Test-Fix-Find-Test Data Example

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

Consider the data given in the first table below. There were 56 total failures and $$T=400\,\!$$. The effectiveness factors of the unique BD modes are given in the second table. Determine the following:

1) Calculate the demonstrated MTBF and failure intensity.

2) Calculate the projected MTBF and failure intensity.

3) What is the rate at which unique BD modes are being generated during this test?

4) If the test continues for an additional 50 hours, what is the minimum number of new unique BD modes expected to be generated?



Solution

1) In order to obtain $${{\widehat{\lambda }}_{CA}}\,\!$$, use the traditional Crow-AMSAA model for test-fix-test to fit all 56 data points, regardless of the failure mode classification to get:
 * $$\begin{align}

\widehat{\beta }= & 0.91026 \\ \widehat{\lambda }= & 0.23969 \end{align}\,\!$$

Thus the achieved or demonstrated failure intensity is estimated by:
 * $$\begin{align}

{{\widehat{\lambda }}_{CA}} = & \widehat{\lambda }\widehat{\beta }{{T}^{\widehat{\beta }-1}} \\ = & 0.23969\times 0.91026\times {{400}^{(0.91026-1)}} \\ = & 0.12744 	\end{align}\,\!$$ The achieved or demonstrated MTBF, $${{M}_{CA}}\,\!$$, is the system reliability attained at the end of test, $$T=400\,\!$$, and is estimated by:


 * $${{\widehat{M}}_{CA}}={{[{{\widehat{\lambda }}_{CA}}]}^{-1}}=7.84708\,\!$$

2) For this data set, $$M=16\,\!$$ and $$T=400\,\!$$.
 * $${{\widehat{\lambda }}_{BD}}=\frac{T}=\frac{32}{400}=0.08\,\!$$


 * $$\overline{d}=\underset{i=1}{\overset{M}{\mathop \sum }}\,{{d}_{i}}/M=0.72125\,\!$$


 * $$\underset{i=1}{\overset{16}{\mathop \sum }}\,(1-{{d}_{i}}){{N}_{i}}/T=0.01955\,\!$$

Calculate maximum likelihood estimates, $$\hat{\beta }\,\!$$ and $$\hat{\lambda }\,\!$$, of the BD modes:
 * $$\begin{align}

{{{\hat{\beta }}}_{BD}}= & 0.74715 \\ {{{\hat{\lambda }}}_{BD}} = & 0.18197 \end{align}\,\!$$

Then:
 * $$\overline{d}\widehat{h}(T|BD)=0.0215\,\!$$

Therefore:
 * $$\begin{align}

{{\widehat{\lambda }}_{EM}} = & {{\widehat{\lambda }}_{CA}}-{{\widehat{\lambda }}_{BD}}+\underset{i=1}{\overset{K}{\mathop \sum }}\,(1-{{d}_{i}})\frac{T}+\overline{d}\widehat{h}(T|BD) \\ = & 0.12744-0.08+0.0196+0.0215 \\ 	 = & 0.08854  	\end{align}\,\!$$

The Crow Extended model projected MTBF is:
 * $$\begin{align}

{{\widehat{M}}_{EM}} = & {{[{{\widehat{\lambda }}_{EM}}]}^{-1}} \\ = & 11.29418 	\end{align}\,\!$$

Consequently, based on the Crow Extended model and the data shown in the tables above, the MTBF grew to 7.85 as a result of the corrective actions for the BC failure modes during the test. The MTBF then jumped to 11.29 after the test as a result of the delayed corrective actions for the BD failure modes. The management strategy can be summarized by the Failure Mode Strategy plot, as shown next.



This pie chart shows that 9.48% of the system's failure intensity has been left in (A modes), 31.81% of the failure intensity due to the BC modes has not been seen yet and 13.40% was removed during the test (BC modes - seen). In addition, 33.23% of the failure intensity due to the BD modes has not been seen yet, 3.37% will remain in the system since the corrective actions will not be completely effective at eliminating the identified failure modes, and 8.72% will be removed after the delayed corrective actions.

3) The rate at which unique BD modes are being generated is equal to $$h{{(T|BD)}^{-1}}\,\!$$, where:


 * $$\begin{align}

h{{(T|BD)}^{-1}} = & \frac{1} \\ = & \frac{T}{M{{\widehat{\beta }}_{BD}}} \\ = & 33.4605 \end{align}\,\!$$

4) Unique BD modes are being generated every 33.4605 hours. If the test continues for another 50 hours, then at least one new unique BD mode would be expected to be seen from this additional testing.

As shown in the next figure, the MTBF of each individual failure mode can be plotted, and the failure modes with the lowest MTBF can be identified. These are the failure modes that cause the majority of the system failures.