# Duane Model Examples

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

## Parameter Estimation Examples

### Graphical Approach

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

1. Given the data in the second and third columns of the above table, the cumulative MTBF, ${{\hat{m}}_{c}}\,\!$, values are calculated in the fourth column. The information in the second and fourth columns are then plotted. The first figure below shows the cumulative MTBF while the second figure below shows the instantaneous MTBF. It can be seen that a straight line represents the MTBF growth very well on log-log scales.
Cumulative MTBF plot
Instantaneous MTBF plot

By changing the x-axis scaling, you are able to extend the line to $T=1\,\!$. You can get the value of $b\,\!$ from the graph by positioning the cursor at the point where the line meets the y-axis. Then read the value of the y-coordinate position at the bottom left corner. In this case, $b\,\!$ is approximately 14 hours. The next figure illustrates this.

Cumulative MTBF plot for $b \approx 14\,\!$ at $T=1\,\!$

Another way of determining $b\,\!$ is to calculate $\alpha \,\!$ by using two points on the fitted straight line and substituting the corresponding ${{\hat{m}}_{c}}\,\!$ and $T\,\!$ values into:

$\alpha =\frac{\ln \left( {{{\hat{m}}}_{{{c}_{2}}}} \right)-\ln \left( {{{\hat{m}}}_{{{c}_{1}}}} \right)}{\ln \left( {{T}_{_{2}}} \right)-\ln \left( {{T}_{_{1}}} \right)}\,\!$

Then substitute this $\alpha \,\!$ and choose a set of values for ${{\hat{m}}_{{{c}_{1}}}}\,\!$ and ${{T}_{_{1}}}\,\!$ into the cumulative MTBF equation, ${{\hat{m}}_{c}}=b{{T}^{\alpha }}\,\!$, and solve for $b\,\!$. The slope of the line, $\alpha \,\!$, may also be found from the linearized form of the cumulative MTBF equation, or :

$\alpha =\frac{\ln \left( {{{\hat{m}}}_{c}} \right)-\ln (b)}{\ln (T)-\ln (1)}\,\!$

Using the cumulative MTBF plot for the example, at ${{T}_{_{1}}}=200\,\!$ hours, ${{\hat{m}}_{{{c}_{1}}}}=100\,\!$ hours, and ${{T}_{_{2}}}=3,500\,\!$ hours, ${{\hat{m}}_{{{c}_{2}}}}=300\,\!$ hours. From the cumulative MTBF plot for $b=14\,\!$ hours when $T=1\,\!$, substituting the first set of values, $b=14\,\!$ hours and $\ln 1=0\,\!$, into the equation yields:

\begin{align} {{\alpha }_{1}}= & \frac{\ln (100)-\ln (14)}{\ln (200)-\ln (1)} \\ = & 0.3711 \end{align}\,\!
2. Substituting the second set of values, $b=14\,\!$ hours and $\ln 1=0,\,\!$ into the equation yields:
\begin{align} {{\alpha }_{2}}= & \frac{\ln (300)-\ln (14)}{\ln (3,500)-\ln (1)} \\ = & 0.3755 \end{align}\,\!
Averaging these two $\alpha \,\!$ values yields a better estimate of $\hat{\alpha }=0.3733\,\!$.
3. Now the equation for the cumulative MTBF growth curve is:
${{\hat{m}}_{c}}=14\cdot {{T}^{\text{ }0.3733}}\,\!$
4. Using the following equation for the instantaneous MTBF, or
${{m}_{i}} = \frac{1}{1-\alpha }{{{\hat{m}}}_{c}},:\ \ \alpha \not{=}1 \,\!$
The equation for the instantaneous MTBF growth curve is:
\begin{align} {{\hat{m}}_{i}}=\frac{1}{1-0.3733} \cdot {{14T}^{0.3733}} \end{align}\,\!
The above equation is plotted in the instantaneous MTBF plot shown in the example and in the cumulative and instantaneous MTBF vs. Time plot. In the following figure, you can see that a parallel shift upward of the cumulative MTBF, ${{\hat{m}}_{c}}\,\!$, line by a distance of $\tfrac{1}{1-\alpha }\,\!$ gives the instantaneous MTBF, or the ${{\hat{m}}_{i}}\,\!$, line.
Cumulative and Instantaneous MTBF vs. Time plot

### Mathematical Approach

#### Example 1

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

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{{{\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}\,\!

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

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)$\ln{T_i}\,\!$ (4)$\ln{T_i}^2\,\!$ (5)$m_c\,\!$ (6)$\ln{m_c}\,\!$ (7)$\ln{m_c}\cdot\ln{T_i}\,\!$
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, ${{m}_{c}}\,\!$, 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{{{\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}\,\!

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

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:

\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{{{\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}\,\!

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}\,\!

### Confidence Bounds Example

Using the values of $\hat{b}\,\!$ and $\hat{\alpha }\,\!$ estimated from the least squares analysis in Least Squares Example 2:

$\hat{b}=1.9453\,\!$
$\hat{\alpha}=0.6133\,\!$

Calculate the 90% confidence bounds for the following:

1. The parameters $\alpha\,\!$ and $b\,\!$.
2. The cumulative and instantaneous failure intensity.
3. The cumulative and instantaneous MTBF.

Solution

1. Use the values of $\hat{b}\,\!$ and $\hat{\alpha }\,\!$ estimated from the least squares analysis. Then:
\begin{align} {{S}_{xx}}&=\left[ \underset{i=1}{\overset{n}{\mathop \sum }}\,{{(\ln {{t}_{i}})}^{2}} \right]-\frac{1}{n}{{\left( \underset{i=1}{\overset{n}{\mathop \sum }}\,\ln ({{t}_{i}}) \right)}^{2}} \\ & = 1400.9084-1301.4545 \\ & = 99.4539 \end{align}\,\!
\begin{align} SE(\hat{\alpha })= & \frac{\sigma }{\sqrt{{{S}_{xx}}}} \\ = & \frac{0.08428}{9.9727} \\ = & 0.008452 \end{align}\,\!
\begin{align} SE(\ln \hat{b})= & \sigma \cdot \sqrt{\frac{\underset{i=1}{\overset{n}{\mathop{\sum }}}\,{{(\ln {{T}_{i}})}^{2}}}{n\cdot {{S}_{xx}}}} \\ = & 0.065960 \end{align}\,\!
Thus, the 90% confidence bounds on parameter $\alpha \,\!$ are:
$C{{B}_{\alpha }}=\hat{\alpha }\pm {{t}_{n-2,\alpha /2}}SE(\hat{\alpha })\,\!$
\begin{align} {{\alpha }_{L}}= & 0.602050 \\ {{\alpha }_{U}}= & 0.624417 \end{align}\,\!
And 90% confidence bounds on parameter $b\,\!$ are:
$C{{B}_{b}}=\hat{b}{{e}^{\pm {{t}_{n-2,\alpha /2}}SE\left[ \ln (\hat{b}) \right]}}\,\!$
\begin{align} {{b}_{L}}= & 1.7831 \\ {{b}_{U}}= & 2.1231 \end{align}\,\!
2. The cumulative failure intensity is:
\begin{align} {{\lambda }_{c}}= & \frac{1}{1.9453}\cdot {{22000}^{-0.6133}} \\ = & 0.00111689 \end{align}\,\!
And the instantaneous failure intensity is equal to:
\begin{align} {{\lambda }_{i}}= & \frac{1}{1.9453}\cdot (1-0.6133)\cdot {{22000}^{-0.6133}} \\ = & 0.00043198 \end{align}\,\!
So, at the 90% confidence level and for $T=22,000\,\!$ hours, the confidence bounds on cumulative failure intensity are:
\begin{align} {{[{{\lambda }_{c}}(t)]}_{L}}= & 0.00106780 \\ {{[{{\lambda }_{c}}(t)]}_{U}}= & 0.00116825 \end{align}\,\!
For the instantaneous failure intensity:
\begin{align} {{[{{\lambda }_{i}}(t)]}_{L}}= & 0.00041299 \\ {{[{{\lambda }_{c}}(t)]}_{U}}= & 0.00045184 \end{align}\,\!
The following figures show the graphs of the cumulative and instantaneous failure intensity. Both are plotted with confidence bounds.
3. The cumulative MTBF is:
\begin{align} {{m}_{c}}(T)= & 1.9453\cdot {{22000}^{0.6133}} \\ = & 895.3395 \end{align}\,\!
And the instantaneous MTBF is:
\begin{align} {{m}_{i}}(T)= & \frac{1.9453}{1-0.6133}\cdot {{22000}^{0.6133}} \\ = & 2314.9369 \end{align}\,\!
So, at 90% confidence level and for $T=22,000\,\!$ hours, the confidence bounds on the cumulative MTBF are:
\begin{align} {{m}_{c}}{{(t)}_{l}}= & 855.9815 \\ {{m}_{c}}{{(t)}_{u}}= & 936.5071 \end{align}\,\!
The confidence bounds for the instantaneous MTBF are:
\begin{align} {{m}_{i}}{{(t)}_{l}}= & 2213.1753 \\ {{m}_{i}}{{(t)}_{u}}= & 2421.3776 \end{align}\,\!
The figure below displays the cumulative MTBF.

The next figure displays the instantaneous MTBF. Both are plotted with confidence bounds.

## Failure Times Data Example

A prototype of a system was tested with design changes incorporated during the test. A total of 12 failures occurred. The data set is given in the following table.

Developmental Test Data
Failure Number Cumulative Test Time(hours)
1 80
2 175
3 265
4 400
5 590
6 1100
7 1650
8 2010
9 2400
10 3380
11 5100
12 6400

Do the following:

1. Estimate the Duane parameters.
2. Plot the cumulative and instantaneous MTBF curves.
3. How many cumulative test and development hours are required to meet an instantaneous MTBF goal of 500 hours?
4. How many cumulative test and development hours are required to meet a cumulative MTBF goal of 500 hours?

Solution

1. The next figure shows the data entered into RGA along with the estimated Duane parameters.
2. The following figure shows the cumulative and instantaneous MTBF curves.
3. The next figure shows the cumulative test and development hours needed for an instantaneous MTBF goal of 500 hours.
4. The figure below shows the cumulative test and development hours needed for a cumulative MTBF goal of 500 hours.

## Discrete Sequential Data Example

Given the sequential success/failure data in the table below, do the following:

1. Estimate the Duane parameters.
2. What is the instantaneous Reliability at the end of the test?
3. How many additional test runs with a one-sided 90% confidence level are required to meet an instantaneous Reliability goal of 80%?
Sequential Data
Run Number Result
1 F
2 F
3 S
4 S
5 S
6 F
7 S
8 F
9 F
10 S
11 S
12 S
13 F
14 S
15 S
16 S
17 S
18 S
19 S
20 S

Solution

1. The following figure shows the data set entered into RGA along with the estimated Duane parameters.
2. The Reliability at the end of the test is equal to 78.22%. Note that this is the DRel that is shown in the control panel in the above figure.
3. The figure below shows the number of test runs with both one-sided confidence bounds at 90% confidence level to achieve an instantaneous Reliability of 80%. Therefore, the number of additional test runs required with a 90% confidence level is equal to $42.2481-20=22.2481\approx 23\,\!$ test runs.