Power Law Model Example

These examples appear in the Reliability Growth and Repairable System Analysis Reference book.

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 $$\widehat{\lambda }\,\!$$ and $$\widehat{\beta }\,\!$$.

Solution

Because the starting time for each system is equal to zero and each system has an equivalent ending time, the general equations for $$\widehat{\beta }\,\!$$ and $$\widehat{\lambda }\,\!$$ reduce to the closed form equations. The maximum likelihood estimates of $$\hat{\beta }\,\!$$ and $$\hat{\lambda }\,\!$$ are then calculated as follows:


 * $$\widehat{\beta }= \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{\mathop{\sum }}}\,\ln (\tfrac{T})} = 0.45300 $$


 * $$\widehat{\lambda }= \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{K{{T}^{\beta }}} =  0.36224 \,\!$$

The system failure intensity function is then estimated by:


 * $$\widehat{u}(t)=\widehat{\lambda }\widehat{\beta }{{t}^{\widehat{\beta }-1}},\text{ }t>0\,\!$$

The next figure is a plot of $$\widehat{u}(t)\,\!$$ 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.