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. 

Bounds on Beta

Fisher Matrix Bounds

The parameter [math]\displaystyle{ \beta \,\! }[/math] must be positive, thus [math]\displaystyle{ \ln \beta \,\! }[/math] is approximately treated as being normally distributed.

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

[math]\displaystyle{ C{{B}_{\beta }}=\hat{\beta }{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{\beta })}/\hat{\beta }}}\,\! }[/math]

[math]\displaystyle{ \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{{{N}_{q}}}{\mathop{\sum }}}\,\ln ({{X}_{i}}{{}_{q}})}\,\! }[/math]

All variance can be calculated using the Fisher Information Matrix.

[math]\displaystyle{ \Lambda \,\! }[/math] is the natural log-likelihood function.

[math]\displaystyle{ \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{{{N}_{q}}}{\mathop \sum }}\,\ln ({{x}_{iq}}) \right]\,\! }[/math]

[math]\displaystyle{ \frac{{{\partial }^{2}}\Lambda }{\partial {{\lambda }^{2}}}=-\frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{{{\lambda }^{2}}}\,\! }[/math]

[math]\displaystyle{ \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]\,\! }[/math]

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

Crow Bounds

Calculate the conditional maximum likelihood estimate of [math]\displaystyle{ \tilde{\beta \,\!}\,\! }[/math] :

[math]\displaystyle{ \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{{{T}_{q}}}{{{X}_{iq}}} \right)}\,\! }[/math]

The Crow 2-sided [math]\displaystyle{ (1-a)\,\! }[/math] 100% confidence bounds on [math]\displaystyle{ \beta \,\! }[/math] are:

[math]\displaystyle{ \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}\,\! }[/math]

Bounds on Lambda

Fisher Matrix Bounds

The parameter [math]\displaystyle{ \lambda \,\! }[/math] must be positive, thus [math]\displaystyle{ \ln \lambda \,\! }[/math] is approximately treated as being normally distributed. These bounds are based on:

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

The approximate confidence bounds on [math]\displaystyle{ \lambda \,\! }[/math] are given as:

[math]\displaystyle{ C{{B}_{\lambda }}=\hat{\lambda }{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{\lambda })}/\hat{\lambda }}}\,\! }[/math]

where [math]\displaystyle{ \hat{\lambda }=\tfrac{n}{T_{K}^{{\hat{\beta }}}}\,\! }[/math].

The variance calculation is the same the equations given in the Bounds on Beta section.

Crow Bounds

Time Terminated

The confidence bounds on [math]\displaystyle{ \lambda \,\! }[/math] for time terminated data are calculated using:

[math]\displaystyle{ \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}\,\! }[/math]

Failure Terminated

The confidence bounds on [math]\displaystyle{ \lambda \,\! }[/math] for failure terminated data are calculated using:

[math]\displaystyle{ \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}\,\! }[/math]

Bounds on Growth Rate

Since the growth rate is equal to [math]\displaystyle{ 1-\beta \,\! }[/math]. the confidence bounds are:

[math]\displaystyle{ \begin{align} Gr.\text{ }Rat{{e}_{L}}= & 1-{{\beta }_{U}} \\ Gr.\text{ }Rat{{e}_{U}}= & 1-{{\beta }_{L}} \end{align}\,\! }[/math]

[math]\displaystyle{ {{\beta }_{L}}\,\! }[/math] and [math]\displaystyle{ {{\beta }_{U}}\,\! }[/math] are obtained using the methods described above in the Bounds on Beta section.

Bounds on Cumulative MTBF

Fisher Matrix Bounds

The cumulative MTBF, [math]\displaystyle{ {{m}_{c}}(t)\,\! }[/math]. must be positive, thus [math]\displaystyle{ \ln {{m}_{c}}(t)\,\! }[/math] is approximately treated as being normally distributed.

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

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

[math]\displaystyle{ CB={{\hat{m}}_{c}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{\hat{m}}_{c}}(t))}/{{\hat{m}}_{c}}(t)}}\,\! }[/math]

where:

[math]\displaystyle{ {{\hat{m}}_{c}}(t)=\frac{1}{\hat{\lambda }}{{t}^{1-\hat{\beta }}}\,\! }[/math]

[math]\displaystyle{ \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}\,\! }[/math]

The variance calculation is the same as the calculations given in the Bounds on Beta section.

[math]\displaystyle{ \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}{{{\hat{\lambda }}^{2}}}{{t}^{1-\hat{\beta }}} \end{align}\,\! }[/math]

Crow Bounds

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

[math]\displaystyle{ C{{(t)}_{L}}=\frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t}\,\! }[/math]

[math]\displaystyle{ C{{(t)}_{u}}=\frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot t}\,\! }[/math]

Then

[math]\displaystyle{ \begin{align} {{[MTB{{F}_{c}}]}_{L}}= & \frac{1}{C{{(t)}_{U}}} \\ {{[MTB{{F}_{c}}]}_{U}}= & \frac{1}{C{{(t)}_{L}}} \end{align}\,\! }[/math]

Bounds on Instantaneous MTBF

Fisher Matrix Bounds

The instantaneous MTBF, [math]\displaystyle{ {{m}_{i}}(t)\,\! }[/math]. must be positive, thus [math]\displaystyle{ \ln {{m}_{i}}(t)\,\! }[/math] is approximately treated as being normally distributed.

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

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

[math]\displaystyle{ CB={{\hat{m}}_{i}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{\hat{m}}_{i}}(t))}/{{\hat{m}}_{i}}(t)}}\,\! }[/math]

where:

[math]\displaystyle{ {{\hat{m}}_{i}}(t)=\frac{1}{\lambda \beta {{t}^{\beta -1}}}\,\! }[/math]

[math]\displaystyle{ \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}\,\! }[/math]

The variance calculation is the same as the calculations given in the Bounds on Beta section.

[math]\displaystyle{ \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}\,\! }[/math]

Crow Bounds

Failure Terminated Data

To calculate the bounds for failure terminated data, consider the following equation:

[math]\displaystyle{ G(\mu |n)=\mathop{}_{0}^{\infty }\frac{{{e}^{-x}}{{x}^{n-2}}}{(n-2)!}\underset{i=0}{\overset{n-1}{\mathop \sum }}\,\frac{1}{i!}{{\left( \frac{\mu }{x} \right)}^{i}}\exp (-\frac{\mu }{x})\,dx\,\! }[/math]

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

[math]\displaystyle{ \begin{align} {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot {{p}_{1}} \\ {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot {{p}_{2}} \end{align}\,\! }[/math]

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

[math]\displaystyle{ \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}\,\! }[/math]

where [math]\displaystyle{ MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}\,\! }[/math].

Time Terminated Data

To calculate the bounds for time terminated data, consider the following equation where [math]\displaystyle{ {{I}_{1}}(.)\,\! }[/math] is the modified Bessel function of order one:

[math]\displaystyle{ H(x|k)=\underset{j=1}{\overset{k}{\mathop \sum }}\,\frac{{{x}^{2j-1}}}{{{2}^{2j-1}}(j-1)!j!{{I}_{1}}(x)}\,\! }[/math]

Find the values [math]\displaystyle{ {{\Pi }_{1}}\,\! }[/math] and [math]\displaystyle{ {{\Pi }_{2}}\,\! }[/math] by finding the solution [math]\displaystyle{ x\,\! }[/math] to [math]\displaystyle{ H(x|k)=\tfrac{\alpha }{2}\,\! }[/math] and [math]\displaystyle{ H(x|k)=1-\tfrac{\alpha }{2}\,\! }[/math] in the cases corresponding to the lower and upper bounds, respectively.

Calculate [math]\displaystyle{ \Pi =\tfrac{{{n}^{2}}}{4{{x}^{2}}}\,\! }[/math] for each case. If using the biased parameters, [math]\displaystyle{ \hat{\beta }\,\! }[/math] and [math]\displaystyle{ \hat{\lambda }\,\! }[/math]. then the upper and lower confidence bounds are:

[math]\displaystyle{ \begin{align} {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot {{\Pi }_{1}} \\ {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot {{\Pi }_{2}} \end{align}\,\! }[/math]

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

[math]\displaystyle{ \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}\,\! }[/math]

where [math]\displaystyle{ MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}\,\! }[/math].

Bounds on Cumulative Failure Intensity

Fisher Matrix Bounds

The cumulative failure intensity, [math]\displaystyle{ {{\lambda }_{c}}(t)\,\! }[/math] must be positive, thus [math]\displaystyle{ \ln {{\lambda }_{c}}(t)\,\! }[/math] is approximately treated as being normally distributed.

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

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

[math]\displaystyle{ CB={{\hat{\lambda }}_{c}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{\hat{\lambda }}_{c}}(t))}/{{\hat{\lambda }}_{c}}(t)}}\,\! }[/math]

where:

[math]\displaystyle{ {{\hat{\lambda }}_{c}}(t)=\hat{\lambda }{{t}^{\hat{\beta }-1}}\,\! }[/math]

and:

[math]\displaystyle{ \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}\,\! }[/math]

The variance calculation is the same as the calculations in the Bounds on Beta section.

[math]\displaystyle{ \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}\,\! }[/math]

Crow Bounds

The Crow cumulative failure intensity confidence bounds are given by:

[math]\displaystyle{ C{{(t)}_{L}}=\frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t}\,\! }[/math]

[math]\displaystyle{ C{{(t)}_{u}}=\frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot t}\,\! }[/math]

Bounds on Instantaneous Failure Intensity

Fisher Matrix Bounds

The instantaneous failure intensity, [math]\displaystyle{ {{\lambda }_{i}}(t)\,\! }[/math]. must be positive, thus [math]\displaystyle{ \ln {{\lambda }_{i}}(t)\,\! }[/math] is approximately treated as being normally distributed.

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

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

[math]\displaystyle{ CB={{\hat{\lambda }}_{i}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{\hat{\lambda }}_{i}}(t))}/{{\hat{\lambda }}_{i}}(t)}}\,\! }[/math]

where [math]\displaystyle{ {{\lambda }_{i}}(t)=\lambda \beta {{t}^{\beta -1}}\,\! }[/math] and:

[math]\displaystyle{ \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}\,\! }[/math]

The variance calculation is the same as the calculations in the Bounds on Beta section.

[math]\displaystyle{ \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}\,\! }[/math]

Crow Bounds

The Crow instantaneous failure intensity confidence bounds are given as:

[math]\displaystyle{ \begin{align} {{[{{\lambda }_{i}}(t)]}_{L}}= & \frac{1}{{{[MTB{{F}_{i}}]}_{U}}} \\ {{[{{\lambda }_{i}}(t)]}_{U}}= & \frac{1}{{{[MTB{{F}_{i}}]}_{L}}} \end{align}\,\! }[/math]

Bounds on Time Given Cumulative MTBF

Fisher Matrix Bounds

The time, [math]\displaystyle{ T\,\! }[/math]. must be positive, thus [math]\displaystyle{ \ln T\,\! }[/math] is approximately treated as being normally distributed.

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

The confidence bounds on the time are given by:

[math]\displaystyle{ CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}}\,\! }[/math]

where:

[math]\displaystyle{ 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 })\,\! }[/math]

The variance calculation is the same as the calculations in the Bounds on Beta section.

[math]\displaystyle{ \hat{T}={{(\lambda \cdot {{m}_{c}})}^{1/(1-\beta )}}\,\! }[/math]

[math]\displaystyle{ \begin{align} \frac{\partial T}{\partial \beta }= & \frac{{{(\lambda \cdot {{m}_{c}})}^{1/(1-\beta )}}\ln (\lambda \cdot {{m}_{c}})}{{{(1-\beta )}^{2}}} \\ \frac{\partial T}{\partial \lambda }= & \frac{{{(\lambda \cdot {{m}_{c}})}^{1/(1-\beta )}}}{\lambda (1-\beta )} \end{align}\,\! }[/math]

Crow Bounds

Step 1: Calculate:

[math]\displaystyle{ \hat{T}={{\left( \frac{{{\lambda }_{c}}(T)}{{\hat{\lambda }}} \right)}^{\tfrac{1}{\beta -1}}}\,\! }[/math]

Step 2: Estimate the number of failures:

[math]\displaystyle{ N(\hat{T})=\hat{\lambda }{{\hat{T}}^{{\hat{\beta }}}}\,\! }[/math]

Step 3: Obtain the confidence bounds on time given the cumulative failure intensity by solving for [math]\displaystyle{ {{t}_{l}}\,\! }[/math] and [math]\displaystyle{ {{t}_{u}}\,\! }[/math] in the following equations:

[math]\displaystyle{ \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}\,\! }[/math]

Bounds on Time Given Instantaneous MTBF

Fisher Matrix Bounds

The time, [math]\displaystyle{ T\,\! }[/math]. must be positive, thus [math]\displaystyle{ \ln T\,\! }[/math] is approximately treated as being normally distributed.

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

The confidence bounds on the time are given by:

[math]\displaystyle{ CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}}\,\! }[/math]

where:

[math]\displaystyle{ 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 })\,\! }[/math]

The variance calculation is the same as the calculations in the Bounds on Beta section.

[math]\displaystyle{ \hat{T}={{(\lambda \beta \cdot MTB{{F}_{i}})}^{1/(1-\beta )}}\,\! }[/math]

[math]\displaystyle{ \begin{align} \frac{\partial T}{\partial \beta }= & {{\left( \lambda \beta \cdot MTB{{F}_{i}} \right)}^{1/(1-\beta )}}[\frac{1}{{{(1-\beta )}^{2}}}\ln (\lambda \beta \cdot MTB{{F}_{i}})+\frac{1}{\beta (1-\beta )}] \\ \frac{\partial T}{\partial \lambda }= & \frac{{{(\lambda \beta \cdot MTB{{F}_{i}})}^{1/(1-\beta )}}}{\lambda (1-\beta )} \end{align}\,\! }[/math]

Crow Bounds

Step 1: Calculate the confidence bounds on the instantaneous MTBF as presented in Section 5.5.2.[Per Kate: THE SECTION 5.5.2 IS ABOUT THE CHI-SQUARED TEST FOR GROUPED DATA. NEED TO ASK SME TO VERIFY THE CORRECT SECTION TO LINK TO.]

Step 2: Calculate the bounds on time as follows.

Failure Terminated Data

[math]\displaystyle{ \hat{T}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{c})}^{1/(1-\beta )}}\,\! }[/math]

So the lower an upper bounds on time are:

[math]\displaystyle{ {{\hat{T}}_{L}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{c}_{1}}})}^{1/(1-\beta )}}\,\! }[/math]

[math]\displaystyle{ {{\hat{T}}_{U}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{c}_{2}}})}^{1/(1-\beta )}}\,\! }[/math]

Time Terminated Data

[math]\displaystyle{ \hat{T}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{\Pi })}^{1/(1-\beta )}}\,\! }[/math]

So the lower and upper bounds on time are:

[math]\displaystyle{ {{\hat{T}}_{L}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{\Pi }_{1}}})}^{1/(1-\beta )}}\,\! }[/math]

[math]\displaystyle{ {{\hat{T}}_{U}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{\Pi }_{2}}})}^{1/(1-\beta )}}\,\! }[/math]

Bounds on Time Given Cumulative Failure Intensity

Fisher Matrix Bounds

The time, [math]\displaystyle{ T\,\! }[/math]. must be positive, thus [math]\displaystyle{ \ln T\,\! }[/math] is approximately treated as being normally distributed.

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

The confidence bounds on the time are given by:

[math]\displaystyle{ CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}}\,\! }[/math]

where:

[math]\displaystyle{ 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 })\,\! }[/math]

The variance calculation is the same as the calculations given in the Bounds on Beta section.

[math]\displaystyle{ \hat{T}={{\left( \frac{{{\lambda }_{c}}(T)}{\lambda } \right)}^{1/(\beta -1)}}\,\! }[/math]

[math]\displaystyle{ \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)}{{{(1-\beta )}^{2}}} \\ \frac{\partial T}{\partial \lambda }= & {{\left( \frac{{{\lambda }_{c}}(T)}{\lambda } \right)}^{1/(\beta -1)}}\frac{1}{\lambda (1-\beta )} \end{align}\,\! }[/math]

Crow Bounds

Step 1: Calculate:

[math]\displaystyle{ \hat{T}={{\left( \frac{{{\lambda }_{c}}(T)}{{\hat{\lambda }}} \right)}^{\tfrac{1}{\beta -1}}}\,\! }[/math]

Step 2: Estimate the number of failures:

[math]\displaystyle{ N(\hat{T})=\hat{\lambda }{{\hat{T}}^{{\hat{\beta }}}}\,\! }[/math]

Step 3: Obtain the confidence bounds on time given the cumulative failure intensity by solving for [math]\displaystyle{ {{t}_{l}}\,\! }[/math] and [math]\displaystyle{ {{t}_{u}}\,\! }[/math] in the following equations:

[math]\displaystyle{ \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}\,\! }[/math]

Bounds on Time Given Instantaneous Failure Intensity

Fisher Matrix Bounds

These bounds are based on:

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

The confidence bounds on the time are given by:

[math]\displaystyle{ CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}}\,\! }[/math]

where:

[math]\displaystyle{ \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}\,\! }[/math]

The variance calculation is the same as the calculations given in the Bounds on Beta section.

[math]\displaystyle{ \hat{T}={{\left( \frac{{{\lambda }_{i}}(T)}{\lambda \cdot \beta } \right)}^{1/(\beta -1)}}\,\! }[/math]

[math]\displaystyle{ \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 })}{{{(\beta -1)}^{2}}}+\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}\,\! }[/math]

Crow Bounds

Step 1: Calculate [math]\displaystyle{ {{\lambda }_{i}}(T)=\tfrac{1}{MTB{{F}_{i}}}\,\! }[/math].

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.

Bounds on Reliability

Fisher Matrix Bounds

These bounds are based on:

[math]\displaystyle{ \log it(\hat{R}(t))\sim N(0,1)\,\! }[/math]

[math]\displaystyle{ \log it(\hat{R}(t))=\ln \left\{ \frac{\hat{R}(t)}{1-\hat{R}(t)} \right\}\,\! }[/math]

The confidence bounds on reliability are given by:

[math]\displaystyle{ 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]}}}\,\! }[/math]

[math]\displaystyle{ 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 })\,\! }[/math]

The variance calculation is the same as the calculations in the Bounds on Beta section.

[math]\displaystyle{ \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}\,\! }[/math]

Crow Bounds

Failure Terminated Data

With failure terminated data, the 100( [math]\displaystyle{ 1-\alpha \,\! }[/math] )% confidence interval for the current reliability at time [math]\displaystyle{ t\,\! }[/math] in a specified mission time [math]\displaystyle{ d\,\! }[/math] is:

[math]\displaystyle{ ({{[\hat{R}(d)]}^{\tfrac{1}{{{p}_{1}}}}},{{[\hat{R}(d)]}^{\tfrac{1}{{{p}_{2}}}}})\,\! }[/math]

where

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

[math]\displaystyle{ {{p}_{1}}\,\! }[/math] and [math]\displaystyle{ {{p}_{2}}\,\! }[/math] can be obtained from the equation for failure terminated data in the Bounds on Instantaneous MTBF section.

Time Terminated Data

With time terminated data, the 100( [math]\displaystyle{ 1-\alpha \,\! }[/math] )% confidence interval for the current reliability at time [math]\displaystyle{ t\,\! }[/math] in a specified mission time [math]\displaystyle{ \tau \,\! }[/math] is:

[math]\displaystyle{ ({{[\hat{R}(d)]}^{\tfrac{1}{{{p}_{1}}}}},{{[\hat{R}(d)]}^{\tfrac{1}{{{p}_{2}}}}})\,\! }[/math]

where:

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

[math]\displaystyle{ {{p}_{1}}\,\! }[/math] and [math]\displaystyle{ {{p}_{2}}\,\! }[/math] can be obtained from the equation for time terminated data in the Bounds on Instantaneous MTBF section.

Bounds on Time Given Reliability and Mission Time

Fisher Matrix Bounds

The time, [math]\displaystyle{ t\,\! }[/math]. must be positive, thus [math]\displaystyle{ \ln t\,\! }[/math] is approximately treated as being normally distributed.

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

The confidence bounds on time are calculated by using:

[math]\displaystyle{ CB=\hat{t}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{t})}/\hat{t}}}\,\! }[/math]

where:

[math]\displaystyle{ 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 })\,\! }[/math]

[math]\displaystyle{ \hat{t}\,\! }[/math] is calculated numerically from:

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

The variance calculations are done by:

[math]\displaystyle{ \begin{align} \frac{\partial t}{\partial \beta }= & \frac{{{{\hat{t}}}^{{\hat{\beta }}}}\ln (\hat{t})-{{(\hat{t}+d)}^{{\hat{\beta }}}}\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{\beta }}}}-{{(\hat{t}+d)}^{{\hat{\beta }}}}}{\hat{\lambda }\hat{\beta }{{(\hat{t}+d)}^{\hat{\beta }-1}}-\hat{\lambda }\hat{\beta }{{{\hat{t}}}^{\hat{\beta }-1}}} \end{align}\,\! }[/math]

Crow Bounds

Failure Terminated Data

Step 1: Calculate [math]\displaystyle{ ({{\hat{R}}_{lower}},{{\hat{R}}_{upper}})=({{R}^{\tfrac{1}{{{p}_{1}}}}},{{R}^{\tfrac{1}{{{p}_{2}}}}})\,\! }[/math].

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

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

Step 4: If [math]\displaystyle{ {{t}_{1}}\lt {{t}_{2}}\,\! }[/math]. then [math]\displaystyle{ {{t}_{lower}}={{t}_{1}}\,\! }[/math] and [math]\displaystyle{ {{t}_{upper}}={{t}_{2}}\,\! }[/math]. If [math]\displaystyle{ {{t}_{1}}\gt {{t}_{2}}\,\! }[/math]. then [math]\displaystyle{ {{t}_{lower}}={{t}_{2}}\,\! }[/math] and [math]\displaystyle{ {{t}_{upper}}={{t}_{1}}\,\! }[/math].

Time Terminated Data
Step 1: Calculate [math]\displaystyle{ ({{\hat{R}}_{lower}},{{\hat{R}}_{upper}})=({{R}^{\tfrac{1}{{{\Pi }_{1}}}}},{{R}^{\tfrac{1}{{{\Pi }_{2}}}}})\,\! }[/math].

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

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

Step 4: If [math]\displaystyle{ {{t}_{1}}\lt {{t}_{2}}\,\! }[/math]. then [math]\displaystyle{ {{t}_{lower}}={{t}_{1}}\,\! }[/math] and [math]\displaystyle{ {{t}_{upper}}={{t}_{2}}\,\! }[/math]. If [math]\displaystyle{ {{t}_{1}}\gt {{t}_{2}}\,\! }[/math]. then [math]\displaystyle{ {{t}_{lower}}={{t}_{2}}\,\! }[/math] and [math]\displaystyle{ {{t}_{upper}}={{t}_{1}}\,\! }[/math].

Bounds on Mission Time Given Reliability and Time

Fisher Matrix Bounds

The mission time, [math]\displaystyle{ d\,\! }[/math]. must be positive, thus [math]\displaystyle{ \ln \left( d \right)\,\! }[/math] is approximately treated as being normally distributed.

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

The confidence bounds on mission time are given by using:

[math]\displaystyle{ CB=\hat{d}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{d})}/\hat{d}}}\,\! }[/math]

where:

[math]\displaystyle{ 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 })\,\! }[/math]

Calculate [math]\displaystyle{ \hat{d}\,\! }[/math] from:

[math]\displaystyle{ \hat{d}={{\left[ {{t}^{{\hat{\beta }}}}-\frac{\ln (R)}{{\hat{\lambda }}} \right]}^{\tfrac{1}{{\hat{\beta }}}}}-t\,\! }[/math]

The variance calculations are done by:

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

Crow Bounds

Failure Terminated Data

Step 1: Calculate [math]\displaystyle{ ({{\hat{R}}_{lower}},{{\hat{R}}_{upper}})=({{R}^{\tfrac{1}{{{p}_{1}}}}},{{R}^{\tfrac{1}{{{p}_{2}}}}})\,\! }[/math].

Step 2: Let [math]\displaystyle{ R={{\hat{R}}_{lower}}\,\! }[/math] and solve for [math]\displaystyle{ {{d}_{1}}\,\! }[/math] such that:

[math]\displaystyle{ {{d}_{1}}={{\left( {{t}^{{\hat{\beta }}}}-\frac{\ln ({{R}_{lower}})}{{\hat{\lambda }}} \right)}^{\tfrac{1}{{\hat{\beta }}}}}-t\,\! }[/math]

Step 3: Let [math]\displaystyle{ R={{\hat{R}}_{upper}}\,\! }[/math] and solve for [math]\displaystyle{ {{d}_{2}}\,\! }[/math] such that:

[math]\displaystyle{ {{d}_{2}}={{\left( {{t}^{{\hat{\beta }}}}-\frac{\ln ({{R}_{upper}})}{{\hat{\lambda }}} \right)}^{\tfrac{1}{{\hat{\beta }}}}}-t\,\! }[/math]

Step 4: If [math]\displaystyle{ {{d}_{1}}\lt {{d}_{2}}\,\! }[/math]. then [math]\displaystyle{ {{d}_{lower}}={{d}_{1}}\,\! }[/math] and [math]\displaystyle{ {{d}_{upper}}={{d}_{2}}\,\! }[/math]. If [math]\displaystyle{ {{d}_{1}}\gt {{d}_{2}}\,\! }[/math]. then [math]\displaystyle{ {{d}_{lower}}={{d}_{2}}\,\! }[/math] and [math]\displaystyle{ {{d}_{upper}}={{d}_{1}}\,\! }[/math].

Time Terminated Data

Step 1: Calculate [math]\displaystyle{ ({{\hat{R}}_{lower}},{{\hat{R}}_{upper}})=({{R}^{\tfrac{1}{{{\Pi }_{1}}}}},{{R}^{\tfrac{1}{{{\Pi }_{2}}}}})\,\! }[/math].

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

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

Step 4: If [math]\displaystyle{ {{d}_{1}}\lt {{d}_{2}}\,\! }[/math]. then [math]\displaystyle{ {{d}_{lower}}={{d}_{1}}\,\! }[/math] and [math]\displaystyle{ {{d}_{upper}}={{d}_{2}}\,\! }[/math]. If [math]\displaystyle{ {{d}_{1}}\gt {{d}_{2}}\,\! }[/math]. then [math]\displaystyle{ {{d}_{lower}}={{d}_{2}}\,\! }[/math] and [math]\displaystyle{ {{d}_{upper}}={{d}_{1}}\,\! }[/math].

Bounds on Cumulative Number of Failures

Fisher Matrix Bounds

The cumulative number of failures, [math]\displaystyle{ N(t)\,\! }[/math]. must be positive, thus [math]\displaystyle{ \ln \left( N(t) \right)\,\! }[/math] is approximately treated as being normally distributed.

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

[math]\displaystyle{ N(t)=\hat{N}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{N}(t))}/\hat{N}(t)}}\,\! }[/math]

where:

[math]\displaystyle{ \hat{N}(t)=\hat{\lambda }{{t}^{\hat{\beta }}}\,\! }[/math]

[math]\displaystyle{ \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}\,\! }[/math]

The variance calculation is the same as the calculations in the Bounds on Beta section.

[math]\displaystyle{ \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}\,\! }[/math]

Crow Bounds

[math]\displaystyle{ \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}\,\! }[/math]

where [math]\displaystyle{ {{\lambda }_{i}}{{(T)}_{L}}\,\! }[/math] and [math]\displaystyle{ {{\lambda }_{i}}{{(T)}_{U}}\,\! }[/math] can be obtained using the equation given in the Bounds on Instantaneous Failure Intensity section.