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===Confidence Bounds for Repairable Systems Analysis===
#REDIRECT [[RGA_Models_for_Repairable_Systems_Analysis#Confidence_Bounds_for_Repairable_Systems_Analysis]]
{{bounds on beta rsa}}
 
{{bounds on lambda rsa}}
 
{{bounds on growth rate rsa}}
 
{{bounds on cumulative mtbf rsa}}
 
{{bounds on instantaneous mtbf rsa}}
 
{{bounds on cumulative failure intensity rsa}}
 
{{bounds on instantaneous failure intensity rsa}}
 
{{bounds on time given cumulative mtbf rsa}}
 
====Bounds on 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>\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:
 
::<math>CB=\widehat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\widehat{T})}/\widehat{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>
=====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>
 
 
So the lower an upper bounds on time are:
 
 
::<math>{{\hat{T}}_{L}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{c}_{1}}})}^{1/(1-\beta )}}</math>
 
 
::<math>{{\hat{T}}_{U}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{c}_{2}}})}^{1/(1-\beta )}}</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:
 
::<math>CB=\widehat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\widehat{T})}/\widehat{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>
 
::<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>
 
 
=====Crow Bounds=====
Step 1: Calculate:
 
 
::<math>\hat{T}={{\left( \frac{{{\lambda }_{c}}(T)}{{\hat{\lambda }}} \right)}^{\tfrac{1}{\beta -1}}}</math>
 
 
Step 2: Estimate the number of failures:
 
 
::<math>N(\hat{T})=\hat{\lambda }{{\hat{T}}^{{\hat{\beta }}}}</math>
 
 
Step 3: Obtain the confidence bounds on time given the cumulative failure intensity by solving for  <math>{{t}_{l}}</math>  and  <math>{{t}_{u}}</math>  in the following equations:
 
::<math>\begin{align}
  & {{t}_{l}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot {{\lambda }_{c}}(T)} \\
& {{t}_{u}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot {{\lambda }_{c}}(T)} 
\end{align}</math>
 
 
====Bounds on Time Given Instantaneous Failure Intensity====
=====Fisher Matrix Bounds=====
These bounds are based on:
 
::<math>\frac{\ln (\widehat{T})-\ln (T)}{\sqrt{Var\left[ \ln (\widehat{T}) \right]}}\sim N(0,1)</math>
 
 
The confidence bounds on the time are given by:
 
 
::<math>CB=\widehat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\widehat{T})}/\widehat{T}}}</math>
 
:where:
 
::<math>\begin{align}
  & 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>\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=====
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:
 
::<math>\log it(\widehat{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>
 
 
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>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 Eqns. (var1), (var2) and (var3).
 
::<math>\begin{align}
  & \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>
 
 
=====Crow Bounds=====
''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>
 
:where
 
::<math>\widehat{R}(\tau )={{e}^{-[\widehat{\lambda }{{(t+\tau )}^{\widehat{\beta }}}-\widehat{\lambda }{{t}^{\widehat{\beta }}}]}}</math>
 
<math>{{p}_{1}}</math> and  <math>{{p}_{2}}</math>  can be obtained from Eqn. (ft).
<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:
 
::<math>\widehat{R}(d)={{e}^{-[\widehat{\lambda }{{(t+d)}^{\widehat{\beta }}}-\widehat{\lambda }{{t}^{\widehat{\beta }}}]}}</math>
 
<math>{{p}_{1}}</math>  and  <math>{{p}_{2}}</math>  can be obtained from Eqn. (tt).
 
====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:
 
::<math>CB=\hat{t}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{t})}/\hat{t}}}</math>
 
: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>\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>
 
The variance calculations are done by:
 
::<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 \lambda }= & \frac{{{{\hat{t}}}^{{\hat{\beta }}}}-{{(\hat{t}+d)}^{{\hat{\beta }}}}}{\hat{\lambda }\hat{\beta }{{(\hat{t}+d)}^{\hat{\beta }-1}}-\hat{\lambda }\hat{\beta }{{{\hat{t}}}^{\hat{\beta }-1}}} 
\end{align}</math>
 
=====Crow Bounds=====
''Failure Terminated Data''
<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>
====Bounds on 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>\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>CB=\hat{d}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{d})}/\hat{d}}}</math>
 
 
:where:
 
 
::<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>
 
 
Calculate  <math>\hat{d}</math>  from:
 
 
::<math>\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>\begin{align}
  & \frac{\partial d}{\partial \beta }= & \left[ \frac{{{t}^{{\hat{\beta }}}}\ln (t)}{{{(t+\hat{d})}^{{\hat{\beta }}}}}-\ln (t+\hat{d}) \right]\cdot \frac{t+\hat{d}}{{\hat{\beta }}} \\
& \frac{\partial d}{\partial \lambda }= & \frac{{{t}^{{\hat{\beta }}}}-{{(t+\hat{d})}^{{\hat{\beta }}}}}{\hat{\lambda }\hat{\beta }{{(t+\hat{d})}^{\hat{\beta }-1}}} 
\end{align}</math>
 
 
=====Crow Bounds=====
''Failure Terminated Data''
<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>
====Bounds on Cumulative Number of Failures====
=====Fisher Matrix Bounds=====
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.
 
::<math>\frac{\ln (\widehat{N}(t))-\ln (N(t))}{\sqrt{Var\left[ \ln \widehat{N}(t) \right]}}\sim N(0,1)</math>
 
 
::<math>N(t)=\widehat{N}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\widehat{N}(t))}/\widehat{N}(t)}}</math>
 
 
:where:
 
::<math>\widehat{N}(t)=\widehat{\lambda }{{t}^{\widehat{\beta }}}</math>
 
<br>
::<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>
 
 
The variance calculation is the same as Eqns. (var1), (var2) and (var3).
 
<br>
::<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>
=====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).
<br>
<br>
=====Example 3=====
Using the data from Example 1, calculate the mission reliability at  <math>t=2000</math>  hours and mission time  <math>d=40</math>  hours  along with the confidence bounds at the 90% confidence level.
<br>
''Solution''
<br>
The maximum likelihood estimates of  <math>\widehat{\lambda }</math>  and  <math>\widehat{\beta }</math>  from Example 1 are:
 
 
::<math>\begin{align}
  & \widehat{\beta }= & 0.45300 \\
& \widehat{\lambda }= & 0.36224 
\end{align}</math>
 
 
From Eq. (reliability), the mission reliability at  <math>t=2000</math>  for mission time  <math>d=40</math>  is:
 
::<math>\begin{align}
  & \widehat{R}(t)= & {{e}^{-\left[ \lambda {{\left( t+d \right)}^{\beta }}-\lambda {{t}^{\beta }} \right]}} \\
& = & 0.90292 
\end{align}</math>
 
 
At the 90% confidence level and  <math>T=2000</math>  hours, the Fisher Matrix confidence bounds for the mission reliability for mission time  <math>d=40</math>  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>\begin{align}
  & {{[\widehat{R}(t)]}_{L}}= & 0.83711 \\
& {{[\widehat{R}(t)]}_{U}}= & 0.94392 
\end{align}</math>
 
 
The Crow confidence bounds for the mission reliability are:
 
::<math>\begin{align}
  & {{[\widehat{R}(t)]}_{L}}= & {{[\widehat{R}(\tau )]}^{\tfrac{1}{{{\Pi }_{1}}}}} \\
& = & {{[0.90292]}^{\tfrac{1}{0.71440}}} \\
& = & 0.86680 \\
& {{[\widehat{R}(t)]}_{U}}= & {{[\widehat{R}(\tau )]}^{\tfrac{1}{{{\Pi }_{2}}}}} \\
& = & {{[0.90292]}^{\tfrac{1}{1.6051}}} \\
& = & 0.93836 
\end{align}</math>
 
 
Figures ConfReliFish and ConfRelCrow show the Fisher Matrix and Crow confidence bounds on mission reliability for mission time  <math>d=40</math> .
 
[[Image:rga13.3.png|thumb|center|300px|Conditional Reliability vs. Time plot with Fisher Matrix confidence bounds.]]
<br>
<br>
[[Image:rga13.4.png|thumb|center|300px|Conditional Reliability vs. Time plot with Crow confidence bounds.]]
 
<br>

Latest revision as of 00:29, 27 August 2012

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