Repairable Systems Analysis Reference Example: Difference between revisions

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Since <math>\,\!S_{1}=S_{2}=S_{3}=</math> 0 and <math>\,\!T_{1}=T_{2}=T_{3}=</math> 200 then the maximum likelihood estimates of <math>\,\!\hat{\beta}</math> and <math>\,\!\hat{\lambda }</math> are given by:
Since <math>\,\!S_{1}=S_{2}=S_{3}=</math> 0 and <math>\,\!T_{1}=T_{2}=T_{3}=</math> 200 then the maximum likelihood estimates of <math>\,\!\hat{\beta}</math> and <math>\,\!\hat{\lambda }</math> are given by:


::<math>\,\!\hat{\beta }=\frac{\sum_{q=1}^{K}N_{q}}{\sum_{q=1}^{K}\sum_{i=1}^{N_{q}}ln \left(\frac{T}{N_{iq}}\right)}</math>
::<math>\begin{align}
\hat{\beta }=&\frac{\sum_{q=1}^{K}N_{q}}{\sum_{q=1}^{K}\sum_{i=1}^{N_{q}}ln \left(\frac{T}{N_{iq}}\right)}\\
 
=&0.6153
\end{align}\,\!</math>

Revision as of 18:24, 6 June 2014

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Repairable Systems Analysis Reference Example

This example compares the results for a repairable systems analysis.


Reference Case

Crow, L.H., Reliability Analysis for Complex Repairable Systems, Reliability and Biometry: Statistical Analysis of Lifelength, pg. 385, 1974.


Data

Simulated Data for 3 Systems with End Time = 200 hours
System 1 System 2 System 3
4.3 0.1 8.4
4.4 5.6 32.4
10.2 18.6 44.7
23.5 19.5 48.4
23.8 24.2 50.6
26.4 26.7 73.6
74 45.1 98.7
77.1 45.8 112.2
92.1 72.7 129.8
197.2 75.7 136
98.6 195.8
120.1
161.8
180.6
190.8


Result

Beta = 0.615, Lambda = 0.461


Results in Weibull++

Since [math]\displaystyle{ \,\!S_{1}=S_{2}=S_{3}= }[/math] 0 and [math]\displaystyle{ \,\!T_{1}=T_{2}=T_{3}= }[/math] 200 then the maximum likelihood estimates of [math]\displaystyle{ \,\!\hat{\beta} }[/math] and [math]\displaystyle{ \,\!\hat{\lambda } }[/math] are given by:

[math]\displaystyle{ \begin{align} \hat{\beta }=&\frac{\sum_{q=1}^{K}N_{q}}{\sum_{q=1}^{K}\sum_{i=1}^{N_{q}}ln \left(\frac{T}{N_{iq}}\right)}\\ =&0.6153 \end{align}\,\! }[/math]
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