Template:Parameter estimation llyd-lipow

Parameter Estimation

When analyzing reliability data in RGA, you have the option to enter the reliability values in percent or in decimal format. However, [math]\displaystyle{ {{\hat{R}}_{\infty }} }[/math] will always be returned in decimal format and not in percent. The estimated parameters in RGA are unitless.

Maximum Likelihood Estimators

For the [math]\displaystyle{ {{k}^{th}} }[/math] stage:

[math]\displaystyle{ {{L}_{k}}=const.\text{ }R_{k}^{{{S}_{k}}}{{(1-{{R}_{k}})}^{{{n}_{k}}-{{S}_{k}}}} }[/math]

And assuming that the results are independent between stages:

[math]\displaystyle{ L=\underset{k=1}{\overset{N}{\mathop \prod }}\,R_{k}^{{{S}_{k}}}{{(1-{{R}_{k}})}^{{{n}_{k}}-{{S}_{k}}}} }[/math]

Then taking the natural log gives:

[math]\displaystyle{ \Lambda =\underset{k=1}{\overset{N}{\mathop \sum }}\,{{S}_{k}}\ln \left( {{R}_{\infty }}-\frac{\alpha }{k} \right)+\underset{k=1}{\overset{N}{\mathop \sum }}\,({{n}_{k}}-{{S}_{k}})\ln \left( 1-{{R}_{\infty }}+\frac{\alpha }{k} \right) }[/math]

Differentiating with respect to [math]\displaystyle{ {{R}_{\infty }} }[/math] and [math]\displaystyle{ \alpha , }[/math] yields:

[math]\displaystyle{ \frac{\partial \Lambda }{\partial {{R}_{\infty }}}=\underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{{{S}_{k}}}{{{R}_{\infty }}-\tfrac{\alpha }{k}}-\underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{{{n}_{k}}-{{S}_{k}}}{1-{{R}_{\infty }}+\tfrac{\alpha }{k}} }[/math]

[math]\displaystyle{ \frac{\partial \Lambda }{\partial \alpha }=-\underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{\tfrac{{{S}_{k}}}{k}}{{{R}_{\infty }}-\tfrac{\alpha }{k}}+\underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{\tfrac{{{n}_{k}}-{{S}_{k}}}{k}}{1-{{R}_{\infty }}+\tfrac{\alpha }{k}} }[/math]

Rearranging Eqns. (R1) and (alpha1) and setting equal to zero gives:

[math]\displaystyle{ \frac{\partial \Lambda }{\partial {{R}_{\infty }}}=\underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{\tfrac{{{S}_{k}}}{{{n}_{k}}}-\left( {{R}_{\infty }}-\tfrac{\alpha }{k} \right)}{\tfrac{1}{{{n}_{k}}}\left( {{R}_{\infty }}-\tfrac{\alpha }{k} \right)\left( 1-{{R}_{\infty }}+\tfrac{\alpha }{k} \right)}=0 }[/math]

[math]\displaystyle{ \frac{\partial \Lambda }{\partial \alpha }=-\underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{\tfrac{1}{k}\tfrac{{{S}_{k}}}{{{n}_{k}}}-\left( {{R}_{\infty }}-\tfrac{\alpha }{k} \right)\tfrac{1}{k}}{\tfrac{1}{{{n}_{k}}}\left( {{R}_{\infty }}-\tfrac{\alpha }{k} \right)\left( 1-{{R}_{\infty }}+\tfrac{\alpha }{k} \right)}=0 }[/math]

Eqns. (R2) and (alpha2) can be solved simultaneously for [math]\displaystyle{ \widehat{\alpha } }[/math] and [math]\displaystyle{ {{\hat{R}}_{\infty }} }[/math] . It should be noted that a closed form solution does not exist for either of the parameters; thus they must be estimated numerically.

Least Squares Estimators

To obtain least squares estimators for [math]\displaystyle{ {{R}_{\infty }} }[/math] and [math]\displaystyle{ \alpha }[/math] , the sum of squares, [math]\displaystyle{ Q }[/math] , of the deviations of the observed success-ratio, [math]\displaystyle{ {{S}_{k}}/{{n}_{k}} }[/math] , is minimized from its expected value, [math]\displaystyle{ {{R}_{\infty }}-\tfrac{\alpha }{k} }[/math] , with respect to the parameters [math]\displaystyle{ {{R}_{\infty }} }[/math] and [math]\displaystyle{ \alpha . }[/math] Therefore, [math]\displaystyle{ Q }[/math] is expressed as:

[math]\displaystyle{ Q=\underset{k=1}{\overset{N}{\mathop \sum }}\,{{\left( \frac{{{S}_{k}}}{{{n}_{k}}}-{{R}_{\infty }}+\frac{\alpha }{k} \right)}^{2}} }[/math]

Taking the derivatives with respect to [math]\displaystyle{ {{R}_{\infty }} }[/math] and [math]\displaystyle{ \alpha }[/math] and setting equal to zero yields:

[math]\displaystyle{ \begin{align} & \frac{\partial Q}{\partial {{R}_{\infty }}}= & -2\underset{k=1}{\overset{N}{\mathop \sum }}\,\left( \frac{{{S}_{k}}}{{{n}_{k}}}-{{R}_{\infty }}+\frac{\alpha }{k} \right)=0 \\ & \frac{\partial Q}{\partial \alpha }= & 2\underset{k=1}{\overset{N}{\mathop \sum }}\,\left( \frac{{{S}_{k}}}{{{n}_{k}}}-{{R}_{\infty }}+\frac{\alpha }{k} \right)\frac{1}{k}=0 \end{align} }[/math]

Solving Eqns. (pqprll) and (pqpall) simultaneously, the least squares estimates of [math]\displaystyle{ {{R}_{\infty }} }[/math] and [math]\displaystyle{ \alpha }[/math] are:

[math]\displaystyle{ {{\hat{R}}_{\infty }}=\frac{\underset{k=1}{\overset{N}{\mathop{\sum }}}\,\tfrac{1}{{{k}^{2}}}\underset{k=1}{\overset{N}{\mathop{\sum }}}\,\tfrac{{{S}_{k}}}{{{n}_{k}}}-\underset{k=1}{\overset{N}{\mathop{\sum }}}\,\tfrac{1}{k}\underset{k=1}{\overset{N}{\mathop{\sum }}}\,\tfrac{{{S}_{k}}}{k{{n}_{k}}}}{N\underset{k=1}{\overset{N}{\mathop{\sum }}}\,\tfrac{1}{{{k}^{2}}}-{{\left( \underset{k=1}{\overset{N}{\mathop{\sum }}}\,\tfrac{1}{k} \right)}^{2}}} }[/math]

or:

[math]\displaystyle{ \text{ }{{\hat{R}}_{\infty }}=\frac{\underset{k=1}{\overset{N}{\mathop{\sum }}}\,\tfrac{1}{{{k}^{2}}}\underset{k=1}{\overset{N}{\mathop{\sum }}}\,{{R}_{k}}-\underset{k=1}{\overset{N}{\mathop{\sum }}}\,\tfrac{1}{k}\underset{k=1}{\overset{N}{\mathop{\sum }}}\,\tfrac{{{R}_{k}}}{k}}{N\underset{k=1}{\overset{N}{\mathop{\sum }}}\,\tfrac{1}{{{k}^{2}}}-{{\left( \underset{k=1}{\overset{N}{\mathop{\sum }}}\,\tfrac{1}{k} \right)}^{2}}} }[/math]

and:

[math]\displaystyle{ \hat{\alpha }=\frac{\underset{k=1}{\overset{N}{\mathop{\sum }}}\,\tfrac{1}{k}\underset{k=1}{\overset{N}{\mathop{\sum }}}\,\tfrac{{{S}_{k}}}{{{n}_{k}}}-N\underset{k=1}{\overset{N}{\mathop{\sum }}}\,\tfrac{{{S}_{k}}}{k{{n}_{k}}}}{N\underset{k=1}{\overset{N}{\mathop{\sum }}}\,\tfrac{1}{{{k}^{2}}}-{{\left( \underset{k=1}{\overset{N}{\mathop{\sum }}}\,\tfrac{1}{k} \right)}^{2}}} }[/math]

or:

[math]\displaystyle{ \hat{\alpha }=\frac{\underset{k=1}{\overset{N}{\mathop{\sum }}}\,\tfrac{1}{k}\underset{k=1}{\overset{N}{\mathop{\sum }}}\,{{R}_{k}}-N\underset{k=1}{\overset{N}{\mathop{\sum }}}\,\tfrac{{{R}_{k}}}{k}}{N\underset{k=1}{\overset{N}{\mathop{\sum }}}\,\tfrac{1}{{{k}^{2}}}-{{\left( \underset{k=1}{\overset{N}{\mathop{\sum }}}\,\tfrac{1}{k} \right)}^{2}}} }[/math]

Example 1
After a 20-stage reliability development test program, 20 groups of success/failure data were obtained and are given in Table 6.1. Do the following:

1) Fit the Lloyd-Lipow model to the data using least squares.

2) Plot the reliabilities predicted by the Lloyd-Lipow model along with the observed reliabilities and compare the results.

Table 6.1 - The test results and reliabilities of each stage calculated from raw data and the predicted reliability

Test Stage Number([math]\displaystyle{ k }[/math])	Number of Tests in Stage([math]\displaystyle{ n_k }[/math])	Number of Successful Tests([math]\displaystyle{ S_k }[/math])	Raw Data Reliability	Lloyd-Lipow Reliability
1	9	6	0.667	0.7002
2	9	5	0.556	0.7369
3	8	7	0.875	0.7552
4	10	6	0.600	0.7662
5	9	7	0.778	0.7736
6	10	8	0.800	0.7788
7	10	7	0.700	0.7827
8	10	6	0.600	0.7858
9	11	7	0.636	0.7882
10	11	9	0.818	0.7902
11	9	9	1.000	0.7919
12	12	10	0.833	0.7933
13	12	9	0.750	0.7945
14	11	8	0.727	0.7956
15	10	7	0.700	0.7965
16	10	8	0.800	0.7973
17	11	10	0.909	0.7980
18	10	9	0.900	0.7987
19	9	8	0.889	0.7992
20	8	7	0.875	0.7998

Solution
From Table 6.1, the least squares estimates are:

[math]\displaystyle{ \begin{align} & \underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{1}{k}= & \underset{k=1}{\overset{20}{\mathop \sum }}\,\frac{1}{k}=3.5977 \\ & \underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{1}{{{k}^{2}}}= & \underset{k=1}{\overset{20}{\mathop \sum }}\,\frac{1}{{{k}^{2}}}=1.5962 \\ & \underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{{{S}_{k}}}{{{n}_{k}}}= & \underset{k=1}{\overset{20}{\mathop \sum }}\,\frac{{{S}_{k}}}{{{n}_{k}}}=15.4131 \end{align} }[/math]

and:

[math]\displaystyle{ \underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{{{S}_{k}}}{k\cdot {{n}_{k}}}=\underset{k=1}{\overset{20}{\mathop \sum }}\,\frac{{{S}_{k}}}{k\cdot {{n}_{k}}}=2.5632 }[/math]

Substituting into Eqns. (ar) and (alph) yields:

[math]\displaystyle{ \begin{align} & {{{\hat{R}}}_{\infty }}= & \frac{(1.5962)(15.413)-(3.5977)(2.5637)}{(20)(1.5962)-{{(3.5977)}^{2}}} \\ & = & 0.8104 \end{align} }[/math]

and:

[math]\displaystyle{ \begin{align} & \hat{\alpha }= & \frac{(3.5977)(15.413)-(20)(2.5637)}{(20)(1.5962)-{{(3.5977)}^{2}}} \\ & = & 0.2207 \end{align} }[/math]

Therefore, the Lloyd-Lipow reliability growth model is as follows, where [math]\displaystyle{ k }[/math] is the test stage.

[math]\displaystyle{ {{R}_{k}}=0.8104-\frac{0.2201}{k} }[/math]

The reliabilities from the raw data and the reliabilities predicted from Eqn. (eq33) are given in the last two columns of Table 6.1. Figure llfig61 shows the plot. Based on the given data, the model cannot do much more than to basically fit a line through the middle of the points.

Comparison of the predicted reliability and the raw data.