Mixed Data  Crow Extended Example
This example appears in the Reliability Growth and Repairable System Analysis Reference.
A oneshot 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.
Mixed Data  
Failures in Interval  Cumulative Trials  Classification  Mode 

1  8  BD  1 
1  8  BD  2 
1  8  BD  3 
0  10  
0  11  
0  12  
1  13  BD  2 
0  14  
0  15  
1  16  BD  4 
0  17  
0  18  
0  19  
1  20  BD  5 
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:
 Calculate the demonstrated reliability at the end of the test.
 Calculate the growth potential reliability.
Solution

Based on the equations presented in CrowAMSAA (NHPP), the parameters of the CrowAMSAA (NHPP) model are estimated as follows:
 [math]\widehat{\beta }=0.8572\,\![/math]
 [math]\widehat{\lambda }=0.4602\,\![/math]
 [math]\hat{\lambda }=\frac{n}{T_{n}^{\hat{\beta }}}\,\![/math]
 [math]\begin{align} \hat{\lambda }= & \frac{n}{{{T}_{n}}} \\ = & \frac{6}{20} \\ = & 0.3 \end{align}\,\![/math]
 [math]\begin{align} {{\lambda }_{i}}(T)=\lambda \beta {{T}^{\beta 1}},\text{with }T\gt 0,\text{ }\lambda \gt 0\text{ and }\beta \gt 0 \end{align}\,\![/math]
 [math]{{Q}_{i}}(20)=\widehat{\lambda }=0.3\,\![/math]
 [math]\begin{align} {{R}_{i}}(20)= & 1{{Q}_{i}}(20) \\ = & 10.3 \\ = & 0.7 \end{align}\,\![/math]
 The growth potential unreliability is:
 [math]\begin{align} {{\widehat{Q}}_{GP}}(T)= & \left( \frac{{{N}_{A}}}{T}+\underset{i=1}{\overset{M}{\mathop \sum }}\,(1{{d}_{i}})\frac{{{N}_{i}}}{T} \right) \\ = & \underset{i=1}{\overset{M}{\mathop \sum }}\,(10.7)\frac{{{N}_{i}}}{T} \\ = & 0.3*(\frac{1+1+1+1+1+1)}{20} \\ = & 0.09 \end{align}\,\![/math]
 [math]\begin{align} {{\widehat{R}}_{GP}}(T) = & 1{{\widehat{Q}}_{GP}}(T) \\ = & 10.09 \\ = & 0.91 \end{align}\,\![/math]