Power Law Model Example
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These examples appear in the Reliability Growth and Repairable System Analysis Reference book.
Parameter Estimation Example
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.
Confidence Bounds Example
Using the data from the power law model example given above, calculate the mission reliability at [math]\displaystyle{ t=2000\,\! }[/math] hours and mission time [math]\displaystyle{ d=40\,\! }[/math] hours along with the confidence bounds at the 90% confidence level.
Solution
The maximum likelihood estimates of [math]\displaystyle{ \widehat{\lambda }\,\! }[/math] and [math]\displaystyle{ \widehat{\beta }\,\! }[/math] from the example are:
- [math]\displaystyle{ \begin{align} \widehat{\beta }= & 0.45300 \\ \widehat{\lambda }= & 0.36224 \end{align}\,\! }[/math]
The mission reliability at [math]\displaystyle{ t=2000\,\! }[/math] for mission time [math]\displaystyle{ d=40\,\! }[/math] is:
- [math]\displaystyle{ \begin{align} \widehat{R}(t)= & {{e}^{-\left[ \lambda {{\left( t+d \right)}^{\beta }}-\lambda {{t}^{\beta }} \right]}} \\ = & 0.90292 \end{align}\,\! }[/math]
At the 90% confidence level and [math]\displaystyle{ T=2000\,\! }[/math] hours, the Fisher matrix confidence bounds for the mission reliability for mission time [math]\displaystyle{ d=40\,\! }[/math] are given by:
- [math]\displaystyle{ CB=\frac{\widehat{R}(t)}{\widehat{R}(t)+(1-\widehat{R}(t)){{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\widehat{R}(t))}/\left[ \widehat{R}(t)(1-\widehat{R}(t)) \right]}}}\,\! }[/math]
- [math]\displaystyle{ \begin{align} {{[\widehat{R}(t)]}_{L}}= & 0.83711 \\ {{[\widehat{R}(t)]}_{U}}= & 0.94392 \end{align}\,\! }[/math]
The Crow confidence bounds for the mission reliability are:
- [math]\displaystyle{ \begin{align} {{[\widehat{R}(t)]}_{L}}= & {{[\widehat{R}(\tau )]}^{\tfrac{1}{{{\Pi }_{1}}}}} \\ = & {{[0.90292]}^{\tfrac{1}{0.71440}}} \\ = & 0.86680 \\ {{[\widehat{R}(t)]}_{U}}= & {{[\widehat{R}(\tau )]}^{\tfrac{1}{{{\Pi }_{2}}}}} \\ = & {{[0.90292]}^{\tfrac{1}{1.6051}}} \\ = & 0.93836 \end{align}\,\! }[/math]
The next two figures show the Fisher matrix and Crow confidence bounds on mission reliability for mission time [math]\displaystyle{ d=40\,\! }[/math].
