Template:Bounds on instantaneous mtbf camsaa-cb

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 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. (variance1) 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

Failure Terminated Data
Consider the following equation:

[math]\displaystyle{ G(\mu |n)=\mathop{}_{0}^{\infty }\frac{{{e}^{-x}}{{x}^{n-2}}}{(n-2)!}\underset{i=0}{\overset{n-1}{\mathop \sum }}\,\frac{1}{i!}{{\left( \frac{\mu }{x} \right)}^{i}}\exp (-\frac{\mu }{x})\,dx }[/math]

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

[math]\displaystyle{ \begin{align} & {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot {{p}_{1}} \\ & {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot {{p}_{2}} \end{align} }[/math]

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

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

where [math]\displaystyle{ MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}} }[/math] .

Time Terminated Data
Consider the following equation where [math]\displaystyle{ {{I}_{1}}(.) }[/math] is the modified Bessel function of order one:

[math]\displaystyle{ H(x|k)=\underset{j=1}{\overset{k}{\mathop \sum }}\,\frac{{{x}^{2j-1}}}{{{2}^{2j-1}}(j-1)!j!{{I}_{1}}(x)} }[/math]

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

[math]\displaystyle{ \begin{align} & {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot {{\Pi }_{1}} \\ & {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot {{\Pi }_{2}} \end{align} }[/math]

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

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

where [math]\displaystyle{ MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}} }[/math] .