Crow-AMSAA Parameter Estimation Example

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

A prototypes of a system was tested 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.

Developmental test data

Row	Time to Event (hr)	[math]\displaystyle{ ln{(T)}\,\! }[/math]
1	2.7	0.99325
2	10.3	2.33214
3	12.5	2.52573
4	30.6	3.42100
5	57.0	4.04305
6	61.3	4.11578
7	80.0	4.38203
8	109.5	4.69592
9	125.0	4.82831
10	128.6	4.85671
11	143.8	4.96842
12	167.9	5.12337
13	229.2	5.43459
14	296.7	5.69272
15	320.6	5.77019
16	328.2	5.79362
17	366.2	5.90318
18	396.7	5.98318
19	421.1	6.04287
20	438.2	6.08268
21	501.2	6.21701
22	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.

Failure rate plot.