Template:Bounds on instantaneous mtbf camsaa-gd

Bounds on Instantaneous MTBF

Fisher Matrix Bounds

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

[math]\displaystyle{ \frac{\ln {{{\hat{m}}}_{i}}(t)-\ln {{m}_{i}}(t)}{\sqrt{Var(\ln {{{\hat{m}}}_{i}}(t)})}\ \tilde{\ }\ N(0,1) }[/math]

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

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

where:

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

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

The variance calculation is the same as Eqn. (variances) and:

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

Crow Bounds

Step 1: Calculate [math]\displaystyle{ P(i)=\tfrac{{{T}_{i}}}{{{T}_{K}}},\,\,i=1,2,\ldots ,K }[/math] .

Step 2: Calculate:

[math]\displaystyle{ A=\underset{i=1}{\overset{K}{\mathop \sum }}\,\frac{{{\left[ P{{(i)}^{{\hat{\beta }}}}\ln P{{(i)}^{{\hat{\beta }}}}-P{{(i-1)}^{\widehat{\beta }}}\ln P{{(i-1)}^{{\hat{\beta }}}} \right]}^{2}}}{\left[ P{{(i)}^{{\hat{\beta }}}}-P{{(i-1)}^{{\hat{\beta }}}} \right]} }[/math]

Step 3: Calculate [math]\displaystyle{ D=\sqrt{\tfrac{1}{A}+1} }[/math] and [math]\displaystyle{ W=\tfrac{({{z}_{1-\alpha /2}})\cdot D}{\sqrt{N}} }[/math] . Thus an approximate 2-sided [math]\displaystyle{ (1-\alpha ) }[/math] 100-percent confidence interval on [math]\displaystyle{ {{\hat{m}}_{i}}(t) }[/math] is:

[math]\displaystyle{ MTB{{F}_{i}}={{\widehat{m}}_{i}}(1\pm W) }[/math]