Test-Find-Test Data Example

This example appears in the Reliability growth reference.

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

 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:


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

 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:


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



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.



 