Test-Find-Test Data Example

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

Consider the data in the first table below. A system was tested for $$T=400\,\!$$ hours. There were a total of $$N=42\,\!$$ failures and all corrective actions will be delayed until after the end of the 400 hour test. Each failure has been designated as either an A failure mode (the cause will not receive a corrective action) or a BD mode (the cause will receive a corrective action). There are $${{N}_{A}}=10\,\!$$ A mode failures and $${{N}_{BD}}=32\,\!$$ BD mode failures. In addition, there are $$M=16\,\!$$ distinct BD failure modes, which means 16 distinct corrective actions will be incorporated into the system at the end of test. The total number of failures for the $${{j}^{th}}\,\!$$ observed distinct BD mode is denoted by $${{N}_{j}}\,\!$$, and the total number of BD failures during the test is $${{N}_{BD}}=\underset{j=1}{\overset{M}{\mathop{\sum }}}\,{{N}_{j}}\,\!$$. These values and effectiveness factors are given in the second table.

Do the following:

1) Determine the projected MTBF and failure intensity.

2) Determine the growth potential MTBF and failure intensity.

3) Determine the demonstrated MTBF and failure intensity.



Solution

1) The maximum likelihood estimates of $${{\beta }_{BD}}\,\!$$ and $${{\lambda }_{BD}}\,\!$$ are determined to be:
 * $$\begin{align}

{{{\hat{\beta }}}_{BD}} = & \frac{M}{\underset{i=1}{\overset{M}{\mathop{\sum }}}\,\ln (\tfrac{T})} \\ = & 0.7970 \\ 	 {{{\hat{\lambda }}}_{BD}} = & 0.1350 \end{align}\,\!$$ The unbiased estimate of $$\beta \,\!$$ is:


 * $$\begin{align}

{{\overline{\beta }}_{BD}} = & \frac{M-1}{M}{{{\hat{\beta }}}_{BD}} \\ = & 0.7472 \end{align}\,\!$$

Based on the test data, $$\overline{d}=\tfrac{1}{M}\underset{i=1}{\overset{M}{\mathop{\sum }}}\,{{d}_{i}}= 0.72125\,\!$$. Therefore, $$B(T)=\overline{d}\tfrac{M{{\overline{\beta }}_{BD}}}{T}=0.0215\,\!$$. The projected failure intensity due to incorporating the 16 corrective actions is:


 * $$\begin{align}

r(T) = & \left( \frac{T}+\underset{i=1}{\overset{M}{\mathop \sum }}\,(1-{{d}_{i}})\frac{T} \right)+\overline{d}\left( \frac{M}{T}{{\overline{\beta }}_{BD}} \right) \\ = & 0.0661 \end{align}\,\!$$

The projected MTBF is:


 * $$M\widehat{T}B{{F}_{P}}={{[r(T)]}^{-1}}=15.127\,\!$$

2) To estimate the maximum reliability that can be attained with this management strategy, use the following calculations.
 * $$\begin{align}

{{N}_{A}}/T=0.0250 \end{align}\,\!$$


 * $$\frac{1}{T}\underset{i=1}{\overset{16}{\mathop \sum }}\,(1-{{d}_{i}}){{N}_{i}}=0.0196\,\!$$

The growth potential failure intensity is estimated by:


 * $$\begin{align}

{{\widehat{r}}_{GP}}(T) = & \left( \frac{T}+\underset{i=1}{\overset{M}{\mathop \sum }}\,(1-{{d}_{i}})\frac{T} \right) \\ = & 0.0250+0.0196 \\  = & 0.0446  \end{align}\,\!$$

The growth potential MTBF is:


 * $$M\widehat{T}B{{F}_{GP}}={{[{{\widehat{r}}_{GP}}]}^{-1}}=22.4467\,\!$$

3) The demonstrated failure intensity and MTBF are estimated by:
 * $$\begin{align}

{{\widehat{\lambda }}_{D}}(T) = & \frac{{{N}_{A}}+{{N}_{BD}}}{T} \\ = & \frac{42}{400} \\ = & 0.1050 	\end{align}\,\!$$
 * $$\begin{align}

M\widehat{T}B{{F}_{D}} = & {{[{{\widehat{\lambda }}_{D}}(T)]}^{-1}} \\ = & 9.5238 	\end{align}\,\!$$

The first chart below shows the demonstrated, projected and growth potential MTBF. The second shows the demonstrated, projected and growth potential failure intensity.