Template:Bounds on time given reliability and mission time rsa

Fisher Matrix Bounds
The time, $$t$$, must be positive, thus  $$\ln t$$  is approximately treated as being normally distributed.


 * $$\frac{\ln (\hat{t})-\ln (t)}{\sqrt{Var\left[ \ln (\hat{t}) \right]}}\sim N(0,1)$$

The confidence bounds on time are calculated by using:


 * $$CB=\hat{t}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{t})}/\hat{t}}}$$


 * where:


 * $$Var(\hat{t})={{\left( \frac{\partial t}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial t}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda })+2\left( \frac{\partial t}{\partial \beta } \right)\left( \frac{\partial t}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda })$$


 * $$\hat{t}$$ is calculated numerically from:


 * $$\widehat{R}(d)={{e}^{-[\widehat{\lambda }{{(\hat{t}+d)}^{\widehat{\beta }}}-\widehat{\lambda }{{{\hat{t}}}^{\widehat{\beta }}}]}}\text{ };\text{ }d\text{ = mission time}$$

The variance calculations are done by:


 * $$\begin{align}

& \frac{\partial t}{\partial \beta }= & \frac{{{{\hat{t}}}^}\ln (\hat{t})-{{(\hat{t}+d)}^}\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{t}+d)}^}}{\hat{\lambda }\hat{\beta }{{(\hat{t}+d)}^{\hat{\beta }-1}}-\hat{\lambda }\hat{\beta }{{{\hat{t}}}^{\hat{\beta }-1}}} \end{align}$$

Crow Bounds
Failure Terminated Data Step 1: Calculate $$({{\hat{R}}_{lower}},{{\hat{R}}_{upper}})=({{R}^{\tfrac{1}}},{{R}^{\tfrac{1}}})$$. Step 2: Let $$R={{\hat{R}}_{lower}}$$  and solve for  $${{t}_{1}}$$  numerically using  $$R={{e}^{-[\widehat{\lambda }{{({{{\hat{t}}}_{1}}+d)}^{\widehat{\beta }}}-\widehat{\lambda }\hat{t}_{1}^{\widehat{\beta }}]}}$$. Step 3: Let $$R={{\hat{R}}_{upper}}$$  and solve for  $${{t}_{2}}$$  numerically using  $$R={{e}^{-[\widehat{\lambda }{{({{{\hat{t}}}_{2}}+d)}^{\widehat{\beta }}}-\widehat{\lambda }\hat{t}_{2}^{\widehat{\beta }}]}}$$. Step 4: If $${{t}_{1}}<{{t}_{2}}$$, then  $${{t}_{lower}}={{t}_{1}}$$  and  $${{t}_{upper}}={{t}_{2}}$$. If $${{t}_{1}}>{{t}_{2}}$$, then  $${{t}_{lower}}={{t}_{2}}$$  and  $${{t}_{upper}}={{t}_{1}}$$. Time Terminated Data Step 1: Calculate $$({{\hat{R}}_{lower}},{{\hat{R}}_{upper}})=({{R}^{\tfrac{1}}},{{R}^{\tfrac{1}}})$$. Step 2: Let $$R={{\hat{R}}_{lower}}$$  and solve for  $${{t}_{1}}$$  numerically using  $$R={{e}^{-[\widehat{\lambda }{{({{{\hat{t}}}_{1}}+d)}^{\widehat{\beta }}}-\widehat{\lambda }\hat{t}_{1}^{\widehat{\beta }}]}}$$. Step 3: Let $$R={{\hat{R}}_{upper}}$$  and solve for  $${{t}_{2}}$$  numerically using  $$R={{e}^{-[\widehat{\lambda }{{({{{\hat{t}}}_{2}}+d)}^{\widehat{\beta }}}-\widehat{\lambda }\hat{t}_{2}^{\widehat{\beta }}]}}$$. Step 4: If $${{t}_{1}}<{{t}_{2}}$$, then  $${{t}_{lower}}={{t}_{1}}$$  and  $${{t}_{upper}}={{t}_{2}}$$. If $${{t}_{1}}>{{t}_{2}}$$, then  $${{t}_{lower}}={{t}_{2}}$$  and  $${{t}_{upper}}={{t}_{1}}$$.