Reliability Data - Modified Gompertz Model - Revision history

Lisa Hacker at 21:31, 18 September 2023

2023-09-18T21:31:21Z

← Older revision		Revision as of 21:31, 18 September 2023
Line 1:		Line 1:
	<noinclude>{{Banner RGA Examples}}		<noinclude>{{Banner RGA Examples}}
	''This example appears in the [~~[Gompertz_Models\|~~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 18:59, 29 May 2014

2014-05-29T18:59:43Z

← Older revision		Revision as of 18:59, 29 May 2014
Line 31:		Line 31:
	\|8\|\| 99.00\|\| 92.76\|\| 99.10\|\| 99.27		\|8\|\| 99.00\|\| 92.76\|\| 99.10\|\| 99.27
	\|}		\|}


	'''Solution'''		'''Solution'''
Line 37:		Line 36:
	To determine the parameters of the modified Gompertz curve, use:		To determine the parameters of the modified Gompertz curve, use:

	::<math>\begin{align}		:<math>\begin{align}
	& {{S}_{1}}(d)= & \underset{i=0}{\overset{2}{\mathop \sum }}\,\ln ({{R}_{oi}}-d) \\		& {{S}_{1}}(d)= & \underset{i=0}{\overset{2}{\mathop \sum }}\,\ln ({{R}_{oi}}-d) \\
	& {{S}_{2}}(d)= & \underset{i=3}{\overset{5}{\mathop \sum }}\,\ln ({{R}_{oi}}-d) \\		& {{S}_{2}}(d)= & \underset{i=3}{\overset{5}{\mathop \sum }}\,\ln ({{R}_{oi}}-d) \\

David J. Groebel at 18:59, 29 May 2014

2014-05-29T18:59:31Z

← Older revision		Revision as of 18:59, 29 May 2014
Line 1:		Line 1:
	<noinclude>{{Banner RGA Examples}}		<noinclude>{{Banner RGA Examples}}
	''This example appears in the [[Gompertz_Models\|Reliability Growth and Repairable System Analysis Reference ~~book~~]]''.		''This example appears in the [[Gompertz_Models\|Reliability Growth and Repairable System Analysis Reference]]''.
	</noinclude>		</noinclude>

Line 43:		Line 43:
	\end{align}\,\!</math>		\end{align}\,\!</math>

	::<math>c(d)={{\left[ \frac{{{S}_{3}}(d)-{{S}_{2}}(d)}{{{S}_{2}}(d)-{{S}_{1}}(d)} \right]}^{\tfrac{1}{3}}}\,\!</math>		:<math>c(d)={{\left[ \frac{{{S}_{3}}(d)-{{S}_{2}}(d)}{{{S}_{2}}(d)-{{S}_{1}}(d)} \right]}^{\tfrac{1}{3}}}\,\!</math>

	::<math>a(d)={{e}^{\left[ \tfrac{1}{3}\left( {{S}_{1}}(d)+\tfrac{{{S}_{2}}(d)-{{S}_{1}}(d)}{1-{{[c(d)]}^{3}}} \right) \right]}}\,\!</math>		:<math>a(d)={{e}^{\left[ \tfrac{1}{3}\left( {{S}_{1}}(d)+\tfrac{{{S}_{2}}(d)-{{S}_{1}}(d)}{1-{{[c(d)]}^{3}}} \right) \right]}}\,\!</math>

	::<math>b(d)={{e}^{\left[ \tfrac{({{S}_{2}}(d)-{{S}_{1}}(d))(c(d)-1)}{{{\left[ 1-{{[c(d)]}^{3}} \right]}^{2}}} \right]}}\,\!</math>		:<math>b(d)={{e}^{\left[ \tfrac{({{S}_{2}}(d)-{{S}_{1}}(d))(c(d)-1)}{{{\left[ 1-{{[c(d)]}^{3}} \right]}^{2}}} \right]}}\,\!</math>

	:and:		and:

	~~::<math>{{R}_{0}}=d+a(d)\cdot b(d)\,\!</math>~~

			:<math>{{R}_{0}}=d+a(d)\cdot b(d)\,\!</math>

	for <math>{{R}_{0}}=31%\,\!</math>, the equation above may be rewritten as:		for <math>{{R}_{0}}=31%\,\!</math>, the equation above may be rewritten as:

	::<math>d-31+a(d)\cdot b(d)=0\,\!</math>		:<math>d-31+a(d)\cdot b(d)=0\,\!</math>


	The equations for parameters <math>c\,\!</math>, <math>a\,\!</math> and <math>b\,\!</math> can now be solved simultaneously. One method for solving these equations numerically is to substitute different values of <math>d\,\!</math>, which must be less than <math>{{R}_{0}}\,\!</math>, into the last equation shown above, and plot the results along the y-axis with the value of <math>d\,\!</math> along the x-axis. The value of <math>d\,\!</math> can then be read from the x-intercept. This can be repeated for greater accuracy using smaller and smaller increments of <math>d\,\!</math>. Once the desired accuracy on <math>d\,\!</math> has been achieved, the value of <math>d\,\!</math> can then be used to solve for <math>a\,\!</math>, <math>b\,\!</math> and <math>c\,\!</math>. For this case, the initial estimates of the parameters are:		The equations for parameters <math>c\,\!</math>, <math>a\,\!</math> and <math>b\,\!</math> can now be solved simultaneously. One method for solving these equations numerically is to substitute different values of <math>d\,\!</math>, which must be less than <math>{{R}_{0}}\,\!</math>, into the last equation shown above, and plot the results along the y-axis with the value of <math>d\,\!</math> along the x-axis. The value of <math>d\,\!</math> can then be read from the x-intercept. This can be repeated for greater accuracy using smaller and smaller increments of <math>d\,\!</math>. Once the desired accuracy on <math>d\,\!</math> has been achieved, the value of <math>d\,\!</math> can then be used to solve for <math>a\,\!</math>, <math>b\,\!</math> and <math>c\,\!</math>. For this case, the initial estimates of the parameters are:
Line 67:		Line 65:
	\widehat{d}= & 30.825		\widehat{d}= & 30.825
	\end{align}\,\!</math>		\end{align}\,\!</math>


	Now, since the initial values have been determined, the Gauss-Newton method can be used. Therefore, substituting <math>{{Y}_{i}}={{R}_{i}},\,\!</math> <math>g_{1}^{(0)}=69.324,\,\!</math> <math>g_{2}^{(0)}=0.002524,\,\!</math> <math>g_{3}^{(0)}=0.46012,\,\!</math> and <math>g_{4}^{(0)}=30.825\,\!</math>, <math>{{Y}^{(0)}},{{D}^{(0)}},\,\!</math> <math>{{\nu }^{(0)}}\,\!</math> become:		Now, since the initial values have been determined, the Gauss-Newton method can be used. Therefore, substituting <math>{{Y}_{i}}={{R}_{i}},\,\!</math> <math>g_{1}^{(0)}=69.324,\,\!</math> <math>g_{2}^{(0)}=0.002524,\,\!</math> <math>g_{3}^{(0)}=0.46012,\,\!</math> and <math>g_{4}^{(0)}=30.825\,\!</math>, <math>{{Y}^{(0)}},{{D}^{(0)}},\,\!</math> <math>{{\nu }^{(0)}}\,\!</math> become:

	::<math>{{Y}^{(0)}}=\left[ \begin{matrix}		:<math>{{Y}^{(0)}}=\left[ \begin{matrix}
	0.000026 \\		0.000026 \\
	0.253873 \\		0.253873 \\
Line 83:		Line 80:
	\end{matrix} \right]\,\!</math>		\end{matrix} \right]\,\!</math>

	::<math>{{D}^{(0)}}=\left[ \begin{matrix}		:<math>{{D}^{(0)}}=\left[ \begin{matrix}
	0.002524 & 69.3240 & 0.0000 & 1 \\		0.002524 & 69.3240 & 0.0000 & 1 \\
	0.063775 & 805.962 & -26.4468 & 1 \\		0.063775 & 805.962 & -26.4468 & 1 \\
Line 95:		Line 92:
	\end{matrix} \right]\,\!</math>		\end{matrix} \right]\,\!</math>

	::<math>{{\nu }^{(0)}}=\left[ \begin{matrix}		:<math>{{\nu }^{(0)}}=\left[ \begin{matrix}
	g_{1}^{(0)} \\		g_{1}^{(0)} \\
	g_{2}^{(0)} \\		g_{2}^{(0)} \\
Line 106:		Line 103:
	30.825 \\		30.825 \\
	\end{matrix} \right]\,\!</math>		\end{matrix} \right]\,\!</math>


	The estimate of the parameters <math>{{\nu }^{(0)}}\,\!</math> is given by:		The estimate of the parameters <math>{{\nu }^{(0)}}\,\!</math> is given by:

	::<math>\begin{align}		:<math>\begin{align}
	{{\widehat{\nu }}^{(0)}}= & {{\left( {{D}^{{{(0)}^{T}}}}{{D}^{(0)}} \right)}^{-1}}{{D}^{{{(0)}^{T}}}}{{Y}^{(0)}} \\		{{\widehat{\nu }}^{(0)}}= & {{\left( {{D}^{{{(0)}^{T}}}}{{D}^{(0)}} \right)}^{-1}}{{D}^{{{(0)}^{T}}}}{{Y}^{(0)}} \\
	= & \left[ \begin{matrix}		= & \left[ \begin{matrix}
Line 119:		Line 115:
	\end{matrix} \right]		\end{matrix} \right]
	\end{align}\,\!</math>		\end{align}\,\!</math>


	The revised estimated regression coefficients in matrix form are given by:		The revised estimated regression coefficients in matrix form are given by:

	::<math>\begin{align}		:<math>\begin{align}
	{{g}^{(1)}}= & {{g}^{(0)}}+{{\widehat{\nu }}^{(0)}}. \\		{{g}^{(1)}}= & {{g}^{(0)}}+{{\widehat{\nu }}^{(0)}}. \\
	= & \left[ \begin{matrix}		= & \left[ \begin{matrix}
Line 143:		Line 138:
	\end{matrix} \right]		\end{matrix} \right]
	\end{align}\,\!</math>		\end{align}\,\!</math>


	With the starting coefficients <math>{{g}^{(0)}}\,\!</math>, <math>Q\,\!</math> is:		With the starting coefficients <math>{{g}^{(0)}}\,\!</math>, <math>Q\,\!</math> is:

	::<math>\begin{align}		:<math>\begin{align}
	{{Q}^{(0)}}= & \underset{i=1}{\overset{N}{\mathop \sum }}\,{{\left( {{Y}_{i}}-f({{T}_{i}},{{g}^{(0)}}) \right)}^{2}} \\		{{Q}^{(0)}}= & \underset{i=1}{\overset{N}{\mathop \sum }}\,{{\left( {{Y}_{i}}-f({{T}_{i}},{{g}^{(0)}}) \right)}^{2}} \\
	= & 2.403672		= & 2.403672
	\end{align}\,\!</math>		\end{align}\,\!</math>


	With the coefficients at the end of the first iteration, <math>{{g}^{(1)}}\,\!</math>, <math>Q\,\!</math> is:		With the coefficients at the end of the first iteration, <math>{{g}^{(1)}}\,\!</math>, <math>Q\,\!</math> is:

	::<math>\begin{align}		:<math>\begin{align}
	{{Q}^{(1)}}= & \underset{i=1}{\overset{N}{\mathop \sum }}\,{{\left[ {{Y}_{i}}-f\left( {{T}_{i}},{{g}^{(1)}} \right) \right]}^{2}} \\		{{Q}^{(1)}}= & \underset{i=1}{\overset{N}{\mathop \sum }}\,{{\left[ {{Y}_{i}}-f\left( {{T}_{i}},{{g}^{(1)}} \right) \right]}^{2}} \\
	= & 2.073964		= & 2.073964
	\end{align}\,\!</math>		\end{align}\,\!</math>


	Therefore:		Therefore:

	::<math>\begin{align}		:<math>\begin{align}
	{{Q}^{(1)}}<{{Q}^{(0)}}		{{Q}^{(1)}}<{{Q}^{(0)}}
	\end{align}\,\!</math>		\end{align}\,\!</math>


	Hence, the Gauss-Newton method works in the right direction. The iterations are continued until the relationship of <math>{{Q}^{(s-1)}}-{{Q}^{(s)}}\simeq 0\,\!</math> has been satisfied. Using RGA, the estimators of the parameters are:		Hence, the Gauss-Newton method works in the right direction. The iterations are continued until the relationship of <math>{{Q}^{(s-1)}}-{{Q}^{(s)}}\simeq 0\,\!</math> has been satisfied. Using RGA, the estimators of the parameters are:

	::<math>\begin{align}		:<math>\begin{align}
	\widehat{a}= & 0.6904 \\		\widehat{a}= & 0.6904 \\
	\widehat{b}= & 0.0020 \\		\widehat{b}= & 0.0020 \\
Line 176:		Line 167:
	\widehat{d}= & 0.3104		\widehat{d}= & 0.3104
	\end{align}\,\!</math>		\end{align}\,\!</math>


	Therefore, the modified Gompertz model is:		Therefore, the modified Gompertz model is:

	::<math>R=0.3104+(0.6904){{(0.0020)}^{{{0.4567}^{T}}}}\,\!</math>		:<math>R=0.3104+(0.6904){{(0.0020)}^{{{0.4567}^{T}}}}\,\!</math>


	Using this equation, the predicted reliability is plotted in the following figure along with the raw data. As you can see, the modified Gompertz curve represents the data very well.		Using this equation, the predicted reliability is plotted in the following figure along with the raw data. As you can see, the modified Gompertz curve represents the data very well.

	[[Image:rga7.5.png\|center\|400px\|Modified Gompertz reliability growth curve.]]		[[Image:rga7.5.png\|center\|400px\|Modified Gompertz reliability growth curve.]]

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

2014-01-21T20:26:59Z

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

New page

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

A reliability growth data set is given in columns 1 and 2 of the following table. Find the modified Gompertz curve that represents the data and plot it comparatively with the raw data.

{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|+'''Development Time vs. Observed Reliability Data and Predicted Reliabilities'''
!Time(months)
!Raw Data Reliability(%)
!Gompertz Reliability(%)
!Logistic Reliability(%)
!Modified Gompertz Reliability(%)
|-
|0|| 31.00|| 25.17|| 22.70|| 31.18
|-
|1|| 35.50|| 38.33|| 38.10|| 35.08
|-
|2|| 49.30|| 51.35|| 56.40|| 49.92
|-
|3|| 70.10|| 62.92|| 73.00|| 69.23
|-
|4|| 83.00|| 72.47|| 85.00|| 83.72
|-
|5|| 92.20|| 79.94|| 93.20|| 92.06
|-
|6|| 96.40|| 85.59|| 96.10|| 96.29
|-
|7|| 98.60|| 89.75|| 98.10|| 98.32
|-
|8|| 99.00|| 92.76|| 99.10|| 99.27
|}

'''Solution'''

To determine the parameters of the modified Gompertz curve, use:

::<math>\begin{align}
& {{S}_{1}}(d)= & \underset{i=0}{\overset{2}{\mathop \sum }}\,\ln ({{R}_{oi}}-d) \\
& {{S}_{2}}(d)= & \underset{i=3}{\overset{5}{\mathop \sum }}\,\ln ({{R}_{oi}}-d) \\
& {{S}_{3}}(d)= & \underset{i=6}{\overset{8}{\mathop \sum }}\,\ln ({{R}_{oi}}-d)
\end{align}\,\!</math>

::<math>c(d)={{\left[ \frac{{{S}_{3}}(d)-{{S}_{2}}(d)}{{{S}_{2}}(d)-{{S}_{1}}(d)} \right]}^{\tfrac{1}{3}}}\,\!</math>

::<math>a(d)={{e}^{\left[ \tfrac{1}{3}\left( {{S}_{1}}(d)+\tfrac{{{S}_{2}}(d)-{{S}_{1}}(d)}{1-{{[c(d)]}^{3}}} \right) \right]}}\,\!</math>

::<math>b(d)={{e}^{\left[ \tfrac{({{S}_{2}}(d)-{{S}_{1}}(d))(c(d)-1)}{{{\left[ 1-{{[c(d)]}^{3}} \right]}^{2}}} \right]}}\,\!</math>

:and:

::<math>{{R}_{0}}=d+a(d)\cdot b(d)\,\!</math>

for <math>{{R}_{0}}=31%\,\!</math>, the equation above may be rewritten as:

::<math>d-31+a(d)\cdot b(d)=0\,\!</math>

The equations for parameters <math>c\,\!</math>, <math>a\,\!</math> and <math>b\,\!</math> can now be solved simultaneously. One method for solving these equations numerically is to substitute different values of <math>d\,\!</math>, which must be less than <math>{{R}_{0}}\,\!</math>, into the last equation shown above, and plot the results along the y-axis with the value of <math>d\,\!</math> along the x-axis. The value of <math>d\,\!</math> can then be read from the x-intercept. This can be repeated for greater accuracy using smaller and smaller increments of <math>d\,\!</math>. Once the desired accuracy on <math>d\,\!</math> has been achieved, the value of <math>d\,\!</math> can then be used to solve for <math>a\,\!</math>, <math>b\,\!</math> and <math>c\,\!</math>. For this case, the initial estimates of the parameters are:

::<math>\begin{align}
\widehat{a}= & 69.324 \\
\widehat{b}= & 0.002524 \\
\widehat{c}= & 0.46012 \\
\widehat{d}= & 30.825
\end{align}\,\!</math>

Now, since the initial values have been determined, the Gauss-Newton method can be used. Therefore, substituting <math>{{Y}_{i}}={{R}_{i}},\,\!</math> <math>g_{1}^{(0)}=69.324,\,\!</math> <math>g_{2}^{(0)}=0.002524,\,\!</math> <math>g_{3}^{(0)}=0.46012,\,\!</math> and <math>g_{4}^{(0)}=30.825\,\!</math>, <math>{{Y}^{(0)}},{{D}^{(0)}},\,\!</math> <math>{{\nu }^{(0)}}\,\!</math> become:

::<math>{{Y}^{(0)}}=\left[ \begin{matrix}
0.000026 \\
0.253873 \\
-1.062940 \\
0.565690 \\
-0.845260 \\
0.096737 \\
0.076450 \\
0.238155 \\
-0.320890 \\
\end{matrix} \right]\,\!</math>

::<math>{{D}^{(0)}}=\left[ \begin{matrix}
0.002524 & 69.3240 & 0.0000 & 1 \\
0.063775 & 805.962 & -26.4468 & 1 \\
0.281835 & 1638.82 & -107.552 & 1 \\
0.558383 & 1493.96 & -147.068 & 1 \\
0.764818 & 941.536 & -123.582 & 1 \\
0.883940 & 500.694 & -82.1487 & 1 \\
0.944818 & 246.246 & -48.4818 & 1 \\
0.974220 & 116.829 & -26.8352 & 1 \\
0.988055 & 54.5185 & -14.3117 & 1 \\
\end{matrix} \right]\,\!</math>

::<math>{{\nu }^{(0)}}=\left[ \begin{matrix}
g_{1}^{(0)} \\
g_{2}^{(0)} \\
g_{3}^{(0)} \\
g_{4}^{(0)} \\
\end{matrix} \right]=\left[ \begin{matrix}
69.324 \\
0.002524 \\
0.46012 \\
30.825 \\
\end{matrix} \right]\,\!</math>

The estimate of the parameters <math>{{\nu }^{(0)}}\,\!</math> is given by:

::<math>\begin{align}
{{\widehat{\nu }}^{(0)}}= & {{\left( {{D}^{{{(0)}^{T}}}}{{D}^{(0)}} \right)}^{-1}}{{D}^{{{(0)}^{T}}}}{{Y}^{(0)}} \\
= & \left[ \begin{matrix}
-0.275569 \\
-0.000549 \\
-0.003202 \\
0.209458 \\
\end{matrix} \right]
\end{align}\,\!</math>

The revised estimated regression coefficients in matrix form are given by:

::<math>\begin{align}
{{g}^{(1)}}= & {{g}^{(0)}}+{{\widehat{\nu }}^{(0)}}. \\
= & \left[ \begin{matrix}
69.324 \\
0.002524 \\
0.46012 \\
30.825 \\
\end{matrix} \right]+\left[ \begin{matrix}
-0.275569 \\
-0.000549 \\
-0.003202 \\
0.209458 \\
\end{matrix} \right] \\
= & \left[ \begin{matrix}
69.0484 \\
0.00198 \\
0.45692 \\
31.0345 \\
\end{matrix} \right]
\end{align}\,\!</math>

With the starting coefficients <math>{{g}^{(0)}}\,\!</math>, <math>Q\,\!</math> is:

::<math>\begin{align}
{{Q}^{(0)}}= & \underset{i=1}{\overset{N}{\mathop \sum }}\,{{\left( {{Y}_{i}}-f({{T}_{i}},{{g}^{(0)}}) \right)}^{2}} \\
= & 2.403672
\end{align}\,\!</math>

With the coefficients at the end of the first iteration, <math>{{g}^{(1)}}\,\!</math>, <math>Q\,\!</math> is:

::<math>\begin{align}
{{Q}^{(1)}}= & \underset{i=1}{\overset{N}{\mathop \sum }}\,{{\left[ {{Y}_{i}}-f\left( {{T}_{i}},{{g}^{(1)}} \right) \right]}^{2}} \\
= & 2.073964
\end{align}\,\!</math>

Therefore:

::<math>\begin{align}
{{Q}^{(1)}}<{{Q}^{(0)}}
\end{align}\,\!</math>

Hence, the Gauss-Newton method works in the right direction. The iterations are continued until the relationship of <math>{{Q}^{(s-1)}}-{{Q}^{(s)}}\simeq 0\,\!</math> has been satisfied. Using RGA, the estimators of the parameters are:

::<math>\begin{align}
\widehat{a}= & 0.6904 \\
\widehat{b}= & 0.0020 \\
\widehat{c}= & 0.4567 \\
\widehat{d}= & 0.3104
\end{align}\,\!</math>

Therefore, the modified Gompertz model is:

::<math>R=0.3104+(0.6904){{(0.0020)}^{{{0.4567}^{T}}}}\,\!</math>

Using this equation, the predicted reliability is plotted in the following figure along with the raw data. As you can see, the modified Gompertz curve represents the data very well.

[[Image:rga7.5.png|center|400px|Modified Gompertz reliability growth curve.]]