution. This is equivalent to fitting a single distribution to describe <math>{{R}_{S}}(t</math> ). In essence, it is like reducing the entire system to a component in order to simplify calculations.
 
For the system in Figure below:
[[Image:5-2.png|thumb|center|300px|<div align="center">Simple two-component system. </div>]]
 

::<math>{{R}_{S}}(t)={{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}</math>

 
To compute an approximate reliability function for this system, <math>{{R}_{A}}(t)\simeq {{R}_{S}}(t)</math> , one would compute <math>n</math> pairs of reliability and time values and then fit a single distribution to the data, or:

 

::<math>\begin{align}
{{R}_{S}}(t= & 10,396.7)=10% \\
{{R}_{S}}(t= & 9,361.9)=20% \\
& ... \\
{{R}_{S}}(t= & 1,053.6)=90%
\end{align}</math>

 

A single distribution, <math>{{R}_{A}}(t)</math> , that approximates <math>{{R}_{S}}(t)</math> can now be computed from these pairs using life data analysis methods. If using the Weibull++ software, one would enter the values as free form data.

===Example 3===
Compute a single Weibull distribution approximation for the system in Example 2.
====Solution to Example 3====
The system in the previous example, shown in Figure "Complex bridge system in Example 2", can be approximated by use of a 2-parameter Weibull distribution with <math>\beta =2.02109</math> and <math>\eta =1123.51</math> . In BlockSim, this is accomplished by representing the entire system as one distribution by going to the Distribution Fit window Figure " Representing a system with a distribution".

 
[[Image:Fig 5.13.PNG|thumb|center|400px|<div align="center"> Fig 5.12 Representing a system with a distribution.</div>]]
 
[[Image:Fig 5.12.PNG|thumb|center|300px|<div align="center"> Fig 5.13 Distribution Fitting window.</div>]]
 
by clicking the Distribution Fit Window, the Distribution Estimator window will appear (Figure "Distribution Fitting window").
 
In this window you can select a distribution to represent the data. BlockSim will then generate a number of system failure times based on the system's reliability function. The system's reliability function can be used to solve for a time value associated with that unreliability value. The distribution of the generated time values can then be fitted to a probability distribution function.
 

Consider a value of <math>F(t)=0.11</math> . Using the system's reliability equation and solving for time, the corresponding time-to-failure for a 0.11 unreliability can be calculated. 
For the system of Example 2, the time for a 0.11 unreliability is 389.786 hours.
 
When enough points have been generated, the selected distribution will be fitted to this data set and the distribution's parameters will be returned. In addition, if ReliaSoft's Weibull++ is installed, the generated data can be viewed/analyzed using a Weibull++ instance, as shown in Figure "Using Weibull++ to calculate distribution parameters".
 
[[Image:Fig 5.14.PNG|thumb|center|400px|<div align="center"> Fig 5.14 Using Weibull++ to calculate distribution parameters.</div>]]
 
It is recommended that the analyst examine the fit to ascertain the applicability of the approximation.

=Duty Cycle=

Components of a system may not operate continuously during a system's mission, or may be subjected to loads greater or lesser than the rated loads during system operation. To model this, a factor called the Duty Cycle ( <math>{{d}_{c}}</math> ) is used. The duty cycle may also be used to account for changes in environmental stress, such as temperature changes, that may effect the operation of a component. The duty cycle is a positive value, with a default value of 1 representing continuous operation at rated load, and any values other than 1 representing other load values with respect to the rated load value (or total operating time). A duty cycle value higher than 1 indicates a load in excess of the rated value. A duty cycle value lower than 1 indicates that the component is operating at a load lower than the rated load or not operating continuously during the system's mission. For instance, a duty cycle of 0.5 may be used for a component that operates only half of the time during the system's mission.
 
The reliability metrics for a component with a duty cycle are calculated as follows. Let <math>{{d}_{c}}</math> represent the duty cycle during a particular mission of the component, <math>t</math> represent the mission time and <math>{t}'</math> represent the accumulated age. Then:
 

::<math>{t}'={{d}_{c}}\times t</math>

 

The reliability equation for the component is:
 

::<math>R({t}')=R({{d}_{c}}\times t)</math>
 

The component ''pdf'' is:
 

::<math>f({t}')=-\frac{d(R({t}'))}{dt}=-\frac{d(R({{d}_{c}}\times t))}{dt}={{d}_{c}}f({{d}_{c}}\times t)</math>
 

The failure rate of the component is:
 

::<math>\lambda ({t}')=\frac{f({t}')}{R({t}')}=\frac{{{d}_{c}}f({{d}_{c}}\times t)}{R({{d}_{c}}\times t)}={{d}_{c}}\lambda ({{d}_{c}}\times t)</math>

===Example 4===
Consider a computer system with three components: a processor, a hard drive and a CD drive in series as shown next. Assume that all three components follow a Weibull failure distribution with the parameters <math>{{\beta }_{1}}=1.5</math> and <math>{{\eta }_{1}}=5000</math> for the processor, <math>{{\beta }_{2}}=2.5</math> and <math>{{\eta }_{2}}=3000</math> for the hard drive, and <math>{{\beta }_{3}}=2</math> and <math>{{\eta }_{3}}=4000</math> for the CD drive. Determine the reliability of the computer system after one year (365 days) of operation, assuming that the CD drive is used only 30% of the time.
 
[[Image:BS5ex4.png|thumb|center|300px|]]

====Solution to Example 4====
The reliability of the processor after 365 days of operation is given by:

::<math>\begin{align}
{{R}_{processor}}(365)= & {{e}^{-{{\left( \tfrac{365}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}} \\
= & {{e}^{-{{\left( \tfrac{365}{5000} \right)}^{1.5}}}} \\
= & 0.9805\text{ or }98.05%
\end{align}</math>

The reliability of the hard drive after 365 days of operation is given by:

::<math>\begin{align}
{{R}_{harddrive}}(365)= & {{e}^{-{{\left( \tfrac{365}{{{\eta }_{2}}} \right)}^{{{\beta }_{2}}}}}} \\
= & {{e}^{-{{\left( \tfrac{365}{3000} \right)}^{2.5}}}} \\
= & 0.9948\text{ or }99.48%
\end{align}</math>

The reliability of the CD drive after 365 days of operation (taking into account the 30% operation using a duty cycle of 0.3) is given by:

::<math>\begin{align}
{{R}_{CDdrive}}(365)= & {{e}^{-{{\left( \tfrac{{{d}_{c}}\times 365}{{{\eta }_{3}}} \right)}^{{{\beta }_{3}}}}}} \\
= & {{e}^{-{{\left( \tfrac{0.3\times 365}{4000} \right)}^{2}}}} \\
= & 0.9993\text{ or }99.93%
\end{align}</math>
 
[[Image:Fig 5.15.PNG|thumb|center|400px|<div align="center"> Fig 5.15 Result for the computer system reliability.</div>]]
 

Thus the reliability of the computer system after 365 days of operation is:

::<math>\begin{align}
{{R}_{s}}(365)= & {{R}_{processor}}(365)\cdot {{R}_{harddrive}}(365)\cdot {{R}_{CDdrive}}(365) \\
= & 0.9805\cdot 0.9948\cdot 0.9993 \\
= & 0.9747\text{ or }97.47%
\end{align}</math>
 
This result can be obtained in BlockSim as shown in Figure "Result for the computer system reliability".
 

=Load Sharing=
As presented in earlier chapters, a reliability block diagram (RBD) allows you to graphically represent how the components within a system are reliability-wise connected. In most cases, independence is assumed across the components within the system. For example, the failure of component A does not affect the failure of component B. However, if a system consists of components that are sharing a load, then the assumption of independence no longer holds true.
 

If one component fails, then the component(s) that are still operating will have to assume the failed unit's portion of the load. Therefore, the reliabilities of the surviving unit(s) will change. Calculating the system reliability is no longer an easy proposition. In the case of load sharing components, the change of the failure distributions of the surviving components must be known in order to determine the system's reliability.
 

To illustrate this, consider the a system of two units connected reliability-wise in parallel (Figure "Two units connected reliability-wise in parallel").
 

[[Image:5-16.png|thumb|center|300px|<div align="center"> Two units connected reliability-wise in parallel.</div>]]
 

Assume that the units must supply an output of 8 volts and that if both units are operational, each unit is to supply 50% of the total output. If one of the units fails, then the surviving unit supplies 100%. Furthermore, assume that having to supply the entire load has a negative impact on the reliability characteristics of the surviving unit. Since the reliability characteristics of the unit change based on whether both or only one is operating, a life distribution along with a life-stress relationship (as discussed in [http://reliawiki.com/index.php/Statistical_Background#A_Brief_Introduction_to_Life-Stress_Relationships|A Brief Introduction to Life-Stress Relationships]) will be needed to model each component.
 

To illustrate the steps needed, we will create the model starting from raw data. Assume that a total of 20 units were tested to failure at 7, 10 and 15 volts. The test data set is presented in the next table.
 

[[Image:5-17.png|thumb|center|300px|]]
 

For this example, Units 1 and 2 are the same component. Therefore, only one set of data was collected. However, it is possible that the load sharing components in a system may not be the same. If that were the case, data would need to be collected for each component.
 

The data set in Table 1 was analyzed using ReliaSoft's ALTA software (as shown in Figure "Using ALTA to calculate component parameters") with the Inverse Power Law as the underlying life-stress relationship and Weibull as the life distribution.
 
 

The estimated model parameters, <math>\beta </math> , <math>K</math> and <math>n</math> , are shown next.
 

::<math>\begin{align}
\beta = & 1.9239 \\
K= & 3.2387\times {{10}^{-7}} \\
n= & 3.4226
\end{align}</math>
 

:Or:
 
::<math>\begin{align}
{{R}_{1}}(t,{{S}_{1}})= & {{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} \\
= & {{e}^{-{{\left( 3.2387\times {{10}^{-7}}S_{1}^{3.4226}t \right)}^{1.9239}}}} \
\end{align}</math>.

 
::<math>{{f}_{1}}(t,{{S}_{1}})=\beta KS_{1}^{n}{{\left( KS_{1}^{n}t \right)}^{\beta -1}}{{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} \ </math>
 

And for this case:

::<math>\begin{align}
{{R}_{1}}(t,{{S}_{1}})= & {{R}_{2}}(t,{{S}_{2}}) \\
{{f}_{1}}(t,{{S}_{1}})= & {{f}_{2}}(t,{{S}_{2}})
\end{align}</math>
 
 
[[Image:Fig 5.17.PNG|thumb|center|500px|<div align="center"> Using ALTA to calculate component parameters.</div>]]
 

Figure "Reliability curves for different voltage output conditions" shows a plot of <math>{{R}_{1}}(t,{{S}_{1}})={{e}^{-{{\left( KS_{1}^{n}t \right)}^{\beta }}}} = {{e}^{-{{\left( 3.2387\times {{10}^{-7}}S_{1}^{3.4226}t \right)}^{1.9239}}}}\ </math>.
 

Now that the failure properties have been determined using the test data, the reliability of the system at some time, <math>t</math> , can be calculated using the following equation:

 
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>

 
:Where:
 

::<math>\begin{align}
{{S}_{1}}= & {{P}_{1}}S \\
{{S}_{2}}= & {{P}_{2}}S
\end{align}</math>

 
[[Image:5-18.png|thumb|center|300px|<div align="center"> Reliability curves for different voltage output conditions. </div>]]
 
[[Image:BS5.19.png|thumb|center|395px| Illustrating <math>{{t}_{e}}</math>]]
 

And:
 

:• <math>S</math> is the total load (or required output). 
:• <math>{{P}_{1}}</math> and <math>{{P}_{2}}</math> are the portion of the total load that each unit supports when both units are operational. In this case, <math>{{P}_{1}}={{P}_{2}}=0.5=50%.</math> 
:• <math>{{S}_{1}}</math> and <math>{{S}_{2}}</math> represent the portions of the load that Unit 1 and Unit 2 must support when both units are operational. 
:• <math>{{t}_{{{1}_{e}}}}</math> is the equivalent operating time for Unit 1 if it had been operating at <math>S</math> instead of <math>{{S}_{1}}</math> . A graphical representation of the equivalent time is shown in Figure 5.19, where the curve marked by L represents the low stress (load) and the curve marked by H represents the high stress (load). 
 

::<math>{{t}_{1e}}</math> can be calculated by:
 
::<math>\begin{align}
{{R}_{1}}(t)= & {{R}_{1}}({{t}_{1e}}) \\
{{e}^{-{{(tKS_{1}^{n})}^{\beta }}}}= & {{e}^{-{{({{t}_{1e}}K{{S}^{n}})}^{\beta }}}} \\
tS_{1}^{n}= & {{t}_{1e}}{{S}^{n}} \\
{{t}_{1e}}= & t{{\left( \frac{{{S}_{1}}}{S} \right)}^{n}},\text{ }{{S}_{1}}={{P}_{1}}S \\
\therefore & {{t}_{1e}}=tP_{1}^{n}
\end{align}</math>

 
::<math>{{t}_{2e}}</math> can be calculated the same way, or:

 
::<math>\begin{align}
{{R}_{2}}(t)= & {{R}_{2}}({{t}_{2e}}) \\
\therefore & {{t}_{2e}}=tP_{2}^{n}
\end{align}</math>
 

In this example, the reliability equations for Unit 1 and Unit 2 are the same since they are the same type of component and demonstrate the same failure properties. In addition, the total output is divided equally between the two units (when both units are operating), so <math>{{t}_{1e}}</math> and <math>{{t}_{2e}}</math> will also be the same.
 
The next step is to determine the reliability of the system after 8,760 hours, <math>R(t=8,760)</math> .
Using
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>
 

the system reliability is found to be:
 

::<math>\begin{align}
R(t=8760)= & 0.8567 \\
= & 85.67%
\end{align}</math>
 

===Load Sharing in BlockSim===
BlockSim uses this formulation when computing reliabilities of units in a load sharing configuration. When using the System Reliability Equation window, BlockSim returns a single token for the reliability of units in a load sharing configuration (as well as in the case of standby redundancy, discussed in the next section). As an example, consider the following RBD with Unit 1 in series with a container that includes two load sharing units.
 
[[Image:BS5.19.2.png|thumb|center|200px|]]
 
BlockSim will return the system equation as:
 

::<math>{{R}_{System}}=+{{R}_{LS}}\cdot {{R}_{1}}</math>
 

Where <math>{{R}_{LS}}</math> implies a form similar to
::<math>\begin{align}
R(t,S)= & {{R}_{1}}(t,{{S}_{1}})\cdot {{R}_{2}}(t,{{S}_{2}}) \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{1}}\left( x,{{S}_{1}} \right)\cdot {{R}_{2}}(x,{{S}_{2}})\cdot \left( \frac{{{R}_{2}}({{t}_{1e}}+(t-x),S)}{{{R}_{2}}({{t}_{1e}},S)} \right)dx \\
& +\underset{o}{\overset{t}{\mathop \int }}\,{{f}_{2}}\left( x,{{S}_{2}} \right)\cdot {{R}_{1}}(x,{{S}_{1}})\cdot \left( \frac{{{R}_{1}}({{t}_{2e}}+(t-x),S)}{{{R}_{1}}({{t}_{2e}},S)} \right)dx
\end{align}</math>
 .
BlockSim allows for <math>k</math> -out-of- <math>n</math> units in a load sharing configuration.

===Example 5===
A component has five possible failure modes, <math>A</math> , <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> , <math>{{B}_{C}}</math> and <math>C</math> , and the <math>B</math> modes are interdependent. The system will fail if mode <math>A</math> occurs, mode <math>C</math> occurs or two out of the three <math>B</math> modes occur.
 

Modes <math>A</math> and <math>C</math> each have a Weibull distribution, with a <math>\beta =2</math> and <math>\eta =10,000</math> and 15,000 respectively. Events <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> and <math>{{B}_{C}}</math> each have an exponential distribution with a mean of 10,000 hours.
 

If any <math>B</math> event occurs (i.e. <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> or <math>{{B}_{C}}</math> ), the remaining <math>B</math> events are more likely to occur. Specifically, the mean times of the remaining <math>B</math> events are halved. Determine the reliability at 1000 hours for this component.
====Solution to Example 5====
The first step is to create the RBD. Modes <math>A</math> , <math>C</math> and a load sharing container with the <math>{{B}_{i}}</math> modes must be drawn in series, as illustrated next.
 
[[Image:BS5.19.3.png|thumb|center|200px|]]
 

The next step is to define the properties for each block, including those for the container. Setting the failure distributions for modes <math>A</math> and <math>C</math> is simple. The more difficult part is setting the properties for the container and the contained blocks. Based on the problem statement, the <math>B</math> modes are in a 2-out-of-3 load sharing redundancy. When all three are working (i.e. when no <math>B</math> mode has occurred), each block has an exponential distribution with <math>\mu=10,000</math>. If one <math>B</math> mode occurs, then the two surviving units have an exponential distribution with <math>\mu =5,000.</math>
 

Assume a Power Life-Stress relationship for the components. Then:
 

::<math>\begin{align}
{{\mu }_{1}}= & \frac{1}{KV_{1}^{n}} \ \\
{{\mu }_{2}}= & \frac{1}{KV_{2}^{n}} \
\end{align}</math>
 

Substituting <math>{{\mu }_{1}}=10,000</math> and <math>{{V}_{1}}=1</math> in <math>{{\mu }_{1}}= \frac{1}{KV_{1}^{n}}</math> and casting it in terms of <math>K</math> yields:
 

::<math>\begin{align}
10,000= & \frac{1}{K} \ \\
K = & \frac{1}{10,000}=0.0001
\end{align}</math>

 
Substituting <math>{{\mu }_{2}}=5,000</math> , <math>{{V}_{2}}=1.5</math> (because if one fails, then each survivor takes on an additional 0.5 units of load) and <math>10,000= \frac{1}{K}</math> for <math>K</math> in <math>{{\mu }_{2}}= \frac{1}{KV_{2}^{n}}</math> yields:

 
::<math>\begin{align}
5,000= & \frac{1}{0.0001\cdot {{(1.5)}^{n}}} \\
0.5= & {{(1.5)}^{-n}} \\
\ln (0.5)= & -n\ln (1.5) \\
n = & 1.7095
\end{align}</math>
 

This also could have been computed in ALTA, as shown in Figure "Calculation performed in ALTA".

The next step is to set the weight proportionality factor. This factor defines the portion of the load that the particular item carries while operating, as well as the load that shifts to the remaining units upon failure of the item. To illustrate, assume three units (1, 2 and 3) are in a load sharing container with weight proportionality factors of 1, 2 and 3 respectively (and a 1-out-of-3 requirement).
 

:• Unit 1 carries <math>\left( \tfrac{1}{1+2+3} \right)=0.166</math> or 16.6% of the total load.
:• Unit 2 carries <math>\left( \tfrac{2}{1+2+3} \right)=0.333</math> or 33.3% of the total load.
:• Unit 3 carries <math>\left( \tfrac{3}{1+2+3} \right)=0.50</math> or 50% of the total load.
 

The actual load on each unit then becomes the product of the entire load defined for the container times the portion carried by that unit. For example, if the container load is 100 lbs, then the portion assigned to Unit 1 will be <math>100\cdot 0.166=16.6</math> lbs.
 

In the current example, all units share the same load and thus have equal weight proportionality factors. Because these factors are relative, if the same number is used for all three items then the results will be the same. Thus, weight proportional factor is set equal to 1 for each item.
 

[[Image:5.20.PNG|thumb|center|500px|<div align="center"> Calculation performed in ALTA.</div>]]
 

 
[[Image:Fig 5.23_2.PNG|thumb|center|400px|<div align="center"> Defining Weight Proportional Factor. </div>]]
 

The last properties that need to be defined are the total load and the 2-out-of-3 redundancy. The total load is dependent on how the parameters were computed. In this case, total load was assumed to be 3 when the parameters were computed (i.e. the load per item was 1 when all worked and 1.5 when two worked). This is defined at the container level, as shown in Figure "Defining Weight Proportional Factor".
When all of the parameters have been specified in BlockSim, the reliability at 1,000 hours can be determined. From the Analytical QCP, this is found to be 93.87%.

=Standby Components=

In the previous section, the case of a system with load sharing components was presented. This is a form of redundancy with dependent components. That is, the failure of one component affects the failure of the other(s). This section presents another form of redundancy: standby redundancy. In standby redundancy the redundant components are set to be under a lighter load condition (or no load) while not needed and under the operating load when they are activated.
 

In standby redundancy the components are set to have two states: an active state and a standby state. Components in standby redundancy have two failure distributions, one for each state. When in the standby state, they have a quiescent (or dormant) failure distribution and when operating, they have an active failure distribution.
 

In the case that both quiescent and active failure distributions are the same, the units are in a simple parallel configuration (also called a hot standby configuration). When the rate of failure of the standby component is lower in quiescent mode than in active mode, that is called a warm standby configuration. When the rate of failure of the standby component is zero in quiescent mode (i.e. the component cannot fail when in standby), that is called a cold standby configuration.

===Simple Standby Configuration===

Consider two components in a standby configuration. Component 1 is the active component with a Weibull failure distribution with parameters <math>\beta </math> = 1.5 and <math>\eta </math> = 1,000. Component 2 is the standby component. When Component 2 is operating, it also has a Weibull failure distribution with <math>\beta </math> = 1.5 and <math>\eta </math> = 1,000. Furthermore, assume the following cases for the quiescent distribution.
 
:• Case 1: The quiescent distribution is the same as the active distribution (hot standby). 
:• Case 2: The quiescent distribution is a Weibull distribution with <math>\beta </math> = 1.5 and <math>\eta </math> = 2000 (warm standby). 
:• Case 3: The component cannot fail in quiescent mode (cold standby). 
 
In this case, the reliability of the system at some time, <math>t</math> , can be obtained using the following equation:
 

::<math>R(t)={{R}_{1}}(t)+\underset{0}{\overset{t}{\mathop \int }}\,{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x)\cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}dx \ </math>
 

Where:
 
:• <math>{{R}_{1}}</math> is the reliability of the active component. 
:• <math>{{f}_{1}}</math> is the <math>pdf</math> of the active component. 
:• <math>{{R}_{2;SB}}</math> is the reliability of the standby component when in standby mode (quiescent reliability). 
:• <math>{{R}_{2;A}}</math> is the reliability of the standby component when in active mode. 
:• <math>{{t}_{e}}</math> is the equivalent operating time for the standby unit if it had been operating at an active mode, such that: 

 
<math>{{R}_{2;SB}}(x)={{R}_{2;A}}({{t}_{e}})</math> 
 
The second equation above can be solved for <math>{{t}_{e}}</math> and substituted into the first equation above.
Figure "Standby container" illustrates the example as entered in BlockSim using a standby container.
 
 
[[Image:5.24.gif|thumb|center|400px|<div align="center"> Standby container.</div>]]

 

The active and standby blocks are within a container, which is used to specify standby redundancy. Since the standby component has two distributions (active and quiescent), the Block Properties window of the standby block has two pages for specifying each one. Figures "Defining the active failure distribution" and "Defining the quiescent failure distribution" illustrate these pages.

 

The system reliability results for 1000 hours are given in the following table:
 
[[Image:5-24.png|thumb|center|400px|]]
 

Note that even though the <math>\beta </math> value for the quiescent distribution is the same as in the active distribution, it is possible that the two can be different. That is, the failure modes present during the quiescent mode could be different from the modes present during the active mode. In that sense, the two distribution types can be different as well (e.g. lognormal when quiescent and Weibull when active).
 

In many cases when considering standby systems, a switching device may also be present that switches from the failed active component to the standby component. The reliability of the switch can also be incorporated into
<math>R(t)={{R}_{1}}(t)+\underset{0}{\overset{t}{\mathop \int }}\,{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x)\cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}dx \ </math>

as presented in the next section.
 

BlockSim's System Reliability Equation window returns a single token for the reliability of units in a standby configuration. This is the same as the load sharing case presented in the previous section.
 

[[Image: Fig 5.25.PNG|thumb|center|400px|<div align="center"> Defining the active failure distribution </div>]]
 
 

[[Image:Fig 5.26.PNG|thumb|center|400px|<div align="center"> Defining the quiescent failure distribution </div>]]
 

===Reliability of Standby Systems with a Switching Device===

In many cases when dealing with standby systems, a switching device is present that will switch to the standby component when the active component fails. Therefore, the failure properties of the switch must also be included in the analysis.
 

[[Image:BS5.26.2.png|thumb|center|300px|]]
 

In most cases when the reliability of a switch is to be included in the analysis, two probabilities can be considered. The first and most common one is the probability of the switch performing the action (i.e. switching) when requested to do so. This is called Switch Probability per Request in BlockSim and is expressed as a static probability (e.g. 90%). The second probability is the quiescent reliability of the switch. This is the reliability of the switch as it ages (e.g. the switch might wear out with age due to corrosion, material degradation, etc.). Thus it is possible for the switch to fail before the active component fails. However, a switch failure does not cause the system to fail, but rather causes the system to fail only if the switch is needed and the switch has failed. For example, if the active component does not fail until the mission end time and the switch fails, then the system does not fail. However, if the active component fails and the switch has also failed, then the system cannot be switched to the standby component and it therefore fails.
 

In analyzing standby components with a switching device, either or both failure probabilities (during the switching or while waiting to switch) can be considered for the switch, since each probability can represent different failure modes. For example, the switch probability per request may represent software-related issues or the probability of detecting the failure of an active component, and the quiescent probability may represent wear-out type failures of the switch.
 

To illustrate the formulation, consider the previous example that assumes perfect switching. To examine the effects of including an imperfect switch, assume that when the active component fails there is a 90% probability that the switch will switch from the active component to the standby component. In addition, assume that the switch can also fail due to a wear-out failure mode described by a Weibull distribution with <math>\beta </math> = 1.7 and <math>\eta </math> = 5000.
 

Therefore, the reliability of the system at some time, <math>t</math> , is given by the following equation.

::<math>\begin{align}
R(t)= & {{R}_{1}}(t) \\
& +\underset{0}{\overset{t}{\mathop \int }}\,\{{{f}_{1}}(x)\cdot {{R}_{2;SB}}(x) \\
& \cdot \frac{{{R}_{2;A}}({{t}_{e}}+t-x)}{{{R}_{2;A}}({{t}_{e}})}\cdot {{R}_{SW;Q}}(x)\cdot {{R}_{SW;REQ}}(x)\}dx
\end{align}</math>

Where:
 
:• <math>{{R}_{1}}</math> is the reliability of the active component. 
:• <math>{{f}_{1}}</math> is the <math>pdf</math> of the active component. 
:• <math>{{R}_{2;SB}}</math> is the reliability of the standby component when in standby mode (quiescent reliability). 
:• <math>{{R}_{2;A}}</math> is the reliability of the standby component when in active mode. 
:• <math>{{R}_{SW;Q}}</math> is the quiescent reliability of the switch. 
:• <math>{{R}_{SW;REQ}}</math> is the switch probability per request. 
:• <math>{{t}_{e}}</math> is the equivalent operating time for the standby unit if it had been operating at an active mode. 
This problem can be solved in BlockSim by including these probabilities in the container's properties, as shown in Figures "Standby container (switch) failure probabilities while attempting to switch" and "Standby container (switch) failure distribution while waiting to switch". In BlockSim, the standby container is acting as the switch.

 

[[Image:Fig 5.28_2.PNG|thumb|center|400px|<div align="center"> Standby container (switch) failure probabilities while attempting to switch </div> ]]
 

 

[[Image:Fig 5.27.PNG|thumb|center|400px|<div align="center"> Standby container (switch) failure distribution while waiting to switch </div>]]
 

 

Note that there are additional properties that can be specified in BlockSim for a switch, such as Switch Restart Probability, No. of Restarts and Switch Delay Time. In many applications, the switch is re-tested (or re-cycled) if it fails to switch the first time. In these cases, it might be possible that it switches in the second or third, or <math>{{n}^{th}}</math> attempt.
 
The Switch Restart Probability specifies each additional attempt's probability of successfully switching and the Finite Restarts specifies the total number of attempts. Note that the Switch Restart Probability specifies the probability of success of each trial (or attempt). The probability of success of <math>n</math> consecutive trials is calculated by BlockSim using the binomial distribution and this probability is then incorporated into equation above. The Switch Delay Time property is related to repairable systems and is considered in BlockSim only when using simulation. When using the analytical solution (i.e. for a non-repairable system), this property is ignored.
 

Solving the analytical solution (as given by Equation above), the following results are obtained.
 

[[Image:5-30.png|thumb|center|400px|]]
 

From the table above, it can be seen that the presence of a switching device has a significant effect on the reliability of a standby system. It is therefore important when modeling standby redundancy to incorporate the switching device reliability properties. It should be noted that this methodology is not the same as treating the switching device as another series component with the standby subsystem. This would be valid only if the failure of the switch resulted in the failure of system (e.g. switch failing open). In equation above, the Switch Probability per Request and quiescent probability are present only in the second term of the equation. Treating these two failure modes as a series configuration with the standby subsystem would imply that they are also present when the active component is functioning (i.e. first term of equation above). This is invalid and would result in the underestimation of the reliability of the system. In other words, these two failure modes become significant only when the active component fails.
 

As an example, and if we consider the warm standby case, the reliability of the system without the switch is 70.57% at 1000 hours. If the system was modeled so that the switching device was in series with the warm standby subsystem, the result would have been:

 

::<math>\begin{align}
{{R}_{S}}(1000)= & {{R}_{Standby}}(1000)\cdot {{R}_{sw,Q(1000)}}\cdot {{R}_{sw,req}} \\
= & 0.7057\cdot 0.9372\cdot 0.9 \\
= & 0.5952
\end{align}</math>

 

In the case where a switch failure mode causes the standby subsystem to fail, then this mode can be modeled as an individual block in series with the standby subsystem.

===Example 6===
Consider a car with four new tires and a full-size spare. Assume the following failure characteristics:
 
:• The tires follow a Weibull distribution with a <math>\beta=4</math> and an <math>\eta =</math> 40,000 miles while on the car due to wear. 
:• The tires also have a probability of failing due to puncture or other causes. For this, assume a constant rate for this occurrence with a probability of 1 every 50,000 miles. 
:• When not on the car (i.e. is a spare), a tire's probability of failing also has a Weibull distribution with a <math>\beta =</math> 2 and <math>\eta =</math> 120,000 miles. 
 
Assume a mission of 1,000 miles. If a tire fails during this trip, it will be replaced with the spare. However, the spare will not be repaired during the trip. In other words, the trip will continue with the spare on the car and if the spare fails the system will fail. Determine the probability of system failure.
====Solution to Example 6====
Active failure distribution for tires:
 
:• Due to wear-out, Weibull <math>\beta =4</math> and <math>\eta =40,000</math> miles. 
:• Due to random puncture, exponential <math>\mu =50,000.</math> 
:• The quiescent failure distribution is a Weibull distribution with <math>\beta =2</math> and <math>\eta =120,000</math> miles. 
 
The block diagram for each tire has two blocks in series, one block representing the wear-out mode and the other the random puncture mode, as shown next:
 
 
[[Image:small5.gif|thumb|center|400px|]]
 

There are five tires, four active and one standby (represented in the diagram by a standby container with a 4-out-of-5 requirement), as shown next:
 
[[Image:BStirecontainer.png|thumb|center|400px|]]

For the standby Wear block, set the active failure and the quiescent distributions, but for the Puncture block, set only the active puncture distribution (because the tire cannot fail due to puncture while stored). Using BlockSim, the probability of system failure is found to be 0.003 or 0.3%.
 

=Note Regarding Numerical Integration Solutions=

Load sharing and standby solutions in BlockSim are performed using numerical integration routines. As with any numerical analysis routine, the solution error depends on the number of iterations performed, the step size chosen and related factors, plus the behavior of the underlying function. By default, BlockSim uses a certain set of preset factors. In general, these defaults are sufficient for most problems. If a higher precision or verification of the precision for a specific problem is required, BlockSim's preset options can be modified and/or the integration error can be assessed using the Integration Parameters... option for each container. For more details, you can refer to the documentation on the Algorithm Setup window in the BlockSim 8 User's help files.

RBDs and Analytical System Reliability

2012-03-19T22:50:59Z

Dingzhou Cao: /* Symbolic Solutions in BlockSim */

{{Template:bsbook|4}}

 
An overall system reliability prediction can be made by looking at the reliabilities of the components that make up the whole system or product. In this chapter, we will examine the methods of performing such calculations. The reliability-wise configuration of components must be determined beforehand. For this reason, we will first look at different component/subsystem configurations, also known as structural properties ([[Appendix_B:_References | Leemis [17]]]). Unless explicitly stated, the components will be assumed to be statistically independent.
=Component Configurations =
 
In order to construct a reliability block diagram, the reliability-wise configuration of the components must be determined. Consequently, the analysis method used for computing the reliability of a system will also depend on the reliability-wise configuration of the components/subsystems. That configuration can be as simple as units arranged in a pure series or parallel configuration. There can also be systems of combined series/parallel configurations or complex systems that cannot be decomposed into groups of series and parallel configurations. The configuration types considered in this reference include:
 
:• Series configuration. 
:• Simple parallel configuration. 
:• Combined (series and parallel) configuration. 
:• Complex configuration. 
:• ''k''-out-of-''n'' parallel configuration. 
:• Configuration with a load sharing container (presented in Chapter 5). 
:• Configuration with a standby container (presented in Chapter 5). 
:• Configuration with inherited subdiagrams. 
:• Configuration with multi blocks. 
:• Configuration with mirrored blocks. 
 
Each of these configurations will be presented, along with analysis methods, in the sections that follow.
 

[[Category:Completed Theoretical Review]]

= Series Systems =

[[Image:BS4im1.png|center|400px|]]
 
In a series configuration, a failure of any component results in the failure of the entire system. In most cases, when considering complete systems at their basic subsystem level, it is found that these are arranged reliability-wise in a series configuration. For example, a personal computer may consist of four basic subsystems: the motherboard, the hard drive, the power supply and the processor. These are reliability-wise in series and a failure of any of these subsystems will cause a system failure. In other words, all of the units in a series system must succeed for the system to succeed.
 
The reliability of the system is the probability that unit 1 succeeds and unit 2 succeeds and all of the other units in the system succeed. So all <math>n</math> units must succeed for the system to succeed. The reliability of the system is then given by:
 

::<math>\begin{align}
R_S = & P({{X}_{1}}\cap {{X}_{2}}\cap ...\cap {{X}_{n}}) \\
= & P({{X}_{1}})P({{X}_{2}}|{{X}_{1}})P({{X}_{3}}|{{X}_{1}}{{X}_{2}}) \cdot\cdot\cdot P({{X}_{n}}|{{X}_{1}}{{X}_{2}}...{{X}_{n-1}})
\end{align}</math>
 
Where:
 
:• <math>{{R}_{s}}</math> = reliability of the system. 
:• <math>{{X}_{i}}</math> = event of unit <math>i</math> being operational. 
:• <math>P({{X}_{i}})</math> = probability that unit <math>i</math> is operational. 
 
In the case where the failure of a component affects the failure rates of other components (i.e. the life distribution characteristics of the other components change when one component fails), then the conditional probabilities in equation above must be considered.
 
However, in the case of independent components, equation above becomes:
 
::<math>{{R}_{s}}=P({{X}_{1}})P({{X}_{2}})...P({{X}_{n}}) </math>
 
:Or:
 
::<math>{{R}_{s}}=\underset{i=1}{\overset{n}{\mathop \prod }}\,P({{X}_{i}})</math>
 
Or, in terms of individual component reliability:
 
::<math>{{R}_{s}}=\underset{i=1}{\overset{n}{\mathop \prod }}\,{{R}_{i}} \ (eqn 2)</math>
 
In other words, for a pure series system, the system reliability is equal to the product of the reliabilities of its constituent components.
{{Example:Analytical-RBD}}

===Effect of Component Reliability in a Series System===
 
[[Image:Chp4image2.png|center|400px|]]
 
In a series configuration, the component with the least reliability has the biggest effect on the system's reliability. There is a saying that a chain is only as strong as its weakest link. This is a good example of the effect of a component in a series system. In a chain, all the rings are in series and if any of the rings break, the system fails. In addition, the weakest link in the chain is the one that will break first. The weakest link dictates the strength of the chain in the same way that the weakest component/subsystem dictates the reliability of a series system. As a result, the reliability of a series system is always less than the reliability of the least reliable component.

===Example 2===
Consider three components arranged reliability-wise in series, where <math>{{R}_{1}}=70%</math> , <math>{{R}_{2}}=80%</math> and <math>{{R}_{3}}=90%</math> (for a given time).
 
In Table "System reliabilities for combinations of component reliabilities", we can examine the effect of each component's reliability on the overall system reliability. The first row of the table shows the given reliability for each component and the corresponding system reliability for these values. In the second row, the reliability of Component 1 is increased by a value of 10% while keeping the reliabilities of the other two components constant. Similarly, by increasing the reliabilities of Components 2 and 3 in rows 3 and 4 by a value of 10%, while keeping the reliabilities of the other components at the given values, we can observe the effect of each component's reliability on the overall system reliability. It is clear that the highest value for the system's reliability was achieved when the reliability of Component 1, which is the least reliable component, was increased by a value of 10%.
 
[[Image:BS4.T1.png|center|500px|System reliabilities for combinations of component reliabilities.]]
 
This conclusion can also be illustrated graphically, as shown in Figure "Effect of component reliability on the overall system reliability
".
 
Note the slight difference in the slopes of the three lines in Figure "Effect of component reliability on the overall system reliability
". The difference in these slopes represents the difference in the effect of each of the components on the overall system reliability. In other words, the system reliability's rate of change with respect to each component's change in reliability is different. This observation will be explored further when the importance measures of components are considered in later chapters. The rate of change of the system's reliability with respect to each of the components is plotted in Figure "Rate of change of series system reliability when increasing the reliability of each component". It can be seen in Figure "Effect of component reliability on the overall system reliability
" that Component 1 has the steepest slope, which indicates that an increase in the reliability of Component 1 will result in a higher increase in the reliability of the system. In other words, Component 1 has a higher ''reliability importance''.
 
[[Image:BS4.1.png|center|500px|Effect of component reliability on the overall system reliability]]
 
[[Image:BS4.2.png|center|500px|Rate of change of series system reliability when increasing the reliability of each component]]
 
 

===Effect of Number of Components in a Series System===
The number of components is another concern in systems with components connected reliability-wise in series. As the number of components connected in series increases, the system's reliability decreases.
 
Figure "Reliability of a system with n statistically independent and identical components arraged reliability-wise in series" illustrates the effect of the number of components arranged reliability-wise in series on the system's reliability for different component reliability values. This figure also demonstrates the dramatic effect that the number of components has on the system's reliability, particularly when the component reliability is low. In other words, in order to achieve a high system reliability, the component reliability must be high also, especially for systems with many components arranged reliability-wise in series.

===Example 3===
Consider a system that consists of a single component. The reliability of the component is 95%, thus the reliability of the system is 95%. What would the reliability of the system be if there were more than one component (with the same individual reliability) in series? Table "System reliability as a function of the number of components" shows the effect on the system's reliability by adding consecutive components (with the same reliability) in series. Figure "Effect of the number of component in a series configuration for two different casese" illustrates the same concept graphically for components with 90% and 95% reliability.
 
[[Image:Table 4.2.png|center|400px|System reliability as a function of the number of components]]

[[Image:BS4.3.png|center|500px|Reliability of a system with n statistically independent and identical components arraged reliability-wise in series]]
 
 
[[Image:BS4.4.png|center|500px|Effect of the number of component in a series configuration for two different casese]]
 
 

= Simple Parallel Systems =

[[Image:chp4image6.png|center|300px|Simple parallel system]]
 
In a simple parallel system, as shown in Figure "Simple parallel system", at least one of the units must succeed for the system to succeed. Units in parallel are also referred to as redundant units. Redundancy is a very important aspect of system design and reliability in that adding redundancy is one of several methods of improving system reliability. It is widely used in the aerospace industry and generally used in mission critical systems. Other example applications include the RAID computer hard drive systems, brake systems and support cables in bridges.
The probability of failure, or unreliability, for a system with <math>n</math> statistically independent parallel components is the probability that unit 1 fails and unit 2 fails and all of the other units in the system fail. So in a parallel system, all <math>n</math> units must fail for the system to fail. Put another way, if unit 1 succeeds or unit 2 succeeds or any of the <math>n</math> units succeeds, then the system succeeds. The unreliability of the system is then given by:
 

::<math>\begin{align}
{{Q}_{s}}= & P({{X}_{1}}\cap {{X}_{2}}\cap ...\cap {{X}_{n}}) \\
= & P({{X}_{1}})P({{X}_{2}}|{{X}_{1}})P({{X}_{3}}|{{X}_{1}}{{X}_{2}})...P({{X}_{n}}|{{X}_{1}}{{X}_{2}}...{{X}_{n-1}}) \ (eqn 3)
\end{align}</math>
 
Where:
 
:• <math>{{Q}_{s}}</math> = unreliability of the system. 
:• <math>{{X}_{i}}</math> = event of failure of unit <math>i</math> . 
:• <math>P({{X}_{i}})</math> = probability of failure of unit <math>i</math>. 
 
In the case where the failure of a component affects the failure rates of other components, then the conditional probabilities in equation above must be considered.
 
However, in the case of independent components, equation above becomes:
 
::<math>{{Q}_{s}}=P({{X}_{1}})P({{X}_{2}})...P({{X}_{n}})</math>
 
:Or:
 
::<math>{{Q}_{s}}=\underset{i=1}{\overset{n}{\mathop \prod }}\,P({{X}_{i}})</math>
 
Or, in terms of component unreliability:
 
::<math>{{Q}_{s}}=\underset{i=1}{\overset{n}{\mathop \prod }}\,{{Q}_{i}}</math>
 
Observe the contrast with the series system, in which the system reliability was the product of the component reliabilities; whereas the parallel system has the overall system unreliability as the product of the component unreliabilities.
 
The reliability of the parallel system is then given by:
 
::<math>\begin{align}
{{R}_{s}}= & 1-Qs=1-({{Q}_{1}}\cdot {{Q}_{2}}\cdot ...\cdot {{Q}_{n}}) \\
= & 1-[(1-{{R}_{1}})\cdot (1-{{R}_{2}})\cdot ...\cdot (1-{{R}_{n}})] \\
= & 1-\underset{i=1}{\overset{n}{\mathop \prod }}\,(1-{{R}_{i}}) \ (eqn 5)
\end{align}</math>
===Example 4===
Consider a system consisting of three subsystems arranged reliability-wise in parallel. Subsystem 1 has a reliability of 99.5%, Subsystem 2 has a reliability of 98.7% and Subsystem 3 has a reliability of 97.3% for a mission of 100 hours. What is the overall reliability of the system for a 100-hour mission?
====Solution to Example 4====
Since the reliabilities of the subsystems are specified for 100 hours, the reliability of the system for a 100-hour mission is:
::<math>\begin{align}
{{R}_{s}}= & 1-(1-0.9950)\cdot(1-0.9870)\cdot(1-0.9730) \\
= & 1-0.000001755 \\
= & 0.999998245 \
\end{align}</math>

===Effect of Component Reliability in a Parallel Configuration===
When we examined a system of components in series, we found that the least reliable component has the biggest effect on the reliability of the system. However, the component with the highest reliability in a parallel configuration has the biggest effect on the system's reliability, since the most reliable component is the one that will most likely fail last. This is a very important property of the parallel configuration, specifically in the design and improvement of systems.
===Example 5===
Consider three components arranged reliability-wise in parallel with .. = 60%, <math>{{R}_{2}}</math> = 70% and <math>{{R}_{3}}</math> = 80% (for a given time). The corresponding reliability for the system is <math>{{R}_{s}}</math> = 97.6%. In Table "System reliability for combinations of component reliabilities", we can examine the effect of each component's reliability on the overall system reliability. The first row of the table shows the given reliability for each component and the corresponding system reliability for these values. In the second row, the reliability of Component 1 is increased by a value of 10% while keeping the reliabilities of the other two components constant. Similarly, by increasing the reliabilities of Components 2 and 3 in the third and fourth rows by a value of 10% while keeping the reliabilities of the other components at the given values, we can observe the effect of each component's reliability on the overall system reliability. It is clear that the highest value for the system's reliability was achieved when the reliability of Component 3, which is the most reliable component, was increased. Once again, this is the opposite of what was encountered with a pure series system.
 
 
[[Image:4.T3.png|center|500px|System reliability for combinations of component reliabilities]]
 
This conclusion can also be illustrated graphically, as shown in Figure "Rate of change of parallel system reliability when increasing the reliability of each component".

====Effect of Number of Components in Parallel====
In the case of the parallel configuration, the number of components has the opposite effect of the one observed for the series configuration. For a parallel configuration, as the number of components/subsystems increases, the system's reliability increases.
 
Figure "Reliability of a system with n statistically independent and identical components arranged reliability-wise in parallel" illustrates that a high system reliability can be achieved with low-reliability components, provided that there are a sufficient number of components in parallel. Note that Figure "Reliability of a system with n statistically independent and identical components arranged reliability-wise in parallel" is the mirror image of Figure "Reliability of a system with n statistically independent and identical components arraged reliability-wise in series", which presents the effect of the number of components in a series configuration.
 
[[Image:BS4.6.png|center|500px|Rate of change of parallel system reliability when increasing the reliability of each component]]
 
 
[[Image:BS4.7.png|center|500px|Reliability of a system with n statistically independent and identical components arranged reliability-wise in parallel]]
 
 

===Example 6===
Consider a system that consists of a single component. The reliability of the component is 60%, thus the reliability of the system is 60%. What would the reliability of the system be if the system were composed of two, four or six such components in parallel?
 
 
[[Image:4.T4.png|center|500px|System reliability as a function of the number of components]]
 
 
Clearly, the reliability of a system can be improved by adding redundancy. However, it must be noted that doing so is usually costly in terms of additional components, additional weight, volume, etc.
 
Reliability optimization and costs are covered in detail in Chapter [[Reliability_Importance_and_Optimized_Reliability_Allocation_(Analytical)|Component Reliability Importance]].
 
[[Image:BS4.8.png|center|500px|Effect of the number of components in a parallel configuration]]
 
 

= Combination of Series and Parallel =

While many smaller systems can be accurately represented by either a simple series or parallel configuration, there may be larger systems that involve both series and parallel configurations in the overall system. Such systems can be analyzed by calculating the reliabilities for the individual series and parallel sections and then combining them in the appropriate manner. Such a methodology is illustrated in the following example.
===Example 7===
Consider a system with three components. Units 1 and 2 are connected in series and Unit 3 is connected in parallel with the first two, as shown in the next figure.
 
[[Image:chp4image10.png|center|400px|]]
 
What is the reliability of the system if <math>{{R}_{1}}</math> = 99.5%, <math>{{R}_{2}}</math> = 98.7% and <math>{{R}_{3}}</math> = 97.3% at 100 hours?
====Solution to Example 7====
First, the reliability of the series segment consisting of Units 1 and 2 is calculated:
 
::<math>\begin{align}
{{R}_{1,2}}= & {{R}_{1}}\cdot {{R}_{2}} \\
{{R}_{1,2}}= & 0.9950\cdot 0.9870 \\
{{R}_{1,2}}= & 0.982065\text{ or }98.2065%
\end{align}</math>
 
The reliability of the overall system is then calculated by treating Units 1 and 2 as one unit with a reliability of 98.2065% connected in parallel with Unit 3. Therefore:
 
::<math>\begin{align}
{{R}_{s}}= & 1-[(1-0.982065)\cdot (1-0.973000)] \\
{{R}_{s}}= & 1-0.000484245 \\
{{R}_{s}}= & 0.999515755 \\
{{R}_{s}}= & 99.95%
\end{align}</math>
 

= k-out-of-n Parallel Configuration =
The <math>k</math> -out-of- <math>n</math> configuration is a special case of parallel redundancy. This type of configuration requires that at least <math>k</math> components succeed out of the total <math>n</math> parallel components for the system to succeed. For example, consider an airplane that has four engines. Furthermore, suppose that the design of the aircraft is such that at least two engines are required to function for the aircraft to remain airborne. This means that the engines are reliability-wise in a <math>k</math> -out-of- <math>n</math> configuration, where <math>k</math> = 2 and <math>n</math> = 4. More specifically, they are in a 2-out-of-4 configuration.
 
Even though we classified the <math>k</math> -out-of- <math>n</math> configuration as a special case of parallel redundancy, it can also be viewed as a general configuration type. As the number of units required to keep the system functioning approaches the total number of units in the system, the system's behavior tends towards that of a series system. If the number of units required is equal to the number of units in the system, it is a series system. In other words, a series system of statistically independent components is an <math>n</math> -out-of- <math>n</math> system and a parallel system of statistically independent components is a <math>1</math> -out-of- <math>n</math> system.
===Reliability of <math>k</math> -out-of- <math>n</math> Independent and Identical Components===
 
[[Image:chp4image11.png|center|500px|<div align="center">2-out-of-4 configuration </div>]]
 
The simplest case of components in a <math>k</math> -out-of- <math>n</math> configuration is when the components are independent and identical. In other words, all the components have the same failure distribution and whenever a failure occurs, the remaining components are not affected. In this case, the reliability of the system with such a configuration can be evaluated using the binomial distribution, or:
 
::<math>{{R}_{s}}(k,n,R)=\underset{r=k}{\overset{n}{\mathop \sum }}\,\left( \begin{matrix}
n \\
r \\
\end{matrix} \right){{R}^{r}}{{(1-R)}^{n-r}} \ </math>
 

Where:
 
:• <math>n</math> is the total number of units in parallel. 
:• <math>k</math> is the minimum number of units required for system success. 
:• <math>R</math> is the reliability of each unit. 

===Example 8===
Consider a system of six pumps of which at least four must function properly for system success. Each pump has an 85% reliability for the mission duration. What is the probability of success of the system for the same mission duration?
====Solution to Example 8====
Using <math>{{R}_{s}}(k,n,R)=\underset{r=k}{\overset{n}{\mathop \sum }}\,\left( \begin{matrix}
n \\
r \\
\end{matrix} \right){{R}^{r}}{{(1-R)}^{n-r}} \ </math> for <math>k=4</math> and <math>n=6</math> :
 
::<math>\begin{align}
{{R}_{s}}= & \underset{r=4}{\overset{6}{\mathop \sum }}\,\left( \begin{matrix}
6 \\
r \\
\end{matrix} \right){{0.85}^{r}}{{(1-0.85)}^{6-r}} \\
= & \left( \begin{matrix}
6 \\
4 \\
\end{matrix} \right){{0.85}^{4}}{{(1-0.85)}^{2}}+\left( \begin{matrix}
6 \\
5 \\
\end{matrix} \right){{0.85}^{5}}{{(1-0.85)}^{1}} \\
& +\left( \begin{matrix}
6 \\
6 \\
\end{matrix} \right){{0.85}^{6}}{{(1-0.85)}^{0}} \\
= & 0.1762+0.3993+0.3771 \\
= & 95.26%
\end{align}</math>
 
One can examine the effect of increasing the number of units required for system success while the total number of units remains constant (in this example, six units). In Figure "2-out-of-4 configuration", the reliability of the <math>k</math> -out-of-6 configuration was plotted versus different numbers of required units.
 
 
[[Image:4.T5.png|center|500px|Reliability for a k-out-of-6 system for different k values]]
 

 
Note that the system configuration becomes a simple parallel configuration for <math>k</math> = 1 and the system is a six-unit series configuration ( <math>{{(0.85)}^{6}}=</math> 0.377) for <math>k</math> = 6.
<math></math>
 
[[Image:BS4.10.png|center|500px|Reliability of a k-out-of-6 configuration for different k values]]
 
 

===Reliability of Nonidentical <math>k</math> -out-of- <math>n</math> Independent Components===
In the case where the <math>k</math> -out-of- <math>n</math> components are not identical, the reliability must be calculated in a different way. One approach, described in detail later in this chapter, is to use the event space method. In this method, all possible operational combinations are considered in order to obtain the system's reliability. The method is illustrated with the following example.
===Example 9===
Three hard drives in a computer system are configured reliability-wise in parallel. At least two of them must function in order for the computer to work properly. Each hard drive is of the same size and speed, but they are made by different manufacturers and have different reliabilities. The reliability of <math>HD\#1</math> is 0.9, <math>HD\#2</math> is 0.88 and <math>HD\#3</math> is 0.85, all at the same mission time.
====Solution to Example 9====
Since at least two hard drives must be functioning at all times, only one failure is allowed. This is a 2-out-of-3 configuration.
 
The following operational combinations are possible for system success:
#All 3 hard drives operate.
#<math>HD\#1</math> fails, while <math>HDs\,\#2</math> and <math>\#3</math> continue to operate.
#<math>HD\#2</math> fails, while <math>HDs\,\#1</math> and <math>\#3</math> continue to operate.
#<math>HD\#3</math> fails, while <math>HDs\,\#1</math> and <math>\#2</math> continue to operate.
 
The probability of success for the system (reliability) can now be expressed as:
 
::<math>\begin{align}
{{P}_{s}}= & {{R}_{1}}{{R}_{2}}{{R}_{3}}+(1-{{R}_{1}}){{R}_{2}}{{R}_{3}}+{{R}_{1}}(1-{{R}_{2}}){{R}_{3}} \\
& +{{R}_{1}}{{R}_{2}}(1-{{R}_{3}})
\end{align}</math>
 
This equation for the reliability of the system can be reduced to:
 
::<math>{{R}_{s}}={{R}_{1}}{{R}_{2}}+{{R}_{2}}{{R}_{3}}+{{R}_{1}}{{R}_{3}}-2{{R}_{1}}{{R}_{2}}{{R}_{3}}</math>
 
:Or:
 
::<math>{{R}_{s}}=95.86%</math>
 
If all three hard drives had the same reliability, <math>R</math> , then the equation for the reliability of the system could be further reduced to:
 
::<math>{{R}_{s}}=3{{R}^{2}}-2{{R}^{3}}</math>
 
Or, using the binomial approach:
 
::<math>\begin{align}
{{R}_{s}}= & \underset{r=2}{\overset{3}{\mathop \sum }}\,\left( \begin{matrix}
3 \\
r \\
\end{matrix} \right){{R}^{r}}{{(1-R)}^{3-r}} \\
= & \left( \begin{matrix}
3 \\
2 \\
\end{matrix} \right){{R}^{2}}(1-R)+\left( \begin{matrix}
3 \\
3 \\
\end{matrix} \right){{R}^{3}}{{(1-R)}^{0}} \\
= & 3\cdot {{R}^{2}}(1-R)+{{R}^{3}} \\
= & 3{{R}^{2}}-2{{R}^{3}}
\end{align}</math>
 
The example can be repeated using BlockSim. The following graphic demonstrates the RBD for the system.
 
<math></math>
 
 
[[Image:Fig_harddrive.PNG|center|400px|]]
 

The RBD is analyzed and the system reliability equation is returned. Figure "System equation results for Example 9" shows the equation returned by BlockSim.
 
[[Image:BS4.11.png|thumb|center|400px|System equation results for Example 9]]
 
Using the Analytical Quick Calculation Pad, the reliability can be calculated to be 0.9586. Figure "Reliability results for Example 9" shows the returned result.
 
Note that you are not required to enter a mission end time for this system into the Analytical QCP because all of the components are static and thus the reliability results are independent of time.
[[Image:Fig 4.12.PNG|thumb|center|500px|<div align="center">Reliability results for Example 9 </div>]]

===Example 10===
Consider the four-engine aircraft discussed previously. If we were to change the problem statement to two out of four engines are required, however no two engines on the same side may fail, then the block diagram would change to the configuration shown in Figure "Block diagram for Example 10".
 
<math></math>
 
Note that this is the same as having two engines in parallel on each wing and then putting the two wings in series.
 

 
[[Image:4.13.png|center|500px|<div align="center">Block diagram for Example 10 </div>]]
 
 

=Consecutive <math>k</math>-out-of-<math>n</math> : <math>F</math> Redundancy=
There are other multiple redundancy types and multiple industry terms. One such example is what is referred to as a consecutive <math>k</math>-out-of-<math>n</math> : <math>F</math> system. To illustrate this configuration type, consider a telecommunications system that consists of a transmitter and receiver with six relay stations to connect them. The relays are situated so that the signal originating from one station can be picked up by the next two stations down the line. For example, a signal from the transmitter can be received by Relay 1 and Relay 2, a signal from Relay 1 can be received by Relay 2 and Relay 3, and so forth. Thus, this arrangement would require two consecutive relays to fail for the system to fail. A diagram of this configuration is shown in Figure "RBD for the consecutive ''k''-out-of-''n'': ''F'' system".
 
<math></math>
 
This type of a configuration is also referred to as a complex system. Complex systems are discussed in the next section.
[[Image:4.14.png|center|600px|<div align="center">RBD for the consecutive ''k''-out-of-''n'': ''F'' system </div>]]

= Complex Systems =
In many cases, it is not easy to recognize which components are in series and which are in parallel in a complex system. The network shown in Figure "Example of a complex system" is a good example of such a complex system.
 
<math></math>
 
 
[[Image:4.15.png|center|500px|<div align="center">Example of a complex system </div>]]
 
The system in Figure "Example of a complex system" cannot be broken down into a group of series and parallel systems. This is primarily due to the fact that component <math>C</math> has two paths leading away from it, whereas <math>B</math> and <math>D</math> have only one. Several methods exist for obtaining the reliability of a complex system including:
 
:• The decomposition method. 
:• The event space method. 
:• The path-tracing method. 
===Decomposition Method===
The decomposition method is an application of the law of total probability. It involves choosing a "key" component and then calculating the reliability of the system twice: once as if the key component failed ( <math>R=0</math> ) and once as if the key component succeeded <math>(R=1)</math> . These two probabilities are then combined to obtain the reliability of the system, since at any given time the key component will be failed or operating. Using probability theory, the equation is:
 
::<math>\begin{align}
{{R}_{s}}= & P(s\cap A)+P(s\cap \overline{A}) \\
= & P(s|A)P(A)+P(s|\overline{A})P(\overline{A})
\end{align}</math>
====Illustration of the Decomposition Method====
Consider three units in series.
 
:• <math>A</math> is the event of Unit 1 success. 
:• <math>B</math> is the event of Unit 2 success. 
:• <math>C</math> is the event of Unit 3 success. 
:• <math>s</math> is the event of system success. 
 
First, select a "key" component for the system. Selecting Unit 1, the probability of success of the system is:
 
::<math>{{R}_{s}}=P(s|A)P(A)+P(s|\overline{A})P(\overline{A})</math>
 
If Unit 1 is good, then:
 
::<math>P(s|A)={{R}_{2}}{{R}_{3}}</math>
 
That is, if Unit 1 is operating, the probability of the success of the system is the probability of Units 2 and 3 succeeding.
 
If Unit 1 fails, then:
 
::<math>P(s|\overline{A})=0</math>
 
That is, if Unit 1 is not operating, the system has failed since a series system requires all of the components to be operating for the system to operate.
Thus the reliability of the system is:
 
::<math>{{R}_{s}}={{R}_{2}}{{R}_{3}}P(A)={{R}_{1}}{{R}_{2}}{{R}_{3}}</math>
 

====Another Illustration of the Decomposition Method====
Consider the following system:
 
[[Image:chp4image14.png|center|300px|]]
 
:• <math>A</math> is the event of Unit 1 success.
:• <math>B</math> is the event of Unit 2 success.
:• <math>C</math> is the event of Unit 3 success.
:• <math>s</math> is the event of system success.
 
Selecting Unit 3 as the `` key'' component, the system reliability is:
 
::<math>{{R}_{s}}=P(s|C)P(C)+P(s|\overline{C})P(\overline{C})</math>
 
If Unit 3 survives, then:
 
::<math>P(s|C)=1</math>
 
That is, since Unit 3 represents half of the parallel section of the system, then as long as it is operating, the entire system operates.
 
If Unit 3 fails, then the system is reduced to:
 
[[Image:chp4image15.png|center|400px|]]
 

::<math>P(s|\overline{C})={{R}_{1}}{{R}_{2}}</math>
 
The reliability of the system is given by:
 
::<math>{{R}_{s}}=P(C)+{{R}_{1}}{{R}_{2}}P(\overline{C})={{R}_{3}}+{{R}_{1}}{{R}_{2}}(1-{{R}_{3}})</math>
 
:Or:
 
::<math>{{R}_{s}}={{R}_{3}}+{{R}_{1}}{{R}_{2}}-{{R}_{1}}{{R}_{2}}{{R}_{3}}</math>

===Event Space Method===
The event space method is an application of the mutually exclusive events axiom. All mutually exclusive events are determined and those that result in system success are considered. The reliability of the system is simply the probability of the union of all mutually exclusive events that yield a system success. Similarly, the unreliability is the probability of the union of all mutually exclusive events that yield a system failure. This is illustrated in the following example.
 
====Illustration of the Event Space Method====
 
Consider the following system, with reliabilities R1, R2 and R3 for a given time.
 
[[Image:chp4image14.png|center|400px|]]
:• <math>A</math> is the event of Unit 1 success. 
:• <math>B</math> is the event of Unit 2 success. 
:• <math>C</math> is the event of Unit 3 success. 
 
The mutually exclusive system events are:
 
::<math>\begin{align}
X1= & ABC-\text{all units succeed}\text{.} \\
X2= & \overline{A}BC-\text{only Unit 1 fails}\text{.} \\
X3= & A\overline{B}C-\text{only Unit 2 fails}\text{.} \\
X4= & AB\overline{C}-\text{only Unit 3 fails}\text{.} \\
X5= & \overline{AB}C-\text{Units 1 and 2 fail}\text{.} \\
X6= & \overline{A}B\overline{C}-\text{Units 1 and 3 fail}\text{.} \\
X7= & A\overline{BC}-\text{Units 2 and 3 fail}\text{.} \\
X8= & \overline{ABC}-\text{all units fail}\text{.}
\end{align}</math>

 
System events <math>{{X}_{6}}</math> , <math>{{X}_{7}}</math> and <math>{{X}_{8}}</math> result in system failure. Thus the probability of failure of the system is:
 
::<math>{{P}_{f}}=P({{X}_{6}}\cup {{X}_{7}}\cup {{X}_{8}})</math>
 
Since events <math>{{X}_{6}}</math> , <math>{{X}_{7}}</math> and <math>{{X}_{8}}</math> are mutually exclusive, then:
 
::<math>{{P}_{f}}=P({{X}_{6}})+P({{X}_{7}})+P({{X}_{8}})</math>
 
:And:
 
::<math>\begin{align}
P({{X}_{6}})= & P(\overline{A}B\overline{C})=(1-{{R}_{1}})({{R}_{2}})(1-{{R}_{3}}) \\
P({{X}_{7}})= & P(A\overline{B}\overline{C})=({{R}_{1}})(1-{{R}_{2}})(1-{{R}_{3}}) \\
P({{X}_{8}})= & P(\overline{A}\overline{B}\overline{C})=(1-{{R}_{1}})(1-{{R}_{2}})(1-{{R}_{3}})
\end{align}</math>
 
Combining terms yields:
 

::<math>{{P}_{f}}=1-{{R}_{1}}{{R}_{2}}-{{R}_{3}}+{{R}_{1}}{{R}_{2}}{{R}_{3}}</math>
 
:Since:
 
::<math>{{R}_{s}}=1-{{P}_{f}}</math>
 
:Then:
 
::<math>{{R}_{s}}={{R}_{1}}{{R}_{2}}+{{R}_{3}}-{{R}_{1}}{{R}_{2}}{{R}_{3}}</math>
 
This is of course the same result as the one obtained previously using the decomposition method.
If <math>{{R}_{1}}</math> = 99.5%, <math>{{R}_{2}}</math> = 98.7% and <math>{{R}_{3}}</math> = 97.3%, then:
 
::<math>\begin{align}
{{R}_{s}}= & 0.995\cdot 0.987+0.973-0.995\cdot 0.987\cdot 0.973 \\
= & 0.999515755
\end{align}</math>
 
:Or:
 
::<math>{{R}_{s}}=99.95%</math>
 

===Path-Tracing Method===
With the path-tracing method, every path from a starting point to an ending point is considered. Since system success involves having at least one path available from one end of the RBD to the other, as long as at least one path from the beginning to the end of the path is available, then the system has not failed. One could consider the RBD to be a plumbing schematic. If a component in the system fails, the "water" can no longer flow through it. As long as there is at least one path for the "water" to flow from the start to the end of the system, the system is successful. This method involves identifying all of the paths the "water" could take and calculating the reliability of the path based on the components that lie along that path. The reliability of the system is simply the probability of the union of these paths. In order to maintain consistency of the analysis, starting and ending blocks for the system must be defined.

===Example 11===
Obtain the reliability equation of the following system.
 
[[Image:BS4ex11.png|center|400px|]]
 
====Solution to Example 11====
The successful paths for this system are:
 

::<math>{{X}_{1}}=ABD\text{ and }{{X}_{2}}=ACD</math>

 

The reliability of the system is simply the probability of the union of these paths:
 

::<math>{{R}_{s}}=P({{X}_{1}}\cup {{X}_{2}})</math>
 

::<math>\begin{align}
P({{X}_{1}}\cup {{X}_{2}})= & P({{X}_{1}})+P({{X}_{2}})-P({{X}_{1}}\cap {{X}_{2}}) \\
= & P(ABD)+P(ACD)-P(ABCD)
\end{align}</math>

 

:Thus:
 

::<math>{{R}_{s}}={{R}_{A}}{{R}_{B}}{{R}_{D}}+{{R}_{A}}{{R}_{C}}{{R}_{D}}-{{R}_{A}}{{R}_{B}}{{R}_{C}}{{R}_{D}}</math>
 

===Example 12===
Obtain the reliability equation of the following system.
 
[[Image:Chp4image14.png|center|400px|]]
 
 

====Solution to Example 12====
Assume starting and ending blocks that cannot fail, as shown next.
 
[[Image:BS.4ex12.2.png|center|500px|]]
 
The paths for this system are:
 
::<math>{{X}_{1}}=1,2\text{ and }{{X}_{2}}=3</math>

 
The probability of success of the system is given by:

::<math>\begin{align}
P({{X}_{1}}\cup {{X}_{2}})= & P({{X}_{1}})+P({{X}_{2}})-P({{X}_{1}}\cap {{X}_{2}}) \\
= & P(1,2)+P(3)-P(1,2,3)
\end{align}</math>
 

:Or:
 

::<math>{{R}_{s}}={{R}_{1}}{{R}_{2}}+{{R}_{3}}-{{R}_{1}}{{R}_{2}}{{R}_{3}}</math>
 

====Starting and Ending Blocks in BlockSim====
Note that BlockSim requires that all diagrams start from a single block and end on a single block. To meet this requirement for this example, we arbitrarily added a starting and an ending block, as shown in Figure "BlockSim representation of the RBD for Example 12". These blocks can be set to a cannot fail condition, or <math>R=1</math> , and thus not affect the outcome. However, when the analysis is performed in BlockSim, the returned equation will include terms for the non-failing blocks, as shown in Figure "BlockSim solution Example 12" and equation.

 

::<math>{{R}_{system}}=({{R}_{S}}\cdot {{R}_{E}}(-{{R}_{1}}\cdot {{R}_{2}}\cdot {{R}_{3}}+{{R}_{1}}\cdot {{R}_{2}}+{{R}_{3}})) \ </math>
 

Note that since <math>{{R}_{S}}={{R}_{E}}=1</math> , the system equation above, can be reduced to:
 

::<math>\begin{align}
{{R}_{system}}= & (1\cdot 1(-{{R}_{1}}\cdot {{R}_{2}}\cdot {{R}_{3}}+{{R}_{1}}\cdot {{R}_{2}}+{{R}_{3}})) \\
= & -{{R}_{1}}\cdot {{R}_{2}}\cdot {{R}_{3}}+{{R}_{1}}\cdot {{R}_{2}}+{{R}_{3}}
\end{align}</math>

 
This is equivalent to <math>{{R}_{s}}={{R}_{1}}{{R}_{2}}+{{R}_{3}}-{{R}_{1}}{{R}_{2}}{{R}_{3}}</math>. The reason that BlockSim includes all items regardless of whether they can fail or not is because BlockSim only recomputes the equation when the system structure has changed. What this means is that the user can alter the failure characteristics of an item without altering the diagram structure. For example, a block that was originally set not to fail can be re-set to a failure distribution and thus it would need to be used in subsequent analyses.
 
[[Image:4.16.png|center|400px|<div align="center">BlockSim representation of the RBD for Example 12 </div>]]
 
 
[[Image:Fig 4.17.PNG|thumb|center|400px|<div align="center">BlockSim solution Example 12 </div>]]
 

===Example 13===
For this example:
 
::<math>a)</math> Determine the reliability equation of the system shown in Figure "Diagram for Example 13" using the decomposition method.
 
::<math>b)</math> Determine the reliability equation of the same system using BlockSim.
====Solution to Example 13====
To obtain the solution:
 
::<math>a)</math> Choose A as the key component, then:
 

::<math>\begin{align}
{{R}_{s}}= & P(s|A)P(A)+P(s|\overline{A})P(\overline{A}) \\
P(s| A)= &{{R}_{B}}{{R}_{F}}\left[ 1-\left( 1-{{R}_{C}} \right)\left( 1-{{R}_{E}} \right) \right] \\
P(s| \overline{A})= &{{R}_{B}}{{R}_{D}}{{R}_{E}}{{R}_{F}} \\
{{R}_{s}}= & \left[ {{R}_{B}}{{R}_{F}}\left[ 1-\left( 1-{{R}_{C}} \right)\left( 1-{{R}_{E}} \right) \right] \right]{{R}_{A}}+\left[ {{R}_{B}}{{R}_{D}}{{R}_{E}}{{R}_{F}} \right](1-{{R}_{A}})
\end{align}</math>

 
::<math>b)</math> Using BlockSim:

 
::<math>{{R}_{s}}={{R}_{B}}\cdot {{R}_{F}}(-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}+{{R}_{A}}\cdot {{R}_{E}}+{{R}_{D}}\cdot {{R}_{E}})</math>
 
[[Image:chp4image21.png|center|400px|<div align="center">Diagram for Example 13. </div>]]
 
[[Image:Fig_4.19.PNG|thumb|center|400px|<div align="center">BlockSim solution for Example 13 </div>]]
 

= Difference Between Physical and Reliability-Wise Arrangement =
Reliability block diagrams are created in order to illustrate the way that components are arranged reliability-wise in a system. So far we have described possible structural properties of a system of components, such as series, parallel, etc. These structural properties, however, refer to the system's state of success or failure based on the states of its components. The physical structural arrangement, even though clearly related to the reliability-wise arrangement, is not necessarily identical to it.
===Example 14===
Consider the following circuit:
 
[[Image:BS3sigma.png|center|400px|]]
 
The equivalent resistance must always be less than <math>1.2\Omega </math> .
 
Draw the reliability block diagram for this circuit.
====Solution to Example 14====
First, let's consider the case where all three resistors operate:
 

::<math>\begin{align}
\frac{1}{{{r}_{eq}}}= & \frac{1}{{{r}_{1}}}+\frac{1}{{{r}_{2}}}+\frac{1}{{{r}_{3}}} \\
= & \frac{1}{3}+\frac{1}{3}+\frac{1}{3} \\
= & 1\Omega
\end{align}</math>

 
Thus, when all components operate, the equivalent resistance is <math>1\Omega </math> , which is less than the maximum resistance of <math>1.2\Omega </math> .
 

Next, consider the case where one of the resistors fails open. In this case, the resistance for the resistor is infinite and the equivalent resistance is:
 

::<math>\frac{1}{{{r}_{eq}}}=\frac{1}{\infty }+\frac{1}{3}+\frac{1}{3}=\frac{2}{3}</math>
 

:Thus:
 

::<math>{{r}_{eq}}=1.5\Omega >1.2\Omega \text{ - System failed}\text{.}</math>

 
If two resistors fail open (e.g. #1 and #2), the equivalent resistance is:
 

::<math>\frac{1}{{{r}_{eq}}}=\frac{1}{\infty }+\frac{1}{\infty }+\frac{1}{3}=\frac{1}{3}</math>

 
:Thus:
 

::<math>{{r}_{eq}}=3\Omega >1.2\Omega \text{ - System failed}\text{.}</math>

 
If all three resistors fail open:
 

::<math>{{r}_{eq}}=\infty >1.2\Omega \text{ - System failed}\text{.}</math>

 
Thus, if <math>{{r}_{1}}</math> , <math>{{r}_{2}}</math> , <math>{{r}_{3}}</math>, or any combination of the three fails, the system fails. Put another way, <math>{{r}_{1}}</math> and <math>{{r}_{2}}</math> and <math>{{r}_{3}}</math> must succeed in order for the system to succeed.
 
The RBD is:

[[Image:chp4image24.png|center|400px|]]
 
In this example it can be seen that even though the three components were physically arranged in parallel, their reliability-wise arrangement is in series.

= Configurations with Load Sharing and Standby Redundancy =
Units in load sharing redundancy exhibit different failure characteristics when one or more fail. In Figure "An RBD with a load sharing container", blocks 1, 2 and 3 are in a load sharing container in BlockSim and have their own failure characteristics. All three must fail for the container to fail. However, as individual items fail, the failure characteristics of the remaining units change since they now have to carry a higher load to compensate for the failed ones. The failure characteristics of each block in a load sharing container are defined using both a life distribution and a life-stress relationship that describe how the life distribution changes as the load changes.
 

Standby redundancy configurations consist of items that are inactive and available to be called into service when/if an active item fails (i.e. the items are on standby). Within BlockSim, a container block with other blocks inside is used to better achieve and streamline the representation and analysis of standby configurations. The container serves a dual purpose. The first is to clearly delineate and define the standby relationships between the active unit(s) and standby unit(s). The second is to serve as the manager of the switching process. For this purpose, the container can be defined with its own probability of successfully activating standby units when needed. Figure "An RBD with a standby container" includes a standby container with three items in standby configuration where one component is active while the other two components are idle. One block within the container must be operating, otherwise the container will fail, leading to a system failure (since the container block is part of a series configuration in this example).
 

Because both of these concepts are better understood when time dependency is considered, they are addressed in more detail in Chapter [[Time-Dependent_System_Reliability_(Analytical)|Time-Dependent System Reliability (Analytical)]].
 
[[Image:Fig_4.20.PNG|center|400px|<div align="center">An RBD with a load sharing container. </div>]]
 
[[Image:Fig_4.21.PNG|center|400px|<div align="center">An RBD with a standby container.</div>]]
 

= Configurations with Inherited Subdiagrams =
Within BlockSim, a subdiagram block inherits some or all of its properties from another block diagram. This allows the analyst to maintain separate diagrams for portions of a system and incorporate those diagrams as components of another diagram. With this technique, it is possible to generate and analyze extremely complex diagrams representing the behavior of many subsystems in a manageable way. In Figure "Illustration of subdiagrams", Subdiagram Block A in the top diagram represents the series configuration of the subsystem reflected in the middle diagram, while Subdiagram Block G in the middle diagram represents the series configuration of the subsubsystem in the bottom diagram. Figure "An RBD of two computer systems in series where each computer system inherits its diagram from another subdiagram" again illustrates this concept.

=Example 15=
For this example, obtain the reliability equation of the system shown in Figure "An RBD of two computer systems in series where each computer system inherits its diagram from another subdiagram".
==Solution to Example 15==
The system reliability equation is:
 

::<math>{{R}_{System}}={{R}_{Computer1}}\cdot {{R}_{Computer2}} \ </math>

 
:Now:
 
::<math>\begin{align}
{{R}_{Computer1}}= & ({{R}_{Power\,Supply}}\cdot {{R}_{Processor}}\cdot {{R}_{HardDrive}} \\
&\cdot(-{{R}_{Fan}}\cdot {{R}_{Fan}}+{{R}_{Fan}}+{{R}_{Fan}})) \
\end{align}</math>
 
Since the structures of the computer systems are the same, <math>{{R}_{Computer1}}={{R}_{Computer2}}</math> , then substituting the first equation above into the second equation above yields:
 
::<math>\begin{align}
{{R}_{System}}= & ({{R}_{Power\,Supply}}\cdot {{R}_{Processor}}) \\
& \cdot {{R}_{HardDrive}}(-R_{Fan}^{2}+2{{R}_{Fan}}){{)}^{2}}
\end{align}</math>

When using BlockSim to compute the equation, the software will return the first equation above for the system and the second equation above for the subdiagram. Even though BlockSim will make these substitutions internally when performing calculations, it does show them in the System Reliability Equation window.
 
[[Image:chp4image27.png|center|500px|Illustration of subdiagrams.]]
 
[[Image:BS4.23.png|center|600px|An RBD of two computer systems in series where each computer system inherits its diagram from another subdiagram]]
 

= Configurations with Multi Blocks =
By using multi blocks within BlockSim, a single block can represent multiple identical blocks in series or in parallel configuration. This technique is simply a way to save time when creating the RBD and to save space within the diagram. Each item represented by a multi block is a separate entity with identical reliability characteristics to the others. However, each item is not rendered individually within the diagram. In other words, if the RBD contains a multi block that represents three identical components in a series configuration, then each of those components fails according to the same failure distribution but each component may fail at different times. Because the items are arranged reliability-wise in series, if one of those components fails, then the multi block fails. It is also possible to define a multi block with multiple identical components arranged reliability-wise in parallel or <math>k</math> -out-of- <math>n</math> redundancy. Figure "Illustrating multi-blocks" demonstrates the use of multi blocks in BlockSim.
 

= Configurations with Mirrored Blocks =
[[Image:Mirror_block_symbol.PNG|center|300px|]]

While multi blocks allow the analyst to represent multiple items with a single block in an RBD, BlockSim's mirrored blocks can be used to represent a single item with more than one block placed in multiple locations within the diagram. Mirrored blocks can be used to simulate bidirectional paths within a diagram. For example, in a reliability block diagram for a communications system where the lines can operate in two directions, the use of mirrored blocks will facilitate realistic simulations for the system maintainability and availability.
 
 
[[Image:chp4image30.png|center|500px|Illustrating multi-blocks.]]

 
It may also be appropriate to use this type of block if the component performs more than one function and the failure to perform each function has a different reliability-wise impact on the system. In mirrored blocks, the duplicate block behaves in the exact same way that the original block does. The failure times and all maintenance events are the same for each duplicate block as for the original block.
 
 
To better illustrate this consider the following block diagram:
 
[[Image:Mirror_block.PNG|center|400px|]]

 
In this diagram <math>Bm</math> is a mirrored block of <math>B</math>. Since <math>B</math> and <math>Bm</math> are identical, this diagram is equivalent to a diagram with <math>A</math>, <math>B</math> and <math>C</math> in series, as shown next:
 
[[Image:EqvMirror.PNG|center|400px|]]
 

==Example 16==
In the diagram shown in Figure "Electircal network diagram.", electricity can flow in both directions. Successful system operation requires at least one output (O1, O2 or O3) to be working.
 
 
Create a block diagram for this system.
===Solution to Example 16===
The bidirectionality of this system can be modeled using mirrored blocks. The diagram is shown in Figure "RBD of Electircal network diagram".
 
 
Blocks 5A, 7A and 1A are duplicates (or mirrored blocks) of 5, 7 and 1 respectively.

{{reliability block diagrams for failure modes}}

= Symbolic Solutions in BlockSim =

 
Several algebraic solutions in BlockSim were used in the prior examples. BlockSim constructs and displays these equations in different ways, depending on the options chosen. As an example, consider the complex system shown in Figure "Complex RBD".
 

When BlockSim constructs the equation internally, it does so in what we will call a symbolic mode. In this mode, portions of the system are segmented. For this example, the symbolic (internal) solution is shown in Figure "BlocksSim's symbolic solution" and composed of the terms shown in Equations <math>{{I}_{7}}</math>, <math>{{D}_{1}}</math>, <math>{{I}_{11}}</math>, <math>{{R}_{System}}</math>.
 
:Or:
 

::<math>\begin{align}
{{I}_{7}}= & -{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}} \\
& +{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}} \\
& -{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}} \\
& +{{R}_{5}}+{{R}_{6}}+{{R}_{4}} \
\end{align}</math>

 
::<math>{{D}_{1}}=+{{R}_{7}}\cdot {{I}_{7}} \ </math>

 
::<math>\begin{align}
{{I}_{11}}= & -{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{D}_{1}} \\
& +{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& -{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{D}_{1}}-{{R}_{2}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& -{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}-{{R}_{2}}\cdot {{R}_{9}}\cdot {{D}_{1}}-{{R}_{9}}\cdot {{R}_{5}}\cdot {{D}_{1}}-{{R}_{9}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& -{{R}_{5}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{9}}+{{R}_{2}}\cdot {{R}_{10}}+{{R}_{9}}\cdot {{D}_{1}} \\
& +{{R}_{5}}\cdot {{D}_{1}}+{{R}_{8}}\cdot {{D}_{1}} \
\end{align}</math>

 
::<math>{{R}_{System}}=+{{R}_{1}}\cdot {{R}_{11}}\cdot {{I}_{11}} \ </math>
 

[[Image:1 thru 11 network.png|thumb|center|500px|Complex RBD]]

[[Image:BS4-30.png|thumb|center|500px|BlocksSim's symbolic solution]]

 

 
These terms use tokens to represent portions of the equation. In the symbolic equation setting, one reads the solution from the bottom up, replacing any occurrences of a particular token with its definition. In this case then, and to obtain a system solution, one begins with <math>{{R}_{System}}</math>. <math>{{R}_{System}}</math> contains the token <math>{{I}_{11}}</math>. This is then substituted into <math>{{R}_{System}}</math> yielding:
 

::<math>\begin{align}
{{R}_{System}}= & +{{R}_{1}}\cdot {{R}_{11}}(-{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& +{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& +{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& -{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{D}_{1}}-{{R}_{2}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& +{{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{8}}\cdot {{D}_{1}}-{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}} \\
& -{{R}_{2}}\cdot {{R}_{9}}\cdot {{D}_{1}}-{{R}_{9}}\cdot {{R}_{5}}\cdot {{D}_{1}}-{{R}_{9}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& -{{R}_{5}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{9}} \\
& +{{R}_{2}}\cdot {{R}_{10}}+{{R}_{9}}\cdot {{D}_{1}}+{{R}_{5}}\cdot {{D}_{1}}+{{R}_{8}}\cdot {{D}_{1}}) \
\end{align}</math>
 

Now equation above contains the token <math>{{D}_{1}}</math>. The next step is to substitute <math>{{D}_{1}}</math> into equation above. Doing so yields <math>{{I}_{7}}</math>, shown next.
 

::<math>\begin{align}
{{R}_{System}}= & +{{R}_{1}}\cdot {{R}_{11}}(-{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\
& +{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\
& +{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\
& +{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\
& -{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\
& -{{R}_{2}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}})+{{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\
& -{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}} \\
& -{{R}_{2}}\cdot {{R}_{9}}\cdot ({{R}_{7}}\cdot {{I}_{7}})-{{R}_{9}}\cdot {{R}_{5}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\
& -{{R}_{9}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}})-{{R}_{5}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}})+{{R}_{2}}\cdot {{R}_{9}} \\
& +{{R}_{2}}\cdot {{R}_{10}}+{{R}_{9}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\
& +{{R}_{5}}\cdot ({{R}_{7}}\cdot {{I}_{7}})+{{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}})) \
\end{align}</math>
 

Equation above contains the token <math>{{D}_{1}}</math>. The last step is then to substitute <math>{{R}_{System}}</math> into equation above:
::<math>{{R}_{System}}=+{{R}_{1}}\cdot {{R}_{11}}(-{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+\ \,{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>+{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>+{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>+{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>-{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>-{{R}_{2}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>+{{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>-{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}-{{R}_{2}}\cdot {{R}_{9}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>-{{R}_{9}}\cdot {{R}_{5}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>-{{R}_{9}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>-{{R}_{5}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>+{{R}_{2}}\cdot {{R}_{9}}+{{R}_{2}}\cdot {{R}_{10}}+{{R}_{9}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>+{{R}_{5}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>+{{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}})))</math>
When the complete equation is chosen, BlockSim's System Reliability Equation window performs these token substitutions automatically. It should be pointed out that the complete equation can get very large. While BlockSim internally can deal with millions of terms in an equation, the System Reliability Equation window will only format and display equations up to 64,000 characters. BlockSim uses a 64K memory buffer for displaying equations.
 

===Use IBS (Identical Block Simplification) Option===

When computing the system equation with the "Use IBS" option selected, BlockSim looks for identical blocks (blocks with the same failure characteristics) and attempts to simplify the equation, if possible. The symbolic solution for the system in the prior case, with the "Use IBS" option selected and setting equal reliability block properties, is:
 
::<math>{{R}_{3}}={{R}_{6}}={{R}_{4}}={{R}_{5}}</math>
 

::<math>{{I}_{7}}=+4{{R}_{3}}-6R_{3}^{2}+4R_{3}^{3}-R_{3}^{4}</math>
 

::<math>{{D}_{1}}=+{{R}_{7}}\cdot {{I}_{7}}</math>
 

::<math>\begin{align}
{{I}_{11}}= & -{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{D}_{1}} \\
& +{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& -{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{D}_{1}}-{{R}_{2}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& -{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}-{{R}_{2}}\cdot {{R}_{9}}\cdot {{D}_{1}}-{{R}_{9}}\cdot {{R}_{5}}\cdot {{D}_{1}} \\
& -{{R}_{9}}\cdot {{R}_{8}}\cdot {{D}_{1}}-{{R}_{5}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{9}}+{{R}_{2}}\cdot {{R}_{10}} \\
& +{{R}_{9}}\cdot {{D}_{1}}+{{R}_{5}}\cdot {{D}_{1}}+{{R}_{8}}\cdot {{D}_{1}}
\end{align}</math>

 
::<math>{{R}_{System}}=+{{R}_{1}}\cdot {{R}_{11}}\cdot {{I}_{11}}</math>

 

[[Image:Fig 4.31.PNG|thumb|center|400px|<div align="center"> BlockSim's complete solution for Complex RBD]]

 
When using IBS, the resulting equation is invalidated if any of the block properties (e.g. failure distributions) have changed since the equation was simplified based on those properties.
 
[[Image:Fig 4.32.PNG|thumb|center|400px|<div align="center"> BlockSim's symbolic solution, with the IBS option selected for Complex RBD]]

RBDs and Analytical System Reliability

2012-03-19T22:48:46Z

Dingzhou Cao: /* Symbolic Solutions in BlockSim */

{{Template:bsbook|4}}

 
An overall system reliability prediction can be made by looking at the reliabilities of the components that make up the whole system or product. In this chapter, we will examine the methods of performing such calculations. The reliability-wise configuration of components must be determined beforehand. For this reason, we will first look at different component/subsystem configurations, also known as structural properties ([[Appendix_B:_References | Leemis [17]]]). Unless explicitly stated, the components will be assumed to be statistically independent.
=Component Configurations =
 
In order to construct a reliability block diagram, the reliability-wise configuration of the components must be determined. Consequently, the analysis method used for computing the reliability of a system will also depend on the reliability-wise configuration of the components/subsystems. That configuration can be as simple as units arranged in a pure series or parallel configuration. There can also be systems of combined series/parallel configurations or complex systems that cannot be decomposed into groups of series and parallel configurations. The configuration types considered in this reference include:
 
:• Series configuration. 
:• Simple parallel configuration. 
:• Combined (series and parallel) configuration. 
:• Complex configuration. 
:• ''k''-out-of-''n'' parallel configuration. 
:• Configuration with a load sharing container (presented in Chapter 5). 
:• Configuration with a standby container (presented in Chapter 5). 
:• Configuration with inherited subdiagrams. 
:• Configuration with multi blocks. 
:• Configuration with mirrored blocks. 
 
Each of these configurations will be presented, along with analysis methods, in the sections that follow.
 

[[Category:Completed Theoretical Review]]

= Series Systems =

[[Image:BS4im1.png|center|400px|]]
 
In a series configuration, a failure of any component results in the failure of the entire system. In most cases, when considering complete systems at their basic subsystem level, it is found that these are arranged reliability-wise in a series configuration. For example, a personal computer may consist of four basic subsystems: the motherboard, the hard drive, the power supply and the processor. These are reliability-wise in series and a failure of any of these subsystems will cause a system failure. In other words, all of the units in a series system must succeed for the system to succeed.
 
The reliability of the system is the probability that unit 1 succeeds and unit 2 succeeds and all of the other units in the system succeed. So all <math>n</math> units must succeed for the system to succeed. The reliability of the system is then given by:
 

::<math>\begin{align}
R_S = & P({{X}_{1}}\cap {{X}_{2}}\cap ...\cap {{X}_{n}}) \\
= & P({{X}_{1}})P({{X}_{2}}|{{X}_{1}})P({{X}_{3}}|{{X}_{1}}{{X}_{2}}) \cdot\cdot\cdot P({{X}_{n}}|{{X}_{1}}{{X}_{2}}...{{X}_{n-1}})
\end{align}</math>
 
Where:
 
:• <math>{{R}_{s}}</math> = reliability of the system. 
:• <math>{{X}_{i}}</math> = event of unit <math>i</math> being operational. 
:• <math>P({{X}_{i}})</math> = probability that unit <math>i</math> is operational. 
 
In the case where the failure of a component affects the failure rates of other components (i.e. the life distribution characteristics of the other components change when one component fails), then the conditional probabilities in equation above must be considered.
 
However, in the case of independent components, equation above becomes:
 
::<math>{{R}_{s}}=P({{X}_{1}})P({{X}_{2}})...P({{X}_{n}}) </math>
 
:Or:
 
::<math>{{R}_{s}}=\underset{i=1}{\overset{n}{\mathop \prod }}\,P({{X}_{i}})</math>
 
Or, in terms of individual component reliability:
 
::<math>{{R}_{s}}=\underset{i=1}{\overset{n}{\mathop \prod }}\,{{R}_{i}} \ (eqn 2)</math>
 
In other words, for a pure series system, the system reliability is equal to the product of the reliabilities of its constituent components.
{{Example:Analytical-RBD}}

===Effect of Component Reliability in a Series System===
 
[[Image:Chp4image2.png|center|400px|]]
 
In a series configuration, the component with the least reliability has the biggest effect on the system's reliability. There is a saying that a chain is only as strong as its weakest link. This is a good example of the effect of a component in a series system. In a chain, all the rings are in series and if any of the rings break, the system fails. In addition, the weakest link in the chain is the one that will break first. The weakest link dictates the strength of the chain in the same way that the weakest component/subsystem dictates the reliability of a series system. As a result, the reliability of a series system is always less than the reliability of the least reliable component.

===Example 2===
Consider three components arranged reliability-wise in series, where <math>{{R}_{1}}=70%</math> , <math>{{R}_{2}}=80%</math> and <math>{{R}_{3}}=90%</math> (for a given time).
 
In Table "System reliabilities for combinations of component reliabilities", we can examine the effect of each component's reliability on the overall system reliability. The first row of the table shows the given reliability for each component and the corresponding system reliability for these values. In the second row, the reliability of Component 1 is increased by a value of 10% while keeping the reliabilities of the other two components constant. Similarly, by increasing the reliabilities of Components 2 and 3 in rows 3 and 4 by a value of 10%, while keeping the reliabilities of the other components at the given values, we can observe the effect of each component's reliability on the overall system reliability. It is clear that the highest value for the system's reliability was achieved when the reliability of Component 1, which is the least reliable component, was increased by a value of 10%.
 
[[Image:BS4.T1.png|center|500px|System reliabilities for combinations of component reliabilities.]]
 
This conclusion can also be illustrated graphically, as shown in Figure "Effect of component reliability on the overall system reliability
".
 
Note the slight difference in the slopes of the three lines in Figure "Effect of component reliability on the overall system reliability
". The difference in these slopes represents the difference in the effect of each of the components on the overall system reliability. In other words, the system reliability's rate of change with respect to each component's change in reliability is different. This observation will be explored further when the importance measures of components are considered in later chapters. The rate of change of the system's reliability with respect to each of the components is plotted in Figure "Rate of change of series system reliability when increasing the reliability of each component". It can be seen in Figure "Effect of component reliability on the overall system reliability
" that Component 1 has the steepest slope, which indicates that an increase in the reliability of Component 1 will result in a higher increase in the reliability of the system. In other words, Component 1 has a higher ''reliability importance''.
 
[[Image:BS4.1.png|center|500px|Effect of component reliability on the overall system reliability]]
 
[[Image:BS4.2.png|center|500px|Rate of change of series system reliability when increasing the reliability of each component]]
 
 

===Effect of Number of Components in a Series System===
The number of components is another concern in systems with components connected reliability-wise in series. As the number of components connected in series increases, the system's reliability decreases.
 
Figure "Reliability of a system with n statistically independent and identical components arraged reliability-wise in series" illustrates the effect of the number of components arranged reliability-wise in series on the system's reliability for different component reliability values. This figure also demonstrates the dramatic effect that the number of components has on the system's reliability, particularly when the component reliability is low. In other words, in order to achieve a high system reliability, the component reliability must be high also, especially for systems with many components arranged reliability-wise in series.

===Example 3===
Consider a system that consists of a single component. The reliability of the component is 95%, thus the reliability of the system is 95%. What would the reliability of the system be if there were more than one component (with the same individual reliability) in series? Table "System reliability as a function of the number of components" shows the effect on the system's reliability by adding consecutive components (with the same reliability) in series. Figure "Effect of the number of component in a series configuration for two different casese" illustrates the same concept graphically for components with 90% and 95% reliability.
 
[[Image:Table 4.2.png|center|400px|System reliability as a function of the number of components]]

[[Image:BS4.3.png|center|500px|Reliability of a system with n statistically independent and identical components arraged reliability-wise in series]]
 
 
[[Image:BS4.4.png|center|500px|Effect of the number of component in a series configuration for two different casese]]
 
 

= Simple Parallel Systems =

[[Image:chp4image6.png|center|300px|Simple parallel system]]
 
In a simple parallel system, as shown in Figure "Simple parallel system", at least one of the units must succeed for the system to succeed. Units in parallel are also referred to as redundant units. Redundancy is a very important aspect of system design and reliability in that adding redundancy is one of several methods of improving system reliability. It is widely used in the aerospace industry and generally used in mission critical systems. Other example applications include the RAID computer hard drive systems, brake systems and support cables in bridges.
The probability of failure, or unreliability, for a system with <math>n</math> statistically independent parallel components is the probability that unit 1 fails and unit 2 fails and all of the other units in the system fail. So in a parallel system, all <math>n</math> units must fail for the system to fail. Put another way, if unit 1 succeeds or unit 2 succeeds or any of the <math>n</math> units succeeds, then the system succeeds. The unreliability of the system is then given by:
 

::<math>\begin{align}
{{Q}_{s}}= & P({{X}_{1}}\cap {{X}_{2}}\cap ...\cap {{X}_{n}}) \\
= & P({{X}_{1}})P({{X}_{2}}|{{X}_{1}})P({{X}_{3}}|{{X}_{1}}{{X}_{2}})...P({{X}_{n}}|{{X}_{1}}{{X}_{2}}...{{X}_{n-1}}) \ (eqn 3)
\end{align}</math>
 
Where:
 
:• <math>{{Q}_{s}}</math> = unreliability of the system. 
:• <math>{{X}_{i}}</math> = event of failure of unit <math>i</math> . 
:• <math>P({{X}_{i}})</math> = probability of failure of unit <math>i</math>. 
 
In the case where the failure of a component affects the failure rates of other components, then the conditional probabilities in equation above must be considered.
 
However, in the case of independent components, equation above becomes:
 
::<math>{{Q}_{s}}=P({{X}_{1}})P({{X}_{2}})...P({{X}_{n}})</math>
 
:Or:
 
::<math>{{Q}_{s}}=\underset{i=1}{\overset{n}{\mathop \prod }}\,P({{X}_{i}})</math>
 
Or, in terms of component unreliability:
 
::<math>{{Q}_{s}}=\underset{i=1}{\overset{n}{\mathop \prod }}\,{{Q}_{i}}</math>
 
Observe the contrast with the series system, in which the system reliability was the product of the component reliabilities; whereas the parallel system has the overall system unreliability as the product of the component unreliabilities.
 
The reliability of the parallel system is then given by:
 
::<math>\begin{align}
{{R}_{s}}= & 1-Qs=1-({{Q}_{1}}\cdot {{Q}_{2}}\cdot ...\cdot {{Q}_{n}}) \\
= & 1-[(1-{{R}_{1}})\cdot (1-{{R}_{2}})\cdot ...\cdot (1-{{R}_{n}})] \\
= & 1-\underset{i=1}{\overset{n}{\mathop \prod }}\,(1-{{R}_{i}}) \ (eqn 5)
\end{align}</math>
===Example 4===
Consider a system consisting of three subsystems arranged reliability-wise in parallel. Subsystem 1 has a reliability of 99.5%, Subsystem 2 has a reliability of 98.7% and Subsystem 3 has a reliability of 97.3% for a mission of 100 hours. What is the overall reliability of the system for a 100-hour mission?
====Solution to Example 4====
Since the reliabilities of the subsystems are specified for 100 hours, the reliability of the system for a 100-hour mission is:
::<math>\begin{align}
{{R}_{s}}= & 1-(1-0.9950)\cdot(1-0.9870)\cdot(1-0.9730) \\
= & 1-0.000001755 \\
= & 0.999998245 \
\end{align}</math>

===Effect of Component Reliability in a Parallel Configuration===
When we examined a system of components in series, we found that the least reliable component has the biggest effect on the reliability of the system. However, the component with the highest reliability in a parallel configuration has the biggest effect on the system's reliability, since the most reliable component is the one that will most likely fail last. This is a very important property of the parallel configuration, specifically in the design and improvement of systems.
===Example 5===
Consider three components arranged reliability-wise in parallel with .. = 60%, <math>{{R}_{2}}</math> = 70% and <math>{{R}_{3}}</math> = 80% (for a given time). The corresponding reliability for the system is <math>{{R}_{s}}</math> = 97.6%. In Table "System reliability for combinations of component reliabilities", we can examine the effect of each component's reliability on the overall system reliability. The first row of the table shows the given reliability for each component and the corresponding system reliability for these values. In the second row, the reliability of Component 1 is increased by a value of 10% while keeping the reliabilities of the other two components constant. Similarly, by increasing the reliabilities of Components 2 and 3 in the third and fourth rows by a value of 10% while keeping the reliabilities of the other components at the given values, we can observe the effect of each component's reliability on the overall system reliability. It is clear that the highest value for the system's reliability was achieved when the reliability of Component 3, which is the most reliable component, was increased. Once again, this is the opposite of what was encountered with a pure series system.
 
 
[[Image:4.T3.png|center|500px|System reliability for combinations of component reliabilities]]
 
This conclusion can also be illustrated graphically, as shown in Figure "Rate of change of parallel system reliability when increasing the reliability of each component".

====Effect of Number of Components in Parallel====
In the case of the parallel configuration, the number of components has the opposite effect of the one observed for the series configuration. For a parallel configuration, as the number of components/subsystems increases, the system's reliability increases.
 
Figure "Reliability of a system with n statistically independent and identical components arranged reliability-wise in parallel" illustrates that a high system reliability can be achieved with low-reliability components, provided that there are a sufficient number of components in parallel. Note that Figure "Reliability of a system with n statistically independent and identical components arranged reliability-wise in parallel" is the mirror image of Figure "Reliability of a system with n statistically independent and identical components arraged reliability-wise in series", which presents the effect of the number of components in a series configuration.
 
[[Image:BS4.6.png|center|500px|Rate of change of parallel system reliability when increasing the reliability of each component]]
 
 
[[Image:BS4.7.png|center|500px|Reliability of a system with n statistically independent and identical components arranged reliability-wise in parallel]]
 
 

===Example 6===
Consider a system that consists of a single component. The reliability of the component is 60%, thus the reliability of the system is 60%. What would the reliability of the system be if the system were composed of two, four or six such components in parallel?
 
 
[[Image:4.T4.png|center|500px|System reliability as a function of the number of components]]
 
 
Clearly, the reliability of a system can be improved by adding redundancy. However, it must be noted that doing so is usually costly in terms of additional components, additional weight, volume, etc.
 
Reliability optimization and costs are covered in detail in Chapter [[Reliability_Importance_and_Optimized_Reliability_Allocation_(Analytical)|Component Reliability Importance]].
 
[[Image:BS4.8.png|center|500px|Effect of the number of components in a parallel configuration]]
 
 

= Combination of Series and Parallel =

While many smaller systems can be accurately represented by either a simple series or parallel configuration, there may be larger systems that involve both series and parallel configurations in the overall system. Such systems can be analyzed by calculating the reliabilities for the individual series and parallel sections and then combining them in the appropriate manner. Such a methodology is illustrated in the following example.
===Example 7===
Consider a system with three components. Units 1 and 2 are connected in series and Unit 3 is connected in parallel with the first two, as shown in the next figure.
 
[[Image:chp4image10.png|center|400px|]]
 
What is the reliability of the system if <math>{{R}_{1}}</math> = 99.5%, <math>{{R}_{2}}</math> = 98.7% and <math>{{R}_{3}}</math> = 97.3% at 100 hours?
====Solution to Example 7====
First, the reliability of the series segment consisting of Units 1 and 2 is calculated:
 
::<math>\begin{align}
{{R}_{1,2}}= & {{R}_{1}}\cdot {{R}_{2}} \\
{{R}_{1,2}}= & 0.9950\cdot 0.9870 \\
{{R}_{1,2}}= & 0.982065\text{ or }98.2065%
\end{align}</math>
 
The reliability of the overall system is then calculated by treating Units 1 and 2 as one unit with a reliability of 98.2065% connected in parallel with Unit 3. Therefore:
 
::<math>\begin{align}
{{R}_{s}}= & 1-[(1-0.982065)\cdot (1-0.973000)] \\
{{R}_{s}}= & 1-0.000484245 \\
{{R}_{s}}= & 0.999515755 \\
{{R}_{s}}= & 99.95%
\end{align}</math>
 

= k-out-of-n Parallel Configuration =
The <math>k</math> -out-of- <math>n</math> configuration is a special case of parallel redundancy. This type of configuration requires that at least <math>k</math> components succeed out of the total <math>n</math> parallel components for the system to succeed. For example, consider an airplane that has four engines. Furthermore, suppose that the design of the aircraft is such that at least two engines are required to function for the aircraft to remain airborne. This means that the engines are reliability-wise in a <math>k</math> -out-of- <math>n</math> configuration, where <math>k</math> = 2 and <math>n</math> = 4. More specifically, they are in a 2-out-of-4 configuration.
 
Even though we classified the <math>k</math> -out-of- <math>n</math> configuration as a special case of parallel redundancy, it can also be viewed as a general configuration type. As the number of units required to keep the system functioning approaches the total number of units in the system, the system's behavior tends towards that of a series system. If the number of units required is equal to the number of units in the system, it is a series system. In other words, a series system of statistically independent components is an <math>n</math> -out-of- <math>n</math> system and a parallel system of statistically independent components is a <math>1</math> -out-of- <math>n</math> system.
===Reliability of <math>k</math> -out-of- <math>n</math> Independent and Identical Components===
 
[[Image:chp4image11.png|center|500px|<div align="center">2-out-of-4 configuration </div>]]
 
The simplest case of components in a <math>k</math> -out-of- <math>n</math> configuration is when the components are independent and identical. In other words, all the components have the same failure distribution and whenever a failure occurs, the remaining components are not affected. In this case, the reliability of the system with such a configuration can be evaluated using the binomial distribution, or:
 
::<math>{{R}_{s}}(k,n,R)=\underset{r=k}{\overset{n}{\mathop \sum }}\,\left( \begin{matrix}
n \\
r \\
\end{matrix} \right){{R}^{r}}{{(1-R)}^{n-r}} \ </math>
 

Where:
 
:• <math>n</math> is the total number of units in parallel. 
:• <math>k</math> is the minimum number of units required for system success. 
:• <math>R</math> is the reliability of each unit. 

===Example 8===
Consider a system of six pumps of which at least four must function properly for system success. Each pump has an 85% reliability for the mission duration. What is the probability of success of the system for the same mission duration?
====Solution to Example 8====
Using <math>{{R}_{s}}(k,n,R)=\underset{r=k}{\overset{n}{\mathop \sum }}\,\left( \begin{matrix}
n \\
r \\
\end{matrix} \right){{R}^{r}}{{(1-R)}^{n-r}} \ </math> for <math>k=4</math> and <math>n=6</math> :
 
::<math>\begin{align}
{{R}_{s}}= & \underset{r=4}{\overset{6}{\mathop \sum }}\,\left( \begin{matrix}
6 \\
r \\
\end{matrix} \right){{0.85}^{r}}{{(1-0.85)}^{6-r}} \\
= & \left( \begin{matrix}
6 \\
4 \\
\end{matrix} \right){{0.85}^{4}}{{(1-0.85)}^{2}}+\left( \begin{matrix}
6 \\
5 \\
\end{matrix} \right){{0.85}^{5}}{{(1-0.85)}^{1}} \\
& +\left( \begin{matrix}
6 \\
6 \\
\end{matrix} \right){{0.85}^{6}}{{(1-0.85)}^{0}} \\
= & 0.1762+0.3993+0.3771 \\
= & 95.26%
\end{align}</math>
 
One can examine the effect of increasing the number of units required for system success while the total number of units remains constant (in this example, six units). In Figure "2-out-of-4 configuration", the reliability of the <math>k</math> -out-of-6 configuration was plotted versus different numbers of required units.
 
 
[[Image:4.T5.png|center|500px|Reliability for a k-out-of-6 system for different k values]]
 

 
Note that the system configuration becomes a simple parallel configuration for <math>k</math> = 1 and the system is a six-unit series configuration ( <math>{{(0.85)}^{6}}=</math> 0.377) for <math>k</math> = 6.
<math></math>
 
[[Image:BS4.10.png|center|500px|Reliability of a k-out-of-6 configuration for different k values]]
 
 

===Reliability of Nonidentical <math>k</math> -out-of- <math>n</math> Independent Components===
In the case where the <math>k</math> -out-of- <math>n</math> components are not identical, the reliability must be calculated in a different way. One approach, described in detail later in this chapter, is to use the event space method. In this method, all possible operational combinations are considered in order to obtain the system's reliability. The method is illustrated with the following example.
===Example 9===
Three hard drives in a computer system are configured reliability-wise in parallel. At least two of them must function in order for the computer to work properly. Each hard drive is of the same size and speed, but they are made by different manufacturers and have different reliabilities. The reliability of <math>HD\#1</math> is 0.9, <math>HD\#2</math> is 0.88 and <math>HD\#3</math> is 0.85, all at the same mission time.
====Solution to Example 9====
Since at least two hard drives must be functioning at all times, only one failure is allowed. This is a 2-out-of-3 configuration.
 
The following operational combinations are possible for system success:
#All 3 hard drives operate.
#<math>HD\#1</math> fails, while <math>HDs\,\#2</math> and <math>\#3</math> continue to operate.
#<math>HD\#2</math> fails, while <math>HDs\,\#1</math> and <math>\#3</math> continue to operate.
#<math>HD\#3</math> fails, while <math>HDs\,\#1</math> and <math>\#2</math> continue to operate.
 
The probability of success for the system (reliability) can now be expressed as:
 
::<math>\begin{align}
{{P}_{s}}= & {{R}_{1}}{{R}_{2}}{{R}_{3}}+(1-{{R}_{1}}){{R}_{2}}{{R}_{3}}+{{R}_{1}}(1-{{R}_{2}}){{R}_{3}} \\
& +{{R}_{1}}{{R}_{2}}(1-{{R}_{3}})
\end{align}</math>
 
This equation for the reliability of the system can be reduced to:
 
::<math>{{R}_{s}}={{R}_{1}}{{R}_{2}}+{{R}_{2}}{{R}_{3}}+{{R}_{1}}{{R}_{3}}-2{{R}_{1}}{{R}_{2}}{{R}_{3}}</math>
 
:Or:
 
::<math>{{R}_{s}}=95.86%</math>
 
If all three hard drives had the same reliability, <math>R</math> , then the equation for the reliability of the system could be further reduced to:
 
::<math>{{R}_{s}}=3{{R}^{2}}-2{{R}^{3}}</math>
 
Or, using the binomial approach:
 
::<math>\begin{align}
{{R}_{s}}= & \underset{r=2}{\overset{3}{\mathop \sum }}\,\left( \begin{matrix}
3 \\
r \\
\end{matrix} \right){{R}^{r}}{{(1-R)}^{3-r}} \\
= & \left( \begin{matrix}
3 \\
2 \\
\end{matrix} \right){{R}^{2}}(1-R)+\left( \begin{matrix}
3 \\
3 \\
\end{matrix} \right){{R}^{3}}{{(1-R)}^{0}} \\
= & 3\cdot {{R}^{2}}(1-R)+{{R}^{3}} \\
= & 3{{R}^{2}}-2{{R}^{3}}
\end{align}</math>
 
The example can be repeated using BlockSim. The following graphic demonstrates the RBD for the system.
 
<math></math>
 
 
[[Image:Fig_harddrive.PNG|center|400px|]]
 

The RBD is analyzed and the system reliability equation is returned. Figure "System equation results for Example 9" shows the equation returned by BlockSim.
 
[[Image:BS4.11.png|thumb|center|400px|System equation results for Example 9]]
 
Using the Analytical Quick Calculation Pad, the reliability can be calculated to be 0.9586. Figure "Reliability results for Example 9" shows the returned result.
 
Note that you are not required to enter a mission end time for this system into the Analytical QCP because all of the components are static and thus the reliability results are independent of time.
[[Image:Fig 4.12.PNG|thumb|center|500px|<div align="center">Reliability results for Example 9 </div>]]

===Example 10===
Consider the four-engine aircraft discussed previously. If we were to change the problem statement to two out of four engines are required, however no two engines on the same side may fail, then the block diagram would change to the configuration shown in Figure "Block diagram for Example 10".
 
<math></math>
 
Note that this is the same as having two engines in parallel on each wing and then putting the two wings in series.
 

 
[[Image:4.13.png|center|500px|<div align="center">Block diagram for Example 10 </div>]]
 
 

=Consecutive <math>k</math>-out-of-<math>n</math> : <math>F</math> Redundancy=
There are other multiple redundancy types and multiple industry terms. One such example is what is referred to as a consecutive <math>k</math>-out-of-<math>n</math> : <math>F</math> system. To illustrate this configuration type, consider a telecommunications system that consists of a transmitter and receiver with six relay stations to connect them. The relays are situated so that the signal originating from one station can be picked up by the next two stations down the line. For example, a signal from the transmitter can be received by Relay 1 and Relay 2, a signal from Relay 1 can be received by Relay 2 and Relay 3, and so forth. Thus, this arrangement would require two consecutive relays to fail for the system to fail. A diagram of this configuration is shown in Figure "RBD for the consecutive ''k''-out-of-''n'': ''F'' system".
 
<math></math>
 
This type of a configuration is also referred to as a complex system. Complex systems are discussed in the next section.
[[Image:4.14.png|center|600px|<div align="center">RBD for the consecutive ''k''-out-of-''n'': ''F'' system </div>]]

= Complex Systems =
In many cases, it is not easy to recognize which components are in series and which are in parallel in a complex system. The network shown in Figure "Example of a complex system" is a good example of such a complex system.
 
<math></math>
 
 
[[Image:4.15.png|center|500px|<div align="center">Example of a complex system </div>]]
 
The system in Figure "Example of a complex system" cannot be broken down into a group of series and parallel systems. This is primarily due to the fact that component <math>C</math> has two paths leading away from it, whereas <math>B</math> and <math>D</math> have only one. Several methods exist for obtaining the reliability of a complex system including:
 
:• The decomposition method. 
:• The event space method. 
:• The path-tracing method. 
===Decomposition Method===
The decomposition method is an application of the law of total probability. It involves choosing a "key" component and then calculating the reliability of the system twice: once as if the key component failed ( <math>R=0</math> ) and once as if the key component succeeded <math>(R=1)</math> . These two probabilities are then combined to obtain the reliability of the system, since at any given time the key component will be failed or operating. Using probability theory, the equation is:
 
::<math>\begin{align}
{{R}_{s}}= & P(s\cap A)+P(s\cap \overline{A}) \\
= & P(s|A)P(A)+P(s|\overline{A})P(\overline{A})
\end{align}</math>
====Illustration of the Decomposition Method====
Consider three units in series.
 
:• <math>A</math> is the event of Unit 1 success. 
:• <math>B</math> is the event of Unit 2 success. 
:• <math>C</math> is the event of Unit 3 success. 
:• <math>s</math> is the event of system success. 
 
First, select a "key" component for the system. Selecting Unit 1, the probability of success of the system is:
 
::<math>{{R}_{s}}=P(s|A)P(A)+P(s|\overline{A})P(\overline{A})</math>
 
If Unit 1 is good, then:
 
::<math>P(s|A)={{R}_{2}}{{R}_{3}}</math>
 
That is, if Unit 1 is operating, the probability of the success of the system is the probability of Units 2 and 3 succeeding.
 
If Unit 1 fails, then:
 
::<math>P(s|\overline{A})=0</math>
 
That is, if Unit 1 is not operating, the system has failed since a series system requires all of the components to be operating for the system to operate.
Thus the reliability of the system is:
 
::<math>{{R}_{s}}={{R}_{2}}{{R}_{3}}P(A)={{R}_{1}}{{R}_{2}}{{R}_{3}}</math>
 

====Another Illustration of the Decomposition Method====
Consider the following system:
 
[[Image:chp4image14.png|center|300px|]]
 
:• <math>A</math> is the event of Unit 1 success.
:• <math>B</math> is the event of Unit 2 success.
:• <math>C</math> is the event of Unit 3 success.
:• <math>s</math> is the event of system success.
 
Selecting Unit 3 as the `` key'' component, the system reliability is:
 
::<math>{{R}_{s}}=P(s|C)P(C)+P(s|\overline{C})P(\overline{C})</math>
 
If Unit 3 survives, then:
 
::<math>P(s|C)=1</math>
 
That is, since Unit 3 represents half of the parallel section of the system, then as long as it is operating, the entire system operates.
 
If Unit 3 fails, then the system is reduced to:
 
[[Image:chp4image15.png|center|400px|]]
 

::<math>P(s|\overline{C})={{R}_{1}}{{R}_{2}}</math>
 
The reliability of the system is given by:
 
::<math>{{R}_{s}}=P(C)+{{R}_{1}}{{R}_{2}}P(\overline{C})={{R}_{3}}+{{R}_{1}}{{R}_{2}}(1-{{R}_{3}})</math>
 
:Or:
 
::<math>{{R}_{s}}={{R}_{3}}+{{R}_{1}}{{R}_{2}}-{{R}_{1}}{{R}_{2}}{{R}_{3}}</math>

===Event Space Method===
The event space method is an application of the mutually exclusive events axiom. All mutually exclusive events are determined and those that result in system success are considered. The reliability of the system is simply the probability of the union of all mutually exclusive events that yield a system success. Similarly, the unreliability is the probability of the union of all mutually exclusive events that yield a system failure. This is illustrated in the following example.
 
====Illustration of the Event Space Method====
 
Consider the following system, with reliabilities R1, R2 and R3 for a given time.
 
[[Image:chp4image14.png|center|400px|]]
:• <math>A</math> is the event of Unit 1 success. 
:• <math>B</math> is the event of Unit 2 success. 
:• <math>C</math> is the event of Unit 3 success. 
 
The mutually exclusive system events are:
 
::<math>\begin{align}
X1= & ABC-\text{all units succeed}\text{.} \\
X2= & \overline{A}BC-\text{only Unit 1 fails}\text{.} \\
X3= & A\overline{B}C-\text{only Unit 2 fails}\text{.} \\
X4= & AB\overline{C}-\text{only Unit 3 fails}\text{.} \\
X5= & \overline{AB}C-\text{Units 1 and 2 fail}\text{.} \\
X6= & \overline{A}B\overline{C}-\text{Units 1 and 3 fail}\text{.} \\
X7= & A\overline{BC}-\text{Units 2 and 3 fail}\text{.} \\
X8= & \overline{ABC}-\text{all units fail}\text{.}
\end{align}</math>

 
System events <math>{{X}_{6}}</math> , <math>{{X}_{7}}</math> and <math>{{X}_{8}}</math> result in system failure. Thus the probability of failure of the system is:
 
::<math>{{P}_{f}}=P({{X}_{6}}\cup {{X}_{7}}\cup {{X}_{8}})</math>
 
Since events <math>{{X}_{6}}</math> , <math>{{X}_{7}}</math> and <math>{{X}_{8}}</math> are mutually exclusive, then:
 
::<math>{{P}_{f}}=P({{X}_{6}})+P({{X}_{7}})+P({{X}_{8}})</math>
 
:And:
 
::<math>\begin{align}
P({{X}_{6}})= & P(\overline{A}B\overline{C})=(1-{{R}_{1}})({{R}_{2}})(1-{{R}_{3}}) \\
P({{X}_{7}})= & P(A\overline{B}\overline{C})=({{R}_{1}})(1-{{R}_{2}})(1-{{R}_{3}}) \\
P({{X}_{8}})= & P(\overline{A}\overline{B}\overline{C})=(1-{{R}_{1}})(1-{{R}_{2}})(1-{{R}_{3}})
\end{align}</math>
 
Combining terms yields:
 

::<math>{{P}_{f}}=1-{{R}_{1}}{{R}_{2}}-{{R}_{3}}+{{R}_{1}}{{R}_{2}}{{R}_{3}}</math>
 
:Since:
 
::<math>{{R}_{s}}=1-{{P}_{f}}</math>
 
:Then:
 
::<math>{{R}_{s}}={{R}_{1}}{{R}_{2}}+{{R}_{3}}-{{R}_{1}}{{R}_{2}}{{R}_{3}}</math>
 
This is of course the same result as the one obtained previously using the decomposition method.
If <math>{{R}_{1}}</math> = 99.5%, <math>{{R}_{2}}</math> = 98.7% and <math>{{R}_{3}}</math> = 97.3%, then:
 
::<math>\begin{align}
{{R}_{s}}= & 0.995\cdot 0.987+0.973-0.995\cdot 0.987\cdot 0.973 \\
= & 0.999515755
\end{align}</math>
 
:Or:
 
::<math>{{R}_{s}}=99.95%</math>
 

===Path-Tracing Method===
With the path-tracing method, every path from a starting point to an ending point is considered. Since system success involves having at least one path available from one end of the RBD to the other, as long as at least one path from the beginning to the end of the path is available, then the system has not failed. One could consider the RBD to be a plumbing schematic. If a component in the system fails, the "water" can no longer flow through it. As long as there is at least one path for the "water" to flow from the start to the end of the system, the system is successful. This method involves identifying all of the paths the "water" could take and calculating the reliability of the path based on the components that lie along that path. The reliability of the system is simply the probability of the union of these paths. In order to maintain consistency of the analysis, starting and ending blocks for the system must be defined.

===Example 11===
Obtain the reliability equation of the following system.
 
[[Image:BS4ex11.png|center|400px|]]
 
====Solution to Example 11====
The successful paths for this system are:
 

::<math>{{X}_{1}}=ABD\text{ and }{{X}_{2}}=ACD</math>

 

The reliability of the system is simply the probability of the union of these paths:
 

::<math>{{R}_{s}}=P({{X}_{1}}\cup {{X}_{2}})</math>
 

::<math>\begin{align}
P({{X}_{1}}\cup {{X}_{2}})= & P({{X}_{1}})+P({{X}_{2}})-P({{X}_{1}}\cap {{X}_{2}}) \\
= & P(ABD)+P(ACD)-P(ABCD)
\end{align}</math>

 

:Thus:
 

::<math>{{R}_{s}}={{R}_{A}}{{R}_{B}}{{R}_{D}}+{{R}_{A}}{{R}_{C}}{{R}_{D}}-{{R}_{A}}{{R}_{B}}{{R}_{C}}{{R}_{D}}</math>
 

===Example 12===
Obtain the reliability equation of the following system.
 
[[Image:Chp4image14.png|center|400px|]]
 
 

====Solution to Example 12====
Assume starting and ending blocks that cannot fail, as shown next.
 
[[Image:BS.4ex12.2.png|center|500px|]]
 
The paths for this system are:
 
::<math>{{X}_{1}}=1,2\text{ and }{{X}_{2}}=3</math>

 
The probability of success of the system is given by:

::<math>\begin{align}
P({{X}_{1}}\cup {{X}_{2}})= & P({{X}_{1}})+P({{X}_{2}})-P({{X}_{1}}\cap {{X}_{2}}) \\
= & P(1,2)+P(3)-P(1,2,3)
\end{align}</math>
 

:Or:
 

::<math>{{R}_{s}}={{R}_{1}}{{R}_{2}}+{{R}_{3}}-{{R}_{1}}{{R}_{2}}{{R}_{3}}</math>
 

====Starting and Ending Blocks in BlockSim====
Note that BlockSim requires that all diagrams start from a single block and end on a single block. To meet this requirement for this example, we arbitrarily added a starting and an ending block, as shown in Figure "BlockSim representation of the RBD for Example 12". These blocks can be set to a cannot fail condition, or <math>R=1</math> , and thus not affect the outcome. However, when the analysis is performed in BlockSim, the returned equation will include terms for the non-failing blocks, as shown in Figure "BlockSim solution Example 12" and equation.

 

::<math>{{R}_{system}}=({{R}_{S}}\cdot {{R}_{E}}(-{{R}_{1}}\cdot {{R}_{2}}\cdot {{R}_{3}}+{{R}_{1}}\cdot {{R}_{2}}+{{R}_{3}})) \ </math>
 

Note that since <math>{{R}_{S}}={{R}_{E}}=1</math> , the system equation above, can be reduced to:
 

::<math>\begin{align}
{{R}_{system}}= & (1\cdot 1(-{{R}_{1}}\cdot {{R}_{2}}\cdot {{R}_{3}}+{{R}_{1}}\cdot {{R}_{2}}+{{R}_{3}})) \\
= & -{{R}_{1}}\cdot {{R}_{2}}\cdot {{R}_{3}}+{{R}_{1}}\cdot {{R}_{2}}+{{R}_{3}}
\end{align}</math>

 
This is equivalent to <math>{{R}_{s}}={{R}_{1}}{{R}_{2}}+{{R}_{3}}-{{R}_{1}}{{R}_{2}}{{R}_{3}}</math>. The reason that BlockSim includes all items regardless of whether they can fail or not is because BlockSim only recomputes the equation when the system structure has changed. What this means is that the user can alter the failure characteristics of an item without altering the diagram structure. For example, a block that was originally set not to fail can be re-set to a failure distribution and thus it would need to be used in subsequent analyses.
 
[[Image:4.16.png|center|400px|<div align="center">BlockSim representation of the RBD for Example 12 </div>]]
 
 
[[Image:Fig 4.17.PNG|thumb|center|400px|<div align="center">BlockSim solution Example 12 </div>]]
 

===Example 13===
For this example:
 
::<math>a)</math> Determine the reliability equation of the system shown in Figure "Diagram for Example 13" using the decomposition method.
 
::<math>b)</math> Determine the reliability equation of the same system using BlockSim.
====Solution to Example 13====
To obtain the solution:
 
::<math>a)</math> Choose A as the key component, then:
 

::<math>\begin{align}
{{R}_{s}}= & P(s|A)P(A)+P(s|\overline{A})P(\overline{A}) \\
P(s| A)= &{{R}_{B}}{{R}_{F}}\left[ 1-\left( 1-{{R}_{C}} \right)\left( 1-{{R}_{E}} \right) \right] \\
P(s| \overline{A})= &{{R}_{B}}{{R}_{D}}{{R}_{E}}{{R}_{F}} \\
{{R}_{s}}= & \left[ {{R}_{B}}{{R}_{F}}\left[ 1-\left( 1-{{R}_{C}} \right)\left( 1-{{R}_{E}} \right) \right] \right]{{R}_{A}}+\left[ {{R}_{B}}{{R}_{D}}{{R}_{E}}{{R}_{F}} \right](1-{{R}_{A}})
\end{align}</math>

 
::<math>b)</math> Using BlockSim:

 
::<math>{{R}_{s}}={{R}_{B}}\cdot {{R}_{F}}(-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}+{{R}_{A}}\cdot {{R}_{E}}+{{R}_{D}}\cdot {{R}_{E}})</math>
 
[[Image:chp4image21.png|center|400px|<div align="center">Diagram for Example 13. </div>]]
 
[[Image:Fig_4.19.PNG|thumb|center|400px|<div align="center">BlockSim solution for Example 13 </div>]]
 

= Difference Between Physical and Reliability-Wise Arrangement =
Reliability block diagrams are created in order to illustrate the way that components are arranged reliability-wise in a system. So far we have described possible structural properties of a system of components, such as series, parallel, etc. These structural properties, however, refer to the system's state of success or failure based on the states of its components. The physical structural arrangement, even though clearly related to the reliability-wise arrangement, is not necessarily identical to it.
===Example 14===
Consider the following circuit:
 
[[Image:BS3sigma.png|center|400px|]]
 
The equivalent resistance must always be less than <math>1.2\Omega </math> .
 
Draw the reliability block diagram for this circuit.
====Solution to Example 14====
First, let's consider the case where all three resistors operate:
 

::<math>\begin{align}
\frac{1}{{{r}_{eq}}}= & \frac{1}{{{r}_{1}}}+\frac{1}{{{r}_{2}}}+\frac{1}{{{r}_{3}}} \\
= & \frac{1}{3}+\frac{1}{3}+\frac{1}{3} \\
= & 1\Omega
\end{align}</math>

 
Thus, when all components operate, the equivalent resistance is <math>1\Omega </math> , which is less than the maximum resistance of <math>1.2\Omega </math> .
 

Next, consider the case where one of the resistors fails open. In this case, the resistance for the resistor is infinite and the equivalent resistance is:
 

::<math>\frac{1}{{{r}_{eq}}}=\frac{1}{\infty }+\frac{1}{3}+\frac{1}{3}=\frac{2}{3}</math>
 

:Thus:
 

::<math>{{r}_{eq}}=1.5\Omega >1.2\Omega \text{ - System failed}\text{.}</math>

 
If two resistors fail open (e.g. #1 and #2), the equivalent resistance is:
 

::<math>\frac{1}{{{r}_{eq}}}=\frac{1}{\infty }+\frac{1}{\infty }+\frac{1}{3}=\frac{1}{3}</math>

 
:Thus:
 

::<math>{{r}_{eq}}=3\Omega >1.2\Omega \text{ - System failed}\text{.}</math>

 
If all three resistors fail open:
 

::<math>{{r}_{eq}}=\infty >1.2\Omega \text{ - System failed}\text{.}</math>

 
Thus, if <math>{{r}_{1}}</math> , <math>{{r}_{2}}</math> , <math>{{r}_{3}}</math>, or any combination of the three fails, the system fails. Put another way, <math>{{r}_{1}}</math> and <math>{{r}_{2}}</math> and <math>{{r}_{3}}</math> must succeed in order for the system to succeed.
 
The RBD is:

[[Image:chp4image24.png|center|400px|]]
 
In this example it can be seen that even though the three components were physically arranged in parallel, their reliability-wise arrangement is in series.

= Configurations with Load Sharing and Standby Redundancy =
Units in load sharing redundancy exhibit different failure characteristics when one or more fail. In Figure "An RBD with a load sharing container", blocks 1, 2 and 3 are in a load sharing container in BlockSim and have their own failure characteristics. All three must fail for the container to fail. However, as individual items fail, the failure characteristics of the remaining units change since they now have to carry a higher load to compensate for the failed ones. The failure characteristics of each block in a load sharing container are defined using both a life distribution and a life-stress relationship that describe how the life distribution changes as the load changes.
 

Standby redundancy configurations consist of items that are inactive and available to be called into service when/if an active item fails (i.e. the items are on standby). Within BlockSim, a container block with other blocks inside is used to better achieve and streamline the representation and analysis of standby configurations. The container serves a dual purpose. The first is to clearly delineate and define the standby relationships between the active unit(s) and standby unit(s). The second is to serve as the manager of the switching process. For this purpose, the container can be defined with its own probability of successfully activating standby units when needed. Figure "An RBD with a standby container" includes a standby container with three items in standby configuration where one component is active while the other two components are idle. One block within the container must be operating, otherwise the container will fail, leading to a system failure (since the container block is part of a series configuration in this example).
 

Because both of these concepts are better understood when time dependency is considered, they are addressed in more detail in Chapter [[Time-Dependent_System_Reliability_(Analytical)|Time-Dependent System Reliability (Analytical)]].
 
[[Image:Fig_4.20.PNG|center|400px|<div align="center">An RBD with a load sharing container. </div>]]
 
[[Image:Fig_4.21.PNG|center|400px|<div align="center">An RBD with a standby container.</div>]]
 

= Configurations with Inherited Subdiagrams =
Within BlockSim, a subdiagram block inherits some or all of its properties from another block diagram. This allows the analyst to maintain separate diagrams for portions of a system and incorporate those diagrams as components of another diagram. With this technique, it is possible to generate and analyze extremely complex diagrams representing the behavior of many subsystems in a manageable way. In Figure "Illustration of subdiagrams", Subdiagram Block A in the top diagram represents the series configuration of the subsystem reflected in the middle diagram, while Subdiagram Block G in the middle diagram represents the series configuration of the subsubsystem in the bottom diagram. Figure "An RBD of two computer systems in series where each computer system inherits its diagram from another subdiagram" again illustrates this concept.

=Example 15=
For this example, obtain the reliability equation of the system shown in Figure "An RBD of two computer systems in series where each computer system inherits its diagram from another subdiagram".
==Solution to Example 15==
The system reliability equation is:
 

::<math>{{R}_{System}}={{R}_{Computer1}}\cdot {{R}_{Computer2}} \ </math>

 
:Now:
 
::<math>\begin{align}
{{R}_{Computer1}}= & ({{R}_{Power\,Supply}}\cdot {{R}_{Processor}}\cdot {{R}_{HardDrive}} \\
&\cdot(-{{R}_{Fan}}\cdot {{R}_{Fan}}+{{R}_{Fan}}+{{R}_{Fan}})) \
\end{align}</math>
 
Since the structures of the computer systems are the same, <math>{{R}_{Computer1}}={{R}_{Computer2}}</math> , then substituting the first equation above into the second equation above yields:
 
::<math>\begin{align}
{{R}_{System}}= & ({{R}_{Power\,Supply}}\cdot {{R}_{Processor}}) \\
& \cdot {{R}_{HardDrive}}(-R_{Fan}^{2}+2{{R}_{Fan}}){{)}^{2}}
\end{align}</math>

When using BlockSim to compute the equation, the software will return the first equation above for the system and the second equation above for the subdiagram. Even though BlockSim will make these substitutions internally when performing calculations, it does show them in the System Reliability Equation window.
 
[[Image:chp4image27.png|center|500px|Illustration of subdiagrams.]]
 
[[Image:BS4.23.png|center|600px|An RBD of two computer systems in series where each computer system inherits its diagram from another subdiagram]]
 

= Configurations with Multi Blocks =
By using multi blocks within BlockSim, a single block can represent multiple identical blocks in series or in parallel configuration. This technique is simply a way to save time when creating the RBD and to save space within the diagram. Each item represented by a multi block is a separate entity with identical reliability characteristics to the others. However, each item is not rendered individually within the diagram. In other words, if the RBD contains a multi block that represents three identical components in a series configuration, then each of those components fails according to the same failure distribution but each component may fail at different times. Because the items are arranged reliability-wise in series, if one of those components fails, then the multi block fails. It is also possible to define a multi block with multiple identical components arranged reliability-wise in parallel or <math>k</math> -out-of- <math>n</math> redundancy. Figure "Illustrating multi-blocks" demonstrates the use of multi blocks in BlockSim.
 

= Configurations with Mirrored Blocks =
[[Image:Mirror_block_symbol.PNG|center|300px|]]

While multi blocks allow the analyst to represent multiple items with a single block in an RBD, BlockSim's mirrored blocks can be used to represent a single item with more than one block placed in multiple locations within the diagram. Mirrored blocks can be used to simulate bidirectional paths within a diagram. For example, in a reliability block diagram for a communications system where the lines can operate in two directions, the use of mirrored blocks will facilitate realistic simulations for the system maintainability and availability.
 
 
[[Image:chp4image30.png|center|500px|Illustrating multi-blocks.]]

 
It may also be appropriate to use this type of block if the component performs more than one function and the failure to perform each function has a different reliability-wise impact on the system. In mirrored blocks, the duplicate block behaves in the exact same way that the original block does. The failure times and all maintenance events are the same for each duplicate block as for the original block.
 
 
To better illustrate this consider the following block diagram:
 
[[Image:Mirror_block.PNG|center|400px|]]

 
In this diagram <math>Bm</math> is a mirrored block of <math>B</math>. Since <math>B</math> and <math>Bm</math> are identical, this diagram is equivalent to a diagram with <math>A</math>, <math>B</math> and <math>C</math> in series, as shown next:
 
[[Image:EqvMirror.PNG|center|400px|]]
 

==Example 16==
In the diagram shown in Figure "Electircal network diagram.", electricity can flow in both directions. Successful system operation requires at least one output (O1, O2 or O3) to be working.
 
 
Create a block diagram for this system.
===Solution to Example 16===
The bidirectionality of this system can be modeled using mirrored blocks. The diagram is shown in Figure "RBD of Electircal network diagram".
 
 
Blocks 5A, 7A and 1A are duplicates (or mirrored blocks) of 5, 7 and 1 respectively.

{{reliability block diagrams for failure modes}}

= Symbolic Solutions in BlockSim =

 
Several algebraic solutions in BlockSim were used in the prior examples. BlockSim constructs and displays these equations in different ways, depending on the options chosen. As an example, consider the complex system shown in Figure "Complex RBD".
 

When BlockSim constructs the equation internally, it does so in what we will call a symbolic mode. In this mode, portions of the system are segmented. For this example, the symbolic (internal) solution is shown in Figure "BlocksSim's symbolic solution" and composed of the terms shown in Equations <math>{{I}_{7}}</math>, <math>{{D}_{1}}</math>, <math>{{I}_{11}}</math>, <math>{{R}_{System}}</math>.
 
:Or:
 

::<math>\begin{align}
{{I}_{7}}= & -{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}} \\
& +{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}} \\
& -{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}} \\
& +{{R}_{5}}+{{R}_{6}}+{{R}_{4}} \
\end{align}</math>

 
::<math>{{D}_{1}}=+{{R}_{7}}\cdot {{I}_{7}} \ </math>

 
::<math>\begin{align}
{{I}_{11}}= & -{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{D}_{1}} \\
& +{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& -{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{D}_{1}}-{{R}_{2}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& -{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}-{{R}_{2}}\cdot {{R}_{9}}\cdot {{D}_{1}}-{{R}_{9}}\cdot {{R}_{5}}\cdot {{D}_{1}}-{{R}_{9}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& -{{R}_{5}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{9}}+{{R}_{2}}\cdot {{R}_{10}}+{{R}_{9}}\cdot {{D}_{1}} \\
& +{{R}_{5}}\cdot {{D}_{1}}+{{R}_{8}}\cdot {{D}_{1}} \
\end{align}</math>

 
::<math>{{R}_{System}}=+{{R}_{1}}\cdot {{R}_{11}}\cdot {{I}_{11}} \ </math>
 

[[Image:1 thru 11 network.png|center|500px|Complex RBD]]

[[Image:BS4-30.png|center|500px|BlocksSim's symbolic solution]]
 

 
These terms use tokens to represent portions of the equation. In the symbolic equation setting, one reads the solution from the bottom up, replacing any occurrences of a particular token with its definition. In this case then, and to obtain a system solution, one begins with <math>{{R}_{System}}</math>. <math>{{R}_{System}}</math> contains the token <math>{{I}_{11}}</math>. This is then substituted into <math>{{R}_{System}}</math> yielding:
 

::<math>\begin{align}
{{R}_{System}}= & +{{R}_{1}}\cdot {{R}_{11}}(-{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& +{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& +{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& -{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{D}_{1}}-{{R}_{2}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& +{{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{8}}\cdot {{D}_{1}}-{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}} \\
& -{{R}_{2}}\cdot {{R}_{9}}\cdot {{D}_{1}}-{{R}_{9}}\cdot {{R}_{5}}\cdot {{D}_{1}}-{{R}_{9}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& -{{R}_{5}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{9}} \\
& +{{R}_{2}}\cdot {{R}_{10}}+{{R}_{9}}\cdot {{D}_{1}}+{{R}_{5}}\cdot {{D}_{1}}+{{R}_{8}}\cdot {{D}_{1}}) \
\end{align}</math>
 

Now equation above contains the token <math>{{D}_{1}}</math>. The next step is to substitute <math>{{D}_{1}}</math> into equation above. Doing so yields <math>{{I}_{7}}</math>, shown next.
 

::<math>\begin{align}
{{R}_{System}}= & +{{R}_{1}}\cdot {{R}_{11}}(-{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\
& +{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\
& +{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\
& +{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\
& -{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\
& -{{R}_{2}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}})+{{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\
& -{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}} \\
& -{{R}_{2}}\cdot {{R}_{9}}\cdot ({{R}_{7}}\cdot {{I}_{7}})-{{R}_{9}}\cdot {{R}_{5}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\
& -{{R}_{9}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}})-{{R}_{5}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}})+{{R}_{2}}\cdot {{R}_{9}} \\
& +{{R}_{2}}\cdot {{R}_{10}}+{{R}_{9}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\
& +{{R}_{5}}\cdot ({{R}_{7}}\cdot {{I}_{7}})+{{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}})) \
\end{align}</math>
 

Equation above contains the token <math>{{D}_{1}}</math>. The last step is then to substitute <math>{{R}_{System}}</math> into equation above:
::<math>{{R}_{System}}=+{{R}_{1}}\cdot {{R}_{11}}(-{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+\ \,{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>+{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>+{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>+{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>-{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>-{{R}_{2}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>+{{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>-{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}-{{R}_{2}}\cdot {{R}_{9}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>-{{R}_{9}}\cdot {{R}_{5}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>-{{R}_{9}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>-{{R}_{5}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>+{{R}_{2}}\cdot {{R}_{9}}+{{R}_{2}}\cdot {{R}_{10}}+{{R}_{9}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>+{{R}_{5}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>+{{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}})))</math>
When the complete equation is chosen, BlockSim's System Reliability Equation window performs these token substitutions automatically. It should be pointed out that the complete equation can get very large. While BlockSim internally can deal with millions of terms in an equation, the System Reliability Equation window will only format and display equations up to 64,000 characters. BlockSim uses a 64K memory buffer for displaying equations.
 

===Use IBS (Identical Block Simplification) Option===

When computing the system equation with the "Use IBS" option selected, BlockSim looks for identical blocks (blocks with the same failure characteristics) and attempts to simplify the equation, if possible. The symbolic solution for the system in the prior case, with the "Use IBS" option selected and setting equal reliability block properties, is:
 
::<math>{{R}_{3}}={{R}_{6}}={{R}_{4}}={{R}_{5}}</math>
 

::<math>{{I}_{7}}=+4{{R}_{3}}-6R_{3}^{2}+4R_{3}^{3}-R_{3}^{4}</math>
 

::<math>{{D}_{1}}=+{{R}_{7}}\cdot {{I}_{7}}</math>
 

::<math>\begin{align}
{{I}_{11}}= & -{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{D}_{1}} \\
& +{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& -{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{D}_{1}}-{{R}_{2}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& -{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}-{{R}_{2}}\cdot {{R}_{9}}\cdot {{D}_{1}}-{{R}_{9}}\cdot {{R}_{5}}\cdot {{D}_{1}} \\
& -{{R}_{9}}\cdot {{R}_{8}}\cdot {{D}_{1}}-{{R}_{5}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{9}}+{{R}_{2}}\cdot {{R}_{10}} \\
& +{{R}_{9}}\cdot {{D}_{1}}+{{R}_{5}}\cdot {{D}_{1}}+{{R}_{8}}\cdot {{D}_{1}}
\end{align}</math>

 
::<math>{{R}_{System}}=+{{R}_{1}}\cdot {{R}_{11}}\cdot {{I}_{11}}</math>

 

[[Image:Fig 4.31.PNG|thumb|center|400px|<div align="center"> BlockSim's complete solution for Complex RBD]]

 
When using IBS, the resulting equation is invalidated if any of the block properties (e.g. failure distributions) have changed since the equation was simplified based on those properties.
 
[[Image:Fig 4.32.PNG|thumb|center|400px|<div align="center"> BlockSim's symbolic solution, with the IBS option selected for Complex RBD]]

RBDs and Analytical System Reliability

2012-03-19T22:47:51Z

Dingzhou Cao: /* Symbolic Solutions in BlockSim */

{{Template:bsbook|4}}

 
An overall system reliability prediction can be made by looking at the reliabilities of the components that make up the whole system or product. In this chapter, we will examine the methods of performing such calculations. The reliability-wise configuration of components must be determined beforehand. For this reason, we will first look at different component/subsystem configurations, also known as structural properties ([[Appendix_B:_References | Leemis [17]]]). Unless explicitly stated, the components will be assumed to be statistically independent.
=Component Configurations =
 
In order to construct a reliability block diagram, the reliability-wise configuration of the components must be determined. Consequently, the analysis method used for computing the reliability of a system will also depend on the reliability-wise configuration of the components/subsystems. That configuration can be as simple as units arranged in a pure series or parallel configuration. There can also be systems of combined series/parallel configurations or complex systems that cannot be decomposed into groups of series and parallel configurations. The configuration types considered in this reference include:
 
:• Series configuration. 
:• Simple parallel configuration. 
:• Combined (series and parallel) configuration. 
:• Complex configuration. 
:• ''k''-out-of-''n'' parallel configuration. 
:• Configuration with a load sharing container (presented in Chapter 5). 
:• Configuration with a standby container (presented in Chapter 5). 
:• Configuration with inherited subdiagrams. 
:• Configuration with multi blocks. 
:• Configuration with mirrored blocks. 
 
Each of these configurations will be presented, along with analysis methods, in the sections that follow.
 

[[Category:Completed Theoretical Review]]

= Series Systems =

[[Image:BS4im1.png|center|400px|]]
 
In a series configuration, a failure of any component results in the failure of the entire system. In most cases, when considering complete systems at their basic subsystem level, it is found that these are arranged reliability-wise in a series configuration. For example, a personal computer may consist of four basic subsystems: the motherboard, the hard drive, the power supply and the processor. These are reliability-wise in series and a failure of any of these subsystems will cause a system failure. In other words, all of the units in a series system must succeed for the system to succeed.
 
The reliability of the system is the probability that unit 1 succeeds and unit 2 succeeds and all of the other units in the system succeed. So all <math>n</math> units must succeed for the system to succeed. The reliability of the system is then given by:
 

::<math>\begin{align}
R_S = & P({{X}_{1}}\cap {{X}_{2}}\cap ...\cap {{X}_{n}}) \\
= & P({{X}_{1}})P({{X}_{2}}|{{X}_{1}})P({{X}_{3}}|{{X}_{1}}{{X}_{2}}) \cdot\cdot\cdot P({{X}_{n}}|{{X}_{1}}{{X}_{2}}...{{X}_{n-1}})
\end{align}</math>
 
Where:
 
:• <math>{{R}_{s}}</math> = reliability of the system. 
:• <math>{{X}_{i}}</math> = event of unit <math>i</math> being operational. 
:• <math>P({{X}_{i}})</math> = probability that unit <math>i</math> is operational. 
 
In the case where the failure of a component affects the failure rates of other components (i.e. the life distribution characteristics of the other components change when one component fails), then the conditional probabilities in equation above must be considered.
 
However, in the case of independent components, equation above becomes:
 
::<math>{{R}_{s}}=P({{X}_{1}})P({{X}_{2}})...P({{X}_{n}}) </math>
 
:Or:
 
::<math>{{R}_{s}}=\underset{i=1}{\overset{n}{\mathop \prod }}\,P({{X}_{i}})</math>
 
Or, in terms of individual component reliability:
 
::<math>{{R}_{s}}=\underset{i=1}{\overset{n}{\mathop \prod }}\,{{R}_{i}} \ (eqn 2)</math>
 
In other words, for a pure series system, the system reliability is equal to the product of the reliabilities of its constituent components.
{{Example:Analytical-RBD}}

===Effect of Component Reliability in a Series System===
 
[[Image:Chp4image2.png|center|400px|]]
 
In a series configuration, the component with the least reliability has the biggest effect on the system's reliability. There is a saying that a chain is only as strong as its weakest link. This is a good example of the effect of a component in a series system. In a chain, all the rings are in series and if any of the rings break, the system fails. In addition, the weakest link in the chain is the one that will break first. The weakest link dictates the strength of the chain in the same way that the weakest component/subsystem dictates the reliability of a series system. As a result, the reliability of a series system is always less than the reliability of the least reliable component.

===Example 2===
Consider three components arranged reliability-wise in series, where <math>{{R}_{1}}=70%</math> , <math>{{R}_{2}}=80%</math> and <math>{{R}_{3}}=90%</math> (for a given time).
 
In Table "System reliabilities for combinations of component reliabilities", we can examine the effect of each component's reliability on the overall system reliability. The first row of the table shows the given reliability for each component and the corresponding system reliability for these values. In the second row, the reliability of Component 1 is increased by a value of 10% while keeping the reliabilities of the other two components constant. Similarly, by increasing the reliabilities of Components 2 and 3 in rows 3 and 4 by a value of 10%, while keeping the reliabilities of the other components at the given values, we can observe the effect of each component's reliability on the overall system reliability. It is clear that the highest value for the system's reliability was achieved when the reliability of Component 1, which is the least reliable component, was increased by a value of 10%.
 
[[Image:BS4.T1.png|center|500px|System reliabilities for combinations of component reliabilities.]]
 
This conclusion can also be illustrated graphically, as shown in Figure "Effect of component reliability on the overall system reliability
".
 
Note the slight difference in the slopes of the three lines in Figure "Effect of component reliability on the overall system reliability
". The difference in these slopes represents the difference in the effect of each of the components on the overall system reliability. In other words, the system reliability's rate of change with respect to each component's change in reliability is different. This observation will be explored further when the importance measures of components are considered in later chapters. The rate of change of the system's reliability with respect to each of the components is plotted in Figure "Rate of change of series system reliability when increasing the reliability of each component". It can be seen in Figure "Effect of component reliability on the overall system reliability
" that Component 1 has the steepest slope, which indicates that an increase in the reliability of Component 1 will result in a higher increase in the reliability of the system. In other words, Component 1 has a higher ''reliability importance''.
 
[[Image:BS4.1.png|center|500px|Effect of component reliability on the overall system reliability]]
 
[[Image:BS4.2.png|center|500px|Rate of change of series system reliability when increasing the reliability of each component]]
 
 

===Effect of Number of Components in a Series System===
The number of components is another concern in systems with components connected reliability-wise in series. As the number of components connected in series increases, the system's reliability decreases.
 
Figure "Reliability of a system with n statistically independent and identical components arraged reliability-wise in series" illustrates the effect of the number of components arranged reliability-wise in series on the system's reliability for different component reliability values. This figure also demonstrates the dramatic effect that the number of components has on the system's reliability, particularly when the component reliability is low. In other words, in order to achieve a high system reliability, the component reliability must be high also, especially for systems with many components arranged reliability-wise in series.

===Example 3===
Consider a system that consists of a single component. The reliability of the component is 95%, thus the reliability of the system is 95%. What would the reliability of the system be if there were more than one component (with the same individual reliability) in series? Table "System reliability as a function of the number of components" shows the effect on the system's reliability by adding consecutive components (with the same reliability) in series. Figure "Effect of the number of component in a series configuration for two different casese" illustrates the same concept graphically for components with 90% and 95% reliability.
 
[[Image:Table 4.2.png|center|400px|System reliability as a function of the number of components]]

[[Image:BS4.3.png|center|500px|Reliability of a system with n statistically independent and identical components arraged reliability-wise in series]]
 
 
[[Image:BS4.4.png|center|500px|Effect of the number of component in a series configuration for two different casese]]
 
 

= Simple Parallel Systems =

[[Image:chp4image6.png|center|300px|Simple parallel system]]
 
In a simple parallel system, as shown in Figure "Simple parallel system", at least one of the units must succeed for the system to succeed. Units in parallel are also referred to as redundant units. Redundancy is a very important aspect of system design and reliability in that adding redundancy is one of several methods of improving system reliability. It is widely used in the aerospace industry and generally used in mission critical systems. Other example applications include the RAID computer hard drive systems, brake systems and support cables in bridges.
The probability of failure, or unreliability, for a system with <math>n</math> statistically independent parallel components is the probability that unit 1 fails and unit 2 fails and all of the other units in the system fail. So in a parallel system, all <math>n</math> units must fail for the system to fail. Put another way, if unit 1 succeeds or unit 2 succeeds or any of the <math>n</math> units succeeds, then the system succeeds. The unreliability of the system is then given by:
 

::<math>\begin{align}
{{Q}_{s}}= & P({{X}_{1}}\cap {{X}_{2}}\cap ...\cap {{X}_{n}}) \\
= & P({{X}_{1}})P({{X}_{2}}|{{X}_{1}})P({{X}_{3}}|{{X}_{1}}{{X}_{2}})...P({{X}_{n}}|{{X}_{1}}{{X}_{2}}...{{X}_{n-1}}) \ (eqn 3)
\end{align}</math>
 
Where:
 
:• <math>{{Q}_{s}}</math> = unreliability of the system. 
:• <math>{{X}_{i}}</math> = event of failure of unit <math>i</math> . 
:• <math>P({{X}_{i}})</math> = probability of failure of unit <math>i</math>. 
 
In the case where the failure of a component affects the failure rates of other components, then the conditional probabilities in equation above must be considered.
 
However, in the case of independent components, equation above becomes:
 
::<math>{{Q}_{s}}=P({{X}_{1}})P({{X}_{2}})...P({{X}_{n}})</math>
 
:Or:
 
::<math>{{Q}_{s}}=\underset{i=1}{\overset{n}{\mathop \prod }}\,P({{X}_{i}})</math>
 
Or, in terms of component unreliability:
 
::<math>{{Q}_{s}}=\underset{i=1}{\overset{n}{\mathop \prod }}\,{{Q}_{i}}</math>
 
Observe the contrast with the series system, in which the system reliability was the product of the component reliabilities; whereas the parallel system has the overall system unreliability as the product of the component unreliabilities.
 
The reliability of the parallel system is then given by:
 
::<math>\begin{align}
{{R}_{s}}= & 1-Qs=1-({{Q}_{1}}\cdot {{Q}_{2}}\cdot ...\cdot {{Q}_{n}}) \\
= & 1-[(1-{{R}_{1}})\cdot (1-{{R}_{2}})\cdot ...\cdot (1-{{R}_{n}})] \\
= & 1-\underset{i=1}{\overset{n}{\mathop \prod }}\,(1-{{R}_{i}}) \ (eqn 5)
\end{align}</math>
===Example 4===
Consider a system consisting of three subsystems arranged reliability-wise in parallel. Subsystem 1 has a reliability of 99.5%, Subsystem 2 has a reliability of 98.7% and Subsystem 3 has a reliability of 97.3% for a mission of 100 hours. What is the overall reliability of the system for a 100-hour mission?
====Solution to Example 4====
Since the reliabilities of the subsystems are specified for 100 hours, the reliability of the system for a 100-hour mission is:
::<math>\begin{align}
{{R}_{s}}= & 1-(1-0.9950)\cdot(1-0.9870)\cdot(1-0.9730) \\
= & 1-0.000001755 \\
= & 0.999998245 \
\end{align}</math>

===Effect of Component Reliability in a Parallel Configuration===
When we examined a system of components in series, we found that the least reliable component has the biggest effect on the reliability of the system. However, the component with the highest reliability in a parallel configuration has the biggest effect on the system's reliability, since the most reliable component is the one that will most likely fail last. This is a very important property of the parallel configuration, specifically in the design and improvement of systems.
===Example 5===
Consider three components arranged reliability-wise in parallel with .. = 60%, <math>{{R}_{2}}</math> = 70% and <math>{{R}_{3}}</math> = 80% (for a given time). The corresponding reliability for the system is <math>{{R}_{s}}</math> = 97.6%. In Table "System reliability for combinations of component reliabilities", we can examine the effect of each component's reliability on the overall system reliability. The first row of the table shows the given reliability for each component and the corresponding system reliability for these values. In the second row, the reliability of Component 1 is increased by a value of 10% while keeping the reliabilities of the other two components constant. Similarly, by increasing the reliabilities of Components 2 and 3 in the third and fourth rows by a value of 10% while keeping the reliabilities of the other components at the given values, we can observe the effect of each component's reliability on the overall system reliability. It is clear that the highest value for the system's reliability was achieved when the reliability of Component 3, which is the most reliable component, was increased. Once again, this is the opposite of what was encountered with a pure series system.
 
 
[[Image:4.T3.png|center|500px|System reliability for combinations of component reliabilities]]
 
This conclusion can also be illustrated graphically, as shown in Figure "Rate of change of parallel system reliability when increasing the reliability of each component".

====Effect of Number of Components in Parallel====
In the case of the parallel configuration, the number of components has the opposite effect of the one observed for the series configuration. For a parallel configuration, as the number of components/subsystems increases, the system's reliability increases.
 
Figure "Reliability of a system with n statistically independent and identical components arranged reliability-wise in parallel" illustrates that a high system reliability can be achieved with low-reliability components, provided that there are a sufficient number of components in parallel. Note that Figure "Reliability of a system with n statistically independent and identical components arranged reliability-wise in parallel" is the mirror image of Figure "Reliability of a system with n statistically independent and identical components arraged reliability-wise in series", which presents the effect of the number of components in a series configuration.
 
[[Image:BS4.6.png|center|500px|Rate of change of parallel system reliability when increasing the reliability of each component]]
 
 
[[Image:BS4.7.png|center|500px|Reliability of a system with n statistically independent and identical components arranged reliability-wise in parallel]]
 
 

===Example 6===
Consider a system that consists of a single component. The reliability of the component is 60%, thus the reliability of the system is 60%. What would the reliability of the system be if the system were composed of two, four or six such components in parallel?
 
 
[[Image:4.T4.png|center|500px|System reliability as a function of the number of components]]
 
 
Clearly, the reliability of a system can be improved by adding redundancy. However, it must be noted that doing so is usually costly in terms of additional components, additional weight, volume, etc.
 
Reliability optimization and costs are covered in detail in Chapter [[Reliability_Importance_and_Optimized_Reliability_Allocation_(Analytical)|Component Reliability Importance]].
 
[[Image:BS4.8.png|center|500px|Effect of the number of components in a parallel configuration]]
 
 

= Combination of Series and Parallel =

While many smaller systems can be accurately represented by either a simple series or parallel configuration, there may be larger systems that involve both series and parallel configurations in the overall system. Such systems can be analyzed by calculating the reliabilities for the individual series and parallel sections and then combining them in the appropriate manner. Such a methodology is illustrated in the following example.
===Example 7===
Consider a system with three components. Units 1 and 2 are connected in series and Unit 3 is connected in parallel with the first two, as shown in the next figure.
 
[[Image:chp4image10.png|center|400px|]]
 
What is the reliability of the system if <math>{{R}_{1}}</math> = 99.5%, <math>{{R}_{2}}</math> = 98.7% and <math>{{R}_{3}}</math> = 97.3% at 100 hours?
====Solution to Example 7====
First, the reliability of the series segment consisting of Units 1 and 2 is calculated:
 
::<math>\begin{align}
{{R}_{1,2}}= & {{R}_{1}}\cdot {{R}_{2}} \\
{{R}_{1,2}}= & 0.9950\cdot 0.9870 \\
{{R}_{1,2}}= & 0.982065\text{ or }98.2065%
\end{align}</math>
 
The reliability of the overall system is then calculated by treating Units 1 and 2 as one unit with a reliability of 98.2065% connected in parallel with Unit 3. Therefore:
 
::<math>\begin{align}
{{R}_{s}}= & 1-[(1-0.982065)\cdot (1-0.973000)] \\
{{R}_{s}}= & 1-0.000484245 \\
{{R}_{s}}= & 0.999515755 \\
{{R}_{s}}= & 99.95%
\end{align}</math>
 

= k-out-of-n Parallel Configuration =
The <math>k</math> -out-of- <math>n</math> configuration is a special case of parallel redundancy. This type of configuration requires that at least <math>k</math> components succeed out of the total <math>n</math> parallel components for the system to succeed. For example, consider an airplane that has four engines. Furthermore, suppose that the design of the aircraft is such that at least two engines are required to function for the aircraft to remain airborne. This means that the engines are reliability-wise in a <math>k</math> -out-of- <math>n</math> configuration, where <math>k</math> = 2 and <math>n</math> = 4. More specifically, they are in a 2-out-of-4 configuration.
 
Even though we classified the <math>k</math> -out-of- <math>n</math> configuration as a special case of parallel redundancy, it can also be viewed as a general configuration type. As the number of units required to keep the system functioning approaches the total number of units in the system, the system's behavior tends towards that of a series system. If the number of units required is equal to the number of units in the system, it is a series system. In other words, a series system of statistically independent components is an <math>n</math> -out-of- <math>n</math> system and a parallel system of statistically independent components is a <math>1</math> -out-of- <math>n</math> system.
===Reliability of <math>k</math> -out-of- <math>n</math> Independent and Identical Components===
 
[[Image:chp4image11.png|center|500px|<div align="center">2-out-of-4 configuration </div>]]
 
The simplest case of components in a <math>k</math> -out-of- <math>n</math> configuration is when the components are independent and identical. In other words, all the components have the same failure distribution and whenever a failure occurs, the remaining components are not affected. In this case, the reliability of the system with such a configuration can be evaluated using the binomial distribution, or:
 
::<math>{{R}_{s}}(k,n,R)=\underset{r=k}{\overset{n}{\mathop \sum }}\,\left( \begin{matrix}
n \\
r \\
\end{matrix} \right){{R}^{r}}{{(1-R)}^{n-r}} \ </math>
 

Where:
 
:• <math>n</math> is the total number of units in parallel. 
:• <math>k</math> is the minimum number of units required for system success. 
:• <math>R</math> is the reliability of each unit. 

===Example 8===
Consider a system of six pumps of which at least four must function properly for system success. Each pump has an 85% reliability for the mission duration. What is the probability of success of the system for the same mission duration?
====Solution to Example 8====
Using <math>{{R}_{s}}(k,n,R)=\underset{r=k}{\overset{n}{\mathop \sum }}\,\left( \begin{matrix}
n \\
r \\
\end{matrix} \right){{R}^{r}}{{(1-R)}^{n-r}} \ </math> for <math>k=4</math> and <math>n=6</math> :
 
::<math>\begin{align}
{{R}_{s}}= & \underset{r=4}{\overset{6}{\mathop \sum }}\,\left( \begin{matrix}
6 \\
r \\
\end{matrix} \right){{0.85}^{r}}{{(1-0.85)}^{6-r}} \\
= & \left( \begin{matrix}
6 \\
4 \\
\end{matrix} \right){{0.85}^{4}}{{(1-0.85)}^{2}}+\left( \begin{matrix}
6 \\
5 \\
\end{matrix} \right){{0.85}^{5}}{{(1-0.85)}^{1}} \\
& +\left( \begin{matrix}
6 \\
6 \\
\end{matrix} \right){{0.85}^{6}}{{(1-0.85)}^{0}} \\
= & 0.1762+0.3993+0.3771 \\
= & 95.26%
\end{align}</math>
 
One can examine the effect of increasing the number of units required for system success while the total number of units remains constant (in this example, six units). In Figure "2-out-of-4 configuration", the reliability of the <math>k</math> -out-of-6 configuration was plotted versus different numbers of required units.
 
 
[[Image:4.T5.png|center|500px|Reliability for a k-out-of-6 system for different k values]]
 

 
Note that the system configuration becomes a simple parallel configuration for <math>k</math> = 1 and the system is a six-unit series configuration ( <math>{{(0.85)}^{6}}=</math> 0.377) for <math>k</math> = 6.
<math></math>
 
[[Image:BS4.10.png|center|500px|Reliability of a k-out-of-6 configuration for different k values]]
 
 

===Reliability of Nonidentical <math>k</math> -out-of- <math>n</math> Independent Components===
In the case where the <math>k</math> -out-of- <math>n</math> components are not identical, the reliability must be calculated in a different way. One approach, described in detail later in this chapter, is to use the event space method. In this method, all possible operational combinations are considered in order to obtain the system's reliability. The method is illustrated with the following example.
===Example 9===
Three hard drives in a computer system are configured reliability-wise in parallel. At least two of them must function in order for the computer to work properly. Each hard drive is of the same size and speed, but they are made by different manufacturers and have different reliabilities. The reliability of <math>HD\#1</math> is 0.9, <math>HD\#2</math> is 0.88 and <math>HD\#3</math> is 0.85, all at the same mission time.
====Solution to Example 9====
Since at least two hard drives must be functioning at all times, only one failure is allowed. This is a 2-out-of-3 configuration.
 
The following operational combinations are possible for system success:
#All 3 hard drives operate.
#<math>HD\#1</math> fails, while <math>HDs\,\#2</math> and <math>\#3</math> continue to operate.
#<math>HD\#2</math> fails, while <math>HDs\,\#1</math> and <math>\#3</math> continue to operate.
#<math>HD\#3</math> fails, while <math>HDs\,\#1</math> and <math>\#2</math> continue to operate.
 
The probability of success for the system (reliability) can now be expressed as:
 
::<math>\begin{align}
{{P}_{s}}= & {{R}_{1}}{{R}_{2}}{{R}_{3}}+(1-{{R}_{1}}){{R}_{2}}{{R}_{3}}+{{R}_{1}}(1-{{R}_{2}}){{R}_{3}} \\
& +{{R}_{1}}{{R}_{2}}(1-{{R}_{3}})
\end{align}</math>
 
This equation for the reliability of the system can be reduced to:
 
::<math>{{R}_{s}}={{R}_{1}}{{R}_{2}}+{{R}_{2}}{{R}_{3}}+{{R}_{1}}{{R}_{3}}-2{{R}_{1}}{{R}_{2}}{{R}_{3}}</math>
 
:Or:
 
::<math>{{R}_{s}}=95.86%</math>
 
If all three hard drives had the same reliability, <math>R</math> , then the equation for the reliability of the system could be further reduced to:
 
::<math>{{R}_{s}}=3{{R}^{2}}-2{{R}^{3}}</math>
 
Or, using the binomial approach:
 
::<math>\begin{align}
{{R}_{s}}= & \underset{r=2}{\overset{3}{\mathop \sum }}\,\left( \begin{matrix}
3 \\
r \\
\end{matrix} \right){{R}^{r}}{{(1-R)}^{3-r}} \\
= & \left( \begin{matrix}
3 \\
2 \\
\end{matrix} \right){{R}^{2}}(1-R)+\left( \begin{matrix}
3 \\
3 \\
\end{matrix} \right){{R}^{3}}{{(1-R)}^{0}} \\
= & 3\cdot {{R}^{2}}(1-R)+{{R}^{3}} \\
= & 3{{R}^{2}}-2{{R}^{3}}
\end{align}</math>
 
The example can be repeated using BlockSim. The following graphic demonstrates the RBD for the system.
 
<math></math>
 
 
[[Image:Fig_harddrive.PNG|center|400px|]]
 

The RBD is analyzed and the system reliability equation is returned. Figure "System equation results for Example 9" shows the equation returned by BlockSim.
 
[[Image:BS4.11.png|thumb|center|400px|System equation results for Example 9]]
 
Using the Analytical Quick Calculation Pad, the reliability can be calculated to be 0.9586. Figure "Reliability results for Example 9" shows the returned result.
 
Note that you are not required to enter a mission end time for this system into the Analytical QCP because all of the components are static and thus the reliability results are independent of time.
[[Image:Fig 4.12.PNG|thumb|center|500px|<div align="center">Reliability results for Example 9 </div>]]

===Example 10===
Consider the four-engine aircraft discussed previously. If we were to change the problem statement to two out of four engines are required, however no two engines on the same side may fail, then the block diagram would change to the configuration shown in Figure "Block diagram for Example 10".
 
<math></math>
 
Note that this is the same as having two engines in parallel on each wing and then putting the two wings in series.
 

 
[[Image:4.13.png|center|500px|<div align="center">Block diagram for Example 10 </div>]]
 
 

=Consecutive <math>k</math>-out-of-<math>n</math> : <math>F</math> Redundancy=
There are other multiple redundancy types and multiple industry terms. One such example is what is referred to as a consecutive <math>k</math>-out-of-<math>n</math> : <math>F</math> system. To illustrate this configuration type, consider a telecommunications system that consists of a transmitter and receiver with six relay stations to connect them. The relays are situated so that the signal originating from one station can be picked up by the next two stations down the line. For example, a signal from the transmitter can be received by Relay 1 and Relay 2, a signal from Relay 1 can be received by Relay 2 and Relay 3, and so forth. Thus, this arrangement would require two consecutive relays to fail for the system to fail. A diagram of this configuration is shown in Figure "RBD for the consecutive ''k''-out-of-''n'': ''F'' system".
 
<math></math>
 
This type of a configuration is also referred to as a complex system. Complex systems are discussed in the next section.
[[Image:4.14.png|center|600px|<div align="center">RBD for the consecutive ''k''-out-of-''n'': ''F'' system </div>]]

= Complex Systems =
In many cases, it is not easy to recognize which components are in series and which are in parallel in a complex system. The network shown in Figure "Example of a complex system" is a good example of such a complex system.
 
<math></math>
 
 
[[Image:4.15.png|center|500px|<div align="center">Example of a complex system </div>]]
 
The system in Figure "Example of a complex system" cannot be broken down into a group of series and parallel systems. This is primarily due to the fact that component <math>C</math> has two paths leading away from it, whereas <math>B</math> and <math>D</math> have only one. Several methods exist for obtaining the reliability of a complex system including:
 
:• The decomposition method. 
:• The event space method. 
:• The path-tracing method. 
===Decomposition Method===
The decomposition method is an application of the law of total probability. It involves choosing a "key" component and then calculating the reliability of the system twice: once as if the key component failed ( <math>R=0</math> ) and once as if the key component succeeded <math>(R=1)</math> . These two probabilities are then combined to obtain the reliability of the system, since at any given time the key component will be failed or operating. Using probability theory, the equation is:
 
::<math>\begin{align}
{{R}_{s}}= & P(s\cap A)+P(s\cap \overline{A}) \\
= & P(s|A)P(A)+P(s|\overline{A})P(\overline{A})
\end{align}</math>
====Illustration of the Decomposition Method====
Consider three units in series.
 
:• <math>A</math> is the event of Unit 1 success. 
:• <math>B</math> is the event of Unit 2 success. 
:• <math>C</math> is the event of Unit 3 success. 
:• <math>s</math> is the event of system success. 
 
First, select a "key" component for the system. Selecting Unit 1, the probability of success of the system is:
 
::<math>{{R}_{s}}=P(s|A)P(A)+P(s|\overline{A})P(\overline{A})</math>
 
If Unit 1 is good, then:
 
::<math>P(s|A)={{R}_{2}}{{R}_{3}}</math>
 
That is, if Unit 1 is operating, the probability of the success of the system is the probability of Units 2 and 3 succeeding.
 
If Unit 1 fails, then:
 
::<math>P(s|\overline{A})=0</math>
 
That is, if Unit 1 is not operating, the system has failed since a series system requires all of the components to be operating for the system to operate.
Thus the reliability of the system is:
 
::<math>{{R}_{s}}={{R}_{2}}{{R}_{3}}P(A)={{R}_{1}}{{R}_{2}}{{R}_{3}}</math>
 

====Another Illustration of the Decomposition Method====
Consider the following system:
 
[[Image:chp4image14.png|center|300px|]]
 
:• <math>A</math> is the event of Unit 1 success.
:• <math>B</math> is the event of Unit 2 success.
:• <math>C</math> is the event of Unit 3 success.
:• <math>s</math> is the event of system success.
 
Selecting Unit 3 as the `` key'' component, the system reliability is:
 
::<math>{{R}_{s}}=P(s|C)P(C)+P(s|\overline{C})P(\overline{C})</math>
 
If Unit 3 survives, then:
 
::<math>P(s|C)=1</math>
 
That is, since Unit 3 represents half of the parallel section of the system, then as long as it is operating, the entire system operates.
 
If Unit 3 fails, then the system is reduced to:
 
[[Image:chp4image15.png|center|400px|]]
 

::<math>P(s|\overline{C})={{R}_{1}}{{R}_{2}}</math>
 
The reliability of the system is given by:
 
::<math>{{R}_{s}}=P(C)+{{R}_{1}}{{R}_{2}}P(\overline{C})={{R}_{3}}+{{R}_{1}}{{R}_{2}}(1-{{R}_{3}})</math>
 
:Or:
 
::<math>{{R}_{s}}={{R}_{3}}+{{R}_{1}}{{R}_{2}}-{{R}_{1}}{{R}_{2}}{{R}_{3}}</math>

===Event Space Method===
The event space method is an application of the mutually exclusive events axiom. All mutually exclusive events are determined and those that result in system success are considered. The reliability of the system is simply the probability of the union of all mutually exclusive events that yield a system success. Similarly, the unreliability is the probability of the union of all mutually exclusive events that yield a system failure. This is illustrated in the following example.
 
====Illustration of the Event Space Method====
 
Consider the following system, with reliabilities R1, R2 and R3 for a given time.
 
[[Image:chp4image14.png|center|400px|]]
:• <math>A</math> is the event of Unit 1 success. 
:• <math>B</math> is the event of Unit 2 success. 
:• <math>C</math> is the event of Unit 3 success. 
 
The mutually exclusive system events are:
 
::<math>\begin{align}
X1= & ABC-\text{all units succeed}\text{.} \\
X2= & \overline{A}BC-\text{only Unit 1 fails}\text{.} \\
X3= & A\overline{B}C-\text{only Unit 2 fails}\text{.} \\
X4= & AB\overline{C}-\text{only Unit 3 fails}\text{.} \\
X5= & \overline{AB}C-\text{Units 1 and 2 fail}\text{.} \\
X6= & \overline{A}B\overline{C}-\text{Units 1 and 3 fail}\text{.} \\
X7= & A\overline{BC}-\text{Units 2 and 3 fail}\text{.} \\
X8= & \overline{ABC}-\text{all units fail}\text{.}
\end{align}</math>

 
System events <math>{{X}_{6}}</math> , <math>{{X}_{7}}</math> and <math>{{X}_{8}}</math> result in system failure. Thus the probability of failure of the system is:
 
::<math>{{P}_{f}}=P({{X}_{6}}\cup {{X}_{7}}\cup {{X}_{8}})</math>
 
Since events <math>{{X}_{6}}</math> , <math>{{X}_{7}}</math> and <math>{{X}_{8}}</math> are mutually exclusive, then:
 
::<math>{{P}_{f}}=P({{X}_{6}})+P({{X}_{7}})+P({{X}_{8}})</math>
 
:And:
 
::<math>\begin{align}
P({{X}_{6}})= & P(\overline{A}B\overline{C})=(1-{{R}_{1}})({{R}_{2}})(1-{{R}_{3}}) \\
P({{X}_{7}})= & P(A\overline{B}\overline{C})=({{R}_{1}})(1-{{R}_{2}})(1-{{R}_{3}}) \\
P({{X}_{8}})= & P(\overline{A}\overline{B}\overline{C})=(1-{{R}_{1}})(1-{{R}_{2}})(1-{{R}_{3}})
\end{align}</math>
 
Combining terms yields:
 

::<math>{{P}_{f}}=1-{{R}_{1}}{{R}_{2}}-{{R}_{3}}+{{R}_{1}}{{R}_{2}}{{R}_{3}}</math>
 
:Since:
 
::<math>{{R}_{s}}=1-{{P}_{f}}</math>
 
:Then:
 
::<math>{{R}_{s}}={{R}_{1}}{{R}_{2}}+{{R}_{3}}-{{R}_{1}}{{R}_{2}}{{R}_{3}}</math>
 
This is of course the same result as the one obtained previously using the decomposition method.
If <math>{{R}_{1}}</math> = 99.5%, <math>{{R}_{2}}</math> = 98.7% and <math>{{R}_{3}}</math> = 97.3%, then:
 
::<math>\begin{align}
{{R}_{s}}= & 0.995\cdot 0.987+0.973-0.995\cdot 0.987\cdot 0.973 \\
= & 0.999515755
\end{align}</math>
 
:Or:
 
::<math>{{R}_{s}}=99.95%</math>
 

===Path-Tracing Method===
With the path-tracing method, every path from a starting point to an ending point is considered. Since system success involves having at least one path available from one end of the RBD to the other, as long as at least one path from the beginning to the end of the path is available, then the system has not failed. One could consider the RBD to be a plumbing schematic. If a component in the system fails, the "water" can no longer flow through it. As long as there is at least one path for the "water" to flow from the start to the end of the system, the system is successful. This method involves identifying all of the paths the "water" could take and calculating the reliability of the path based on the components that lie along that path. The reliability of the system is simply the probability of the union of these paths. In order to maintain consistency of the analysis, starting and ending blocks for the system must be defined.

===Example 11===
Obtain the reliability equation of the following system.
 
[[Image:BS4ex11.png|center|400px|]]
 
====Solution to Example 11====
The successful paths for this system are:
 

::<math>{{X}_{1}}=ABD\text{ and }{{X}_{2}}=ACD</math>

 

The reliability of the system is simply the probability of the union of these paths:
 

::<math>{{R}_{s}}=P({{X}_{1}}\cup {{X}_{2}})</math>
 

::<math>\begin{align}
P({{X}_{1}}\cup {{X}_{2}})= & P({{X}_{1}})+P({{X}_{2}})-P({{X}_{1}}\cap {{X}_{2}}) \\
= & P(ABD)+P(ACD)-P(ABCD)
\end{align}</math>

 

:Thus:
 

::<math>{{R}_{s}}={{R}_{A}}{{R}_{B}}{{R}_{D}}+{{R}_{A}}{{R}_{C}}{{R}_{D}}-{{R}_{A}}{{R}_{B}}{{R}_{C}}{{R}_{D}}</math>
 

===Example 12===
Obtain the reliability equation of the following system.
 
[[Image:Chp4image14.png|center|400px|]]
 
 

====Solution to Example 12====
Assume starting and ending blocks that cannot fail, as shown next.
 
[[Image:BS.4ex12.2.png|center|500px|]]
 
The paths for this system are:
 
::<math>{{X}_{1}}=1,2\text{ and }{{X}_{2}}=3</math>

 
The probability of success of the system is given by:

::<math>\begin{align}
P({{X}_{1}}\cup {{X}_{2}})= & P({{X}_{1}})+P({{X}_{2}})-P({{X}_{1}}\cap {{X}_{2}}) \\
= & P(1,2)+P(3)-P(1,2,3)
\end{align}</math>
 

:Or:
 

::<math>{{R}_{s}}={{R}_{1}}{{R}_{2}}+{{R}_{3}}-{{R}_{1}}{{R}_{2}}{{R}_{3}}</math>
 

====Starting and Ending Blocks in BlockSim====
Note that BlockSim requires that all diagrams start from a single block and end on a single block. To meet this requirement for this example, we arbitrarily added a starting and an ending block, as shown in Figure "BlockSim representation of the RBD for Example 12". These blocks can be set to a cannot fail condition, or <math>R=1</math> , and thus not affect the outcome. However, when the analysis is performed in BlockSim, the returned equation will include terms for the non-failing blocks, as shown in Figure "BlockSim solution Example 12" and equation.

 

::<math>{{R}_{system}}=({{R}_{S}}\cdot {{R}_{E}}(-{{R}_{1}}\cdot {{R}_{2}}\cdot {{R}_{3}}+{{R}_{1}}\cdot {{R}_{2}}+{{R}_{3}})) \ </math>
 

Note that since <math>{{R}_{S}}={{R}_{E}}=1</math> , the system equation above, can be reduced to:
 

::<math>\begin{align}
{{R}_{system}}= & (1\cdot 1(-{{R}_{1}}\cdot {{R}_{2}}\cdot {{R}_{3}}+{{R}_{1}}\cdot {{R}_{2}}+{{R}_{3}})) \\
= & -{{R}_{1}}\cdot {{R}_{2}}\cdot {{R}_{3}}+{{R}_{1}}\cdot {{R}_{2}}+{{R}_{3}}
\end{align}</math>

 
This is equivalent to <math>{{R}_{s}}={{R}_{1}}{{R}_{2}}+{{R}_{3}}-{{R}_{1}}{{R}_{2}}{{R}_{3}}</math>. The reason that BlockSim includes all items regardless of whether they can fail or not is because BlockSim only recomputes the equation when the system structure has changed. What this means is that the user can alter the failure characteristics of an item without altering the diagram structure. For example, a block that was originally set not to fail can be re-set to a failure distribution and thus it would need to be used in subsequent analyses.
 
[[Image:4.16.png|center|400px|<div align="center">BlockSim representation of the RBD for Example 12 </div>]]
 
 
[[Image:Fig 4.17.PNG|thumb|center|400px|<div align="center">BlockSim solution Example 12 </div>]]
 

===Example 13===
For this example:
 
::<math>a)</math> Determine the reliability equation of the system shown in Figure "Diagram for Example 13" using the decomposition method.
 
::<math>b)</math> Determine the reliability equation of the same system using BlockSim.
====Solution to Example 13====
To obtain the solution:
 
::<math>a)</math> Choose A as the key component, then:
 

::<math>\begin{align}
{{R}_{s}}= & P(s|A)P(A)+P(s|\overline{A})P(\overline{A}) \\
P(s| A)= &{{R}_{B}}{{R}_{F}}\left[ 1-\left( 1-{{R}_{C}} \right)\left( 1-{{R}_{E}} \right) \right] \\
P(s| \overline{A})= &{{R}_{B}}{{R}_{D}}{{R}_{E}}{{R}_{F}} \\
{{R}_{s}}= & \left[ {{R}_{B}}{{R}_{F}}\left[ 1-\left( 1-{{R}_{C}} \right)\left( 1-{{R}_{E}} \right) \right] \right]{{R}_{A}}+\left[ {{R}_{B}}{{R}_{D}}{{R}_{E}}{{R}_{F}} \right](1-{{R}_{A}})
\end{align}</math>

 
::<math>b)</math> Using BlockSim:

 
::<math>{{R}_{s}}={{R}_{B}}\cdot {{R}_{F}}(-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}+{{R}_{A}}\cdot {{R}_{E}}+{{R}_{D}}\cdot {{R}_{E}})</math>
 
[[Image:chp4image21.png|center|400px|<div align="center">Diagram for Example 13. </div>]]
 
[[Image:Fig_4.19.PNG|thumb|center|400px|<div align="center">BlockSim solution for Example 13 </div>]]
 

= Difference Between Physical and Reliability-Wise Arrangement =
Reliability block diagrams are created in order to illustrate the way that components are arranged reliability-wise in a system. So far we have described possible structural properties of a system of components, such as series, parallel, etc. These structural properties, however, refer to the system's state of success or failure based on the states of its components. The physical structural arrangement, even though clearly related to the reliability-wise arrangement, is not necessarily identical to it.
===Example 14===
Consider the following circuit:
 
[[Image:BS3sigma.png|center|400px|]]
 
The equivalent resistance must always be less than <math>1.2\Omega </math> .
 
Draw the reliability block diagram for this circuit.
====Solution to Example 14====
First, let's consider the case where all three resistors operate:
 

::<math>\begin{align}
\frac{1}{{{r}_{eq}}}= & \frac{1}{{{r}_{1}}}+\frac{1}{{{r}_{2}}}+\frac{1}{{{r}_{3}}} \\
= & \frac{1}{3}+\frac{1}{3}+\frac{1}{3} \\
= & 1\Omega
\end{align}</math>

 
Thus, when all components operate, the equivalent resistance is <math>1\Omega </math> , which is less than the maximum resistance of <math>1.2\Omega </math> .
 

Next, consider the case where one of the resistors fails open. In this case, the resistance for the resistor is infinite and the equivalent resistance is:
 

::<math>\frac{1}{{{r}_{eq}}}=\frac{1}{\infty }+\frac{1}{3}+\frac{1}{3}=\frac{2}{3}</math>
 

:Thus:
 

::<math>{{r}_{eq}}=1.5\Omega >1.2\Omega \text{ - System failed}\text{.}</math>

 
If two resistors fail open (e.g. #1 and #2), the equivalent resistance is:
 

::<math>\frac{1}{{{r}_{eq}}}=\frac{1}{\infty }+\frac{1}{\infty }+\frac{1}{3}=\frac{1}{3}</math>

 
:Thus:
 

::<math>{{r}_{eq}}=3\Omega >1.2\Omega \text{ - System failed}\text{.}</math>

 
If all three resistors fail open:
 

::<math>{{r}_{eq}}=\infty >1.2\Omega \text{ - System failed}\text{.}</math>

 
Thus, if <math>{{r}_{1}}</math> , <math>{{r}_{2}}</math> , <math>{{r}_{3}}</math>, or any combination of the three fails, the system fails. Put another way, <math>{{r}_{1}}</math> and <math>{{r}_{2}}</math> and <math>{{r}_{3}}</math> must succeed in order for the system to succeed.
 
The RBD is:

[[Image:chp4image24.png|center|400px|]]
 
In this example it can be seen that even though the three components were physically arranged in parallel, their reliability-wise arrangement is in series.

= Configurations with Load Sharing and Standby Redundancy =
Units in load sharing redundancy exhibit different failure characteristics when one or more fail. In Figure "An RBD with a load sharing container", blocks 1, 2 and 3 are in a load sharing container in BlockSim and have their own failure characteristics. All three must fail for the container to fail. However, as individual items fail, the failure characteristics of the remaining units change since they now have to carry a higher load to compensate for the failed ones. The failure characteristics of each block in a load sharing container are defined using both a life distribution and a life-stress relationship that describe how the life distribution changes as the load changes.
 

Standby redundancy configurations consist of items that are inactive and available to be called into service when/if an active item fails (i.e. the items are on standby). Within BlockSim, a container block with other blocks inside is used to better achieve and streamline the representation and analysis of standby configurations. The container serves a dual purpose. The first is to clearly delineate and define the standby relationships between the active unit(s) and standby unit(s). The second is to serve as the manager of the switching process. For this purpose, the container can be defined with its own probability of successfully activating standby units when needed. Figure "An RBD with a standby container" includes a standby container with three items in standby configuration where one component is active while the other two components are idle. One block within the container must be operating, otherwise the container will fail, leading to a system failure (since the container block is part of a series configuration in this example).
 

Because both of these concepts are better understood when time dependency is considered, they are addressed in more detail in Chapter [[Time-Dependent_System_Reliability_(Analytical)|Time-Dependent System Reliability (Analytical)]].
 
[[Image:Fig_4.20.PNG|center|400px|<div align="center">An RBD with a load sharing container. </div>]]
 
[[Image:Fig_4.21.PNG|center|400px|<div align="center">An RBD with a standby container.</div>]]
 

= Configurations with Inherited Subdiagrams =
Within BlockSim, a subdiagram block inherits some or all of its properties from another block diagram. This allows the analyst to maintain separate diagrams for portions of a system and incorporate those diagrams as components of another diagram. With this technique, it is possible to generate and analyze extremely complex diagrams representing the behavior of many subsystems in a manageable way. In Figure "Illustration of subdiagrams", Subdiagram Block A in the top diagram represents the series configuration of the subsystem reflected in the middle diagram, while Subdiagram Block G in the middle diagram represents the series configuration of the subsubsystem in the bottom diagram. Figure "An RBD of two computer systems in series where each computer system inherits its diagram from another subdiagram" again illustrates this concept.

=Example 15=
For this example, obtain the reliability equation of the system shown in Figure "An RBD of two computer systems in series where each computer system inherits its diagram from another subdiagram".
==Solution to Example 15==
The system reliability equation is:
 

::<math>{{R}_{System}}={{R}_{Computer1}}\cdot {{R}_{Computer2}} \ </math>

 
:Now:
 
::<math>\begin{align}
{{R}_{Computer1}}= & ({{R}_{Power\,Supply}}\cdot {{R}_{Processor}}\cdot {{R}_{HardDrive}} \\
&\cdot(-{{R}_{Fan}}\cdot {{R}_{Fan}}+{{R}_{Fan}}+{{R}_{Fan}})) \
\end{align}</math>
 
Since the structures of the computer systems are the same, <math>{{R}_{Computer1}}={{R}_{Computer2}}</math> , then substituting the first equation above into the second equation above yields:
 
::<math>\begin{align}
{{R}_{System}}= & ({{R}_{Power\,Supply}}\cdot {{R}_{Processor}}) \\
& \cdot {{R}_{HardDrive}}(-R_{Fan}^{2}+2{{R}_{Fan}}){{)}^{2}}
\end{align}</math>

When using BlockSim to compute the equation, the software will return the first equation above for the system and the second equation above for the subdiagram. Even though BlockSim will make these substitutions internally when performing calculations, it does show them in the System Reliability Equation window.
 
[[Image:chp4image27.png|center|500px|Illustration of subdiagrams.]]
 
[[Image:BS4.23.png|center|600px|An RBD of two computer systems in series where each computer system inherits its diagram from another subdiagram]]
 

= Configurations with Multi Blocks =
By using multi blocks within BlockSim, a single block can represent multiple identical blocks in series or in parallel configuration. This technique is simply a way to save time when creating the RBD and to save space within the diagram. Each item represented by a multi block is a separate entity with identical reliability characteristics to the others. However, each item is not rendered individually within the diagram. In other words, if the RBD contains a multi block that represents three identical components in a series configuration, then each of those components fails according to the same failure distribution but each component may fail at different times. Because the items are arranged reliability-wise in series, if one of those components fails, then the multi block fails. It is also possible to define a multi block with multiple identical components arranged reliability-wise in parallel or <math>k</math> -out-of- <math>n</math> redundancy. Figure "Illustrating multi-blocks" demonstrates the use of multi blocks in BlockSim.
 

= Configurations with Mirrored Blocks =
[[Image:Mirror_block_symbol.PNG|center|300px|]]

While multi blocks allow the analyst to represent multiple items with a single block in an RBD, BlockSim's mirrored blocks can be used to represent a single item with more than one block placed in multiple locations within the diagram. Mirrored blocks can be used to simulate bidirectional paths within a diagram. For example, in a reliability block diagram for a communications system where the lines can operate in two directions, the use of mirrored blocks will facilitate realistic simulations for the system maintainability and availability.
 
 
[[Image:chp4image30.png|center|500px|Illustrating multi-blocks.]]

 
It may also be appropriate to use this type of block if the component performs more than one function and the failure to perform each function has a different reliability-wise impact on the system. In mirrored blocks, the duplicate block behaves in the exact same way that the original block does. The failure times and all maintenance events are the same for each duplicate block as for the original block.
 
 
To better illustrate this consider the following block diagram:
 
[[Image:Mirror_block.PNG|center|400px|]]

 
In this diagram <math>Bm</math> is a mirrored block of <math>B</math>. Since <math>B</math> and <math>Bm</math> are identical, this diagram is equivalent to a diagram with <math>A</math>, <math>B</math> and <math>C</math> in series, as shown next:
 
[[Image:EqvMirror.PNG|center|400px|]]
 

==Example 16==
In the diagram shown in Figure "Electircal network diagram.", electricity can flow in both directions. Successful system operation requires at least one output (O1, O2 or O3) to be working.
 
 
Create a block diagram for this system.
===Solution to Example 16===
The bidirectionality of this system can be modeled using mirrored blocks. The diagram is shown in Figure "RBD of Electircal network diagram".
 
 
Blocks 5A, 7A and 1A are duplicates (or mirrored blocks) of 5, 7 and 1 respectively.

{{reliability block diagrams for failure modes}}

= Symbolic Solutions in BlockSim =

 
Several algebraic solutions in BlockSim were used in the prior examples. BlockSim constructs and displays these equations in different ways, depending on the options chosen. As an example, consider the complex system shown in Figure "Complex RBD".
 

When BlockSim constructs the equation internally, it does so in what we will call a symbolic mode. In this mode, portions of the system are segmented. For this example, the symbolic (internal) solution is shown in Figure "BlocksSim's symbolic solution" and composed of the terms shown in Equations <math>{{I}_{7}}</math>, <math>{{D}_{1}}</math>, <math>{{I}_{11}}</math>, <math>{{R}_{System}}</math>.
 
:Or:
 

::<math>\begin{align}
{{I}_{7}}= & -{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}} \\
& +{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}} \\
& -{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}} \\
& +{{R}_{5}}+{{R}_{6}}+{{R}_{4}} \
\end{align}</math>

 
::<math>{{D}_{1}}=+{{R}_{7}}\cdot {{I}_{7}} \ </math>

 
::<math>\begin{align}
{{I}_{11}}= & -{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{D}_{1}} \\
& +{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& -{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{D}_{1}}-{{R}_{2}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& -{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}-{{R}_{2}}\cdot {{R}_{9}}\cdot {{D}_{1}}-{{R}_{9}}\cdot {{R}_{5}}\cdot {{D}_{1}}-{{R}_{9}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& -{{R}_{5}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{9}}+{{R}_{2}}\cdot {{R}_{10}}+{{R}_{9}}\cdot {{D}_{1}} \\
& +{{R}_{5}}\cdot {{D}_{1}}+{{R}_{8}}\cdot {{D}_{1}} \
\end{align}</math>

 
::<math>{{R}_{System}}=+{{R}_{1}}\cdot {{R}_{11}}\cdot {{I}_{11}} \ </math>
 

[[Image:1 thru 11 network.png|center|500px|Complex RBD]]

[[Image:BS4-30.png|center|500px|BlocksSim's symbolic solution]]
 
[[Image:Fig 4.30.png|center|400px|<div align="center"> BlocksSim's symbolic solution]]
 
 
These terms use tokens to represent portions of the equation. In the symbolic equation setting, one reads the solution from the bottom up, replacing any occurrences of a particular token with its definition. In this case then, and to obtain a system solution, one begins with <math>{{R}_{System}}</math>. <math>{{R}_{System}}</math> contains the token <math>{{I}_{11}}</math>. This is then substituted into <math>{{R}_{System}}</math> yielding:
 

::<math>\begin{align}
{{R}_{System}}= & +{{R}_{1}}\cdot {{R}_{11}}(-{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& +{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& +{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& -{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{D}_{1}}-{{R}_{2}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& +{{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{8}}\cdot {{D}_{1}}-{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}} \\
& -{{R}_{2}}\cdot {{R}_{9}}\cdot {{D}_{1}}-{{R}_{9}}\cdot {{R}_{5}}\cdot {{D}_{1}}-{{R}_{9}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& -{{R}_{5}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{9}} \\
& +{{R}_{2}}\cdot {{R}_{10}}+{{R}_{9}}\cdot {{D}_{1}}+{{R}_{5}}\cdot {{D}_{1}}+{{R}_{8}}\cdot {{D}_{1}}) \
\end{align}</math>
 

Now equation above contains the token <math>{{D}_{1}}</math>. The next step is to substitute <math>{{D}_{1}}</math> into equation above. Doing so yields <math>{{I}_{7}}</math>, shown next.
 

::<math>\begin{align}
{{R}_{System}}= & +{{R}_{1}}\cdot {{R}_{11}}(-{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\
& +{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\
& +{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\
& +{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\
& -{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\
& -{{R}_{2}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}})+{{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\
& -{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}} \\
& -{{R}_{2}}\cdot {{R}_{9}}\cdot ({{R}_{7}}\cdot {{I}_{7}})-{{R}_{9}}\cdot {{R}_{5}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\
& -{{R}_{9}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}})-{{R}_{5}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}})+{{R}_{2}}\cdot {{R}_{9}} \\
& +{{R}_{2}}\cdot {{R}_{10}}+{{R}_{9}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\
& +{{R}_{5}}\cdot ({{R}_{7}}\cdot {{I}_{7}})+{{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}})) \
\end{align}</math>
 

Equation above contains the token <math>{{D}_{1}}</math>. The last step is then to substitute <math>{{R}_{System}}</math> into equation above:
::<math>{{R}_{System}}=+{{R}_{1}}\cdot {{R}_{11}}(-{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+\ \,{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>+{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>+{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>+{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>-{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>-{{R}_{2}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>+{{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>-{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}-{{R}_{2}}\cdot {{R}_{9}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>-{{R}_{9}}\cdot {{R}_{5}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>-{{R}_{9}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>-{{R}_{5}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>+{{R}_{2}}\cdot {{R}_{9}}+{{R}_{2}}\cdot {{R}_{10}}+{{R}_{9}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>+{{R}_{5}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))</math> <math>+{{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}})))</math>
When the complete equation is chosen, BlockSim's System Reliability Equation window performs these token substitutions automatically. It should be pointed out that the complete equation can get very large. While BlockSim internally can deal with millions of terms in an equation, the System Reliability Equation window will only format and display equations up to 64,000 characters. BlockSim uses a 64K memory buffer for displaying equations.
 

===Use IBS (Identical Block Simplification) Option===

When computing the system equation with the "Use IBS" option selected, BlockSim looks for identical blocks (blocks with the same failure characteristics) and attempts to simplify the equation, if possible. The symbolic solution for the system in the prior case, with the "Use IBS" option selected and setting equal reliability block properties, is:
 
::<math>{{R}_{3}}={{R}_{6}}={{R}_{4}}={{R}_{5}}</math>
 

::<math>{{I}_{7}}=+4{{R}_{3}}-6R_{3}^{2}+4R_{3}^{3}-R_{3}^{4}</math>
 

::<math>{{D}_{1}}=+{{R}_{7}}\cdot {{I}_{7}}</math>
 

::<math>\begin{align}
{{I}_{11}}= & -{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{D}_{1}} \\
& +{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& -{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{D}_{1}}-{{R}_{2}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\
& -{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}-{{R}_{2}}\cdot {{R}_{9}}\cdot {{D}_{1}}-{{R}_{9}}\cdot {{R}_{5}}\cdot {{D}_{1}} \\
& -{{R}_{9}}\cdot {{R}_{8}}\cdot {{D}_{1}}-{{R}_{5}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{9}}+{{R}_{2}}\cdot {{R}_{10}} \\
& +{{R}_{9}}\cdot {{D}_{1}}+{{R}_{5}}\cdot {{D}_{1}}+{{R}_{8}}\cdot {{D}_{1}}
\end{align}</math>

 
::<math>{{R}_{System}}=+{{R}_{1}}\cdot {{R}_{11}}\cdot {{I}_{11}}</math>

 

[[Image:Fig 4.31.PNG|thumb|center|400px|<div align="center"> BlockSim's complete solution for Complex RBD]]

 
When using IBS, the resulting equation is invalidated if any of the block properties (e.g. failure distributions) have changed since the equation was simplified based on those properties.
 
[[Image:Fig 4.32.PNG|thumb|center|400px|<div align="center"> BlockSim's symbolic solution, with the IBS option selected for Complex RBD]]

File:BS4-30.png

2012-03-19T22:44:50Z

Dingzhou Cao:

Template:Xfmea Control Plan Example

2012-03-19T21:10:20Z

Dingzhou Cao: Created page with 'Let's look at a PFMEA of the "Front Door L.H." in a automobile.'

Let's look at a PFMEA of the "Front Door L.H." in a automobile.

Template:Risk Discovery Analysis Example for XFMEA

2012-03-19T20:41:37Z

Dingzhou Cao:

Let’s examine the new Risk Discovery analysis feature in Xfmea through a fictional example. Suppose that a technology company is developing a new complex product. It is a multi-function printer that prints, copies, scans and has several other features, such as scan to e-mail, document management, etc.

Figure below shows the system hierarchy, with some subsystems fully expanded down to the component level and others collapsed at the subsystem level.
[[Image:Printer System for FMEA Example.png|thumb|center|300px|]]

As a best practice, the team first conducts an FMEA at the top (system) level in order to address any potential problems related to interactions and interfaces. Starting the analysis at lower levels has been proven to be a practice that can lead to omission of interactions between subsystems. In most complex systems, interactions will account for more than 50% of the total failure modes that the system will experience. In such cases, it is essential to start with a top-down approach.

Once the top (system) level FMEA has been completed, the next question that the FMEA team faces is which subsystems and components to focus on. This is where the Risk Discovery feature in Xfmea is used.

The team adds a Risk Discovery analysis for each of the main subsystems of the product: the printing system, the copying system, the scanning system and the document management system. Xfmea supports two configurable methods for this type of analysis: a set of yes/no questions (where a yes answer indicates that the assembly/component poses a risk that warrants more detailed analysis) or a set of predefined rating scales (which can be used to calculate an overall rating for each assembly/component). For this example, we will assume that the organization has selected to use the Questions tab with the sample questions that are shipped by default with the software. These questions address the issues that would typically be considered with a change point analysis approach (new technology, new application, historical problems, supplier capability), together with other relevant issues (safety, regulatory requirements and mission criticality).
[[Image:Printer System for FMEA Example 1.png|thumb|center|500px|]]

If the rating scales method is used, for each factor, a risk rating scale is assigned to it, as shown in the figure below.
[[Image:Printer System for FMEA Example 2.png|thumb|center|500px|]]

File:Printer System for FMEA Example 2.png

2012-03-19T20:41:25Z

Dingzhou Cao:

Template:Risk Discovery Analysis Example for XFMEA

2012-03-19T20:37:33Z

Dingzhou Cao:

Template:Risk Discovery Analysis Example for XFMEA

2012-03-19T20:37:19Z

Dingzhou Cao:

Let’s examine the new Risk Discovery analysis feature in Xfmea through a fictional example. Suppose that a technology company is developing a new complex product. It is a multi-function printer that prints, copies, scans and has several other features, such as scan to e-mail, document management, etc.

Figure below shows the system hierarchy, with some subsystems fully expanded down to the component level and others collapsed at the subsystem level.
[[Image:Printer System for FMEA Example.png|thumb|center|400px|]]

As a best practice, the team first conducts an FMEA at the top (system) level in order to address any potential problems related to interactions and interfaces. Starting the analysis at lower levels has been proven to be a practice that can lead to omission of interactions between subsystems. In most complex systems, interactions will account for more than 50% of the total failure modes that the system will experience. In such cases, it is essential to start with a top-down approach.

Once the top (system) level FMEA has been completed, the next question that the FMEA team faces is which subsystems and components to focus on. This is where the Risk Discovery feature in Xfmea is used.

The team adds a Risk Discovery analysis for each of the main subsystems of the product: the printing system, the copying system, the scanning system and the document management system. Xfmea supports two configurable methods for this type of analysis: a set of yes/no questions (where a yes answer indicates that the assembly/component poses a risk that warrants more detailed analysis) or a set of predefined rating scales (which can be used to calculate an overall rating for each assembly/component). For this example, we will assume that the organization has selected to use the Questions tab with the sample questions that are shipped by default with the software. These questions address the issues that would typically be considered with a change point analysis approach (new technology, new application, historical problems, supplier capability), together with other relevant issues (safety, regulatory requirements and mission criticality).
[[Image:Printer System for FMEA Example 1.png|thumb|center|400px|]]

File:Printer System for FMEA Example 1.png

2012-03-19T20:37:01Z

Dingzhou Cao:

Template:Risk Discovery Analysis Example for XFMEA

2012-03-19T19:02:52Z

Dingzhou Cao:

Let’s examine the new Risk Discovery analysis feature in Xfmea through a fictional example. Suppose that a technology company is developing a new complex product. It is a multi-function printer that prints, copies, scans and has several other features, such as scan to e-mail, document management, etc.

Figure below shows the system hierarchy, with some subsystems fully expanded down to the component level and others collapsed at the subsystem level.
[[Image:Printer System for FMEA Example.png|thumb|center|400px|]]

As a best practice, the team first conducts an FMEA at the top (system) level in order to address any potential problems related to interactions and interfaces. Starting the analysis at lower levels has been proven to be a practice that can lead to omission of interactions between subsystems. In most complex systems, interactions will account for more than 50% of the total failure modes that the system will experience. In such cases, it is essential to start with a top-down approach.

Once the top (system) level FMEA has been completed, the next question that the FMEA team faces is which subsystems and components to focus on. This is where the Risk Discovery feature in Xfmea is used.

The team adds a Risk Discovery analysis for each of the main subsystems of the product: the printing system, the copying system, the scanning system and the document management system. Xfmea supports two configurable methods for this type of analysis: a set of yes/no questions (where a yes answer indicates that the assembly/component poses a risk that warrants more detailed analysis) or a set of predefined rating scales (which can be used to calculate an overall rating for each assembly/component). For this example, we will assume that the organization has selected to use the Questions tab with the sample questions that are shipped by default with the software. These questions address the issues that would typically be considered with a change point analysis approach (new technology, new application, historical problems, supplier capability), together with other relevant issues (safety, regulatory requirements and mission criticality).

When the team uses the questions to discuss each subsystem, they determine that the highest risk assemblies are the printing system followed by the document management system. The copying and scanning systems are highly leveraged from previous designs and as a result they were graded lower in terms of risk. The two assemblies that will receive subsystem level FMEA are marked by selecting the "Mark item for more detailed analysis" check box and they are then highlighted in the system configuration display. Now the team has a clear path of where to focus their FMEA efforts first. Figure 2 shows the analysis for the printing system and Figure 3 shows the system configuration with two subsystems highlighted for more detailed analysis.

Template:Risk Discovery Analysis Example for XFMEA

2012-03-19T19:00:28Z

Dingzhou Cao:

Template:Risk Discovery Analysis Example for XFMEA

2012-03-19T18:57:31Z

Dingzhou Cao: Created page with 'Let’s examine the new Risk Discovery analysis feature in Xfmea through a fictional example. Suppose that a technology company is developing a new complex product. It is a multi…'

File:Printer System for FMEA Example.png

2012-03-19T18:57:23Z

Dingzhou Cao:

Fault Tree Diagrams and System Analysis

2012-03-19T17:31:05Z

Dingzhou Cao: /* Fault Trees and Simulation */

{{Template:bsbook|10}}

BlockSim allows system modeling using both Reliability Block Diagrams (RBDs) and Fault Trees. This chapter introduces basic fault tree analysis and points out the similarities (and differences) between RBDs and fault tree diagrams. Principles, methods and concepts discussed in previous chapters are used.
=Fault Tree and Reliability Block Diagrams=
Fault trees and reliability block diagrams are both symbolic analytical logic techniques that can be applied to analyze system reliability and related characteristics. Although the symbols and structures of the two diagram types differ, most of the logical constructs in a fault tree diagram (FTD) can also be modeled with a reliability block diagram (RBD). This chapter presents a brief introduction to fault tree analysis concepts and illustrates the similarities between fault tree diagrams and reliability block diagrams.
 
===Fault Tree Analysis: Brief Introduction===
 
Bell Telephone Laboratories developed the concept of fault tree analysis in 1962 for the U.S. Air Force for use with the Minuteman system. It was later adopted and extensively applied by the Boeing Company. A fault tree diagram follows a top-down structure and represents a graphical model of the pathways within a system that can lead to a foreseeable, undesirable loss event (or a failure). The pathways interconnect contributory events and conditions using standard logic symbols (AND, OR, etc.).
 

Fault tree diagrams consist of gates and events connected with lines. The AND and OR gates are the two most commonly used gates in a fault tree. To illustrate the use of these gates, consider two events (called ``input events'') that can lead to another event (called the ``output event''). If the occurrence of either input event causes the output event to occur, then these input events are connected using an OR gate. Alternatively, if both input events must occur in order for the output event to occur, then they are connected by an AND gate. Figure “Fault tree where the occurrence of either ''A'' or ''B'' can cause system failure” shows a simple fault tree diagram in which either <math>A</math> or <math>B</math> must occur in order for the output event to occur. In this diagram, the two events are connected to an OR gate. If the output event is system failure and the two input events are component failures, then this fault tree indicates that the failure of <math>A</math> or <math>B</math> causes the system to fail. The RBD equivalent for this configuration is a simple series system with two blocks, <math>A</math> and <math>B</math> , as shown in Figure “he RBD representation of the fault tree”.
 

[[Image:1.png|thumb|center|500px|Fault tree where the occurrence of either ''A'' or ''B'' can cause system failure.]]
 
 
[[Image:10.2.gif|thumb|center|200px|The RBD representation of the fault tree.]]
 
[[Category:Completed Theoretical Review]]

=Basic Gates=
Gates are the logic symbols that interconnect contributory events and conditions in a fault tree diagram. The AND and OR gates described above, as well as a Voting OR gate in which the output event occurs if a certain number of the input events occur (i.e. <math>k</math> -out-of- <math>n</math> redundancy), are the most basic types of gates in classical fault tree analysis. These gates are explicitly provided for in BlockSim and are described in this section along with their BlockSim implementations. Additional gates are introduced in the following sections.
 

===AND Gate===
 
[[Image:I10.1.gif|thumb|center|400px|]]
<math></math>

In an AND gate, the output event occurs if all input events occur. In system reliability terms, this implies that all components must fail (input) in order for the system to fail (output). When using RBDs, the equivalent is a simple parallel configuration.
 
====Example====
 
Consider a system with two components, <math>A</math> and <math>B</math> . The system fails if both <math>A</math> and <math>B</math> fail. Draw the fault tree and reliability block diagram for the system. The next two figures show both the FTD and RBD representations.
 
<center>[[Image:I10.2.gif|thumb|center|300px|]][[Image:I10.3.gif|thumb|center|300px|]]</center>
 
 

The reliability equation for either configuration is:

 

::<math>{{R}_{System}}={{R}_{A}}+{{R}_{B}}-{{R}_{A}}\cdot {{R}_{B}}</math>
 
Figure below show the analytic equation from BlockSim 8.

[[Image:AND gate.png|thumb|center|400px|]]

===OR Gate===
 
In an OR gate, the output event occurs if at least one of the input events occurs. In system reliability terms, this implies that if any component fails (input) then the system will fail (output). When using RBDs, the equivalent is a series configuration.
 
 
[[Image:I10.4.gif|thumb|center|400px|]]
 

{{BlockSim Analytical Fault Tree Diagram Example}}

===Voting OR Gate===
 
[[Image:I10.7.gif|thumb|center|400px|]]
 

In a Voting OR gate, the output event occurs if <math>k</math> or more of the input events occur. In system reliability terms, this implies that if any <math>k</math> -out-of- <math>n</math> components fail (input) then the system will fail (output).
 

The equivalent RBD construct is a node and is similar to a <math>k</math> -out-of- <math>n</math> parallel configuration with a distinct difference, as discussed next. To illustrate this difference, consider a fault tree diagram with a <math>2</math> -out-of- <math>4</math> Voting OR gate, as shown in Figure “Illustration of a 2-out-or-4 Voting OR gate”. In this diagram, the system will fail if any two of the blocks below fail. Equivalently, this can be represented by the RBD shown in Figure “Equivalent representation of the 2-out-of-4 Voting OR gate” using a <math>3</math> -out-of- <math>4</math> node.
 

In this configuration, the system will not fail if three out of four components are operating, but will fail if more than one fails. In other words, the fault tree considers <math>k</math> -out-of- <math>n</math> failures for the system failure while the RBD considers <math>k</math> -out-of- <math>n</math> successes for system success.

[[Image:10.3.gif|thumb|center|400px|Illustration of a 2-out-or-4 Voting OR gate.]]
 
[[Image:10.4.gif|thumb|center|400px|Equivalent representation of the 2-out-of-4 Voting OR gate.]]

 

====Expanding the Classical Voting OR Gate====
 
Classical Voting OR gates have no properties and cannot fail or be repaired (i.e. they cannot be an event themselves). In BlockSim, Voting OR gates behave like nodes in an RBD; thus, they can also fail and be repaired just like any other event. By default, when a Voting OR gate is inserted into an FTD within BlockSim, the gate is set so that it cannot fail (classical definition). However, this property can be modified to allow for additional flexibility.
 

====Example====
 
Consider a system with three components, <math>A,</math> <math>B</math> and <math>C</math> . The system fails if any two components fail. Draw the fault tree and reliability block diagram for the system. The next two figures show both the FTD and RBD representations.

 
<center>[[Image:I10.8.gif|thumb|center|300px|]][[Image:I10.9.gif|thumb|center|300px|]]</center>

 
The reliability equation for either configuration is:

 
::<math>{{R}_{System}}=-2\cdot {{R}_{A}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{A}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{C}}+{{R}_{B}}\cdot {{R}_{C}}\ </math>
 

Equation above assumes a classical Voting OR gate (i.e. the voting gate itself cannot fail). If the gate can fail then the equation is modified as follows:
 

::<math>{{R}_{System}}={{R}_{Voting}}\left( -2\cdot {{R}_{A}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{A}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{C}}+{{R}_{B}}\cdot {{R}_{C}} \right)</math>
 

Note that while both the gate and the node are 2-out-of-3, they represent different circumstances. The Voting OR gate in the fault tree indicates that if two components fail then the system will fail; while the node in the reliability block diagram indicates that if at least two components succeed then the system will succeed.
 

===Combining Basic Gates===
 
As in reliability block diagrams where different configuration types can be combined in the same diagram, fault tree analysis gates can also be combined to create more complex representations. As an example, consider the fault tree diagram shown in figures below.
 
 
[[Image:10.5.gif|thumb|center|500px|A sample FTD using different gates.]]
 
 
[[Image:10.6.gif|thumb|center|300px|RBD representation of the FTD shown in figure above.]]
 

===Discussion of Basic Gates and Events===
 
A fault tree diagram is always drawn in a top-down manner with lowest item being a basic event block. Classical fault tree gates have no properties (i.e. they cannot fail).
 

=New BlockSim Gates=

In addition to the gates defined above, other gates exist in classical FTA. These additional gates (e.g. Sequence Enforcing, Priority AND, etc.) are usually used to describe more complex redundancy configurations and are described in later sections. First, we will introduce two new advanced gates that can be used to append to and/or replace classical fault tree gates.
These two new gates are the Load Sharing and Standby gates. Classical fault trees (or any other fault tree standard to our knowledge) do not allow for load sharing redundancy (or event dependency). To overcome this limitation, and to provide fault trees with the same flexibility as BlockSim's RBDs, we will define a Load Sharing gate in this section. Additionally, traditional fault trees do not provide the full capability to model standby redundancy configurations (including the quiescent failure distribution), although basic standby can be represented in traditional fault tree diagrams using a Priority AND gate or a Sequence Enforcing gate, discussed in later sections.
 

===Load Sharing Gate===
 
[[Image:I10.10.gif|thumb|center|400px|]]
 

A Load Sharing gate behaves just like BlockSim's Load Sharing containers for RBDs. Load Sharing containers were discussed in [[Time-Dependent_System_Reliability_(Analytical)|Time-Dependent System Reliability (Analytical)]] and [[RBDs_and_Analytical_System_Reliability|RBDs and Analytical System Reliability]]. Events leading into a Load Sharing gate have distributions and life-stress relationships, just like contained blocks. Furthermore, the gate defines the load and the number required to cause the output event (i.e. the Load Sharing gate is defined with a <math>k</math> -out-of- <math>n</math> vote ). In BlockSim, no additional gates are allowed below a Load Sharing gate.
 

====Example====
 
A component has five possible failure modes, <math>A</math> , <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> , <math>{{B}_{C}}</math> and <math>C</math> , and the <math>B</math> modes are interdependent. The system will fail if mode <math>A</math> occurs, mode <math>C</math> occurs or two out of the three <math>B</math> modes occur. Modes <math>A</math> and <math>C</math> have a Weibull distribution with <math>\beta =2</math> and <math>\eta =10,000</math> and <math>15,000</math> respectively. Events <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> and <math>{{B}_{C}}</math> have an exponential distribution with a mean of <math>10,000</math> hours. If any <math>B</math> event occurs (i.e. <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> or <math>{{B}_{C}}</math> ), the remaining <math>B</math> events are more likely to occur. Specifically, the mean times of the remaining <math>B</math> events are halved. Determine the reliability at <math>1,000</math> hours for this component.
 
====Solution to Example====
 
The first step is to create the fault tree as shown in figure below. Note that both an OR gate and a Load Sharing gate are used.
 
[[Image:10.7.gif|thumb|center|400px| Fault tree for the example illustrating a Load Sharing gate.]]

 
The next step is to define the properties for each event block and the Load Sharing gate. Setting the failure distributions for modes <math>A</math> and <math>C</math> is simple. 
[[Image:10.8.gif|thumb|center|400px| Load sharing parameters.]]
 

The more difficult part is setting the properties of the Load Sharing gate (which are the same as an RBD container) and the dependent load sharing events (which are the same as the contained blocks in an RBD). Based on the problem statement, the <math>B</math> modes are in a 2-out-of-3 load sharing redundancy. When all three are working (i.e. when no <math>B</math> mode has occurred), each block has an exponential distribution with <math>\mu =10,000</math> . If one <math>B</math> mode occurs, then the two surviving units have an exponential distribution with <math>\mu =5,000.</math>
 
Assume an inverse power life-stress relationship for the components. Then:

 
::<math>{{\mu }_{1}}= \frac{1}{KV_{1}^{n}}\ </math>

::<math>{{\mu }_{2}}= \frac{1}{KV_{2}^{n}}\ </math>
 

Substituting <math>{{\mu }_{1}}=10,000</math> and <math>{{V}_{1}}=1</math> in <math>{{\mu}_{1}}= \frac{1}{KV_{1}^{n}}\ </math> and casting it in terms of <math>K</math> yields:
 

::<math>\begin{align}
10,000= & \frac{1}{K} \\
K= & \frac{1}{10,000}=0.0001
\end{align}</math>

 

Substituting <math>{{\mu }_{2}}=5,000</math> , <math>{{V}_{2}}=1.5</math> (because if one fails, then each survivor takes on an additional 0.5 units of load) and <math>10,000=\frac{1}{K} </math> for <math>K</math> in <math>{{\mu }_{2}}= \frac{1}{KV_{2}^{n}}\ </math> yields:

 

::<math>\begin{align}
5,000= & \frac{1}{0.0001\cdot {{(1.5)}^{n}}} \\
0.5= & {{(1.5)}^{-n}} \\
\ln (0.5)= & -n\ln (1.5) \\
n= & 1.7095
\end{align}</math>

 

This also could have been computed in ReliaSoft's ALTA software or with the Load & Life Parameter Experimenter in BlockSim. This was done in Chapter 5.
 

At this point, the parameters for the load sharing units have been computed and can be set, as shown in Figure "Load sharing parameters". (Note: when define the IPL-Exponential model, we just need to specify the value for K and n, the value for Use Stress is not a issue here, leave it as default number 10 or any number will be good.)

 

The next step is to set the weight proportionality factor. This factor defines the portion of the load that the particular item carries while operating, as well as the load that shifts to the remaining units upon failure of the item. To illustrate, assume three units (1, 2 and 3) are in a load sharing redundancy, represented in the fault tree diagram by a Load Sharing gate, with weight proportionality factors of 1, 2 and 3 respectively (and a 3-out-of-3 requirement).
 
:• Unit 1 carries <math>\left( \tfrac{1}{1+2+3} \right)=0.166</math> or 16.6% of the total load.
:• Unit 2 carries <math>\left( \tfrac{2}{1+2+3} \right)=0.333</math> or 33.3% of the total load.
:• Unit 3 carries <math>\left( \tfrac{3}{1+2+3} \right)=0.50</math> or 50% of the total load.
 
 
The actual load on each unit then becomes the product of the entire load defined for the gate multiplied by the portion carried by that unit. For example, if the load is 100 lbs, then the portion assigned to Unit 1 will be <math>100\cdot 0.166=16.6</math> lbs.

In the current example, all units share the same load; thus, they have equal weight proportionality factors. Because these factors are relative, if the same number is used for all three items then the results will be the same. For simplicity, we will set the factor equal to 1 for each item.
 

Once the properties have been specified in BlockSim, the reliability at 1000 hours can be determined. From the Analytical QCP, this is found to be 93.87%.

===Standby Gate===
 
[[Image:SB.gif|thumb|center|400px|]]
 

A Standby gate behaves just like a standby container in BlockSim's RBDs. Standby containers were discussed in [[Time-Dependent_System_Reliability_(Analytical)|Time-Dependent System Reliability (Analytical)]] and [[RBDs_and_Analytical_System_Reliability|RBDs and Analytical System Reliability]]. Events leading into a Standby gate have active and quiescent failure distributions, just like contained blocks. Furthermore, the gate acts as the switch, can fail and can also define the number of active blocks whose failure would cause system failure (i.e. the Active Vote Number required ). In BlockSim, no additional gates are allowed below a Standby gate.
 
====Example====
 
Consider a system with two units, <math>A</math> and <math>B</math> , in a standby configuration. Unit <math>A</math> is active and unit <math>B</math> is in a ``warm'' standby configuration. Furthermore, assume perfect switching (i.e. the switch cannot fail and the switch occurs instantly). Units <math>A</math> and <math>B</math> have the following failure properties:
 
:• Block <math>A</math> (Active):
::o Failure Distribution: Weibull; <math>\beta </math> =1.5; <math>\eta </math> = <math>1,000</math> hours.
:• Block <math>B</math> (Standby):
::o Energized failure distribution: Weibull; <math>\beta </math> =1.5; <math>\eta </math> = <math>1,000</math> hours.
::o Quiescent failure distribution: Weibull; <math>\beta </math> =1.5; <math>\eta </math> = <math>2,000</math> hours.
 
Determine the reliability of the system for <math>500</math> hours.

 

====Example Solution====
 
The fault tree diagram for this configuration is shown next and <math>R(t=500)=94.26%</math> .
 
[[Image:I10.11.gif|thumb|center|400px|]]
 

=Additional Classical Gates and Their Equivalents in BlockSim=
===Sequence Enforcing Gate===
 
Various graphical symbols have been used to represent a Sequence Enforcing gate. It is a variation of an AND gate in which each item must happen in sequence. In other words, events are constrained to occur in a specific sequence and the output event occurs if all input events occur in that specified sequence. This is identical to a cold standby redundant configuration (i.e. <math>k</math> units in standby with no quiescent failure distribution and no switch failure probability). BlockSim does not explicitly provide a Sequence Enforcing gate; however, it can be easily modeled using the more advanced Standby gate, described previously.
 
===Inhibit Gate===
 
[[Image:I10.12.gif|thumb|center|400px|]]
 

In an Inhibit gate, the output event occurs if all input events occur and an additional conditional event occurs. It is an AND gate with an additional event. In reality, an Inhibit gate provides no additional modeling capabilities but is used to illustrate the fact that an additional event must also occur. As an example, consider the case where events <math>A</math> and <math>B</math> must occur as well as a third event <math>C</math> (the so-called conditional event) in order for the system to fail. One can represent this in a fault tree by using an AND gate with three events, <math>A</math> , <math>B</math> and <math>C</math> , as shown in Figure "Using an AND gate to represent an inhibit relationship". Classical fault tree diagrams have the conditional event drawn to the side and the gate drawn as a hexagon, as shown Figure "Traditional use of an Inhibit gate". It should be noted that both representations are equivalent from an analysis standpoint.
 

 
[[Image:10.10.gif|thumb|center|400px|Using an AND gate to represent an inhibit relationship.]]
 
 
[[Image:10.11.gif|thumb|center|400px|Traditional use of an Inhibit gate.]]
 
 
[[Image:10.12.gif|thumb|center|400px|Including the conditional event inside the Inhibit gate.]]
 
 
BlockSim explicitly provides an Inhibit gate. This gate functions just like an AND gate with the exception that failure/repair characteristics can be assigned to the gate itself. This allows the construction shown in Figure "Traditional use of an Inhibit gate"(if the gate itself is set to not fail). Additionally, one could encapsulate event <math>C</math> inside the gate (since the gate can have properties), as shown Figure "Including the conditional event inside the Inhibit gate". Note that all three figures can be represented using a single RBD with events <math>A</math> , <math>B</math> and <math>C</math> in parallel.
 

===Priority AND Gate===
 
[[Image:I10.13.gif|thumb|center|400px|]]
 
 

With a Priority AND gate, the output event occurs if all input events occur in a specific sequence. This is an AND gate that requires that all events occur in a specific sequence. At first, this may seem identical to the Sequence Enforcing gate discussed earlier. However, it differs from this gate in the fact that events can occur out of sequence (i.e. are not constrained to occur in a specific sequence) but the output event only occurs if the sequence is followed. To better illustrate this, consider the case of two motors in standby configuration with motor <math>A</math> being the primary motor and motor <math>B</math> in standby. If motor <math>A</math> fails, then the switch (which can also fail) activates motor <math>B</math> . Then the system will fail if motor <math>A</math> fails and the switch fails to switch, or if the switch succeeds but motor <math>B</math> fails subsequent to the switching action. In this scenario, the events must occur in the order noted; however, it is possible for the switch or motor <math>B</math> to fail (in a quiescent mode) without causing a system failure, if <math>A</math> never fails. BlockSim does not explicitly provide a Priority AND gate. However, like the Sequence Enforcing gate, it can be easily modeled using the more advanced Standby gate.
 

===Transfer Gate===
[[Image:I10.14.gif|thumb|center|400px|]]
 
Transfer in/out gates are used to indicate a transfer/continuation of one fault tree to another. In classical fault trees, the Transfer gate is generally used to signify the continuation of a tree on a separate sheet. This is the same as a subdiagram block in an RBD. BlockSim does not explicitly provide a Transfer gate. However, it does allow for subdiagrams (or sub-trees), which provide for greater flexibility. Additionally, a subdiagram in a BlockSim fault tree can be an RBD and vice versa. BlockSim uses the more intuitive folder symbol to represent subdiagrams.
 
[[Image:i10.15.gif|thumb|center|400px|]]
 
As an example, consider the fault tree of the robot manipulator shown in Figure figrobotall. Figure robotsubdiagram illustrates the same fault tree with the use of subdiagrams (Transfer gates). The referenced subdiagrams are shown in subsequent figures. Note that this is using multiple levels of indenture (i.e. the subdiagram has subdiagrams and so forth).
 
The RBD representation of the fault tree shown in Figure "A" is given in Figure "H". This same RBD could have been represented using subdiagrams, as shown in Figure "I". In this figure, which is the RBD representation of Figure "B", the subdiagrams in the RBD link to the fault trees of Figures "D" and "C" and their sub-trees.
 
[[Image:BS10.13.png|thumb|center|400px|A: A sample fault tree for a robot manipulator, showing all items in a single tree.]]
 
[[Image:10.14.gif|thumb|center|400px|B: The fault tree of Figure 10.12 using subdiagrams (Transfer gates). The subdiagrams are shown in Figures "C" and "D".]]
 
[[Image:10.15.gif|thumb|center|400px|C: The fault tree of the robot arm mechanism. This subdiagram is referenced in Figure "B".]]
 
[[Image:10.16.gif|thumb|center|400px|D: The fault tree for the arm jams/collides event. This subdiagram is referenced in Figure "B". It also includes a subdiagram continuation to Figure "E".]]
 
[[Image:10.17.gif|thumb|center|400px|E: The brake shutdown event referenced from Figure "D". it also includes a subdiagram continuation to Figure "F".]]
 
[[Image:10.18.gif|thumb|center|400px|F: The watchdog ESD fails event referenced from Figure "F". It also includes a subdiagram continuation to Figure "G"]]
 
[[Image:10.19.gif|thumb|center|400px|G: The communication fails event referenced from Figure "G".]]
 
[[Image:10.20.gif|thumb|center|400px|H: This is the RBD equivalent of the complete fault tree of Figure "A".]]
 
[[Image:10.21.gif|thumb|center|400px|I: The RBD representation of Figure "B" with the subdiagrams in the RBD linked to the fault trees of Figures "C" and "D" and their sub-trees.]]
 

===XOR Gate===
 
[[Image:I10.16.gif|thumb|center|400px|]]
 

In an XOR gate, the output event occurs if exactly one input event occurs. This is similar to an OR gate with the exception that if more than one input event occurs then the output event does not occur. For example, if there are two input events then the XOR gate indicates that the output event occurs if only one of the input events occurs but not if zero or both of these events occur. From a system reliability perspective, this would imply that a two-component system would function even if both components had failed. Furthermore, when dealing with time-varying failure distributions, and if system components do not operate through failure, a failure occurrence of both components at the exact same time ( <math>dt)</math> is an unreachable state; thus an OR gate would suffice. For these reasons, an RBD equivalent of an XOR gate is not presented here and BlockSim does not explicitly provide an XOR gate.
 

=Event Classifications=
Traditional fault trees use different shapes to represent different events. Unlike gates, however, different events in a fault tree are not treated differently from an analytical perspective. Rather, the event shapes are used to convey additional information visually. BlockSim includes some of the main event symbols from classical fault tree analysis and provides utilities for changing the graphical look of a block to illustrate a different type of event. Some of these event classifications are given next. From a properties perspective, all events defined in BlockSim can have fixed probabilities, failure distributions, repair distributions, crews, spares, etc. In other words, fault tree event blocks can have all the properties that an RBD block can have. This is an enhancement and a significant expansion over traditional fault trees, which generally include just a fixed probability of occurrence and/or a constant failure rate.
 
===Basic Event===
 
[[Image:5.png|thumb|center|300px|]]
 
A basic event (or failure event) is identical to an RBD block and has been traditionally represented by a circle.
 

===Undeveloped Event===
 
[[Image:diamond.gif|thumb|center|300px|]]
 
An undeveloped event has the same properties as a basic event with the exception that it is graphically rendered as a diamond. The diamond representation graphically illustrates that this event could have been expanded into a separate fault tree but was not. In other words, the analyst uses a different symbol to convey that the event could have been developed (broken down) further but he/she has chosen not to do so for the analysis.
 

===House Event===
[[Image:pentagon.gif|thumb|center|300px|]]
 

A house event is an event that can be set to occur or not occur (i.e. it usually has a fixed probability of 0 or 1). It is usually used to turn paths on or off or to make paths of a tree functional or non-functional. Furthermore, the terms failed house and working house have been used to signify probabilities of 0 and 1 respectively. In BlockSim, a house shape is available for an event and a house event has the same properties as a basic event, keeping in mind that an event can be set to Cannot Fail or Failed from the block properties window.
 

===Conditional Event===
[[Image:oval.gif|thumb|center|300px|]]

A conditional event is represented by an ellipse and specifies a condition. Again, it has all the properties of a basic event. It can be applied to any gate. As an example, event <math>C</math> in the first figure below would be the conditional event and it would be represented more applicably by an ellipse than a circle, as shown in the second figure below.
 
 
[[Image:10.10.gif|thumb|center|400px|Using an AND gate to represent an inhibit relationship.]]
[[Image:10.22.gif|thumb|center|400px|Using an ellipse attached to an inhubit gate (with no gate properties) to show the conditional event. This is mathematically equivalent to figure above.]]
 

=Comparing Fault Trees and RBDs=
The most fundamental difference between fault tree diagrams and reliability block diagrams is that you work in the success space in an RBD while you work in the failure space in a fault tree. In other words, the RBD considers success combinations while the fault tree considers failure combinations. In addition, fault trees have traditionally been used to analyze fixed probabilities (i.e. each event that comprises the tree has a fixed probability of occurring) while RBDs may include time-varying distributions for the success (reliability equation) and other properties, such as repair/restoration distributions. In general (and with some specific exceptions), a fault tree can be easily converted to an RBD. However, it is generally more difficult to convert an RBD into a fault tree, especially if one allows for highly complex configurations.
 
As you can see from the discussion to this point, an RBD equivalent exists for most of the constructs that are supported by classical fault tree analysis. With these constructs, you can perform the same powerful system analysis, including simulation, regardless of how you choose to represent the system thus erasing the distinction between fault trees and reliability block diagrams.
 
===Example Using Both RBDs and Fault Trees===
 
Assume that a component can fail due to six independent primary failure modes: <math>A</math> , <math>B</math> , <math>C</math> , <math>D</math> , <math>E</math> and <math>F</math> . Some of these primary modes can be broken down further into the events that can cause them, or sub-modes. Furthermore, assume that once a mode occurs, the event also occurs and the mode does not go away. Specifically:
 
:• The component fails if mode <math>A</math> , <math>B</math> or <math>C</math> occurs.
:• If mode <math>D</math> , <math>E</math> or <math>F</math> occurs alone, the component does not fail; however, the component will fail if any two (or more) of these modes occur (i.e. <math>D</math> and <math>E</math> ; <math>D</math> and <math>F</math> ; <math>E</math> and <math>F</math> ).
:• Modes <math>D</math> , <math>E</math> and <math>F</math> have a constant rate of occurrence (exponential distribution) with mean times of occurrence of <math>200,000</math> , <math>175,000</math> and <math>500,000</math> hours, respectively.
:• The rates of occurrence for modes <math>A</math> , <math>B</math> and <math>C</math> depend on their sub-modes.
 
Do the following:

::#Determine the reliability of the component after 1 year ( <math>8,760</math> hours).
::#Determine the B10 life of the component.
::#Determine the mean time to failure ( <math>MTTF</math> ) of the component.
::#Rank the modes in order of importance at 1 year.
::#Re-calculate results 1, 2 and 3 assuming mode <math>B</math> is eliminated.

To begin the analysis, modes <math>A</math> , <math>B</math> and <math>C</math> can be broken down further based on specific events (sub-modes), as defined next.
 
 
'''Mode <math>A</math>'''
 
 
There are five independent events (sub-modes) associated with mode <math>A</math> : events <math>S1</math> , <math>S2</math> , <math>T1</math> , <math>T2</math> and <math>Y</math> . It is assumed that events <math>S1</math> and <math>S2</math> each have a constant rate of occurrence with a probability of occurrence in a single year ( <math>8760</math> hours) of <math>1</math> in <math>10,000</math> and <math>1</math> in <math>20,000</math> , respectively. Events <math>T1</math> and <math>T2</math> are more likely to occur in an older component than a newer one (i.e. they have an increasing rate of occurrence) and have a probability of occurrence of <math>1</math> in <math>10,000</math> and <math>1</math> in <math>20,000</math> , respectively, in a single year and <math>1</math> in <math>1,000</math> and <math>1</math> in <math>3,000</math> , respectively, after two years. Event <math>Y</math> also has a constant rate of occurrence with a probability of occurrence of <math>1</math> in <math>1,000</math> in a single year. There are three possible ways for mode <math>A</math> to manifest itself:
 
::#Events <math>S1</math> and <math>S2</math> both occur.
::#Event <math>T1</math> or <math>T2</math> occurs.
::#Event <math>Y</math> and either event <math>S1</math> or event <math>S2</math> occur (i.e. events <math>Y</math> and <math>S1</math> or events <math>Y</math> and <math>S2</math> occur).

'''RBD Solution for Mode <math>A</math>'''

The RBD that satisfies the conditions for mode <math>A</math> is shown in figure below.

[[Image:6.png|thumb|center|400px|Reliability block diagram for mode A.]]
 

[[Image:7.png|thumb|center|400px|Fault tree for mode ''A''.]]
 

Each mode is identified in the RBD. Furthermore, two additional items are included: a starting block ( <math>NF</math> ) and an end node (2/2). The starting block and the end node are set so they cannot fail and, therefore, will not affect the results. The end node is used to define a 2-out-of-2 configuration (i.e. both paths leading into the node must work).
 
 
'''Fault Tree Solution for Mode <math>A</math>'''
 
 
The fault tree for mode <math>A</math> is shown in Figure "Fault tree for mode ''A''". Each mode is identified as an event in the fault tree. Figure below shows an alternative representation for mode <math>A</math> using mirrored events for <math>S1</math> and <math>S2</math> . Further discussion on mirrored events is provided in the next section.
[[Image:8.png|thumb|center|400px|An alternative representation of the fault tree for mode ''A'' using mirrored events.]]
 
 
'''Mode <math>A</math> Discussion'''
 
 
The system reliability equation for this configuration (regardless of how it is drawn) is:

 
::<math>\begin{align}
R(t)= & -2{{R}_{T2}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{S1}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{S2}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{T1}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{Y}} \\
& +{{R}_{T2}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{S1}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{S2}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{T1}} \\
& +{{R}_{T2}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{S1}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{T1}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{Y}} \\
& +{{R}_{T2}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{S2}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{T1}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{Y}}
\end{align}</math>
 

 
Based on the given probabilities, distribution parameters are computed for each block (either RBD block or the fault tree event block). One way is to compute them using the Parameter Experimenter, as shown in figure below. In this figure and for <math>S1</math> , the probability is <math>1</math> in <math>10,000</math> in one year ( <math>8760</math> hours), thus the exponential failure rate is 1.1416e-8. This can be repeated for <math>S2</math> and <math>Y</math> .
 

[[Image:10.26.gif|thumb|center|400px|BlockSim's Parameter Experimenter.]]
 
Events <math>T1</math> and <math>T2</math> need to be modeled using a life distribution that does not have a constant failure rate. Using BlockSim's Parameter Experimenter and selecting the Weibull distribution, the parameter values for events <math>T1</math> and <math>T2</math> are shown in Figures below, respectively.
 
[[Image:10.27.gif|thumb|center|400px|Parameter vealues for event ''T''1.]]
 

[[Image:10.28.gif|thumb|center|400px|Parameter values for event ''T''2.]]
 
'''Mode <math>B</math>'''
 
 
There are three dependent events associated with mode <math>B</math> : events <math>BA</math> , <math>BB</math> and <math>BC</math> .
 
:• Two out of the three events must occur for mode <math>B</math> to occur.
::o Events <math>BA</math> , <math>BB</math> and <math>BC</math> all have an exponential distribution with a mean of <math>50,000</math> hours.
::o The events are dependent (i.e. if <math>BA</math> , <math>BB</math> or <math>BC</math> occurs, then the remaining events are more likely to occur). Specifically, when one event occurs, the <math>MTTF</math> of the remaining events is halved.
 

This is basically a load sharing configuration. The reliability function for each block will change depending on the other events. Therefore, the reliability of each block is not only dependent on time, but also on the stress (load) that the block experiences.
 
 
'''RBD Solution for Mode <math>B</math>'''
 
 
The reliability block diagram for mode <math>B</math> is shown in figure below.
[[Image:10.29.gif|thumb|center|400px|Reliability block diagram for mode ''B''.]]
 

 
'''Fault Tree Solution for Mode <math>B</math>'''
 
The fault tree for mode <math>B</math> is shown in figure below. A Load Sharing gate is used.
 
[[Image:10.30.gif|thumb|center|400px|Fault tree diagram for mode ''B''(using a Load Sharing gate unique to BlockSim).]]
::<math></math>

 

 
'''Mode <math>B</math> Discussion'''
 
 
To describe the dependency, a Load Sharing gate and dependent event blocks are used. Since the failure rate is assumed to be constant, an exponential distribution is used. Furthermore, for simplicity, an Arrhenius life-stress relationship is used with the parameters B=2.0794 and C=6250.
 

 
'''Mode <math>C</math>'''
 
 
There are two sequential events associated with mode <math>C</math> : <math>CA</math> and <math>CB</math> .
 
:• Both events must occur for mode <math>C</math> to occur.
:• Event <math>CB</math> will only occur if event <math>CA</math> has occurred.
:• If event <math>CA</math> has not occurred, then event <math>CB</math> will not occur.
:• Events <math>CA</math> and <math>CB</math> both occur based on a Weibull distribution.
:• For event <math>CA</math> , <math>\beta </math> =2 and <math>\eta </math> =30,000 hours.
:• For event <math>CB</math> , <math>\beta </math> =2 and <math>\eta </math> =10,000 hours.
 
 
'''RBD Solution for Mode <math>C</math>'''
 
 
To model this, you can think of a scenario similar to standby redundancy. Basically, if <math>CA</math> occurs then <math>CB</math> gets initiated. A Standby container can be used to model this, as shown in figure below.
 

[[Image:10.32.gif|thumb|center|400px|Standby container for mode C.]]
 

 
In this case, event <math>CA</math> is set as the active component and <math>CB</math> as the standby. If event <math>CA</math> occurs, <math>CB</math> will be initiated. For this analysis, a perfect switch is assumed. The properties are set in BlockSim as follows:
 
Contained Items
 
:• <math>CA</math> : Active failure distribution, Weibull distribution ( <math>\beta </math> =2, <math>\eta </math> =30,000).
:• <math>CA</math> : Quiescent failure distribution: None, cannot fail or age in this mode.
:• <math>CB</math> : Active failure distribution, Weibull distribution ( <math>\beta </math> =2, <math>\eta </math> =10,000).
:• <math>CB</math> : Quiescent failure distribution: None, cannot fail or age in this mode.
 
Switch
 
:• Active Switching: Always works (100% reliability) and instant switch (no delays).
:• Quiescent Switch failure distribution: None, cannot fail or age in this mode.
 

'''Fault Tree Solution for Mode <math>C</math>'''
The fault tree for mode <math>C</math> is shown in figure below. Note that the sequence is enforced by the Standby gate (used as a Sequence Enforcing gate).

[[Image:10.31.png|thumb|center|400px|Standby (Sequence Enforcing) gate for model ''C'']]
 
 

'''Mode <math>C</math> Discussion'''
 
The failure distribution settings for event <math>CA</math> are shown in figure below.
[[Image:10.34.gif|thumb|center|400px|Failure distribution settings for event ''C'' ''A''.]]
 

The failure distribution properties for event <math>CB</math> are set in the same manner.
 

'''Modes <math>D</math> , <math>E</math> and <math>F</math>'''
 
Modes <math>D</math> , <math>E</math> and <math>F</math> can all be represented using the exponential distribution. The failure distribution properties for modes <math>D</math> , <math>E</math> and <math>F</math> are:
 
:• <math>D</math> : <math>MTTF</math> = 200,000 hours.
:• <math>E</math> : <math>MTTF</math> = 175,000 hours.
:• <math>F</math> : <math>MTTF</math> = 500,000 hours.
 
'''The Entire Component'''
 
The last step is to set up the model for the component based on the primary modes ( <math>A</math> , <math>B</math> , <math>C</math> , <math>D</math> , <math>E</math> and <math>F</math> ). Modes <math>A</math> , <math>B</math> and <math>C</math> can each be represented by single blocks that encapsulate the subdiagrams already created. The RBD in figure below represents the primary failure modes for the component while the fault tree in Figure "Fault tree of the component" illustrates the same. The node represented by 2/3 in the RBD indicates a 2-out-of-3 configuration. The Voting OR gate in the fault tree accomplishes the same. Subdiagrams are used in both configurations for the sub-modes.
 
[[Image:10.35.gif|thumb|center|400px|RBD of the component.]]
 

Once the diagrams have been created, the reliability equation for the system can be obtained, as follows:
 
[[Image:10.36.gif|thumb|center|400px|Fault tree of the component.]]
 

::<math>\begin{align}
R{{(t)}_{System}}= & R{{(t)}_{A}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{B}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{F}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{D}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{C}} \\
& +R{{(t)}_{A}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{B}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{F}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{C}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{E}} \\
& +R{{(t)}_{A}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{B}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{D}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{C}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{E}} \\
& -2(R{{(t)}_{A}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{B}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{F}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{D}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{C}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{E}})
\end{align}</math>
 
Where <math>R{{(t)}_{A}}</math> , <math>R{{(t)}_{B}}</math> and <math>R{{(t)}_{C}}</math> are the reliability equations corresponding to the sub-modes.
 
'''Analysis'''
 

The questions posed earlier can be answered using BlockSim. Regardless of the approach used (i.e. RBD or FTA), the answers are the same.
 
::1. The reliability of the component at 1 year (8,760 hours) can be calculated using the Analytical Quick Calculation Pad (QCP) or by viewing the reliability vs. time plot, as displayed in Figure "Reliability vs. time plot for the component". <math>R(t=8760)=86.4975%</math> .
::2. Using the Analytical QCP, the B10 life of the component is estimated to be 7,373.94 hours.
::3. Using the Analytical QCP, the mean life of the component is estimated to be 21,659.68 hours.
::4. The ranking of the modes after 1 year can be shown via the Static Reliability Importance plot, as shown in Figure below.
::5. Re-computing the results for 1, 2 and 3 assuming mode <math>B</math> is removed:
:::a) <math>R(t=8760)=98.72%.</math>
:::b) <math>B10</math> = 16,928.38 hours.
:::c) <math>MTTF</math> = 34,552.89 hours.

[[Image:10.37.gif|thumb|center|400px|Reliability vs. time plot for the component.]]
 

[[Image:10.38.gif|thumb|center|400px|Static reliability importance for each of the modes at ''t''=8,760 hours.]]
 

'''Discussion'''
 
There are multiple options for modeling systems with fault trees and RBDs in BlockSim. The first figure below shows the complete fault tree for the component without using subdiagrams (Transfer gates) while the second figure below illustrates a hybrid analysis utilizing an RBD for the component and fault trees as the subdiagrams. The results are the same regardless of the option chosen.
 
[[Image:10.37.png|thumb|center|600px|Fault tree for the component without using subdiagrams(Transfer gates)]]

[[Image:10.40.gif|thumb|center|400px|A hybrid solution using an RBD for the component and fault trees as subdiagrams.]]
 

=Using Mirrored Blocks to Represent Complex RBDs as FTDs=
A fault tree cannot normally represent a complex RBD. As an example, consider the RBD shown in figure below.
 
[[Image:10.41.gif|thumb|center|500px|A complex RBD that cannot be represented by a fault tree unless duplicate events are utilized.]]
 

A fault tree representation for this RBD is shown in the first figure below. Note that the same event is used more than once in the fault tree diagram. To correctly analyze this, the duplicate events need to be set up as "mirrored" events to the parent event. In other words, the same event is represented in two locations in the fault tree diagram. It should be pointed out that the RBD in the second figure below is also equivalent to the RBD of figure above and the fault tree of the first figure below.
 
[[Image:10.42.gif|thumb|center|500px|A fault tree representation using mirrored blocks (events) of the complex RBD.]]
 

[[Image:10.43.gif|thumb|center|500px|An RBD using mirrored blocks that is equivalent to both the RBD and FTD.]]
 

=Fault Trees and Simulation=

The slightly modified constructs in BlockSim erase the distinction between RBDs and fault trees. Given this, any analysis that is possible in a BlockSim RBD (including throughput analysis) is also available when using fault trees.
 
{{Template:FT and RBD Simulation Example}}

=Additional Fault Tree Topics=

===Minimal Cut Sets===
 
Traditional solution of fault trees involves the determination of so-called minimal cut sets. Minimal cut sets are all the unique combinations of component failures that can cause system failure. Specifically, a cut set is said to be a minimal cut set if, when any basic event is removed from the set, the remaining events collectively are no longer a cut set [[Appendix_B:_References | Kececioglu [10]]]. As an example, consider the fault tree shown in Figure "Minimal cut set example". The system will fail if {1, 2, 3 and 4 fail} or {1, 2 and 3 fail} or {1, 2 and 4 fail}.
 

All of these are cut sets. However, the one including all components is not a minimal cut set because, if 3 and 4 are removed, the remaining events are also a cut set. Therefore, the minimal cut sets for this configuration are {1, 2 , 3} or {1, 2, 4}. This may be more evident by examining the RBD equivalent of the first figure below, as shown in the second figure below.
 

[[Image:9.png|thumb|center|400px|Minimal cut set example.]]
 

[[Image:10.png|thumb|center|400px|RBD of the fault tree shown in figure above.]]
 

BlockSim does not use the cut sets methodology when analyzing fault trees. However, interested users can obtain these cut sets for both fault trees and block diagrams with the command available in the Tools menu.

Fault Tree Diagrams and System Analysis

2012-03-19T17:29:06Z

Dingzhou Cao: /* Example */

{{Template:bsbook|10}}

BlockSim allows system modeling using both Reliability Block Diagrams (RBDs) and Fault Trees. This chapter introduces basic fault tree analysis and points out the similarities (and differences) between RBDs and fault tree diagrams. Principles, methods and concepts discussed in previous chapters are used.
=Fault Tree and Reliability Block Diagrams=
Fault trees and reliability block diagrams are both symbolic analytical logic techniques that can be applied to analyze system reliability and related characteristics. Although the symbols and structures of the two diagram types differ, most of the logical constructs in a fault tree diagram (FTD) can also be modeled with a reliability block diagram (RBD). This chapter presents a brief introduction to fault tree analysis concepts and illustrates the similarities between fault tree diagrams and reliability block diagrams.
 
===Fault Tree Analysis: Brief Introduction===
 
Bell Telephone Laboratories developed the concept of fault tree analysis in 1962 for the U.S. Air Force for use with the Minuteman system. It was later adopted and extensively applied by the Boeing Company. A fault tree diagram follows a top-down structure and represents a graphical model of the pathways within a system that can lead to a foreseeable, undesirable loss event (or a failure). The pathways interconnect contributory events and conditions using standard logic symbols (AND, OR, etc.).
 

Fault tree diagrams consist of gates and events connected with lines. The AND and OR gates are the two most commonly used gates in a fault tree. To illustrate the use of these gates, consider two events (called ``input events'') that can lead to another event (called the ``output event''). If the occurrence of either input event causes the output event to occur, then these input events are connected using an OR gate. Alternatively, if both input events must occur in order for the output event to occur, then they are connected by an AND gate. Figure “Fault tree where the occurrence of either ''A'' or ''B'' can cause system failure” shows a simple fault tree diagram in which either <math>A</math> or <math>B</math> must occur in order for the output event to occur. In this diagram, the two events are connected to an OR gate. If the output event is system failure and the two input events are component failures, then this fault tree indicates that the failure of <math>A</math> or <math>B</math> causes the system to fail. The RBD equivalent for this configuration is a simple series system with two blocks, <math>A</math> and <math>B</math> , as shown in Figure “he RBD representation of the fault tree”.
 

[[Image:1.png|thumb|center|500px|Fault tree where the occurrence of either ''A'' or ''B'' can cause system failure.]]
 
 
[[Image:10.2.gif|thumb|center|200px|The RBD representation of the fault tree.]]
 
[[Category:Completed Theoretical Review]]

=Basic Gates=
Gates are the logic symbols that interconnect contributory events and conditions in a fault tree diagram. The AND and OR gates described above, as well as a Voting OR gate in which the output event occurs if a certain number of the input events occur (i.e. <math>k</math> -out-of- <math>n</math> redundancy), are the most basic types of gates in classical fault tree analysis. These gates are explicitly provided for in BlockSim and are described in this section along with their BlockSim implementations. Additional gates are introduced in the following sections.
 

===AND Gate===
 
[[Image:I10.1.gif|thumb|center|400px|]]
<math></math>

In an AND gate, the output event occurs if all input events occur. In system reliability terms, this implies that all components must fail (input) in order for the system to fail (output). When using RBDs, the equivalent is a simple parallel configuration.
 
====Example====
 
Consider a system with two components, <math>A</math> and <math>B</math> . The system fails if both <math>A</math> and <math>B</math> fail. Draw the fault tree and reliability block diagram for the system. The next two figures show both the FTD and RBD representations.
 
<center>[[Image:I10.2.gif|thumb|center|300px|]][[Image:I10.3.gif|thumb|center|300px|]]</center>
 
 

The reliability equation for either configuration is:

 

::<math>{{R}_{System}}={{R}_{A}}+{{R}_{B}}-{{R}_{A}}\cdot {{R}_{B}}</math>
 
Figure below show the analytic equation from BlockSim 8.

[[Image:AND gate.png|thumb|center|400px|]]

===OR Gate===
 
In an OR gate, the output event occurs if at least one of the input events occurs. In system reliability terms, this implies that if any component fails (input) then the system will fail (output). When using RBDs, the equivalent is a series configuration.
 
 
[[Image:I10.4.gif|thumb|center|400px|]]
 

{{BlockSim Analytical Fault Tree Diagram Example}}

===Voting OR Gate===
 
[[Image:I10.7.gif|thumb|center|400px|]]
 

In a Voting OR gate, the output event occurs if <math>k</math> or more of the input events occur. In system reliability terms, this implies that if any <math>k</math> -out-of- <math>n</math> components fail (input) then the system will fail (output).
 

The equivalent RBD construct is a node and is similar to a <math>k</math> -out-of- <math>n</math> parallel configuration with a distinct difference, as discussed next. To illustrate this difference, consider a fault tree diagram with a <math>2</math> -out-of- <math>4</math> Voting OR gate, as shown in Figure “Illustration of a 2-out-or-4 Voting OR gate”. In this diagram, the system will fail if any two of the blocks below fail. Equivalently, this can be represented by the RBD shown in Figure “Equivalent representation of the 2-out-of-4 Voting OR gate” using a <math>3</math> -out-of- <math>4</math> node.
 

In this configuration, the system will not fail if three out of four components are operating, but will fail if more than one fails. In other words, the fault tree considers <math>k</math> -out-of- <math>n</math> failures for the system failure while the RBD considers <math>k</math> -out-of- <math>n</math> successes for system success.

[[Image:10.3.gif|thumb|center|400px|Illustration of a 2-out-or-4 Voting OR gate.]]
 
[[Image:10.4.gif|thumb|center|400px|Equivalent representation of the 2-out-of-4 Voting OR gate.]]

 

====Expanding the Classical Voting OR Gate====
 
Classical Voting OR gates have no properties and cannot fail or be repaired (i.e. they cannot be an event themselves). In BlockSim, Voting OR gates behave like nodes in an RBD; thus, they can also fail and be repaired just like any other event. By default, when a Voting OR gate is inserted into an FTD within BlockSim, the gate is set so that it cannot fail (classical definition). However, this property can be modified to allow for additional flexibility.
 

====Example====
 
Consider a system with three components, <math>A,</math> <math>B</math> and <math>C</math> . The system fails if any two components fail. Draw the fault tree and reliability block diagram for the system. The next two figures show both the FTD and RBD representations.

 
<center>[[Image:I10.8.gif|thumb|center|300px|]][[Image:I10.9.gif|thumb|center|300px|]]</center>

 
The reliability equation for either configuration is:

 
::<math>{{R}_{System}}=-2\cdot {{R}_{A}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{A}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{C}}+{{R}_{B}}\cdot {{R}_{C}}\ </math>
 

Equation above assumes a classical Voting OR gate (i.e. the voting gate itself cannot fail). If the gate can fail then the equation is modified as follows:
 

::<math>{{R}_{System}}={{R}_{Voting}}\left( -2\cdot {{R}_{A}}\cdot {{R}_{B}}\cdot {{R}_{C}}+{{R}_{A}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{C}}+{{R}_{B}}\cdot {{R}_{C}} \right)</math>
 

Note that while both the gate and the node are 2-out-of-3, they represent different circumstances. The Voting OR gate in the fault tree indicates that if two components fail then the system will fail; while the node in the reliability block diagram indicates that if at least two components succeed then the system will succeed.
 

===Combining Basic Gates===
 
As in reliability block diagrams where different configuration types can be combined in the same diagram, fault tree analysis gates can also be combined to create more complex representations. As an example, consider the fault tree diagram shown in figures below.
 
 
[[Image:10.5.gif|thumb|center|500px|A sample FTD using different gates.]]
 
 
[[Image:10.6.gif|thumb|center|300px|RBD representation of the FTD shown in figure above.]]
 

===Discussion of Basic Gates and Events===
 
A fault tree diagram is always drawn in a top-down manner with lowest item being a basic event block. Classical fault tree gates have no properties (i.e. they cannot fail).
 

=New BlockSim Gates=

In addition to the gates defined above, other gates exist in classical FTA. These additional gates (e.g. Sequence Enforcing, Priority AND, etc.) are usually used to describe more complex redundancy configurations and are described in later sections. First, we will introduce two new advanced gates that can be used to append to and/or replace classical fault tree gates.
These two new gates are the Load Sharing and Standby gates. Classical fault trees (or any other fault tree standard to our knowledge) do not allow for load sharing redundancy (or event dependency). To overcome this limitation, and to provide fault trees with the same flexibility as BlockSim's RBDs, we will define a Load Sharing gate in this section. Additionally, traditional fault trees do not provide the full capability to model standby redundancy configurations (including the quiescent failure distribution), although basic standby can be represented in traditional fault tree diagrams using a Priority AND gate or a Sequence Enforcing gate, discussed in later sections.
 

===Load Sharing Gate===
 
[[Image:I10.10.gif|thumb|center|400px|]]
 

A Load Sharing gate behaves just like BlockSim's Load Sharing containers for RBDs. Load Sharing containers were discussed in [[Time-Dependent_System_Reliability_(Analytical)|Time-Dependent System Reliability (Analytical)]] and [[RBDs_and_Analytical_System_Reliability|RBDs and Analytical System Reliability]]. Events leading into a Load Sharing gate have distributions and life-stress relationships, just like contained blocks. Furthermore, the gate defines the load and the number required to cause the output event (i.e. the Load Sharing gate is defined with a <math>k</math> -out-of- <math>n</math> vote ). In BlockSim, no additional gates are allowed below a Load Sharing gate.
 

====Example====
 
A component has five possible failure modes, <math>A</math> , <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> , <math>{{B}_{C}}</math> and <math>C</math> , and the <math>B</math> modes are interdependent. The system will fail if mode <math>A</math> occurs, mode <math>C</math> occurs or two out of the three <math>B</math> modes occur. Modes <math>A</math> and <math>C</math> have a Weibull distribution with <math>\beta =2</math> and <math>\eta =10,000</math> and <math>15,000</math> respectively. Events <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> and <math>{{B}_{C}}</math> have an exponential distribution with a mean of <math>10,000</math> hours. If any <math>B</math> event occurs (i.e. <math>{{B}_{A}}</math> , <math>{{B}_{B}}</math> or <math>{{B}_{C}}</math> ), the remaining <math>B</math> events are more likely to occur. Specifically, the mean times of the remaining <math>B</math> events are halved. Determine the reliability at <math>1,000</math> hours for this component.
 
====Solution to Example====
 
The first step is to create the fault tree as shown in figure below. Note that both an OR gate and a Load Sharing gate are used.
 
[[Image:10.7.gif|thumb|center|400px| Fault tree for the example illustrating a Load Sharing gate.]]

 
The next step is to define the properties for each event block and the Load Sharing gate. Setting the failure distributions for modes <math>A</math> and <math>C</math> is simple. 
[[Image:10.8.gif|thumb|center|400px| Load sharing parameters.]]
 

The more difficult part is setting the properties of the Load Sharing gate (which are the same as an RBD container) and the dependent load sharing events (which are the same as the contained blocks in an RBD). Based on the problem statement, the <math>B</math> modes are in a 2-out-of-3 load sharing redundancy. When all three are working (i.e. when no <math>B</math> mode has occurred), each block has an exponential distribution with <math>\mu =10,000</math> . If one <math>B</math> mode occurs, then the two surviving units have an exponential distribution with <math>\mu =5,000.</math>
 
Assume an inverse power life-stress relationship for the components. Then:

 
::<math>{{\mu }_{1}}= \frac{1}{KV_{1}^{n}}\ </math>

::<math>{{\mu }_{2}}= \frac{1}{KV_{2}^{n}}\ </math>
 

Substituting <math>{{\mu }_{1}}=10,000</math> and <math>{{V}_{1}}=1</math> in <math>{{\mu}_{1}}= \frac{1}{KV_{1}^{n}}\ </math> and casting it in terms of <math>K</math> yields:
 

::<math>\begin{align}
10,000= & \frac{1}{K} \\
K= & \frac{1}{10,000}=0.0001
\end{align}</math>

 

Substituting <math>{{\mu }_{2}}=5,000</math> , <math>{{V}_{2}}=1.5</math> (because if one fails, then each survivor takes on an additional 0.5 units of load) and <math>10,000=\frac{1}{K} </math> for <math>K</math> in <math>{{\mu }_{2}}= \frac{1}{KV_{2}^{n}}\ </math> yields:

 

::<math>\begin{align}
5,000= & \frac{1}{0.0001\cdot {{(1.5)}^{n}}} \\
0.5= & {{(1.5)}^{-n}} \\
\ln (0.5)= & -n\ln (1.5) \\
n= & 1.7095
\end{align}</math>

 

This also could have been computed in ReliaSoft's ALTA software or with the Load & Life Parameter Experimenter in BlockSim. This was done in Chapter 5.
 

At this point, the parameters for the load sharing units have been computed and can be set, as shown in Figure "Load sharing parameters". (Note: when define the IPL-Exponential model, we just need to specify the value for K and n, the value for Use Stress is not a issue here, leave it as default number 10 or any number will be good.)

 

The next step is to set the weight proportionality factor. This factor defines the portion of the load that the particular item carries while operating, as well as the load that shifts to the remaining units upon failure of the item. To illustrate, assume three units (1, 2 and 3) are in a load sharing redundancy, represented in the fault tree diagram by a Load Sharing gate, with weight proportionality factors of 1, 2 and 3 respectively (and a 3-out-of-3 requirement).
 
:• Unit 1 carries <math>\left( \tfrac{1}{1+2+3} \right)=0.166</math> or 16.6% of the total load.
:• Unit 2 carries <math>\left( \tfrac{2}{1+2+3} \right)=0.333</math> or 33.3% of the total load.
:• Unit 3 carries <math>\left( \tfrac{3}{1+2+3} \right)=0.50</math> or 50% of the total load.
 
 
The actual load on each unit then becomes the product of the entire load defined for the gate multiplied by the portion carried by that unit. For example, if the load is 100 lbs, then the portion assigned to Unit 1 will be <math>100\cdot 0.166=16.6</math> lbs.

In the current example, all units share the same load; thus, they have equal weight proportionality factors. Because these factors are relative, if the same number is used for all three items then the results will be the same. For simplicity, we will set the factor equal to 1 for each item.
 

Once the properties have been specified in BlockSim, the reliability at 1000 hours can be determined. From the Analytical QCP, this is found to be 93.87%.

===Standby Gate===
 
[[Image:SB.gif|thumb|center|400px|]]
 

A Standby gate behaves just like a standby container in BlockSim's RBDs. Standby containers were discussed in [[Time-Dependent_System_Reliability_(Analytical)|Time-Dependent System Reliability (Analytical)]] and [[RBDs_and_Analytical_System_Reliability|RBDs and Analytical System Reliability]]. Events leading into a Standby gate have active and quiescent failure distributions, just like contained blocks. Furthermore, the gate acts as the switch, can fail and can also define the number of active blocks whose failure would cause system failure (i.e. the Active Vote Number required ). In BlockSim, no additional gates are allowed below a Standby gate.
 
====Example====
 
Consider a system with two units, <math>A</math> and <math>B</math> , in a standby configuration. Unit <math>A</math> is active and unit <math>B</math> is in a ``warm'' standby configuration. Furthermore, assume perfect switching (i.e. the switch cannot fail and the switch occurs instantly). Units <math>A</math> and <math>B</math> have the following failure properties:
 
:• Block <math>A</math> (Active):
::o Failure Distribution: Weibull; <math>\beta </math> =1.5; <math>\eta </math> = <math>1,000</math> hours.
:• Block <math>B</math> (Standby):
::o Energized failure distribution: Weibull; <math>\beta </math> =1.5; <math>\eta </math> = <math>1,000</math> hours.
::o Quiescent failure distribution: Weibull; <math>\beta </math> =1.5; <math>\eta </math> = <math>2,000</math> hours.
 
Determine the reliability of the system for <math>500</math> hours.

 

====Example Solution====
 
The fault tree diagram for this configuration is shown next and <math>R(t=500)=94.26%</math> .
 
[[Image:I10.11.gif|thumb|center|400px|]]
 

=Additional Classical Gates and Their Equivalents in BlockSim=
===Sequence Enforcing Gate===
 
Various graphical symbols have been used to represent a Sequence Enforcing gate. It is a variation of an AND gate in which each item must happen in sequence. In other words, events are constrained to occur in a specific sequence and the output event occurs if all input events occur in that specified sequence. This is identical to a cold standby redundant configuration (i.e. <math>k</math> units in standby with no quiescent failure distribution and no switch failure probability). BlockSim does not explicitly provide a Sequence Enforcing gate; however, it can be easily modeled using the more advanced Standby gate, described previously.
 
===Inhibit Gate===
 
[[Image:I10.12.gif|thumb|center|400px|]]
 

In an Inhibit gate, the output event occurs if all input events occur and an additional conditional event occurs. It is an AND gate with an additional event. In reality, an Inhibit gate provides no additional modeling capabilities but is used to illustrate the fact that an additional event must also occur. As an example, consider the case where events <math>A</math> and <math>B</math> must occur as well as a third event <math>C</math> (the so-called conditional event) in order for the system to fail. One can represent this in a fault tree by using an AND gate with three events, <math>A</math> , <math>B</math> and <math>C</math> , as shown in Figure "Using an AND gate to represent an inhibit relationship". Classical fault tree diagrams have the conditional event drawn to the side and the gate drawn as a hexagon, as shown Figure "Traditional use of an Inhibit gate". It should be noted that both representations are equivalent from an analysis standpoint.
 

 
[[Image:10.10.gif|thumb|center|400px|Using an AND gate to represent an inhibit relationship.]]
 
 
[[Image:10.11.gif|thumb|center|400px|Traditional use of an Inhibit gate.]]
 
 
[[Image:10.12.gif|thumb|center|400px|Including the conditional event inside the Inhibit gate.]]
 
 
BlockSim explicitly provides an Inhibit gate. This gate functions just like an AND gate with the exception that failure/repair characteristics can be assigned to the gate itself. This allows the construction shown in Figure "Traditional use of an Inhibit gate"(if the gate itself is set to not fail). Additionally, one could encapsulate event <math>C</math> inside the gate (since the gate can have properties), as shown Figure "Including the conditional event inside the Inhibit gate". Note that all three figures can be represented using a single RBD with events <math>A</math> , <math>B</math> and <math>C</math> in parallel.
 

===Priority AND Gate===
 
[[Image:I10.13.gif|thumb|center|400px|]]
 
 

With a Priority AND gate, the output event occurs if all input events occur in a specific sequence. This is an AND gate that requires that all events occur in a specific sequence. At first, this may seem identical to the Sequence Enforcing gate discussed earlier. However, it differs from this gate in the fact that events can occur out of sequence (i.e. are not constrained to occur in a specific sequence) but the output event only occurs if the sequence is followed. To better illustrate this, consider the case of two motors in standby configuration with motor <math>A</math> being the primary motor and motor <math>B</math> in standby. If motor <math>A</math> fails, then the switch (which can also fail) activates motor <math>B</math> . Then the system will fail if motor <math>A</math> fails and the switch fails to switch, or if the switch succeeds but motor <math>B</math> fails subsequent to the switching action. In this scenario, the events must occur in the order noted; however, it is possible for the switch or motor <math>B</math> to fail (in a quiescent mode) without causing a system failure, if <math>A</math> never fails. BlockSim does not explicitly provide a Priority AND gate. However, like the Sequence Enforcing gate, it can be easily modeled using the more advanced Standby gate.
 

===Transfer Gate===
[[Image:I10.14.gif|thumb|center|400px|]]
 
Transfer in/out gates are used to indicate a transfer/continuation of one fault tree to another. In classical fault trees, the Transfer gate is generally used to signify the continuation of a tree on a separate sheet. This is the same as a subdiagram block in an RBD. BlockSim does not explicitly provide a Transfer gate. However, it does allow for subdiagrams (or sub-trees), which provide for greater flexibility. Additionally, a subdiagram in a BlockSim fault tree can be an RBD and vice versa. BlockSim uses the more intuitive folder symbol to represent subdiagrams.
 
[[Image:i10.15.gif|thumb|center|400px|]]
 
As an example, consider the fault tree of the robot manipulator shown in Figure figrobotall. Figure robotsubdiagram illustrates the same fault tree with the use of subdiagrams (Transfer gates). The referenced subdiagrams are shown in subsequent figures. Note that this is using multiple levels of indenture (i.e. the subdiagram has subdiagrams and so forth).
 
The RBD representation of the fault tree shown in Figure "A" is given in Figure "H". This same RBD could have been represented using subdiagrams, as shown in Figure "I". In this figure, which is the RBD representation of Figure "B", the subdiagrams in the RBD link to the fault trees of Figures "D" and "C" and their sub-trees.
 
[[Image:BS10.13.png|thumb|center|400px|A: A sample fault tree for a robot manipulator, showing all items in a single tree.]]
 
[[Image:10.14.gif|thumb|center|400px|B: The fault tree of Figure 10.12 using subdiagrams (Transfer gates). The subdiagrams are shown in Figures "C" and "D".]]
 
[[Image:10.15.gif|thumb|center|400px|C: The fault tree of the robot arm mechanism. This subdiagram is referenced in Figure "B".]]
 
[[Image:10.16.gif|thumb|center|400px|D: The fault tree for the arm jams/collides event. This subdiagram is referenced in Figure "B". It also includes a subdiagram continuation to Figure "E".]]
 
[[Image:10.17.gif|thumb|center|400px|E: The brake shutdown event referenced from Figure "D". it also includes a subdiagram continuation to Figure "F".]]
 
[[Image:10.18.gif|thumb|center|400px|F: The watchdog ESD fails event referenced from Figure "F". It also includes a subdiagram continuation to Figure "G"]]
 
[[Image:10.19.gif|thumb|center|400px|G: The communication fails event referenced from Figure "G".]]
 
[[Image:10.20.gif|thumb|center|400px|H: This is the RBD equivalent of the complete fault tree of Figure "A".]]
 
[[Image:10.21.gif|thumb|center|400px|I: The RBD representation of Figure "B" with the subdiagrams in the RBD linked to the fault trees of Figures "C" and "D" and their sub-trees.]]
 

===XOR Gate===
 
[[Image:I10.16.gif|thumb|center|400px|]]
 

In an XOR gate, the output event occurs if exactly one input event occurs. This is similar to an OR gate with the exception that if more than one input event occurs then the output event does not occur. For example, if there are two input events then the XOR gate indicates that the output event occurs if only one of the input events occurs but not if zero or both of these events occur. From a system reliability perspective, this would imply that a two-component system would function even if both components had failed. Furthermore, when dealing with time-varying failure distributions, and if system components do not operate through failure, a failure occurrence of both components at the exact same time ( <math>dt)</math> is an unreachable state; thus an OR gate would suffice. For these reasons, an RBD equivalent of an XOR gate is not presented here and BlockSim does not explicitly provide an XOR gate.
 

=Event Classifications=
Traditional fault trees use different shapes to represent different events. Unlike gates, however, different events in a fault tree are not treated differently from an analytical perspective. Rather, the event shapes are used to convey additional information visually. BlockSim includes some of the main event symbols from classical fault tree analysis and provides utilities for changing the graphical look of a block to illustrate a different type of event. Some of these event classifications are given next. From a properties perspective, all events defined in BlockSim can have fixed probabilities, failure distributions, repair distributions, crews, spares, etc. In other words, fault tree event blocks can have all the properties that an RBD block can have. This is an enhancement and a significant expansion over traditional fault trees, which generally include just a fixed probability of occurrence and/or a constant failure rate.
 
===Basic Event===
 
[[Image:5.png|thumb|center|300px|]]
 
A basic event (or failure event) is identical to an RBD block and has been traditionally represented by a circle.
 

===Undeveloped Event===
 
[[Image:diamond.gif|thumb|center|300px|]]
 
An undeveloped event has the same properties as a basic event with the exception that it is graphically rendered as a diamond. The diamond representation graphically illustrates that this event could have been expanded into a separate fault tree but was not. In other words, the analyst uses a different symbol to convey that the event could have been developed (broken down) further but he/she has chosen not to do so for the analysis.
 

===House Event===
[[Image:pentagon.gif|thumb|center|300px|]]
 

A house event is an event that can be set to occur or not occur (i.e. it usually has a fixed probability of 0 or 1). It is usually used to turn paths on or off or to make paths of a tree functional or non-functional. Furthermore, the terms failed house and working house have been used to signify probabilities of 0 and 1 respectively. In BlockSim, a house shape is available for an event and a house event has the same properties as a basic event, keeping in mind that an event can be set to Cannot Fail or Failed from the block properties window.
 

===Conditional Event===
[[Image:oval.gif|thumb|center|300px|]]

A conditional event is represented by an ellipse and specifies a condition. Again, it has all the properties of a basic event. It can be applied to any gate. As an example, event <math>C</math> in the first figure below would be the conditional event and it would be represented more applicably by an ellipse than a circle, as shown in the second figure below.
 
 
[[Image:10.10.gif|thumb|center|400px|Using an AND gate to represent an inhibit relationship.]]
[[Image:10.22.gif|thumb|center|400px|Using an ellipse attached to an inhubit gate (with no gate properties) to show the conditional event. This is mathematically equivalent to figure above.]]
 

=Comparing Fault Trees and RBDs=
The most fundamental difference between fault tree diagrams and reliability block diagrams is that you work in the success space in an RBD while you work in the failure space in a fault tree. In other words, the RBD considers success combinations while the fault tree considers failure combinations. In addition, fault trees have traditionally been used to analyze fixed probabilities (i.e. each event that comprises the tree has a fixed probability of occurring) while RBDs may include time-varying distributions for the success (reliability equation) and other properties, such as repair/restoration distributions. In general (and with some specific exceptions), a fault tree can be easily converted to an RBD. However, it is generally more difficult to convert an RBD into a fault tree, especially if one allows for highly complex configurations.
 
As you can see from the discussion to this point, an RBD equivalent exists for most of the constructs that are supported by classical fault tree analysis. With these constructs, you can perform the same powerful system analysis, including simulation, regardless of how you choose to represent the system thus erasing the distinction between fault trees and reliability block diagrams.
 
===Example Using Both RBDs and Fault Trees===
 
Assume that a component can fail due to six independent primary failure modes: <math>A</math> , <math>B</math> , <math>C</math> , <math>D</math> , <math>E</math> and <math>F</math> . Some of these primary modes can be broken down further into the events that can cause them, or sub-modes. Furthermore, assume that once a mode occurs, the event also occurs and the mode does not go away. Specifically:
 
:• The component fails if mode <math>A</math> , <math>B</math> or <math>C</math> occurs.
:• If mode <math>D</math> , <math>E</math> or <math>F</math> occurs alone, the component does not fail; however, the component will fail if any two (or more) of these modes occur (i.e. <math>D</math> and <math>E</math> ; <math>D</math> and <math>F</math> ; <math>E</math> and <math>F</math> ).
:• Modes <math>D</math> , <math>E</math> and <math>F</math> have a constant rate of occurrence (exponential distribution) with mean times of occurrence of <math>200,000</math> , <math>175,000</math> and <math>500,000</math> hours, respectively.
:• The rates of occurrence for modes <math>A</math> , <math>B</math> and <math>C</math> depend on their sub-modes.
 
Do the following:

::#Determine the reliability of the component after 1 year ( <math>8,760</math> hours).
::#Determine the B10 life of the component.
::#Determine the mean time to failure ( <math>MTTF</math> ) of the component.
::#Rank the modes in order of importance at 1 year.
::#Re-calculate results 1, 2 and 3 assuming mode <math>B</math> is eliminated.

To begin the analysis, modes <math>A</math> , <math>B</math> and <math>C</math> can be broken down further based on specific events (sub-modes), as defined next.
 
 
'''Mode <math>A</math>'''
 
 
There are five independent events (sub-modes) associated with mode <math>A</math> : events <math>S1</math> , <math>S2</math> , <math>T1</math> , <math>T2</math> and <math>Y</math> . It is assumed that events <math>S1</math> and <math>S2</math> each have a constant rate of occurrence with a probability of occurrence in a single year ( <math>8760</math> hours) of <math>1</math> in <math>10,000</math> and <math>1</math> in <math>20,000</math> , respectively. Events <math>T1</math> and <math>T2</math> are more likely to occur in an older component than a newer one (i.e. they have an increasing rate of occurrence) and have a probability of occurrence of <math>1</math> in <math>10,000</math> and <math>1</math> in <math>20,000</math> , respectively, in a single year and <math>1</math> in <math>1,000</math> and <math>1</math> in <math>3,000</math> , respectively, after two years. Event <math>Y</math> also has a constant rate of occurrence with a probability of occurrence of <math>1</math> in <math>1,000</math> in a single year. There are three possible ways for mode <math>A</math> to manifest itself:
 
::#Events <math>S1</math> and <math>S2</math> both occur.
::#Event <math>T1</math> or <math>T2</math> occurs.
::#Event <math>Y</math> and either event <math>S1</math> or event <math>S2</math> occur (i.e. events <math>Y</math> and <math>S1</math> or events <math>Y</math> and <math>S2</math> occur).

'''RBD Solution for Mode <math>A</math>'''

The RBD that satisfies the conditions for mode <math>A</math> is shown in figure below.

[[Image:6.png|thumb|center|400px|Reliability block diagram for mode A.]]
 

[[Image:7.png|thumb|center|400px|Fault tree for mode ''A''.]]
 

Each mode is identified in the RBD. Furthermore, two additional items are included: a starting block ( <math>NF</math> ) and an end node (2/2). The starting block and the end node are set so they cannot fail and, therefore, will not affect the results. The end node is used to define a 2-out-of-2 configuration (i.e. both paths leading into the node must work).
 
 
'''Fault Tree Solution for Mode <math>A</math>'''
 
 
The fault tree for mode <math>A</math> is shown in Figure "Fault tree for mode ''A''". Each mode is identified as an event in the fault tree. Figure below shows an alternative representation for mode <math>A</math> using mirrored events for <math>S1</math> and <math>S2</math> . Further discussion on mirrored events is provided in the next section.
[[Image:8.png|thumb|center|400px|An alternative representation of the fault tree for mode ''A'' using mirrored events.]]
 
 
'''Mode <math>A</math> Discussion'''
 
 
The system reliability equation for this configuration (regardless of how it is drawn) is:

 
::<math>\begin{align}
R(t)= & -2{{R}_{T2}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{S1}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{S2}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{T1}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{Y}} \\
& +{{R}_{T2}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{S1}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{S2}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{T1}} \\
& +{{R}_{T2}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{S1}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{T1}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{Y}} \\
& +{{R}_{T2}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{S2}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{T1}}\text{ }\!\!\cdot\!\!\text{ }{{R}_{Y}}
\end{align}</math>
 

 
Based on the given probabilities, distribution parameters are computed for each block (either RBD block or the fault tree event block). One way is to compute them using the Parameter Experimenter, as shown in figure below. In this figure and for <math>S1</math> , the probability is <math>1</math> in <math>10,000</math> in one year ( <math>8760</math> hours), thus the exponential failure rate is 1.1416e-8. This can be repeated for <math>S2</math> and <math>Y</math> .
 

[[Image:10.26.gif|thumb|center|400px|BlockSim's Parameter Experimenter.]]
 
Events <math>T1</math> and <math>T2</math> need to be modeled using a life distribution that does not have a constant failure rate. Using BlockSim's Parameter Experimenter and selecting the Weibull distribution, the parameter values for events <math>T1</math> and <math>T2</math> are shown in Figures below, respectively.
 
[[Image:10.27.gif|thumb|center|400px|Parameter vealues for event ''T''1.]]
 

[[Image:10.28.gif|thumb|center|400px|Parameter values for event ''T''2.]]
 
'''Mode <math>B</math>'''
 
 
There are three dependent events associated with mode <math>B</math> : events <math>BA</math> , <math>BB</math> and <math>BC</math> .
 
:• Two out of the three events must occur for mode <math>B</math> to occur.
::o Events <math>BA</math> , <math>BB</math> and <math>BC</math> all have an exponential distribution with a mean of <math>50,000</math> hours.
::o The events are dependent (i.e. if <math>BA</math> , <math>BB</math> or <math>BC</math> occurs, then the remaining events are more likely to occur). Specifically, when one event occurs, the <math>MTTF</math> of the remaining events is halved.
 

This is basically a load sharing configuration. The reliability function for each block will change depending on the other events. Therefore, the reliability of each block is not only dependent on time, but also on the stress (load) that the block experiences.
 
 
'''RBD Solution for Mode <math>B</math>'''
 
 
The reliability block diagram for mode <math>B</math> is shown in figure below.
[[Image:10.29.gif|thumb|center|400px|Reliability block diagram for mode ''B''.]]
 

 
'''Fault Tree Solution for Mode <math>B</math>'''
 
The fault tree for mode <math>B</math> is shown in figure below. A Load Sharing gate is used.
 
[[Image:10.30.gif|thumb|center|400px|Fault tree diagram for mode ''B''(using a Load Sharing gate unique to BlockSim).]]
::<math></math>

 

 
'''Mode <math>B</math> Discussion'''
 
 
To describe the dependency, a Load Sharing gate and dependent event blocks are used. Since the failure rate is assumed to be constant, an exponential distribution is used. Furthermore, for simplicity, an Arrhenius life-stress relationship is used with the parameters B=2.0794 and C=6250.
 

 
'''Mode <math>C</math>'''
 
 
There are two sequential events associated with mode <math>C</math> : <math>CA</math> and <math>CB</math> .
 
:• Both events must occur for mode <math>C</math> to occur.
:• Event <math>CB</math> will only occur if event <math>CA</math> has occurred.
:• If event <math>CA</math> has not occurred, then event <math>CB</math> will not occur.
:• Events <math>CA</math> and <math>CB</math> both occur based on a Weibull distribution.
:• For event <math>CA</math> , <math>\beta </math> =2 and <math>\eta </math> =30,000 hours.
:• For event <math>CB</math> , <math>\beta </math> =2 and <math>\eta </math> =10,000 hours.
 
 
'''RBD Solution for Mode <math>C</math>'''
 
 
To model this, you can think of a scenario similar to standby redundancy. Basically, if <math>CA</math> occurs then <math>CB</math> gets initiated. A Standby container can be used to model this, as shown in figure below.
 

[[Image:10.32.gif|thumb|center|400px|Standby container for mode C.]]
 

 
In this case, event <math>CA</math> is set as the active component and <math>CB</math> as the standby. If event <math>CA</math> occurs, <math>CB</math> will be initiated. For this analysis, a perfect switch is assumed. The properties are set in BlockSim as follows:
 
Contained Items
 
:• <math>CA</math> : Active failure distribution, Weibull distribution ( <math>\beta </math> =2, <math>\eta </math> =30,000).
:• <math>CA</math> : Quiescent failure distribution: None, cannot fail or age in this mode.
:• <math>CB</math> : Active failure distribution, Weibull distribution ( <math>\beta </math> =2, <math>\eta </math> =10,000).
:• <math>CB</math> : Quiescent failure distribution: None, cannot fail or age in this mode.
 
Switch
 
:• Active Switching: Always works (100% reliability) and instant switch (no delays).
:• Quiescent Switch failure distribution: None, cannot fail or age in this mode.
 

'''Fault Tree Solution for Mode <math>C</math>'''
The fault tree for mode <math>C</math> is shown in figure below. Note that the sequence is enforced by the Standby gate (used as a Sequence Enforcing gate).

[[Image:10.31.png|thumb|center|400px|Standby (Sequence Enforcing) gate for model ''C'']]
 
 

'''Mode <math>C</math> Discussion'''
 
The failure distribution settings for event <math>CA</math> are shown in figure below.
[[Image:10.34.gif|thumb|center|400px|Failure distribution settings for event ''C'' ''A''.]]
 

The failure distribution properties for event <math>CB</math> are set in the same manner.
 

'''Modes <math>D</math> , <math>E</math> and <math>F</math>'''
 
Modes <math>D</math> , <math>E</math> and <math>F</math> can all be represented using the exponential distribution. The failure distribution properties for modes <math>D</math> , <math>E</math> and <math>F</math> are:
 
:• <math>D</math> : <math>MTTF</math> = 200,000 hours.
:• <math>E</math> : <math>MTTF</math> = 175,000 hours.
:• <math>F</math> : <math>MTTF</math> = 500,000 hours.
 
'''The Entire Component'''
 
The last step is to set up the model for the component based on the primary modes ( <math>A</math> , <math>B</math> , <math>C</math> , <math>D</math> , <math>E</math> and <math>F</math> ). Modes <math>A</math> , <math>B</math> and <math>C</math> can each be represented by single blocks that encapsulate the subdiagrams already created. The RBD in figure below represents the primary failure modes for the component while the fault tree in Figure "Fault tree of the component" illustrates the same. The node represented by 2/3 in the RBD indicates a 2-out-of-3 configuration. The Voting OR gate in the fault tree accomplishes the same. Subdiagrams are used in both configurations for the sub-modes.
 
[[Image:10.35.gif|thumb|center|400px|RBD of the component.]]
 

Once the diagrams have been created, the reliability equation for the system can be obtained, as follows:
 
[[Image:10.36.gif|thumb|center|400px|Fault tree of the component.]]
 

::<math>\begin{align}
R{{(t)}_{System}}= & R{{(t)}_{A}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{B}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{F}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{D}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{C}} \\
& +R{{(t)}_{A}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{B}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{F}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{C}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{E}} \\
& +R{{(t)}_{A}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{B}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{D}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{C}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{E}} \\
& -2(R{{(t)}_{A}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{B}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{F}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{D}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{C}}\text{ }\!\!\cdot\!\!\text{ }R{{(t)}_{E}})
\end{align}</math>
 
Where <math>R{{(t)}_{A}}</math> , <math>R{{(t)}_{B}}</math> and <math>R{{(t)}_{C}}</math> are the reliability equations corresponding to the sub-modes.
 
'''Analysis'''
 

The questions posed earlier can be answered using BlockSim. Regardless of the approach used (i.e. RBD or FTA), the answers are the same.
 
::1. The reliability of the component at 1 year (8,760 hours) can be calculated using the Analytical Quick Calculation Pad (QCP) or by viewing the reliability vs. time plot, as displayed in Figure "Reliability vs. time plot for the component". <math>R(t=8760)=86.4975%</math> .
::2. Using the Analytical QCP, the B10 life of the component is estimated to be 7,373.94 hours.
::3. Using the Analytical QCP, the mean life of the component is estimated to be 21,659.68 hours.
::4. The ranking of the modes after 1 year can be shown via the Static Reliability Importance plot, as shown in Figure below.
::5. Re-computing the results for 1, 2 and 3 assuming mode <math>B</math> is removed:
:::a) <math>R(t=8760)=98.72%.</math>
:::b) <math>B10</math> = 16,928.38 hours.
:::c) <math>MTTF</math> = 34,552.89 hours.

[[Image:10.37.gif|thumb|center|400px|Reliability vs. time plot for the component.]]
 

[[Image:10.38.gif|thumb|center|400px|Static reliability importance for each of the modes at ''t''=8,760 hours.]]
 

'''Discussion'''
 
There are multiple options for modeling systems with fault trees and RBDs in BlockSim. The first figure below shows the complete fault tree for the component without using subdiagrams (Transfer gates) while the second figure below illustrates a hybrid analysis utilizing an RBD for the component and fault trees as the subdiagrams. The results are the same regardless of the option chosen.
 
[[Image:10.37.png|thumb|center|600px|Fault tree for the component without using subdiagrams(Transfer gates)]]

[[Image:10.40.gif|thumb|center|400px|A hybrid solution using an RBD for the component and fault trees as subdiagrams.]]
 

=Using Mirrored Blocks to Represent Complex RBDs as FTDs=
A fault tree cannot normally represent a complex RBD. As an example, consider the RBD shown in figure below.
 
[[Image:10.41.gif|thumb|center|500px|A complex RBD that cannot be represented by a fault tree unless duplicate events are utilized.]]
 

A fault tree representation for this RBD is shown in the first figure below. Note that the same event is used more than once in the fault tree diagram. To correctly analyze this, the duplicate events need to be set up as "mirrored" events to the parent event. In other words, the same event is represented in two locations in the fault tree diagram. It should be pointed out that the RBD in the second figure below is also equivalent to the RBD of figure above and the fault tree of the first figure below.
 
[[Image:10.42.gif|thumb|center|500px|A fault tree representation using mirrored blocks (events) of the complex RBD.]]
 

[[Image:10.43.gif|thumb|center|500px|An RBD using mirrored blocks that is equivalent to both the RBD and FTD.]]
 

=Fault Trees and Simulation=

The slightly modified constructs in BlockSim erase the distinction between RBDs and fault trees. Given this, any analysis that is possible in a BlockSim RBD (including throughput analysis) is also available when using fault trees.
 
===Example===
 
Consider the RBD shown in the first figure below and its equivalent fault tree representation, as shown in the second figure below.
 
[[Image:10.44.gif|thumb|center|400px|RBD for a repairable system.]]
 
[[Image:10.45.gif|thumb|center|400px|Fault tree equivalent of the repairable system shown in figure above.]]
 

 
Furthermore, assume the following basic failure and repair properties for each block and event:
 
:• Block <math>A</math> :
::o Failure Distribution: Weibull; <math>\beta </math> =1.5; <math>\eta </math> =1,000.
::o Corrective Distribution: Weibull; <math>\beta </math> =1.5; <math>\eta </math> =100.
:• Block <math>B</math> :
::o Failure Distribution: Exponential; <math>\mu </math> =10,000.
::o Corrective Distribution: Weibull; <math>\beta </math> =1.5; <math>\eta </math> =20.
:• Block <math>C</math> :
::o Failure Distribution: Normal; <math>\mu </math> =1,000; <math>\sigma </math> =200.
::o Corrective Distribution: Normal; <math>\mu </math> =6; <math>\sigma </math> =2.
:• Block <math>D</math> :
::o Failure Distribution: Weibull; <math>\beta </math> =1.5; <math>\eta </math> =10,000.
::o Corrective Distribution: Exponential; <math>\mu </math> =10.
:• Block <math>E</math> :
::o Failure Distribution: Weibull; <math>\beta </math> =3; <math>\eta </math> =1,000.
::o Corrective Distribution: Weibull; <math>\beta </math> =1.5; <math>\eta </math> =20.
:• Block <math>F</math> :
::o Failure Distribution: Weibull; <math>\beta </math> =1.5; <math>\eta </math> =5,000.
::o Corrective Distribution: Weibull; <math>\beta </math> =1.5; <math>\eta </math> =100.
:• Block <math>G</math> :
::o Failure Distribution: Exponential; <math>\mu </math> =100,000.
::o Corrective Distribution: Weibull; <math>\beta </math> =1.5; <math>\eta </math> =10.
:• Block <math>H</math> :
::o Failure Distribution: Normal; <math>\mu </math> =5,000; <math>\sigma </math> =50.
::o Corrective Distribution: Normal; <math>\mu </math> =10; <math>\sigma </math> =2.
 
A sample table of simulation results is given next for up to <math>t=1,000</math> , using <math>2,000</math> simulations for each diagram and an identical seed.
 
[[Image:FT and RBD example.png|thumb|center|400px|]]
As expected, the results are equivalent (within an expected difference due to simulation) regardless of the diagram type used. It should be pointed out that even though the same seed was used by both diagrams, the results are not always expected to be identical because the order in which the blocks are read from a fault tree diagram during the simulation may differ from the order in which they are read in the RBD; thus using a different random number stream for each block (e.g. block <math>G</math> in the RBD may receive a different sequence of random numbers than event block <math>G</math> in the FT).
 

=Additional Fault Tree Topics=

===Minimal Cut Sets===
 
Traditional solution of fault trees involves the determination of so-called minimal cut sets. Minimal cut sets are all the unique combinations of component failures that can cause system failure. Specifically, a cut set is said to be a minimal cut set if, when any basic event is removed from the set, the remaining events collectively are no longer a cut set [[Appendix_B:_References | Kececioglu [10]]]. As an example, consider the fault tree shown in Figure "Minimal cut set example". The system will fail if {1, 2, 3 and 4 fail} or {1, 2 and 3 fail} or {1, 2 and 4 fail}.
 

All of these are cut sets. However, the one including all components is not a minimal cut set because, if 3 and 4 are removed, the remaining events are also a cut set. Therefore, the minimal cut sets for this configuration are {1, 2 , 3} or {1, 2, 4}. This may be more evident by examining the RBD equivalent of the first figure below, as shown in the second figure below.
 

[[Image:9.png|thumb|center|400px|Minimal cut set example.]]
 

[[Image:10.png|thumb|center|400px|RBD of the fault tree shown in figure above.]]
 

BlockSim does not use the cut sets methodology when analyzing fault trees. However, interested users can obtain these cut sets for both fault trees and block diagrams with the command available in the Tools menu.

File:FT and RBD example.png

2012-03-19T17:28:24Z

Dingzhou Cao:

Template:BlockSim Analytical FRED Report Example

2012-03-19T17:11:11Z

Dingzhou Cao:

You can use the analytical results and maintainability/availability simulation results for your BlockSim diagrams and fault trees to generate Failure Reporting, Evaluating and Display (FRED) Reports. These reports provide a graphical demonstration of the reliability and maintainability/availability characteristics of the components in a system and help to identify the components that may require improvement.

Analytical FRED reports illustrate the reliability characteristics of a system’s components in a flexible, color-coded format.

For example, consider a analytical diagram like this:
[[Image:Anlytical FRED example.png|thumb|center|246px|]]

Block A with Weibull(Beta=1.5, Eta=1000), Block B with Weibull(Beta=1.5, Eta=700), Block C with Weibull(Beta=1.5, Eta=600) and Block D with Weibull(Beta=1.5, Eta=900).

The FRED reports for this diagram looks like this:
[[Image:Anlytical FRED example 1.png|thumb|center|546px|]]

The pentagon-shaped block in the first row represents the reliability results for the first diagram included in the analytical FRED report.

* t displays the time at which the reliability is reported for the system and its blocks and at which time the importance factor is reported for the blocks. You set the time in the At Time field in the control panel.

* R displays the reliability of the block at time t.

* Imp displays the reliability importance of the block at time t.

While two blocks may have the same reliability, their importance factors may differ because of their positions within the RBD. For example, a block in a series configuration may have a greater importance factor than a block in a parallel configuration, because any failure of the block in the series configuration will bring the system down.

Template:BlockSim Analytical FRED Report Example

2012-03-19T16:58:59Z

Dingzhou Cao:

Template:BlockSim Analytical FRED Report Example

2012-03-19T16:58:49Z

Dingzhou Cao:

Template:BlockSim Analytical FRED Report Example

2012-03-19T16:58:32Z

Dingzhou Cao:

File:Anlytical FRED example 1.png

2012-03-19T16:58:16Z

Dingzhou Cao:

Template:BlockSim Analytical FRED Report Example

2012-03-19T16:55:07Z

Dingzhou Cao:

Template:BlockSim Analytical FRED Report Example

2012-03-19T16:53:25Z

Dingzhou Cao:

Template:BlockSim Analytical FRED Report Example

2012-03-19T16:53:17Z

Dingzhou Cao:

File:Anlytical FRED example.png

2012-03-19T16:51:58Z

Dingzhou Cao:

Template:BlockSim Analytical FRED Report Example

2012-03-19T16:49:31Z

Dingzhou Cao:

Template:BlockSim Analytical FRED Report Example

2012-03-19T16:44:49Z

Dingzhou Cao:

Template:BlockSim Analytical FRED Report Example

2012-03-19T16:44:04Z

Dingzhou Cao: Created page with 'You can use the analytical results and maintainability/availability simulation results for your BlockSim diagrams and fault trees to generate Failure Reporting, Evaluating and Di…'

Template:Workbook Example

2012-03-19T16:22:36Z

Dingzhou Cao:

Using ReliaSoft’s analysis workbook and Function Wizard, you can generate many different types of results.

For example, suppose that you have the reliability data for two components from different suppliers. You can create a table that shows the reliability for each at a given time.

* First create the folios (in Weibull++ or ALTA) or diagrams (in BlockSim), add some data and calculate them.
* Add an Analysis Workbook to your project and do not select a default data source.
* Associate the two analyses to the spreadsheet.
* Update the column headings in the report (i.e., Time in column A, Reliability Data Source 1 in column B and Reliability Data Source 2 in column C).
* In cell A1 enter the first time value and in cell A2 use a function like =A1+100 to increment the time value used in that cell. Fill in cells A3 through A12 in a similar manner. When finished, your window would look like this:
[[Image:Report example1 WorkBook.png|thumb|center|346px|]]

* Click in cell B1 and open the Function Wizard. Select the RELIABILITY function (in Weibull++ or BlockSim) or the RELIABILITY_S function (in ALTA), use cell A1 for the time value and select the first associated data source. Then drag the column down to cell B12.

: Tip:In Weibull++ you can use the RELIABILITY function to generate both standard and conditional calculations. For standard calculations, you would enter a value in the Age field and leave the Add Time field empty. For conditional calculations, you would enter the starting time in the Age field and the additional time in the Add Time field. In ALTA and BlockSim you can use the CONDRELIABILITY_S function to generate conditional calculations in a similar manner.

* Starting in cell C1, repeat for the second associated data source. When finished, your window would look like this: [[Image:Report example2 WorkBook.png|thumb|center|346px|]]

By looking at the data in the table you can see which design has the better reliability at a given time.

File:Report example2 WorkBook.png

2012-03-19T16:22:06Z

Dingzhou Cao:

Template:Workbook Example

2012-03-19T16:21:49Z

Dingzhou Cao:

Using ReliaSoft’s analysis workbook and Function Wizard, you can generate many different types of results.

For example, suppose that you have the reliability data for two components from different suppliers. You can create a table that shows the reliability for each at a given time.

* First create the folios (in Weibull++ or ALTA) or diagrams (in BlockSim), add some data and calculate them.
* Add an Analysis Workbook to your project and do not select a default data source.
* Associate the two analyses to the spreadsheet.
* Update the column headings in the report (i.e., Time in column A, Reliability Data Source 1 in column B and Reliability Data Source 2 in column C).
* In cell A1 enter the first time value and in cell A2 use a function like =A1+100 to increment the time value used in that cell. Fill in cells A3 through A12 in a similar manner. When finished, your window would look like this:
[[Image:Report example1 WorkBook.png|thumb|center|346px|]]

* Click in cell B1 and open the Function Wizard. Select the RELIABILITY function (in Weibull++ or BlockSim) or the RELIABILITY_S function (in ALTA), use cell A1 for the time value and select the first associated data source. Then drag the column down to cell B12.

: Tip:In Weibull++ you can use the RELIABILITY function to generate both standard and conditional calculations. For standard calculations, you would enter a value in the Age field and leave the Add Time field empty. For conditional calculations, you would enter the starting time in the Age field and the additional time in the Add Time field. In ALTA and BlockSim you can use the CONDRELIABILITY_S function to generate conditional calculations in a similar manner.

* Starting in cell C1, repeat for the second associated data source. When finished, your window would look like this: Figure

By looking at the data in the table you can see which design has the better reliability at a given time.

File:Report example1 WorkBook.png

2012-03-19T16:19:55Z

Dingzhou Cao:

Template:Workbook Example

2012-03-19T16:18:30Z

Dingzhou Cao:

Using ReliaSoft’s analysis workbook and Function Wizard, you can generate many different types of results.

For example, suppose that you have the reliability data for two components from different suppliers. You can create a table that shows the reliability for each at a given time.

* First create the folios (in Weibull++ or ALTA) or diagrams (in BlockSim), add some data and calculate them.
* Add an Analysis Workbook to your project and do not select a default data source.
* Associate the two analyses to the spreadsheet.
* Update the column headings in the report (i.e., Time in column A, Reliability Data Source 1 in column B and Reliability Data Source 2 in column C).
* In cell A1 enter the first time value and in cell A2 use a function like =A1+100 to increment the time value used in that cell. Fill in cells A3 through A12 in a similar manner. When finished, your window would look like this: Figure
* Click in cell B1 and open the Function Wizard. Select the RELIABILITY function (in Weibull++ or BlockSim) or the RELIABILITY_S function (in ALTA), use cell A1 for the time value and select the first associated data source. Then drag the column down to cell B12.

: Tip:In Weibull++ you can use the RELIABILITY function to generate both standard and conditional calculations. For standard calculations, you would enter a value in the Age field and leave the Add Time field empty. For conditional calculations, you would enter the starting time in the Age field and the additional time in the Add Time field. In ALTA and BlockSim you can use the CONDRELIABILITY_S function to generate conditional calculations in a similar manner.

* Starting in cell C1, repeat for the second associated data source. When finished, your window would look like this: Figure

By looking at the data in the table you can see which design has the better reliability at a given time.

Template:Workbook Example

2012-03-19T16:18:10Z

Dingzhou Cao:

Using ReliaSoft’s analysis workbook and Function Wizard, you can generate many different types of results.

For example, suppose that you have the reliability data for two components from different suppliers. You can create a table that shows the reliability for each at a given time.

* First create the folios (in Weibull++ or ALTA) or diagrams (in BlockSim), add some data and calculate them.
# Add an Analysis Workbook to your project and do not select a default data source.
# Associate the two analyses to the spreadsheet.
# Update the column headings in the report (i.e., Time in column A, Reliability Data Source 1 in column B and Reliability Data Source 2 in column C).
# In cell A1 enter the first time value and in cell A2 use a function like =A1+100 to increment the time value used in that cell. Fill in cells A3 through A12 in a similar manner. When finished, your window would look like this: Figure
# Click in cell B1 and open the Function Wizard. Select the RELIABILITY function (in Weibull++ or BlockSim) or the RELIABILITY_S function (in ALTA), use cell A1 for the time value and select the first associated data source. Then drag the column down to cell B12.

: Tip:In Weibull++ you can use the RELIABILITY function to generate both standard and conditional calculations. For standard calculations, you would enter a value in the Age field and leave the Add Time field empty. For conditional calculations, you would enter the starting time in the Age field and the additional time in the Add Time field. In ALTA and BlockSim you can use the CONDRELIABILITY_S function to generate conditional calculations in a similar manner.

# Starting in cell C1, repeat for the second associated data source. When finished, your window would look like this: Figure
By looking at the data in the table you can see which design has the better reliability at a given time.