Template:Bounds on time given reliability and mission time rsa: Difference between revisions

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(Created page with '====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 …')
 
 
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====Bounds on Time Given Reliability and Mission Time====
#REDIRECT [[RGA_Models_for_Repairable_Systems_Analysis#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>

Latest revision as of 00:38, 27 August 2012

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