Crow-AMSAA Model - Grouped Data Example

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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]