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| ::<math>\begin{align} | | ::<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)}\\ | | \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 | | =&0.6153 |
| \end{align}\,\!</math> | | \end{align}\,\!</math> |
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Repairable Systems Analysis Reference Example
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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]
- [math]\displaystyle{ \begin{align}
\hat{\lambda }=&\frac{\sum_{q=1}^{K}N_{q}}{KT^{\hat{\beta }}}\\
\\
=&0.4605
\end{align}\,\! }[/math]
