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{{template:RGA BOOK|E|Confidence Bounds for Repairable Systems Analysis}}
{{template:RGA BOOK|Appendix E|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 [[RGA Models for Repairable Systems Analysis|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. 
In this appendix, we will present the two methods used in the RGA software to estimate the confidence bounds for [[Repairable Systems Analysis|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. 


<PER LISA: ASK SME HOW THIS COMPARES TO THE APPENDIX FOR THE CONFIDENCE BOUNDS ON THE CROW EXTENDED MODEL. FOR EXAMPLES, SHOULD ANY IDENTICAL SECTIONS BE PLACED INTO TEMPLATES?>
===Beta===<!-- THIS SECTION HEADER IS LINKED FROM ANOTHER SECTION IN THIS
PAGE. IF YOU RENAME THE SECTION, YOU MUST UPDATE THE LINK(S). -->
====Fisher Matrix Bounds====
The parameter <math>\beta \,\!</math> must be positive, thus <math>\ln \beta \,\!</math> is approximately treated as being normally distributed.


====Bounds on Beta====
:<math>\frac{\ln (\hat{\beta })-\ln (\beta )}{\sqrt{Var\left[ \ln (\hat{\beta }) \right]}}\ \tilde{\ }\ N(0,1)\,\!</math>
=====Fisher Matrix Bounds=====
The parameter  <math>\beta \,\!</math>  must be positive, thus  <math>\ln \beta \,\!</math> is approximately treated as being normally distributed.


::<math>\frac{\ln (\widehat{\beta })-\ln (\beta )}{\sqrt{Var\left[ \ln (\widehat{\beta }) \right]}}\ \tilde{\ }\ N(0,1)</math>
:<math>C{{B}_{\beta }}=\hat{\beta }{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{\beta })}/\hat{\beta }}}\,\!</math>


::<math>C{{B}_{\beta }}=\widehat{\beta }{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\widehat{\beta })}/\widehat{\beta }}}</math>
:<math>\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>


::<math>\widehat{\beta }=\frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\widehat{\lambda }\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\left[ (T_{q}^{\widehat{\beta }}\ln ({{T}_{q}})-S_{q}^{\widehat{\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>\Lambda \,\!</math> is the natural log-likelihood function.


All variance can be calculated using the Fisher Information Matrix.
:<math>\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>\Lambda \,\!</math> is the natural log-likelihood function.
:<math>\frac{{{\partial }^{2}}\Lambda }{\partial {{\lambda }^{2}}}=-\frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{{{\lambda }^{2}}}\,\!</math>


::<math>\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>\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>\frac{{{\partial }^{2}}\Lambda }{\partial {{\lambda }^{2}}}=-\frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{{{\lambda }^{2}}}</math>
:<math>\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>


::<math>\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>
====Crow Bounds====
Calculate the conditional maximum likelihood estimate of <math>\tilde{\beta \,\!}\,\!</math> :


::<math>\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>
:<math>\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>


=====Crow Bounds=====
The Crow 2-sided <math>(1-a)\,\!</math> 100% confidence bounds on <math>\beta \,\!</math> are:  
Calculate the conditional maximum likelihood estimate of  <math>\tilde{\beta \,\!}</math> :


::<math>\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>
:<math>\begin{align}
 
The Crow 2-sided  <math>(1-a)\,\!</math> 100-percent confidence bounds on  <math>\beta \,\!</math>  are:
 
::<math>\begin{align}
   {{\beta }_{L}}= & \tilde{\beta }\frac{\chi _{\tfrac{\alpha }{2},2M}^{2}}{2M} \\  
   {{\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}   
   {{\beta }_{U}}= & \tilde{\beta }\frac{\chi _{1-\tfrac{\alpha }{2},2M}^{2}}{2M}   
\end{align}</math>
\end{align}\,\!</math>
 
===Lambda===
====Fisher Matrix Bounds====
The parameter <math>\lambda \,\!</math> must be positive, thus <math>\ln \lambda \,\!</math> is approximately treated as being normally distributed. These bounds are based on:


====Bounds on Lambda====
:<math>\frac{\ln (\hat{\lambda })-\ln (\lambda )}{\sqrt{Var\left[ \ln (\hat{\lambda }) \right]}}\ \tilde{\ }\ N(0,1)\,\!</math>
=====Fisher Matrix Bounds=====
The parameter  <math>\lambda </math>  must be positive, thus  <math>\ln \lambda </math> is approximately treated as being normally distributed. These bounds are based on:


The approximate confidence bounds on <math>\lambda \,\!</math> are given as:


::<math>\frac{\ln (\widehat{\lambda })-\ln (\lambda )}{\sqrt{Var\left[ \ln (\widehat{\lambda }) \right]}}\ \tilde{\ }\ N(0,1)</math>
:<math>C{{B}_{\lambda }}=\hat{\lambda }{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{\lambda })}/\hat{\lambda }}}\,\!</math>


<br>
where <math>\hat{\lambda }=\tfrac{n}{T_{K}^{{\hat{\beta }}}}\,\!</math>.
The approximate confidence bounds on  <math>\lambda </math> are given as:


The variance calculation is the same the equations given in the confidence bounds on [[Confidence_Bounds_for_Repairable_Systems_Analysis#Beta|Beta]].


::<math>C{{B}_{\lambda }}=\widehat{\lambda }{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\widehat{\lambda })}/\widehat{\lambda }}}</math>
====Crow Bounds====


'''Failure Terminated'''


where  <math>\widehat{\lambda }=\tfrac{n}{T_{K}^{{\hat{\beta }}}}</math> .
The confidence bounds on <math>\lambda \,\!</math> for failure terminated data are calculated using:
The variance calculation is the same as Eqns. (var1), (var2) and (var3).
<br>
<br>
=====Crow Bounds=====
''Time Terminated''
<br>
The confidence bounds on <math>\lambda </math>   for time terminated data are calculated using:


:<math>\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>


::<math>\begin{align}
where:
  & {{\lambda }_{L}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot \underset{q=1}{\overset{K}{\mathop{\sum }}}\,T_{q}^{^{\beta }}} \\  
*<math>N\,\!</math> = total number of failures.
& {{\lambda }_{u}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot \underset{q=1}{\overset{K}{\mathop{\sum }}}\,T_{q}^{^{\beta }}}
*<math>K\,\!</math> = number of systems.
\end{align}</math>
*<math>{{T}_{q}}\,\!</math> = end time for the <math>{{q}^{th}}</math> system.


'''Time Terminated'''


The confidence bounds on <math>\lambda \,\!</math> for time terminated data are calculated using:


''Failure Terminated''
:<math>\begin{align}
<br>
  {{\lambda }_{L}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot \underset{q=1}{\overset{K}{\mathop{\sum }}}\,T_{q}^{^{\beta }}} \\
The confidence bounds on <math>\lambda </math> for failure terminated data are calculated using:
  {{\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>


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


::<math>\begin{align}
===Cumulative Number of Failures===
  & {{\lambda }_{L}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot \underset{q=1}{\overset{K}{\mathop{\sum }}}\,T_{q}^{^{\beta }}} \\
====Fisher Matrix Bounds====
& {{\lambda }_{u}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N}^{2}}{2\cdot \underset{q=1}{\overset{K}{\mathop{\sum }}}\,T_{q}^{^{\beta }}} 
The cumulative number of failures, <math>N(t)\,\!</math>. must be positive, thus <math>\ln \left( N(t) \right)\,\!</math> is approximately treated as being normally distributed.
\end{align}</math>


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


====Bounds on Growth Rate====
:<math>N(t)=\hat{N}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{N}(t))}/\hat{N}(t)}}\,\!</math>
Since the growth rate is equal to  <math>1-\beta </math> , the confidence bounds are:


::<math>\begin{align}
where:  
  & Gr.\text{ }Rat{{e}_{L}}= & 1-{{\beta }_{U}} \\
& Gr.\text{ }Rat{{e}_{U}}= & 1-{{\beta }_{L}} 
\end{align}</math>


If Fisher Matrix confidence bounds are used then  <math>{{\beta }_{L}}</math>  and  <math>{{\beta }_{U}}</math>  are obtained from Eqn. (betafc). If Crow bounds are used then  <math>{{\beta }_{L}}</math>  and  <math>{{\beta }_{U}}</math>  are obtained from Eqn. (betacc).
:<math>\hat{N}(t)=\hat{\lambda }{{t}^{\hat{\beta }}}\,\!</math>
<br>
<br>


:<math>\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>


====Bounds on Cumulative MTBF====
The variance calculation is the same as the calculations in the confidence bounds on [[Confidence_Bounds_for_Repairable_Systems_Analysis#Beta|Beta]].
=====Fisher Matrix Bounds=====
The cumulative MTBF,  <math>{{m}_{c}}(t)</math> , must be positive, thus  <math>\ln {{m}_{c}}(t)</math>  is approximately treated as being normally distributed.  


::<math>\frac{\ln ({{\widehat{m}}_{c}}(t))-\ln ({{m}_{c}}(t))}{\sqrt{Var\left[ \ln ({{\widehat{m}}_{c}}(t)) \right]}}\ \tilde{\ }\ N(0,1)</math>
:<math>\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>


The approximate confidence bounds on the cumulative MTBF are then estimated from:
====Crow Bounds====
The 2-sided confidence bounds on the cumulative number of failures are given by:


:<math>N{{\left( t \right)}_{L}}=\frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot S}\,\!</math>


::<math>CB={{\widehat{m}}_{c}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{\widehat{m}}_{c}}(t))}/{{\widehat{m}}_{c}}(t)}}</math>
:<math>N{{\left( t \right)}_{U}}=\frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot S}\,\!</math>


:where:  
where:


::<math>{{\widehat{m}}_{c}}(t)=\frac{1}{\widehat{\lambda }}{{t}^{1-\widehat{\beta }}}</math>
*<math>N\,\!</math> = total number of failures across all systems. This is not the number of failures up to time <math>t\,\!</math>.
*<math>S=\frac{\left( \frac{N}{{\hat{\lambda }}} \right)}{{{t}^{{\hat{\beta }}}}}\,\!</math>
*<math>t\,\!</math> = time at which calculations are being conducted.


===Cumulative Failure Intensity===
====Fisher Matrix Bounds====
The cumulative failure intensity, <math>{{\lambda }_{c}}(t)\,\!</math> must be positive, thus <math>\ln {{\lambda }_{c}}(t)\,\!</math> is approximately treated as being normally distributed.


::<math>\begin{align}
:<math>\frac{\ln ({{\hat{\lambda }}_{c}}(t))-\ln ({{\lambda }_{c}}(t))}{\sqrt{Var\left[ \ln ({{\hat{\lambda }}_{c}}(t)) \right]}}\ \tilde{\ }\ N(0,1)\,\!</math>
  & Var({{\widehat{m}}_{c}}(t))= & {{\left( \frac{\partial {{m}_{c}}(t)}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial {{m}_{c}}(t)}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda }) \\
&  & +2\left( \frac{\partial {{m}_{c}}(t)}{\partial \beta } \right)\left( \frac{\partial {{m}_{c}}(t)}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda })\,
\end{align}</math>


The variance calculation is the same as Eqns. (var1), (var2) and (var3).
The approximate confidence bounds on the cumulative failure intensity are then estimated using:


::<math>\begin{align}
:<math>CB={{\hat{\lambda }}_{c}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{\hat{\lambda }}_{c}}(t))}/{{\hat{\lambda }}_{c}}(t)}}\,\!</math>
  & \frac{\partial {{m}_{c}}(t)}{\partial \beta }= & -\frac{1}{\widehat{\lambda }}{{t}^{1-\widehat{\beta }}}\ln (t) \\
& \frac{\partial {{m}_{c}}(t)}{\partial \lambda }= & -\frac{1}{{{\widehat{\lambda }}^{2}}}{{t}^{1-\widehat{\beta }}} 
\end{align}</math>


where:


=====Crow Bounds=====
:<math>{{\hat{\lambda }}_{c}}(t)=\hat{\lambda }{{t}^{\hat{\beta }-1}}\,\!</math>
To calculate the Crow confidence bounds on cumulative MTBF, first calculate the Crow cumulative failure intensity confidence bounds:


::<math>C{{(t)}_{L}}=\frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t}</math>
and:  


:<math>\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>


::<math>C{{(t)}_{u}}=\frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot t}</math>
The variance calculation is the same as the calculations in the confidence bounds on [[Confidence_Bounds_for_Repairable_Systems_Analysis#Beta|Beta]].


:Then
:<math>\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>


::<math>\begin{align}
====Crow Bounds====
  & {{[MTB{{F}_{c}}]}_{L}}= & \frac{1}{C{{(t)}_{U}}} \\
The 2-sided confidence bounds on the cumulative failure intensity are given by:
& {{[MTB{{F}_{c}}]}_{U}}= & \frac{1}{C{{(t)}_{L}}} 
\end{align}</math>


:<math>CFI_L=\frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t \cdot S}\,\!</math>


====Bounds on Instantaneous MTBF====
:<math>CFI_U=\frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot t \cdot S}\,\!</math>
=====Fisher Matrix Bounds=====
The instantaneous MTBF,  <math>{{m}_{i}}(t)</math> , must be positive, thus  <math>\ln {{m}_{i}}(t)</math> is approximately treated as being normally distributed.


::<math>\frac{\ln ({{\widehat{m}}_{i}}(t))-\ln ({{m}_{i}}(t))}{\sqrt{Var\left[ \ln ({{\widehat{m}}_{i}}(t)) \right]}}\ \tilde{\ }\ N(0,1)</math>
where:


*<math>N\,\!</math> = total number of failures across all systems. This is not the number of failures up to time <math>t\,\!</math>.
*<math>S=\frac{\left( \frac{N}{{\hat{\lambda }}} \right)}{{{t}^{{\hat{\beta }}}}}\,\!</math>
*<math>t\,\!</math> = time at which calculations are being conducted.


The approximate confidence bounds on the instantaneous MTBF are then estimated from:
===Cumulative MTBF===
====Fisher Matrix Bounds====
The cumulative MTBF, <math>{{m}_{c}}(t)\,\!</math>. must be positive, thus <math>\ln {{m}_{c}}(t)\,\!</math> is approximately treated as being normally distributed.


::<math>CB={{\widehat{m}}_{i}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{\widehat{m}}_{i}}(t))}/{{\widehat{m}}_{i}}(t)}}</math>
:<math>\frac{\ln ({{\hat{m}}_{c}}(t))-\ln ({{m}_{c}}(t))}{\sqrt{Var\left[ \ln ({{\hat{m}}_{c}}(t)) \right]}}\ \tilde{\ }\ N(0,1)\,\!</math>


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


::<math>{{\widehat{m}}_{i}}(t)=\frac{1}{\lambda \beta {{t}^{\beta -1}}}</math>
:<math>CB={{\hat{m}}_{c}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{\hat{m}}_{c}}(t))}/{{\hat{m}}_{c}}(t)}}\,\!</math>


where:  
::<math>\begin{align}
  & Var({{\widehat{m}}_{i}}(t))= & {{\left( \frac{\partial {{m}_{i}}(t)}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial {{m}_{i}}(t)}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda }) \\
&  & +2\left( \frac{\partial {{m}_{i}}(t)}{\partial \beta } \right)\left( \frac{\partial {{m}_{i}}(t)}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda }) 
\end{align}</math>


:<math>{{\hat{m}}_{c}}(t)=\frac{1}{\hat{\lambda }}{{t}^{1-\hat{\beta }}}\,\!</math>


The variance calculation is the same as (var1), (var2) and (var3).
:<math>\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>


::<math>\begin{align}
The variance calculation is the same as the calculations given in the confidence bounds on [[Confidence_Bounds_for_Repairable_Systems_Analysis#Beta|Beta]].
  & \frac{\partial {{m}_{i}}(t)}{\partial \beta }= & -\frac{1}{\widehat{\lambda }{{\widehat{\beta }}^{2}}}{{t}^{1-\widehat{\beta }}}-\frac{1}{\widehat{\lambda }\widehat{\beta }}{{t}^{1-\widehat{\beta }}}\ln (t) \\
& \frac{\partial {{m}_{i}}(t)}{\partial \lambda }= & -\frac{1}{{{\widehat{\lambda }}^{2}}\widehat{\beta }}{{t}^{1-\widehat{\beta }}} 
\end{align}</math>


:<math>\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=====
====Crow Bounds====
''Failure Terminated Data''
The 2-sided confidence bounds on the cumulative MTBF <math>(CMTBF)\,\!</math> are given by:
<br>
To calculate the bounds for failure terminated data, consider the following equation:  


::<math>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>
:<math>\begin{align}
& CMTBF_{L}=\frac{1}{CFI_{U}} \\  
& CMTBF_{U}=\frac{1}{CFI_{L}}
\end{align}\,\!</math>


Find the values  <math>{{p}_{1}}</math>  and  <math>{{p}_{2}}</math>  by finding the solution  <math>c</math>  to  <math>G({{n}^{2}}/c|n)=\xi </math>  for  <math>\xi =\tfrac{\alpha }{2}</math> and <math>\xi =1-\tfrac{\alpha }{2}</math> , respectively. If using the biased parameters,  <math>\hat{\beta }</math>  and  <math>\hat{\lambda }</math> , then the upper and lower confidence bounds are:
where <math>CFI_L\,\!</math> and <math>CFI_U\,\!</math> are calculated using the process for the confidence bounds on [[Confidence Bounds for Repairable Systems Analysis#Crow_Bounds_4|cumulative failure intensity]].


::<math>\begin{align}
===Instantaneous MTBF===<!-- THIS SECTION HEADER IS LINKED FROM ANOTHER SECTION IN THIS
  & {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot {{p}_{1}} \\  
PAGE. IF YOU RENAME THE SECTION, YOU MUST UPDATE THE LINK(S). -->
& {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot {{p}_{2}} 
====Fisher Matrix Bounds====
\end{align}</math>
The instantaneous MTBF, <math>{{m}_{i}}(t)\,\!</math>. must be positive, thus <math>\ln {{m}_{i}}(t)\,\!</math> is approximately treated as being normally distributed.


where  <math>MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}</math> . If using the unbiased parameters,  <math>\bar{\beta }</math>  and  <math>\bar{\lambda }</math> , then the upper and lower confidence bounds are:
:<math>\frac{\ln ({{\hat{m}}_{i}}(t))-\ln ({{m}_{i}}(t))}{\sqrt{Var\left[ \ln ({{\hat{m}}_{i}}(t)) \right]}}\ \tilde{\ }\ N(0,1)\,\!</math>


::<math>\begin{align}
The approximate confidence bounds on the instantaneous MTBF are then estimated from:  
  & {{[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>MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}</math> .
:<math>CB={{\hat{m}}_{i}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{\hat{m}}_{i}}(t))}/{{\hat{m}}_{i}}(t)}}\,\!</math>
<br>
<br>
''Time Terminated Data''
<br>
To calculate the bounds for time terminated data, consider the following equation where  <math>{{I}_{1}}(.)</math> is the modified Bessel function of order one:


::<math>H(x|k)=\underset{j=1}{\overset{k}{\mathop \sum }}\,\frac{{{x}^{2j-1}}}{{{2}^{2j-1}}(j-1)!j!{{I}_{1}}(x)}</math>
where:  


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


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


where  <math>MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}</math> . If using the unbiased parameters,  <math>\bar{\beta }</math>  and  <math>\bar{\lambda }</math> , then the upper and lower confidence bounds are:
The variance calculation is the same as the calculations given in the confidence bounds on [[Confidence_Bounds_for_Repairable_Systems_Analysis#Beta|Beta]].


::<math>\begin{align}
:<math>\begin{align}
  & {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot \left( \frac{N-1}{N} \right)\cdot {{\Pi }_{1}} \\  
  \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) \\  
& {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot \left( \frac{N-1}{N} \right)\cdot {{\Pi }_{2}}   
  \frac{\partial {{m}_{i}}(t)}{\partial \lambda }= & -\frac{1}{{{\hat{\lambda }}^{2}}\hat{\beta }}{{t}^{1-\hat{\beta }}}   
\end{align}</math>
\end{align}\,\!</math>


where  <math>MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}</math> .
====Crow Bounds====
'''Failure Terminated'''


For failure terminated data and the 2-sided confidence bounds on instantaneous MTBF <math>(IMTBF)\,\!</math>, consider the following equation:


====Bounds on Cumulative Failure Intensity====
:<math>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>
=====Fisher Matrix Bounds=====
The cumulative failure intensity,  <math>{{\lambda }_{c}}(t)</math>  must be positive, thus  <math>\ln {{\lambda }_{c}}(t)</math> is approximately treated as being normally distributed.


::<math>\frac{\ln ({{\widehat{\lambda }}_{c}}(t))-\ln ({{\lambda }_{c}}(t))}{\sqrt{Var\left[ \ln ({{\widehat{\lambda }}_{c}}(t)) \right]}}\ \tilde{\ }\ N(0,1)</math>
Find the values <math>{{p}_{1}}\,\!</math> and <math>{{p}_{2}}\,\!</math> by finding the solution
<math>G\left( \left. \frac{{{n}^{2}}}{c} \right|n \right)=\frac{\alpha }{2}</math> and <math>G\left( \left. \frac{{{n}^{2}}}{c} \right|n \right)=1-\frac{\alpha }{2}</math> for the lower and upper bounds, respectively.


The approximate confidence bounds on the cumulative failure intensity are then estimated using:  
If using the biased parameters, <math>\hat{\beta }\,\!</math> and <math>\hat{\lambda }\,\!</math>, then the upper and lower confidence bounds are:


::<math>CB={{\widehat{\lambda }}_{c}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{\widehat{\lambda }}_{c}}(t))}/{{\widehat{\lambda }}_{c}}(t)}}</math>
:<math>\begin{align}
  {{IMTBF}_{L}}= & IMTBF\cdot {{p}_{1}} \\
  {{IMTBF}_{U}}= & IMTBF\cdot {{p}_{2}}
\end{align}\,\!</math>


:where:
where <math>IMTBF=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}\,\!</math>.


::<math>{{\widehat{\lambda }}_{c}}(t)=\widehat{\lambda }{{t}^{\widehat{\beta }-1}}</math>
If using the unbiased parameters, <math>\bar{\beta }\,\!</math> and <math>\bar{\lambda }\,\!</math>, then the upper and lower confidence bounds are:


:and:
:<math>\begin{align}
  {{IMTBF}_{L}}= & IMTBF\cdot \left( \frac{N-2}{N} \right)\cdot {{p}_{1}} \\
  {{IMTBF}_{U}}= & IMTBF\cdot \left( \frac{N-2}{N} \right)\cdot {{p}_{2}} 
\end{align}\,\!</math>


::<math>\begin{align}
where <math>IMTBF=\tfrac{1}{\bar{\lambda }\bar{\beta }{{t}^{\bar{\beta }-1}}}\,\!</math>.
  & Var({{\widehat{\lambda }}_{c}}(t))= & {{\left( \frac{\partial {{\lambda }_{c}}(t)}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial {{\lambda }_{c}}(t)}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda }) \\
&  & +2\left( \frac{\partial {{\lambda }_{c}}(t)}{\partial \beta } \right)\left( \frac{\partial {{\lambda }_{c}}(t)}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda }) 
\end{align}</math>


'''Time Terminated'''


The variance calculation is the same as Eqns. (var1), (var2) and (var3):  
Consider the following equation where <math>{{I}_{1}}(.)\,\!</math> is the modified Bessel function of order one:  


::<math>\begin{align}
:<math>H(x|k)=\underset{j=1}{\overset{k}{\mathop \sum }}\,\frac{{{x}^{2j-1}}}{{{2}^{2j-1}}(j-1)!j!{{I}_{1}}(x)}\,\!</math>
  & \frac{\partial {{\lambda }_{c}}(t)}{\partial \beta }= & \widehat{\lambda }{{t}^{\widehat{\beta }-1}}\ln (t) \\
& \frac{\partial {{\lambda }_{c}}(t)}{\partial \lambda }= & {{t}^{\widehat{\beta }-1}}
\end{align}</math>


<br>
Find the values <math>{{\Pi }_{1}}\,\!</math> and <math>{{\Pi }_{2}}\,\!</math> by finding the solution <math>x\,\!</math> to <math>H(x|k)=\tfrac{\alpha }{2}\,\!</math> and <math>H(x|k)=1-\tfrac{\alpha }{2}\,\!</math> in the cases corresponding to the lower and upper bounds, respectively. Calculate <math>\Pi =\tfrac{4{{n}^{2}}}{{{x}^{2}}}\,\!</math> for each case.
=====Crow Bounds=====
The Crow cumulative failure intensity confidence bounds are given by:


::<math>C{{(t)}_{L}}=\frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t}</math>
If using the biased parameters, <math>\hat{\beta }\,\!</math> and <math>\hat{\lambda }\,\!</math>, then the upper and lower confidence bounds are:


:<math>\begin{align}
  {{IMTBF}_{L}}= & IMTBF\cdot {{\Pi }_{1}} \\
  {{IMTBF}_{U}}= & IMTBF\cdot {{\Pi }_{2}} 
\end{align}\,\!</math>


::<math>C{{(t)}_{u}}=\frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot t}</math>
where <math>IMTBF=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}\,\!</math>.


If using the unbiased parameters, <math>\bar{\beta }\,\!</math> and <math>\bar{\lambda }\,\!</math>, then the upper and lower confidence bounds are:


====Bounds on Instantaneous Failure Intensity====
:<math>\begin{align}
=====Fisher Matrix Bounds=====
  {{IMTBF}_{L}}= & IMTBF\cdot \left( \frac{N-1}{N} \right)\cdot {{\Pi }_{1}} \\
The instantaneous failure intensity,  <math>{{\lambda }_{i}}(t)</math> , must be positive, thus  <math>\ln {{\lambda }_{i}}(t)</math> is approximately treated as being normally distributed.
  {{IMTBF}_{U}}= & IMTBF\cdot \left( \frac{N-1}{N} \right)\cdot {{\Pi }_{2}} 
\end{align}\,\!</math>


::<math>\frac{\ln ({{\widehat{\lambda }}_{i}}(t))-\ln ({{\lambda }_{i}}(t))}{\sqrt{Var\left[ \ln ({{\widehat{\lambda }}_{i}}(t)) \right]}}\sim N(0,1)</math>
where <math>IMTBF=\tfrac{1}{\bar{\lambda }\bar{\beta }{{t}^{\bar{\beta }-1}}}\,\!</math>.


<br>
===Instantaneous Failure Intensity===<!-- THIS SECTION HEADER IS LINKED FROM ANOTHER SECTION IN THIS PAGE. IF YOU RENAME THE SECTION, YOU MUST UPDATE THE LINK(S). -->
The approximate confidence bounds on the instantaneous failure intensity are then estimated from:
====Fisher Matrix Bounds====
The instantaneous failure intensity, <math>{{\lambda }_{i}}(t)\,\!</math>. must be positive, thus <math>\ln {{\lambda }_{i}}(t)\,\!</math> is approximately treated as being normally distributed.


::<math>CB={{\widehat{\lambda }}_{i}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{\widehat{\lambda }}_{i}}(t))}/{{\widehat{\lambda }}_{i}}(t)}}</math>
:<math>\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:


where  <math>{{\lambda }_{i}}(t)=\lambda \beta {{t}^{\beta -1}}</math> and:
:<math>CB={{\hat{\lambda }}_{i}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{\hat{\lambda }}_{i}}(t))}/{{\hat{\lambda }}_{i}}(t)}}\,\!</math>


::<math>\begin{align}
where <math>{{\lambda }_{i}}(t)=\lambda \beta {{t}^{\beta -1}}\,\!</math> and:
  & Var({{\widehat{\lambda }}_{i}}(t))= & {{\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda }) \\
&  & +2\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \beta } \right)\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda }) 
\end{align}</math>


<br>
:<math>\begin{align}
The variance calculation is the same as Eqns. (var1), (var2) and (var3):
  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>


::<math>\begin{align}
The variance calculation is the same as the calculations in the confidence bounds on [[Confidence_Bounds_for_Repairable_Systems_Analysis#Beta|Beta]].
  & \frac{\partial {{\lambda }_{i}}(t)}{\partial \beta }= & \hat{\lambda }{{t}^{\widehat{\beta }-1}}+\hat{\lambda }\hat{\beta }{{t}^{\widehat{\beta }-1}}\ln (t) \\
& \frac{\partial {{\lambda }_{i}}(t)}{\partial \lambda }= & \widehat{\beta }{{t}^{\widehat{\beta }-1}} 
\end{align}</math>


:<math>\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=====
====Crow Bounds====
The Crow instantaneous failure intensity confidence bounds are given as:  
The 2-sided confidence bounds on the instantaneous failure intensity <math>(IFI)\,\!</math> are given by:


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


where <math>IMTB{{F}_{L}}\,\!</math> and <math>IMTB{{F}_{U}}\,\!</math> are calculated using the process presented for the confidence bounds on the [[Confidence Bounds for Repairable Systems Analysis#Instantaneous_MTBF|instantaneous MTBF]].


====Bounds on Time Given Cumulative MTBF====
===Time Given Cumulative Failure Intensity===
=====Fisher Matrix Bounds=====
====Fisher Matrix Bounds====
The time, <math>T</math> , must be positive, thus <math>\ln T</math> is approximately treated as being normally distributed.  
The time, <math>T\,\!</math>. must be positive, thus <math>\ln T\,\!</math> is approximately treated as being normally distributed.  


::<math>\frac{\ln (\widehat{T})-\ln (T)}{\sqrt{Var\left[ \ln (\widehat{T}) \right]}}\ \tilde{\ }\ N(0,1)</math>
:<math>\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:  
The confidence bounds on the time are given by:  


::<math>CB=\widehat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\widehat{T})}/\widehat{T}}}</math>
:<math>CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}}\,\!</math>
 
:where:
 
::<math>Var(\widehat{T})={{\left( \frac{\partial T}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial T}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda })+2\left( \frac{\partial T}{\partial \beta } \right)\left( \frac{\partial T}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda })</math>
 
The variance calculation is the same as Eqns. (var1), (var2) and (var3).
 
::<math>\widehat{T}={{(\lambda \cdot {{m}_{c}})}^{1/(1-\beta )}}</math>


where:


::<math>\begin{align}
:<math>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>
  & \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>


The variance calculation is the same as the calculations given in the confidence bounds on [[Confidence_Bounds_for_Repairable_Systems_Analysis#Beta|Beta]].


=====Crow Bounds=====
:<math>\hat{T}={{\left( \frac{{{\lambda }_{c}}(T)}{\lambda } \right)}^{1/(\beta -1)}}\,\!</math>
Step 1: Calculate:


::<math>\hat{T}={{\left( \frac{{{\lambda }_{c}}(T)}{{\hat{\lambda }}} \right)}^{\tfrac{1}{\beta -1}}}</math>
:<math>\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>


Step 2: Estimate the number of failures:
====Crow Bounds====
The 2-sided confidence bounds on time given cumulative failure intensity <math>(CFI)\,\!</math> are given by:


::<math>N(\hat{T})=\hat{\lambda }{{\hat{T}}^{{\hat{\beta }}}}</math>
:<math>\hat{t}={{\left( \frac{CFI}{{\hat{\lambda }}} \right)}^{\tfrac{1}{\hat{\beta }-1}}}\,\!</math>


Step 3: Obtain the confidence bounds on time given the cumulative failure intensity by solving for  <math>{{t}_{l}}</math> and  <math>{{t}_{u}}</math>  in the following equations:  
Then estimate, the number of failures, <math>N\,\!</math>, such that:


::<math>\begin{align}
:<math>N=\hat{\lambda }{{\hat{t}}^{{\hat{\beta }}}}\,\!</math>
  & {{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>


<br>
The lower and upper confidence bounds on time are then estimated using:


:<math>\begin{align}
  {{t}_{L}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot CFI} \\
  {{t}_{U}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot CFI} 
\end{align}\,\!</math>


====Bounds on Time Given Instantaneous MTBF====
===Time Given Cumulative MTBF===
=====Fisher Matrix Bounds=====
====Fisher Matrix Bounds====
The time, <math>T</math> , must be positive, thus <math>\ln T</math> is approximately treated as being normally distributed.  
The time, <math>T\,\!</math>. must be positive, thus <math>\ln T\,\!</math> is approximately treated as being normally distributed.  


::<math>\frac{\ln (\widehat{T})-\ln (T)}{\sqrt{Var\left[ \ln (\widehat{T}) \right]}}\ \tilde{\ }\ N(0,1)</math>
:<math>\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:  
The confidence bounds on the time are given by:  


::<math>CB=\widehat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\widehat{T})}/\widehat{T}}}</math>
:<math>CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}}\,\!</math>
 
:where:
 
::<math>Var(\widehat{T})={{\left( \frac{\partial T}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial T}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda })+2\left( \frac{\partial T}{\partial \beta } \right)\left( \frac{\partial T}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda })</math>
 
The variance calculation is the same as Eqns. (var1), (var2) and (var3).
 
 
::<math>\widehat{T}={{(\lambda \beta \cdot MTB{{F}_{i}})}^{1/(1-\beta )}}</math>
 
 
::<math>\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>


<br>
where:  
=====Crow Bounds=====
Step 1: Calculate the confidence bounds on the instantaneous MTBF as presented in Section 5.5.2.
<br>
Step 2: Calculate the bounds on time as follows.
<br>
<br>
''Failure Terminated Data''


::<math>\hat{T}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{c})}^{1/(1-\beta )}}</math>
:<math>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 confidence bounds on [[Confidence_Bounds_for_Repairable_Systems_Analysis#Beta|Beta]].


So the lower an upper bounds on time are:
:<math>\hat{T}={{(\lambda \cdot {{m}_{c}})}^{1/(1-\beta )}}\,\!</math>


:<math>\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>


::<math>{{\hat{T}}_{L}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{c}_{1}}})}^{1/(1-\beta )}}</math>
====Crow Bounds====
The 2-sided confidence bounds on time given cumulative MTBF <math>(CMTBF)\,\!</math> are estimated using the process for the confidence bounds on [[Confidence Bounds for Repairable Systems Analysis#Time_Given_Cumulative_Failure_Intensity|time given cumulative failure intensity]] <math>(CFI)\,\!</math> where <math>CFI=\frac{1}{CMTBF}\,\!</math>.


===Time Given Instantaneous MTBF===
====Fisher Matrix Bounds====
The time, <math>T\,\!</math>. must be positive, thus <math>\ln T\,\!</math> is approximately treated as being normally distributed.


::<math>{{\hat{T}}_{U}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{c}_{2}}})}^{1/(1-\beta )}}</math>
:<math>\frac{\ln (\hat{T})-\ln (T)}{\sqrt{Var\left[ \ln (\hat{T}) \right]}}\ \tilde{\ }\ N(0,1)\,\!</math>
 
 
''Time Terminated Data''
 
::<math>\hat{T}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{\Pi })}^{1/(1-\beta )}}</math>
 
 
So the lower and upper bounds on time are:
 
 
::<math>{{\hat{T}}_{L}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{\Pi }_{1}}})}^{1/(1-\beta )}}</math>
 
 
::<math>{{\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>T</math> , must be positive, thus  <math>\ln T</math>  is approximately treated as being normally distributed.
 
::<math>\frac{\ln (\widehat{T})-\ln (T)}{\sqrt{Var\left[ \ln \widehat{T} \right]}}\ \tilde{\ }\ N(0,1)</math>


The confidence bounds on the time are given by:  
The confidence bounds on the time are given by:  


::<math>CB=\widehat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\widehat{T})}/\widehat{T}}}</math>
:<math>CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}}\,\!</math>
 
:where:
 
::<math>Var(\widehat{T})={{\left( \frac{\partial T}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial T}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda })+2\left( \frac{\partial T}{\partial \beta } \right)\left( \frac{\partial T}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda })</math>
 
The variance calculation is the same as Eqns. (var1), (var2) and (var3):


::<math>\widehat{T}={{\left( \frac{{{\lambda }_{c}}(T)}{\lambda } \right)}^{1/(\beta -1)}}</math>
where:  


:<math>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>\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>


The variance calculation is the same as the calculations in the confidence bounds on [[Confidence_Bounds_for_Repairable_Systems_Analysis#Beta|Beta]].


=====Crow Bounds=====
:<math>\hat{T}={{(\lambda \beta \cdot MTB{{F}_{i}})}^{1/(1-\beta )}}\,\!</math>
Step 1: Calculate:


:<math>\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>


::<math>\hat{T}={{\left( \frac{{{\lambda }_{c}}(T)}{{\hat{\lambda }}} \right)}^{\tfrac{1}{\beta -1}}}</math>
====Crow Bounds====


'''Failure Terminated'''


Step 2: Estimate the number of failures:
Calculate the constants <math>p_1\,\!</math> and <math>p_2\,\!</math> using procedures described for the confidence bounds on [[Confidence Bounds for Repairable Systems Analysis#Instantaneous_MTBF|instantaneous MTBF]]. The lower and upper confidence bounds on time are then given by:


:<math>{{\hat{t}}_{L}}={{\left( \frac{\lambda \beta \cdot IMTBF}{{{p}_{1}}} \right)}^{\tfrac{1}{1-\beta }}}</math>


::<math>N(\hat{T})=\hat{\lambda }{{\hat{T}}^{{\hat{\beta }}}}</math>
:<math>{{\hat{t}}_{U}}={{\left( \frac{\lambda \beta \cdot IMTBF}{{{p}_{2}}} \right)}^{\tfrac{1}{1-\beta }}}</math>


'''Time Terminated'''


Step 3: Obtain the confidence bounds on time given the cumulative failure intensity by solving for  <math>{{t}_{l}}</math> and <math>{{t}_{u}}</math> in the following equations:  
Calculate the constants <math>{{\Pi }_{1}}\,\!</math> and <math>{{\Pi }_{2}}\,\!</math> using procedures described for the confidence bounds on [[Confidence Bounds for Repairable Systems Analysis#Instantaneous_MTBF|instantaneous MTBF]]. The lower and upper confidence bounds on time are then given by:


::<math>\begin{align}
:<math>{{\hat{t}}_{L}}={{\left( \frac{\lambda \beta \cdot IMTBF}{{{\Pi }_{1}}} \right)}^{\tfrac{1}{1-\beta }}}\,\!</math>
  & {{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>


:<math>{{\hat{t}}_{U}}={{\left( \frac{\lambda \beta \cdot IMTBF}{{{\Pi }_{2}}} \right)}^{\tfrac{1}{1-\beta }}}\,\!</math>


====Bounds on Time Given Instantaneous Failure Intensity====
===Time Given Instantaneous Failure Intensity===
=====Fisher Matrix Bounds=====
====Fisher Matrix Bounds====
These bounds are based on:  
These bounds are based on:  


::<math>\frac{\ln (\widehat{T})-\ln (T)}{\sqrt{Var\left[ \ln (\widehat{T}) \right]}}\sim N(0,1)</math>
:<math>\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:
The confidence bounds on the time are given by:


:<math>CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}}\,\!</math>


::<math>CB=\widehat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\widehat{T})}/\widehat{T}}}</math>
where:  


:where:
:<math>\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>


::<math>\begin{align}
The variance calculation is the same as the calculations given in the confidence bounds on [[Confidence_Bounds_for_Repairable_Systems_Analysis#Beta|Beta]].
  & Var(\widehat{T})= & {{\left( \frac{\partial T}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial T}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda }) \\
&  & +2\left( \frac{\partial T}{\partial \beta } \right)\left( \frac{\partial T}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda }) 
\end{align}</math>


The variance calculation is the same as Eqns. (var1), (var2) and (var3).
:<math>\hat{T}={{\left( \frac{{{\lambda }_{i}}(T)}{\lambda \cdot \beta } \right)}^{1/(\beta -1)}}\,\!</math>


::<math>\widehat{T}={{\left( \frac{{{\lambda }_{i}}(T)}{\lambda \cdot \beta } \right)}^{1/(\beta -1)}}</math>
:<math>\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====
The 2-sided confidence bounds on time given instantaneous failure intensity <math>(IFI)\,\!</math> are estimated using the process for the confidence bounds on [[Confidence Bounds for Repairable Systems Analysis#Time_Given_Instantaneous_MTBF|time given instantaneous MTBF]] where <math>IMTBF=\frac{1}{IFI}\,\!</math>.


::<math>\begin{align}
===Reliability===
  & \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 )}] \\
====Fisher Matrix Bounds====
& \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>{{\lambda }_{i}}(T)=\tfrac{1}{MTB{{F}_{i}}}</math> .
<br>
Step 2: Use the equations from 13.1.7.9 to calculate the bounds on time given the instantaneous failure intensity.
<br>
<br>
 
 
====Bounds on Reliability====
=====Fisher Matrix Bounds=====
These bounds are based on:  
These bounds are based on:  


::<math>\log it(\widehat{R}(t))\sim N(0,1)</math>
:<math>\log it(\hat{R}(t))\sim N(0,1)\,\!</math>
 
 
::<math>\log it(\widehat{R}(t))=\ln \left\{ \frac{\widehat{R}(t)}{1-\widehat{R}(t)} \right\}</math>


:<math>\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:  
The confidence bounds on reliability are given by:  


::<math>CB=\frac{\widehat{R}(t)}{\widehat{R}(t)+(1-\widehat{R}(t)){{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\widehat{R}(t))}/\left[ \widehat{R}(t)(1-\widehat{R}(t)) \right]}}}</math>
:<math>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>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>


::<math>Var(\widehat{R}(t))={{\left( \frac{\partial R}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial R}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda })+2\left( \frac{\partial R}{\partial \beta } \right)\left( \frac{\partial R}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda })</math>
The variance calculation is the same as the calculations in the confidence bounds on [[Confidence_Bounds_for_Repairable_Systems_Analysis#Beta|Beta]].


:<math>\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>


The variance calculation is the same as Eqns. (var1), (var2) and (var3).
====Crow Bounds====
'''Failure Terminated'''


::<math>\begin{align}
For failure terminated data, the 100( <math>1-\alpha \,\!</math> )% confidence interval on the current reliability at time <math>t\,\!</math> for a specified mission duration <math>d\,\!</math> is:
  & \frac{\partial R}{\partial \beta }= & {{e}^{-[\widehat{\lambda }{{(t+d)}^{\widehat{\beta }}}-\widehat{\lambda }{{t}^{\widehat{\beta }}}]}}[\lambda {{t}^{\widehat{\beta }}}\ln (t)-\lambda {{(t+d)}^{\widehat{\beta }}}\ln (t+d)] \\
& \frac{\partial R}{\partial \lambda }= & {{e}^{-[\widehat{\lambda }{{(t+d)}^{\widehat{\beta }}}-\widehat{\lambda }{{t}^{\widehat{\beta }}}]}}[{{t}^{\widehat{\beta }}}-{{(t+d)}^{\widehat{\beta }}}] 
\end{align}</math>


:<math>\left( {{\left[ \hat{R}\left( d \right) \right]}^{\tfrac{1}{{{p}_{1}}}}},{{\left[ \hat{R}\left( d \right) \right]}^{\tfrac{1}{{{p}_{2}}}}} \right)\,\!</math>


=====Crow Bounds=====
where:
''Failure Terminated Data''
<br>
With failure terminated data, the 100( <math>1-\alpha </math> )% confidence interval for the current reliability at time  <math>t</math>  in a specified mission time  <math>d</math>  is:  


::<math>({{[\widehat{R}(d)]}^{\tfrac{1}{{{p}_{1}}}}},{{[\hat{R}(d)]}^{\tfrac{1}{{{p}_{2}}}}})</math>
*<math>\hat{R}\left( d \right)={{e}^{-\left[ \hat{\lambda }{{\left( t+d \right)}^{{\hat{\beta }}}}-\hat{\lambda }{{t}^{{\hat{\beta }}}} \right]}}\,\!</math>
*<math>p_1\,\!</math> and <math>p_2\,\!</math> are obtained from the confidence bounds on [[Confidence Bounds for Repairable Systems Analysis#Instantaneous_MTBF|instantaneous MTBF]] for failure terminated data.


:where
'''Time Terminated'''


::<math>\widehat{R}(\tau )={{e}^{-[\widehat{\lambda }{{(t+\tau )}^{\widehat{\beta }}}-\widehat{\lambda }{{t}^{\widehat{\beta }}}]}}</math>
For time terminated data, the 100( <math>1-\alpha \,\!</math> )% confidence interval on the current reliability at time <math>t\,\!</math> for a specified mission duration <math>d\,\!</math> is:


<math>{{p}_{1}}</math> and  <math>{{p}_{2}}</math>  can be obtained from Eqn. (ft).
:<math>\left( {{\left[ \hat{R}\left( d \right) \right]}^{\tfrac{1}{{{\Pi }_{1}}}}},{{\left[ \hat{R}\left( d \right) \right]}^{\tfrac{1}{{{\Pi }_{2}}}}} \right)\,\!</math>  
<br>
<br>
''Time Terminated Data''
<br>
With time terminated data, the 100( <math>1-\alpha </math> )% confidence interval for the current reliability at time  <math>t</math>  in a specified mission time  <math>\tau </math> is:


::<math>({{[\widehat{R}(d)]}^{\tfrac{1}{{{p}_{1}}}}},{{[\hat{R}(d)]}^{\tfrac{1}{{{p}_{2}}}}})</math>
where:  


:where:
*<math>\hat{R}\left( d \right)={{e}^{-\left[ \hat{\lambda }{{\left( t+d \right)}^{{\hat{\beta }}}}-\hat{\lambda }{{t}^{{\hat{\beta }}}} \right]}}\,\!</math>
* <math>{{\Pi }_{1}}\,\!</math> and <math>{{\Pi }_{2}}\,\!</math> are obtained from the confidence bounds on [[Confidence Bounds for Repairable Systems Analysis#Instantaneous_MTBF|instantaneous MTBF]] for time terminated data.


::<math>\widehat{R}(d)={{e}^{-[\widehat{\lambda }{{(t+d)}^{\widehat{\beta }}}-\widehat{\lambda }{{t}^{\widehat{\beta }}}]}}</math>
===Time Given Reliability and Mission Time===
====Fisher Matrix Bounds====
The time, <math>t\,\!</math>. must be positive, thus <math>\ln t\,\!</math> is approximately treated as being normally distributed.


<math>{{p}_{1}}</math>  and  <math>{{p}_{2}}</math>  can be obtained from Eqn. (tt).
:<math>\frac{\ln (\hat{t})-\ln (t)}{\sqrt{Var\left[ \ln (\hat{t}) \right]}}\sim N(0,1)\,\!</math>
 
 
====Bounds on Time Given Reliability and Mission Time====
=====Fisher Matrix Bounds=====
The time,  <math>t</math> , must be positive, thus  <math>\ln t</math>  is approximately treated as being normally distributed.
 
::<math>\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:
The confidence bounds on time are calculated by using:


::<math>CB=\hat{t}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{t})}/\hat{t}}}</math>
:<math>CB=\hat{t}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{t})}/\hat{t}}}\,\!</math>


:where:
where:


::<math>Var(\hat{t})={{\left( \frac{\partial t}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial t}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda })+2\left( \frac{\partial t}{\partial \beta } \right)\left( \frac{\partial t}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda })</math>
:<math>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>\hat{t}</math> is calculated numerically from:
<math>\hat{t}\,\!</math> is calculated numerically from:


::<math>\widehat{R}(d)={{e}^{-[\widehat{\lambda }{{(\hat{t}+d)}^{\widehat{\beta }}}-\widehat{\lambda }{{{\hat{t}}}^{\widehat{\beta }}}]}}\text{ };\text{ }d\text{ = mission time}</math>
:<math>\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:
The variance calculations are done by:


::<math>\begin{align}
:<math>\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 \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}}}   
  \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>
\end{align}\,\!</math>


=====Crow Bounds=====
====Crow Bounds====
''Failure Terminated Data''
'''Failure Terminated'''
<br>
Step 1: Calculate  <math>({{\hat{R}}_{lower}},{{\hat{R}}_{upper}})=({{R}^{\tfrac{1}{{{p}_{1}}}}},{{R}^{\tfrac{1}{{{p}_{2}}}}})</math> .
<br>
Step 2: Let  <math>R={{\hat{R}}_{lower}}</math>  and solve for  <math>{{t}_{1}}</math>  numerically using  <math>R={{e}^{-[\widehat{\lambda }{{({{{\hat{t}}}_{1}}+d)}^{\widehat{\beta }}}-\widehat{\lambda }\hat{t}_{1}^{\widehat{\beta }}]}}</math> .
<br>
Step 3: Let  <math>R={{\hat{R}}_{upper}}</math>  and solve for  <math>{{t}_{2}}</math>  numerically using  <math>R={{e}^{-[\widehat{\lambda }{{({{{\hat{t}}}_{2}}+d)}^{\widehat{\beta }}}-\widehat{\lambda }\hat{t}_{2}^{\widehat{\beta }}]}}</math> .
<br>
Step 4: If  <math>{{t}_{1}}<{{t}_{2}}</math> , then  <math>{{t}_{lower}}={{t}_{1}}</math>  and  <math>{{t}_{upper}}={{t}_{2}}</math> . If  <math>{{t}_{1}}>{{t}_{2}}</math> , then  <math>{{t}_{lower}}={{t}_{2}}</math>  and  <math>{{t}_{upper}}={{t}_{1}}</math> .
<br>
<br>
''Time Terminated Data''
<br>
Step 1: Calculate  <math>({{\hat{R}}_{lower}},{{\hat{R}}_{upper}})=({{R}^{\tfrac{1}{{{\Pi }_{1}}}}},{{R}^{\tfrac{1}{{{\Pi }_{2}}}}})</math> .
<br>
Step 2: Let  <math>R={{\hat{R}}_{lower}}</math>  and solve for  <math>{{t}_{1}}</math>  numerically using  <math>R={{e}^{-[\widehat{\lambda }{{({{{\hat{t}}}_{1}}+d)}^{\widehat{\beta }}}-\widehat{\lambda }\hat{t}_{1}^{\widehat{\beta }}]}}</math> .
<br>
Step 3: Let  <math>R={{\hat{R}}_{upper}}</math>  and solve for  <math>{{t}_{2}}</math>  numerically using  <math>R={{e}^{-[\widehat{\lambda }{{({{{\hat{t}}}_{2}}+d)}^{\widehat{\beta }}}-\widehat{\lambda }\hat{t}_{2}^{\widehat{\beta }}]}}</math> .
<br>
Step 4: If  <math>{{t}_{1}}<{{t}_{2}}</math> , then  <math>{{t}_{lower}}={{t}_{1}}</math>  and  <math>{{t}_{upper}}={{t}_{2}}</math> . If  <math>{{t}_{1}}>{{t}_{2}}</math> , then  <math>{{t}_{lower}}={{t}_{2}}</math>  and  <math>{{t}_{upper}}={{t}_{1}}</math> .
<br>
<br>


For failure terminated data, the 2-sided confidence bounds on time given reliability and mission time estimated by calculating:


====Bounds on Mission Time Given Reliability and Time====
:<math>\left( {{{\hat{R}}}_{L}},{{{\hat{R}}}_{U}} \right)=\left( {{R}^{\tfrac{1}{{{p}_{1}}}}},{{R}^{\tfrac{1}{{{p}_{2}}}}}\, \right)\,\!</math>
=====Fisher Matrix Bounds=====
The mission time,  <math>d</math> , must be positive, thus  <math>\ln \left( d \right)</math> is approximately treated as being normally distributed.


::<math>\frac{\ln (\hat{d})-\ln (d)}{\sqrt{Var\left[ \ln (\hat{d}) \right]}}\sim N(0,1)</math>
where <math>p_1\,\!</math> and <math>p_2\,\!</math> are obtained from the confidence bounds on [[Confidence Bounds for Repairable Systems Analysis#Instantaneous_MTBF|instantaneous MTBF]] for failure terminated data.


Let <math>R={{\hat{R}}_{L}}\,\!</math> and solve numerically for <math>{{t}_{1}}\,\!</math> using <math>R={{e}^{-[\hat{\lambda }{{({{{\hat{t}}}_{1}}+d)}^{\hat{\beta }}}-\hat{\lambda }\hat{t}_{1}^{\hat{\beta }}]}}\,\!</math>.


The confidence bounds on mission time are given by using:
Let <math>R={{\hat{R}}_{U}}\,\!</math> and solve numerically for <math>{{t}_{2}}\,\!</math> using <math>R={{e}^{-[\hat{\lambda }{{({{{\hat{t}}}_{2}}+d)}^{\hat{\beta }}}-\hat{\lambda }\hat{t}_{2}^{\hat{\beta }}]}}\,\!</math>.


If <math>{{t}_{1}}<{{t}_{2}}\,\!</math> then <math>{{t}_{L}}={{t}_{1}}\,\!</math> and <math>{{t}_{U}}={{t}_{2}}\,\!</math>. If <math>{{t}_{1}}>{{t}_{2}}\,\!</math> then <math>{{t}_{L}}={{t}_{2}}\,\!</math> and <math>{{t}_{U}}={{t}_{1}}\,\!</math>.


::<math>CB=\hat{d}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{d})}/\hat{d}}}</math>
'''Time Terminated'''


For time terminated data, the 2-sided confidence bounds on time given reliability and mission time estimated by calculating:


:where:
:<math>\left( {{{\hat{R}}}_{L}},{{{\hat{R}}}_{U}} \right)=\left( {{R}^{\tfrac{1}{{{\Pi }_{1}}}}},{{R}^{\tfrac{1}{{{\Pi }_{2}}}}}\, \right)</math>.


where <math>\Pi_1\,\!</math> and <math>\Pi_2\,\!</math> are obtained from the confidence bounds on [[Confidence Bounds for Repairable Systems Analysis#Instantaneous_MTBF|instantaneous MTBF]] for time terminated data.


::<math>Var(\hat{d})={{\left( \frac{\partial d}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial d}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda })+2\left( \frac{\partial td}{\partial \beta } \right)\left( \frac{\partial d}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda })</math>
Let <math>R={{\hat{R}}_{L}}\,\!</math> and solve numerically for <math>{{t}_{1}}\,\!</math> using <math>R={{e}^{-[\hat{\lambda }{{({{{\hat{t}}}_{1}}+d)}^{\hat{\beta }}}-\hat{\lambda }\hat{t}_{1}^{\hat{\beta }}]}}\,\!</math>.


Let <math>R={{\hat{R}}_{U}}\,\!</math> and solve numerically for <math>{{t}_{2}}\,\!</math> using <math>R={{e}^{-[\hat{\lambda }{{({{{\hat{t}}}_{2}}+d)}^{\hat{\beta }}}-\hat{\lambda }\hat{t}_{2}^{\hat{\beta }}]}}\,\!</math>.


Calculate  <math>\hat{d}</math> from:
If <math>{{t}_{1}}<{{t}_{2}}\,\!</math>. then <math>{{t}_{L}}={{t}_{1}}\,\!</math> and <math>{{t}_{U}}={{t}_{2}}\,\!</math>. If <math>{{t}_{1}}>{{t}_{2}}\,\!</math>. then <math>{{t}_{L}}={{t}_{2}}\,\!</math> and <math>{{t}_{U}}={{t}_{1}}\,\!</math>.


===Mission Time Given Reliability and Time===
====Fisher Matrix Bounds====
The mission time, <math>d\,\!</math>. must be positive, thus <math>\ln \left( d \right)\,\!</math> is approximately treated as being normally distributed.


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


The variance calculations are done by:
:<math>CB=\hat{d}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{d})}/\hat{d}}}\,\!</math>


where:


::<math>\begin{align}
:<math>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>
  & \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>


Calculate <math>\hat{d}\,\!</math> from:


=====Crow Bounds=====
:<math>\hat{d}={{\left[ {{t}^{{\hat{\beta }}}}-\frac{\ln (R)}{{\hat{\lambda }}} \right]}^{\tfrac{1}{{\hat{\beta }}}}}-t\,\!</math>
''Failure Terminated Data''
<br>
Step 1: Calculate  <math>({{\hat{R}}_{lower}},{{\hat{R}}_{upper}})=({{R}^{\tfrac{1}{{{p}_{1}}}}},{{R}^{\tfrac{1}{{{p}_{2}}}}})</math> .
<br>
Step 2: Let  <math>R={{\hat{R}}_{lower}}</math>  and solve for  <math>{{d}_{1}}</math>  such that:
 
 
::<math>{{d}_{1}}={{\left( {{t}^{{\hat{\beta }}}}-\frac{\ln ({{R}_{lower}})}{{\hat{\lambda }}} \right)}^{\tfrac{1}{{\hat{\beta }}}}}-t</math>
 
 
Step 3: Let  <math>R={{\hat{R}}_{upper}}</math>  and solve for  <math>{{d}_{2}}</math>  such that:
 
 
::<math>{{d}_{2}}={{\left( {{t}^{{\hat{\beta }}}}-\frac{\ln ({{R}_{upper}})}{{\hat{\lambda }}} \right)}^{\tfrac{1}{{\hat{\beta }}}}}-t</math>
 
 
Step 4: If  <math>{{d}_{1}}<{{d}_{2}}</math> , then  <math>{{d}_{lower}}={{d}_{1}}</math>  and  <math>{{d}_{upper}}={{d}_{2}}</math> . If  <math>{{d}_{1}}>{{d}_{2}}</math> , then  <math>{{d}_{lower}}={{d}_{2}}</math>  and  <math>{{d}_{upper}}={{d}_{1}}</math> .
<br>
<br>
''Time Terminated Data''
<br>
Step 1: Calculate  <math>({{\hat{R}}_{lower}},{{\hat{R}}_{upper}})=({{R}^{\tfrac{1}{{{\Pi }_{1}}}}},{{R}^{\tfrac{1}{{{\Pi }_{2}}}}})</math> .
<br>
Step 2: Let  <math>R={{\hat{R}}_{lower}}</math>  and solve for  <math>{{d}_{1}}</math>  using Eqn. (CBR1).
<br>
Step 3: Let  <math>R={{\hat{R}}_{upper}}</math>  and solve for  <math>{{d}_{2}}</math>  using Eqn. (CBR2).
<br>
Step 4: If  <math>{{d}_{1}}<{{d}_{2}}</math> , then  <math>{{d}_{lower}}={{d}_{1}}</math>  and  <math>{{d}_{upper}}={{d}_{2}}</math> . If  <math>{{d}_{1}}>{{d}_{2}}</math> , then  <math>{{d}_{lower}}={{d}_{2}}</math>  and  <math>{{d}_{upper}}={{d}_{1}}</math> .
<br>
<br>


The variance calculations are done by:


====Bounds on Cumulative Number of Failures====
:<math>\begin{align}
=====Fisher Matrix Bounds=====
  \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 }}} \\
The cumulative number of failures,  <math>N(t)</math> , must be positive, thus  <math>\ln \left( N(t) \right)</math> is approximately treated as being normally distributed.
  \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>


::<math>\frac{\ln (\widehat{N}(t))-\ln (N(t))}{\sqrt{Var\left[ \ln \widehat{N}(t) \right]}}\sim N(0,1)</math>
====Crow Bounds====
'''Failure Terminated'''


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


::<math>N(t)=\widehat{N}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\widehat{N}(t))}/\widehat{N}(t)}}</math>
Step 2: Let <math>R={{\hat{R}}_{lower}}\,\!</math> and solve for <math>{{d}_{1}}\,\!</math> such that:


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


:where:  
Step 3: Let <math>R={{\hat{R}}_{upper}}\,\!</math> and solve for <math>{{d}_{2}}\,\!</math> such that:


::<math>\widehat{N}(t)=\widehat{\lambda }{{t}^{\widehat{\beta }}}</math>
:<math>{{d}_{2}}={{\left( {{t}^{{\hat{\beta }}}}-\frac{\ln ({{R}_{upper}})}{{\hat{\lambda }}} \right)}^{\tfrac{1}{{\hat{\beta }}}}}-t\,\!</math>


<br>
Step 4: If <math>{{d}_{1}}<{{d}_{2}}\,\!</math>. then <math>{{d}_{lower}}={{d}_{1}}\,\!</math> and <math>{{d}_{upper}}={{d}_{2}}\,\!</math>. If <math>{{d}_{1}}>{{d}_{2}}\,\!</math>. then <math>{{d}_{lower}}={{d}_{2}}\,\!</math> and <math>{{d}_{upper}}={{d}_{1}}\,\!</math>.
::<math>\begin{align}
  & Var(\widehat{N}(t))= & {{\left( \frac{\partial N(t)}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial N(t)}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda }) \\
&  & +2\left( \frac{\partial N(t)}{\partial \beta } \right)\left( \frac{\partial N(t)}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda }) 
\end{align}</math>


'''Time Terminated'''


The variance calculation is the same as Eqns. (var1), (var2) and (var3).  
Step 1: Calculate <math>({{\hat{R}}_{lower}},{{\hat{R}}_{upper}})=({{R}^{\tfrac{1}{{{\Pi }_{1}}}}},{{R}^{\tfrac{1}{{{\Pi }_{2}}}}})\,\!</math>.


<br>
Step 2: Let <math>R={{\hat{R}}_{lower}}\,\!</math> and solve for <math>{{d}_{1}}\,\!</math> using the same equation given for the failure terminated data.
::<math>\begin{align}
  & \frac{\partial N(t)}{\partial \beta }= & \hat{\lambda }{{t}^{\widehat{\beta }}}\ln (t) \\
& \frac{\partial N(t)}{\partial \lambda }= & t\widehat{\beta } 
\end{align}</math>


<br>
Step 3: Let <math>R={{\hat{R}}_{upper}}\,\!</math> and solve for <math>{{d}_{2}}\,\!</math> using the same equation given for the failure terminated data.
=====Crow Bounds=====
::<math>\begin{array}{*{35}{l}}
  {{N}_{L}}(T)=\tfrac{T}{\widehat{\beta }}{{\lambda }_{i}}{{(T)}_{L}} \\
  {{N}_{U}}(T)=\tfrac{T}{\widehat{\beta }}{{\lambda }_{i}}{{(T)}_{U}}  \\
\end{array}</math>


where  <math>{{\lambda }_{i}}{{(T)}_{L}}</math> and <math>{{\lambda }_{i}}{{(T)}_{U}}</math> can be obtained from Eqn. (inr).
Step 4: If <math>{{d}_{1}}<{{d}_{2}}\,\!</math>. then <math>{{d}_{lower}}={{d}_{1}}\,\!</math> and <math>{{d}_{upper}}={{d}_{2}}\,\!</math>. If <math>{{d}_{1}}>{{d}_{2}}\,\!</math>. then <math>{{d}_{lower}}={{d}_{2}}\,\!</math> and <math>{{d}_{upper}}={{d}_{1}}\,\!</math>.

Latest revision as of 20:42, 18 September 2023

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Chapter Appendix E: Confidence Bounds for Repairable Systems Analysis


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Chapter Appendix E  
Confidence Bounds for Repairable Systems Analysis  

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Available Software:
RGA

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More Resources:
RGA examples

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. 

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]

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 confidence bounds on Beta.

Crow Bounds

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]

where:

  • [math]\displaystyle{ N\,\! }[/math] = total number of failures.
  • [math]\displaystyle{ K\,\! }[/math] = number of systems.
  • [math]\displaystyle{ {{T}_{q}}\,\! }[/math] = end time for the [math]\displaystyle{ {{q}^{th}} }[/math] system.

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]

where:

  • [math]\displaystyle{ N\,\! }[/math] = total number of failures.
  • [math]\displaystyle{ K\,\! }[/math] = number of systems.
  • [math]\displaystyle{ {{T}_{q}}\,\! }[/math] = end time for the [math]\displaystyle{ {{q}^{th}} }[/math] system.

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 confidence bounds on Beta.

[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

The 2-sided confidence bounds on the cumulative number of failures are given by:

[math]\displaystyle{ N{{\left( t \right)}_{L}}=\frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot S}\,\! }[/math]
[math]\displaystyle{ N{{\left( t \right)}_{U}}=\frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot S}\,\! }[/math]

where:

  • [math]\displaystyle{ N\,\! }[/math] = total number of failures across all systems. This is not the number of failures up to time [math]\displaystyle{ t\,\! }[/math].
  • [math]\displaystyle{ S=\frac{\left( \frac{N}{{\hat{\lambda }}} \right)}{{{t}^{{\hat{\beta }}}}}\,\! }[/math]
  • [math]\displaystyle{ t\,\! }[/math] = time at which calculations are being conducted.

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 confidence bounds on Beta.

[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 2-sided confidence bounds on the cumulative failure intensity are given by:

[math]\displaystyle{ CFI_L=\frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t \cdot S}\,\! }[/math]
[math]\displaystyle{ CFI_U=\frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot t \cdot S}\,\! }[/math]

where:

  • [math]\displaystyle{ N\,\! }[/math] = total number of failures across all systems. This is not the number of failures up to time [math]\displaystyle{ t\,\! }[/math].
  • [math]\displaystyle{ S=\frac{\left( \frac{N}{{\hat{\lambda }}} \right)}{{{t}^{{\hat{\beta }}}}}\,\! }[/math]
  • [math]\displaystyle{ t\,\! }[/math] = time at which calculations are being conducted.

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 confidence bounds on Beta.

[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

The 2-sided confidence bounds on the cumulative MTBF [math]\displaystyle{ (CMTBF)\,\! }[/math] are given by:

[math]\displaystyle{ \begin{align} & CMTBF_{L}=\frac{1}{CFI_{U}} \\ & CMTBF_{U}=\frac{1}{CFI_{L}} \end{align}\,\! }[/math]

where [math]\displaystyle{ CFI_L\,\! }[/math] and [math]\displaystyle{ CFI_U\,\! }[/math] are calculated using the process for the confidence bounds on cumulative failure intensity.

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 confidence bounds on Beta.

[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

For failure terminated data and the 2-sided confidence bounds on instantaneous MTBF [math]\displaystyle{ (IMTBF)\,\! }[/math], 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{ G\left( \left. \frac{{{n}^{2}}}{c} \right|n \right)=\frac{\alpha }{2} }[/math] and [math]\displaystyle{ G\left( \left. \frac{{{n}^{2}}}{c} \right|n \right)=1-\frac{\alpha }{2} }[/math] for the lower and upper bounds, 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} {{IMTBF}_{L}}= & IMTBF\cdot {{p}_{1}} \\ {{IMTBF}_{U}}= & IMTBF\cdot {{p}_{2}} \end{align}\,\! }[/math]

where [math]\displaystyle{ IMTBF=\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} {{IMTBF}_{L}}= & IMTBF\cdot \left( \frac{N-2}{N} \right)\cdot {{p}_{1}} \\ {{IMTBF}_{U}}= & IMTBF\cdot \left( \frac{N-2}{N} \right)\cdot {{p}_{2}} \end{align}\,\! }[/math]

where [math]\displaystyle{ IMTBF=\tfrac{1}{\bar{\lambda }\bar{\beta }{{t}^{\bar{\beta }-1}}}\,\! }[/math].

Time Terminated

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{4{{n}^{2}}}{{{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} {{IMTBF}_{L}}= & IMTBF\cdot {{\Pi }_{1}} \\ {{IMTBF}_{U}}= & IMTBF\cdot {{\Pi }_{2}} \end{align}\,\! }[/math]

where [math]\displaystyle{ IMTBF=\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} {{IMTBF}_{L}}= & IMTBF\cdot \left( \frac{N-1}{N} \right)\cdot {{\Pi }_{1}} \\ {{IMTBF}_{U}}= & IMTBF\cdot \left( \frac{N-1}{N} \right)\cdot {{\Pi }_{2}} \end{align}\,\! }[/math]

where [math]\displaystyle{ IMTBF=\tfrac{1}{\bar{\lambda }\bar{\beta }{{t}^{\bar{\beta }-1}}}\,\! }[/math].

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 confidence bounds on Beta.

[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 2-sided confidence bounds on the instantaneous failure intensity [math]\displaystyle{ (IFI)\,\! }[/math] are given by:

[math]\displaystyle{ \begin{align} {IFI_{L}}= & \frac{1}{{IMTBF}_{U}} \\ {IFI_{U}}= & \frac{1}{{IMTBF}_{L}} \end{align}\,\! }[/math]

where [math]\displaystyle{ IMTB{{F}_{L}}\,\! }[/math] and [math]\displaystyle{ IMTB{{F}_{U}}\,\! }[/math] are calculated using the process presented for the confidence bounds on the instantaneous MTBF.

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 confidence bounds on Beta.

[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

The 2-sided confidence bounds on time given cumulative failure intensity [math]\displaystyle{ (CFI)\,\! }[/math] are given by:

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

Then estimate, the number of failures, [math]\displaystyle{ N\,\! }[/math], such that:

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

The lower and upper confidence bounds on time are then estimated using:

[math]\displaystyle{ \begin{align} {{t}_{L}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot CFI} \\ {{t}_{U}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot CFI} \end{align}\,\! }[/math]

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 confidence bounds on Beta.

[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

The 2-sided confidence bounds on time given cumulative MTBF [math]\displaystyle{ (CMTBF)\,\! }[/math] are estimated using the process for the confidence bounds on time given cumulative failure intensity [math]\displaystyle{ (CFI)\,\! }[/math] where [math]\displaystyle{ CFI=\frac{1}{CMTBF}\,\! }[/math].

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 confidence bounds on Beta.

[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

Failure Terminated

Calculate the constants [math]\displaystyle{ p_1\,\! }[/math] and [math]\displaystyle{ p_2\,\! }[/math] using procedures described for the confidence bounds on instantaneous MTBF. The lower and upper confidence bounds on time are then given by:

[math]\displaystyle{ {{\hat{t}}_{L}}={{\left( \frac{\lambda \beta \cdot IMTBF}{{{p}_{1}}} \right)}^{\tfrac{1}{1-\beta }}} }[/math]
[math]\displaystyle{ {{\hat{t}}_{U}}={{\left( \frac{\lambda \beta \cdot IMTBF}{{{p}_{2}}} \right)}^{\tfrac{1}{1-\beta }}} }[/math]

Time Terminated

Calculate the constants [math]\displaystyle{ {{\Pi }_{1}}\,\! }[/math] and [math]\displaystyle{ {{\Pi }_{2}}\,\! }[/math] using procedures described for the confidence bounds on instantaneous MTBF. The lower and upper confidence bounds on time are then given by:

[math]\displaystyle{ {{\hat{t}}_{L}}={{\left( \frac{\lambda \beta \cdot IMTBF}{{{\Pi }_{1}}} \right)}^{\tfrac{1}{1-\beta }}}\,\! }[/math]
[math]\displaystyle{ {{\hat{t}}_{U}}={{\left( \frac{\lambda \beta \cdot IMTBF}{{{\Pi }_{2}}} \right)}^{\tfrac{1}{1-\beta }}}\,\! }[/math]

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 confidence bounds on Beta.

[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

The 2-sided confidence bounds on time given instantaneous failure intensity [math]\displaystyle{ (IFI)\,\! }[/math] are estimated using the process for the confidence bounds on time given instantaneous MTBF where [math]\displaystyle{ IMTBF=\frac{1}{IFI}\,\! }[/math].

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 confidence bounds on Beta.

[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

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

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

where:

  • [math]\displaystyle{ \hat{R}\left( d \right)={{e}^{-\left[ \hat{\lambda }{{\left( t+d \right)}^{{\hat{\beta }}}}-\hat{\lambda }{{t}^{{\hat{\beta }}}} \right]}}\,\! }[/math]
  • [math]\displaystyle{ p_1\,\! }[/math] and [math]\displaystyle{ p_2\,\! }[/math] are obtained from the confidence bounds on instantaneous MTBF for failure terminated data.

Time Terminated

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

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

where:

  • [math]\displaystyle{ \hat{R}\left( d \right)={{e}^{-\left[ \hat{\lambda }{{\left( t+d \right)}^{{\hat{\beta }}}}-\hat{\lambda }{{t}^{{\hat{\beta }}}} \right]}}\,\! }[/math]
  • [math]\displaystyle{ {{\Pi }_{1}}\,\! }[/math] and [math]\displaystyle{ {{\Pi }_{2}}\,\! }[/math] are obtained from the confidence bounds on instantaneous MTBF for time terminated data.

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

For failure terminated data, the 2-sided confidence bounds on time given reliability and mission time estimated by calculating:

[math]\displaystyle{ \left( {{{\hat{R}}}_{L}},{{{\hat{R}}}_{U}} \right)=\left( {{R}^{\tfrac{1}{{{p}_{1}}}}},{{R}^{\tfrac{1}{{{p}_{2}}}}}\, \right)\,\! }[/math]

where [math]\displaystyle{ p_1\,\! }[/math] and [math]\displaystyle{ p_2\,\! }[/math] are obtained from the confidence bounds on instantaneous MTBF for failure terminated data.

Let [math]\displaystyle{ R={{\hat{R}}_{L}}\,\! }[/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].

Let [math]\displaystyle{ R={{\hat{R}}_{U}}\,\! }[/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].

If [math]\displaystyle{ {{t}_{1}}\lt {{t}_{2}}\,\! }[/math] then [math]\displaystyle{ {{t}_{L}}={{t}_{1}}\,\! }[/math] and [math]\displaystyle{ {{t}_{U}}={{t}_{2}}\,\! }[/math]. If [math]\displaystyle{ {{t}_{1}}\gt {{t}_{2}}\,\! }[/math] then [math]\displaystyle{ {{t}_{L}}={{t}_{2}}\,\! }[/math] and [math]\displaystyle{ {{t}_{U}}={{t}_{1}}\,\! }[/math].

Time Terminated

For time terminated data, the 2-sided confidence bounds on time given reliability and mission time estimated by calculating:

[math]\displaystyle{ \left( {{{\hat{R}}}_{L}},{{{\hat{R}}}_{U}} \right)=\left( {{R}^{\tfrac{1}{{{\Pi }_{1}}}}},{{R}^{\tfrac{1}{{{\Pi }_{2}}}}}\, \right) }[/math].

where [math]\displaystyle{ \Pi_1\,\! }[/math] and [math]\displaystyle{ \Pi_2\,\! }[/math] are obtained from the confidence bounds on instantaneous MTBF for time terminated data.

Let [math]\displaystyle{ R={{\hat{R}}_{L}}\,\! }[/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].

Let [math]\displaystyle{ R={{\hat{R}}_{U}}\,\! }[/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].

If [math]\displaystyle{ {{t}_{1}}\lt {{t}_{2}}\,\! }[/math]. then [math]\displaystyle{ {{t}_{L}}={{t}_{1}}\,\! }[/math] and [math]\displaystyle{ {{t}_{U}}={{t}_{2}}\,\! }[/math]. If [math]\displaystyle{ {{t}_{1}}\gt {{t}_{2}}\,\! }[/math]. then [math]\displaystyle{ {{t}_{L}}={{t}_{2}}\,\! }[/math] and [math]\displaystyle{ {{t}_{U}}={{t}_{1}}\,\! }[/math].

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

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

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].

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