Time-Dependent System Reliability for Components in Parallel - Revision history

Lisa Hacker at 20:57, 18 September 2023

2023-09-18T20:57:16Z

← Older revision		Revision as of 20:57, 18 September 2023
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	<noinclude>{{Banner BlockSim Examples}}		<noinclude>{{Banner BlockSim Examples}}
	''This example appears in the [~~[Time-Dependent_System_Reliability_(Analytical)#Examples\|~~System ~~Analysis Reference book]~~]''.		''This example appears in the [https://help.reliasoft.com/reference/system_analysis System analysis reference]''.

Sharon Honecker: changed R(t,T) to R(t|T)

2017-02-08T20:19:56Z

changed R(t,T) to R(t|T)

← Older revision		Revision as of 20:19, 8 February 2017
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	Lastly, the conditional reliability can be obtained using <math>R(T,t)=\frac{R(T+t)}{R(T)}\,\!</math> and <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}\,\!</math>, or:		Lastly, the conditional reliability can be obtained using <math>R(t\|T)=\frac{R(T+t)}{R(T)}\,\!</math> and <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}\,\!</math>, or:

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

Sharon Honecker at 20:17, 8 February 2017

2017-02-08T20:17:45Z

← Older revision		Revision as of 20:17, 8 February 2017
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	[[Image:BS5.7.png\|center\|650px\|<div align="center"> ''pdf'' plot for the system.</div>\|link=]]		[[Image:BS5.7.png\|center\|650px\|<div align="center"> ''pdf'' plot for the system.</div>\|link=]]

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

	::<math>\begin{align}		::<math>\begin{align}

Richard House at 20:49, 5 January 2016

2016-01-05T20:49:42Z

← Older revision		Revision as of 20:49, 5 January 2016
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	The <math>MTTF\,\!</math> of the system is obtained by integrating the system's reliability function given by <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}\,\!</math> from time zero to infinity, as given by <math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ \,\!</math>. Using BlockSim's Analytical QCP, an <math>MTTF\,\!</math> of 1007.8 hours is calculated, as shown in the figure below.		The <math>MTTF\,\!</math> of the system is obtained by integrating the system's reliability function given by <math>{{R}_{s}}(t)=2\cdot {{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}-5\cdot {{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+2\cdot {{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}\,\!</math> from time zero to infinity, as given by <math>MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \ \,\!</math>. Using BlockSim's Analytical QCP, an <math>MTTF\,\!</math> of 1007.8 hours is calculated, as shown in the figure below.

	[[Image:Fig 5.9.PNG\|center\|450px\|<div align="center"> MTTF of the system. </div>]]		[[Image:Fig 5.9.PNG\|center\|450px\|<div align="center"> MTTF of the system. </div>\|link=]]


	The warranty time can be obtained by solving <math>{{R}_{s}}(t)\,\!</math> with respect to time for a system reliability <math>{{R}_{s}}=0.9\,\!</math>. Using the Analytical QCP and selecting the '''Reliable Life''' option, a time of 372.72 hours is obtained, as shown in the following figure.		The warranty time can be obtained by solving <math>{{R}_{s}}(t)\,\!</math> with respect to time for a system reliability <math>{{R}_{s}}=0.9\,\!</math>. Using the Analytical QCP and selecting the '''Reliable Life''' option, a time of 372.72 hours is obtained, as shown in the following figure.

	[[Image:Fig 5.10.PNG\|center\|450px\|<div align="center"> Time at which ''R''=0.9 or 90% for the system.</div>]]		[[Image:Fig 5.10.PNG\|center\|450px\|<div align="center"> Time at which ''R''=0.9 or 90% for the system.</div>\|link=]]


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	This can be calculated using BlockSim's Analytical QCP, as shown below.		This can be calculated using BlockSim's Analytical QCP, as shown below.

	[[Image:Fig 5.11.PNG\|center\|450px\|<div align="center"> Conditional reliability calculation for the system.</div>]]		[[Image:Fig 5.11.PNG\|center\|450px\|<div align="center"> Conditional reliability calculation for the system.</div>\|link=]]

Richard House at 20:31, 5 January 2016

2016-01-05T20:31:30Z

← Older revision		Revision as of 20:31, 5 January 2016
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	Consider the system shown next.		Consider the system shown next.
	<!-- THE DATA SET IN THIS EXAMPLE IS ALSO USED IN ANOTHER EXAMPLE. IF YOU EDIT THE DATA SET ON THIS PAGE, YOU MUST ALSO EDIT THE DATA SET AND RESULTS IN THE PAGE: Example Using a Distribution to Approximate the CDF -->		<!-- THE DATA SET IN THIS EXAMPLE IS ALSO USED IN ANOTHER EXAMPLE. IF YOU EDIT THE DATA SET ON THIS PAGE, YOU MUST ALSO EDIT THE DATA SET AND RESULTS IN THE PAGE: Example Using a Distribution to Approximate the CDF -->
	[[Image:BS5.5.png\|center\|~~700px~~\|<div align="center"> Complex bridge system in Example 2. </div>\|link=]]		[[Image:BS5.5.png\|center\|600px\|<div align="center"> Complex bridge system in Example 2. </div>\|link=]]

	Components <math>A\,\!</math> through <math>E\,\!</math> are Weibull distributed with <math>\beta =1.2\,\!</math> and <math>\eta =1230\,\!</math> hours. The starting and ending blocks cannot fail.		Components <math>A\,\!</math> through <math>E\,\!</math> are Weibull distributed with <math>\beta =1.2\,\!</math> and <math>\eta =1230\,\!</math> hours. The starting and ending blocks cannot fail.

Richard House at 20:31, 5 January 2016

2016-01-05T20:31:06Z

← Older revision		Revision as of 20:31, 5 January 2016
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	The corresponding plot is given in the following figure.		The corresponding plot is given in the following figure.

	[[Image:BS5.6.png\|center\|~~700px~~\|<div align="center"> Reliability plot for the system. </div>\|link=]]		[[Image:BS5.6.png\|center\|650px\|<div align="center"> Reliability plot for the system. </div>\|link=]]

	In order to obtain the system's ''pdf'', the derivative of the reliability equation given above is taken with respect to time, resulting in:		In order to obtain the system's ''pdf'', the derivative of the reliability equation given above is taken with respect to time, resulting in:
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	The ''pdf'' can now be plotted for different time values, <math>t\,\!</math>, as shown in the following figure.		The ''pdf'' can now be plotted for different time values, <math>t\,\!</math>, as shown in the following figure.

	[[Image:BS5.7.png\|center\|~~700px~~\|<div align="center"> ''pdf'' plot for the system.</div>\|link=]]		[[Image:BS5.7.png\|center\|650px\|<div align="center"> ''pdf'' plot for the system.</div>\|link=]]

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

	[[Image:BS5.8.png\|center\|~~700px~~\|<div align="center"> Failure rate for the system.</div>\|link=]]		[[Image:BS5.8.png\|center\|650px\|<div align="center"> Failure rate for the system.</div>\|link=]]

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

Richard House at 23:03, 4 January 2016

2016-01-04T23:03:40Z

← Older revision		Revision as of 23:03, 4 January 2016
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	The corresponding plot is given in the following figure.		The corresponding plot is given in the following figure.

	[[Image:BS5.6.png\|center\|~~300px~~\|<div align="center"> Reliability plot for the system. </div>]]		[[Image:BS5.6.png\|center\|700px\|<div align="center"> Reliability plot for the system. </div>\|link=]]

	In order to obtain the system's ''pdf'', the derivative of the reliability equation given above is taken with respect to time, resulting in:		In order to obtain the system's ''pdf'', the derivative of the reliability equation given above is taken with respect to time, resulting in:
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	The ''pdf'' can now be plotted for different time values, <math>t\,\!</math>, as shown in the following figure.		The ''pdf'' can now be plotted for different time values, <math>t\,\!</math>, as shown in the following figure.

	[[Image:BS5.7.png\|center\|~~300px~~\|<div align="center"> ''pdf'' plot for the system.</div>]]		[[Image:BS5.7.png\|center\|700px\|<div align="center"> ''pdf'' plot for the system.</div>\|link=]]

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

	[[Image:BS5.8.png\|center\|~~300px~~\|<div align="center"> Failure rate for the system.</div>]]		[[Image:BS5.8.png\|center\|700px\|<div align="center"> Failure rate for the system.</div>\|link=]]

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

Richard House at 22:58, 4 January 2016

2016-01-04T22:58:49Z

← Older revision		Revision as of 22:58, 4 January 2016
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	Consider the system shown next.		Consider the system shown next.
	<!-- THE DATA SET IN THIS EXAMPLE IS ALSO USED IN ANOTHER EXAMPLE. IF YOU EDIT THE DATA SET ON THIS PAGE, YOU MUST ALSO EDIT THE DATA SET AND RESULTS IN THE PAGE: Example Using a Distribution to Approximate the CDF -->		<!-- THE DATA SET IN THIS EXAMPLE IS ALSO USED IN ANOTHER EXAMPLE. IF YOU EDIT THE DATA SET ON THIS PAGE, YOU MUST ALSO EDIT THE DATA SET AND RESULTS IN THE PAGE: Example Using a Distribution to Approximate the CDF -->
	[[Image:BS5.5.png\|center\|~~450px~~\|<div align="center"> Complex bridge system in Example 2. </div>]]		[[Image:BS5.5.png\|center\|700px\|<div align="center"> Complex bridge system in Example 2. </div>\|link=]]

	Components <math>A\,\!</math> through <math>E\,\!</math> are Weibull distributed with <math>\beta =1.2\,\!</math> and <math>\eta =1230\,\!</math> hours. The starting and ending blocks cannot fail.		Components <math>A\,\!</math> through <math>E\,\!</math> are Weibull distributed with <math>\beta =1.2\,\!</math> and <math>\eta =1230\,\!</math> hours. The starting and ending blocks cannot fail.

Richard House at 21:18, 17 September 2012

2012-09-17T21:18:25Z

Kate Racaza at 22:17, 20 August 2012

2012-08-20T22:17:06Z

← Older revision		Revision as of 22:17, 20 August 2012
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	Note that since the starting and ending blocks cannot fail, <math>{{R}_{Start}}=1\,\!</math> and <math>{{R}_{End}}=1,\,\!</math> equation above can be reduced to:		Note that since the starting and ending blocks cannot fail, <math>{{R}_{Start}}=1\,\!</math> and <math>{{R}_{End}}=1,\,\!</math> the equation above can be reduced to:

	::<math>\begin{align}		::<math>\begin{align}
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	[[Image:BS5.7.png\|center\|300px\|<div align="center"> ''pdf'' plot for the system.</div>]]		[[Image:BS5.7.png\|center\|300px\|<div align="center"> ''pdf'' plot for the system.</div>]]

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

	::<math>\begin{align}		::<math>\begin{align}
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	The warranty time can be obtained by solving <math>{{R}_{s}}(t)\,\!</math> with respect to time for a system reliability <math>{{R}_{s}}=0.9\,\!</math> . Using the Analytical QCP and selecting the ~~Warranty Time~~ option, a time of 372.72 hours is obtained, as shown in the following figure.		The warranty time can be obtained by solving <math>{{R}_{s}}(t)\,\!</math> with respect to time for a system reliability <math>{{R}_{s}}=0.9\,\!</math>. Using the Analytical QCP and selecting the '''Reliable Life''' option, a time of 372.72 hours is obtained, as shown in the following figure.

	[[Image:Fig 5.10.PNG\|center\|450px\|<div align="center"> Time at which ''R''=0.9 or 90% for the system.</div>]]		[[Image:Fig 5.10.PNG\|center\|450px\|<div align="center"> Time at which ''R''=0.9 or 90% for the system.</div>]]

← Older revision		Revision as of 21:18, 17 September 2012
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	[[Image:BS5.5.png\|center\|450px\|<div align="center"> Complex bridge system in Example 2. </div>]]		[[Image:BS5.5.png\|center\|450px\|<div align="center"> Complex bridge system in Example 2. </div>]]

	Components <math>A\,\!</math> through <math>E\,\!</math> are Weibull distributed with <math>\beta =1.2\,\!</math> and <math>\eta =1230\,\!</math> hours. The starting and ending blocks cannot fail.		Components <math>A\,\!</math> through <math>E\,\!</math> are Weibull distributed with <math>\beta =1.2\,\!</math> and <math>\eta =1230\,\!</math> hours. The starting and ending blocks cannot fail.

	Determine the following:		Determine the following:
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	& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\		& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
	& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}))		& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}))
	\end{align}</math>		\end{align}\,\!</math>


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

	::<math>\begin{align}		::<math>\begin{align}
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	& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\		& -{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}} \\
	& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}		& +{{R}_{D}}\cdot {{R}_{C}}\cdot {{R}_{B}}+{{R}_{A}}\cdot {{R}_{D}}+{{R}_{B}}\cdot {{R}_{E}}
	\end{align}</math>		\end{align}\,\!</math>


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

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


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


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	::<math>\begin{align}		::<math>\begin{align}
	{{R}_{s}}(t)=2R{{(t)}^{2}}+2R{{(t)}^{3}}-5R{{(t)}^{4}}+2R{{(t)}^{5}}		{{R}_{s}}(t)=2R{{(t)}^{2}}+2R{{(t)}^{3}}-5R{{(t)}^{4}}+2R{{(t)}^{5}}
	\end{align}</math>		\end{align}\,\!</math>

	Or, in terms of the failure distribution:		Or, in terms of the failure distribution:

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

	The corresponding plot is given in the following figure.		The corresponding plot is given in the following figure.
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	{{f}_{s}}(t)= & 4\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+6\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \\		{{f}_{s}}(t)= & 4\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-2{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+6\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-3{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}} \\
	& -20\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+10\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}		& -20\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-4{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}+10\cdot \frac{\beta }{\eta }{{\left( \frac{t}{\eta } \right)}^{\beta -1}}{{e}^{-5{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}
	\end{align}</math>		\end{align}\,\!</math>

	The ''pdf'' can now be plotted for different time values, <math>t\,\!</math> , as shown in the following figure.		The ''pdf'' can now be plotted for different time values, <math>t\,\!</math>, as shown in the following figure.

	[[Image:BS5.7.png\|center\|300px\|<div align="center"> ''pdf'' plot for the system.</div>]]		[[Image:BS5.7.png\|center\|300px\|<div align="center"> ''pdf'' plot for the system.</div>]]

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

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

	The corresponding plot is given below.		The corresponding plot is given below.
Line 95:		Line 95:
	[[Image:BS5.8.png\|center\|300px\|<div align="center"> Failure rate for the system.</div>]]		[[Image:BS5.8.png\|center\|300px\|<div align="center"> Failure rate for the system.</div>]]

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

	[[Image:Fig 5.9.PNG\|center\|450px\|<div align="center"> MTTF of the system. </div>]]		[[Image:Fig 5.9.PNG\|center\|450px\|<div align="center"> MTTF of the system. </div>]]


	The warranty time can be obtained by solving <math>{{R}_{s}}(t)\,\!</math> with respect to time for a system reliability <math>{{R}_{s}}=0.9\,\!</math>. Using the Analytical QCP and selecting the '''Reliable Life''' option, a time of 372.72 hours is obtained, as shown in the following figure.		The warranty time can be obtained by solving <math>{{R}_{s}}(t)\,\!</math> with respect to time for a system reliability <math>{{R}_{s}}=0.9\,\!</math>. Using the Analytical QCP and selecting the '''Reliable Life''' option, a time of 372.72 hours is obtained, as shown in the following figure.

	[[Image:Fig 5.10.PNG\|center\|450px\|<div align="center"> Time at which ''R''=0.9 or 90% for the system.</div>]]		[[Image:Fig 5.10.PNG\|center\|450px\|<div align="center"> Time at which ''R''=0.9 or 90% for the system.</div>]]


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

	::<math>\begin{align}		::<math>\begin{align}
Line 111:		Line 111:
	= & \frac{0.883825}{0.975321} \\		= & \frac{0.883825}{0.975321} \\
	= & 0.906189		= & 0.906189
	\end{align}</math>		\end{align}\,\!</math>

	This can be calculated using BlockSim's Analytical QCP, as shown below.		This can be calculated using BlockSim's Analytical QCP, as shown below.

	[[Image:Fig 5.11.PNG\|center\|450px\|<div align="center"> Conditional reliability calculation for the system.</div>]]		[[Image:Fig 5.11.PNG\|center\|450px\|<div align="center"> Conditional reliability calculation for the system.</div>]]