Template:Bounds on cumulative mtbf rsa

Bounds on Cumulative MTBF

Fisher Matrix Bounds

The cumulative MTBF, [math]\displaystyle{ {{m}_{c}}(t) }[/math] , must be positive, thus [math]\displaystyle{ \ln {{m}_{c}}(t) }[/math] is approximately treated as being normally distributed.

[math]\displaystyle{ \frac{\ln ({{\widehat{m}}_{c}}(t))-\ln ({{m}_{c}}(t))}{\sqrt{Var\left[ \ln ({{\widehat{m}}_{c}}(t)) \right]}}\ \tilde{\ }\ N(0,1) }[/math]

The approximate confidence bounds on the cumulative MTBF are then estimated from:

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

where:

[math]\displaystyle{ {{\widehat{m}}_{c}}(t)=\frac{1}{\widehat{\lambda }}{{t}^{1-\widehat{\beta }}} }[/math]

[math]\displaystyle{ \begin{align} & Var({{\widehat{m}}_{c}}(t))= & {{\left( \frac{\partial {{m}_{c}}(t)}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial {{m}_{c}}(t)}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda }) \\ & & +2\left( \frac{\partial {{m}_{c}}(t)}{\partial \beta } \right)\left( \frac{\partial {{m}_{c}}(t)}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda })\, \end{align} }[/math]

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

[math]\displaystyle{ \begin{align} & \frac{\partial {{m}_{c}}(t)}{\partial \beta }= & -\frac{1}{\widehat{\lambda }}{{t}^{1-\widehat{\beta }}}\ln (t) \\ & \frac{\partial {{m}_{c}}(t)}{\partial \lambda }= & -\frac{1}{{{\widehat{\lambda }}^{2}}}{{t}^{1-\widehat{\beta }}} \end{align} }[/math]

Crow Bounds

To calculate the Crow confidence bounds on cumulative MTBF, first calculate the Crow cumulative failure intensity confidence bounds:

[math]\displaystyle{ C{{(t)}_{L}}=\frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t} }[/math]

[math]\displaystyle{ C{{(t)}_{u}}=\frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot t} }[/math]

Then

[math]\displaystyle{ \begin{align} & {{[MTB{{F}_{c}}]}_{L}}= & \frac{1}{C{{(t)}_{U}}} \\ & {{[MTB{{F}_{c}}]}_{U}}= & \frac{1}{C{{(t)}_{L}}} \end{align} }[/math]