Template:Confidence bounds camsaa

Confidence Bounds
This section presents the methods used in the RGA software to estimate the confidence bounds for the Crow-AMSAA model when applied to developmental testing data. RGA provides two methods to estimate the confidence bounds. The Fisher Matrix (FM) method, which is commonly employed in the reliability field, is based on the Fisher information matrix. The Crow Bounds (Crow) method has been developed by Dr. Crow.

Fisher Matrix Bounds
The cumulative MTBF, $${{m}_{c}}(t)$$, must be positive, thus  $$\ln {{m}_{c}}(t)$$  is treated as being normally distributed as well.


 * $$\frac{\ln {{{\hat{m}}}_{c}}(t)-\ln {{m}_{c}}(t)}{\sqrt{Var(\ln {{{\hat{m}}}_{c}}(t)})}\ \tilde{\ }\ N(0,1)$$

The approximate confidence bounds on the cumulative MTBF are then estimated from:


 * $$CB={{\hat{m}}_{c}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{c}}(t))}/{{{\hat{m}}}_{c}}(t)}}$$


 * where:


 * $${{\hat{m}}_{c}}(t)=\frac{1}{{t}^{1-\hat{\beta }}}$$


 * $$\begin{align}

& Var({{{\hat{m}}}_{c}}(t))= & {{\left( \frac{\partial {{m}_{c}}(t)}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial {{m}_{c}}(t)}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial {{m}_{c}}(t)}{\partial \beta } \right)\left( \frac{\partial {{m}_{c}}(t)}{\partial \lambda } \right)cov(\hat{\beta },\hat{\lambda })\, \end{align}$$ The variance calculation is the same as Eqn. (variance1) and:


 * $$\begin{align}

& \frac{\partial {{m}_{c}}(t)}{\partial \beta }= & -\frac{1}{{t}^{1-\hat{\beta }}}\ln t \\ & \frac{\partial {{m}_{c}}(t)}{\partial \lambda }= & -\frac{1}{{t}^{1-\hat{\beta }}} \end{align}$$

Crow Bounds
To calculate the Crow confidence bounds on cumulative MTBF, first calculate the Crow cumulative failure intensity confidence bounds:


 * $$C{{(t)}_{L}}=\frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t}$$


 * $$C{{(t)}_{U}}=\frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot t}$$


 * Then:


 * $$\begin{align}

& {{[MTB{{F}_{c}}]}_{L}}= & \frac{1}{C{{(t)}_{U}}} \\ & {{[MTB{{F}_{c}}]}_{U}}= & \frac{1}{C{{(t)}_{L}}} \end{align}$$

Fisher Matrix Bounds
The instantaneous MTBF, $${{m}_{i}}(t)$$, must be positive, thus  $$\ln {{m}_{i}}(t)$$  is treated as being normally distributed as well.


 * $$\frac{\ln {{{\hat{m}}}_{i}}(t)-\ln {{m}_{i}}(t)}{\sqrt{Var(\ln {{{\hat{m}}}_{i}}(t)})}\ \tilde{\ }\ N(0,1)$$

The approximate confidence bounds on the instantaneous MTBF are then estimated from:


 * $$CB={{\hat{m}}_{i}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{i}}(t))}/{{{\hat{m}}}_{i}}(t)}}$$


 * where:


 * $${{\hat{m}}_{i}}(t)=\frac{1}{\lambda \beta {{t}^{\beta -1}}}$$


 * $$\begin{align}

& Var({{{\hat{m}}}_{i}}(t))= & {{\left( \frac{\partial {{m}_{i}}(t)}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial {{m}_{i}}(t)}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial {{m}_{i}}(t)}{\partial \beta } \right)\left( \frac{\partial {{m}_{i}}(t)}{\partial \lambda } \right)cov(\hat{\beta },\hat{\lambda }). \end{align}$$

The variance calculation is the same as Eqn. (variance1) and:


 * $$\begin{align}

& \frac{\partial {{m}_{i}}(t)}{\partial \beta }= & -\frac{1}{\hat{\lambda }{{{\hat{\beta }}}^{2}}}{{t}^{1-\hat{\beta }}}-\frac{1}{\hat{\lambda }\hat{\beta }}{{t}^{1-\hat{\beta }}}\ln t \\ & \frac{\partial {{m}_{i}}(t)}{\partial \lambda }= & -\frac{1}{{{{\hat{\lambda }}}^{2}}\hat{\beta }}{{t}^{1-\hat{\beta }}} \end{align}$$

Crow Bounds
Failure Terminated Data Consider the following equation:


 * $$G(\mu |n)=\mathop{}_{0}^{\infty }\frac{(n-2)!}\underset{i=0}{\overset{n-1}{\mathop \sum }}\,\frac{1}{i!}{{\left( \frac{\mu }{x} \right)}^{i}}\exp (-\frac{\mu }{x})\,dx$$

Find the values $${{p}_{1}}$$  and  $${{p}_{2}}$$  by finding the solution  $$c$$  to  $$G({{n}^{2}}/c|n)=\xi $$  for  $$\xi =\tfrac{\alpha }{2}$$  and  $$\xi =1-\tfrac{\alpha }{2}$$, respectively. If using the biased parameters, $$\hat{\beta }$$  and  $$\hat{\lambda }$$, then the upper and lower confidence bounds are:


 * $$\begin{align}

& {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot {{p}_{1}} \\ & {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot {{p}_{2}} \end{align}$$

where $$MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}$$. If using the unbiased parameters, $$\bar{\beta }$$  and  $$\bar{\lambda }$$, then the upper and lower confidence bounds are:


 * $$\begin{align}

& {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot \left( \frac{N-2}{N} \right)\cdot {{p}_{1}} \\ & {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot \left( \frac{N-2}{N} \right)\cdot {{p}_{2}} \end{align}$$

where $$MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}$$. Time Terminated Data Consider the following equation where $${{I}_{1}}(.)$$  is the modified Bessel function of order one:


 * $$H(x|k)=\underset{j=1}{\overset{k}{\mathop \sum }}\,\frac{{{2}^{2j-1}}(j-1)!j!{{I}_{1}}(x)}$$

Find the values $${{\Pi }_{1}}$$  and  $${{\Pi }_{2}}$$  by finding the solution  $$x$$  to  $$H(x|k)=\tfrac{\alpha }{2}$$  and  $$H(x|k)=1-\tfrac{\alpha }{2}$$  in the cases corresponding to the lower and upper bounds, respectively. Calculate $$\Pi =\tfrac{4{{n}^{2}}}$$  for each case. If using the biased parameters, $$\hat{\beta }$$  and  $$\hat{\lambda }$$, then the upper and lower confidence bounds are:


 * $$\begin{align}

& {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot {{\Pi }_{1}} \\ & {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot {{\Pi }_{2}} \end{align}$$

where $$MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}$$. If using the unbiased parameters, $$\bar{\beta }$$  and  $$\bar{\lambda }$$, then the upper and lower confidence bounds are:


 * $$\begin{align}

& {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot \left( \frac{N-1}{N} \right)\cdot {{\Pi }_{1}} \\ & {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot \left( \frac{N-1}{N} \right)\cdot {{\Pi }_{2}} \end{align}$$

where $$MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}$$.

Fisher Matrix Bounds
The cumulative failure intensity, $${{\lambda }_{c}}(t)$$, must be positive, thus  $$\ln {{\lambda }_{c}}(t)$$  is treated as being normally distributed.


 * $$\frac{\ln {{{\hat{\lambda }}}_{c}}(t)-\ln {{\lambda }_{c}}(t)}{\sqrt{Var(\ln {{{\hat{\lambda }}}_{c}}(t)})}\ \tilde{\ }\ N(0,1)$$

The approximate confidence bounds on the cumulative failure intensity are then estimated from:


 * $$CB={{\hat{\lambda }}_{c}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{{\hat{\lambda }}}_{c}}(t))}/{{{\hat{\lambda }}}_{c}}(t)}}$$


 * where:


 * $${{\hat{\lambda }}_{c}}(t)=\hat{\lambda }{{t}^{\hat{\beta }-1}}$$


 * and:


 * $$\begin{align}

& Var({{{\hat{\lambda }}}_{c}}(t))= & {{\left( \frac{\partial {{\lambda }_{c}}(t)}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial {{\lambda }_{c}}(t)}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial {{\lambda }_{c}}(t)}{\partial \beta } \right)\left( \frac{\partial {{\lambda }_{c}}(t)}{\partial \lambda } \right)cov(\hat{\beta },\,\,\,\hat{\lambda }) \end{align}$$

The variance calculation is the same as Eqn. (variance1) and:


 * $$\begin{align}

& \frac{\partial {{\lambda }_{c}}(t)}{\partial \beta }= & \hat{\lambda }{{t}^{\hat{\beta }-1}}\ln t \\ & \frac{\partial {{\lambda }_{c}}(t)}{\partial \lambda }= & {{t}^{\hat{\beta }-1}} \end{align}$$

Crow Bounds
The Crow cumulative failure intensity confidence bounds are given as:


 * $$\begin{align}

& C{{(t)}_{L}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t} \\ & C{{(t)}_{U}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot t} \end{align}$$

Fisher Matrix Bounds
The instantaneous failure intensity, $${{\lambda }_{i}}(t)$$, must be positive, thus  $$\ln {{\lambda }_{i}}(t)$$  is treated as being normally distributed.


 * $$\frac{\ln {{{\hat{\lambda }}}_{i}}(t)-\ln {{\lambda }_{i}}(t)}{\sqrt{Var(\ln {{{\hat{\lambda }}}_{i}}(t)})}\text{ }\tilde{\ }\text{ }N(0,1)$$

The approximate confidence bounds on the instantaneous failure intensity are then estimated from:


 * $$CB={{\hat{\lambda }}_{i}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{{\hat{\lambda }}}_{i}}(t))}/{{{\hat{\lambda }}}_{i}}(t)}}$$


 * where


 * $${{\lambda }_{i}}(t)=\lambda \beta {{t}^{\beta -1}}$$


 * $$\begin{align}

& Var({{{\hat{\lambda }}}_{i}}(t))= & {{\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \beta } \right)\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \lambda } \right)cov(\hat{\beta },\,\,\,\hat{\lambda }) \end{align}$$

The variance calculation is the same as Eqn. (variance1) and:


 * $$\begin{align}

& \frac{\partial {{\lambda }_{i}}(t)}{\partial \beta }= & \hat{\lambda }{{t}^{\hat{\beta }-1}}+\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}\ln t \\ & \frac{\partial {{\lambda }_{i}}(t)}{\partial \lambda }= & \hat{\beta }{{t}^{\hat{\beta }-1}} \end{align}$$

Crow Bounds
The Crow instantaneous failure intensity confidence bounds are given as:


 * $$\begin{align}

& {{\lambda }_{i}}{{(t)}_{L}}= & \frac{1} \\ & {{\lambda }_{i}}{{(t)}_{U}}= & \frac{1} \end{align}$$

Fisher Matrix Bounds
The time, $$T$$, must be positive, thus  $$\ln T$$  is treated as being normally distributed.


 * $$\frac{\ln \hat{T}-\ln T}{\sqrt{Var(\ln \hat{T}})}\ \tilde{\ }\ N(0,1)$$

Confidence bounds on the time are given by:
 * $$CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}}$$


 * where:


 * $$\begin{align}

& Var(\hat{T})= & {{\left( \frac{\partial T}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial T}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial T}{\partial \beta } \right)\left( \frac{\partial T}{\partial \lambda } \right)cov(\hat{\beta },\,\,\,\hat{\lambda }) \end{align}$$ The variance calculation is the same as Eqn. (variance1) and:
 * $$\begin{align}

& \frac{\partial T}{\partial \beta }= & \frac{-{{\left( \tfrac{{{\lambda }_{c}}(T)}{\lambda } \right)}^{1/(\beta -1)}}\ln \left( \tfrac{{{\lambda }_{c}}(T)}{\lambda } \right)} \\ & \frac{\partial T}{\partial \lambda }= & {{\left( \frac{{{\lambda }_{c}}(T)}{\lambda } \right)}^{1/(\beta -1)}}\frac{1}{\lambda (1-\beta )} \end{align}$$

Crow Bounds

 * Step 1: Calculate:


 * $$\hat{T}={{\left( \frac{{{\lambda }_{c}}(T)}{{\hat{\lambda }}} \right)}^{\tfrac{1}{\beta -1}}}$$


 * Step 2: Estimate the number of failures:


 * $$N(\hat{T})=\hat{\lambda }{{\hat{T}}^}$$


 * Step 3: Obtain the confidence bounds on time given the cumulative failure intensity by solving for $${{t}_{l}}$$  and  $${{t}_{u}}$$  in the following equations:


 * $$\begin{align}

& {{t}_{l}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot {{\lambda }_{c}}(T)} \\ & {{t}_{u}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot {{\lambda }_{c}}(T)} \end{align}$$

Fisher Matrix Bounds
The time, $$T$$, must be positive, thus  $$\ln T$$  is treated as being normally distributed.
 * $$\frac{\ln \hat{T}-\ln T}{\sqrt{Var(\ln \hat{T}})}\ \tilde{\ }\ N(0,1)$$

Confidence bounds on the time are given by:
 * $$CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}}$$
 * where:
 * $$\begin{align}

& Var(\hat{T})= & {{\left( \frac{\partial T}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial T}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial T}{\partial \beta } \right)\left( \frac{\partial T}{\partial \lambda } \right)cov(\hat{\beta },\,\,\,\hat{\lambda }) \end{align}$$ The variance calculation is the same as Eqn. (variance1) and:
 * $$\hat{T}={{(\lambda \cdot {{m}_{c}})}^{1/(1-\beta )}}$$


 * $$\begin{align}

& \frac{\partial T}{\partial \beta }= & \frac{{{(\lambda \cdot \,{{m}_{c}})}^{1/(1-\beta )}}\ln (\lambda \cdot \text{ }{{m}_{c}})} \\ & \frac{\partial T}{\partial \lambda }= & \frac{\lambda (1-\beta )} \end{align}$$

Crow Bounds

 * Step 1: Calculate $${{\lambda }_{c}}(T)=\tfrac{1}{MTB{{F}_{c}}}$$.
 * Step 2: Use the equations from 5.2.8.2 to calculate the bounds on time given the cumulative failure intensity.

Fisher Matrix Bounds
The time, $$T$$, must be positive, thus  $$\ln T$$  is treated as being normally distributed.


 * $$\frac{\ln \hat{T}-\ln T}{\sqrt{Var(\ln \hat{T}})}\ \tilde{\ }\ N(0,1)$$

Confidence bounds on the time are given by:


 * $$CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}}$$


 * where:


 * $$\begin{align}

& Var(\hat{T})= & {{\left( \frac{\partial T}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial T}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial T}{\partial \beta } \right)\left( \frac{\partial T}{\partial \lambda } \right)cov(\hat{\beta },\,\,\,\hat{\lambda }) \end{align}$$

The variance calculation is the same as Eqn. (variance1) and:


 * $$\hat{T}={{(\lambda \beta \cdot MTB{{F}_{i}})}^{1/(1-\beta )}}$$


 * $$\begin{align}

& \frac{\partial T}{\partial \beta }= & {{\left( \lambda \beta \cdot MTB{{F}_{i}} \right)}^{1/(1-\beta )}}\left[ \frac{1}\ln (\lambda \beta \cdot MTB{{F}_{i}})+\frac{1}{\beta (1-\beta )} \right] \\ & \frac{\partial T}{\partial \lambda }= & \frac{\lambda (1-\beta )} \end{align}$$

Crow Bounds

 * Step 1: Calculate the confidence bounds on the instantaneous MTBF as presented in Section 5.5.2.
 * Step 2: Calculate the bounds on time as follows.

Failure Terminated Data

 * $$\hat{T}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{c})}^{1/(1-\beta )}}$$

So the lower an upper bounds on time are:


 * $${{\hat{T}}_{L}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{c}_{1}}})}^{1/(1-\beta )}}$$


 * $${{\hat{T}}_{U}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{c}_{2}}})}^{1/(1-\beta )}}$$

Time Terminated Data


 * $$\hat{T}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{\Pi })}^{1/(1-\beta )}}$$

So the lower and upper bounds on time are:


 * $${{\hat{T}}_{L}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{\Pi }_{1}}})}^{1/(1-\beta )}}$$


 * $${{\hat{T}}_{U}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{\Pi }_{2}}})}^{1/(1-\beta )}}$$

Fisher Matrix Bounds
The time, $$T$$, must be positive, thus  $$\ln T$$  is treated as being normally distributed.


 * $$\frac{\ln \hat{T}-\ln T}{\sqrt{Var(\ln \hat{T}})}\ \tilde{\ }\ N(0,1)$$

Confidence bounds on the time are given by:


 * $$CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}}$$


 * where:


 * $$\begin{align}

& Var(\hat{T})= & {{\left( \frac{\partial T}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial T}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial T}{\partial \beta } \right)\left( \frac{\partial T}{\partial \lambda } \right)cov(\hat{\beta },\,\,\,\hat{\lambda }) \end{align}$$

The variance calculation is the same as Eqn. (variance1) and:


 * $$\hat{T}={{\left( \frac{{{\lambda }_{i}}(T)}{\lambda \beta } \right)}^{1/(\beta -1)}}$$


 * $$\begin{align}

& \frac{\partial T}{\partial \beta }= & {{\left( \frac{{{\lambda }_{i}}(T)}{\lambda \beta } \right)}^{1/(\beta -1)}}\left[ -\frac{\ln (\tfrac{{{\lambda }_{i}}(T)}{\lambda \beta })}+\frac{1}{\beta (1-\beta )} \right] \\ & \frac{\partial T}{\partial \lambda }= & {{\left( \frac{{{\lambda }_{i}}(T)}{\lambda \beta } \right)}^{1/(\beta -1)}}\frac{1}{\lambda (1-\beta )} \end{align}$$

Crow Bounds

 * Step 1: Calculate $$MTB{{F}_{i}}=\tfrac{1}{{{\lambda }_{i}}(T)}$$.
 * Step 2: Use the equations from 5.2.10.2 to calculate the bounds on time given the instantaneous failure intensity.

Fisher Matrix Bounds
The cumulative number of failures, $$N(t)$$, must be positive, thus  $$\ln N(t)$$  is treated as being normally distributed.


 * $$\frac{\ln \hat{N}(t)-\ln N(t)}{\sqrt{Var(\ln \hat{N}(t)})}\ \tilde{\ }\ N(0,1)$$


 * $$N(t)=\hat{N}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{N}(t))}/\hat{N}(t)}}$$


 * where:


 * $$\hat{N}(t)=\hat{\lambda }{{t}^}$$


 * $$\begin{align}

& Var(\hat{N}(t))= & {{\left( \frac{\partial \hat{N}(t)}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial \hat{N}(t)}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial \hat{N}(t)}{\partial \beta } \right)\left( \frac{\partial \hat{N}(t)}{\partial \lambda } \right)cov(\hat{\beta },\hat{\lambda }) \end{align}$$

The variance calculation is the same as Eqn. (variance1) and:


 * $$\begin{align}

& \frac{\partial \hat{N}(t)}{\partial \beta }= & \hat{\lambda }{{t}^}\ln t \\ & \frac{\partial \hat{N}(t)}{\partial \lambda }= & {{t}^} \end{align}$$

Crow Bounds
The Crow cumulative number of failure confidence bounds are:


 * $$\begin{align}

& {{N}_{L}}(T)= & \frac{T}{{\lambda }_{i}}{{(T)}_{L}} \\ & {{N}_{U}}(T)= & \frac{T}{{\lambda }_{i}}{{(T)}_{U}} \end{align}$$

where $${{\lambda }_{i}}{{(T)}_{L}}$$  and  $${{\lambda }_{i}}{{(T)}_{U}}$$  can be obtained from Eqn. (amsaac14). Example 2 Calculate the 90% 2-sided confidence bounds on the cumulative and instantaneous failure intensity for the data from Example 1 given in Table 5.1.

Solution Fisher Matrix Bounds Using $$\widehat{\beta }$$  and  $$\widehat{\lambda }$$  estimated in Example 1, Eqns. (lambda2partial), (beta2partial) and (lambdabeta2partial) are:


 * $$\begin{align}

& \frac{{{\partial }^{2}}\Lambda }{\partial {{\lambda }^{2}}}= & -\frac{22}=-122.43 \\ & \frac{{{\partial }^{2}}\Lambda }{\partial {{\beta }^{2}}}= & -\frac{22}-0.4239\cdot {{620}^{0.6142}}{{(\ln 620)}^{2}}=-967.68 \\ & \frac{{{\partial }^{2}}\Lambda }{\partial \lambda \partial \beta }= & -{{620}^{0.6142}}\ln 620=-333.64 \end{align}$$

The Fisher Matrix then becomes: For $$T=620$$  hr, the partial derivatives of the cumulative and instantaneous failure intensities are:


 * $$\begin{align}

& \frac{\partial {{\lambda }_{c}}(T)}{\partial \beta }= & \widehat{\lambda }{{T}^{\widehat{\beta }-1}}\ln (T) \\ & = & 0.4239\cdot {{620}^{-0.3858}}\ln 620 \\ & = & 0.22811336 \\ & \frac{\partial {{\lambda }_{c}}(T)}{\partial \lambda }= & {{T}^{\widehat{\beta }-1}} \\ & = & {{620}^{-0.3858}} \\ & = & 0.083694185  \end{align}$$


 * $$\begin{align}

& \frac{\partial {{\lambda }_{i}}(T)}{\partial \beta }= & \widehat{\lambda }{{T}^{\widehat{\beta }-1}}+\widehat{\lambda }\widehat{\beta }{{T}^{\widehat{\beta }-1}}\ln T \\ & = & 0.4239\cdot {{620}^{-0.3858}}+0.4239\cdot 0.6142\cdot {{620}^{-0.3858}}\ln 620 \\ & = & 0.17558519 \end{align}$$
 * $$\begin{align}

& \frac{\partial {{\lambda }_{i}}(T)}{\partial \lambda }= & \widehat{\beta }{{T}^{\widehat{\beta }-1}} \\ & = & 0.6142\cdot {{620}^{-0.3858}} \\ & = & 0.051404969 \end{align}$$

Therefore, the variances become:

The cumulative and instantaneous failure intensities at $$T=620$$  hr are:
 * $$\begin{align}

& {{\lambda }_{c}}(T)= & 0.03548 \\ & {{\lambda }_{i}}(T)= & 0.02179 \end{align}$$

So, at the 90% confidence level and for $$T=620$$  hr, the Fisher Matrix confidence bounds for the cumulative failure intensity are:


 * $$\begin{align}

& {{[{{\lambda }_{c}}(T)]}_{L}}= & 0.02499 \\ & {{[{{\lambda }_{c}}(T)]}_{U}}= & 0.05039 \end{align}$$

The confidence bounds for the instantaneous failure intensity are:


 * $$\begin{align}

& {{[{{\lambda }_{i}}(T)]}_{L}}= & 0.01327 \\ & {{[{{\lambda }_{i}}(T)]}_{U}}= & 0.03579 \end{align}$$

Figures 4fig82 and 4fig83 display plots of the Fisher Matrix confidence bounds for the cumulative and instantaneous failure intensity, respectively. Crow Bounds The Crow confidence bounds for the cumulative failure intensity at the 90% confidence level and for $$T=620$$  hr are:


 * $$\begin{align}

& {{[{{\lambda }_{c}}(T)]}_{L}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t} \\ & = & \frac{29.787476}{2*620} \\ & = & 0.02402 \\ & {{[{{\lambda }_{c}}(T)]}_{U}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot t} \\ & = & \frac{62.8296}{2*620} \\ & = & 0.05067 \end{align}$$

The Crow confidence bounds for the instantaneous failure intensity at the 90% confidence level and for $$T=620$$  hr are:


 * $$\begin{align}

& {{[{{\lambda }_{i}}(t)]}_{L}}= & \frac{1} \\ & = & \frac{1}{MTB{{F}_{i}}\cdot U} \\ & = & 0.01179 \end{align}$$


 * $$\begin{align}

& {{[{{\lambda }_{i}}(t)]}_{U}}= & \frac{1} \\ & = & \frac{1}{MTB{{F}_{i}}\cdot L} \\ & = & 0.03253 \end{align}$$

Figures 4fig84 and 4fig85 display plots of the Crow confidence bounds for the cumulative and instantaneous failure intensity, respectively.
 * $$\begin{align}

& Var(\widehat{\lambda })= & 0.13519969 \\ & Var(\widehat{\beta })= & 0.017105343 \\ & Cov(\widehat{\beta },\widehat{\lambda })= & -0.046614609 \end{align}$$ Example 3 Calculate the confidence bounds on the cumulative and instantaneous MTBF for the data in Table 5.1. Solution Fisher Matrix Bounds From the previous example:

And for $$T=620$$  hr, the partial derivatives of the cumulative and instantaneous MTBF are:


 * $$\begin{align}

& \frac{\partial {{m}_{c}}(T)}{\partial \beta }= & -\frac{1}{\widehat{\lambda }}{{T}^{1-\widehat{\beta }}}\ln T \\ & = & -\frac{1}{0.4239}{{620}^{0.3858}}\ln 620 \\ & = & -181.23135 \\ & \frac{\partial {{m}_{c}}(T)}{\partial \lambda }= & -\frac{1}{{T}^{1-\widehat{\beta }}} \\ & = & -\frac{1}{{620}^{0.3858}} \\ & = & -66.493299 \\ & \frac{\partial {{m}_{i}}(T)}{\partial \beta }= & -\frac{1}{\widehat{\lambda }{{\widehat{\beta }}^{2}}}{{T}^{1-\beta }}-\frac{1}{\widehat{\lambda }\widehat{\beta }}{{T}^{1-\widehat{\beta }}}\ln T \\ & = & -\frac{1}{0.4239\cdot {{0.6142}^{2}}}{{620}^{0.3858}}-\frac{1}{0.4239\cdot 0.6142}{{620}^{0.3858}}\ln 620 \\ & = & -369.78634 \\ & \frac{\partial {{m}_{i}}(T)}{\partial \lambda }= & -\frac{1}{{{\widehat{\lambda }}^{2}}\widehat{\beta }}{{T}^{1-\widehat{\beta }}} \\ & = & -\frac{1}{{{0.4239}^{2}}\cdot 0.6142}\cdot {{620}^{0.3858}} \\ & = & -108.26001 \end{align}$$

Therefore, the variances become:


 * $$\begin{align}

& Var({{\widehat{m}}_{c}}(T))= & {{\left( -181.23135 \right)}^{2}}\cdot 0.017105343+{{\left( -66.493299 \right)}^{2}}\cdot 0.13519969 \\ & & -2\cdot \left( -181.23135 \right)\cdot \left( -66.493299 \right)\cdot 0.046614609 \\ & = & 36.113376 \end{align}$$


 * $$\begin{align}

& Var({{\widehat{m}}_{i}}(T))= & {{\left( -369.78634 \right)}^{2}}\cdot 0.017105343+{{\left( -108.26001 \right)}^{2}}\cdot 0.13519969 \\ & & -2\cdot \left( -369.78634 \right)\cdot \left( -108.26001 \right)\cdot 0.046614609 \\ & = & 191.33709 \end{align}$$

So, at 90% confidence level and $$T=620$$  hr, the Fisher Matrix confidence bounds are:


 * $$\begin{align}

& {{[{{m}_{c}}(T)]}_{L}}= & {{{\hat{m}}}_{c}}(t){{e}^{-{{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{c}}(t))}/{{{\hat{m}}}_{c}}(t)}} \\ & = & 19.84581 \\ & {{[{{m}_{c}}(T)]}_{U}}= & {{{\hat{m}}}_{c}}(t){{e}^{{{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{c}}(t))}/{{{\hat{m}}}_{c}}(t)}} \\ & = & 40.01927 \end{align}$$
 * $$\begin{align}

& {{[{{m}_{i}}(T)]}_{L}}= & {{{\hat{m}}}_{i}}(t){{e}^{-{{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{i}}(t))}/{{{\hat{m}}}_{i}}(t)}} \\ & = & 27.94261 \\ & {{[{{m}_{i}}(T)]}_{U}}= & {{{\hat{m}}}_{i}}(t){{e}^{{{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{i}}(t))}/{{{\hat{m}}}_{i}}(t)}} \\ & = & 75.34193 \end{align}$$

Figures 4fig86 and 4fig87 show plots of the Fisher Matrix confidence bounds for the cumulative and instantaneous MTBFs. Crow Bounds The Crow confidence bounds for the cumulative MTBF and the instantaneous MTBF at the 90% confidence level and for $$T=620$$  hr are:


 * $$\begin{align}

& {{[{{m}_{c}}(T)]}_{L}}= & \frac{1} \\ & = & 20.5023 \\ & {{[{{m}_{c}}(T)]}_{U}}= & \frac{1} \\ & = & 41.6282 \end{align}$$


 * $$\begin{align}

& {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot {{\Pi }_{1}} \\ & = & 30.7445 \\ & {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot {{\Pi }_{2}} \\ & = & 84.7972 \end{align}$$

Figures 4fig88 and 4fig89 show plots of the Crow confidence bounds for the cumulative and instantaneous MTBF. Confidence bounds can also be obtained on the parameters $$\widehat{\beta }$$  and  $$\widehat{\lambda }$$. For Fisher Matrix confidence bounds:


 * $$\begin{align}

& {{\beta }_{L}}= & \hat{\beta }{{e}^{{{z}_{\alpha }}\sqrt{Var(\hat{\beta })}/\hat{\beta }}} \\ & = & 0.4325 \\ & {{\beta }_{U}}= & \hat{\beta }{{e}^{-{{z}_{\alpha }}\sqrt{Var(\hat{\beta })}/\hat{\beta }}} \\ & = & 0.8722 \end{align}$$
 * and:


 * $$\begin{align}

& {{\lambda }_{L}}= & \hat{\lambda }{{e}^{{{z}_{\alpha }}\sqrt{Var(\hat{\lambda })}/\hat{\lambda }}} \\ & = & 0.1016 \\ & {{\lambda }_{U}}= & \hat{\lambda }{{e}^{-{{z}_{\alpha }}\sqrt{Var(\hat{\lambda })}/\hat{\lambda }}} \\ & = & 1.7691 \end{align}$$ For Crow confidence bounds:


 * $$\begin{align}

& {{\beta }_{L}}= & 0.4527 \\ & {{\beta }_{U}}= & 0.9350 \end{align}$$
 * and:


 * $$\begin{align}

& {{\lambda }_{L}}= & 0.2870 \\ & {{\lambda }_{U}}= & 0.5827 \end{align}$$