Lloyd-Lipow Least Squares Example

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This example appears in the Reliability Growth and Repairable System Analysis Reference.

After a 20-stage reliability development test program, 20 groups of success/failure data were obtained and are given in the table below. 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.

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

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]

   Using these estimates to obtain [math]\displaystyle{ \hat{R_{\infty}}\,\! }[/math] and [math]\displaystyle{ \hat{\alpha}\,\! }[/math]yields:

   [math]\displaystyle{ \begin{align} \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}}} \\ = & \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{\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}}}\\ = & \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 the Lloyd-Lipow reliability growth model are given in the last two columns of the table. The figure below 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.