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| ===Bounds on Instantaneous MTBF===
| | #REDIRECT [[Crow-AMSAA_-_NHPP#Bounds_on_Instantaneous_MTBF]] |
| ====Fisher Matrix Bounds====
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| The instantaneous MTBF, <math>{{m}_{i}}(t)</math> , must be positive, thus <math>\ln {{m}_{i}}(t)</math> is treated as being normally distributed as well.
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| ::<math>\frac{\ln {{{\hat{m}}}_{i}}(t)-\ln {{m}_{i}}(t)}{\sqrt{Var(\ln {{{\hat{m}}}_{i}}(t)})}\ \tilde{\ }\ N(0,1)</math>
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| The approximate confidence bounds on the instantaneous MTBF are then estimated from:
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| ::<math>CB={{\hat{m}}_{i}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{i}}(t))}/{{{\hat{m}}}_{i}}(t)}}</math>
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| <br>
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| :where:
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| <br>
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| ::<math>{{\hat{m}}_{i}}(t)=\frac{1}{\lambda \beta {{t}^{\beta -1}}}</math>
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| ::<math>\begin{align}
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| & 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 }) \\
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| & & +2\left( \frac{\partial {{m}_{i}}(t)}{\partial \beta } \right)\left( \frac{\partial {{m}_{i}}(t)}{\partial \lambda } \right)cov(\hat{\beta },\hat{\lambda }).
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| \end{align}</math>
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| The variance calculation is the same as Eqn. (variance1) and:
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| ::<math>\begin{align}
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| & \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 \\
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| & \frac{\partial {{m}_{i}}(t)}{\partial \lambda }= & -\frac{1}{{{{\hat{\lambda }}}^{2}}\hat{\beta }}{{t}^{1-\hat{\beta }}}
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| \end{align}</math>
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| ====Crow Bounds====
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| '''Failure Terminated Data'''
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| <br>
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| Consider the following equation:
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| ::<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>
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| 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:
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| ::<math>\begin{align}
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| & {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot {{p}_{1}} \\
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| & {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot {{p}_{2}}
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| \end{align}</math>
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| 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:
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| ::<math>\begin{align}
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| & {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot \left( \frac{N-2}{N} \right)\cdot {{p}_{1}} \\
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| & {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot \left( \frac{N-2}{N} \right)\cdot {{p}_{2}}
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| \end{align}</math>
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| where <math>MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}</math> .
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| <br>
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| <br>
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| '''Time Terminated Data'''
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| <br>
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| Consider the following equation where <math>{{I}_{1}}(.)</math> is the modified Bessel function of order one:
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| ::<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>
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| 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. If using the biased parameters, <math>\hat{\beta }</math> and <math>\hat{\lambda }</math> , then the upper and lower confidence bounds are:
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| ::<math>\begin{align}
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| & {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot {{\Pi }_{1}} \\
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| & {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot {{\Pi }_{2}}
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| \end{align}</math>
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| 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:
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| ::<math>\begin{align}
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| & {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot \left( \frac{N-1}{N} \right)\cdot {{\Pi }_{1}} \\
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| & {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot \left( \frac{N-1}{N} \right)\cdot {{\Pi }_{2}}
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| \end{align}</math>
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| where <math>MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}</math> .
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