Grouped per Configuration Data - Logistic Model - Revision history

Lisa Hacker at 21:29, 18 September 2023

2023-09-18T21:29:36Z

← 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 [~~[Logistic\|~~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 16:57, 20 June 2014

2014-06-20T16:57:21Z

← Older revision		Revision as of 16:57, 20 June 2014
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	The figure below shows the Reliability vs. Time plot.		The figure below shows the Reliability vs. Time plot.

	[[Image:rga8.5.png\|center\|~~400px\|Logistic Reliability vs. Time plot displaying the intervals.~~]]		[[Image:rga8.5.png\|center\|450px]]

David J. Groebel at 18:19, 29 May 2014

2014-05-29T18:19:32Z

← Older revision		Revision as of 18:19, 29 May 2014
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	<noinclude>{{Banner RGA Examples}}		<noinclude>{{Banner RGA Examples}}
	''This example appears in the [[Logistic\|Reliability Growth and Repairable System Analysis Reference ~~book~~]]''.		''This example appears in the [[Logistic\|Reliability Growth and Repairable System Analysis Reference]]''.
	</noinclude>		</noinclude>

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	\|10\|\| 1\|\| 8\|\| 0.9000		\|10\|\| 1\|\| 8\|\| 0.9000
	\|}		\|}


	'''Solution'''		'''Solution'''
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	The observed reliability is <math>1-\tfrac{\#\text{ of failures}}{\#\text{ of units}}\,\!</math> and the last column of the table above shows the values for each group. With <math>N=9\,\!</math>, the least square estimator <math>\overline{Y}</math> becomes:		The observed reliability is <math>1-\tfrac{\#\text{ of failures}}{\#\text{ of units}}\,\!</math> and the last column of the table above shows the values for each group. With <math>N=9\,\!</math>, the least square estimator <math>\overline{Y}</math> becomes:

	::<math>\begin{align}		:<math>\begin{align}
	\bar{Y}= & \frac{1}{9}\underset{i=0}{\overset{8}{\mathop \sum }}\,\ln \left( \frac{1}{{{R}_{i}}}-1 \right) \\		\bar{Y}= & \frac{1}{9}\underset{i=0}{\overset{8}{\mathop \sum }}\,\ln \left( \frac{1}{{{R}_{i}}}-1 \right) \\
	= & -1.4036		= & -1.4036
	\end{align}\,\!</math>		\end{align}\,\!</math>


	and:		and:

	::<math>\begin{align}		:<math>\begin{align}
	\bar{T}= & \frac{1}{9}\underset{i=0}{\overset{8}{\mathop \sum }}\,{{T}_{i}} \\		\bar{T}= & \frac{1}{9}\underset{i=0}{\overset{8}{\mathop \sum }}\,{{T}_{i}} \\
	= & 4 \\		= & 4 \\
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	\underset{i=0}{\overset{8}{\mathop \sum }}\,{{T}_{i}}{{Y}_{i}}= & -68.33		\underset{i=0}{\overset{8}{\mathop \sum }}\,{{T}_{i}}{{Y}_{i}}= & -68.33
	\end{align}\,\!</math>		\end{align}\,\!</math>


	Now from the least squares estimators, <math>\hat{b_{i}}\,\!</math> and <math>\hat{b_{0}}\,\!</math>, we have:		Now from the least squares estimators, <math>\hat{b_{i}}\,\!</math> and <math>\hat{b_{0}}\,\!</math>, we have:

	::<math>\begin{align}		:<math>\begin{align}
	{{{\hat{b}}}_{1}}= & \frac{\underset{i=0}{\overset{8}{\mathop{\sum }}}\,{{T}_{i}}{{Y}_{i}}-N\cdot \bar{T}\cdot \bar{Y}}{\underset{i=0}{\overset{8}{\mathop{\sum }}}\,T_{i}^{2}-N\cdot {{{\bar{T}}}^{2}}} \\		{{{\hat{b}}}_{1}}= & \frac{\underset{i=0}{\overset{8}{\mathop{\sum }}}\,{{T}_{i}}{{Y}_{i}}-N\cdot \bar{T}\cdot \bar{Y}}{\underset{i=0}{\overset{8}{\mathop{\sum }}}\,T_{i}^{2}-N\cdot {{{\bar{T}}}^{2}}} \\
	= & \frac{-68.33-9\cdot 4\cdot \left( -1.4036 \right)}{204-9\cdot {{4}^{2}}} \\		= & \frac{-68.33-9\cdot 4\cdot \left( -1.4036 \right)}{204-9\cdot {{4}^{2}}} \\
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	= & -0.2168		= & -0.2168
	\end{align}\,\!</math>		\end{align}\,\!</math>


	Therefore:		Therefore:

	::<math>\begin{align}		:<math>\begin{align}
	\widehat{b}= & {{e}^{-0.2168}} \\		\widehat{b}= & {{e}^{-0.2168}} \\
	= & 0.8051 \\		= & 0.8051 \\
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	= & 0.2967		= & 0.2967
	\end{align}\,\!</math>		\end{align}\,\!</math>


	The Logistic reliability model that best fits the data is given by:		The Logistic reliability model that best fits the data is given by:

	::<math>R=\frac{1}{1+0.8051\cdot \ \,{{e}^{-0.2967T}}}\,\!</math>		:<math>R=\frac{1}{1+0.8051\cdot \ \,{{e}^{-0.2967T}}}\,\!</math>


	The figure below shows the Reliability vs. Time plot.		The figure below shows the Reliability vs. Time plot.

	[[Image:rga8.5.png\|center\|400px\|Logistic Reliability vs. Time plot displaying the intervals.]]		[[Image:rga8.5.png\|center\|400px\|Logistic Reliability vs. Time plot displaying the intervals.]]

Kate Racaza: Created page with '{{Banner RGA Examples}} ''This example appears in the Reliability Growth and Repairable System Analysis Reference book''. Some equipment und…'

2014-01-21T20:52:22Z

Created page with '<noinclude>{{Banner RGA Examples}} ''This example appears in the Reliability Growth and Repairable System Analysis Reference book''. </noinclude> Some equipment und…'

New page

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

Some equipment underwent testing in different stages. The testing may have been performed in subsequent days, weeks or months with an unequal number of units tested every day. Each group was tested and several failures occurred. The data set is given in columns 1 and 2 of the following table. Find the Logistic model that best fits the data, and plot it along with the reliability observed from the raw data.

{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|+'''Grouped per Configuration Data'''
!Number of Units
!Number of Failures
!<math>T_i\,\!</math>
!Observed Reliability
|-
|10|| 5|| 0|| 0.5000
|-
|8|| 3|| 1|| 0.6250
|-
|9|| 3|| 2|| 0.6667
|-
|9|| 2|| 3|| 0.7778
|-
|10|| 2|| 4|| 0.8000
|-
|10|| 1|| 5|| 0.9000
|-
|10|| 1|| 6|| 0.9000
|-
|10|| 1|| 7|| 0.9000
|-
|10|| 1|| 8|| 0.9000
|}

'''Solution'''

The observed reliability is <math>1-\tfrac{\#\text{ of failures}}{\#\text{ of units}}\,\!</math> and the last column of the table above shows the values for each group. With <math>N=9\,\!</math>, the least square estimator <math>\overline{Y}</math> becomes:

::<math>\begin{align}
\bar{Y}= & \frac{1}{9}\underset{i=0}{\overset{8}{\mathop \sum }}\,\ln \left( \frac{1}{{{R}_{i}}}-1 \right) \\
= & -1.4036
\end{align}\,\!</math>

and:

::<math>\begin{align}
\bar{T}= & \frac{1}{9}\underset{i=0}{\overset{8}{\mathop \sum }}\,{{T}_{i}} \\
= & 4 \\
\underset{i=0}{\overset{8}{\mathop \sum }}\,T_{i}^{2}= & 204 \\
\underset{i=0}{\overset{8}{\mathop \sum }}\,{{T}_{i}}{{Y}_{i}}= & -68.33
\end{align}\,\!</math>

Now from the least squares estimators, <math>\hat{b_{i}}\,\!</math> and <math>\hat{b_{0}}\,\!</math>, we have:

::<math>\begin{align}
{{{\hat{b}}}_{1}}= & \frac{\underset{i=0}{\overset{8}{\mathop{\sum }}}\,{{T}_{i}}{{Y}_{i}}-N\cdot \bar{T}\cdot \bar{Y}}{\underset{i=0}{\overset{8}{\mathop{\sum }}}\,T_{i}^{2}-N\cdot {{{\bar{T}}}^{2}}} \\
= & \frac{-68.33-9\cdot 4\cdot \left( -1.4036 \right)}{204-9\cdot {{4}^{2}}} \\
= & -0.2967 \\
& \\
{{{\hat{b}}}_{0}}= & \bar{Y}-{{{\hat{b}}}_{1}}\bar{T} \\
= & \left( -1.4036 \right)-\left( -0.2967 \right)\cdot 4.0 \\
= & -0.2168
\end{align}\,\!</math>

Therefore:

::<math>\begin{align}
\widehat{b}= & {{e}^{-0.2168}} \\
= & 0.8051 \\
\widehat{k}= & -(-0.2967) \\
= & 0.2967
\end{align}\,\!</math>

The Logistic reliability model that best fits the data is given by:

::<math>R=\frac{1}{1+0.8051\cdot \ \,{{e}^{-0.2967T}}}\,\!</math>

The figure below shows the Reliability vs. Time plot.

[[Image:rga8.5.png|center|400px|Logistic Reliability vs. Time plot displaying the intervals.]]