Duane Linear Regression Examples
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These examples appear in the Reliability growth reference.
The following examples demonstrate how to estimate the parameters of the Duane model using a mathematical approach.
Example 1
A complex system's reliability growth is being monitored and the data set is given in the table below.
| Point Number | Cumulative Test Time(hours) | Cumulative Failures | Cumulative MTBF(hours) | Instantaneous MTBF(hours) |
|---|---|---|---|---|
| 1 | 200 | 2 | 100.0 | 100 |
| 2 | 400 | 3 | 133.0 | 200 |
| 3 | 600 | 4 | 150.0 | 200 |
| 4 | 3,000 | 11 | 273.0 | 342.8 |
Do the following:
- Plot the cumulative MTBF growth curve.
- Write the equation of this growth curve.
- Write the equation of the instantaneous MTBF growth model.
- Plot the instantaneous MTBF growth curve.
Solution
From the data table:
- [math]\displaystyle{ \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}\,\! }[/math]
Obtain the value of [math]\displaystyle{ \hat{\alpha}\,\! }[/math] from the least squares analysis, or:
- [math]\displaystyle{ \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{{{\left( \underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln ({{T}_{i}}) \right)}^{2}}}{n}} \\ & = \frac{130.66-\tfrac{25.693\cdot 20.116}{4}}{168.99-\tfrac{{{25.693}^{2}}}{4}} \\ & = 0.3671 \end{align}\,\! }[/math]
Obtain the value [math]\displaystyle{ \hat{b}\,\! }[/math] from the least squares analysis, or:
- [math]\displaystyle{ \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}\,\! }[/math]
Therefore, the cumulative MTBF becomes:
- [math]\displaystyle{ \begin{align} \hat{m_{c}}&=bT^{\alpha } \\ &=14.456\cdot {{T}^{0.3671}} \\ \end{align}\,\! }[/math]
The equation for the instantaneous MTBF growth curve is:
- [math]\displaystyle{ \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}\,\! }[/math]
Example 2
For the data given in columns 1 and 2 of the following table, estimate the Duane parameters using least squares.
| (1)Failure Number | (2)Failure Time(hours) | (3)[math]\displaystyle{ \ln{T_i}\,\! }[/math] | (4)[math]\displaystyle{ \ln{T_i}^2\,\! }[/math] | (5)[math]\displaystyle{ m_c\,\! }[/math] | (6)[math]\displaystyle{ \ln{m_c}\,\! }[/math] | (7)[math]\displaystyle{ \ln{m_c}\cdot\ln{T_i}\,\! }[/math] |
|---|---|---|---|---|---|---|
| 1 | 9.2 | 2.219 | 4.925 | 9.200 | 2.219 | 4.925 |
| 2 | 25 | 3.219 | 10.361 | 12.500 | 2.526 | 8.130 |
| 3 | 61.5 | 4.119 | 16.966 | 20.500 | 3.020 | 12.441 |
| 4 | 260 | 5.561 | 30.921 | 65.000 | 4.174 | 23.212 |
| 5 | 300 | 5.704 | 32.533 | 60.000 | 4.094 | 23.353 |
| 6 | 710 | 6.565 | 43.103 | 118.333 | 4.774 | 31.339 |
| 7 | 916 | 6.820 | 46.513 | 130.857 | 4.874 | 33.241 |
| 8 | 1010 | 6.918 | 47.855 | 126.250 | 4.838 | 33.470 |
| 9 | 1220 | 7.107 | 50.504 | 135.556 | 4.909 | 34.889 |
| 10 | 2530 | 7.836 | 61.402 | 253.000 | 5.533 | 43.359 |
| 11 | 3350 | 8.117 | 65.881 | 304.545 | 5.719 | 46.418 |
| 12 | 4200 | 8.343 | 69.603 | 350.000 | 5.858 | 48.872 |
| 13 | 4410 | 8.392 | 70.419 | 339.231 | 5.827 | 48.895 |
| 14 | 4990 | 8.515 | 72.508 | 356.429 | 5.876 | 50.036 |
| 15 | 5570 | 8.625 | 74.393 | 371.333 | 5.917 | 51.036 |
| 16 | 8310 | 9.025 | 81.455 | 519.375 | 6.253 | 56.431 |
| 17 | 8530 | 9.051 | 81.927 | 501.765 | 6.218 | 56.282 |
| 18 | 9200 | 9.127 | 83.301 | 511.111 | 6.237 | 56.921 |
| 19 | 10500 | 9.259 | 85.731 | 552.632 | 6.315 | 58.469 |
| 20 | 12100 | 9.401 | 88.378 | 605.000 | 6.405 | 60.215 |
| 21 | 13400 | 9.503 | 90.307 | 638.095 | 6.458 | 61.375 |
| 22 | 14600 | 9.589 | 91.945 | 663.636 | 6.498 | 62.305 |
| 23 | 22000 | 9.999 | 99.976 | 956.522 | 6.863 | 68.625 |
| Sum = | 173.013 | 1400.908 | 7600.870 | 121.406 | 974.242 |
Solution
To estimate the parameters using least squares, the values in columns 3, 4, 5, 6 and 7 are calculated. The cumulative MTBF, [math]\displaystyle{ {{m}_{c}}\,\! }[/math], is calculated by dividing the failure time by the failure number. The value of [math]\displaystyle{ \hat{\alpha }\,\! }[/math] is:
- [math]\displaystyle{ \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{{{\left( \underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln ({{T}_{i}}) \right)}^{2}}}{n}} \\ & = \frac{974.242-\tfrac{173.013\cdot 121.406}{23}}{1400.908-\tfrac{{{(173.013)}^{2}}}{23}} \\ & = 0.6133 \end{align}\,\! }[/math]
The estimator of [math]\displaystyle{ b\,\! }[/math] is estimated to be:
- [math]\displaystyle{ \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}\,\! }[/math]
Therefore, the cumulative MTBF becomes:
- [math]\displaystyle{ \begin{align} \hat{m_{c}}&= bT^{\alpha } \\ & =1.9453\cdot {{T}^{0.613}} \\ \end{align}\,\! }[/math]
Using the equation for the instantaneous MTBF growth curve,
- [math]\displaystyle{ \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}\,\! }[/math]
Example 3
For the data given in the following table, estimate the Duane parameters using least squares.
| Run Number | Failed Unit | Test Time 1 | Test Time 2 | Cumulative Time |
|---|---|---|---|---|
| 1 | 1 | 0.2 | 2.0 | 2.2 |
| 2 | 2 | 1.7 | 2.9 | 4.6 |
| 3 | 2 | 4.5 | 5.2 | 9.7 |
| 4 | 2 | 5.8 | 9.1 | 14.9 |
| 5 | 2 | 17.3 | 9.2 | 26.5 |
| 6 | 2 | 29.3 | 24.1 | 53.4 |
| 7 | 1 | 36.5 | 61.1 | 97.6 |
| 8 | 2 | 46.3 | 69.6 | 115.9 |
| 9 | 1 | 63.6 | 78.1 | 141.7 |
| 10 | 2 | 64.4 | 85.4 | 149.8 |
| 11 | 1 | 74.3 | 93.6 | 167.9 |
| 12 | 1 | 106.6 | 103 | 209.6 |
| 13 | 2 | 195.2 | 117 | 312.2 |
| 14 | 2 | 235.1 | 134.3 | 369.4 |
| 15 | 1 | 248.7 | 150.2 | 398.9 |
| 16 | 2 | 256.8 | 164.6 | 421.4 |
| 17 | 2 | 261.1 | 174.3 | 435.4 |
| 18 | 2 | 299.4 | 193.2 | 492.6 |
| 19 | 1 | 305.3 | 234.2 | 539.5 |
| 20 | 1 | 326.9 | 257.3 | 584.2 |
| 21 | 1 | 339.2 | 290.2 | 629.4 |
| 22 | 1 | 366.1 | 293.1 | 659.2 |
| 23 | 2 | 466.4 | 316.4 | 782.8 |
| 24 | 1 | 504 | 373.2 | 877.2 |
| 25 | 1 | 510 | 375.1 | 885.1 |
| 26 | 2 | 543.2 | 386.1 | 929.3 |
| 27 | 2 | 635.4 | 453.3 | 1088.7 |
| 28 | 1 | 641.2 | 485.8 | 1127 |
| 29 | 2 | 755.8 | 573.6 | 1329.4 |
Solution
The solution to this example follows the same procedure as the previous example. Therefore, from the table shown above:
- [math]\displaystyle{ \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}\,\! }[/math]
For least squares, the value of [math]\displaystyle{ \alpha \,\! }[/math] is:
- [math]\displaystyle{ \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{{{\left( \underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln ({{T}_{i}}) \right)}^{2}}}{n}} \\ & = \frac{483.154-\tfrac{154.151\cdot 82.884}{29}}{902.592-\tfrac{{{(154.151)}^{2}}}{29}} \\ & = 0.5115 \end{align}\,\! }[/math]
The value of the estimator [math]\displaystyle{ b\,\! }[/math] is:
- [math]\displaystyle{ \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}\,\! }[/math]
Therefore, the cumulative MTBF is:
- [math]\displaystyle{ \begin{align} \hat{m_{c}}&=bT^{\alpha } & = 1.1495\cdot {{T}^{0.5115}} \end{align}\,\! }[/math]
Using the equation for the instantaneous MTBF growth,
- [math]\displaystyle{ \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}\,\! }[/math]