Template:Bounds on mission time given reliability and time rsa

Fisher Matrix Bounds
The mission time, $$d$$, must be positive, thus  $$\ln \left( d \right)$$  is approximately treated as being normally distributed.


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

The confidence bounds on mission time are given by using:


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


 * where:


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

Calculate $$\hat{d}$$  from:


 * $$\hat{d}={{\left[ {{t}^}-\frac{\ln (R)}{{\hat{\lambda }}} \right]}^{\tfrac{1}}}-t$$

The variance calculations are done by:


 * $$\begin{align}

& \frac{\partial d}{\partial \beta }= & \left[ \frac{{{t}^}\ln (t)}-\ln (t+\hat{d}) \right]\cdot \frac{t+\hat{d}} \\ & \frac{\partial d}{\partial \lambda }= & \frac{{{t}^}-{{(t+\hat{d})}^}}{\hat{\lambda }\hat{\beta }{{(t+\hat{d})}^{\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  $${{d}_{1}}$$  such that:


 * $${{d}_{1}}={{\left( {{t}^}-\frac{\ln ({{R}_{lower}})}{{\hat{\lambda }}} \right)}^{\tfrac{1}}}-t$$

Step 3: Let $$R={{\hat{R}}_{upper}}$$  and solve for  $${{d}_{2}}$$  such that:


 * $${{d}_{2}}={{\left( {{t}^}-\frac{\ln ({{R}_{upper}})}{{\hat{\lambda }}} \right)}^{\tfrac{1}}}-t$$

Step 4: If $${{d}_{1}}<{{d}_{2}}$$, then  $${{d}_{lower}}={{d}_{1}}$$  and  $${{d}_{upper}}={{d}_{2}}$$. If $${{d}_{1}}>{{d}_{2}}$$, then  $${{d}_{lower}}={{d}_{2}}$$  and  $${{d}_{upper}}={{d}_{1}}$$. 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  $${{d}_{1}}$$  using Eqn. (CBR1). Step 3: Let $$R={{\hat{R}}_{upper}}$$  and solve for  $${{d}_{2}}$$  using Eqn. (CBR2). Step 4: If $${{d}_{1}}<{{d}_{2}}$$, then  $${{d}_{lower}}={{d}_{1}}$$  and  $${{d}_{upper}}={{d}_{2}}$$. If $${{d}_{1}}>{{d}_{2}}$$, then  $${{d}_{lower}}={{d}_{2}}$$  and  $${{d}_{upper}}={{d}_{1}}$$.