Duane Linear Regression Examples

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

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

Least Squares Example 1

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

Cumulative Test Hours and the Corresponding Observed Failures for the Complex System
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:

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:

[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} \widehat{\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} \widehat{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]


Least Squares Example 2

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

Failure times data
(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{ \widehat{\alpha }\,\! }[/math] is:

[math]\displaystyle{ \begin{align} \widehat{\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} \widehat{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]
Duane plot for Example 3.


Least Squares Example 3

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

Multiple systems (known operating times) data}
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} \widehat{\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} \widehat{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]