Lloyd-Lipow Model Examples
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This example appears in the Reliability Growth and Repairable System Analysis Reference book.
Parameter Esimation
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:
- Fit the Lloyd-Lipow model to the data using least squares.
- Plot the reliabilities predicted by the Lloyd-Lipow model along with the observed reliabilities, and compare the results.
| 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
- 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]
- [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]
- [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]
- [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]
- [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.
Confidence Bounds
Consider the success/failure data given in the following table. Solve for the Lloyd-Lipow parameters using least squares analysis, and plot the Lloyd-Lipow reliability with 2-sided confidence bounds at the 90% confidence level.
| Test Stage Number([math]\displaystyle{ k\,\! }[/math]) | Result | Number of Tests([math]\displaystyle{ n_k\,\! }[/math]> | Successful Tests([math]\displaystyle{ S_k=R_i\,\! }[/math]) |
|---|---|---|---|
| 1 | F | 1 | 0 |
| 2 | F | 1 | 0 |
| 3 | F | 1 | 0 |
| 4 | S | 1 | 0.2500 |
| 5 | F | 1 | 0.2000 |
| 6 | F | 1 | 0.1667 |
| 7 | S | 1 | 0.2857 |
| 8 | S | 1 | 0.3750 |
| 9 | S | 1 | 0.4444 |
| 10 | S | 1 | 0.5000 |
| 11 | S | 1 | 0.5455 |
| 12 | S | 1 | 0.5833 |
| 13 | S | 1 | 0.6154 |
| 14 | S | 1 | 0.6429 |
| 15 | S | 1 | 0.6667 |
| 16 | S | 1 | 0.6875 |
| 17 | F | 1 | 0.6471 |
| 18 | S | 1 | 0.6667 |
| 19 | F | 1 | 0.6316 |
| 20 | S | 1 | 0.6500 |
| 21 | S | 1 | 0.6667 |
| 22 | S | 1 | 0.6818 |
Solution
Note that the data set contains three consecutive failures at the beginning of the test. These failures will be ignored throughout the analysis because it is considered that the test starts when the reliability is not equal to zero or one. The number of data points is now reduced to 19. Also, note that the only time that the first three first failures are considered is to calculate the observed reliability in the test. For example, given this data set, the observed reliability at stage 4 is [math]\displaystyle{ 1/4=0.25\,\! }[/math]. This is considered to be the reliability at stage 1.
From the table, the least squares estimates can be calculated as follows:
- [math]\displaystyle{ \begin{align} \underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{1}{k}= & \underset{k=1}{\overset{19}{\mathop \sum }}\,\frac{1}{k}=3.54774 \\ \underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{1}{{{k}^{2}}}= & \underset{k=1}{\overset{19}{\mathop \sum }}\,\frac{1}{{{k}^{2}}}=1.5936 \\ \underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{{{S}_{k}}}{{{n}_{k}}}= & \underset{k=1}{\overset{19}{\mathop \sum }}\,\frac{{{S}_{k}}}{{{n}_{k}}}=9.907 \end{align}\,\! }[/math]
and:
- [math]\displaystyle{ \underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{{{S}_{k}}}{k\cdot {{n}_{k}}}=\underset{k=1}{\overset{19}{\mathop \sum }}\,\frac{{{S}_{k}}}{k\cdot {{n}_{k}}}=1.3002\,\! }[/math]
Using these estimates to obtain [math]\displaystyle{ \hat{R}_{\infty}\,\! }[/math] and [math]\displaystyle{ \hat{\alpha}\,\! }[/math] yields:
- [math]\displaystyle{ \begin{align} {{{\hat{R}}}_{\infty }} = & \frac{(1.5936)(9.907)-(3.5477)(1.3002)}{(19)(1.5936)-{{(3.5477)}^{2}}} \\ = & 0.6316 \end{align}\,\! }[/math]
and:
- [math]\displaystyle{ \begin{align} \hat{\alpha } = & \frac{(3.5477)(9.907)-(19)(1.3002)}{(19)(1.5936)-{{(3.5477)}^{2}}} \\ = & 0.5902 \end{align}\,\! }[/math]
Therefore, the Lloyd-Lipow reliability growth model is as follows, where [math]\displaystyle{ k\,\! }[/math] is the number of the test stage.
- [math]\displaystyle{ {{R}_{k}}=0.6316-\frac{0.5902}{k}\,\! }[/math]
Using the data from the table:
- [math]\displaystyle{ \begin{align} \frac{{{\partial }^{2}}\Lambda }{\partial R_{\infty }^{2}} = & -176.847-40.500=-217.347 \\ \frac{{{\partial }^{2}}\Lambda }{\partial {{\alpha }^{2}}} = & -146.763-2.1274=-148.891 \\ \frac{{{\partial }^{2}}\Lambda }{\partial {{R}_{\infty }}\partial \alpha } = & 149.909-6.5660=143.343 \end{align}\,\! }[/math]
The variances can be calculated using the Fisher Matrix:
- [math]\displaystyle{ \begin{align} {{\left[ \begin{matrix} 217.347 & -143.343 \\ -143.343 & 148.891 \\ \end{matrix} \right]}^{-1}}= & \left[ \begin{matrix} Var({{\widehat{R}}_{\infty }}) & Cov({{\widehat{R}}_{\infty }},\widehat{\alpha }) \\ Cov({{\widehat{R}}_{\infty }},\widehat{\alpha }) & Var(\widehat{\alpha }) \\ \end{matrix} \right] \\ = & \left[ \begin{matrix} 0.0126033 & 0.0121335 \\ 0.0121335 & 0.0183977 \\ \end{matrix} \right] \end{align}\,\! }[/math]
The variance of [math]\displaystyle{ {{R}_{k}}\,\! }[/math] is therefore:
- [math]\displaystyle{ Var({{\widehat{R}}_{k}})=0.0126031+\frac{1}{{{k}^{2}}}\cdot 0.0183977-\frac{2}{k}\cdot 0.0121335\,\! }[/math]
The confidence bounds on reliability can now can be calculated. The associated confidence bounds on reliability at the 90% confidence level are plotted in the following figure, with the predicted reliability, [math]\displaystyle{ {{R}_{k}}\,\! }[/math].
