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

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(Created page with '====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)</m…')
 
 
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====Bounds on Mission Time Given Reliability and Time====
#REDIRECT [[RGA_Models_for_Repairable_Systems_Analysis#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> .
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<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>
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Latest revision as of 00:39, 27 August 2012

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