Lloyd-Lipow Confidence Bounds Example - Revision history

Lisa Hacker at 21:29, 18 September 2023

2023-09-18T21:29:12Z

← Older revision		Revision as of 21:29, 18 September 2023
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	<noinclude>{{Banner RGA Examples}}		<noinclude>{{Banner RGA Examples}}
	''This example appears in the [~~[Lloyd-Lipow\|~~Reliability ~~Growth and Repairable System Analysis Reference]~~]''.		''This example appears in the [https://help.reliasoft.com/reference/reliability_growth_and_repairable_system_analysis Reliability growth reference]''.
	</noinclude>		</noinclude>

David J. Groebel at 19:43, 29 May 2014

2014-05-29T19:43:23Z

← Older revision		Revision as of 19:43, 29 May 2014
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	<noinclude>{{Banner RGA Examples}}		<noinclude>{{Banner RGA Examples}}
	''This example appears in the [[Lloyd-Lipow\|Reliability Growth and Repairable System Analysis Reference ~~book~~]]''.		''This example appears in the [[Lloyd-Lipow\|Reliability Growth and Repairable System Analysis Reference]]''.
	</noinclude>		</noinclude>

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	\|}		\|}


	'''Solution'''		'''Solution'''

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	From the table, the least squares estimates can be calculated as follows:		From the table, the least squares estimates can be calculated as follows:

	::<math>\begin{align}		:<math>\begin{align}
	\underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{1}{k}= & \underset{k=1}{\overset{19}{\mathop \sum }}\,\frac{1}{k}=3.54774 \\		\underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{1}{k}= & \underset{k=1}{\overset{19}{\mathop \sum }}\,\frac{1}{k}=3.54774 \\
	\underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{1}{{{k}^{2}}}= & \underset{k=1}{\overset{19}{\mathop \sum }}\,\frac{1}{{{k}^{2}}}=1.5936 \\		\underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{1}{{{k}^{2}}}= & \underset{k=1}{\overset{19}{\mathop \sum }}\,\frac{1}{{{k}^{2}}}=1.5936 \\
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	\end{align}\,\!</math>		\end{align}\,\!</math>

			and:

	~~:and:~~		:<math>\underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{{{S}_{k}}}{k\cdot {{n}_{k}}}=\underset{k=1}{\overset{19}{\mathop \sum }}\,\frac{{{S}_{k}}}{k\cdot {{n}_{k}}}=1.3002\,\!</math>

	::<math>\underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{{{S}_{k}}}{k\cdot {{n}_{k}}}=\underset{k=1}{\overset{19}{\mathop \sum }}\,\frac{{{S}_{k}}}{k\cdot {{n}_{k}}}=1.3002\,\!</math>


	Using these estimates to obtain <math>\hat{R}_{\infty}\,\!</math> and <math>\hat{\alpha}\,\!</math> yields:		Using these estimates to obtain <math>\hat{R}_{\infty}\,\!</math> and <math>\hat{\alpha}\,\!</math> yields:

			:<math>\begin{align}
	::<math>\begin{align}
	{{{\hat{R}}}_{\infty }} = & \frac{(1.5936)(9.907)-(3.5477)(1.3002)}{(19)(1.5936)-{{(3.5477)}^{2}}} \\		{{{\hat{R}}}_{\infty }} = & \frac{(1.5936)(9.907)-(3.5477)(1.3002)}{(19)(1.5936)-{{(3.5477)}^{2}}} \\
	= & 0.6316		= & 0.6316
	\end{align}\,\!</math>		\end{align}\,\!</math>

	:and:		and:

	::<math>\begin{align}		:<math>\begin{align}
	\hat{\alpha } = & \frac{(3.5477)(9.907)-(19)(1.3002)}{(19)(1.5936)-{{(3.5477)}^{2}}} \\		\hat{\alpha } = & \frac{(3.5477)(9.907)-(19)(1.3002)}{(19)(1.5936)-{{(3.5477)}^{2}}} \\
	= & 0.5902		= & 0.5902
	\end{align}\,\!</math>		\end{align}\,\!</math>


	Therefore, the Lloyd-Lipow reliability growth model is as follows, where <math>k\,\!</math> is the number of the test stage.		Therefore, the Lloyd-Lipow reliability growth model is as follows, where <math>k\,\!</math> is the number of the test stage.

	::<math>{{R}_{k}}=0.6316-\frac{0.5902}{k}\,\!</math>		:<math>{{R}_{k}}=0.6316-\frac{0.5902}{k}\,\!</math>


	Using the data from the table:		Using the data from the table:
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	\frac{{{\partial }^{2}}\Lambda }{\partial {{R}_{\infty }}\partial \alpha } = & 149.909-6.5660=143.343		\frac{{{\partial }^{2}}\Lambda }{\partial {{R}_{\infty }}\partial \alpha } = & 149.909-6.5660=143.343
	\end{align}\,\!</math>		\end{align}\,\!</math>


	The variances can be calculated using the Fisher Matrix:		The variances can be calculated using the Fisher Matrix:

	::<math>\begin{align}		:<math>\begin{align}
	{{\left[ \begin{matrix}		{{\left[ \begin{matrix}
	217.347 & -143.343 \\		217.347 & -143.343 \\
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	\end{matrix} \right]		\end{matrix} \right]
	\end{align}\,\!</math>		\end{align}\,\!</math>


	The variance of <math>{{R}_{k}}\,\!</math> is therefore:		The variance of <math>{{R}_{k}}\,\!</math> is therefore:

	::<math>Var({{\widehat{R}}_{k}})=0.0126031+\frac{1}{{{k}^{2}}}\cdot 0.0183977-\frac{2}{k}\cdot 0.0121335\,\!</math>		:<math>Var({{\widehat{R}}_{k}})=0.0126031+\frac{1}{{{k}^{2}}}\cdot 0.0183977-\frac{2}{k}\cdot 0.0121335\,\!</math>

	The confidence bounds on reliability can now can be calculated. The associated confidence bounds on reliability at the 90% confidence level are plotted in the following figure, with the predicted reliability, <math>{{R}_{k}}\,\!</math>.		The confidence bounds on reliability can now can be calculated. The associated confidence bounds on reliability at the 90% confidence level are plotted in the following figure, with the predicted reliability, <math>{{R}_{k}}\,\!</math>.

	[[Image:rga6.3.png\|center\|~~400px~~\|Reliability vs. Time Plot with 90% confidence bounds.]]		[[Image:rga6.3.png\|center\|450px\|Reliability vs. Time Plot with 90% confidence bounds.]]

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<noinclude>{{Banner RGA Examples}}
''This example appears in the [[Lloyd-Lipow|Reliability Growth and Repairable System Analysis Reference book]]''.
</noinclude>

Consider the success/failure data given in the following table. Solve for the Lloyd-Lipow parameters using least squares analysis, and plot the Lloyd-Lipow reliability with 2-sided confidence bounds at the 90% confidence level.

{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|+'''Success/Failure Data for a Variable Number of Tests Performed in Each Test Stage'''
!Test Stage Number(<math>k\,\!</math>)
!Result
!Number of Tests(<math>n_k\,\!</math>>
!Successful Tests(<math>S_k=R_i\,\!</math>)
|-
|1|| F|| 1|| 0
|-
|2|| F|| 1|| 0
|-
|3|| F|| 1|| 0
|-
|4|| S|| 1|| 0.2500
|-
|5|| F|| 1|| 0.2000
|-
|6|| F|| 1|| 0.1667
|-
|7|| S|| 1|| 0.2857
|-
|8|| S|| 1|| 0.3750
|-
|9|| S|| 1|| 0.4444
|-
|10|| S|| 1|| 0.5000
|-
|11|| S|| 1|| 0.5455
|-
|12|| S|| 1|| 0.5833
|-
|13|| S|| 1|| 0.6154
|-
|14|| S|| 1|| 0.6429
|-
|15|| S|| 1|| 0.6667
|-
|16|| S|| 1|| 0.6875
|-
|17|| F|| 1|| 0.6471
|-
|18|| S|| 1|| 0.6667
|-
|19|| F|| 1|| 0.6316
|-
|20|| S|| 1|| 0.6500
|-
|21|| S|| 1|| 0.6667
|-
|22|| S|| 1|| 0.6818
|}

'''Solution'''

Note that the data set contains three consecutive failures at the beginning of the test. These failures will be ignored throughout the analysis because it is considered that the test starts when the reliability is not equal to zero or one. The number of data points is now reduced to 19. Also, note that the only time that the first three first failures are considered is to calculate the observed reliability in the test. For example, given this data set, the observed reliability at stage 4 is <math>1/4=0.25\,\!</math>. This is considered to be the reliability at stage 1.

From the table, the least squares estimates can be calculated as follows:

::<math>\begin{align}
\underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{1}{k}= & \underset{k=1}{\overset{19}{\mathop \sum }}\,\frac{1}{k}=3.54774 \\
\underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{1}{{{k}^{2}}}= & \underset{k=1}{\overset{19}{\mathop \sum }}\,\frac{1}{{{k}^{2}}}=1.5936 \\
\underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{{{S}_{k}}}{{{n}_{k}}}= & \underset{k=1}{\overset{19}{\mathop \sum }}\,\frac{{{S}_{k}}}{{{n}_{k}}}=9.907
\end{align}\,\!</math>

:and:

::<math>\underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{{{S}_{k}}}{k\cdot {{n}_{k}}}=\underset{k=1}{\overset{19}{\mathop \sum }}\,\frac{{{S}_{k}}}{k\cdot {{n}_{k}}}=1.3002\,\!</math>

Using these estimates to obtain <math>\hat{R}_{\infty}\,\!</math> and <math>\hat{\alpha}\,\!</math> yields:

::<math>\begin{align}
{{{\hat{R}}}_{\infty }} = & \frac{(1.5936)(9.907)-(3.5477)(1.3002)}{(19)(1.5936)-{{(3.5477)}^{2}}} \\
= & 0.6316
\end{align}\,\!</math>

:and:

::<math>\begin{align}
\hat{\alpha } = & \frac{(3.5477)(9.907)-(19)(1.3002)}{(19)(1.5936)-{{(3.5477)}^{2}}} \\
= & 0.5902
\end{align}\,\!</math>

Therefore, the Lloyd-Lipow reliability growth model is as follows, where <math>k\,\!</math> is the number of the test stage.

::<math>{{R}_{k}}=0.6316-\frac{0.5902}{k}\,\!</math>

Using the data from the table:

::<math>\begin{align}
\frac{{{\partial }^{2}}\Lambda }{\partial R_{\infty }^{2}} = & -176.847-40.500=-217.347 \\
\frac{{{\partial }^{2}}\Lambda }{\partial {{\alpha }^{2}}} = & -146.763-2.1274=-148.891 \\
\frac{{{\partial }^{2}}\Lambda }{\partial {{R}_{\infty }}\partial \alpha } = & 149.909-6.5660=143.343
\end{align}\,\!</math>

The variances can be calculated using the Fisher Matrix:

::<math>\begin{align}
{{\left[ \begin{matrix}
217.347 & -143.343 \\
-143.343 & 148.891 \\
\end{matrix} \right]}^{-1}}= & \left[ \begin{matrix}
Var({{\widehat{R}}_{\infty }}) & Cov({{\widehat{R}}_{\infty }},\widehat{\alpha }) \\
Cov({{\widehat{R}}_{\infty }},\widehat{\alpha }) & Var(\widehat{\alpha }) \\
\end{matrix} \right] \\
= & \left[ \begin{matrix}
0.0126033 & 0.0121335 \\
0.0121335 & 0.0183977 \\
\end{matrix} \right]
\end{align}\,\!</math>

The variance of <math>{{R}_{k}}\,\!</math> is therefore:

::<math>Var({{\widehat{R}}_{k}})=0.0126031+\frac{1}{{{k}^{2}}}\cdot 0.0183977-\frac{2}{k}\cdot 0.0121335\,\!</math>

The confidence bounds on reliability can now can be calculated. The associated confidence bounds on reliability at the 90% confidence level are plotted in the following figure, with the predicted reliability, <math>{{R}_{k}}\,\!</math>.

[[Image:rga6.3.png|center|400px|Reliability vs. Time Plot with 90% confidence bounds.]]