Template:Bounds on cumulative number of failures camsaa-cb

Bounds on Cumulative Number of Failures

Fisher Matrix Bounds

The cumulative number of failures, [math]\displaystyle{ N(t) }[/math] , must be positive, thus [math]\displaystyle{ \ln N(t) }[/math] is treated as being normally distributed.

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

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

where:

[math]\displaystyle{ \hat{N}(t)=\hat{\lambda }{{t}^{{\hat{\beta }}}} }[/math]

[math]\displaystyle{ \begin{align} & Var(\hat{N}(t))= & {{\left( \frac{\partial \hat{N}(t)}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial \hat{N}(t)}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial \hat{N}(t)}{\partial \beta } \right)\left( \frac{\partial \hat{N}(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 \hat{N}(t)}{\partial \beta }= & \hat{\lambda }{{t}^{{\hat{\beta }}}}\ln t \\ & \frac{\partial \hat{N}(t)}{\partial \lambda }= & {{t}^{{\hat{\beta }}}} \end{align} }[/math]

Crow Bounds

The Crow cumulative number of failure confidence bounds are:

[math]\displaystyle{ \begin{align} & {{N}_{L}}(T)= & \frac{T}{{\hat{\beta }}}{{\lambda }_{i}}{{(T)}_{L}} \\ & {{N}_{U}}(T)= & \frac{T}{{\hat{\beta }}}{{\lambda }_{i}}{{(T)}_{U}} \end{align} }[/math]

where [math]\displaystyle{ {{\lambda }_{i}}{{(T)}_{L}} }[/math] and [math]\displaystyle{ {{\lambda }_{i}}{{(T)}_{U}} }[/math] can be obtained from Eqn. (amsaac14).

Example 2
Calculate the 90% 2-sided confidence bounds on the cumulative and instantaneous failure intensity for the data from Example 1 given in Table 5.1.

Solution Fisher Matrix Bounds
Using [math]\displaystyle{ \widehat{\beta } }[/math] and [math]\displaystyle{ \widehat{\lambda } }[/math] estimated in Example 1, Eqns. (lambda2partial), (beta2partial) and (lambdabeta2partial) are:

[math]\displaystyle{ \begin{align} & \frac{{{\partial }^{2}}\Lambda }{\partial {{\lambda }^{2}}}= & -\frac{22}{{{0.4239}^{2}}}=-122.43 \\ & \frac{{{\partial }^{2}}\Lambda }{\partial {{\beta }^{2}}}= & -\frac{22}{{{0.6142}^{2}}}-0.4239\cdot {{620}^{0.6142}}{{(\ln 620)}^{2}}=-967.68 \\ & \frac{{{\partial }^{2}}\Lambda }{\partial \lambda \partial \beta }= & -{{620}^{0.6142}}\ln 620=-333.64 \end{align} }[/math]

The Fisher Matrix then becomes:

For [math]\displaystyle{ T=620 }[/math] hr, the partial derivatives of the cumulative and instantaneous failure intensities are:

[math]\displaystyle{ \begin{align} & \frac{\partial {{\lambda }_{c}}(T)}{\partial \beta }= & \widehat{\lambda }{{T}^{\widehat{\beta }-1}}\ln (T) \\ & = & 0.4239\cdot {{620}^{-0.3858}}\ln 620 \\ & = & 0.22811336 \\ & \frac{\partial {{\lambda }_{c}}(T)}{\partial \lambda }= & {{T}^{\widehat{\beta }-1}} \\ & = & {{620}^{-0.3858}} \\ & = & 0.083694185 \end{align} }[/math]

[math]\displaystyle{ \begin{align} & \frac{\partial {{\lambda }_{i}}(T)}{\partial \beta }= & \widehat{\lambda }{{T}^{\widehat{\beta }-1}}+\widehat{\lambda }\widehat{\beta }{{T}^{\widehat{\beta }-1}}\ln T \\ & = & 0.4239\cdot {{620}^{-0.3858}}+0.4239\cdot 0.6142\cdot {{620}^{-0.3858}}\ln 620 \\ & = & 0.17558519 \end{align} }[/math]

[math]\displaystyle{ \begin{align} & \frac{\partial {{\lambda }_{i}}(T)}{\partial \lambda }= & \widehat{\beta }{{T}^{\widehat{\beta }-1}} \\ & = & 0.6142\cdot {{620}^{-0.3858}} \\ & = & 0.051404969 \end{align} }[/math]

Therefore, the variances become:

The cumulative and instantaneous failure intensities at [math]\displaystyle{ T=620 }[/math] hr are:

[math]\displaystyle{ \begin{align} & {{\lambda }_{c}}(T)= & 0.03548 \\ & {{\lambda }_{i}}(T)= & 0.02179 \end{align} }[/math]

So, at the 90% confidence level and for [math]\displaystyle{ T=620 }[/math] hr, the Fisher Matrix confidence bounds for the cumulative failure intensity are:

[math]\displaystyle{ \begin{align} & {{[{{\lambda }_{c}}(T)]}_{L}}= & 0.02499 \\ & {{[{{\lambda }_{c}}(T)]}_{U}}= & 0.05039 \end{align} }[/math]

The confidence bounds for the instantaneous failure intensity are:

[math]\displaystyle{ \begin{align} & {{[{{\lambda }_{i}}(T)]}_{L}}= & 0.01327 \\ & {{[{{\lambda }_{i}}(T)]}_{U}}= & 0.03579 \end{align} }[/math]

Figures 4fig82 and 4fig83 display plots of the Fisher Matrix confidence bounds for the cumulative and instantaneous failure intensity, respectively.

Cumulative failure intensity with 2-sided 90% Fisher Matrix confidence bounds.

Instantaneous failure intensity with 2-sided 90% Fisher Matrix confidence bounds.

Crow Bounds
The Crow confidence bounds for the cumulative failure intensity at the 90% confidence level and for [math]\displaystyle{ T=620 }[/math] hr are:

[math]\displaystyle{ \begin{align} & {{[{{\lambda }_{c}}(T)]}_{L}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t} \\ & = & \frac{29.787476}{2*620} \\ & = & 0.02402 \\ & {{[{{\lambda }_{c}}(T)]}_{U}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot t} \\ & = & \frac{62.8296}{2*620} \\ & = & 0.05067 \end{align} }[/math]

The Crow confidence bounds for the instantaneous failure intensity at the 90% confidence level and for [math]\displaystyle{ T=620 }[/math] hr are:

[math]\displaystyle{ \begin{align} & {{[{{\lambda }_{i}}(t)]}_{L}}= & \frac{1}{{{[MTB{{F}_{i}}]}_{U}}} \\ & = & \frac{1}{MTB{{F}_{i}}\cdot U} \\ & = & 0.01179 \end{align} }[/math]

[math]\displaystyle{ \begin{align} & {{[{{\lambda }_{i}}(t)]}_{U}}= & \frac{1}{{{[MTB{{F}_{i}}]}_{L}}} \\ & = & \frac{1}{MTB{{F}_{i}}\cdot L} \\ & = & 0.03253 \end{align} }[/math]

Figures 4fig84 and 4fig85 display plots of the Crow confidence bounds for the cumulative and instantaneous failure intensity, respectively.

Cumulative failure intensity with 2-sided 90% Crow confidence bounds.

Instantaneous failure intensity with 2-sided 90% Crow confidence bounds.

[math]\displaystyle{ \begin{align} & Var(\widehat{\lambda })= & 0.13519969 \\ & Var(\widehat{\beta })= & 0.017105343 \\ & Cov(\widehat{\beta },\widehat{\lambda })= & -0.046614609 \end{align} }[/math]

Example 3
Calculate the confidence bounds on the cumulative and instantaneous MTBF for the data in Table 5.1.
Solution
Fisher Matrix Bounds
From the previous example:

And for [math]\displaystyle{ T=620 }[/math] hr, the partial derivatives of the cumulative and instantaneous MTBF are:

[math]\displaystyle{ \begin{align} & \frac{\partial {{m}_{c}}(T)}{\partial \beta }= & -\frac{1}{\widehat{\lambda }}{{T}^{1-\widehat{\beta }}}\ln T \\ & = & -\frac{1}{0.4239}{{620}^{0.3858}}\ln 620 \\ & = & -181.23135 \\ & \frac{\partial {{m}_{c}}(T)}{\partial \lambda }= & -\frac{1}{{{\widehat{\lambda }}^{2}}}{{T}^{1-\widehat{\beta }}} \\ & = & -\frac{1}{{{0.4239}^{2}}}{{620}^{0.3858}} \\ & = & -66.493299 \\ & \frac{\partial {{m}_{i}}(T)}{\partial \beta }= & -\frac{1}{\widehat{\lambda }{{\widehat{\beta }}^{2}}}{{T}^{1-\beta }}-\frac{1}{\widehat{\lambda }\widehat{\beta }}{{T}^{1-\widehat{\beta }}}\ln T \\ & = & -\frac{1}{0.4239\cdot {{0.6142}^{2}}}{{620}^{0.3858}}-\frac{1}{0.4239\cdot 0.6142}{{620}^{0.3858}}\ln 620 \\ & = & -369.78634 \\ & \frac{\partial {{m}_{i}}(T)}{\partial \lambda }= & -\frac{1}{{{\widehat{\lambda }}^{2}}\widehat{\beta }}{{T}^{1-\widehat{\beta }}} \\ & = & -\frac{1}{{{0.4239}^{2}}\cdot 0.6142}\cdot {{620}^{0.3858}} \\ & = & -108.26001 \end{align} }[/math]

Therefore, the variances become:

[math]\displaystyle{ \begin{align} & Var({{\widehat{m}}_{c}}(T))= & {{\left( -181.23135 \right)}^{2}}\cdot 0.017105343+{{\left( -66.493299 \right)}^{2}}\cdot 0.13519969 \\ & & -2\cdot \left( -181.23135 \right)\cdot \left( -66.493299 \right)\cdot 0.046614609 \\ & = & 36.113376 \end{align} }[/math]

[math]\displaystyle{ \begin{align} & Var({{\widehat{m}}_{i}}(T))= & {{\left( -369.78634 \right)}^{2}}\cdot 0.017105343+{{\left( -108.26001 \right)}^{2}}\cdot 0.13519969 \\ & & -2\cdot \left( -369.78634 \right)\cdot \left( -108.26001 \right)\cdot 0.046614609 \\ & = & 191.33709 \end{align} }[/math]

So, at 90% confidence level and [math]\displaystyle{ T=620 }[/math] hr, the Fisher Matrix confidence bounds are:

[math]\displaystyle{ \begin{align} & {{[{{m}_{c}}(T)]}_{L}}= & {{{\hat{m}}}_{c}}(t){{e}^{-{{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{c}}(t))}/{{{\hat{m}}}_{c}}(t)}} \\ & = & 19.84581 \\ & {{[{{m}_{c}}(T)]}_{U}}= & {{{\hat{m}}}_{c}}(t){{e}^{{{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{c}}(t))}/{{{\hat{m}}}_{c}}(t)}} \\ & = & 40.01927 \end{align} }[/math]

[math]\displaystyle{ \begin{align} & {{[{{m}_{i}}(T)]}_{L}}= & {{{\hat{m}}}_{i}}(t){{e}^{-{{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{i}}(t))}/{{{\hat{m}}}_{i}}(t)}} \\ & = & 27.94261 \\ & {{[{{m}_{i}}(T)]}_{U}}= & {{{\hat{m}}}_{i}}(t){{e}^{{{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{i}}(t))}/{{{\hat{m}}}_{i}}(t)}} \\ & = & 75.34193 \end{align} }[/math]

Figures 4fig86 and 4fig87 show plots of the Fisher Matrix confidence bounds for the cumulative and instantaneous MTBFs.

Cumulative MTBF with 2-sided 90% Fisher Matrix confidence bounds.

Instantaneous MTBF with 2-sided Fisher Matrix confidence bounds.

Crow Bounds
The Crow confidence bounds for the cumulative MTBF and the instantaneous MTBF at the 90% confidence level and for [math]\displaystyle{ T=620 }[/math] hr are:

[math]\displaystyle{ \begin{align} & {{[{{m}_{c}}(T)]}_{L}}= & \frac{1}{{{[{{\lambda }_{c}}(T)]}_{U}}} \\ & = & 20.5023 \\ & {{[{{m}_{c}}(T)]}_{U}}= & \frac{1}{{{[{{\lambda }_{c}}(T)]}_{L}}} \\ & = & 41.6282 \end{align} }[/math]

[math]\displaystyle{ \begin{align} & {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot {{\Pi }_{1}} \\ & = & 30.7445 \\ & {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot {{\Pi }_{2}} \\ & = & 84.7972 \end{align} }[/math]

Figures 4fig88 and 4fig89 show plots of the Crow confidence bounds for the cumulative and instantaneous MTBF.

Cumulative MTBF with 2-sided 90% Crow confidence bounds.

Instantaneous MTBF with 2-sided 90% Crow confidence bounds.

Confidence bounds can also be obtained on the parameters [math]\displaystyle{ \widehat{\beta } }[/math] and [math]\displaystyle{ \widehat{\lambda } }[/math] . For Fisher Matrix confidence bounds:

[math]\displaystyle{ \begin{align} & {{\beta }_{L}}= & \hat{\beta }{{e}^{{{z}_{\alpha }}\sqrt{Var(\hat{\beta })}/\hat{\beta }}} \\ & = & 0.4325 \\ & {{\beta }_{U}}= & \hat{\beta }{{e}^{-{{z}_{\alpha }}\sqrt{Var(\hat{\beta })}/\hat{\beta }}} \\ & = & 0.8722 \end{align} }[/math]

and:

[math]\displaystyle{ \begin{align} & {{\lambda }_{L}}= & \hat{\lambda }{{e}^{{{z}_{\alpha }}\sqrt{Var(\hat{\lambda })}/\hat{\lambda }}} \\ & = & 0.1016 \\ & {{\lambda }_{U}}= & \hat{\lambda }{{e}^{-{{z}_{\alpha }}\sqrt{Var(\hat{\lambda })}/\hat{\lambda }}} \\ & = & 1.7691 \end{align} }[/math]

For Crow confidence bounds:

[math]\displaystyle{ \begin{align} & {{\beta }_{L}}= & 0.4527 \\ & {{\beta }_{U}}= & 0.9350 \end{align} }[/math]

and:

[math]\displaystyle{ \begin{align} & {{\lambda }_{L}}= & 0.2870 \\ & {{\lambda }_{U}}= & 0.5827 \end{align} }[/math]