Template:Parameter estimation camsaa

Parameter Estimation

Maximum Likelihood Estimators

The probability density function ( [math]\displaystyle{ pdf }[/math] ) of the [math]\displaystyle{ {{i}^{th}} }[/math] event given that the [math]\displaystyle{ {{(i-1)}^{th}} }[/math] event occurred at [math]\displaystyle{ {{T}_{i-1}} }[/math] is:

[math]\displaystyle{ f({{T}_{i}}|{{T}_{i-1}})=\frac{\beta }{\eta }{{\left( \frac{{{T}_{i}}}{\eta } \right)}^{\beta -1}}\cdot {{e}^{-\tfrac{1}{{{\eta }^{\beta }}}\left( T_{i}^{\beta }-T_{i-1}^{\beta } \right)}} }[/math]

The likelihood function is:

[math]\displaystyle{ L={{\lambda }^{n}}{{\beta }^{n}}{{e}^{-\lambda {{T}^{*\beta }}}}\underset{i=1}{\overset{n}{\mathop \prod }}\,T_{i}^{\beta -1} }[/math]

where [math]\displaystyle{ {{T}^{*}} }[/math] is the termination time and is given by:

[math]\displaystyle{ {{T}^{*}}=\left\{ \begin{matrix} {{T}_{n}}\text{ if the test is failure terminated} \\ T\gt {{T}_{n}}\text{ if the test is time terminated} \\ \end{matrix} \right\} }[/math]

Taking the natural log on both sides:

[math]\displaystyle{ \Lambda =n\ln \lambda +n\ln \beta -\lambda {{T}^{*\beta }}+(\beta -1)\underset{i=1}{\overset{n}{\mathop \sum }}\,\ln {{T}_{i}} }[/math]

And differentiating with respect to [math]\displaystyle{ \lambda }[/math] yields:

[math]\displaystyle{ \frac{\partial \Lambda }{\partial \lambda }=\frac{n}{\lambda }-{{T}^{*\beta }} }[/math]

Set equal to zero and solve for [math]\displaystyle{ \lambda }[/math] :

[math]\displaystyle{ \widehat{\lambda }=\frac{n}{{{T}^{*\beta }}} }[/math]

Now differentiate Eqn. (amsaa4) with respect to [math]\displaystyle{ \beta }[/math] :

[math]\displaystyle{ \frac{\partial \Lambda }{\partial \beta }=\frac{n}{\beta }-\lambda {{T}^{*\beta }}\ln {{T}^{*}}+\underset{i=1}{\overset{n}{\mathop \sum }}\,\ln {{T}_{i}} }[/math]

Set equal to zero and solve for [math]\displaystyle{ \beta }[/math] :

[math]\displaystyle{ \widehat{\beta }=\frac{n}{n\ln {{T}^{*}}-\underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln {{T}_{i}}} }[/math]

Biasing and Unbiasing of Beta

Eqn. (6) returns the biased estimate of [math]\displaystyle{ \beta }[/math] . The unbiased estimate of [math]\displaystyle{ \beta }[/math] can be calculated by using the following relationships. For time terminated data (meaning that the test ends after a specified number of failures):

[math]\displaystyle{ \bar{\beta }=\frac{N-1}{N}\hat{\beta } }[/math]

For failure terminated data (meaning that the test ends after a specified test time):

[math]\displaystyle{ \bar{\beta }=\frac{N-2}{N}\hat{\beta } }[/math]

Example 1
Two prototypes of a system were tested simultaneously with design changes incorporated during the test. Table 5.1 presents the data collected over the entire test. Find the Crow-AMSAA parameters and the intensity function using maximum likelihood estimators.

Table 5.1 - Developmental test data for two identical systems

Failure Number	Failed Unit	Test Time Unit 1(hr)	Test Time Unit 2(hr)	Total Test Time(hr)	[math]\displaystyle{ ln{(T)} }[/math]
1	1	1.0	1.7	2.7	0.99325
2	1	7.3	3.0	10.3	2.33214
3	2	8.7	3.8	12.5	2.52573
4	2	23.3	7.3	30.6	3.42100
5	2	46.4	10.6	57.0	4.04305
6	1	50.1	11.2	61.3	4.11578
7	1	57.8	22.2	80.0	4.38203
8	2	82.1	27.4	109.5	4.69592
9	2	86.6	38.4	125.0	4.82831
10	1	87.0	41.6	128.6	4.85671
11	2	98.7	45.1	143.8	4.96842
12	1	102.2	65.7	167.9	5.12337
13	1	139.2	90.0	229.2	5.43459
14	1	166.6	130.1	296.7	5.69272
15	2	180.8	139.8	320.6	5.77019
16	1	181.3	146.9	328.2	5.79362
17	2	207.9	158.3	366.2	5.90318
18	2	209.8	186.9	396.7	5.98318
19	2	226.9	194.2	421.1	6.04287
20	1	232.2	206.0	438.2	6.08268
21	2	267.5	233.7	501.2	6.21701
22	2	330.1	289.9	620.0	6.42972

Solution
For the failure terminated test, using Eqn. (amsaa6):

[math]\displaystyle{ \widehat{\beta }=\frac{22}{22\ln 620-\underset{i=1}{\overset{22}{\mathop{\sum }}}\,\ln {{T}_{i}}} }[/math]

where:

[math]\displaystyle{ \underset{i=1}{\overset{22}{\mathop \sum }}\,\ln {{T}_{i}}=105.6355 }[/math]

Then:

[math]\displaystyle{ \widehat{\beta }=\frac{22}{22\ln 620-105.6355}=0.6142 }[/math]

From Eqn. (amsaa5):

[math]\displaystyle{ \widehat{\lambda }=\frac{22}{{{620}^{0.6142}}}=0.4239 }[/math]

Therefore, [math]\displaystyle{ {{\lambda }_{i}}(T) }[/math] becomes:

[math]\displaystyle{ \begin{align} & {{\widehat{\lambda }}_{i}}(T)= & 0.4239\cdot 0.6142\cdot {{620}^{-0.3858}} \\ & = & 0.0217906\frac{\text{failures}}{\text{hr}} \end{align} }[/math]

Figure 4fig81 shows the plot of the failure rate. If no further changes are made, the estimated MTBF is [math]\displaystyle{ \tfrac{1}{0.0217906} }[/math] or 46 hr.

File:Rga5.1.png

Failure rate plot for Example 5-1 using Maximum Likelihood Estimation.