Template:Bounds on mission time given reliability and time rsa

Bounds on Mission Time Given Reliability and Time

Fisher Matrix Bounds

The mission time, [math]\displaystyle{ d }[/math] , must be positive, thus [math]\displaystyle{ \ln \left( d \right) }[/math] is approximately treated as being normally distributed.

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

where:

[math]\displaystyle{ 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]\displaystyle{ \hat{d} }[/math] from:

[math]\displaystyle{ \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]\displaystyle{ \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
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{ {{d}_{1}} }[/math] such that:

[math]\displaystyle{ {{d}_{1}}={{\left( {{t}^{{\hat{\beta }}}}-\frac{\ln ({{R}_{lower}})}{{\hat{\lambda }}} \right)}^{\tfrac{1}{{\hat{\beta }}}}}-t }[/math]

Step 3: Let [math]\displaystyle{ R={{\hat{R}}_{upper}} }[/math] and solve for [math]\displaystyle{ {{d}_{2}} }[/math] such that:

[math]\displaystyle{ {{d}_{2}}={{\left( {{t}^{{\hat{\beta }}}}-\frac{\ln ({{R}_{upper}})}{{\hat{\lambda }}} \right)}^{\tfrac{1}{{\hat{\beta }}}}}-t }[/math]

Step 4: If [math]\displaystyle{ {{d}_{1}}\lt {{d}_{2}} }[/math] , then [math]\displaystyle{ {{d}_{lower}}={{d}_{1}} }[/math] and [math]\displaystyle{ {{d}_{upper}}={{d}_{2}} }[/math] . If [math]\displaystyle{ {{d}_{1}}\gt {{d}_{2}} }[/math] , then [math]\displaystyle{ {{d}_{lower}}={{d}_{2}} }[/math] and [math]\displaystyle{ {{d}_{upper}}={{d}_{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{ {{d}_{1}} }[/math] using Eqn. (CBR1).
Step 3: Let [math]\displaystyle{ R={{\hat{R}}_{upper}} }[/math] and solve for [math]\displaystyle{ {{d}_{2}} }[/math] using Eqn. (CBR2).
Step 4: If [math]\displaystyle{ {{d}_{1}}\lt {{d}_{2}} }[/math] , then [math]\displaystyle{ {{d}_{lower}}={{d}_{1}} }[/math] and [math]\displaystyle{ {{d}_{upper}}={{d}_{2}} }[/math] . If [math]\displaystyle{ {{d}_{1}}\gt {{d}_{2}} }[/math] , then [math]\displaystyle{ {{d}_{lower}}={{d}_{2}} }[/math] and [math]\displaystyle{ {{d}_{upper}}={{d}_{1}} }[/math] .