Template:Bounds on time given cumulative mtbf rsa

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


 * $$\frac{\ln (\widehat{T})-\ln (T)}{\sqrt{Var\left[ \ln (\widehat{T}) \right]}}\ \tilde{\ }\ N(0,1)$$

The confidence bounds on the time are given by:


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


 * where:


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

The variance calculation is the same as Eqns. (var1), (var2) and (var3).


 * $$\widehat{T}={{(\lambda \cdot {{m}_{c}})}^{1/(1-\beta )}}$$


 * $$\begin{align}

& \frac{\partial T}{\partial \beta }= & \frac{{{(\lambda \cdot {{m}_{c}})}^{1/(1-\beta )}}\ln (\lambda \cdot {{m}_{c}})} \\ & \frac{\partial T}{\partial \lambda }= & \frac{\lambda (1-\beta )} \end{align}$$

Crow Bounds
Step 1: Calculate:


 * $$\hat{T}={{\left( \frac{{{\lambda }_{c}}(T)}{{\hat{\lambda }}} \right)}^{\tfrac{1}{\beta -1}}}$$

Step 2: Estimate the number of failures:


 * $$N(\hat{T})=\hat{\lambda }{{\hat{T}}^}$$

Step 3: Obtain the confidence bounds on time given the cumulative failure intensity by solving for $${{t}_{l}}$$  and  $${{t}_{u}}$$  in the following equations:


 * $$\begin{align}

& {{t}_{l}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot {{\lambda }_{c}}(T)} \\ & {{t}_{u}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot {{\lambda }_{c}}(T)} \end{align}$$