Lloyd-Lipow Least Squares Example

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

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.

Solution

The least squares estimates are:


 * $$\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}= & \underset{k=1}{\overset{20}{\mathop \sum }}\,\frac{1}=1.5962 \\ \underset{k=1}{\overset{N}{\mathop \sum }}\,\frac= & \underset{k=1}{\overset{20}{\mathop \sum }}\,\frac=15.4131 \end{align}\,\!$$


 * and:


 * $$\underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{k\cdot {{n}_{k}}}=\underset{k=1}{\overset{20}{\mathop \sum }}\,\frac{k\cdot {{n}_{k}}}=2.5632\,\!$$

Using these estimates to obtain $$\hat{R_{\infty}}\,\!$$ and $$\hat{\alpha}\,\!$$yields:


 * $$\begin{align}

\text{ }{{\hat{R}}_{\infty }}= &\frac{\underset{k=1}{\overset{N}{\mathop{\sum }}}\,\tfrac{1}\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{k}}{N\underset{k=1}{\overset{N}{\mathop{\sum }}}\,\tfrac{1}-{{\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}\,\!$$


 * and:


 * $$\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{k}}{N\underset{k=1}{\overset{N}{\mathop{\sum }}}\,\tfrac{1}-{{\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}\,\!$$

Therefore, the Lloyd-Lipow reliability growth model is as follows, where $$k\,\!$$ is the test stage.


 * $${{R}_{k}}=0.8104-\frac{0.2201}{k}\,\!$$

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.