Power Law Model Parameter Estimation Example: Difference between revisions
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<noinclude>{{Banner RGA Examples}}{{Navigation box}} | <noinclude>{{Banner RGA Examples}}{{Navigation box}} | ||
''These examples appear in the [[Repairable_Systems_Analysis|Reliability Growth and Repairable System Analysis Reference | ''These examples appear in the [[Repairable_Systems_Analysis|Reliability Growth and Repairable System Analysis Reference]]''. | ||
</noinclude> | </noinclude> | ||
For the data in the following table, the starting time for each system is equal to 0 and the ending time for each system is 2,000 hours. Calculate the maximum likelihood estimates <math>\widehat{\lambda }\,\!</math> and <math>\widehat{\beta }\,\!</math>. | For the data in the following table, the starting time for each system is equal to 0 and the ending time for each system is 2,000 hours. Calculate the maximum likelihood estimates <math>\widehat{\lambda }\,\!</math> and <math>\widehat{\beta }\,\!</math>. | ||
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Because the starting time for each system is equal to zero and each system has an equivalent ending time, the general equations for <math>\widehat{\beta }\,\!</math> and <math>\widehat{\lambda }\,\!</math> reduce to the closed form equations. The maximum likelihood estimates of <math>\hat{\beta }\,\!</math> and <math>\hat{\lambda }\,\!</math> are then calculated as follows: | Because the starting time for each system is equal to zero and each system has an equivalent ending time, the general equations for <math>\widehat{\beta }\,\!</math> and <math>\widehat{\lambda }\,\!</math> reduce to the closed form equations. The maximum likelihood estimates of <math>\hat{\beta }\,\!</math> and <math>\hat{\lambda }\,\!</math> are then calculated as follows: | ||
:<math>\widehat{\beta }= \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln (\tfrac{T}{{{X}_{iq}}})} = 0.45300 </math> | |||
:<math>\widehat{\lambda }= \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{K{{T}^{\beta }}} = 0.36224 \,\!</math> | |||
The system failure intensity function is then estimated by: | The system failure intensity function is then estimated by: | ||
:<math>\widehat{u}(t)=\widehat{\lambda }\widehat{\beta }{{t}^{\widehat{\beta }-1}},\text{ }t>0\,\!</math> | |||
The next figure is a plot of <math>\widehat{u}(t)\,\!</math> over the period (0, 3000). Clearly, the estimated failure intensity function is most representative over the range of the data and any extrapolation should be viewed with the usual caution. | The next figure is a plot of <math>\widehat{u}(t)\,\!</math> over the period (0, 3000). Clearly, the estimated failure intensity function is most representative over the range of the data and any extrapolation should be viewed with the usual caution. | ||
[[File:Instantaneous Failure Intensity vs. Time plot.png|450px | [[File:Instantaneous Failure Intensity vs. Time plot.png|450px|center]] | ||
Revision as of 20:03, 2 June 2014
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These examples appear in the Reliability Growth and Repairable System Analysis Reference.
For the data in the following table, the starting time for each system is equal to 0 and the ending time for each system is 2,000 hours. Calculate the maximum likelihood estimates [math]\displaystyle{ \widehat{\lambda }\,\! }[/math] and [math]\displaystyle{ \widehat{\beta }\,\! }[/math].
| Repairable system failure data | ||
| System 1 ( [math]\displaystyle{ {{X}_{i1}}\,\! }[/math] ) | System 2 ( [math]\displaystyle{ {{X}_{i2}}\,\! }[/math] ) | System 3 ( [math]\displaystyle{ {{X}_{i3}}\,\! }[/math] ) |
|---|---|---|
| 1.2 | 1.4 | 0.3 |
| 55.6 | 35.0 | 32.6 |
| 72.7 | 46.8 | 33.4 |
| 111.9 | 65.9 | 241.7 |
| 121.9 | 181.1 | 396.2 |
| 303.6 | 712.6 | 444.4 |
| 326.9 | 1005.7 | 480.8 |
| 1568.4 | 1029.9 | 588.9 |
| 1913.5 | 1675.7 | 1043.9 |
| 1787.5 | 1136.1 | |
| 1867.0 | 1288.1 | |
| 1408.1 | ||
| 1439.4 | ||
| 1604.8 | ||
| [math]\displaystyle{ {{N}_{1}}=9\,\! }[/math] | [math]\displaystyle{ {{N}_{2}}=11\,\! }[/math] | [math]\displaystyle{ {{N}_{3}}=14\,\! }[/math] |
Solution
Because the starting time for each system is equal to zero and each system has an equivalent ending time, the general equations for [math]\displaystyle{ \widehat{\beta }\,\! }[/math] and [math]\displaystyle{ \widehat{\lambda }\,\! }[/math] reduce to the closed form equations. The maximum likelihood estimates of [math]\displaystyle{ \hat{\beta }\,\! }[/math] and [math]\displaystyle{ \hat{\lambda }\,\! }[/math] are then calculated as follows:
- [math]\displaystyle{ \widehat{\beta }= \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln (\tfrac{T}{{{X}_{iq}}})} = 0.45300 }[/math]
- [math]\displaystyle{ \widehat{\lambda }= \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{K{{T}^{\beta }}} = 0.36224 \,\! }[/math]
The system failure intensity function is then estimated by:
- [math]\displaystyle{ \widehat{u}(t)=\widehat{\lambda }\widehat{\beta }{{t}^{\widehat{\beta }-1}},\text{ }t\gt 0\,\! }[/math]
The next figure is a plot of [math]\displaystyle{ \widehat{u}(t)\,\! }[/math] over the period (0, 3000). Clearly, the estimated failure intensity function is most representative over the range of the data and any extrapolation should be viewed with the usual caution.
