Crow-AMSAA Parameter Estimation Example

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

A prototype of a system was tested with design changes incorporated during the test. The following table presents the data collected over the entire test. Find the Crow-AMSAA parameters and the intensity function using maximum likelihood estimators.

Developmental test data Solution

For the failure terminated test, $${\beta}\,\!$$ is:


 * $$\begin{align}

\widehat{\beta }&=\frac{n}{n\ln {{T}^{*}}-\underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln {{T}_{i}}} \\ &=\frac{22}{22\ln 620-\underset{i=1}{\overset{22}{\mathop{\sum }}}\,\ln {{T}_{i}}} \\ \end{align}\,\!$$

where:


 * $$\underset{i=1}{\overset{22}{\mathop \sum }}\,\ln {{T}_{i}}=105.6355\,\!$$

Then:


 * $$\widehat{\beta }=\frac{22}{22\ln 620-105.6355}=0.6142\,\!$$

And for $${\lambda}\,\!$$ :


 * $$\begin{align}

\widehat{\lambda }&=\frac{n} \\ & =\frac{22}=0.4239 \\ \end{align}\,\!$$

Therefore, $${{\lambda }_{i}}(T)\,\!$$ becomes:


 * $$\begin{align}

{{\widehat{\lambda }}_{i}}(T)= & 0.4239\cdot 0.6142\cdot {{620}^{-0.3858}} \\ = & 0.0217906\frac{\text{failures}}{\text{hr}} \end{align}\,\!$$

The next figure shows the plot of the failure rate. If no further changes are made, the estimated MTBF is $$\tfrac{1}{0.0217906}\,\!$$ or 46 hours.



Confidence Bounds on Cumulative and Instantaneous Failure Intensity
Using the values of $$\hat{\beta }\,\!$$ and $$\hat{\lambda }\,\!$$ estimated in the example given above, 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:


 * $$\begin{align}

\frac{{{\partial }^{2}}\Lambda }{\partial {{\lambda }^{2}}} = & -\frac{22}=-122.43 \\ \frac{{{\partial }^{2}}\Lambda }{\partial {{\beta }^{2}}} = & -\frac{22}-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}\,\!$$

The Fisher Matrix then becomes:


 * $$\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}\,\!$$

For $$T=620\,\!$$ hours, the partial derivatives of the cumulative and instantaneous failure intensities are:


 * $$\begin{align}

\frac{\partial {{\lambda }_{c}}(T)}{\partial \beta }= & \hat{\lambda }{{T}^{\hat{\beta }-1}}\ln (T) \\ = & 0.4239\cdot {{620}^{-0.3858}}\ln 620 \\ = & 0.22811336 \\   \frac{\partial {{\lambda }_{c}}(T)}{\partial \lambda }=  & {{T}^{\hat{\beta }-1}} \\ = & {{620}^{-0.3858}} \\   =  & 0.083694185  \end{align}\,\!$$


 * $$\begin{align}

\frac{\partial {{\lambda }_{i}}(T)}{\partial \beta }= & \hat{\lambda }{{T}^{\hat{\beta }-1}}+\hat{\lambda }\hat{\beta }{{T}^{\hat{\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}\,\!$$


 * $$\begin{align}

\frac{\partial {{\lambda }_{i}}(T)}{\partial \lambda }= & \hat{\beta }{{T}^{\hat{\beta }-1}} \\ = & 0.6142\cdot {{620}^{-0.3858}} \\ = & 0.051404969  \end{align}\,\!$$

Therefore, the variances become:


 * $$\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}\,\!$$

The cumulative and instantaneous failure intensities at $$T=620\,\!$$ hours are:


 * $$\begin{align}

{{\lambda }_{c}}(T)= & 0.03548 \\ {{\lambda }_{i}}(T)= & 0.02179 \end{align}\,\!$$

So, at the 90% confidence level and for $$T=620\,\!$$ hours, the Fisher Matrix confidence bounds for the cumulative failure intensity are:


 * $$\begin{align}

{{[{{\lambda }_{c}}(T)]}_{L}}= & 0.02499 \\ {{[{{\lambda }_{c}}(T)]}_{U}}= & 0.05039 \end{align}\,\!$$

The confidence bounds for the instantaneous failure intensity are:


 * $$\begin{align}

{{[{{\lambda }_{i}}(T)]}_{L}}= & 0.01327 \\ {{[{{\lambda }_{i}}(T)]}_{U}}= & 0.03579 \end{align}\,\!$$

The following figures display plots of the Fisher Matrix confidence bounds for the cumulative and instantaneous failure intensity, respectively.





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 $$T=620\,\!$$ hours are:


 * $$\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}\,\!$$

The Crow confidence bounds for the instantaneous failure intensity at the 90% confidence level and for $$T=620\,\!$$ hours are calculated by first estimating the bounds on the instantaneous MTBF. Once these are calculated, take the inverse as shown below. Details on the confidence bounds for instantaneous MTBF are presented here.


 * $$\begin{align}

{{[{{\lambda }_{i}}(t)]}_{L}} = & \frac{1} \\ = & \frac{1}{MTB{{F}_{i}}\cdot U} \\ = & 0.01179 \end{align}\,\!$$


 * $$\begin{align}

{{[{{\lambda }_{i}}(t)]}_{U}}= & \frac{1} \\ = & \frac{1}{MTB{{F}_{i}}\cdot L} \\ = & 0.03253  \end{align}\,\!$$

The following figures display plots of the Crow confidence bounds for the cumulative and instantaneous failure intensity, respectively.





Confidence Bounds on Cumulative and Instantaneous MTBF
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:


 * $$\begin{align}

Var(\hat{\lambda }) = & 0.13519969 \\ Var(\hat{\beta }) = & 0.017105343 \\ Cov(\hat{\beta },\hat{\lambda }) = & -0.046614609 \end{align}\,\!$$

And for $$T=620\,\!$$ hours, the partial derivatives of the cumulative and instantaneous MTBF are:


 * $$\begin{align}

\frac{\partial {{m}_{c}}(T)}{\partial \beta }= & -\frac{1}{\hat{\lambda }}{{T}^{1-\hat{\beta }}}\ln T \\ = & -\frac{1}{0.4239}{{620}^{0.3858}}\ln 620 \\ = & -181.23135 \\  \frac{\partial {{m}_{c}}(T)}{\partial \lambda } = & -\frac{1}{{T}^{1-\hat{\beta }}} \\ = & -\frac{1}{{620}^{0.3858}} \\ = & -66.493299 \\  \frac{\partial {{m}_{i}}(T)}{\partial \beta } = & -\frac{1}{\hat{\lambda }{{\hat{\beta }}^{2}}}{{T}^{1-\beta }}-\frac{1}{\hat{\lambda }\hat{\beta }}{{T}^{1-\hat{\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}{{{\hat{\lambda }}^{2}}\hat{\beta }}{{T}^{1-\hat{\beta }}} \\ = & -\frac{1}{{{0.4239}^{2}}\cdot 0.6142}\cdot {{620}^{0.3858}} \\ = & -108.26001 \end{align}\,\!$$

Therefore, the variances become:


 * $$\begin{align}

Var({{\hat{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}\,\!$$


 * $$\begin{align}

Var({{\hat{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}\,\!$$

So, at 90% confidence level and $$T=620\,\!$$ hours, the Fisher Matrix confidence bounds are:


 * $$\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}\,\!$$


 * $$\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}\,\!$$

The following two figures show plots of the Fisher Matrix confidence bounds for the cumulative and instantaneous MTBFs.





Crow Bounds

The Crow confidence bounds for the cumulative MTBF and the instantaneous MTBF at the 90% confidence level and for $$T=620\,\!$$ hours are:


 * $$\begin{align}

{{[{{m}_{c}}(T)]}_{L}} = & \frac{1} \\ = & 20.5023 \\  {{[{{m}_{c}}(T)]}_{U}}  = & \frac{1} \\ = & 41.6282 \end{align}\,\!$$


 * $$\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}\,\!$$

The figures below show plots of the Crow confidence bounds for the cumulative and instantaneous MTBF.



Confidence bounds can also be obtained on the parameters $$\hat{\beta }\,\!$$ and $$\hat{\lambda }\,\!$$. For Fisher Matrix confidence bounds:


 * $$\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}\,\!$$

and:


 * $$\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}\,\!$$

For Crow confidence bounds:


 * $$\begin{align}

{{\beta }_{L}}= & 0.4527 \\ {{\beta }_{U}}= & 0.9350 \end{align}\,\!$$

and:


 * $$\begin{align}

{{\lambda }_{L}}= & 0.2870 \\ {{\lambda }_{U}}= & 0.5827 \end{align}\,\!$$