Power Law Model Parameter Estimation Example

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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 [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.