Template:Bounds on time given cumulative mtbf rsa: Difference between revisions

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(Created page with '====Bounds on Time Given Cumulative MTBF==== =====Fisher Matrix Bounds===== The time, <math>T</math> , must be positive, thus <math>\ln T</math> is approximately treated as be…')
 
 
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====Bounds on Time Given Cumulative MTBF====
#REDIRECT [[RGA_Models_for_Repairable_Systems_Analysis#Bounds_on_Time_Given_Cumulative_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 \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>
 
 
=====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>
 
<br>

Latest revision as of 00:35, 27 August 2012

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