Crow-AMSAA Model - Grouped Data Example: Difference between revisions

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''This example appears in the [[Crow-AMSAA (NHPP)|Reliability Growth and Repairable System Analysis Reference]]''.
''This example appears in the [https://help.reliasoft.com/reference/reliability_growth_and_repairable_system_analysis Reliability growth reference]''.
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Latest revision as of 21:20, 18 September 2023

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This example appears in the Reliability growth reference.


Consider the grouped failure times data given in the following table. Solve for the Crow-AMSAA parameters using MLE.

Grouped Failure Times Data
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]