Template:Confidence bounds gompz

Confidence Bounds
The approximate reliability confidence bounds under the Modified Gompertz model can be obtained using nonlinear regression. Additionally, the reliability is always between $$0$$  and  $$1$$. In order to keep the endpoints of the confidence interval, the logit transformation can be used to obtain the confidence bounds on reliability.


 * $$CB=\frac{{{{\hat{R}}}_{i}}+(1-{{{\hat{R}}}_{i}}){{e}^{\pm {{z}_{\alpha }}{{{\hat{\sigma }}}_{R}}/\left[ {{{\hat{R}}}_{i}}(1-{{{\hat{R}}}_{i}}) \right]}}}$$


 * $${{\hat{\sigma }}^{2}}=\frac{SSE}{n-p}$$

where $$p$$  is the total number of groups (in this case 4) and  $$n$$  is the total number of items in each group.

Example 3

A reliability growth data set is given in Table 7.4, columns 1 and 2. Find the Modified Gompertz curve that represents the data and plot it comparatively with the raw data.

Solution To determine the parameters of the Modified Gompertz curve, use:


 * $$\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}$$


 * $$c(d)={{\left[ \frac{{{S}_{3}}(d)-{{S}_{2}}(d)}{{{S}_{2}}(d)-{{S}_{1}}(d)} \right]}^{\tfrac{1}{3}}}$$


 * $$a(d)={{e}^{\left[ \tfrac{1}{3}\left( {{S}_{1}}(d)+\tfrac{{{S}_{2}}(d)-{{S}_{1}}(d)}{1-{{[c(d)]}^{3}}} \right) \right]}}$$


 * $$b(d)={{e}^{\left[ \tfrac{({{S}_{2}}(d)-{{S}_{1}}(d))(c(d)-1)} \right]}}$$


 * and:


 * $${{R}_{0}}=d+a(d)\cdot b(d)$$

where $${{R}_{0}}=31%$$. Then, Eqn. (eq27) may be rewritten as:
 * $$d-31+a(d)\cdot b(d)=0$$

Eqns. (eq24), (eq25), (eq26) and (eq28) can now be solved simultaneously. One method for solving these equations numerically is to substitute different values of $$d$$, which must be less than  $${{R}_{0}}$$ , into Eqn. (eq28) and plot along the y-axis with the value of $$d$$  along the x-axis. The value of $$d$$  can then be read from the x-intercept. This can be repeated for greater accuracy using smaller and smaller increments of $$d$$. Once the desired accuracy on $$d$$  has been achieved, the value of  $$d$$  can then be substituted into Eqns. (eq24), (eq25) and (eq26). Now $$a$$,  $$b$$  and  $$c$$  can be calculated. For this case, the initial estimates of the parameters are:


 * $$\begin{align}

& \widehat{a}= & 69.324 \\ & \widehat{b}= & 0.002524 \\ & \widehat{c}= & 0.46012 \\ & \widehat{d}= & 30.825 \end{align}$$

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


 * $${{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]$$


 * $${{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]$$


 * $${{\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]$$ The estimate of the parameters $${{\nu }^{(0)}}$$  is given by:


 * $$\begin{align}

& {{\widehat{\nu }}^{(0)}}= & {{\left( {{D}^}{{D}^{(0)}} \right)}^{-1}}{{D}^}{{Y}^{(0)}} \\ & = & \left[ \begin{matrix} -0.275569 \\   -0.000549  \\   -0.003202  \\   0.209458  \\ \end{matrix} \right] \end{align}$$ The revised estimated regression coefficients in matrix form are given by:
 * $$\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}$$ With the starting coefficients $${{g}^{(0)}}$$,  $$Q$$  is:
 * $$\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}$$ With the coefficients at the end of the first iteration, $${{g}^{(1)}}$$,  $$Q$$  is:
 * $$\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}$$
 * Therefore:


 * $${{Q}^{(1)}}<{{Q}^{(0)}}$$

Hence, the Gauss-Newton method works in the right direction. The iterations are continued until the relationship of Eqn. (critir) has been satisfied. Using RGA, the estimators of the parameters are:
 * $$\begin{align}

& \widehat{a}= & 0.6904 \\ & \widehat{b}= & 0.0020 \\ & \widehat{c}= & 0.4567 \\ & \widehat{d}= & 0.3104 \end{align}$$ Therefore, the Modified Gompertz model is:
 * $$R=0.3104+(0.6904){{(0.0020)}^}$$

Using Eqn. (eq29), the predicted reliability is plotted in the following figure along with the raw data. It can be seen from the plot in Figure MGomp1 that the Modified Gompertz curve represents the data very well.