Repairable Systems Analysis
The previous chapters presented analysis methods for data obtained during developmental testing. However, data from systems in the field can also be analyzed in RGA. This type of data is called fielded systems data and is analogous to warranty data. Fielded systems can be categorized into two basic types: one-time or nonrepairable systems and reusable or repairable systems. In the latter case, under continuous operation, the system is repaired, but not replaced after each failure. For example, if a water pump in a vehicle fails, the water pump is replaced and the vehicle is repaired. Two types of analysis are presented in this chapter. The first is repairable systems analysis where the reliability of a system can be tracked and quantified based on data from multiple systems in the field. The second is fleet analysis where data from multiple systems in the field can be collected and analyzed so that reliability metrics for the fleet as a whole can be quantified.
Background
Most complex systems, such as automobiles, communication systems, aircraft, printers, medical diagnostics systems, helicopters, etc., are repaired and not replaced when they fail. When these systems are fielded or subjected to a customer use environment, it is often of considerable interest to determine the reliability and other performance characteristics under these conditions. Areas of interest may include assessing the expected number of failures during the warranty period, maintaining a minimum mission reliability, evaluating the rate of wearout, determining when to replace or overhaul a system and minimizing life cycle costs. In general, a lifetime distribution, such as the Weibull distribution, cannot be used to address these issues. In order to address the reliability characteristics of complex repairable systems, a process is often used instead of a distribution. The most popular process model is the Power Law model. This model is popular for several reasons. One is that it has a very practical foundation in terms of minimal repair. This is the situation when the repair of a failed system is just enough to get the system operational again. Second, if the time to first failure follows the Weibull distribution, then each succeeding failure is governed by the Power Law model in the case of minimal repair. From this point of view, the Power Law model is an extension of the Weibull distribution.
Sometimes, the Crow Extended model, which was introduced in a previous chapter for analyzing developmental data, is also applied for fielded repairable systems. Applying the Crow Extended model on repairable system data allows analysts to project the system MTBF after reliability-related issues are addressed during the field operation. Projections are calculated based on the mode classifications (A, BC and BD). The calculation procedure is the same as the one for the developmental data.and is not repeated in this chapter.
The previous chapters presented analysis methods for data obtained during developmental testing. However, data from systems in the field can also be analyzed in RGA. This type of data is called fielded systems data and is analogous to warranty data. Fielded systems can be categorized into two basic types: one-time or nonrepairable systems and reusable or repairable systems. In the latter case, under continuous operation, the system is repaired, but not replaced after each failure. For example, if a water pump in a vehicle fails, the water pump is replaced and the vehicle is repaired. Two types of analysis are presented in this chapter. The first is repairable systems analysis where the reliability of a system can be tracked and quantified based on data from multiple systems in the field. The second is fleet analysis where data from multiple systems in the field can be collected and analyzed so that reliability metrics for the fleet as a whole can be quantified.
Background
Most complex systems, such as automobiles, communication systems, aircraft, printers, medical diagnostics systems, helicopters, etc., are repaired and not replaced when they fail. When these systems are fielded or subjected to a customer use environment, it is often of considerable interest to determine the reliability and other performance characteristics under these conditions. Areas of interest may include assessing the expected number of failures during the warranty period, maintaining a minimum mission reliability, evaluating the rate of wearout, determining when to replace or overhaul a system and minimizing life cycle costs. In general, a lifetime distribution, such as the Weibull distribution, cannot be used to address these issues. In order to address the reliability characteristics of complex repairable systems, a process is often used instead of a distribution. The most popular process model is the Power Law model. This model is popular for several reasons. One is that it has a very practical foundation in terms of minimal repair. This is the situation when the repair of a failed system is just enough to get the system operational again. Second, if the time to first failure follows the Weibull distribution, then each succeeding failure is governed by the Power Law model in the case of minimal repair. From this point of view, the Power Law model is an extension of the Weibull distribution.
Sometimes, the Crow Extended model, which was introduced in a previous chapter for analyzing developmental data, is also applied for fielded repairable systems. Applying the Crow Extended model on repairable system data allows analysts to project the system MTBF after reliability-related issues are addressed during the field operation. Projections are calculated based on the mode classifications (A, BC and BD). The calculation procedure is the same as the one for the developmental data.and is not repeated in this chapter.
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Parameter Estimation
Suppose that the number of systems under study is [math]\displaystyle{ K }[/math] and the [math]\displaystyle{ {{q}^{th}} }[/math] system is observed continuously from time [math]\displaystyle{ {{S}_{q}} }[/math] to time [math]\displaystyle{ {{T}_{q}} }[/math] , [math]\displaystyle{ q=1,2,\ldots ,K }[/math] . During the period [math]\displaystyle{ [{{S}_{q}},{{T}_{q}}] }[/math] , let [math]\displaystyle{ {{N}_{q}} }[/math] be the number of failures experienced by the [math]\displaystyle{ {{q}^{th}} }[/math] system and let [math]\displaystyle{ {{X}_{i,q}} }[/math] be the age of this system at the [math]\displaystyle{ {{i}^{th}} }[/math] occurrence of failure, [math]\displaystyle{ i=1,2,\ldots ,{{N}_{q}} }[/math] . It is also possible that the times [math]\displaystyle{ {{S}_{q}} }[/math] and [math]\displaystyle{ {{T}_{q}} }[/math] may be observed failure times for the [math]\displaystyle{ {{q}^{th}} }[/math] system. If [math]\displaystyle{ {{X}_{{{N}_{q}},q}}={{T}_{q}} }[/math] then the data on the [math]\displaystyle{ {{q}^{th}} }[/math] system is said to be failure terminated and [math]\displaystyle{ {{T}_{q}} }[/math] is a random variable with [math]\displaystyle{ {{N}_{q}} }[/math] fixed. If [math]\displaystyle{ {{X}_{{{N}_{q}},q}}\lt {{T}_{q}} }[/math] then the data on the [math]\displaystyle{ {{q}^{th}} }[/math] system is said to be time terminated with [math]\displaystyle{ {{N}_{q}} }[/math] a random variable. The maximum likelihood estimates of [math]\displaystyle{ \lambda }[/math] and [math]\displaystyle{ \beta }[/math] are values satisfying the Eqns. (lambdaPowerLaw) and (BetaPowerLaw).
- [math]\displaystyle{ \begin{align} & \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\left( T_{q}^{\widehat{\beta }}-S_{q}^{\widehat{\beta }} \right)} \\ & \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\widehat{\lambda }\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\left[ T_{q}^{\widehat{\beta }}\ln ({{T}_{q}})-S_{q}^{\widehat{\beta }}\ln ({{S}_{q}}) \right]-\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln ({{X}_{i,q}})} \end{align} }[/math]
where [math]\displaystyle{ 0\ln 0 }[/math] is defined to be 0. In general, these equations cannot be solved explicitly for [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta }, }[/math] but must be solved by iterative procedures. Once [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta } }[/math] have been estimated, the maximum likelihood estimate of the intensity function is given by:
- [math]\displaystyle{ \widehat{u}(t)=\widehat{\lambda }\widehat{\beta }{{t}^{\widehat{\beta }-1}} }[/math]
If [math]\displaystyle{ {{S}_{1}}={{S}_{2}}=\ldots ={{S}_{q}}=0 }[/math] and [math]\displaystyle{ {{T}_{1}}={{T}_{2}}=\ldots ={{T}_{q}} }[/math] [math]\displaystyle{ \,(q=1,2,\ldots ,K) }[/math] then the maximum likelihood estimates [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta } }[/math] are in closed form.
- [math]\displaystyle{ \begin{align} & \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{K{{T}^{\beta }}} \\ & \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln (\tfrac{T}{{{X}_{iq}}})} \end{align} }[/math]
The following examples illustrate these estimation procedures.
Example 1
For the data in Table 13.1, the starting time for each system is equal to [math]\displaystyle{ 0 }[/math] and the ending time for each system is 2000 hours. Calculate the maximum likelihood estimates [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta } }[/math] .
| Table 13.1 - Repairable system failure data | ||
| System 1 ( [math]\displaystyle{ {{X}_{i1}} }[/math] ) | System 2 ( [math]\displaystyle{ {{X}_{i2}} }[/math] ) | System 3 ( [math]\displaystyle{ {{X}_{i3}} }[/math] ) |
|---|---|---|
| 1.2 | 1.4 | 0.3 |
| 55.6 | 35.0 | 32.6 |
| 72.7 | 46.8 | 33.4 |
| 111.9 | 65.9 | 241.7 |
| 121.9 | 181.1 | 396.2 |
| 303.6 | 712.6 | 444.4 |
| 326.9 | 1005.7 | 480.8 |
| 1568.4 | 1029.9 | 588.9 |
| 1913.5 | 1675.7 | 1043.9 |
| 1787.5 | 1136.1 | |
| 1867.0 | 1288.1 | |
| 1408.1 | ||
| 1439.4 | ||
| 1604.8 | ||
| [math]\displaystyle{ {{N}_{1}}=9 }[/math] | [math]\displaystyle{ {{N}_{2}}=11 }[/math] | [math]\displaystyle{ {{N}_{3}}=14 }[/math] |
Solution
Since the starting time for each system is equal to zero and each system has an equivalent ending time, the general Eqns. (lambdaPowerLaw) and (BetaPowerLaw) reduce to the closed form Eqns. (sample1) and (sample2). The maximum likelihood estimates of [math]\displaystyle{ \hat{\beta } }[/math] and [math]\displaystyle{ \hat{\lambda } }[/math] are then calculated as follows:
- [math]\displaystyle{ \begin{align} & \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln (\tfrac{T}{{{X}_{iq}}})} \\ & = & 0.45300 \end{align} }[/math]
- [math]\displaystyle{ \begin{align} & \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{K{{T}^{\beta }}} \\ & = & 0.36224 \end{align} }[/math]
The system failure intensity function is then estimated by:
- [math]\displaystyle{ \widehat{u}(t)=\widehat{\lambda }\widehat{\beta }{{t}^{\widehat{\beta }-1}},\text{ }t\gt 0 }[/math]
Figure wpp intensity is a plot of [math]\displaystyle{ \widehat{u}(t) }[/math] over the period (0, 3000). Clearly, the estimated failure intensity function is most representative over the range of the data and any extrapolation should be viewed with the usual caution.
Goodness-of-Fit Tests for Repairable System Analysis
It is generally desirable to test the compatibility of a model and data by a statistical goodness-of-fit test. A parametric Cramér-von Mises goodness-of-fit test is used for the multiple system and repairable system Power Law model, as proposed by Crow in [17]. This goodness-of-fit test is appropriate whenever the start time for each system is 0 and the failure data is complete over the continuous interval [math]\displaystyle{ [0,{{T}_{q}}] }[/math] with no gaps in the data. The Chi-Squared test is a goodness-of-fit test that can be applied under more general circumstances. In addition, the Common Beta Hypothesis test also can be used to compare the intensity functions of the individual systems by comparing the [math]\displaystyle{ {{\beta }_{q}} }[/math] values of each system. Lastly, the Laplace Trend test checks for trends within the data. Due to their general applicatoin, the Common Beta Hypothesis test and the Laplace Trend test are both presented in Appendix B. The Cramér-von Mises and Chi-Squared goodness-of-fit tests are illustrated next.
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Economical Life Model
One consideration in reducing the cost to maintain repairable systems is to establish an overhaul policy that will minimize the total life cost of the system. However, an overhaul policy makes sense only if [math]\displaystyle{ \beta \gt 1 }[/math] . It does not make sense to implement an overhaul policy if [math]\displaystyle{ \beta \lt 1 }[/math] since wearout is not present. If you assume that there is a point at which it is cheaper to overhaul a system than to continue repairs, what is the overhaul time that will minimize the total life cycle cost while considering repair cost and the cost of overhaul?
Denote [math]\displaystyle{ {{C}_{1}} }[/math] as the average repair cost (unscheduled), [math]\displaystyle{ {{C}_{2}} }[/math] as the replacement or overhaul cost and [math]\displaystyle{ {{C}_{3}} }[/math] as the average cost of scheduled maintenance. Scheduled maintenance is performed for every [math]\displaystyle{ S }[/math] miles or time interval. In addition, let [math]\displaystyle{ {{N}_{1}} }[/math] be the number of failures in [math]\displaystyle{ [0,t] }[/math] and let [math]\displaystyle{ {{N}_{2}} }[/math] be the number of replacements in [math]\displaystyle{ [0,t] }[/math] . Suppose that replacement or overhaul occurs at times [math]\displaystyle{ T }[/math] , [math]\displaystyle{ 2T }[/math] , [math]\displaystyle{ 3T }[/math] . The problem is to select the optimum overhaul time [math]\displaystyle{ T={{T}_{0}} }[/math] so as to minimize the long term average system cost (unscheduled maintenance, replacement cost and scheduled maintenance). Since [math]\displaystyle{ \beta \gt 1 }[/math] , the average system cost is minimized when the system is overhauled (or replaced) at time [math]\displaystyle{ {{T}_{0}} }[/math] such that the instantaneous maintenance cost equals the average system cost.
The total system cost between overhaul or replacement is:
- [math]\displaystyle{ TSC(T)={{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\frac{T}{S} }[/math]
So the average system cost is:
- [math]\displaystyle{ C(T)=\frac{{{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\tfrac{T}{S}}{T} }[/math]
The instantaneous maintenance cost at time [math]\displaystyle{ T }[/math] is equal to:
- [math]\displaystyle{ IMC(T)={{C}_{1}}\lambda \beta {{T}^{\beta -1}}+\frac{{{C}_{3}}}{S} }[/math]
The following equation holds at optimum overhaul time [math]\displaystyle{ {{T}_{0}} }[/math] :
- [math]\displaystyle{ \begin{align} & {{C}_{1}}\lambda \beta T_{0}^{\beta -1}+\frac{{{C}_{3}}}{S}= & \frac{{{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\tfrac{T}{S}}{T} \\ & = & \frac{{{C}_{1}}\lambda T_{0}^{\beta }+{{C}_{2}}+{{C}_{3}}\tfrac{{{T}_{0}}}{S}}{{{T}_{0}}} \end{align} }[/math]
- Therefore:
- [math]\displaystyle{ {{T}_{0}}={{\left[ \frac{{{C}_{2}}}{\lambda (\beta -1){{C}_{1}}} \right]}^{1/\beta }} }[/math]
When there is no scheduled maintenance, Eqn. (ecolm) becomes:
- [math]\displaystyle{ {{C}_{1}}\lambda \beta T_{0}^{\beta -1}=\frac{{{C}_{1}}\lambda T_{0}^{\beta }+{{C}_{2}}}{{{T}_{0}}} }[/math]
The optimum overhaul time, [math]\displaystyle{ {{T}_{0}} }[/math] , is the same as Eqn. (optimt), so for periodic maintenance scheduled every [math]\displaystyle{ S }[/math] miles, the replacement or overhaul time is the same as for the unscheduled and replacement or overhaul cost model.
Examples
Example 6 (repairable system data)
This case study is based on the data given in the article Graphical Analysis of Repair Data by Dr. Wayne Nelson [23]. The data in Table 13.10 represents repair data on an automatic transmission from a sample of 34 cars. For each car, the data set shows mileage at the time of each transmission repair, along with the latest mileage. The + indicates the latest mileage observed without failure. Car 1, for example, had a repair at 7068 miles and was observed until 26,744 miles. Do the following:
- 1) Estimate the parameters of the Power Law model.
- 2) Estimate the number of warranty claims for a 36,000 mile warranty policy for an estimated fleet of 35,000 vehicles.
| Table 13.10 - Automatic transmission data | ||||
| Car | Mileage | Car | Mileage | |
|---|---|---|---|---|
| 1 | 7068, 26744+ | 18 | 17955+ | |
| 2 | 28, 13809+ | 19 | 19507+ | |
| 3 | 48, 1440, 29834+ | 20 | 24177+ | |
| 4 | 530, 25660+ | 21 | 22854+ | |
| 5 | 21762+ | 22 | 17844+ | |
| 6 | 14235+ | 23 | 22637+ | |
| 7 | 1388, 18228+ | 24 | 375, 19607+ | |
| 8 | 21401+ | 25 | 19403+ | |
| 9 | 21876+ | 26 | 20997+ | |
| 10 | 5094, 18228+ | 27 | 19175+ | |
| 11 | 21691+ | 28 | 20425+ | |
| 12 | 20890+ | 29 | 22149+ | |
| 13 | 22486+ | 30 | 21144+ | |
| 14 | 19321+ | 31 | 21237+ | |
| 15 | 21585+ | 32 | 14281+ | |
| 16 | 18676+ | 33 | 8250, 21974+ | |
| 17 | 23520+ | 34 | 19250, 21888+ | |
Solution to Example 6
- 1) The estimated Power Law parameters are shown in Figure Repair3.
- 2) The expected number of failures at 36,000 miles can be estimated using the QCP as shown in Figure Repair4. The model predicts that 0.3559 failures per system will occur by 36,000 miles. This means that for a fleet of 35,000 vehicles, the expected warranty claims are 0.3559 * 35,000 = 12,456.
[math]\displaystyle{ }[/math]
[math]\displaystyle{ }[/math]
Example 7 (repairable system data)
Field data have been collected for a system that begins its wearout phase at time zero. The start time for each system is equal to zero and the end time for each system is 10,000 miles. Each system is scheduled to undergo an overhaul after a certain number of miles. It has been determined that the cost of an overhaul is four times more expensive than a repair. Table 13.11 presents the data. Do the following:
- 1) Estimate the parameters of the Power Law model.
- 2) Determine the optimum overhaul interval.
- 3) If [math]\displaystyle{ \beta \lt 1 }[/math] , would it be cost-effective to implement an overhaul policy?
| Table 13.11 - Field data | ||
| System 1 | System 2 | System 3 |
|---|---|---|
| 1006.3 | 722.7 | 619.1 |
| 2261.2 | 1950.9 | 1519.1 |
| 2367 | 3259.6 | 2956.6 |
| 2615.5 | 4733.9 | 3114.8 |
| 2848.1 | 5105.1 | 3657.9 |
| 4073 | 5624.1 | 4268.9 |
| 5708.1 | 5806.3 | 6690.2 |
| 6464.1 | 5855.6 | 6803.1 |
| 6519.7 | 6325.2 | 7323.9 |
| 6799.1 | 6999.4 | 7501.4 |
| 7342.9 | 7084.4 | 7641.2 |
| 7736 | 7105.9 | 7851.6 |
| 8246.1 | 7290.9 | 8147.6 |
| 7614.2 | 8221.9 | |
| 8332.1 | 9560.5 | |
| 8368.5 | 9575.4 | |
| 8947.9 | ||
| 9012.3 | ||
| 9135.9 | ||
| 9147.5 | ||
| 9601 | ||
Solution to Example 7
- 1) Figure Repair5 shows the estimated Power Law parameters.
- 2) The QCP can be used to calculate the optimum overhaul interval as shown in Figure Repair6.
- 3) Since [math]\displaystyle{ \beta \lt 1 }[/math] then the systems are not wearing out and it would not be cost-effective to implement an overhaul policy. An overhaul policy makes sense only if the systems are wearing out. Otherwise, an overhauled unit would have the same probability of failing as a unit that was not overhauled.
[math]\displaystyle{ }[/math]
Example 8 (repairable system data)
Failures and fixes of two repairable systems in the field are recorded. Both systems start from time 0. System 1 ends at time = 504 and system 2 ends at time = 541. All the BD modes are fixed at the end of the test. A fixed effectiveness factor equal to 0.6 is used. Answer the following questions:
- 1) Estimate the parameters of the Crow Extended model.
- 2) Calculate the projected MTBF after the delayed fixes.
- 3) What is the expected number of failures at time 1,000, if no fixes were performed for the future failures?
Solution to Example 8
- 1) Figure CrowExtendedRepair shows the estimated Crow Extended parameters.
- 2) Figure CrowExtendedMTBF shows the projected MTBF at time = 541 (i.e. the age of the oldest system).
- 3) Figure CrowExtendedNumofFailure shows the expected number of failures at time = 1,000.
[math]\displaystyle{ }[/math]
The previous chapters presented analysis methods for data obtained during developmental testing. However, data from systems in the field can also be analyzed in RGA. This type of data is called fielded systems data and is analogous to warranty data. Fielded systems can be categorized into two basic types: one-time or nonrepairable systems and reusable or repairable systems. In the latter case, under continuous operation, the system is repaired, but not replaced after each failure. For example, if a water pump in a vehicle fails, the water pump is replaced and the vehicle is repaired. Two types of analysis are presented in this chapter. The first is repairable systems analysis where the reliability of a system can be tracked and quantified based on data from multiple systems in the field. The second is fleet analysis where data from multiple systems in the field can be collected and analyzed so that reliability metrics for the fleet as a whole can be quantified.
Background
Most complex systems, such as automobiles, communication systems, aircraft, printers, medical diagnostics systems, helicopters, etc., are repaired and not replaced when they fail. When these systems are fielded or subjected to a customer use environment, it is often of considerable interest to determine the reliability and other performance characteristics under these conditions. Areas of interest may include assessing the expected number of failures during the warranty period, maintaining a minimum mission reliability, evaluating the rate of wearout, determining when to replace or overhaul a system and minimizing life cycle costs. In general, a lifetime distribution, such as the Weibull distribution, cannot be used to address these issues. In order to address the reliability characteristics of complex repairable systems, a process is often used instead of a distribution. The most popular process model is the Power Law model. This model is popular for several reasons. One is that it has a very practical foundation in terms of minimal repair. This is the situation when the repair of a failed system is just enough to get the system operational again. Second, if the time to first failure follows the Weibull distribution, then each succeeding failure is governed by the Power Law model in the case of minimal repair. From this point of view, the Power Law model is an extension of the Weibull distribution.
Sometimes, the Crow Extended model, which was introduced in a previous chapter for analyzing developmental data, is also applied for fielded repairable systems. Applying the Crow Extended model on repairable system data allows analysts to project the system MTBF after reliability-related issues are addressed during the field operation. Projections are calculated based on the mode classifications (A, BC and BD). The calculation procedure is the same as the one for the developmental data.and is not repeated in this chapter.
Template loop detected: Template:Distribution ex rsa
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Template loop detected: Template:Using power law model rsa
Parameter Estimation
Suppose that the number of systems under study is [math]\displaystyle{ K }[/math] and the [math]\displaystyle{ {{q}^{th}} }[/math] system is observed continuously from time [math]\displaystyle{ {{S}_{q}} }[/math] to time [math]\displaystyle{ {{T}_{q}} }[/math] , [math]\displaystyle{ q=1,2,\ldots ,K }[/math] . During the period [math]\displaystyle{ [{{S}_{q}},{{T}_{q}}] }[/math] , let [math]\displaystyle{ {{N}_{q}} }[/math] be the number of failures experienced by the [math]\displaystyle{ {{q}^{th}} }[/math] system and let [math]\displaystyle{ {{X}_{i,q}} }[/math] be the age of this system at the [math]\displaystyle{ {{i}^{th}} }[/math] occurrence of failure, [math]\displaystyle{ i=1,2,\ldots ,{{N}_{q}} }[/math] . It is also possible that the times [math]\displaystyle{ {{S}_{q}} }[/math] and [math]\displaystyle{ {{T}_{q}} }[/math] may be observed failure times for the [math]\displaystyle{ {{q}^{th}} }[/math] system. If [math]\displaystyle{ {{X}_{{{N}_{q}},q}}={{T}_{q}} }[/math] then the data on the [math]\displaystyle{ {{q}^{th}} }[/math] system is said to be failure terminated and [math]\displaystyle{ {{T}_{q}} }[/math] is a random variable with [math]\displaystyle{ {{N}_{q}} }[/math] fixed. If [math]\displaystyle{ {{X}_{{{N}_{q}},q}}\lt {{T}_{q}} }[/math] then the data on the [math]\displaystyle{ {{q}^{th}} }[/math] system is said to be time terminated with [math]\displaystyle{ {{N}_{q}} }[/math] a random variable. The maximum likelihood estimates of [math]\displaystyle{ \lambda }[/math] and [math]\displaystyle{ \beta }[/math] are values satisfying the Eqns. (lambdaPowerLaw) and (BetaPowerLaw).
- [math]\displaystyle{ \begin{align} & \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\left( T_{q}^{\widehat{\beta }}-S_{q}^{\widehat{\beta }} \right)} \\ & \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\widehat{\lambda }\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\left[ T_{q}^{\widehat{\beta }}\ln ({{T}_{q}})-S_{q}^{\widehat{\beta }}\ln ({{S}_{q}}) \right]-\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln ({{X}_{i,q}})} \end{align} }[/math]
where [math]\displaystyle{ 0\ln 0 }[/math] is defined to be 0. In general, these equations cannot be solved explicitly for [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta }, }[/math] but must be solved by iterative procedures. Once [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta } }[/math] have been estimated, the maximum likelihood estimate of the intensity function is given by:
- [math]\displaystyle{ \widehat{u}(t)=\widehat{\lambda }\widehat{\beta }{{t}^{\widehat{\beta }-1}} }[/math]
If [math]\displaystyle{ {{S}_{1}}={{S}_{2}}=\ldots ={{S}_{q}}=0 }[/math] and [math]\displaystyle{ {{T}_{1}}={{T}_{2}}=\ldots ={{T}_{q}} }[/math] [math]\displaystyle{ \,(q=1,2,\ldots ,K) }[/math] then the maximum likelihood estimates [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta } }[/math] are in closed form.
- [math]\displaystyle{ \begin{align} & \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{K{{T}^{\beta }}} \\ & \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln (\tfrac{T}{{{X}_{iq}}})} \end{align} }[/math]
The following examples illustrate these estimation procedures.
Example 1
For the data in Table 13.1, the starting time for each system is equal to [math]\displaystyle{ 0 }[/math] and the ending time for each system is 2000 hours. Calculate the maximum likelihood estimates [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta } }[/math] .
| Table 13.1 - Repairable system failure data | ||
| System 1 ( [math]\displaystyle{ {{X}_{i1}} }[/math] ) | System 2 ( [math]\displaystyle{ {{X}_{i2}} }[/math] ) | System 3 ( [math]\displaystyle{ {{X}_{i3}} }[/math] ) |
|---|---|---|
| 1.2 | 1.4 | 0.3 |
| 55.6 | 35.0 | 32.6 |
| 72.7 | 46.8 | 33.4 |
| 111.9 | 65.9 | 241.7 |
| 121.9 | 181.1 | 396.2 |
| 303.6 | 712.6 | 444.4 |
| 326.9 | 1005.7 | 480.8 |
| 1568.4 | 1029.9 | 588.9 |
| 1913.5 | 1675.7 | 1043.9 |
| 1787.5 | 1136.1 | |
| 1867.0 | 1288.1 | |
| 1408.1 | ||
| 1439.4 | ||
| 1604.8 | ||
| [math]\displaystyle{ {{N}_{1}}=9 }[/math] | [math]\displaystyle{ {{N}_{2}}=11 }[/math] | [math]\displaystyle{ {{N}_{3}}=14 }[/math] |
Solution
Since the starting time for each system is equal to zero and each system has an equivalent ending time, the general Eqns. (lambdaPowerLaw) and (BetaPowerLaw) reduce to the closed form Eqns. (sample1) and (sample2). The maximum likelihood estimates of [math]\displaystyle{ \hat{\beta } }[/math] and [math]\displaystyle{ \hat{\lambda } }[/math] are then calculated as follows:
- [math]\displaystyle{ \begin{align} & \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln (\tfrac{T}{{{X}_{iq}}})} \\ & = & 0.45300 \end{align} }[/math]
- [math]\displaystyle{ \begin{align} & \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{K{{T}^{\beta }}} \\ & = & 0.36224 \end{align} }[/math]
The system failure intensity function is then estimated by:
- [math]\displaystyle{ \widehat{u}(t)=\widehat{\lambda }\widehat{\beta }{{t}^{\widehat{\beta }-1}},\text{ }t\gt 0 }[/math]
Figure wpp intensity is a plot of [math]\displaystyle{ \widehat{u}(t) }[/math] over the period (0, 3000). Clearly, the estimated failure intensity function is most representative over the range of the data and any extrapolation should be viewed with the usual caution.
Goodness-of-Fit Tests for Repairable System Analysis
It is generally desirable to test the compatibility of a model and data by a statistical goodness-of-fit test. A parametric Cramér-von Mises goodness-of-fit test is used for the multiple system and repairable system Power Law model, as proposed by Crow in [17]. This goodness-of-fit test is appropriate whenever the start time for each system is 0 and the failure data is complete over the continuous interval [math]\displaystyle{ [0,{{T}_{q}}] }[/math] with no gaps in the data. The Chi-Squared test is a goodness-of-fit test that can be applied under more general circumstances. In addition, the Common Beta Hypothesis test also can be used to compare the intensity functions of the individual systems by comparing the [math]\displaystyle{ {{\beta }_{q}} }[/math] values of each system. Lastly, the Laplace Trend test checks for trends within the data. Due to their general applicatoin, the Common Beta Hypothesis test and the Laplace Trend test are both presented in Appendix B. The Cramér-von Mises and Chi-Squared goodness-of-fit tests are illustrated next.
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Economical Life Model
One consideration in reducing the cost to maintain repairable systems is to establish an overhaul policy that will minimize the total life cost of the system. However, an overhaul policy makes sense only if [math]\displaystyle{ \beta \gt 1 }[/math] . It does not make sense to implement an overhaul policy if [math]\displaystyle{ \beta \lt 1 }[/math] since wearout is not present. If you assume that there is a point at which it is cheaper to overhaul a system than to continue repairs, what is the overhaul time that will minimize the total life cycle cost while considering repair cost and the cost of overhaul?
Denote [math]\displaystyle{ {{C}_{1}} }[/math] as the average repair cost (unscheduled), [math]\displaystyle{ {{C}_{2}} }[/math] as the replacement or overhaul cost and [math]\displaystyle{ {{C}_{3}} }[/math] as the average cost of scheduled maintenance. Scheduled maintenance is performed for every [math]\displaystyle{ S }[/math] miles or time interval. In addition, let [math]\displaystyle{ {{N}_{1}} }[/math] be the number of failures in [math]\displaystyle{ [0,t] }[/math] and let [math]\displaystyle{ {{N}_{2}} }[/math] be the number of replacements in [math]\displaystyle{ [0,t] }[/math] . Suppose that replacement or overhaul occurs at times [math]\displaystyle{ T }[/math] , [math]\displaystyle{ 2T }[/math] , [math]\displaystyle{ 3T }[/math] . The problem is to select the optimum overhaul time [math]\displaystyle{ T={{T}_{0}} }[/math] so as to minimize the long term average system cost (unscheduled maintenance, replacement cost and scheduled maintenance). Since [math]\displaystyle{ \beta \gt 1 }[/math] , the average system cost is minimized when the system is overhauled (or replaced) at time [math]\displaystyle{ {{T}_{0}} }[/math] such that the instantaneous maintenance cost equals the average system cost.
The total system cost between overhaul or replacement is:
- [math]\displaystyle{ TSC(T)={{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\frac{T}{S} }[/math]
So the average system cost is:
- [math]\displaystyle{ C(T)=\frac{{{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\tfrac{T}{S}}{T} }[/math]
The instantaneous maintenance cost at time [math]\displaystyle{ T }[/math] is equal to:
- [math]\displaystyle{ IMC(T)={{C}_{1}}\lambda \beta {{T}^{\beta -1}}+\frac{{{C}_{3}}}{S} }[/math]
The following equation holds at optimum overhaul time [math]\displaystyle{ {{T}_{0}} }[/math] :
- [math]\displaystyle{ \begin{align} & {{C}_{1}}\lambda \beta T_{0}^{\beta -1}+\frac{{{C}_{3}}}{S}= & \frac{{{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\tfrac{T}{S}}{T} \\ & = & \frac{{{C}_{1}}\lambda T_{0}^{\beta }+{{C}_{2}}+{{C}_{3}}\tfrac{{{T}_{0}}}{S}}{{{T}_{0}}} \end{align} }[/math]
- Therefore:
- [math]\displaystyle{ {{T}_{0}}={{\left[ \frac{{{C}_{2}}}{\lambda (\beta -1){{C}_{1}}} \right]}^{1/\beta }} }[/math]
When there is no scheduled maintenance, Eqn. (ecolm) becomes:
- [math]\displaystyle{ {{C}_{1}}\lambda \beta T_{0}^{\beta -1}=\frac{{{C}_{1}}\lambda T_{0}^{\beta }+{{C}_{2}}}{{{T}_{0}}} }[/math]
The optimum overhaul time, [math]\displaystyle{ {{T}_{0}} }[/math] , is the same as Eqn. (optimt), so for periodic maintenance scheduled every [math]\displaystyle{ S }[/math] miles, the replacement or overhaul time is the same as for the unscheduled and replacement or overhaul cost model.
Examples
Example 6 (repairable system data)
This case study is based on the data given in the article Graphical Analysis of Repair Data by Dr. Wayne Nelson [23]. The data in Table 13.10 represents repair data on an automatic transmission from a sample of 34 cars. For each car, the data set shows mileage at the time of each transmission repair, along with the latest mileage. The + indicates the latest mileage observed without failure. Car 1, for example, had a repair at 7068 miles and was observed until 26,744 miles. Do the following:
- 1) Estimate the parameters of the Power Law model.
- 2) Estimate the number of warranty claims for a 36,000 mile warranty policy for an estimated fleet of 35,000 vehicles.
| Table 13.10 - Automatic transmission data | ||||
| Car | Mileage | Car | Mileage | |
|---|---|---|---|---|
| 1 | 7068, 26744+ | 18 | 17955+ | |
| 2 | 28, 13809+ | 19 | 19507+ | |
| 3 | 48, 1440, 29834+ | 20 | 24177+ | |
| 4 | 530, 25660+ | 21 | 22854+ | |
| 5 | 21762+ | 22 | 17844+ | |
| 6 | 14235+ | 23 | 22637+ | |
| 7 | 1388, 18228+ | 24 | 375, 19607+ | |
| 8 | 21401+ | 25 | 19403+ | |
| 9 | 21876+ | 26 | 20997+ | |
| 10 | 5094, 18228+ | 27 | 19175+ | |
| 11 | 21691+ | 28 | 20425+ | |
| 12 | 20890+ | 29 | 22149+ | |
| 13 | 22486+ | 30 | 21144+ | |
| 14 | 19321+ | 31 | 21237+ | |
| 15 | 21585+ | 32 | 14281+ | |
| 16 | 18676+ | 33 | 8250, 21974+ | |
| 17 | 23520+ | 34 | 19250, 21888+ | |
Solution to Example 6
- 1) The estimated Power Law parameters are shown in Figure Repair3.
- 2) The expected number of failures at 36,000 miles can be estimated using the QCP as shown in Figure Repair4. The model predicts that 0.3559 failures per system will occur by 36,000 miles. This means that for a fleet of 35,000 vehicles, the expected warranty claims are 0.3559 * 35,000 = 12,456.
[math]\displaystyle{ }[/math]
[math]\displaystyle{ }[/math]
Example 7 (repairable system data)
Field data have been collected for a system that begins its wearout phase at time zero. The start time for each system is equal to zero and the end time for each system is 10,000 miles. Each system is scheduled to undergo an overhaul after a certain number of miles. It has been determined that the cost of an overhaul is four times more expensive than a repair. Table 13.11 presents the data. Do the following:
- 1) Estimate the parameters of the Power Law model.
- 2) Determine the optimum overhaul interval.
- 3) If [math]\displaystyle{ \beta \lt 1 }[/math] , would it be cost-effective to implement an overhaul policy?
| Table 13.11 - Field data | ||
| System 1 | System 2 | System 3 |
|---|---|---|
| 1006.3 | 722.7 | 619.1 |
| 2261.2 | 1950.9 | 1519.1 |
| 2367 | 3259.6 | 2956.6 |
| 2615.5 | 4733.9 | 3114.8 |
| 2848.1 | 5105.1 | 3657.9 |
| 4073 | 5624.1 | 4268.9 |
| 5708.1 | 5806.3 | 6690.2 |
| 6464.1 | 5855.6 | 6803.1 |
| 6519.7 | 6325.2 | 7323.9 |
| 6799.1 | 6999.4 | 7501.4 |
| 7342.9 | 7084.4 | 7641.2 |
| 7736 | 7105.9 | 7851.6 |
| 8246.1 | 7290.9 | 8147.6 |
| 7614.2 | 8221.9 | |
| 8332.1 | 9560.5 | |
| 8368.5 | 9575.4 | |
| 8947.9 | ||
| 9012.3 | ||
| 9135.9 | ||
| 9147.5 | ||
| 9601 | ||
Solution to Example 7
- 1) Figure Repair5 shows the estimated Power Law parameters.
- 2) The QCP can be used to calculate the optimum overhaul interval as shown in Figure Repair6.
- 3) Since [math]\displaystyle{ \beta \lt 1 }[/math] then the systems are not wearing out and it would not be cost-effective to implement an overhaul policy. An overhaul policy makes sense only if the systems are wearing out. Otherwise, an overhauled unit would have the same probability of failing as a unit that was not overhauled.
[math]\displaystyle{ }[/math]
Example 8 (repairable system data)
Failures and fixes of two repairable systems in the field are recorded. Both systems start from time 0. System 1 ends at time = 504 and system 2 ends at time = 541. All the BD modes are fixed at the end of the test. A fixed effectiveness factor equal to 0.6 is used. Answer the following questions:
- 1) Estimate the parameters of the Crow Extended model.
- 2) Calculate the projected MTBF after the delayed fixes.
- 3) What is the expected number of failures at time 1,000, if no fixes were performed for the future failures?
Solution to Example 8
- 1) Figure CrowExtendedRepair shows the estimated Crow Extended parameters.
- 2) Figure CrowExtendedMTBF shows the projected MTBF at time = 541 (i.e. the age of the oldest system).
- 3) Figure CrowExtendedNumofFailure shows the expected number of failures at time = 1,000.
[math]\displaystyle{ }[/math]
The previous chapters presented analysis methods for data obtained during developmental testing. However, data from systems in the field can also be analyzed in RGA. This type of data is called fielded systems data and is analogous to warranty data. Fielded systems can be categorized into two basic types: one-time or nonrepairable systems and reusable or repairable systems. In the latter case, under continuous operation, the system is repaired, but not replaced after each failure. For example, if a water pump in a vehicle fails, the water pump is replaced and the vehicle is repaired. Two types of analysis are presented in this chapter. The first is repairable systems analysis where the reliability of a system can be tracked and quantified based on data from multiple systems in the field. The second is fleet analysis where data from multiple systems in the field can be collected and analyzed so that reliability metrics for the fleet as a whole can be quantified.
Background
Most complex systems, such as automobiles, communication systems, aircraft, printers, medical diagnostics systems, helicopters, etc., are repaired and not replaced when they fail. When these systems are fielded or subjected to a customer use environment, it is often of considerable interest to determine the reliability and other performance characteristics under these conditions. Areas of interest may include assessing the expected number of failures during the warranty period, maintaining a minimum mission reliability, evaluating the rate of wearout, determining when to replace or overhaul a system and minimizing life cycle costs. In general, a lifetime distribution, such as the Weibull distribution, cannot be used to address these issues. In order to address the reliability characteristics of complex repairable systems, a process is often used instead of a distribution. The most popular process model is the Power Law model. This model is popular for several reasons. One is that it has a very practical foundation in terms of minimal repair. This is the situation when the repair of a failed system is just enough to get the system operational again. Second, if the time to first failure follows the Weibull distribution, then each succeeding failure is governed by the Power Law model in the case of minimal repair. From this point of view, the Power Law model is an extension of the Weibull distribution.
Sometimes, the Crow Extended model, which was introduced in a previous chapter for analyzing developmental data, is also applied for fielded repairable systems. Applying the Crow Extended model on repairable system data allows analysts to project the system MTBF after reliability-related issues are addressed during the field operation. Projections are calculated based on the mode classifications (A, BC and BD). The calculation procedure is the same as the one for the developmental data.and is not repeated in this chapter.
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Parameter Estimation
Suppose that the number of systems under study is [math]\displaystyle{ K }[/math] and the [math]\displaystyle{ {{q}^{th}} }[/math] system is observed continuously from time [math]\displaystyle{ {{S}_{q}} }[/math] to time [math]\displaystyle{ {{T}_{q}} }[/math] , [math]\displaystyle{ q=1,2,\ldots ,K }[/math] . During the period [math]\displaystyle{ [{{S}_{q}},{{T}_{q}}] }[/math] , let [math]\displaystyle{ {{N}_{q}} }[/math] be the number of failures experienced by the [math]\displaystyle{ {{q}^{th}} }[/math] system and let [math]\displaystyle{ {{X}_{i,q}} }[/math] be the age of this system at the [math]\displaystyle{ {{i}^{th}} }[/math] occurrence of failure, [math]\displaystyle{ i=1,2,\ldots ,{{N}_{q}} }[/math] . It is also possible that the times [math]\displaystyle{ {{S}_{q}} }[/math] and [math]\displaystyle{ {{T}_{q}} }[/math] may be observed failure times for the [math]\displaystyle{ {{q}^{th}} }[/math] system. If [math]\displaystyle{ {{X}_{{{N}_{q}},q}}={{T}_{q}} }[/math] then the data on the [math]\displaystyle{ {{q}^{th}} }[/math] system is said to be failure terminated and [math]\displaystyle{ {{T}_{q}} }[/math] is a random variable with [math]\displaystyle{ {{N}_{q}} }[/math] fixed. If [math]\displaystyle{ {{X}_{{{N}_{q}},q}}\lt {{T}_{q}} }[/math] then the data on the [math]\displaystyle{ {{q}^{th}} }[/math] system is said to be time terminated with [math]\displaystyle{ {{N}_{q}} }[/math] a random variable. The maximum likelihood estimates of [math]\displaystyle{ \lambda }[/math] and [math]\displaystyle{ \beta }[/math] are values satisfying the Eqns. (lambdaPowerLaw) and (BetaPowerLaw).
- [math]\displaystyle{ \begin{align} & \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\left( T_{q}^{\widehat{\beta }}-S_{q}^{\widehat{\beta }} \right)} \\ & \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\widehat{\lambda }\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\left[ T_{q}^{\widehat{\beta }}\ln ({{T}_{q}})-S_{q}^{\widehat{\beta }}\ln ({{S}_{q}}) \right]-\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln ({{X}_{i,q}})} \end{align} }[/math]
where [math]\displaystyle{ 0\ln 0 }[/math] is defined to be 0. In general, these equations cannot be solved explicitly for [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta }, }[/math] but must be solved by iterative procedures. Once [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta } }[/math] have been estimated, the maximum likelihood estimate of the intensity function is given by:
- [math]\displaystyle{ \widehat{u}(t)=\widehat{\lambda }\widehat{\beta }{{t}^{\widehat{\beta }-1}} }[/math]
If [math]\displaystyle{ {{S}_{1}}={{S}_{2}}=\ldots ={{S}_{q}}=0 }[/math] and [math]\displaystyle{ {{T}_{1}}={{T}_{2}}=\ldots ={{T}_{q}} }[/math] [math]\displaystyle{ \,(q=1,2,\ldots ,K) }[/math] then the maximum likelihood estimates [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta } }[/math] are in closed form.
- [math]\displaystyle{ \begin{align} & \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{K{{T}^{\beta }}} \\ & \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln (\tfrac{T}{{{X}_{iq}}})} \end{align} }[/math]
The following examples illustrate these estimation procedures.
Example 1
For the data in Table 13.1, the starting time for each system is equal to [math]\displaystyle{ 0 }[/math] and the ending time for each system is 2000 hours. Calculate the maximum likelihood estimates [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta } }[/math] .
| Table 13.1 - Repairable system failure data | ||
| System 1 ( [math]\displaystyle{ {{X}_{i1}} }[/math] ) | System 2 ( [math]\displaystyle{ {{X}_{i2}} }[/math] ) | System 3 ( [math]\displaystyle{ {{X}_{i3}} }[/math] ) |
|---|---|---|
| 1.2 | 1.4 | 0.3 |
| 55.6 | 35.0 | 32.6 |
| 72.7 | 46.8 | 33.4 |
| 111.9 | 65.9 | 241.7 |
| 121.9 | 181.1 | 396.2 |
| 303.6 | 712.6 | 444.4 |
| 326.9 | 1005.7 | 480.8 |
| 1568.4 | 1029.9 | 588.9 |
| 1913.5 | 1675.7 | 1043.9 |
| 1787.5 | 1136.1 | |
| 1867.0 | 1288.1 | |
| 1408.1 | ||
| 1439.4 | ||
| 1604.8 | ||
| [math]\displaystyle{ {{N}_{1}}=9 }[/math] | [math]\displaystyle{ {{N}_{2}}=11 }[/math] | [math]\displaystyle{ {{N}_{3}}=14 }[/math] |
Solution
Since the starting time for each system is equal to zero and each system has an equivalent ending time, the general Eqns. (lambdaPowerLaw) and (BetaPowerLaw) reduce to the closed form Eqns. (sample1) and (sample2). The maximum likelihood estimates of [math]\displaystyle{ \hat{\beta } }[/math] and [math]\displaystyle{ \hat{\lambda } }[/math] are then calculated as follows:
- [math]\displaystyle{ \begin{align} & \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln (\tfrac{T}{{{X}_{iq}}})} \\ & = & 0.45300 \end{align} }[/math]
- [math]\displaystyle{ \begin{align} & \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{K{{T}^{\beta }}} \\ & = & 0.36224 \end{align} }[/math]
The system failure intensity function is then estimated by:
- [math]\displaystyle{ \widehat{u}(t)=\widehat{\lambda }\widehat{\beta }{{t}^{\widehat{\beta }-1}},\text{ }t\gt 0 }[/math]
Figure wpp intensity is a plot of [math]\displaystyle{ \widehat{u}(t) }[/math] over the period (0, 3000). Clearly, the estimated failure intensity function is most representative over the range of the data and any extrapolation should be viewed with the usual caution.
Goodness-of-Fit Tests for Repairable System Analysis
It is generally desirable to test the compatibility of a model and data by a statistical goodness-of-fit test. A parametric Cramér-von Mises goodness-of-fit test is used for the multiple system and repairable system Power Law model, as proposed by Crow in [17]. This goodness-of-fit test is appropriate whenever the start time for each system is 0 and the failure data is complete over the continuous interval [math]\displaystyle{ [0,{{T}_{q}}] }[/math] with no gaps in the data. The Chi-Squared test is a goodness-of-fit test that can be applied under more general circumstances. In addition, the Common Beta Hypothesis test also can be used to compare the intensity functions of the individual systems by comparing the [math]\displaystyle{ {{\beta }_{q}} }[/math] values of each system. Lastly, the Laplace Trend test checks for trends within the data. Due to their general applicatoin, the Common Beta Hypothesis test and the Laplace Trend test are both presented in Appendix B. The Cramér-von Mises and Chi-Squared goodness-of-fit tests are illustrated next.
Template loop detected: Template:Cramer-con mises test rsa
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Economical Life Model
One consideration in reducing the cost to maintain repairable systems is to establish an overhaul policy that will minimize the total life cost of the system. However, an overhaul policy makes sense only if [math]\displaystyle{ \beta \gt 1 }[/math] . It does not make sense to implement an overhaul policy if [math]\displaystyle{ \beta \lt 1 }[/math] since wearout is not present. If you assume that there is a point at which it is cheaper to overhaul a system than to continue repairs, what is the overhaul time that will minimize the total life cycle cost while considering repair cost and the cost of overhaul?
Denote [math]\displaystyle{ {{C}_{1}} }[/math] as the average repair cost (unscheduled), [math]\displaystyle{ {{C}_{2}} }[/math] as the replacement or overhaul cost and [math]\displaystyle{ {{C}_{3}} }[/math] as the average cost of scheduled maintenance. Scheduled maintenance is performed for every [math]\displaystyle{ S }[/math] miles or time interval. In addition, let [math]\displaystyle{ {{N}_{1}} }[/math] be the number of failures in [math]\displaystyle{ [0,t] }[/math] and let [math]\displaystyle{ {{N}_{2}} }[/math] be the number of replacements in [math]\displaystyle{ [0,t] }[/math] . Suppose that replacement or overhaul occurs at times [math]\displaystyle{ T }[/math] , [math]\displaystyle{ 2T }[/math] , [math]\displaystyle{ 3T }[/math] . The problem is to select the optimum overhaul time [math]\displaystyle{ T={{T}_{0}} }[/math] so as to minimize the long term average system cost (unscheduled maintenance, replacement cost and scheduled maintenance). Since [math]\displaystyle{ \beta \gt 1 }[/math] , the average system cost is minimized when the system is overhauled (or replaced) at time [math]\displaystyle{ {{T}_{0}} }[/math] such that the instantaneous maintenance cost equals the average system cost.
The total system cost between overhaul or replacement is:
- [math]\displaystyle{ TSC(T)={{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\frac{T}{S} }[/math]
So the average system cost is:
- [math]\displaystyle{ C(T)=\frac{{{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\tfrac{T}{S}}{T} }[/math]
The instantaneous maintenance cost at time [math]\displaystyle{ T }[/math] is equal to:
- [math]\displaystyle{ IMC(T)={{C}_{1}}\lambda \beta {{T}^{\beta -1}}+\frac{{{C}_{3}}}{S} }[/math]
The following equation holds at optimum overhaul time [math]\displaystyle{ {{T}_{0}} }[/math] :
- [math]\displaystyle{ \begin{align} & {{C}_{1}}\lambda \beta T_{0}^{\beta -1}+\frac{{{C}_{3}}}{S}= & \frac{{{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\tfrac{T}{S}}{T} \\ & = & \frac{{{C}_{1}}\lambda T_{0}^{\beta }+{{C}_{2}}+{{C}_{3}}\tfrac{{{T}_{0}}}{S}}{{{T}_{0}}} \end{align} }[/math]
- Therefore:
- [math]\displaystyle{ {{T}_{0}}={{\left[ \frac{{{C}_{2}}}{\lambda (\beta -1){{C}_{1}}} \right]}^{1/\beta }} }[/math]
When there is no scheduled maintenance, Eqn. (ecolm) becomes:
- [math]\displaystyle{ {{C}_{1}}\lambda \beta T_{0}^{\beta -1}=\frac{{{C}_{1}}\lambda T_{0}^{\beta }+{{C}_{2}}}{{{T}_{0}}} }[/math]
The optimum overhaul time, [math]\displaystyle{ {{T}_{0}} }[/math] , is the same as Eqn. (optimt), so for periodic maintenance scheduled every [math]\displaystyle{ S }[/math] miles, the replacement or overhaul time is the same as for the unscheduled and replacement or overhaul cost model.
Examples
Example 6 (repairable system data)
This case study is based on the data given in the article Graphical Analysis of Repair Data by Dr. Wayne Nelson [23]. The data in Table 13.10 represents repair data on an automatic transmission from a sample of 34 cars. For each car, the data set shows mileage at the time of each transmission repair, along with the latest mileage. The + indicates the latest mileage observed without failure. Car 1, for example, had a repair at 7068 miles and was observed until 26,744 miles. Do the following:
- 1) Estimate the parameters of the Power Law model.
- 2) Estimate the number of warranty claims for a 36,000 mile warranty policy for an estimated fleet of 35,000 vehicles.
| Table 13.10 - Automatic transmission data | ||||
| Car | Mileage | Car | Mileage | |
|---|---|---|---|---|
| 1 | 7068, 26744+ | 18 | 17955+ | |
| 2 | 28, 13809+ | 19 | 19507+ | |
| 3 | 48, 1440, 29834+ | 20 | 24177+ | |
| 4 | 530, 25660+ | 21 | 22854+ | |
| 5 | 21762+ | 22 | 17844+ | |
| 6 | 14235+ | 23 | 22637+ | |
| 7 | 1388, 18228+ | 24 | 375, 19607+ | |
| 8 | 21401+ | 25 | 19403+ | |
| 9 | 21876+ | 26 | 20997+ | |
| 10 | 5094, 18228+ | 27 | 19175+ | |
| 11 | 21691+ | 28 | 20425+ | |
| 12 | 20890+ | 29 | 22149+ | |
| 13 | 22486+ | 30 | 21144+ | |
| 14 | 19321+ | 31 | 21237+ | |
| 15 | 21585+ | 32 | 14281+ | |
| 16 | 18676+ | 33 | 8250, 21974+ | |
| 17 | 23520+ | 34 | 19250, 21888+ | |
Solution to Example 6
- 1) The estimated Power Law parameters are shown in Figure Repair3.
- 2) The expected number of failures at 36,000 miles can be estimated using the QCP as shown in Figure Repair4. The model predicts that 0.3559 failures per system will occur by 36,000 miles. This means that for a fleet of 35,000 vehicles, the expected warranty claims are 0.3559 * 35,000 = 12,456.
[math]\displaystyle{ }[/math]
[math]\displaystyle{ }[/math]
Example 7 (repairable system data)
Field data have been collected for a system that begins its wearout phase at time zero. The start time for each system is equal to zero and the end time for each system is 10,000 miles. Each system is scheduled to undergo an overhaul after a certain number of miles. It has been determined that the cost of an overhaul is four times more expensive than a repair. Table 13.11 presents the data. Do the following:
- 1) Estimate the parameters of the Power Law model.
- 2) Determine the optimum overhaul interval.
- 3) If [math]\displaystyle{ \beta \lt 1 }[/math] , would it be cost-effective to implement an overhaul policy?
| Table 13.11 - Field data | ||
| System 1 | System 2 | System 3 |
|---|---|---|
| 1006.3 | 722.7 | 619.1 |
| 2261.2 | 1950.9 | 1519.1 |
| 2367 | 3259.6 | 2956.6 |
| 2615.5 | 4733.9 | 3114.8 |
| 2848.1 | 5105.1 | 3657.9 |
| 4073 | 5624.1 | 4268.9 |
| 5708.1 | 5806.3 | 6690.2 |
| 6464.1 | 5855.6 | 6803.1 |
| 6519.7 | 6325.2 | 7323.9 |
| 6799.1 | 6999.4 | 7501.4 |
| 7342.9 | 7084.4 | 7641.2 |
| 7736 | 7105.9 | 7851.6 |
| 8246.1 | 7290.9 | 8147.6 |
| 7614.2 | 8221.9 | |
| 8332.1 | 9560.5 | |
| 8368.5 | 9575.4 | |
| 8947.9 | ||
| 9012.3 | ||
| 9135.9 | ||
| 9147.5 | ||
| 9601 | ||
Solution to Example 7
- 1) Figure Repair5 shows the estimated Power Law parameters.
- 2) The QCP can be used to calculate the optimum overhaul interval as shown in Figure Repair6.
- 3) Since [math]\displaystyle{ \beta \lt 1 }[/math] then the systems are not wearing out and it would not be cost-effective to implement an overhaul policy. An overhaul policy makes sense only if the systems are wearing out. Otherwise, an overhauled unit would have the same probability of failing as a unit that was not overhauled.
[math]\displaystyle{ }[/math]
Example 8 (repairable system data)
Failures and fixes of two repairable systems in the field are recorded. Both systems start from time 0. System 1 ends at time = 504 and system 2 ends at time = 541. All the BD modes are fixed at the end of the test. A fixed effectiveness factor equal to 0.6 is used. Answer the following questions:
- 1) Estimate the parameters of the Crow Extended model.
- 2) Calculate the projected MTBF after the delayed fixes.
- 3) What is the expected number of failures at time 1,000, if no fixes were performed for the future failures?
Solution to Example 8
- 1) Figure CrowExtendedRepair shows the estimated Crow Extended parameters.
- 2) Figure CrowExtendedMTBF shows the projected MTBF at time = 541 (i.e. the age of the oldest system).
- 3) Figure CrowExtendedNumofFailure shows the expected number of failures at time = 1,000.
[math]\displaystyle{ }[/math]
Parameter Estimation
Suppose that the number of systems under study is [math]\displaystyle{ K }[/math] and the [math]\displaystyle{ {{q}^{th}} }[/math] system is observed continuously from time [math]\displaystyle{ {{S}_{q}} }[/math] to time [math]\displaystyle{ {{T}_{q}} }[/math] , [math]\displaystyle{ q=1,2,\ldots ,K }[/math] . During the period [math]\displaystyle{ [{{S}_{q}},{{T}_{q}}] }[/math] , let [math]\displaystyle{ {{N}_{q}} }[/math] be the number of failures experienced by the [math]\displaystyle{ {{q}^{th}} }[/math] system and let [math]\displaystyle{ {{X}_{i,q}} }[/math] be the age of this system at the [math]\displaystyle{ {{i}^{th}} }[/math] occurrence of failure, [math]\displaystyle{ i=1,2,\ldots ,{{N}_{q}} }[/math] . It is also possible that the times [math]\displaystyle{ {{S}_{q}} }[/math] and [math]\displaystyle{ {{T}_{q}} }[/math] may be observed failure times for the [math]\displaystyle{ {{q}^{th}} }[/math] system. If [math]\displaystyle{ {{X}_{{{N}_{q}},q}}={{T}_{q}} }[/math] then the data on the [math]\displaystyle{ {{q}^{th}} }[/math] system is said to be failure terminated and [math]\displaystyle{ {{T}_{q}} }[/math] is a random variable with [math]\displaystyle{ {{N}_{q}} }[/math] fixed. If [math]\displaystyle{ {{X}_{{{N}_{q}},q}}\lt {{T}_{q}} }[/math] then the data on the [math]\displaystyle{ {{q}^{th}} }[/math] system is said to be time terminated with [math]\displaystyle{ {{N}_{q}} }[/math] a random variable. The maximum likelihood estimates of [math]\displaystyle{ \lambda }[/math] and [math]\displaystyle{ \beta }[/math] are values satisfying the Eqns. (lambdaPowerLaw) and (BetaPowerLaw).
- [math]\displaystyle{ \begin{align} & \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\left( T_{q}^{\widehat{\beta }}-S_{q}^{\widehat{\beta }} \right)} \\ & \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\widehat{\lambda }\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\left[ T_{q}^{\widehat{\beta }}\ln ({{T}_{q}})-S_{q}^{\widehat{\beta }}\ln ({{S}_{q}}) \right]-\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln ({{X}_{i,q}})} \end{align} }[/math]
where [math]\displaystyle{ 0\ln 0 }[/math] is defined to be 0. In general, these equations cannot be solved explicitly for [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta }, }[/math] but must be solved by iterative procedures. Once [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta } }[/math] have been estimated, the maximum likelihood estimate of the intensity function is given by:
- [math]\displaystyle{ \widehat{u}(t)=\widehat{\lambda }\widehat{\beta }{{t}^{\widehat{\beta }-1}} }[/math]
If [math]\displaystyle{ {{S}_{1}}={{S}_{2}}=\ldots ={{S}_{q}}=0 }[/math] and [math]\displaystyle{ {{T}_{1}}={{T}_{2}}=\ldots ={{T}_{q}} }[/math] [math]\displaystyle{ \,(q=1,2,\ldots ,K) }[/math] then the maximum likelihood estimates [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta } }[/math] are in closed form.
- [math]\displaystyle{ \begin{align} & \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{K{{T}^{\beta }}} \\ & \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln (\tfrac{T}{{{X}_{iq}}})} \end{align} }[/math]
The following examples illustrate these estimation procedures.
Example 1
For the data in Table 13.1, the starting time for each system is equal to [math]\displaystyle{ 0 }[/math] and the ending time for each system is 2000 hours. Calculate the maximum likelihood estimates [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta } }[/math] .
| Table 13.1 - Repairable system failure data | ||
| System 1 ( [math]\displaystyle{ {{X}_{i1}} }[/math] ) | System 2 ( [math]\displaystyle{ {{X}_{i2}} }[/math] ) | System 3 ( [math]\displaystyle{ {{X}_{i3}} }[/math] ) |
|---|---|---|
| 1.2 | 1.4 | 0.3 |
| 55.6 | 35.0 | 32.6 |
| 72.7 | 46.8 | 33.4 |
| 111.9 | 65.9 | 241.7 |
| 121.9 | 181.1 | 396.2 |
| 303.6 | 712.6 | 444.4 |
| 326.9 | 1005.7 | 480.8 |
| 1568.4 | 1029.9 | 588.9 |
| 1913.5 | 1675.7 | 1043.9 |
| 1787.5 | 1136.1 | |
| 1867.0 | 1288.1 | |
| 1408.1 | ||
| 1439.4 | ||
| 1604.8 | ||
| [math]\displaystyle{ {{N}_{1}}=9 }[/math] | [math]\displaystyle{ {{N}_{2}}=11 }[/math] | [math]\displaystyle{ {{N}_{3}}=14 }[/math] |
Solution
Since the starting time for each system is equal to zero and each system has an equivalent ending time, the general Eqns. (lambdaPowerLaw) and (BetaPowerLaw) reduce to the closed form Eqns. (sample1) and (sample2). The maximum likelihood estimates of [math]\displaystyle{ \hat{\beta } }[/math] and [math]\displaystyle{ \hat{\lambda } }[/math] are then calculated as follows:
- [math]\displaystyle{ \begin{align} & \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln (\tfrac{T}{{{X}_{iq}}})} \\ & = & 0.45300 \end{align} }[/math]
- [math]\displaystyle{ \begin{align} & \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{K{{T}^{\beta }}} \\ & = & 0.36224 \end{align} }[/math]
The system failure intensity function is then estimated by:
- [math]\displaystyle{ \widehat{u}(t)=\widehat{\lambda }\widehat{\beta }{{t}^{\widehat{\beta }-1}},\text{ }t\gt 0 }[/math]
Figure wpp intensity is a plot of [math]\displaystyle{ \widehat{u}(t) }[/math] over the period (0, 3000). Clearly, the estimated failure intensity function is most representative over the range of the data and any extrapolation should be viewed with the usual caution.
Goodness-of-Fit Tests for Repairable System Analysis
It is generally desirable to test the compatibility of a model and data by a statistical goodness-of-fit test. A parametric Cramér-von Mises goodness-of-fit test is used for the multiple system and repairable system Power Law model, as proposed by Crow in [17]. This goodness-of-fit test is appropriate whenever the start time for each system is 0 and the failure data is complete over the continuous interval [math]\displaystyle{ [0,{{T}_{q}}] }[/math] with no gaps in the data. The Chi-Squared test is a goodness-of-fit test that can be applied under more general circumstances. In addition, the Common Beta Hypothesis test also can be used to compare the intensity functions of the individual systems by comparing the [math]\displaystyle{ {{\beta }_{q}} }[/math] values of each system. Lastly, the Laplace Trend test checks for trends within the data. Due to their general applicatoin, the Common Beta Hypothesis test and the Laplace Trend test are both presented in Appendix B. The Cramér-von Mises and Chi-Squared goodness-of-fit tests are illustrated next.
The previous chapters presented analysis methods for data obtained during developmental testing. However, data from systems in the field can also be analyzed in RGA. This type of data is called fielded systems data and is analogous to warranty data. Fielded systems can be categorized into two basic types: one-time or nonrepairable systems and reusable or repairable systems. In the latter case, under continuous operation, the system is repaired, but not replaced after each failure. For example, if a water pump in a vehicle fails, the water pump is replaced and the vehicle is repaired. Two types of analysis are presented in this chapter. The first is repairable systems analysis where the reliability of a system can be tracked and quantified based on data from multiple systems in the field. The second is fleet analysis where data from multiple systems in the field can be collected and analyzed so that reliability metrics for the fleet as a whole can be quantified.
Background
Most complex systems, such as automobiles, communication systems, aircraft, printers, medical diagnostics systems, helicopters, etc., are repaired and not replaced when they fail. When these systems are fielded or subjected to a customer use environment, it is often of considerable interest to determine the reliability and other performance characteristics under these conditions. Areas of interest may include assessing the expected number of failures during the warranty period, maintaining a minimum mission reliability, evaluating the rate of wearout, determining when to replace or overhaul a system and minimizing life cycle costs. In general, a lifetime distribution, such as the Weibull distribution, cannot be used to address these issues. In order to address the reliability characteristics of complex repairable systems, a process is often used instead of a distribution. The most popular process model is the Power Law model. This model is popular for several reasons. One is that it has a very practical foundation in terms of minimal repair. This is the situation when the repair of a failed system is just enough to get the system operational again. Second, if the time to first failure follows the Weibull distribution, then each succeeding failure is governed by the Power Law model in the case of minimal repair. From this point of view, the Power Law model is an extension of the Weibull distribution.
Sometimes, the Crow Extended model, which was introduced in a previous chapter for analyzing developmental data, is also applied for fielded repairable systems. Applying the Crow Extended model on repairable system data allows analysts to project the system MTBF after reliability-related issues are addressed during the field operation. Projections are calculated based on the mode classifications (A, BC and BD). The calculation procedure is the same as the one for the developmental data.and is not repeated in this chapter.
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Parameter Estimation
Suppose that the number of systems under study is [math]\displaystyle{ K }[/math] and the [math]\displaystyle{ {{q}^{th}} }[/math] system is observed continuously from time [math]\displaystyle{ {{S}_{q}} }[/math] to time [math]\displaystyle{ {{T}_{q}} }[/math] , [math]\displaystyle{ q=1,2,\ldots ,K }[/math] . During the period [math]\displaystyle{ [{{S}_{q}},{{T}_{q}}] }[/math] , let [math]\displaystyle{ {{N}_{q}} }[/math] be the number of failures experienced by the [math]\displaystyle{ {{q}^{th}} }[/math] system and let [math]\displaystyle{ {{X}_{i,q}} }[/math] be the age of this system at the [math]\displaystyle{ {{i}^{th}} }[/math] occurrence of failure, [math]\displaystyle{ i=1,2,\ldots ,{{N}_{q}} }[/math] . It is also possible that the times [math]\displaystyle{ {{S}_{q}} }[/math] and [math]\displaystyle{ {{T}_{q}} }[/math] may be observed failure times for the [math]\displaystyle{ {{q}^{th}} }[/math] system. If [math]\displaystyle{ {{X}_{{{N}_{q}},q}}={{T}_{q}} }[/math] then the data on the [math]\displaystyle{ {{q}^{th}} }[/math] system is said to be failure terminated and [math]\displaystyle{ {{T}_{q}} }[/math] is a random variable with [math]\displaystyle{ {{N}_{q}} }[/math] fixed. If [math]\displaystyle{ {{X}_{{{N}_{q}},q}}\lt {{T}_{q}} }[/math] then the data on the [math]\displaystyle{ {{q}^{th}} }[/math] system is said to be time terminated with [math]\displaystyle{ {{N}_{q}} }[/math] a random variable. The maximum likelihood estimates of [math]\displaystyle{ \lambda }[/math] and [math]\displaystyle{ \beta }[/math] are values satisfying the Eqns. (lambdaPowerLaw) and (BetaPowerLaw).
- [math]\displaystyle{ \begin{align} & \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\left( T_{q}^{\widehat{\beta }}-S_{q}^{\widehat{\beta }} \right)} \\ & \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\widehat{\lambda }\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\left[ T_{q}^{\widehat{\beta }}\ln ({{T}_{q}})-S_{q}^{\widehat{\beta }}\ln ({{S}_{q}}) \right]-\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln ({{X}_{i,q}})} \end{align} }[/math]
where [math]\displaystyle{ 0\ln 0 }[/math] is defined to be 0. In general, these equations cannot be solved explicitly for [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta }, }[/math] but must be solved by iterative procedures. Once [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta } }[/math] have been estimated, the maximum likelihood estimate of the intensity function is given by:
- [math]\displaystyle{ \widehat{u}(t)=\widehat{\lambda }\widehat{\beta }{{t}^{\widehat{\beta }-1}} }[/math]
If [math]\displaystyle{ {{S}_{1}}={{S}_{2}}=\ldots ={{S}_{q}}=0 }[/math] and [math]\displaystyle{ {{T}_{1}}={{T}_{2}}=\ldots ={{T}_{q}} }[/math] [math]\displaystyle{ \,(q=1,2,\ldots ,K) }[/math] then the maximum likelihood estimates [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta } }[/math] are in closed form.
- [math]\displaystyle{ \begin{align} & \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{K{{T}^{\beta }}} \\ & \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln (\tfrac{T}{{{X}_{iq}}})} \end{align} }[/math]
The following examples illustrate these estimation procedures.
Example 1
For the data in Table 13.1, the starting time for each system is equal to [math]\displaystyle{ 0 }[/math] and the ending time for each system is 2000 hours. Calculate the maximum likelihood estimates [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta } }[/math] .
| Table 13.1 - Repairable system failure data | ||
| System 1 ( [math]\displaystyle{ {{X}_{i1}} }[/math] ) | System 2 ( [math]\displaystyle{ {{X}_{i2}} }[/math] ) | System 3 ( [math]\displaystyle{ {{X}_{i3}} }[/math] ) |
|---|---|---|
| 1.2 | 1.4 | 0.3 |
| 55.6 | 35.0 | 32.6 |
| 72.7 | 46.8 | 33.4 |
| 111.9 | 65.9 | 241.7 |
| 121.9 | 181.1 | 396.2 |
| 303.6 | 712.6 | 444.4 |
| 326.9 | 1005.7 | 480.8 |
| 1568.4 | 1029.9 | 588.9 |
| 1913.5 | 1675.7 | 1043.9 |
| 1787.5 | 1136.1 | |
| 1867.0 | 1288.1 | |
| 1408.1 | ||
| 1439.4 | ||
| 1604.8 | ||
| [math]\displaystyle{ {{N}_{1}}=9 }[/math] | [math]\displaystyle{ {{N}_{2}}=11 }[/math] | [math]\displaystyle{ {{N}_{3}}=14 }[/math] |
Solution
Since the starting time for each system is equal to zero and each system has an equivalent ending time, the general Eqns. (lambdaPowerLaw) and (BetaPowerLaw) reduce to the closed form Eqns. (sample1) and (sample2). The maximum likelihood estimates of [math]\displaystyle{ \hat{\beta } }[/math] and [math]\displaystyle{ \hat{\lambda } }[/math] are then calculated as follows:
- [math]\displaystyle{ \begin{align} & \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln (\tfrac{T}{{{X}_{iq}}})} \\ & = & 0.45300 \end{align} }[/math]
- [math]\displaystyle{ \begin{align} & \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{K{{T}^{\beta }}} \\ & = & 0.36224 \end{align} }[/math]
The system failure intensity function is then estimated by:
- [math]\displaystyle{ \widehat{u}(t)=\widehat{\lambda }\widehat{\beta }{{t}^{\widehat{\beta }-1}},\text{ }t\gt 0 }[/math]
Figure wpp intensity is a plot of [math]\displaystyle{ \widehat{u}(t) }[/math] over the period (0, 3000). Clearly, the estimated failure intensity function is most representative over the range of the data and any extrapolation should be viewed with the usual caution.
Goodness-of-Fit Tests for Repairable System Analysis
It is generally desirable to test the compatibility of a model and data by a statistical goodness-of-fit test. A parametric Cramér-von Mises goodness-of-fit test is used for the multiple system and repairable system Power Law model, as proposed by Crow in [17]. This goodness-of-fit test is appropriate whenever the start time for each system is 0 and the failure data is complete over the continuous interval [math]\displaystyle{ [0,{{T}_{q}}] }[/math] with no gaps in the data. The Chi-Squared test is a goodness-of-fit test that can be applied under more general circumstances. In addition, the Common Beta Hypothesis test also can be used to compare the intensity functions of the individual systems by comparing the [math]\displaystyle{ {{\beta }_{q}} }[/math] values of each system. Lastly, the Laplace Trend test checks for trends within the data. Due to their general applicatoin, the Common Beta Hypothesis test and the Laplace Trend test are both presented in Appendix B. The Cramér-von Mises and Chi-Squared goodness-of-fit tests are illustrated next.
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Economical Life Model
One consideration in reducing the cost to maintain repairable systems is to establish an overhaul policy that will minimize the total life cost of the system. However, an overhaul policy makes sense only if [math]\displaystyle{ \beta \gt 1 }[/math] . It does not make sense to implement an overhaul policy if [math]\displaystyle{ \beta \lt 1 }[/math] since wearout is not present. If you assume that there is a point at which it is cheaper to overhaul a system than to continue repairs, what is the overhaul time that will minimize the total life cycle cost while considering repair cost and the cost of overhaul?
Denote [math]\displaystyle{ {{C}_{1}} }[/math] as the average repair cost (unscheduled), [math]\displaystyle{ {{C}_{2}} }[/math] as the replacement or overhaul cost and [math]\displaystyle{ {{C}_{3}} }[/math] as the average cost of scheduled maintenance. Scheduled maintenance is performed for every [math]\displaystyle{ S }[/math] miles or time interval. In addition, let [math]\displaystyle{ {{N}_{1}} }[/math] be the number of failures in [math]\displaystyle{ [0,t] }[/math] and let [math]\displaystyle{ {{N}_{2}} }[/math] be the number of replacements in [math]\displaystyle{ [0,t] }[/math] . Suppose that replacement or overhaul occurs at times [math]\displaystyle{ T }[/math] , [math]\displaystyle{ 2T }[/math] , [math]\displaystyle{ 3T }[/math] . The problem is to select the optimum overhaul time [math]\displaystyle{ T={{T}_{0}} }[/math] so as to minimize the long term average system cost (unscheduled maintenance, replacement cost and scheduled maintenance). Since [math]\displaystyle{ \beta \gt 1 }[/math] , the average system cost is minimized when the system is overhauled (or replaced) at time [math]\displaystyle{ {{T}_{0}} }[/math] such that the instantaneous maintenance cost equals the average system cost.
The total system cost between overhaul or replacement is:
- [math]\displaystyle{ TSC(T)={{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\frac{T}{S} }[/math]
So the average system cost is:
- [math]\displaystyle{ C(T)=\frac{{{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\tfrac{T}{S}}{T} }[/math]
The instantaneous maintenance cost at time [math]\displaystyle{ T }[/math] is equal to:
- [math]\displaystyle{ IMC(T)={{C}_{1}}\lambda \beta {{T}^{\beta -1}}+\frac{{{C}_{3}}}{S} }[/math]
The following equation holds at optimum overhaul time [math]\displaystyle{ {{T}_{0}} }[/math] :
- [math]\displaystyle{ \begin{align} & {{C}_{1}}\lambda \beta T_{0}^{\beta -1}+\frac{{{C}_{3}}}{S}= & \frac{{{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\tfrac{T}{S}}{T} \\ & = & \frac{{{C}_{1}}\lambda T_{0}^{\beta }+{{C}_{2}}+{{C}_{3}}\tfrac{{{T}_{0}}}{S}}{{{T}_{0}}} \end{align} }[/math]
- Therefore:
- [math]\displaystyle{ {{T}_{0}}={{\left[ \frac{{{C}_{2}}}{\lambda (\beta -1){{C}_{1}}} \right]}^{1/\beta }} }[/math]
When there is no scheduled maintenance, Eqn. (ecolm) becomes:
- [math]\displaystyle{ {{C}_{1}}\lambda \beta T_{0}^{\beta -1}=\frac{{{C}_{1}}\lambda T_{0}^{\beta }+{{C}_{2}}}{{{T}_{0}}} }[/math]
The optimum overhaul time, [math]\displaystyle{ {{T}_{0}} }[/math] , is the same as Eqn. (optimt), so for periodic maintenance scheduled every [math]\displaystyle{ S }[/math] miles, the replacement or overhaul time is the same as for the unscheduled and replacement or overhaul cost model.
Examples
Example 6 (repairable system data)
This case study is based on the data given in the article Graphical Analysis of Repair Data by Dr. Wayne Nelson [23]. The data in Table 13.10 represents repair data on an automatic transmission from a sample of 34 cars. For each car, the data set shows mileage at the time of each transmission repair, along with the latest mileage. The + indicates the latest mileage observed without failure. Car 1, for example, had a repair at 7068 miles and was observed until 26,744 miles. Do the following:
- 1) Estimate the parameters of the Power Law model.
- 2) Estimate the number of warranty claims for a 36,000 mile warranty policy for an estimated fleet of 35,000 vehicles.
| Table 13.10 - Automatic transmission data | ||||
| Car | Mileage | Car | Mileage | |
|---|---|---|---|---|
| 1 | 7068, 26744+ | 18 | 17955+ | |
| 2 | 28, 13809+ | 19 | 19507+ | |
| 3 | 48, 1440, 29834+ | 20 | 24177+ | |
| 4 | 530, 25660+ | 21 | 22854+ | |
| 5 | 21762+ | 22 | 17844+ | |
| 6 | 14235+ | 23 | 22637+ | |
| 7 | 1388, 18228+ | 24 | 375, 19607+ | |
| 8 | 21401+ | 25 | 19403+ | |
| 9 | 21876+ | 26 | 20997+ | |
| 10 | 5094, 18228+ | 27 | 19175+ | |
| 11 | 21691+ | 28 | 20425+ | |
| 12 | 20890+ | 29 | 22149+ | |
| 13 | 22486+ | 30 | 21144+ | |
| 14 | 19321+ | 31 | 21237+ | |
| 15 | 21585+ | 32 | 14281+ | |
| 16 | 18676+ | 33 | 8250, 21974+ | |
| 17 | 23520+ | 34 | 19250, 21888+ | |
Solution to Example 6
- 1) The estimated Power Law parameters are shown in Figure Repair3.
- 2) The expected number of failures at 36,000 miles can be estimated using the QCP as shown in Figure Repair4. The model predicts that 0.3559 failures per system will occur by 36,000 miles. This means that for a fleet of 35,000 vehicles, the expected warranty claims are 0.3559 * 35,000 = 12,456.
[math]\displaystyle{ }[/math]
[math]\displaystyle{ }[/math]
Example 7 (repairable system data)
Field data have been collected for a system that begins its wearout phase at time zero. The start time for each system is equal to zero and the end time for each system is 10,000 miles. Each system is scheduled to undergo an overhaul after a certain number of miles. It has been determined that the cost of an overhaul is four times more expensive than a repair. Table 13.11 presents the data. Do the following:
- 1) Estimate the parameters of the Power Law model.
- 2) Determine the optimum overhaul interval.
- 3) If [math]\displaystyle{ \beta \lt 1 }[/math] , would it be cost-effective to implement an overhaul policy?
| Table 13.11 - Field data | ||
| System 1 | System 2 | System 3 |
|---|---|---|
| 1006.3 | 722.7 | 619.1 |
| 2261.2 | 1950.9 | 1519.1 |
| 2367 | 3259.6 | 2956.6 |
| 2615.5 | 4733.9 | 3114.8 |
| 2848.1 | 5105.1 | 3657.9 |
| 4073 | 5624.1 | 4268.9 |
| 5708.1 | 5806.3 | 6690.2 |
| 6464.1 | 5855.6 | 6803.1 |
| 6519.7 | 6325.2 | 7323.9 |
| 6799.1 | 6999.4 | 7501.4 |
| 7342.9 | 7084.4 | 7641.2 |
| 7736 | 7105.9 | 7851.6 |
| 8246.1 | 7290.9 | 8147.6 |
| 7614.2 | 8221.9 | |
| 8332.1 | 9560.5 | |
| 8368.5 | 9575.4 | |
| 8947.9 | ||
| 9012.3 | ||
| 9135.9 | ||
| 9147.5 | ||
| 9601 | ||
Solution to Example 7
- 1) Figure Repair5 shows the estimated Power Law parameters.
- 2) The QCP can be used to calculate the optimum overhaul interval as shown in Figure Repair6.
- 3) Since [math]\displaystyle{ \beta \lt 1 }[/math] then the systems are not wearing out and it would not be cost-effective to implement an overhaul policy. An overhaul policy makes sense only if the systems are wearing out. Otherwise, an overhauled unit would have the same probability of failing as a unit that was not overhauled.
[math]\displaystyle{ }[/math]
Example 8 (repairable system data)
Failures and fixes of two repairable systems in the field are recorded. Both systems start from time 0. System 1 ends at time = 504 and system 2 ends at time = 541. All the BD modes are fixed at the end of the test. A fixed effectiveness factor equal to 0.6 is used. Answer the following questions:
- 1) Estimate the parameters of the Crow Extended model.
- 2) Calculate the projected MTBF after the delayed fixes.
- 3) What is the expected number of failures at time 1,000, if no fixes were performed for the future failures?
Solution to Example 8
- 1) Figure CrowExtendedRepair shows the estimated Crow Extended parameters.
- 2) Figure CrowExtendedMTBF shows the projected MTBF at time = 541 (i.e. the age of the oldest system).
- 3) Figure CrowExtendedNumofFailure shows the expected number of failures at time = 1,000.
[math]\displaystyle{ }[/math]
The previous chapters presented analysis methods for data obtained during developmental testing. However, data from systems in the field can also be analyzed in RGA. This type of data is called fielded systems data and is analogous to warranty data. Fielded systems can be categorized into two basic types: one-time or nonrepairable systems and reusable or repairable systems. In the latter case, under continuous operation, the system is repaired, but not replaced after each failure. For example, if a water pump in a vehicle fails, the water pump is replaced and the vehicle is repaired. Two types of analysis are presented in this chapter. The first is repairable systems analysis where the reliability of a system can be tracked and quantified based on data from multiple systems in the field. The second is fleet analysis where data from multiple systems in the field can be collected and analyzed so that reliability metrics for the fleet as a whole can be quantified.
Background
Most complex systems, such as automobiles, communication systems, aircraft, printers, medical diagnostics systems, helicopters, etc., are repaired and not replaced when they fail. When these systems are fielded or subjected to a customer use environment, it is often of considerable interest to determine the reliability and other performance characteristics under these conditions. Areas of interest may include assessing the expected number of failures during the warranty period, maintaining a minimum mission reliability, evaluating the rate of wearout, determining when to replace or overhaul a system and minimizing life cycle costs. In general, a lifetime distribution, such as the Weibull distribution, cannot be used to address these issues. In order to address the reliability characteristics of complex repairable systems, a process is often used instead of a distribution. The most popular process model is the Power Law model. This model is popular for several reasons. One is that it has a very practical foundation in terms of minimal repair. This is the situation when the repair of a failed system is just enough to get the system operational again. Second, if the time to first failure follows the Weibull distribution, then each succeeding failure is governed by the Power Law model in the case of minimal repair. From this point of view, the Power Law model is an extension of the Weibull distribution.
Sometimes, the Crow Extended model, which was introduced in a previous chapter for analyzing developmental data, is also applied for fielded repairable systems. Applying the Crow Extended model on repairable system data allows analysts to project the system MTBF after reliability-related issues are addressed during the field operation. Projections are calculated based on the mode classifications (A, BC and BD). The calculation procedure is the same as the one for the developmental data.and is not repeated in this chapter.
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Parameter Estimation
Suppose that the number of systems under study is [math]\displaystyle{ K }[/math] and the [math]\displaystyle{ {{q}^{th}} }[/math] system is observed continuously from time [math]\displaystyle{ {{S}_{q}} }[/math] to time [math]\displaystyle{ {{T}_{q}} }[/math] , [math]\displaystyle{ q=1,2,\ldots ,K }[/math] . During the period [math]\displaystyle{ [{{S}_{q}},{{T}_{q}}] }[/math] , let [math]\displaystyle{ {{N}_{q}} }[/math] be the number of failures experienced by the [math]\displaystyle{ {{q}^{th}} }[/math] system and let [math]\displaystyle{ {{X}_{i,q}} }[/math] be the age of this system at the [math]\displaystyle{ {{i}^{th}} }[/math] occurrence of failure, [math]\displaystyle{ i=1,2,\ldots ,{{N}_{q}} }[/math] . It is also possible that the times [math]\displaystyle{ {{S}_{q}} }[/math] and [math]\displaystyle{ {{T}_{q}} }[/math] may be observed failure times for the [math]\displaystyle{ {{q}^{th}} }[/math] system. If [math]\displaystyle{ {{X}_{{{N}_{q}},q}}={{T}_{q}} }[/math] then the data on the [math]\displaystyle{ {{q}^{th}} }[/math] system is said to be failure terminated and [math]\displaystyle{ {{T}_{q}} }[/math] is a random variable with [math]\displaystyle{ {{N}_{q}} }[/math] fixed. If [math]\displaystyle{ {{X}_{{{N}_{q}},q}}\lt {{T}_{q}} }[/math] then the data on the [math]\displaystyle{ {{q}^{th}} }[/math] system is said to be time terminated with [math]\displaystyle{ {{N}_{q}} }[/math] a random variable. The maximum likelihood estimates of [math]\displaystyle{ \lambda }[/math] and [math]\displaystyle{ \beta }[/math] are values satisfying the Eqns. (lambdaPowerLaw) and (BetaPowerLaw).
- [math]\displaystyle{ \begin{align} & \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\left( T_{q}^{\widehat{\beta }}-S_{q}^{\widehat{\beta }} \right)} \\ & \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\widehat{\lambda }\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\left[ T_{q}^{\widehat{\beta }}\ln ({{T}_{q}})-S_{q}^{\widehat{\beta }}\ln ({{S}_{q}}) \right]-\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln ({{X}_{i,q}})} \end{align} }[/math]
where [math]\displaystyle{ 0\ln 0 }[/math] is defined to be 0. In general, these equations cannot be solved explicitly for [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta }, }[/math] but must be solved by iterative procedures. Once [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta } }[/math] have been estimated, the maximum likelihood estimate of the intensity function is given by:
- [math]\displaystyle{ \widehat{u}(t)=\widehat{\lambda }\widehat{\beta }{{t}^{\widehat{\beta }-1}} }[/math]
If [math]\displaystyle{ {{S}_{1}}={{S}_{2}}=\ldots ={{S}_{q}}=0 }[/math] and [math]\displaystyle{ {{T}_{1}}={{T}_{2}}=\ldots ={{T}_{q}} }[/math] [math]\displaystyle{ \,(q=1,2,\ldots ,K) }[/math] then the maximum likelihood estimates [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta } }[/math] are in closed form.
- [math]\displaystyle{ \begin{align} & \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{K{{T}^{\beta }}} \\ & \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln (\tfrac{T}{{{X}_{iq}}})} \end{align} }[/math]
The following examples illustrate these estimation procedures.
Example 1
For the data in Table 13.1, the starting time for each system is equal to [math]\displaystyle{ 0 }[/math] and the ending time for each system is 2000 hours. Calculate the maximum likelihood estimates [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta } }[/math] .
| Table 13.1 - Repairable system failure data | ||
| System 1 ( [math]\displaystyle{ {{X}_{i1}} }[/math] ) | System 2 ( [math]\displaystyle{ {{X}_{i2}} }[/math] ) | System 3 ( [math]\displaystyle{ {{X}_{i3}} }[/math] ) |
|---|---|---|
| 1.2 | 1.4 | 0.3 |
| 55.6 | 35.0 | 32.6 |
| 72.7 | 46.8 | 33.4 |
| 111.9 | 65.9 | 241.7 |
| 121.9 | 181.1 | 396.2 |
| 303.6 | 712.6 | 444.4 |
| 326.9 | 1005.7 | 480.8 |
| 1568.4 | 1029.9 | 588.9 |
| 1913.5 | 1675.7 | 1043.9 |
| 1787.5 | 1136.1 | |
| 1867.0 | 1288.1 | |
| 1408.1 | ||
| 1439.4 | ||
| 1604.8 | ||
| [math]\displaystyle{ {{N}_{1}}=9 }[/math] | [math]\displaystyle{ {{N}_{2}}=11 }[/math] | [math]\displaystyle{ {{N}_{3}}=14 }[/math] |
Solution
Since the starting time for each system is equal to zero and each system has an equivalent ending time, the general Eqns. (lambdaPowerLaw) and (BetaPowerLaw) reduce to the closed form Eqns. (sample1) and (sample2). The maximum likelihood estimates of [math]\displaystyle{ \hat{\beta } }[/math] and [math]\displaystyle{ \hat{\lambda } }[/math] are then calculated as follows:
- [math]\displaystyle{ \begin{align} & \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln (\tfrac{T}{{{X}_{iq}}})} \\ & = & 0.45300 \end{align} }[/math]
- [math]\displaystyle{ \begin{align} & \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{K{{T}^{\beta }}} \\ & = & 0.36224 \end{align} }[/math]
The system failure intensity function is then estimated by:
- [math]\displaystyle{ \widehat{u}(t)=\widehat{\lambda }\widehat{\beta }{{t}^{\widehat{\beta }-1}},\text{ }t\gt 0 }[/math]
Figure wpp intensity is a plot of [math]\displaystyle{ \widehat{u}(t) }[/math] over the period (0, 3000). Clearly, the estimated failure intensity function is most representative over the range of the data and any extrapolation should be viewed with the usual caution.
Goodness-of-Fit Tests for Repairable System Analysis
It is generally desirable to test the compatibility of a model and data by a statistical goodness-of-fit test. A parametric Cramér-von Mises goodness-of-fit test is used for the multiple system and repairable system Power Law model, as proposed by Crow in [17]. This goodness-of-fit test is appropriate whenever the start time for each system is 0 and the failure data is complete over the continuous interval [math]\displaystyle{ [0,{{T}_{q}}] }[/math] with no gaps in the data. The Chi-Squared test is a goodness-of-fit test that can be applied under more general circumstances. In addition, the Common Beta Hypothesis test also can be used to compare the intensity functions of the individual systems by comparing the [math]\displaystyle{ {{\beta }_{q}} }[/math] values of each system. Lastly, the Laplace Trend test checks for trends within the data. Due to their general applicatoin, the Common Beta Hypothesis test and the Laplace Trend test are both presented in Appendix B. The Cramér-von Mises and Chi-Squared goodness-of-fit tests are illustrated next.
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Economical Life Model
One consideration in reducing the cost to maintain repairable systems is to establish an overhaul policy that will minimize the total life cost of the system. However, an overhaul policy makes sense only if [math]\displaystyle{ \beta \gt 1 }[/math] . It does not make sense to implement an overhaul policy if [math]\displaystyle{ \beta \lt 1 }[/math] since wearout is not present. If you assume that there is a point at which it is cheaper to overhaul a system than to continue repairs, what is the overhaul time that will minimize the total life cycle cost while considering repair cost and the cost of overhaul?
Denote [math]\displaystyle{ {{C}_{1}} }[/math] as the average repair cost (unscheduled), [math]\displaystyle{ {{C}_{2}} }[/math] as the replacement or overhaul cost and [math]\displaystyle{ {{C}_{3}} }[/math] as the average cost of scheduled maintenance. Scheduled maintenance is performed for every [math]\displaystyle{ S }[/math] miles or time interval. In addition, let [math]\displaystyle{ {{N}_{1}} }[/math] be the number of failures in [math]\displaystyle{ [0,t] }[/math] and let [math]\displaystyle{ {{N}_{2}} }[/math] be the number of replacements in [math]\displaystyle{ [0,t] }[/math] . Suppose that replacement or overhaul occurs at times [math]\displaystyle{ T }[/math] , [math]\displaystyle{ 2T }[/math] , [math]\displaystyle{ 3T }[/math] . The problem is to select the optimum overhaul time [math]\displaystyle{ T={{T}_{0}} }[/math] so as to minimize the long term average system cost (unscheduled maintenance, replacement cost and scheduled maintenance). Since [math]\displaystyle{ \beta \gt 1 }[/math] , the average system cost is minimized when the system is overhauled (or replaced) at time [math]\displaystyle{ {{T}_{0}} }[/math] such that the instantaneous maintenance cost equals the average system cost.
The total system cost between overhaul or replacement is:
- [math]\displaystyle{ TSC(T)={{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\frac{T}{S} }[/math]
So the average system cost is:
- [math]\displaystyle{ C(T)=\frac{{{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\tfrac{T}{S}}{T} }[/math]
The instantaneous maintenance cost at time [math]\displaystyle{ T }[/math] is equal to:
- [math]\displaystyle{ IMC(T)={{C}_{1}}\lambda \beta {{T}^{\beta -1}}+\frac{{{C}_{3}}}{S} }[/math]
The following equation holds at optimum overhaul time [math]\displaystyle{ {{T}_{0}} }[/math] :
- [math]\displaystyle{ \begin{align} & {{C}_{1}}\lambda \beta T_{0}^{\beta -1}+\frac{{{C}_{3}}}{S}= & \frac{{{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\tfrac{T}{S}}{T} \\ & = & \frac{{{C}_{1}}\lambda T_{0}^{\beta }+{{C}_{2}}+{{C}_{3}}\tfrac{{{T}_{0}}}{S}}{{{T}_{0}}} \end{align} }[/math]
- Therefore:
- [math]\displaystyle{ {{T}_{0}}={{\left[ \frac{{{C}_{2}}}{\lambda (\beta -1){{C}_{1}}} \right]}^{1/\beta }} }[/math]
When there is no scheduled maintenance, Eqn. (ecolm) becomes:
- [math]\displaystyle{ {{C}_{1}}\lambda \beta T_{0}^{\beta -1}=\frac{{{C}_{1}}\lambda T_{0}^{\beta }+{{C}_{2}}}{{{T}_{0}}} }[/math]
The optimum overhaul time, [math]\displaystyle{ {{T}_{0}} }[/math] , is the same as Eqn. (optimt), so for periodic maintenance scheduled every [math]\displaystyle{ S }[/math] miles, the replacement or overhaul time is the same as for the unscheduled and replacement or overhaul cost model.
Examples
Example 6 (repairable system data)
This case study is based on the data given in the article Graphical Analysis of Repair Data by Dr. Wayne Nelson [23]. The data in Table 13.10 represents repair data on an automatic transmission from a sample of 34 cars. For each car, the data set shows mileage at the time of each transmission repair, along with the latest mileage. The + indicates the latest mileage observed without failure. Car 1, for example, had a repair at 7068 miles and was observed until 26,744 miles. Do the following:
- 1) Estimate the parameters of the Power Law model.
- 2) Estimate the number of warranty claims for a 36,000 mile warranty policy for an estimated fleet of 35,000 vehicles.
| Table 13.10 - Automatic transmission data | ||||
| Car | Mileage | Car | Mileage | |
|---|---|---|---|---|
| 1 | 7068, 26744+ | 18 | 17955+ | |
| 2 | 28, 13809+ | 19 | 19507+ | |
| 3 | 48, 1440, 29834+ | 20 | 24177+ | |
| 4 | 530, 25660+ | 21 | 22854+ | |
| 5 | 21762+ | 22 | 17844+ | |
| 6 | 14235+ | 23 | 22637+ | |
| 7 | 1388, 18228+ | 24 | 375, 19607+ | |
| 8 | 21401+ | 25 | 19403+ | |
| 9 | 21876+ | 26 | 20997+ | |
| 10 | 5094, 18228+ | 27 | 19175+ | |
| 11 | 21691+ | 28 | 20425+ | |
| 12 | 20890+ | 29 | 22149+ | |
| 13 | 22486+ | 30 | 21144+ | |
| 14 | 19321+ | 31 | 21237+ | |
| 15 | 21585+ | 32 | 14281+ | |
| 16 | 18676+ | 33 | 8250, 21974+ | |
| 17 | 23520+ | 34 | 19250, 21888+ | |
Solution to Example 6
- 1) The estimated Power Law parameters are shown in Figure Repair3.
- 2) The expected number of failures at 36,000 miles can be estimated using the QCP as shown in Figure Repair4. The model predicts that 0.3559 failures per system will occur by 36,000 miles. This means that for a fleet of 35,000 vehicles, the expected warranty claims are 0.3559 * 35,000 = 12,456.
[math]\displaystyle{ }[/math]
[math]\displaystyle{ }[/math]
Example 7 (repairable system data)
Field data have been collected for a system that begins its wearout phase at time zero. The start time for each system is equal to zero and the end time for each system is 10,000 miles. Each system is scheduled to undergo an overhaul after a certain number of miles. It has been determined that the cost of an overhaul is four times more expensive than a repair. Table 13.11 presents the data. Do the following:
- 1) Estimate the parameters of the Power Law model.
- 2) Determine the optimum overhaul interval.
- 3) If [math]\displaystyle{ \beta \lt 1 }[/math] , would it be cost-effective to implement an overhaul policy?
| Table 13.11 - Field data | ||
| System 1 | System 2 | System 3 |
|---|---|---|
| 1006.3 | 722.7 | 619.1 |
| 2261.2 | 1950.9 | 1519.1 |
| 2367 | 3259.6 | 2956.6 |
| 2615.5 | 4733.9 | 3114.8 |
| 2848.1 | 5105.1 | 3657.9 |
| 4073 | 5624.1 | 4268.9 |
| 5708.1 | 5806.3 | 6690.2 |
| 6464.1 | 5855.6 | 6803.1 |
| 6519.7 | 6325.2 | 7323.9 |
| 6799.1 | 6999.4 | 7501.4 |
| 7342.9 | 7084.4 | 7641.2 |
| 7736 | 7105.9 | 7851.6 |
| 8246.1 | 7290.9 | 8147.6 |
| 7614.2 | 8221.9 | |
| 8332.1 | 9560.5 | |
| 8368.5 | 9575.4 | |
| 8947.9 | ||
| 9012.3 | ||
| 9135.9 | ||
| 9147.5 | ||
| 9601 | ||
Solution to Example 7
- 1) Figure Repair5 shows the estimated Power Law parameters.
- 2) The QCP can be used to calculate the optimum overhaul interval as shown in Figure Repair6.
- 3) Since [math]\displaystyle{ \beta \lt 1 }[/math] then the systems are not wearing out and it would not be cost-effective to implement an overhaul policy. An overhaul policy makes sense only if the systems are wearing out. Otherwise, an overhauled unit would have the same probability of failing as a unit that was not overhauled.
[math]\displaystyle{ }[/math]
Example 8 (repairable system data)
Failures and fixes of two repairable systems in the field are recorded. Both systems start from time 0. System 1 ends at time = 504 and system 2 ends at time = 541. All the BD modes are fixed at the end of the test. A fixed effectiveness factor equal to 0.6 is used. Answer the following questions:
- 1) Estimate the parameters of the Crow Extended model.
- 2) Calculate the projected MTBF after the delayed fixes.
- 3) What is the expected number of failures at time 1,000, if no fixes were performed for the future failures?
Solution to Example 8
- 1) Figure CrowExtendedRepair shows the estimated Crow Extended parameters.
- 2) Figure CrowExtendedMTBF shows the projected MTBF at time = 541 (i.e. the age of the oldest system).
- 3) Figure CrowExtendedNumofFailure shows the expected number of failures at time = 1,000.
[math]\displaystyle{ }[/math]
The previous chapters presented analysis methods for data obtained during developmental testing. However, data from systems in the field can also be analyzed in RGA. This type of data is called fielded systems data and is analogous to warranty data. Fielded systems can be categorized into two basic types: one-time or nonrepairable systems and reusable or repairable systems. In the latter case, under continuous operation, the system is repaired, but not replaced after each failure. For example, if a water pump in a vehicle fails, the water pump is replaced and the vehicle is repaired. Two types of analysis are presented in this chapter. The first is repairable systems analysis where the reliability of a system can be tracked and quantified based on data from multiple systems in the field. The second is fleet analysis where data from multiple systems in the field can be collected and analyzed so that reliability metrics for the fleet as a whole can be quantified.
Background
Most complex systems, such as automobiles, communication systems, aircraft, printers, medical diagnostics systems, helicopters, etc., are repaired and not replaced when they fail. When these systems are fielded or subjected to a customer use environment, it is often of considerable interest to determine the reliability and other performance characteristics under these conditions. Areas of interest may include assessing the expected number of failures during the warranty period, maintaining a minimum mission reliability, evaluating the rate of wearout, determining when to replace or overhaul a system and minimizing life cycle costs. In general, a lifetime distribution, such as the Weibull distribution, cannot be used to address these issues. In order to address the reliability characteristics of complex repairable systems, a process is often used instead of a distribution. The most popular process model is the Power Law model. This model is popular for several reasons. One is that it has a very practical foundation in terms of minimal repair. This is the situation when the repair of a failed system is just enough to get the system operational again. Second, if the time to first failure follows the Weibull distribution, then each succeeding failure is governed by the Power Law model in the case of minimal repair. From this point of view, the Power Law model is an extension of the Weibull distribution.
Sometimes, the Crow Extended model, which was introduced in a previous chapter for analyzing developmental data, is also applied for fielded repairable systems. Applying the Crow Extended model on repairable system data allows analysts to project the system MTBF after reliability-related issues are addressed during the field operation. Projections are calculated based on the mode classifications (A, BC and BD). The calculation procedure is the same as the one for the developmental data.and is not repeated in this chapter.
Template loop detected: Template:Distribution ex rsa
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Template loop detected: Template:Using power law model rsa
Parameter Estimation
Suppose that the number of systems under study is [math]\displaystyle{ K }[/math] and the [math]\displaystyle{ {{q}^{th}} }[/math] system is observed continuously from time [math]\displaystyle{ {{S}_{q}} }[/math] to time [math]\displaystyle{ {{T}_{q}} }[/math] , [math]\displaystyle{ q=1,2,\ldots ,K }[/math] . During the period [math]\displaystyle{ [{{S}_{q}},{{T}_{q}}] }[/math] , let [math]\displaystyle{ {{N}_{q}} }[/math] be the number of failures experienced by the [math]\displaystyle{ {{q}^{th}} }[/math] system and let [math]\displaystyle{ {{X}_{i,q}} }[/math] be the age of this system at the [math]\displaystyle{ {{i}^{th}} }[/math] occurrence of failure, [math]\displaystyle{ i=1,2,\ldots ,{{N}_{q}} }[/math] . It is also possible that the times [math]\displaystyle{ {{S}_{q}} }[/math] and [math]\displaystyle{ {{T}_{q}} }[/math] may be observed failure times for the [math]\displaystyle{ {{q}^{th}} }[/math] system. If [math]\displaystyle{ {{X}_{{{N}_{q}},q}}={{T}_{q}} }[/math] then the data on the [math]\displaystyle{ {{q}^{th}} }[/math] system is said to be failure terminated and [math]\displaystyle{ {{T}_{q}} }[/math] is a random variable with [math]\displaystyle{ {{N}_{q}} }[/math] fixed. If [math]\displaystyle{ {{X}_{{{N}_{q}},q}}\lt {{T}_{q}} }[/math] then the data on the [math]\displaystyle{ {{q}^{th}} }[/math] system is said to be time terminated with [math]\displaystyle{ {{N}_{q}} }[/math] a random variable. The maximum likelihood estimates of [math]\displaystyle{ \lambda }[/math] and [math]\displaystyle{ \beta }[/math] are values satisfying the Eqns. (lambdaPowerLaw) and (BetaPowerLaw).
- [math]\displaystyle{ \begin{align} & \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\left( T_{q}^{\widehat{\beta }}-S_{q}^{\widehat{\beta }} \right)} \\ & \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\widehat{\lambda }\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\left[ T_{q}^{\widehat{\beta }}\ln ({{T}_{q}})-S_{q}^{\widehat{\beta }}\ln ({{S}_{q}}) \right]-\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln ({{X}_{i,q}})} \end{align} }[/math]
where [math]\displaystyle{ 0\ln 0 }[/math] is defined to be 0. In general, these equations cannot be solved explicitly for [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta }, }[/math] but must be solved by iterative procedures. Once [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta } }[/math] have been estimated, the maximum likelihood estimate of the intensity function is given by:
- [math]\displaystyle{ \widehat{u}(t)=\widehat{\lambda }\widehat{\beta }{{t}^{\widehat{\beta }-1}} }[/math]
If [math]\displaystyle{ {{S}_{1}}={{S}_{2}}=\ldots ={{S}_{q}}=0 }[/math] and [math]\displaystyle{ {{T}_{1}}={{T}_{2}}=\ldots ={{T}_{q}} }[/math] [math]\displaystyle{ \,(q=1,2,\ldots ,K) }[/math] then the maximum likelihood estimates [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta } }[/math] are in closed form.
- [math]\displaystyle{ \begin{align} & \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{K{{T}^{\beta }}} \\ & \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln (\tfrac{T}{{{X}_{iq}}})} \end{align} }[/math]
The following examples illustrate these estimation procedures.
Example 1
For the data in Table 13.1, the starting time for each system is equal to [math]\displaystyle{ 0 }[/math] and the ending time for each system is 2000 hours. Calculate the maximum likelihood estimates [math]\displaystyle{ \widehat{\lambda } }[/math] and [math]\displaystyle{ \widehat{\beta } }[/math] .
| Table 13.1 - Repairable system failure data | ||
| System 1 ( [math]\displaystyle{ {{X}_{i1}} }[/math] ) | System 2 ( [math]\displaystyle{ {{X}_{i2}} }[/math] ) | System 3 ( [math]\displaystyle{ {{X}_{i3}} }[/math] ) |
|---|---|---|
| 1.2 | 1.4 | 0.3 |
| 55.6 | 35.0 | 32.6 |
| 72.7 | 46.8 | 33.4 |
| 111.9 | 65.9 | 241.7 |
| 121.9 | 181.1 | 396.2 |
| 303.6 | 712.6 | 444.4 |
| 326.9 | 1005.7 | 480.8 |
| 1568.4 | 1029.9 | 588.9 |
| 1913.5 | 1675.7 | 1043.9 |
| 1787.5 | 1136.1 | |
| 1867.0 | 1288.1 | |
| 1408.1 | ||
| 1439.4 | ||
| 1604.8 | ||
| [math]\displaystyle{ {{N}_{1}}=9 }[/math] | [math]\displaystyle{ {{N}_{2}}=11 }[/math] | [math]\displaystyle{ {{N}_{3}}=14 }[/math] |
Solution
Since the starting time for each system is equal to zero and each system has an equivalent ending time, the general Eqns. (lambdaPowerLaw) and (BetaPowerLaw) reduce to the closed form Eqns. (sample1) and (sample2). The maximum likelihood estimates of [math]\displaystyle{ \hat{\beta } }[/math] and [math]\displaystyle{ \hat{\lambda } }[/math] are then calculated as follows:
- [math]\displaystyle{ \begin{align} & \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln (\tfrac{T}{{{X}_{iq}}})} \\ & = & 0.45300 \end{align} }[/math]
- [math]\displaystyle{ \begin{align} & \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{K{{T}^{\beta }}} \\ & = & 0.36224 \end{align} }[/math]
The system failure intensity function is then estimated by:
- [math]\displaystyle{ \widehat{u}(t)=\widehat{\lambda }\widehat{\beta }{{t}^{\widehat{\beta }-1}},\text{ }t\gt 0 }[/math]
Figure wpp intensity is a plot of [math]\displaystyle{ \widehat{u}(t) }[/math] over the period (0, 3000). Clearly, the estimated failure intensity function is most representative over the range of the data and any extrapolation should be viewed with the usual caution.
Goodness-of-Fit Tests for Repairable System Analysis
It is generally desirable to test the compatibility of a model and data by a statistical goodness-of-fit test. A parametric Cramér-von Mises goodness-of-fit test is used for the multiple system and repairable system Power Law model, as proposed by Crow in [17]. This goodness-of-fit test is appropriate whenever the start time for each system is 0 and the failure data is complete over the continuous interval [math]\displaystyle{ [0,{{T}_{q}}] }[/math] with no gaps in the data. The Chi-Squared test is a goodness-of-fit test that can be applied under more general circumstances. In addition, the Common Beta Hypothesis test also can be used to compare the intensity functions of the individual systems by comparing the [math]\displaystyle{ {{\beta }_{q}} }[/math] values of each system. Lastly, the Laplace Trend test checks for trends within the data. Due to their general applicatoin, the Common Beta Hypothesis test and the Laplace Trend test are both presented in Appendix B. The Cramér-von Mises and Chi-Squared goodness-of-fit tests are illustrated next.
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Template loop detected: Template:Confidence bounds for rsa
Economical Life Model
One consideration in reducing the cost to maintain repairable systems is to establish an overhaul policy that will minimize the total life cost of the system. However, an overhaul policy makes sense only if [math]\displaystyle{ \beta \gt 1 }[/math] . It does not make sense to implement an overhaul policy if [math]\displaystyle{ \beta \lt 1 }[/math] since wearout is not present. If you assume that there is a point at which it is cheaper to overhaul a system than to continue repairs, what is the overhaul time that will minimize the total life cycle cost while considering repair cost and the cost of overhaul?
Denote [math]\displaystyle{ {{C}_{1}} }[/math] as the average repair cost (unscheduled), [math]\displaystyle{ {{C}_{2}} }[/math] as the replacement or overhaul cost and [math]\displaystyle{ {{C}_{3}} }[/math] as the average cost of scheduled maintenance. Scheduled maintenance is performed for every [math]\displaystyle{ S }[/math] miles or time interval. In addition, let [math]\displaystyle{ {{N}_{1}} }[/math] be the number of failures in [math]\displaystyle{ [0,t] }[/math] and let [math]\displaystyle{ {{N}_{2}} }[/math] be the number of replacements in [math]\displaystyle{ [0,t] }[/math] . Suppose that replacement or overhaul occurs at times [math]\displaystyle{ T }[/math] , [math]\displaystyle{ 2T }[/math] , [math]\displaystyle{ 3T }[/math] . The problem is to select the optimum overhaul time [math]\displaystyle{ T={{T}_{0}} }[/math] so as to minimize the long term average system cost (unscheduled maintenance, replacement cost and scheduled maintenance). Since [math]\displaystyle{ \beta \gt 1 }[/math] , the average system cost is minimized when the system is overhauled (or replaced) at time [math]\displaystyle{ {{T}_{0}} }[/math] such that the instantaneous maintenance cost equals the average system cost.
The total system cost between overhaul or replacement is:
- [math]\displaystyle{ TSC(T)={{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\frac{T}{S} }[/math]
So the average system cost is:
- [math]\displaystyle{ C(T)=\frac{{{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\tfrac{T}{S}}{T} }[/math]
The instantaneous maintenance cost at time [math]\displaystyle{ T }[/math] is equal to:
- [math]\displaystyle{ IMC(T)={{C}_{1}}\lambda \beta {{T}^{\beta -1}}+\frac{{{C}_{3}}}{S} }[/math]
The following equation holds at optimum overhaul time [math]\displaystyle{ {{T}_{0}} }[/math] :
- [math]\displaystyle{ \begin{align} & {{C}_{1}}\lambda \beta T_{0}^{\beta -1}+\frac{{{C}_{3}}}{S}= & \frac{{{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\tfrac{T}{S}}{T} \\ & = & \frac{{{C}_{1}}\lambda T_{0}^{\beta }+{{C}_{2}}+{{C}_{3}}\tfrac{{{T}_{0}}}{S}}{{{T}_{0}}} \end{align} }[/math]
- Therefore:
- [math]\displaystyle{ {{T}_{0}}={{\left[ \frac{{{C}_{2}}}{\lambda (\beta -1){{C}_{1}}} \right]}^{1/\beta }} }[/math]
When there is no scheduled maintenance, Eqn. (ecolm) becomes:
- [math]\displaystyle{ {{C}_{1}}\lambda \beta T_{0}^{\beta -1}=\frac{{{C}_{1}}\lambda T_{0}^{\beta }+{{C}_{2}}}{{{T}_{0}}} }[/math]
The optimum overhaul time, [math]\displaystyle{ {{T}_{0}} }[/math] , is the same as Eqn. (optimt), so for periodic maintenance scheduled every [math]\displaystyle{ S }[/math] miles, the replacement or overhaul time is the same as for the unscheduled and replacement or overhaul cost model.
Examples
Example 6 (repairable system data)
This case study is based on the data given in the article Graphical Analysis of Repair Data by Dr. Wayne Nelson [23]. The data in Table 13.10 represents repair data on an automatic transmission from a sample of 34 cars. For each car, the data set shows mileage at the time of each transmission repair, along with the latest mileage. The + indicates the latest mileage observed without failure. Car 1, for example, had a repair at 7068 miles and was observed until 26,744 miles. Do the following:
- 1) Estimate the parameters of the Power Law model.
- 2) Estimate the number of warranty claims for a 36,000 mile warranty policy for an estimated fleet of 35,000 vehicles.
| Table 13.10 - Automatic transmission data | ||||
| Car | Mileage | Car | Mileage | |
|---|---|---|---|---|
| 1 | 7068, 26744+ | 18 | 17955+ | |
| 2 | 28, 13809+ | 19 | 19507+ | |
| 3 | 48, 1440, 29834+ | 20 | 24177+ | |
| 4 | 530, 25660+ | 21 | 22854+ | |
| 5 | 21762+ | 22 | 17844+ | |
| 6 | 14235+ | 23 | 22637+ | |
| 7 | 1388, 18228+ | 24 | 375, 19607+ | |
| 8 | 21401+ | 25 | 19403+ | |
| 9 | 21876+ | 26 | 20997+ | |
| 10 | 5094, 18228+ | 27 | 19175+ | |
| 11 | 21691+ | 28 | 20425+ | |
| 12 | 20890+ | 29 | 22149+ | |
| 13 | 22486+ | 30 | 21144+ | |
| 14 | 19321+ | 31 | 21237+ | |
| 15 | 21585+ | 32 | 14281+ | |
| 16 | 18676+ | 33 | 8250, 21974+ | |
| 17 | 23520+ | 34 | 19250, 21888+ | |
Solution to Example 6
- 1) The estimated Power Law parameters are shown in Figure Repair3.
- 2) The expected number of failures at 36,000 miles can be estimated using the QCP as shown in Figure Repair4. The model predicts that 0.3559 failures per system will occur by 36,000 miles. This means that for a fleet of 35,000 vehicles, the expected warranty claims are 0.3559 * 35,000 = 12,456.
[math]\displaystyle{ }[/math]
[math]\displaystyle{ }[/math]
Example 7 (repairable system data)
Field data have been collected for a system that begins its wearout phase at time zero. The start time for each system is equal to zero and the end time for each system is 10,000 miles. Each system is scheduled to undergo an overhaul after a certain number of miles. It has been determined that the cost of an overhaul is four times more expensive than a repair. Table 13.11 presents the data. Do the following:
- 1) Estimate the parameters of the Power Law model.
- 2) Determine the optimum overhaul interval.
- 3) If [math]\displaystyle{ \beta \lt 1 }[/math] , would it be cost-effective to implement an overhaul policy?
| Table 13.11 - Field data | ||
| System 1 | System 2 | System 3 |
|---|---|---|
| 1006.3 | 722.7 | 619.1 |
| 2261.2 | 1950.9 | 1519.1 |
| 2367 | 3259.6 | 2956.6 |
| 2615.5 | 4733.9 | 3114.8 |
| 2848.1 | 5105.1 | 3657.9 |
| 4073 | 5624.1 | 4268.9 |
| 5708.1 | 5806.3 | 6690.2 |
| 6464.1 | 5855.6 | 6803.1 |
| 6519.7 | 6325.2 | 7323.9 |
| 6799.1 | 6999.4 | 7501.4 |
| 7342.9 | 7084.4 | 7641.2 |
| 7736 | 7105.9 | 7851.6 |
| 8246.1 | 7290.9 | 8147.6 |
| 7614.2 | 8221.9 | |
| 8332.1 | 9560.5 | |
| 8368.5 | 9575.4 | |
| 8947.9 | ||
| 9012.3 | ||
| 9135.9 | ||
| 9147.5 | ||
| 9601 | ||
Solution to Example 7
- 1) Figure Repair5 shows the estimated Power Law parameters.
- 2) The QCP can be used to calculate the optimum overhaul interval as shown in Figure Repair6.
- 3) Since [math]\displaystyle{ \beta \lt 1 }[/math] then the systems are not wearing out and it would not be cost-effective to implement an overhaul policy. An overhaul policy makes sense only if the systems are wearing out. Otherwise, an overhauled unit would have the same probability of failing as a unit that was not overhauled.
[math]\displaystyle{ }[/math]
Example 8 (repairable system data)
Failures and fixes of two repairable systems in the field are recorded. Both systems start from time 0. System 1 ends at time = 504 and system 2 ends at time = 541. All the BD modes are fixed at the end of the test. A fixed effectiveness factor equal to 0.6 is used. Answer the following questions:
- 1) Estimate the parameters of the Crow Extended model.
- 2) Calculate the projected MTBF after the delayed fixes.
- 3) What is the expected number of failures at time 1,000, if no fixes were performed for the future failures?
Solution to Example 8
- 1) Figure CrowExtendedRepair shows the estimated Crow Extended parameters.
- 2) Figure CrowExtendedMTBF shows the projected MTBF at time = 541 (i.e. the age of the oldest system).
- 3) Figure CrowExtendedNumofFailure shows the expected number of failures at time = 1,000.
[math]\displaystyle{ }[/math]
Economical Life Model
One consideration in reducing the cost to maintain repairable systems is to establish an overhaul policy that will minimize the total life cost of the system. However, an overhaul policy makes sense only if [math]\displaystyle{ \beta \gt 1 }[/math] . It does not make sense to implement an overhaul policy if [math]\displaystyle{ \beta \lt 1 }[/math] since wearout is not present. If you assume that there is a point at which it is cheaper to overhaul a system than to continue repairs, what is the overhaul time that will minimize the total life cycle cost while considering repair cost and the cost of overhaul?
Denote [math]\displaystyle{ {{C}_{1}} }[/math] as the average repair cost (unscheduled), [math]\displaystyle{ {{C}_{2}} }[/math] as the replacement or overhaul cost and [math]\displaystyle{ {{C}_{3}} }[/math] as the average cost of scheduled maintenance. Scheduled maintenance is performed for every [math]\displaystyle{ S }[/math] miles or time interval. In addition, let [math]\displaystyle{ {{N}_{1}} }[/math] be the number of failures in [math]\displaystyle{ [0,t] }[/math] and let [math]\displaystyle{ {{N}_{2}} }[/math] be the number of replacements in [math]\displaystyle{ [0,t] }[/math] . Suppose that replacement or overhaul occurs at times [math]\displaystyle{ T }[/math] , [math]\displaystyle{ 2T }[/math] , [math]\displaystyle{ 3T }[/math] . The problem is to select the optimum overhaul time [math]\displaystyle{ T={{T}_{0}} }[/math] so as to minimize the long term average system cost (unscheduled maintenance, replacement cost and scheduled maintenance). Since [math]\displaystyle{ \beta \gt 1 }[/math] , the average system cost is minimized when the system is overhauled (or replaced) at time [math]\displaystyle{ {{T}_{0}} }[/math] such that the instantaneous maintenance cost equals the average system cost.
The total system cost between overhaul or replacement is:
- [math]\displaystyle{ TSC(T)={{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\frac{T}{S} }[/math]
So the average system cost is:
- [math]\displaystyle{ C(T)=\frac{{{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\tfrac{T}{S}}{T} }[/math]
The instantaneous maintenance cost at time [math]\displaystyle{ T }[/math] is equal to:
- [math]\displaystyle{ IMC(T)={{C}_{1}}\lambda \beta {{T}^{\beta -1}}+\frac{{{C}_{3}}}{S} }[/math]
The following equation holds at optimum overhaul time [math]\displaystyle{ {{T}_{0}} }[/math] :
- [math]\displaystyle{ \begin{align} & {{C}_{1}}\lambda \beta T_{0}^{\beta -1}+\frac{{{C}_{3}}}{S}= & \frac{{{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\tfrac{T}{S}}{T} \\ & = & \frac{{{C}_{1}}\lambda T_{0}^{\beta }+{{C}_{2}}+{{C}_{3}}\tfrac{{{T}_{0}}}{S}}{{{T}_{0}}} \end{align} }[/math]
- Therefore:
- [math]\displaystyle{ {{T}_{0}}={{\left[ \frac{{{C}_{2}}}{\lambda (\beta -1){{C}_{1}}} \right]}^{1/\beta }} }[/math]
When there is no scheduled maintenance, Eqn. (ecolm) becomes:
- [math]\displaystyle{ {{C}_{1}}\lambda \beta T_{0}^{\beta -1}=\frac{{{C}_{1}}\lambda T_{0}^{\beta }+{{C}_{2}}}{{{T}_{0}}} }[/math]
The optimum overhaul time, [math]\displaystyle{ {{T}_{0}} }[/math] , is the same as Eqn. (optimt), so for periodic maintenance scheduled every [math]\displaystyle{ S }[/math] miles, the replacement or overhaul time is the same as for the unscheduled and replacement or overhaul cost model.
Examples
Example 6 (repairable system data)
This case study is based on the data given in the article Graphical Analysis of Repair Data by Dr. Wayne Nelson [23]. The data in Table 13.10 represents repair data on an automatic transmission from a sample of 34 cars. For each car, the data set shows mileage at the time of each transmission repair, along with the latest mileage. The + indicates the latest mileage observed without failure. Car 1, for example, had a repair at 7068 miles and was observed until 26,744 miles. Do the following:
- 1) Estimate the parameters of the Power Law model.
- 2) Estimate the number of warranty claims for a 36,000 mile warranty policy for an estimated fleet of 35,000 vehicles.
| Table 13.10 - Automatic transmission data | ||||
| Car | Mileage | Car | Mileage | |
|---|---|---|---|---|
| 1 | 7068, 26744+ | 18 | 17955+ | |
| 2 | 28, 13809+ | 19 | 19507+ | |
| 3 | 48, 1440, 29834+ | 20 | 24177+ | |
| 4 | 530, 25660+ | 21 | 22854+ | |
| 5 | 21762+ | 22 | 17844+ | |
| 6 | 14235+ | 23 | 22637+ | |
| 7 | 1388, 18228+ | 24 | 375, 19607+ | |
| 8 | 21401+ | 25 | 19403+ | |
| 9 | 21876+ | 26 | 20997+ | |
| 10 | 5094, 18228+ | 27 | 19175+ | |
| 11 | 21691+ | 28 | 20425+ | |
| 12 | 20890+ | 29 | 22149+ | |
| 13 | 22486+ | 30 | 21144+ | |
| 14 | 19321+ | 31 | 21237+ | |
| 15 | 21585+ | 32 | 14281+ | |
| 16 | 18676+ | 33 | 8250, 21974+ | |
| 17 | 23520+ | 34 | 19250, 21888+ | |
Solution to Example 6
- 1) The estimated Power Law parameters are shown in Figure Repair3.
- 2) The expected number of failures at 36,000 miles can be estimated using the QCP as shown in Figure Repair4. The model predicts that 0.3559 failures per system will occur by 36,000 miles. This means that for a fleet of 35,000 vehicles, the expected warranty claims are 0.3559 * 35,000 = 12,456.
[math]\displaystyle{ }[/math]
[math]\displaystyle{ }[/math]
Example 7 (repairable system data)
Field data have been collected for a system that begins its wearout phase at time zero. The start time for each system is equal to zero and the end time for each system is 10,000 miles. Each system is scheduled to undergo an overhaul after a certain number of miles. It has been determined that the cost of an overhaul is four times more expensive than a repair. Table 13.11 presents the data. Do the following:
- 1) Estimate the parameters of the Power Law model.
- 2) Determine the optimum overhaul interval.
- 3) If [math]\displaystyle{ \beta \lt 1 }[/math] , would it be cost-effective to implement an overhaul policy?
| Table 13.11 - Field data | ||
| System 1 | System 2 | System 3 |
|---|---|---|
| 1006.3 | 722.7 | 619.1 |
| 2261.2 | 1950.9 | 1519.1 |
| 2367 | 3259.6 | 2956.6 |
| 2615.5 | 4733.9 | 3114.8 |
| 2848.1 | 5105.1 | 3657.9 |
| 4073 | 5624.1 | 4268.9 |
| 5708.1 | 5806.3 | 6690.2 |
| 6464.1 | 5855.6 | 6803.1 |
| 6519.7 | 6325.2 | 7323.9 |
| 6799.1 | 6999.4 | 7501.4 |
| 7342.9 | 7084.4 | 7641.2 |
| 7736 | 7105.9 | 7851.6 |
| 8246.1 | 7290.9 | 8147.6 |
| 7614.2 | 8221.9 | |
| 8332.1 | 9560.5 | |
| 8368.5 | 9575.4 | |
| 8947.9 | ||
| 9012.3 | ||
| 9135.9 | ||
| 9147.5 | ||
| 9601 | ||
Solution to Example 7
- 1) Figure Repair5 shows the estimated Power Law parameters.
- 2) The QCP can be used to calculate the optimum overhaul interval as shown in Figure Repair6.
- 3) Since [math]\displaystyle{ \beta \lt 1 }[/math] then the systems are not wearing out and it would not be cost-effective to implement an overhaul policy. An overhaul policy makes sense only if the systems are wearing out. Otherwise, an overhauled unit would have the same probability of failing as a unit that was not overhauled.
[math]\displaystyle{ }[/math]
Example 8 (repairable system data)
Failures and fixes of two repairable systems in the field are recorded. Both systems start from time 0. System 1 ends at time = 504 and system 2 ends at time = 541. All the BD modes are fixed at the end of the test. A fixed effectiveness factor equal to 0.6 is used. Answer the following questions:
- 1) Estimate the parameters of the Crow Extended model.
- 2) Calculate the projected MTBF after the delayed fixes.
- 3) What is the expected number of failures at time 1,000, if no fixes were performed for the future failures?
Solution to Example 8
- 1) Figure CrowExtendedRepair shows the estimated Crow Extended parameters.
- 2) Figure CrowExtendedMTBF shows the projected MTBF at time = 541 (i.e. the age of the oldest system).
- 3) Figure CrowExtendedNumofFailure shows the expected number of failures at time = 1,000.
[math]\displaystyle{ }[/math]