Mixed Data - Crow Extended Example

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

A one-shot system underwent reliability growth testing for a total of 20 trials. The test was performed as a combination of groups of units with the same configuration and individual trials. The following table shows the data set. The Failures in Interval column specifies the number of failures that occurred in each interval and the Cumulative Trials column specifies the cumulative number of trials at the end of that interval. In other words, the first three rows contain the data from the first trial, in which 8 units with the same configuration were tested and 3 failures (with different failure modes) were observed. The next row contains data from the second trial, in which 2 units with the same configuration were tested and no failures occurred. And so on.

The table also gives the classifications of the failure modes. There are 5 BD modes. The average effectiveness factor for the BD modes is 0.7. Do the following:
 * 1)	Calculate the demonstrated reliability at the end of the test.


 * 2)	Calculate the growth potential reliability.

Solution

Based on the equations presented in Crow-AMSAA (NHPP), the parameters of the Crow-AMSAA (NHPP) model are estimated as follows:


 * $$\widehat{\beta }=0.8572\,\!$$

and:


 * $$\widehat{\lambda }=0.4602\,\!$$

However, because there are only A or BD modes for mixed data, there is no growth during the test. In other words, the hypothesis for the $$\hat{\beta }\,\!$$ parameter is that $$\hat{\beta }=1.\,\!$$

From the Crow-AMSAA (NHPP) model, we know that:


 * $$\hat{\lambda }=\frac{n}{T_{n}^{\hat{\beta }}}\,\!$$

or, if $$\hat{\beta }=1\,\!$$, this becomes:


 * $$\begin{align}

\hat{\lambda }= & \frac{n} \\ = & \frac{6}{20} \\ = & 0.3 \end{align}\,\!$$

As we have seen, the Crow-AMSAA instantaneous failure intensity, $${{\lambda }_{i}}(T)\,\!$$, is defined as:


 * $$\begin{align}

{{\lambda }_{i}}(T)=\lambda \beta {{T}^{\beta -1}},\text{with }T>0,\text{ }\lambda >0\text{ and }\beta >0 \end{align}\,\!$$

Using trials instead of time, and accommodating for $$\hat{\beta }=1\,\!$$, we can calculate the instantaneous failure probability at the end of the test, or $$T=20\,\!$$ :


 * $${{Q}_{i}}(20)=\widehat{\lambda }=0.3\,\!$$

So the instantaneous reliability at the end of the test, or demonstrated reliability, is:


 * $$\begin{align}

{{R}_{i}}(20)= & 1-{{Q}_{i}}(20) \\ = & 1-0.3 \\  = & 0.7  \end{align}\,\!$$

The next figure shows the data sheet as calculated in the RGA software.



The growth potential unreliability is:


 * $$\begin{align}

{{\widehat{Q}}_{GP}}(T)= & \left( \frac{T}+\underset{i=1}{\overset{M}{\mathop \sum }}\,(1-{{d}_{i}})\frac{T} \right) \\ = & \underset{i=1}{\overset{M}{\mathop \sum }}\,(1-0.7)\frac{T} \\ = & 0.3*(\frac{1+1+1+1+1+1)}{20} \\ = & 0.09 \end{align}\,\!$$

So the growth potential reliability is:


 * $$\begin{align}

{{\widehat{R}}_{GP}}(T) = & 1-{{\widehat{Q}}_{GP}}(T) \\ = & 1-0.09 \\  = & 0.91  \end{align}\,\!$$

The figures below show the calculation of the growth potential reliability for the mixed data using the RGA software's QCP, followed by the growth potential plot.