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]

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]