Power Law Model Parameter Estimation Example

These examples appear in the Reliability growth 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 $$\widehat{\lambda }\,\!$$ 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.