View source for Reliability Data - Modified Gompertz Model

<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.]]