Crow-AMSAA Confidence Bounds Example

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This example appears in the Reliability Growth and Repairable System Analysis Reference book.


Using the values of [math]\displaystyle{ \widehat{\beta }\,\! }[/math] and [math]\displaystyle{ \widehat{\lambda }\,\! }[/math] estimated in the Failure Times - Example 1, calculate the 90% 2-sided confidence bounds on the cumulative and instantaneous failure intensity.


Solution

Fisher Matrix Bounds

The partial derivatives for the Fisher Matrix confidence bounds 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:

[math]\displaystyle{ \begin{align} \begin{bmatrix}122.43 & 333.64\\ 333.64 & 967.68\end{bmatrix}^{-1} & = \begin{bmatrix}Var(\hat{\lambda}) & Cov(\hat{\beta},\hat{\lambda})\\ Cov(\hat{\beta},\hat{\lambda}) & Var(\hat{\beta})\end{bmatrix} \\ & = \begin{bmatrix} 0.13519969 & -0.046614609\\ -0.046614609 & 0.017105343 \end{bmatrix} \end{align}\,\! }[/math]


For [math]\displaystyle{ T=620\,\! }[/math] hours, 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:

[math]\displaystyle{ \begin{align} Var(\hat{\lambda_{c}}(T)) & = 0.22811336^{2}\cdot 0.017105343\ + 0.083694185^{2} \cdot 0.13519969\ -2\cdot 0.22811336\cdot 0.083694185\cdot 0.046614609 \\ & = 0.00005721408 \\ Var(\hat{\lambda_{i}}(T)) & = 0.17558519^{2}\cdot 0.01715343\ + 0.051404969^{2}\cdot 0.13519969\ -2\cdot 0.17558519\cdot 0.051404969\cdot 0.046614609 \\ &= 0.0000431393 \end{align}\,\! }[/math]


The cumulative and instantaneous failure intensities at [math]\displaystyle{ T=620\,\! }[/math] hours 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] hours, 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]


The following figures 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

Given that the data is failure terminated, the Crow confidence bounds for the cumulative failure intensity at the 90% confidence level and for [math]\displaystyle{ T=620\,\! }[/math] hours 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\cdot t} \\ = & \frac{60.48089}{2*620} \\ = & 0.048775 \end{align}\,\! }[/math]

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

The following figures 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.


Failure Times - Example 3

Calculate the confidence bounds on the cumulative and instantaneous MTBF for the data from the example given above.


Solution

Fisher Matrix Bounds

From the previous example:

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


And for [math]\displaystyle{ T=620\,\! }[/math] hours, 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] hours, 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]


The following two figures 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] hours 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]


The figures below 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]