Crow-AMSAA Parameter Estimation Example: Difference between revisions

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<noinclude>{{Banner RGA Examples}}
<noinclude>{{Banner RGA Examples}}
''This example appears in the [[Crow-AMSAA (NHPP)|Reliability Growth and Repairable System Analysis Reference book]]''.
''This example appears in the [https://help.reliasoft.com/reference/reliability_growth_and_repairable_system_analysis Reliability growth reference]''.
</noinclude>
</noinclude>


A prototype 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.
A prototype 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.


<center>'''Developmental test data''' </center>
<center>'''Developmental Test Data''' </center>
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{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
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The next figure shows the plot of the failure rate. If no further changes are made, the estimated MTBF is <math>\tfrac{1}{0.0217906}\,\!</math> or 46 hours.
The next figure shows the plot of the failure rate. If no further changes are made, the estimated MTBF is <math>\tfrac{1}{0.0217906}\,\!</math> or 46 hours.


[[Image:FIvsTimeExample1.png|center|500px|Failure Intensity vs. Time plot]]
[[Image:FIvsTimeExample1.png|center|450px]]

Latest revision as of 21:21, 18 September 2023

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This example appears in the Reliability growth reference.


A prototype 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.

FIvsTimeExample1.png