Crow-AMSAA Parameter Estimation Example

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

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.