Duane Linear Regression Examples

These examples appear in the Reliability Growth and Repairable System Analysis Reference.

The following examples demonstrate how to estimate the parameters of the Duane model using a mathematical approach.

Example 1
====Example 1====

Using the same data set from the graphical approach example, estimate the parameters of the MTBF model using least squares.

A complex system's reliability growth is being monitored and the data set is given in the table below.

Do the following:


 * 1) Plot the cumulative MTBF growth curve.
 * 2) Write the equation of this growth curve.
 * 3) Write the equation of the instantaneous MTBF growth model.
 * 4) Plot the instantaneous MTBF growth curve.

Solution

From the data table:


 * $$\begin{align}

\underset{i=1}{\overset{n}{\mathop \sum }}\,\ln ({{T}_{i}})&= & 25.693 \\ \underset{i=1}{\overset{n}{\mathop \sum }}\,\ln ({{T}_{i}})\ln ({{m}_{ci}})&= & 130.66 \\ \underset{i=1}{\overset{n}{\mathop \sum }}\,\ln ({{m}_{ci}})&= & 20.116 \\ \underset{i=1}{\overset{n}{\mathop \sum }}\,{{\left[ \ln ({{T}_{i}}) \right]}^{2}}&= & 168.99 \end{align}\,\!$$

Obtain the value of $$\hat{\alpha}\,\!$$ from the least squares analysis, or:
 * $$\begin{align}

\hat{\alpha }&=\frac{\underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln ({{T}_{i}})\ln ({{m}_{ci}})-\tfrac{\underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln ({{T}_{i}})\underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln ({{m}_{ci}})}{n}}{\underset{i=1}{\overset{n}{\mathop{\sum }}}\,{{\left[ \ln ({{T}_{i}}) \right]}^{2}}-\tfrac{n}} \\ & = \frac{130.66-\tfrac{25.693\cdot 20.116}{4}}{168.99-\tfrac{4}} \\ & = 0.3671  \end{align}\,\!$$

Obtain the value $$\hat{b}\,\!$$ from the least squares analysis, or:


 * $$\begin{align}

\hat{b}&={{e}^{\tfrac{1}{n}\left[ \underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln ({{m}_{ci}})-\alpha \underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln ({{T}_{i}}) \right]}} \\ & = {{e}^{\tfrac{1}{4}(20.116-0.3671\cdot 25.693)}} \\ & = 14.456  \end{align}\,\!$$

Therefore, the cumulative MTBF becomes:


 * $$\begin{align}

\hat{m_{c}}&=bT^{\alpha } \\ &=14.456\cdot {{T}^{0.3671}} \\ \end{align}\,\!$$

The equation for the instantaneous MTBF growth curve is:


 * $$\begin{align}

{{\hat{m}}_{i}}&=\frac{1}{1-\alpha }{{{\hat{m}}}_{c}},:\ \ \alpha \not{=}1 \\ &=\frac{1}{1-0.3671}(14.456){{T}^{0.3671}} \\ \end{align}\,\!$$

==Example 2== ====Example 2==== For the data given in columns 1 and 2 of the following table, estimate the Duane parameters using least squares.

Solution

To estimate the parameters using least squares, the values in columns 3, 4, 5, 6 and 7 are calculated. The cumulative MTBF, $${{m}_{c}}\,\!$$, is calculated by dividing the failure time by the failure number. The value of $$\hat{\alpha }\,\!$$ is:


 * $$\begin{align}

\hat{\alpha }&=\frac{\underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln ({{T}_{i}})\ln ({{m}_{ci}})-\tfrac{\underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln ({{T}_{i}})\underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln ({{m}_{ci}})}{n}}{\underset{i=1}{\overset{n}{\mathop{\sum }}}\,{{\left[ \ln ({{T}_{i}}) \right]}^{2}}-\tfrac{n}} \\ & = \frac{974.242-\tfrac{173.013\cdot 121.406}{23}}{1400.908-\tfrac{23}} \\ & = 0.6133 \end{align}\,\!$$

The estimator of $$b\,\!$$ is estimated to be:


 * $$\begin{align}

\hat{b}&={{e}^{\tfrac{1}{n}\left[ \underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln ({{m}_{ci}})-\alpha \underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln ({{T}_{i}}) \right]}} \\ & = {{e}^{\tfrac{1}{23}(121.406-0.6133\cdot 173.013)}} \\ & = 1.9453 \end{align}\,\!$$

Therefore, the cumulative MTBF becomes:


 * $$\begin{align}

\hat{m_{c}}&= bT^{\alpha } \\ & =1.9453\cdot {{T}^{0.613}} \\ \end{align}\,\!$$

Using the equation for the instantaneous MTBF growth curve,


 * $$\begin{align}

{{\hat{m}}_{i}}&=\frac{1}{1-\alpha }{{{\hat{m}}}_{c}},:\ \ \alpha \not{=}1 \\ & =\frac{1}{1-0.613}(1.945){{T}^{0.613}} \\ \end{align}\,\!$$



==Example 3== ====Example 3====

For the data given in the following table, estimate the Duane parameters using least squares.

Solution

The solution to this example follows the same procedure as the previous example. Therefore, from the table shown above:


 * $$\begin{align}

\underset{i=1}{\overset{29}{\mathop \sum }}\,\ln ({{T}_{i}})= & 154.151 \\ \underset{i=1}{\overset{29}{\mathop \sum }}\,\ln {{({{T}_{i}})}^{2}}= & 902.592 \\ \underset{i=1}{\overset{29}{\mathop \sum }}\,\ln ({{m}_{c}})= & 82.884 \\ \underset{i=1}{\overset{29}{\mathop \sum }}\,\ln ({{T}_{i}})\cdot \ln ({{m}_{c}})= & 483.154 \end{align}\,\!$$

For least squares, the value of $$\alpha \,\!$$ is:


 * $$\begin{align}

\hat{\alpha }&=\frac{\underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln ({{T}_{i}})\ln ({{m}_{ci}})-\tfrac{\underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln ({{T}_{i}})\underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln ({{m}_{ci}})}{n}}{\underset{i=1}{\overset{n}{\mathop{\sum }}}\,{{\left[ \ln ({{T}_{i}}) \right]}^{2}}-\tfrac{n}} \\ & = \frac{483.154-\tfrac{154.151\cdot 82.884}{29}}{902.592-\tfrac{29}} \\ & = 0.5115 \end{align}\,\!$$

The value of the estimator $$b\,\!$$ is:


 * $$\begin{align}

\hat{b}&={{e}^{\tfrac{1}{n}\left[ \underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln ({{m}_{ci}})-\alpha \underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln ({{T}_{i}}) \right]}} \\ & = {{e}^{\tfrac{1}{29}(82.884-0.5115\cdot 154.151)}} \\ & = 1.1495 \end{align}\,\!$$

Therefore, the cumulative MTBF is:


 * $$\begin{align}

\hat{m_{c}}&=bT^{\alpha } & = 1.1495\cdot {{T}^{0.5115}} \end{align}\,\!$$

Using the equation for the instantaneous MTBF growth,


 * $$\begin{align}

{{\hat{m}}_{i}}&=\frac{1}{1-\alpha }{{{\hat{m}}}_{c}},:\ \ \alpha \not{=}1 \\ & =\frac{1}{1-0.5115}(1.1495){{T}^{0.5115}} \end{align}\,\!$$