Crow-AMSAA Parameter Estimation Example
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This example appears in the Reliability Growth and Repairable System Analysis Reference book.
Two prototypes of a system were tested simultaneously with design changes incorporated during the test. The following table presents the data collected over the entire test. Find the Crow-AMSAA parameters and the intensity function using maximum likelihood estimators.
| Failure Number | Failed Unit | Test Time Unit 1(hours) | Test Time Unit 2(hours) | Total Test Time(hours) | [math]\displaystyle{ ln{(T)}\,\! }[/math] |
|---|---|---|---|---|---|
| 1 | 1 | 1.0 | 1.7 | 2.7 | 0.99325 |
| 2 | 1 | 7.3 | 3.0 | 10.3 | 2.33214 |
| 3 | 2 | 8.7 | 3.8 | 12.5 | 2.52573 |
| 4 | 2 | 23.3 | 7.3 | 30.6 | 3.42100 |
| 5 | 2 | 46.4 | 10.6 | 57.0 | 4.04305 |
| 6 | 1 | 50.1 | 11.2 | 61.3 | 4.11578 |
| 7 | 1 | 57.8 | 22.2 | 80.0 | 4.38203 |
| 8 | 2 | 82.1 | 27.4 | 109.5 | 4.69592 |
| 9 | 2 | 86.6 | 38.4 | 125.0 | 4.82831 |
| 10 | 1 | 87.0 | 41.6 | 128.6 | 4.85671 |
| 11 | 2 | 98.7 | 45.1 | 143.8 | 4.96842 |
| 12 | 1 | 102.2 | 65.7 | 167.9 | 5.12337 |
| 13 | 1 | 139.2 | 90.0 | 229.2 | 5.43459 |
| 14 | 1 | 166.6 | 130.1 | 296.7 | 5.69272 |
| 15 | 2 | 180.8 | 139.8 | 320.6 | 5.77019 |
| 16 | 1 | 181.3 | 146.9 | 328.2 | 5.79362 |
| 17 | 2 | 207.9 | 158.3 | 366.2 | 5.90318 |
| 18 | 2 | 209.8 | 186.9 | 396.7 | 5.98318 |
| 19 | 2 | 226.9 | 194.2 | 421.1 | 6.04287 |
| 20 | 1 | 232.2 | 206.0 | 438.2 | 6.08268 |
| 21 | 2 | 267.5 | 233.7 | 501.2 | 6.21701 |
| 22 | 2 | 330.1 | 289.9 | 620.0 | 6.42972 |
Solution
For the failure terminated test, [math]\displaystyle{ {\beta}\,\! }[/math] is:
- [math]\displaystyle{ \begin{align} \widehat{\beta }&=\frac{n}{n\ln {{T}^{*}}-\underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln {{T}_{i}}} \\ &=\frac{22}{22\ln 620-\underset{i=1}{\overset{22}{\mathop{\sum }}}\,\ln {{T}_{i}}} \\ \end{align}\,\! }[/math]
- where:
- [math]\displaystyle{ \underset{i=1}{\overset{22}{\mathop \sum }}\,\ln {{T}_{i}}=105.6355\,\! }[/math]
- Then:
- [math]\displaystyle{ \widehat{\beta }=\frac{22}{22\ln 620-105.6355}=0.6142\,\! }[/math]
And for [math]\displaystyle{ {\lambda}\,\! }[/math] :
- [math]\displaystyle{ \begin{align} \widehat{\lambda }&=\frac{n}{{{T}^{*\beta }}} \\ & =\frac{22}{{{620}^{0.6142}}}=0.4239 \\ \end{align}\,\! }[/math]
Therefore, [math]\displaystyle{ {{\lambda }_{i}}(T)\,\! }[/math] becomes:
- [math]\displaystyle{ \begin{align} {{\widehat{\lambda }}_{i}}(T)= & 0.4239\cdot 0.6142\cdot {{620}^{-0.3858}} \\ = & 0.0217906\frac{\text{failures}}{\text{hr}} \end{align}\,\! }[/math]
The next figure shows the plot of the failure rate. If no further changes are made, the estimated MTBF is [math]\displaystyle{ \tfrac{1}{0.0217906}\,\! }[/math] or 46 hours.
