Template:Bounds on time given instantaneous mtbf rsa

Bounds on Time Given Instantaneous MTBF

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

The confidence bounds on the time are given by:

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

where:

[math]\displaystyle{ 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 }) }[/math]

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

[math]\displaystyle{ \widehat{T}={{(\lambda \beta \cdot MTB{{F}_{i}})}^{1/(1-\beta )}} }[/math]

[math]\displaystyle{ \begin{align} & \frac{\partial T}{\partial \beta }= & {{\left( \lambda \beta \cdot MTB{{F}_{i}} \right)}^{1/(1-\beta )}}[\frac{1}{{{(1-\beta )}^{2}}}\ln (\lambda \beta \cdot MTB{{F}_{i}})+\frac{1}{\beta (1-\beta )}] \\ & \frac{\partial T}{\partial \lambda }= & \frac{{{(\lambda \beta \cdot MTB{{F}_{i}})}^{1/(1-\beta )}}}{\lambda (1-\beta )} \end{align} }[/math]

Crow Bounds

Step 1: Calculate the confidence bounds on the instantaneous MTBF as presented in Section 5.5.2.
Step 2: Calculate the bounds on time as follows.

Failure Terminated Data

[math]\displaystyle{ \hat{T}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{c})}^{1/(1-\beta )}} }[/math]

So the lower an upper bounds on time are:

[math]\displaystyle{ {{\hat{T}}_{L}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{c}_{1}}})}^{1/(1-\beta )}} }[/math]

[math]\displaystyle{ {{\hat{T}}_{U}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{c}_{2}}})}^{1/(1-\beta )}} }[/math]

Time Terminated Data

[math]\displaystyle{ \hat{T}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{\Pi })}^{1/(1-\beta )}} }[/math]

So the lower and upper bounds on time are:

[math]\displaystyle{ {{\hat{T}}_{L}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{\Pi }_{1}}})}^{1/(1-\beta )}} }[/math]

[math]\displaystyle{ {{\hat{T}}_{U}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{\Pi }_{2}}})}^{1/(1-\beta )}} }[/math]