Crow-AMSAA Model - Grouped Data Example
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This example appears in the Reliability Growth and Repairable System Analysis Reference book.
Consider the grouped failure times data given in the following table. Solve for the Crow-AMSAA parameters using MLE.
| Run Number | Cumulative Failures | End Time(hours) | [math]\displaystyle{ \ln{(T_i)}\,\! }[/math] | [math]\displaystyle{ \ln{(T_i)^2}\,\! }[/math] | [math]\displaystyle{ \ln{(\theta_i)}\,\! }[/math] | [math]\displaystyle{ \ln{(T_i)}\cdot\ln{(\theta_i)}\,\! }[/math] |
|---|---|---|---|---|---|---|
| 1 | 2 | 200 | 5.298 | 28.072 | 0.693 | 3.673 |
| 2 | 3 | 400 | 5.991 | 35.898 | 1.099 | 6.582 |
| 3 | 4 | 600 | 6.397 | 40.921 | 1.386 | 8.868 |
| 4 | 11 | 3000 | 8.006 | 64.102 | 2.398 | 19.198 |
| Sum = | 25.693 | 168.992 | 5.576 | 38.321 |
Solution
Using RGA, the value of [math]\displaystyle{ \hat{\beta }\,\! }[/math], which must be solved numerically, is 0.6315. Using this value, the estimator of [math]\displaystyle{ \lambda \,\! }[/math] is:
- [math]\displaystyle{ \begin{align} \hat{\lambda } = & \frac{11}{3,{{000}^{0.6315}}} \\ = & 0.0701 \end{align}\,\! }[/math]
Therefore, the intensity function becomes:
- [math]\displaystyle{ \hat{\rho }(T)=0.0701\cdot 0.6315\cdot {{T}^{-0.3685}}\,\! }[/math]