Template:Bounds on instantaneous mtbf rsa

Fisher Matrix Bounds
The instantaneous MTBF, $${{m}_{i}}(t)$$, must be positive, thus  $$\ln {{m}_{i}}(t)$$  is approximately treated as being normally distributed.


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

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


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


 * where:


 * $${{\widehat{m}}_{i}}(t)=\frac{1}{\lambda \beta {{t}^{\beta -1}}}$$


 * $$\begin{align}

& Var({{\widehat{m}}_{i}}(t))= & {{\left( \frac{\partial {{m}_{i}}(t)}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial {{m}_{i}}(t)}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda }) \\ & & +2\left( \frac{\partial {{m}_{i}}(t)}{\partial \beta } \right)\left( \frac{\partial {{m}_{i}}(t)}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda }) \end{align}$$

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


 * $$\begin{align}

& \frac{\partial {{m}_{i}}(t)}{\partial \beta }= & -\frac{1}{\widehat{\lambda }{{\widehat{\beta }}^{2}}}{{t}^{1-\widehat{\beta }}}-\frac{1}{\widehat{\lambda }\widehat{\beta }}{{t}^{1-\widehat{\beta }}}\ln (t) \\ & \frac{\partial {{m}_{i}}(t)}{\partial \lambda }= & -\frac{1}{{{\widehat{\lambda }}^{2}}\widehat{\beta }}{{t}^{1-\widehat{\beta }}} \end{align}$$

Crow Bounds
Failure Terminated Data To calculate the bounds for failure terminated data, consider the following equation:


 * $$G(\mu |n)=\mathop{}_{0}^{\infty }\frac{(n-2)!}\underset{i=0}{\overset{n-1}{\mathop \sum }}\,\frac{1}{i!}{{\left( \frac{\mu }{x} \right)}^{i}}\exp (-\frac{\mu }{x})\,dx$$

Find the values $${{p}_{1}}$$  and  $${{p}_{2}}$$  by finding the solution  $$c$$  to  $$G({{n}^{2}}/c|n)=\xi $$  for  $$\xi =\tfrac{\alpha }{2}$$  and  $$\xi =1-\tfrac{\alpha }{2}$$, respectively. If using the biased parameters, $$\hat{\beta }$$  and  $$\hat{\lambda }$$, then the upper and lower confidence bounds are:


 * $$\begin{align}

& {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot {{p}_{1}} \\ & {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot {{p}_{2}} \end{align}$$

where $$MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}$$. If using the unbiased parameters, $$\bar{\beta }$$  and  $$\bar{\lambda }$$, then the upper and lower confidence bounds are:


 * $$\begin{align}

& {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot \left( \frac{N-2}{N} \right)\cdot {{p}_{1}} \\ & {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot \left( \frac{N-2}{N} \right)\cdot {{p}_{2}} \end{align}$$

where $$MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}$$. Time Terminated Data To calculate the bounds for time terminated data, consider the following equation where $${{I}_{1}}(.)$$  is the modified Bessel function of order one:


 * $$H(x|k)=\underset{j=1}{\overset{k}{\mathop \sum }}\,\frac{{{2}^{2j-1}}(j-1)!j!{{I}_{1}}(x)}$$

Find the values $${{\Pi }_{1}}$$  and  $${{\Pi }_{2}}$$  by finding the solution  $$x$$  to  $$H(x|k)=\tfrac{\alpha }{2}$$  and  $$H(x|k)=1-\tfrac{\alpha }{2}$$  in the cases corresponding to the lower and upper bounds, respectively. Calculate $$\Pi =\tfrac{4{{x}^{2}}}$$  for each case. If using the biased parameters, $$\hat{\beta }$$  and  $$\hat{\lambda }$$, then the upper and lower confidence bounds are:


 * $$\begin{align}

& {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot {{\Pi }_{1}} \\ & {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot {{\Pi }_{2}} \end{align}$$

where $$MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}$$. If using the unbiased parameters, $$\bar{\beta }$$  and  $$\bar{\lambda }$$, then the upper and lower confidence bounds are:


 * $$\begin{align}

& {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot \left( \frac{N-1}{N} \right)\cdot {{\Pi }_{1}} \\ & {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot \left( \frac{N-1}{N} \right)\cdot {{\Pi }_{2}} \end{align}$$

where $$MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}$$.