Concurrent Operating Times - Crow-AMSAA (NHPP) Example

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

Six systems were subjected to a reliability growth test, and a total of 82 failures were observed. Given the data in the table below, which presents the start/end times and times-to-failure for each system, do the following:


 * 1) Estimate the parameters of the Crow-AMSAA model using maximum likelihood estimation.
 * 2) Determine how many additional failures would be generated if testing continues until 3,000 hours.

Solution

 To estimate the parameters $$\hat{\beta }\,\!$$ and $$\hat{\lambda}\,\!$$, the equivalent single system (ESS) must first be determined. The ESS is given below:

Equivalent Single System

Given the ESS data, the value of $$\hat{\beta }\,\!$$ is calculated using:


 * $$\hat{\beta }=\frac{n}{n\ln {{T}^{*}}-\underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln {{T}_{i}}}\,\!$$
 * $$\hat{\beta }=0.8939\,\!$$

where $$n\,\!$$ is the number of failures and $$T^*\,\!$$ is the termination time. The termination time is the sum of end times for each of the systems which equals 2909.

$$\hat{\lambda}\,\!$$ is estimated with:


 * $$\hat{\lambda }=\frac{n}{{{T}^{*}}^{\beta }}$$
 * $$\hat{\lambda }=0.0657\,\!$$

The next figure shows the parameters estimated using RGA.



 The number of failures can be estimated using the Quick Calculation Pad, as shown next. The estimated number of failures at 3,000 hours is equal to 84.2892 and 82 failures were observed during testing. Therefore, the number of additional failures generated if testing continues until 3,000 hours is equal to $$84.2892-82=2.2892\approx 3\,\!$$



 