Confidence Bounds for Repairable Systems Analysis

In this appendix, we will present the two methods used in the RGA software to estimate the confidence bounds for Repairable Systems Analysis. The Fisher Matrix approach is based on the Fisher Information Matrix and is commonly employed in the reliability field. The Crow bounds were developed by Dr. Larry Crow.

Fisher Matrix Bounds
The parameter $$\beta \,\!$$ must be positive, thus $$\ln \beta \,\!$$ is approximately treated as being normally distributed.


 * $$\frac{\ln (\hat{\beta })-\ln (\beta )}{\sqrt{Var\left[ \ln (\hat{\beta }) \right]}}\ \tilde{\ }\ N(0,1)\,\!$$


 * $$C{{B}_{\beta }}=\hat{\beta }{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{\beta })}/\hat{\beta }}}\,\!$$


 * $$\hat{\beta }=\frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\hat{\lambda }\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\left[ (T_{q}^{\hat{\beta }}\ln ({{T}_{q}})-S_{q}^{\hat{\beta }}\ln ({{S}_{q}}) \right]-\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{\mathop{\sum }}}\,\ln ({{X}_{i}}{{}_{q}})}\,\!$$

All variance can be calculated using the Fisher Information Matrix. $$\Lambda \,\!$$ is the natural log-likelihood function.


 * $$\Lambda =\underset{q=1}{\overset{K}{\mathop \sum }}\,\left[ {{N}_{q}}(\ln (\lambda )+\ln (\beta ))-\lambda (T_{q}^{\beta }-S_{q}^{\beta })+(\beta -1)\underset{i=1}{\overset{\mathop \sum }}\,\ln ({{x}_{iq}}) \right]\,\!$$


 * $$\frac{{{\partial }^{2}}\Lambda }{\partial {{\lambda }^{2}}}=-\frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}\,\!$$


 * $$\frac{{{\partial }^{2}}\Lambda }{\partial \lambda \partial \beta }=-\underset{q=1}{\overset{K}{\mathop \sum }}\,\left[ T_{q}^{\beta }\ln ({{T}_{q}})-S_{q}^{\beta }\ln ({{S}_{q}}) \right]\,\!$$


 * $$\frac{{{\partial }^{2}}\Lambda }{\partial {{\beta }^{2}}}=-\frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}-\lambda \underset{q=1}{\overset{K}{\mathop \sum }}\,\left[ T_{q}^{\beta }{{(\ln ({{T}_{q}}))}^{2}}-S_{q}^{\beta }{{(\ln ({{S}_{q}}))}^{2}} \right]\,\!$$

Crow Bounds
Calculate the conditional maximum likelihood estimate of $$\tilde{\beta \,\!}\,\!$$ :


 * $$\tilde{\beta }=\frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{M}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{M}{\mathop{\sum }}}\,\ln \left( \tfrac \right)}\,\!$$

The Crow 2-sided $$(1-a)\,\!$$ 100% confidence bounds on $$\beta \,\!$$ are:


 * $$\begin{align}

{{\beta }_{L}}= & \tilde{\beta }\frac{\chi _{\tfrac{\alpha }{2},2M}^{2}}{2M} \\ {{\beta }_{U}}= & \tilde{\beta }\frac{\chi _{1-\tfrac{\alpha }{2},2M}^{2}}{2M} \end{align}\,\!$$

Fisher Matrix Bounds
The parameter $$\lambda \,\!$$ must be positive, thus $$\ln \lambda \,\!$$ is approximately treated as being normally distributed. These bounds are based on:


 * $$\frac{\ln (\hat{\lambda })-\ln (\lambda )}{\sqrt{Var\left[ \ln (\hat{\lambda }) \right]}}\ \tilde{\ }\ N(0,1)\,\!$$

The approximate confidence bounds on $$\lambda \,\!$$ are given as:


 * $$C{{B}_{\lambda }}=\hat{\lambda }{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{\lambda })}/\hat{\lambda }}}\,\!$$

where $$\hat{\lambda }=\tfrac{n}{T_{K}^}\,\!$$.

The variance calculation is the same the equations given in the confidence bounds on Beta.

Crow Bounds
Failure Terminated

The confidence bounds on $$\lambda \,\!$$ for failure terminated data are calculated using:


 * $$\begin{align}

{{\lambda }_{L}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot \underset{q=1}{\overset{K}{\mathop{\sum }}}\,T_{q}^{^{\beta }}} \\ {{\lambda }_{U}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N}^{2}}{2\cdot \underset{q=1}{\overset{K}{\mathop{\sum }}}\,T_{q}^{^{\beta }}} \end{align}\,\!$$

where:
 * $$N\,\!$$ = total number of failures.
 * $$K\,\!$$ = number of systems.
 * $${{T}_{q}}\,\!$$ = end time for the $${{q}^{th}}$$ system.

Time Terminated

The confidence bounds on $$\lambda \,\!$$ for time terminated data are calculated using:


 * $$\begin{align}

{{\lambda }_{L}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot \underset{q=1}{\overset{K}{\mathop{\sum }}}\,T_{q}^{^{\beta }}} \\ {{\lambda }_{U}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot \underset{q=1}{\overset{K}{\mathop{\sum }}}\,T_{q}^{^{\beta }}} \end{align}\,\!$$

where:
 * $$N\,\!$$ = total number of failures.
 * $$K\,\!$$ = number of systems.
 * $${{T}_{q}}\,\!$$ = end time for the $${{q}^{th}}$$ system.

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


 * $$\frac{\ln (\hat{N}(t))-\ln (N(t))}{\sqrt{Var\left[ \ln \hat{N}(t) \right]}}\sim N(0,1)\,\!$$


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

where:


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


 * $$\begin{align}

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

The variance calculation is the same as the calculations in the confidence bounds on Beta.


 * $$\begin{align}

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

Crow Bounds

 * $$\begin{array}{*{35}{l}}

{{N}_{L}}(T)=\tfrac{T}{\hat{\beta }}{{\lambda }_{i}}{{(T)}_{L}} \\ {{N}_{U}}(T)=\tfrac{T}{\hat{\beta }}{{\lambda }_{i}}{{(T)}_{U}} \\ \end{array}\,\!$$

where $${{\lambda }_{i}}{{(T)}_{L}}\,\!$$ and $${{\lambda }_{i}}{{(T)}_{U}}\,\!$$ can be obtained using the equations given for the confidence bounds on Instantaneous Failure Intensity.

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


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

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


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

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 the calculations in the confidence bounds on Beta.


 * $$\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 by:


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


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

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


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

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:


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


 * $$\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 the calculations given in the confidence bounds on Beta.


 * $$\begin{align}

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

Crow Bounds
The 2-sided confidence bounds on the cumulative MTBF $$(CMTBF)\,\!$$ are given by:


 * $$\begin{align}

& CMTBF_{L}=\frac{1}{CFI_{U}} \\ & CMTBF_{U}=\frac{1}{CFI_{L}} \end{align}\,\!$$

where $$CFI_L\,\!$$ and $$CFI_U\,\!$$ are calculated using the process for the confidence bounds on cumulative failure intensity.

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


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

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 the calculations given in the confidence bounds on Beta.


 * $$\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

To calculate the bounds for 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

To calculate the bounds for 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 $$\Pi =\tfrac{4{{x}^{2}}}\,\!$$ 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 instantaneous failure intensity, $${{\lambda }_{i}}(t)\,\!$$. must be positive, thus $$\ln {{\lambda }_{i}}(t)\,\!$$ is approximately treated as being normally distributed.


 * $$\frac{\ln ({{\hat{\lambda }}_{i}}(t))-\ln ({{\lambda }_{i}}(t))}{\sqrt{Var\left[ \ln ({{\hat{\lambda }}_{i}}(t)) \right]}}\sim N(0,1)\,\!$$

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 $${{\lambda }_{i}}(t)=\lambda \beta {{t}^{\beta -1}}\,\!$$ and:


 * $$\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 the calculations in the confidence bounds on Beta.


 * $$\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 approximately treated as being normally distributed.


 * $$\frac{\ln (\hat{T})-\ln (T)}{\sqrt{Var\left[ \ln (\hat{T}) \right]}}\ \tilde{\ }\ N(0,1)\,\!$$

The confidence bounds on the time are given by:


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

where:


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

The variance calculation is the same as the calculations in the confidence bounds on Beta.


 * $$\hat{T}={{(\lambda \cdot {{m}_{c}})}^{1/(1-\beta )}}\,\!$$


 * $$\begin{align}

\frac{\partial T}{\partial \beta }= & \frac{{{(\lambda \cdot {{m}_{c}})}^{1/(1-\beta )}}\ln (\lambda \cdot {{m}_{c}})} \\ \frac{\partial T}{\partial \lambda }= & \frac{\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 approximately treated as being normally distributed.


 * $$\frac{\ln (\hat{T})-\ln (T)}{\sqrt{Var\left[ \ln (\hat{T}) \right]}}\ \tilde{\ }\ N(0,1)\,\!$$

The confidence bounds on the time are given by:


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

where:


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

The variance calculation is the same as the calculations in the confidence bounds on Beta.


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


 * $$\begin{align}

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

Crow Bounds
Step 1: Calculate the confidence bounds on the instantaneous MTBF.

Step 2: Calculate the bounds on time as follows.

Failure Terminated


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

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


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

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 approximately treated as being normally distributed.


 * $$\frac{\ln (\hat{T})-\ln (T)}{\sqrt{Var\left[ \ln \hat{T} \right]}}\ \tilde{\ }\ N(0,1)\,\!$$

The confidence bounds on the time are given by:


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

where:


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

The variance calculation is the same as the calculations given in the confidence bounds on Beta.


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


 * $$\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
These bounds are based on:


 * $$\frac{\ln (\hat{T})-\ln (T)}{\sqrt{Var\left[ \ln (\hat{T}) \right]}}\sim N(0,1)\,\!$$

The 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 the calculations given in the confidence bounds on Beta.


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


 * $$\begin{align}

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

Crow Bounds
Step 1: Calculate $${{\lambda }_{i}}(T)=\tfrac{1}{MTB{{F}_{i}}}\,\!$$.

Step 2: Use the equations in the Bounds on Time Given Instantaneous MTBF section to calculate the bounds on time given the instantaneous failure intensity.

Fisher Matrix Bounds
These bounds are based on:


 * $$\log it(\hat{R}(t))\sim N(0,1)\,\!$$


 * $$\log it(\hat{R}(t))=\ln \left\{ \frac{\hat{R}(t)}{1-\hat{R}(t)} \right\}\,\!$$

The confidence bounds on reliability are given by:


 * $$CB=\frac{\hat{R}(t)}{\hat{R}(t)+(1-\hat{R}(t)){{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{R}(t))}/\left[ \hat{R}(t)(1-\hat{R}(t)) \right]}}}\,\!$$


 * $$Var(\hat{R}(t))={{\left( \frac{\partial R}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial R}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda })+2\left( \frac{\partial R}{\partial \beta } \right)\left( \frac{\partial R}{\partial \lambda } \right)cov(\hat{\beta },\hat{\lambda })\,\!$$

The variance calculation is the same as the calculations in the confidence bounds on Beta.


 * $$\begin{align}

\frac{\partial R}{\partial \beta }= & {{e}^{-[\hat{\lambda }{{(t+d)}^{\hat{\beta }}}-\hat{\lambda }{{t}^{\hat{\beta }}}]}}[\lambda {{t}^{\hat{\beta }}}\ln (t)-\lambda {{(t+d)}^{\hat{\beta }}}\ln (t+d)] \\ \frac{\partial R}{\partial \lambda }= & {{e}^{-[\hat{\lambda }{{(t+d)}^{\hat{\beta }}}-\hat{\lambda }{{t}^{\hat{\beta }}}]}}[{{t}^{\hat{\beta }}}-{{(t+d)}^{\hat{\beta }}}] \end{align}\,\!$$

Crow Bounds
Failure Terminated

With failure terminated data, the 100( $$1-\alpha \,\!$$ )% confidence interval for the current reliability at time $$t\,\!$$ in a specified mission time $$d\,\!$$ is:


 * $$({{[\hat{R}(d)]}^{\tfrac{1}}},{{[\hat{R}(d)]}^{\tfrac{1}}})\,\!$$

where


 * $$\hat{R}(\tau )={{e}^{-[\hat{\lambda }{{(t+\tau )}^{\hat{\beta }}}-\hat{\lambda }{{t}^{\hat{\beta }}}]}}\,\!$$

$${{p}_{1}}\,\!$$ and $${{p}_{2}}\,\!$$ can be obtained from the equations for failure terminated data for the confidence bounds on Instantaneous MTBF.

Time Terminated

With time terminated data, the 100( $$1-\alpha \,\!$$ )% confidence interval for the current reliability at time $$t\,\!$$ in a specified mission time $$\tau \,\!$$ is:


 * $$({{[\hat{R}(d)]}^{\tfrac{1}}},{{[\hat{R}(d)]}^{\tfrac{1}}})\,\!$$

where:


 * $$\hat{R}(d)={{e}^{-[\hat{\lambda }{{(t+d)}^{\hat{\beta }}}-\hat{\lambda }{{t}^{\hat{\beta }}}]}}\,\!$$

$${{p}_{1}}\,\!$$ and $${{p}_{2}}\,\!$$ can be obtained from the equation for time terminated data for the confidence bounds on Instantaneous MTBF.

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


 * $$\frac{\ln (\hat{t})-\ln (t)}{\sqrt{Var\left[ \ln (\hat{t}) \right]}}\sim N(0,1)\,\!$$

The confidence bounds on time are calculated by using:


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

where:


 * $$Var(\hat{t})={{\left( \frac{\partial t}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial t}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda })+2\left( \frac{\partial t}{\partial \beta } \right)\left( \frac{\partial t}{\partial \lambda } \right)cov(\hat{\beta },\hat{\lambda })\,\!$$

$$\hat{t}\,\!$$ is calculated numerically from:


 * $$\hat{R}(d)={{e}^{-[\hat{\lambda }{{(\hat{t}+d)}^{\hat{\beta }}}-\hat{\lambda }{{{\hat{t}}}^{\hat{\beta }}}]}}\text{ };\text{ }d\text{ = mission time}\,\!$$

The variance calculations are done by:


 * $$\begin{align}

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

Crow Bounds
Failure Terminated

Step 1: Calculate $$({{\hat{R}}_{lower}},{{\hat{R}}_{upper}})=({{R}^{\tfrac{1}}},{{R}^{\tfrac{1}}})\,\!$$.

Step 2: Let $$R={{\hat{R}}_{lower}}\,\!$$ and solve numerically for $${{t}_{1}}\,\!$$ using $$R={{e}^{-[\hat{\lambda }{{({{{\hat{t}}}_{1}}+d)}^{\hat{\beta }}}-\hat{\lambda }\hat{t}_{1}^{\hat{\beta }}]}}\,\!$$.

Step 3: Let $$R={{\hat{R}}_{upper}}\,\!$$ and solve numerically for $${{t}_{2}}\,\!$$ using $$R={{e}^{-[\hat{\lambda }{{({{{\hat{t}}}_{2}}+d)}^{\hat{\beta }}}-\hat{\lambda }\hat{t}_{2}^{\hat{\beta }}]}}\,\!$$.

Step 4: If $${{t}_{1}}<{{t}_{2}}\,\!$$. then $${{t}_{lower}}={{t}_{1}}\,\!$$ and $${{t}_{upper}}={{t}_{2}}\,\!$$. If $${{t}_{1}}>{{t}_{2}}\,\!$$. then $${{t}_{lower}}={{t}_{2}}\,\!$$ and $${{t}_{upper}}={{t}_{1}}\,\!$$.

Time Terminated

Step 1: Calculate $$({{\hat{R}}_{lower}},{{\hat{R}}_{upper}})=({{R}^{\tfrac{1}}},{{R}^{\tfrac{1}}})\,\!$$.

Step 2: Let $$R={{\hat{R}}_{lower}}\,\!$$ and solve numerically for $${{t}_{1}}\,\!$$ using $$R={{e}^{-[\hat{\lambda }{{({{{\hat{t}}}_{1}}+d)}^{\hat{\beta }}}-\hat{\lambda }\hat{t}_{1}^{\hat{\beta }}]}}\,\!$$.

Step 3: Let $$R={{\hat{R}}_{upper}}\,\!$$ and solve numerically for $${{t}_{2}}\,\!$$ using $$R={{e}^{-[\hat{\lambda }{{({{{\hat{t}}}_{2}}+d)}^{\hat{\beta }}}-\hat{\lambda }\hat{t}_{2}^{\hat{\beta }}]}}\,\!$$.

Step 4: If $${{t}_{1}}<{{t}_{2}}\,\!$$. then $${{t}_{lower}}={{t}_{1}}\,\!$$ and $${{t}_{upper}}={{t}_{2}}\,\!$$. If $${{t}_{1}}>{{t}_{2}}\,\!$$. then $${{t}_{lower}}={{t}_{2}}\,\!$$ and $${{t}_{upper}}={{t}_{1}}\,\!$$.

Fisher Matrix Bounds
The mission time, $$d\,\!$$. must be positive, thus $$\ln \left( d \right)\,\!$$ is approximately treated as being normally distributed.


 * $$\frac{\ln (\hat{d})-\ln (d)}{\sqrt{Var\left[ \ln (\hat{d}) \right]}}\sim N(0,1)\,\!$$

The confidence bounds on mission time are given by using:


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

where:


 * $$Var(\hat{d})={{\left( \frac{\partial d}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial d}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda })+2\left( \frac{\partial td}{\partial \beta } \right)\left( \frac{\partial d}{\partial \lambda } \right)cov(\hat{\beta },\hat{\lambda })\,\!$$

Calculate $$\hat{d}\,\!$$ from:


 * $$\hat{d}={{\left[ {{t}^}-\frac{\ln (R)}{{\hat{\lambda }}} \right]}^{\tfrac{1}}}-t\,\!$$

The variance calculations are done by:


 * $$\begin{align}

\frac{\partial d}{\partial \beta }= & \left[ \frac{{{t}^}\ln (t)}-\ln (t+\hat{d}) \right]\cdot \frac{t+\hat{d}} \\ \frac{\partial d}{\partial \lambda }= & \frac{{{t}^}-{{(t+\hat{d})}^}}{\hat{\lambda }\hat{\beta }{{(t+\hat{d})}^{\hat{\beta }-1}}} \end{align}\,\!$$

Crow Bounds
Failure Terminated

Step 1: Calculate $$({{\hat{R}}_{lower}},{{\hat{R}}_{upper}})=({{R}^{\tfrac{1}}},{{R}^{\tfrac{1}}})\,\!$$.

Step 2: Let $$R={{\hat{R}}_{lower}}\,\!$$ and solve for $${{d}_{1}}\,\!$$ such that:


 * $${{d}_{1}}={{\left( {{t}^}-\frac{\ln ({{R}_{lower}})}{{\hat{\lambda }}} \right)}^{\tfrac{1}}}-t\,\!$$

Step 3: Let $$R={{\hat{R}}_{upper}}\,\!$$ and solve for $${{d}_{2}}\,\!$$ such that:


 * $${{d}_{2}}={{\left( {{t}^}-\frac{\ln ({{R}_{upper}})}{{\hat{\lambda }}} \right)}^{\tfrac{1}}}-t\,\!$$

Step 4: If $${{d}_{1}}<{{d}_{2}}\,\!$$. then $${{d}_{lower}}={{d}_{1}}\,\!$$ and $${{d}_{upper}}={{d}_{2}}\,\!$$. If $${{d}_{1}}>{{d}_{2}}\,\!$$. then $${{d}_{lower}}={{d}_{2}}\,\!$$ and $${{d}_{upper}}={{d}_{1}}\,\!$$.

Time Terminated

Step 1: Calculate $$({{\hat{R}}_{lower}},{{\hat{R}}_{upper}})=({{R}^{\tfrac{1}}},{{R}^{\tfrac{1}}})\,\!$$.

Step 2: Let $$R={{\hat{R}}_{lower}}\,\!$$ and solve for $${{d}_{1}}\,\!$$ using the same equation given for the failure terminated data.

Step 3: Let $$R={{\hat{R}}_{upper}}\,\!$$ and solve for $${{d}_{2}}\,\!$$ using the same equation given for the failure terminated data.

Step 4: If $${{d}_{1}}<{{d}_{2}}\,\!$$. then $${{d}_{lower}}={{d}_{1}}\,\!$$ and $${{d}_{upper}}={{d}_{2}}\,\!$$. If $${{d}_{1}}>{{d}_{2}}\,\!$$. then $${{d}_{lower}}={{d}_{2}}\,\!$$ and $${{d}_{upper}}={{d}_{1}}\,\!$$.