Duane Model Examples

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These examples appear in the Reliability growth reference.

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, [math]\displaystyle{ {{\hat{m}}_{c}}\,\! }[/math], 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 [math]\displaystyle{ T=1\,\! }[/math]. You can get the value of [math]\displaystyle{ b\,\! }[/math] 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, [math]\displaystyle{ b\,\! }[/math] is approximately 14 hours. The next figure illustrates this.

   
   Cumulative MTBF plot for [math]\displaystyle{ b \approx 14\,\! }[/math] at [math]\displaystyle{ T=1\,\! }[/math]

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

   [math]\displaystyle{ \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)}\,\! }[/math]

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

   [math]\displaystyle{ \alpha =\frac{\ln \left( {{{\hat{m}}}_{c}} \right)-\ln (b)}{\ln (T)-\ln (1)}\,\! }[/math]

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

   [math]\displaystyle{ \begin{align} {{\alpha }_{1}}= & \frac{\ln (100)-\ln (14)}{\ln (200)-\ln (1)} \\ = & 0.3711 \end{align}\,\! }[/math]

2. Substituting the second set of values, [math]\displaystyle{ b=14\,\! }[/math] hours and [math]\displaystyle{ \ln 1=0,\,\! }[/math] into the equation yields:

   [math]\displaystyle{ \begin{align} {{\alpha }_{2}}= & \frac{\ln (300)-\ln (14)}{\ln (3,500)-\ln (1)} \\ = & 0.3755 \end{align}\,\! }[/math]

   Averaging these two [math]\displaystyle{ \alpha \,\! }[/math] values yields a better estimate of [math]\displaystyle{ \hat{\alpha }=0.3733\,\! }[/math].
3. Now the equation for the cumulative MTBF growth curve is:

   [math]\displaystyle{ {{\hat{m}}_{c}}=14\cdot {{T}^{\text{ }0.3733}}\,\! }[/math]

4. Using the following equation for the instantaneous MTBF, or

   [math]\displaystyle{ {{m}_{i}} = \frac{1}{1-\alpha }{{{\hat{m}}}_{c}},:\ \ \alpha \not{=}1 \,\! }[/math]

   The equation for the instantaneous MTBF growth curve is:

   [math]\displaystyle{ \begin{align} {{\hat{m}}_{i}}=\frac{1}{1-0.3733} \cdot {{14T}^{0.3733}} \end{align}\,\! }[/math]

   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, [math]\displaystyle{ {{\hat{m}}_{c}}\,\! }[/math], line by a distance of [math]\displaystyle{ \tfrac{1}{1-\alpha }\,\! }[/math] gives the instantaneous MTBF, or the [math]\displaystyle{ {{\hat{m}}_{i}}\,\! }[/math], 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:

[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.

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{ \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.

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} \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]

Confidence Bounds Example

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

[math]\displaystyle{ \hat{b}=1.9453\,\! }[/math]

[math]\displaystyle{ \hat{\alpha}=0.6133\,\! }[/math]

Calculate the 90% confidence bounds for the following:

1. The parameters [math]\displaystyle{ \alpha\,\! }[/math] and [math]\displaystyle{ b\,\! }[/math].
2. The cumulative and instantaneous failure intensity.
3. The cumulative and instantaneous MTBF.

Solution

1. Use the values of [math]\displaystyle{ \hat{b}\,\! }[/math] and [math]\displaystyle{ \hat{\alpha }\,\! }[/math] estimated from the least squares analysis. Then:

   [math]\displaystyle{ \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}\,\! }[/math]

   [math]\displaystyle{ \begin{align} SE(\hat{\alpha })= & \frac{\sigma }{\sqrt{{{S}_{xx}}}} \\ = & \frac{0.08428}{9.9727} \\ = & 0.008452 \end{align}\,\! }[/math]

   [math]\displaystyle{ \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}\,\! }[/math]

   Thus, the 90% confidence bounds on parameter [math]\displaystyle{ \alpha \,\! }[/math] are:

   [math]\displaystyle{ C{{B}_{\alpha }}=\hat{\alpha }\pm {{t}_{n-2,\alpha /2}}SE(\hat{\alpha })\,\! }[/math]

   [math]\displaystyle{ \begin{align} {{\alpha }_{L}}= & 0.602050 \\ {{\alpha }_{U}}= & 0.624417 \end{align}\,\! }[/math]

   And 90% confidence bounds on parameter [math]\displaystyle{ b\,\! }[/math] are:

   [math]\displaystyle{ C{{B}_{b}}=\hat{b}{{e}^{\pm {{t}_{n-2,\alpha /2}}SE\left[ \ln (\hat{b}) \right]}}\,\! }[/math]

   [math]\displaystyle{ \begin{align} {{b}_{L}}= & 1.7831 \\ {{b}_{U}}= & 2.1231 \end{align}\,\! }[/math]

2. The cumulative failure intensity is:

   [math]\displaystyle{ \begin{align} {{\lambda }_{c}}= & \frac{1}{1.9453}\cdot {{22000}^{-0.6133}} \\ = & 0.00111689 \end{align}\,\! }[/math]

   And the instantaneous failure intensity is equal to:

   [math]\displaystyle{ \begin{align} {{\lambda }_{i}}= & \frac{1}{1.9453}\cdot (1-0.6133)\cdot {{22000}^{-0.6133}} \\ = & 0.00043198 \end{align}\,\! }[/math]

   So, at the 90% confidence level and for [math]\displaystyle{ T=22,000\,\! }[/math] hours, the confidence bounds on cumulative failure intensity are:

   [math]\displaystyle{ \begin{align} {{[{{\lambda }_{c}}(t)]}_{L}}= & 0.00106780 \\ {{[{{\lambda }_{c}}(t)]}_{U}}= & 0.00116825 \end{align}\,\! }[/math]

   For the instantaneous failure intensity:

   [math]\displaystyle{ \begin{align} {{[{{\lambda }_{i}}(t)]}_{L}}= & 0.00041299 \\ {{[{{\lambda }_{c}}(t)]}_{U}}= & 0.00045184 \end{align}\,\! }[/math]

   The following figures show the graphs of the cumulative and instantaneous failure intensity. Both are plotted with confidence bounds.
   
   
3. The cumulative MTBF is:

   [math]\displaystyle{ \begin{align} {{m}_{c}}(T)= & 1.9453\cdot {{22000}^{0.6133}} \\ = & 895.3395 \end{align}\,\! }[/math]

   And the instantaneous MTBF is:

   [math]\displaystyle{ \begin{align} {{m}_{i}}(T)= & \frac{1.9453}{1-0.6133}\cdot {{22000}^{0.6133}} \\ = & 2314.9369 \end{align}\,\! }[/math]

   So, at 90% confidence level and for [math]\displaystyle{ T=22,000\,\! }[/math] hours, the confidence bounds on the cumulative MTBF are:

   [math]\displaystyle{ \begin{align} {{m}_{c}}{{(t)}_{l}}= & 855.9815 \\ {{m}_{c}}{{(t)}_{u}}= & 936.5071 \end{align}\,\! }[/math]

   The confidence bounds for the instantaneous MTBF are:

   [math]\displaystyle{ \begin{align} {{m}_{i}}{{(t)}_{l}}= & 2213.1753 \\ {{m}_{i}}{{(t)}_{u}}= & 2421.3776 \end{align}\,\! }[/math]

   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 [math]\displaystyle{ 42.2481-20=22.2481\approx 23\,\! }[/math] test runs.