Template:Bounds on time given reliability and mission time rsa

Bounds on Time Given Reliability and Mission Time

Fisher Matrix Bounds

The time, [math]\displaystyle{ t }[/math] , must be positive, thus [math]\displaystyle{ \ln t }[/math] is approximately treated as being normally distributed.

[math]\displaystyle{ \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]\displaystyle{ CB=\hat{t}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{t})}/\hat{t}}} }[/math]

where:

[math]\displaystyle{ 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]\displaystyle{ \hat{t} }[/math] is calculated numerically from:

[math]\displaystyle{ \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]\displaystyle{ \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
Step 1: Calculate [math]\displaystyle{ ({{\hat{R}}_{lower}},{{\hat{R}}_{upper}})=({{R}^{\tfrac{1}{{{p}_{1}}}}},{{R}^{\tfrac{1}{{{p}_{2}}}}}) }[/math] .
Step 2: Let [math]\displaystyle{ R={{\hat{R}}_{lower}} }[/math] and solve for [math]\displaystyle{ {{t}_{1}} }[/math] numerically using [math]\displaystyle{ R={{e}^{-[\widehat{\lambda }{{({{{\hat{t}}}_{1}}+d)}^{\widehat{\beta }}}-\widehat{\lambda }\hat{t}_{1}^{\widehat{\beta }}]}} }[/math] .
Step 3: Let [math]\displaystyle{ R={{\hat{R}}_{upper}} }[/math] and solve for [math]\displaystyle{ {{t}_{2}} }[/math] numerically using [math]\displaystyle{ R={{e}^{-[\widehat{\lambda }{{({{{\hat{t}}}_{2}}+d)}^{\widehat{\beta }}}-\widehat{\lambda }\hat{t}_{2}^{\widehat{\beta }}]}} }[/math] .
Step 4: If [math]\displaystyle{ {{t}_{1}}\lt {{t}_{2}} }[/math] , then [math]\displaystyle{ {{t}_{lower}}={{t}_{1}} }[/math] and [math]\displaystyle{ {{t}_{upper}}={{t}_{2}} }[/math] . If [math]\displaystyle{ {{t}_{1}}\gt {{t}_{2}} }[/math] , then [math]\displaystyle{ {{t}_{lower}}={{t}_{2}} }[/math] and [math]\displaystyle{ {{t}_{upper}}={{t}_{1}} }[/math] .

Time Terminated Data
Step 1: Calculate [math]\displaystyle{ ({{\hat{R}}_{lower}},{{\hat{R}}_{upper}})=({{R}^{\tfrac{1}{{{\Pi }_{1}}}}},{{R}^{\tfrac{1}{{{\Pi }_{2}}}}}) }[/math] .
Step 2: Let [math]\displaystyle{ R={{\hat{R}}_{lower}} }[/math] and solve for [math]\displaystyle{ {{t}_{1}} }[/math] numerically using [math]\displaystyle{ R={{e}^{-[\widehat{\lambda }{{({{{\hat{t}}}_{1}}+d)}^{\widehat{\beta }}}-\widehat{\lambda }\hat{t}_{1}^{\widehat{\beta }}]}} }[/math] .
Step 3: Let [math]\displaystyle{ R={{\hat{R}}_{upper}} }[/math] and solve for [math]\displaystyle{ {{t}_{2}} }[/math] numerically using [math]\displaystyle{ R={{e}^{-[\widehat{\lambda }{{({{{\hat{t}}}_{2}}+d)}^{\widehat{\beta }}}-\widehat{\lambda }\hat{t}_{2}^{\widehat{\beta }}]}} }[/math] .
Step 4: If [math]\displaystyle{ {{t}_{1}}\lt {{t}_{2}} }[/math] , then [math]\displaystyle{ {{t}_{lower}}={{t}_{1}} }[/math] and [math]\displaystyle{ {{t}_{upper}}={{t}_{2}} }[/math] . If [math]\displaystyle{ {{t}_{1}}\gt {{t}_{2}} }[/math] , then [math]\displaystyle{ {{t}_{lower}}={{t}_{2}} }[/math] and [math]\displaystyle{ {{t}_{upper}}={{t}_{1}} }[/math] .