Sequential Data - Logistic Model - Revision history

Lisa Hacker at 21:23, 18 September 2023

2023-09-18T21:23:47Z

← Older revision		Revision as of 21:23, 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 17:01, 20 June 2014

2014-06-20T17:01:40Z

← Older revision		Revision as of 17:01, 20 June 2014
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	The following figure shows the Reliability vs. Time plot.		The following figure shows the Reliability vs. Time plot.

	[[Image:rga8.4.png\|center\|~~400px\|Logistic Reliability vs. Time plot.~~]]		[[Image:rga8.4.png\|center\|450px]]

David J. Groebel at 18:25, 29 May 2014

2014-05-29T18:25:00Z

← Older revision		Revision as of 18:25, 29 May 2014
Line 1:		Line 1:
	<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>

Line 46:		Line 46:
	Therefore:		Therefore:

	::<math>\begin{align}		:<math>\begin{align}
	\bar{Y}= & \frac{1}{N}\underset{i=0}{\overset{N-1}{\mathop \sum }}\,{{Y}_{i}} \\		\bar{Y}= & \frac{1}{N}\underset{i=0}{\overset{N-1}{\mathop \sum }}\,{{Y}_{i}} \\
	= & \frac{1}{14}\underset{i=0}{\overset{13}{\mathop \sum }}\,\ln \left( \frac{1}{{{R}_{i}}}-1 \right) \\		= & \frac{1}{14}\underset{i=0}{\overset{13}{\mathop \sum }}\,\ln \left( \frac{1}{{{R}_{i}}}-1 \right) \\
	= & -0.43163		= & -0.43163
	\end{align}\,\!</math>		\end{align}\,\!</math>


	and:		and:

	::<math>\begin{align}		:<math>\begin{align}
	\bar{T}= & \frac{1}{14}\underset{i=0}{\overset{13}{\mathop \sum }}\,{{T}_{i}} \\		\bar{T}= & \frac{1}{14}\underset{i=0}{\overset{13}{\mathop \sum }}\,{{T}_{i}} \\
	= & 6.5 \\		= & 6.5 \\
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	\underset{i=0}{\overset{13}{\mathop \sum }}\,{{T}_{i}}{{Y}_{i}}= & -61.69		\underset{i=0}{\overset{13}{\mathop \sum }}\,{{T}_{i}}{{Y}_{i}}= & -61.69
	\end{align}\,\!</math>		\end{align}\,\!</math>


	Now, from the least squares estimators, the values are:		Now, from the least squares estimators, the values are:

	::<math>\begin{align}		:<math>\begin{align}
	{{{\hat{b}}}_{1}}&= \frac{\underset{i=0}{\overset{N-1}{\mathop{\sum }}}\,{{T}_{i}}{{Y}_{i}}-N\cdot \bar{T}\cdot \bar{Y}}{\underset{i=0}{\overset{N-1}{\mathop{\sum }}}\,T_{i}^{2}-N\cdot {{{\bar{T}}}^{2}}} \\ \\		{{{\hat{b}}}_{1}}&= \frac{\underset{i=0}{\overset{N-1}{\mathop{\sum }}}\,{{T}_{i}}{{Y}_{i}}-N\cdot \bar{T}\cdot \bar{Y}}{\underset{i=0}{\overset{N-1}{\mathop{\sum }}}\,T_{i}^{2}-N\cdot {{{\bar{T}}}^{2}}} \\ \\
	&= \frac{-61.69-14 \cdot 6.5 \cdot (-.43163)}{819.0-14 \cdot 6.5^{2}} \\		&= \frac{-61.69-14 \cdot 6.5 \cdot (-.43163)}{819.0-14 \cdot 6.5^{2}} \\
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	&= 0.2087		&= 0.2087
	\end{align}\,\!</math>		\end{align}\,\!</math>


	Therefore:		Therefore:

	::<math>\begin{align}		:<math>\begin{align}
	\widehat{b}= & {{e}^{0.2087}} \\		\widehat{b}= & {{e}^{0.2087}} \\
	= & 1.2321 \\		= & 1.2321 \\
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	= & 0.0985		= & 0.0985
	\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+1.2321\cdot \ \,{{e}^{-0.0985T}}}\,\!</math>		:<math>R=\frac{1}{1+1.2321\cdot \ \,{{e}^{-0.0985T}}}\,\!</math>


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

	[[Image:rga8.4.png\|center\|400px\|Logistic Reliability vs. Time plot.]]		[[Image:rga8.4.png\|center\|400px\|Logistic Reliability vs. Time plot.]]

Kate Racaza: Created page with '{{Banner RGA Examples}} ''This example appears in the Reliability Growth and Repairable System Analysis Reference book''. A prototype was te…'

2014-01-21T20:50:33Z

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

New page

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

A prototype was tested under a success/failure pattern. The test consisted of 15 runs. The following table presents the data from the test. Find the Logistic model that best fits the data set, and plot it along with the reliability observed from the raw data.

{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|+'''Sequential Success/Failure Data with Observed Reliability Values'''
!Time
!Result
!Observed Reliability
|-
|0|| F|| 0.5000
|-
|1|| F|| 0.3333
|-
|2|| S|| 0.5000
|-
|3|| S|| 0.6000
|-
|4|| F|| 0.5000
|-
|5|| S|| 0.5714
|-
|6|| S|| 0.6250
|-
|7|| S|| 0.6667
|-
|8|| S|| 0.7000
|-
|9|| F|| 0.6364
|-
|10|| S|| 0.6667
|-
|11|| S|| 0.6923
|-
|12|| S|| 0.7143
|-
|13|| S|| 0.7333
|}

'''Solution'''

The first run is ignored because it was a success, and the reliability at that point was 100%. This failure 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 test essentially begins at time 1, and is now considered as time 0 with <math>N=14\,\!</math>. The observed reliability is shown in the last column of the table. Keep in mind that the observed reliability values still account for the initial suspension.

Therefore:

::<math>\begin{align}
\bar{Y}= & \frac{1}{N}\underset{i=0}{\overset{N-1}{\mathop \sum }}\,{{Y}_{i}} \\
= & \frac{1}{14}\underset{i=0}{\overset{13}{\mathop \sum }}\,\ln \left( \frac{1}{{{R}_{i}}}-1 \right) \\
= & -0.43163
\end{align}\,\!</math>

and:

::<math>\begin{align}
\bar{T}= & \frac{1}{14}\underset{i=0}{\overset{13}{\mathop \sum }}\,{{T}_{i}} \\
= & 6.5 \\
\underset{i=0}{\overset{13}{\mathop \sum }}\,T_{i}^{2}= & 819.0 \\
\underset{i=0}{\overset{13}{\mathop \sum }}\,{{T}_{i}}{{Y}_{i}}= & -61.69
\end{align}\,\!</math>

Now, from the least squares estimators, the values are:

::<math>\begin{align}
{{{\hat{b}}}_{1}}&= \frac{\underset{i=0}{\overset{N-1}{\mathop{\sum }}}\,{{T}_{i}}{{Y}_{i}}-N\cdot \bar{T}\cdot \bar{Y}}{\underset{i=0}{\overset{N-1}{\mathop{\sum }}}\,T_{i}^{2}-N\cdot {{{\bar{T}}}^{2}}} \\ \\
&= \frac{-61.69-14 \cdot 6.5 \cdot (-.43163)}{819.0-14 \cdot 6.5^{2}} \\
&= -0.0985 \\ \\

\hat{b_{0}}&= \overline{Y} - \hat{b_{1}}\overline{T} \\
&= (-.043163)-(-0.0985) \cdot 6.5 \\
&= 0.2087
\end{align}\,\!</math>

Therefore:

::<math>\begin{align}
\widehat{b}= & {{e}^{0.2087}} \\
= & 1.2321 \\
\widehat{k}= & -(-0.0985) \\
= & 0.0985
\end{align}\,\!</math>

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

::<math>R=\frac{1}{1+1.2321\cdot \ \,{{e}^{-0.0985T}}}\,\!</math>

The following figure shows the Reliability vs. Time plot.

[[Image:rga8.4.png|center|400px|Logistic Reliability vs. Time plot.]]