Weibull Distribution - Examples

=General Examples=

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Example 7 In this example, we will determine the median rank value used for plotting the sixth failure from a sample size of ten. This will be used to illustrate two of the built-in functions in Weibull++'s Quick Statistical Reference.

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Solution to Example 7 First, open the ‘‘‘Quick Statistical Reference‘‘‘ by clicking its icon.

F'I'G'U'R'E'H'E'R'EN = 10,j = 6,m = 2(10 − 6 + 1) = 10 or by selecting ‘‘‘Quick Statistical Reference‘‘‘\ from the ‘‘‘Tools‘‘‘ menu. In this example N = 10, &lt;span class="texhtml" /&gt;&lt;span class="texhtml" /&gt;&lt;span class="texhtml" /&gt;, and $$ n=2\times 6=12 $$. Thus, from the &lt;span class="texhtml" /&gt;-distribution rank equation:

$$ MR={\frac{1}{1+\left( {\frac{10-6+1}{6}}\right) F_{0.50;10;12}}} $$ Calculate the value of F0.50:10:12 by using the ‘‘‘Inverse ‘‘‘ F ‘‘‘-Distribution Values‘‘‘ option from the ‘‘‘Quick Statistical Reference‘‘‘, or F0.50;10;12 = 0.9886 as shown next:

F'I'G'U'R'E'H'E'R'E Consequently:

$$ MR={\frac{1}{1+{\frac{5}{6}}\times 0.9886}}=0.5483=54.83\ $$ Another method is to use the Median Ranks option directly, which yields MR(%)=54.8305%, as shown next:

F'I'G'U'R'E'H'E'R'E 

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Example 8 What is the unreliability of the units in Example 1 for a mission duration of 30 hours, starting the mission at age zero? To replicate the results in this reference with Weibull++,\ choose RRX\ (Rank Regression on X) as the calculation method.

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Solution to Example 8 First, use Weibull++ to obtain the parameters using RRX. Then, we investigate several methods of solution for this problem. The first, and more laborious, method is to extract the information directly from the plot. You may do this with either the screen plot in ‘‘‘RS Draw ‘‘‘ or the printed copy of the plot. (When extracting information from the screen plot in ‘‘‘RS Draw‘‘‘, note that the translated axis position of your mouse is always shown on the bottom right corner.) $$ FIGURE HERE $$ Using this first method, enter either the screen plot or the printed plot with $$ T=30 $$ hours, go up vertically to the straight line fitted to the data, then go horizontally to the ordinate, and read off $$ MR=\hat{Q}(T=30 \hbox{ hr})\simeq 23% $$. Then, a good estimate of the sought unreliability is 23%. (Also, the reliability estimate is $$ 1.0-0.23=0.77 $$ or $$ 77% $$) The second method involves the use of the ‘‘‘Quick Calculation Pad‘‘‘ ‘‘‘(QCP)‘‘‘‘‘.‘‘ To activate the ‘‘‘Quick Calculation Pad‘‘‘ ‘‘\ ‘‘either click its icon on the ‘‘‘Data Analysis ‘‘‘toolbar, or ‘‘‘Data Folio Control Panel‘‘‘ $$ FIGURE HERE $$ Click ‘‘‘Calculate‘‘‘ to get the result. The result is shown in the figure above, or $$ \hat{Q}(T)=22.98% $$ Note that the results in QCP vary according to the parameter estimation method used. The above results are obtained using RRX.

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Example 9 What is the reliability for a new mission of $$ t=10 $$ hours duration, starting the new mission at the age of $$ T=30 $$ hours, for the same data as Example 7?

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Solution to Example 9 The conditional reliability is given by: $$ R(t|T)={\frac{R(T+t)}{R(T)}} $$ or: $$ \hat{R}(10\hbox{ hr}|30\hbox{ hr})={\frac{\hat{R}(30\hbox{ hr}+10\hbox{ hr})}{\hat{R}(30\hbox { hr})}}={\frac{\hat{R}(40\hbox{ hr})}{\hat{R}(30 \hbox{ hr})}} $$ Again, the ‘‘‘Quick Calculation Pad‘‘‘ can provide this result directly and more accurately than the plot. $$ FIGURE HERE $$

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Example 10 What is the longest mission that this equipment should undertake for a reliability of $$ 90\ $$?

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Solution to Example 10 Using the ‘‘‘QCP‘‘‘ again, choose ‘‘‘Warranty (Time) Information‘‘‘ and enter the ‘‘‘Required Reliability‘‘‘, $$ 0.90 $$, and click ‘‘‘Calculate‘‘‘. The result is $$ 15.993 $$ hours. $$ FIGURE HERE $$

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Example 11 Assume that ten identical units ($$ N=10 $$) are being reliability tested at the same application and operation stress levels. Six of these units fail during this test after operating the following numbers of hours, $$ T_{j} $$: 150, 105, 83, 123, 64 and 46. The test is stopped at the sixth failure. Find the parameters of the Weibull $$ pdf $$ that represents these data.

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Solution to Example 11 Open a new Data Folio choosing ‘‘‘Times-to-failure data‘‘‘, ‘‘‘My data set contains suspensions (right censored data)‘‘‘ and ‘‘‘I want to enter data in groups‘‘‘. $$ FIGURE HERE $$ Enter the data in the appropriate columns.\ Note that there are four suspensions, as only six of the ten units were tested to failure (the next figure shows the data as entered). Use the three-parameter Weibull and MLE for the calculations. $$ FIGURE HERE $$ Plot the data. $$ FIGURE HERE $$ Note that the original data points, on the curved line, were adjusted by subtracting 30.92 hours to yield a straight line as shown above.

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Example 12 Suppose that we have run an experiment with eight units being tested and the following is a table of their last inspection times and times-to-failure:

Analyze the data using several different parameter estimation techniques and compare the results.

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Solution to Example 12 This data set can be entered into Weibull++ by opening a new ‘‘‘Data Folio‘‘‘ and choosing ‘‘‘Times-to-failure‘‘‘ and ‘‘‘My data set contains interval and/or left censored data‘‘‘. $$ FIGURE HERE $$

The data is entered as follows, $$ FIGURE HERE $$

The computed parameters using\ maximum likelihood are: $$ \hat{\beta }=5.76 $$ $$ \hat{\eta }=44.68 $$

using RRX or rank regression on X:  $$ \hat{\beta }=5.70 $$ $$ \hat{\eta }=44.54 $$

and using RRY or rank regression on Y:  $$ \hat{\beta }=5.41 $$ $$ \hat{\eta }=44.76 $$

The plot of the MLE solution with the two-sided 90% confidence bounds is: $$ FIGURE HERE $$

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Example 13 ACME company manufactures widgets, and is currently engaged in reliability testing a new widget design. Nineteen units are being reliability tested, but due to the tremendous demand for widgets, units are removed from the test whenever the production cannot cover the demand. The test is terminated at the 67$$ ^{th} $$ day when the last widget is removed from the test. Table 6.6 contains the collected data.

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Solution to Example 13 In this example, we see that the number of failures is less than the number of suspensions. This is a very common situation, since reliability tests are often terminated before all units fail due to financial or time constraints. Further, some suspensions will be recorded when a failure occurs that is not due to a legitimate failure mode, such as operator error. \ In cases such as this, a suspension is recorded, since the unit under test cannot be said to have had a legitimate failure. This data set can be entered into Weibull++ using ‘‘‘Times-to-failure‘‘‘ and ‘‘‘My data set contains suspensions (right censored data)‘‘‘. $$ FIGURE HERE $$

We will use the two-parameter Weibull to solve this problem. The parameters using maximum likelihood are: $$ \hat{\beta }=1.145, \hat{\eta }=65.97 $$

using RRX: $$ \hat{\beta }=0.914, \hat{\eta }=79.38 $$

and using RRY: $$ \hat{\beta }=0.895, \hat{\eta }=82.02 $$

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Example 14 Suppose we want to model a left censored, right censored, interval, and complete data set, consisting of 274 units under test of which 185 units fail. Table 6.8 contains the data.

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Solution to Example 14 This data set can be entered into Weibull++ by selecting the ‘‘‘ Times-to-failure‘‘‘ and ‘‘‘My data set contains suspensions (right censored data)‘‘‘,’‘‘\ My data set contains interval and/or left censored data‘‘‘ and ‘‘‘I\ want to enter data in groups‘‘‘ options. $$ FIGURE HERE $$ Since standard ranking methods for dealing with these different data types are inadequate, we will want to use the ReliaSoft ranking method. \ This option is the default in Weibull++ when dealing with interval data. The computed parameters using MLE are: $$ \hat{\beta }=0.748, \hat{\eta }=44.38 $$

using RRX: $$ \hat{\beta }=1.057, \hat{\eta }=36.29 $$

and using RRY: $$ \hat{\beta }=0.998, \hat{\eta }=37.16 $$

The plot with the two-sided 90% confidence bounds for the rank regression on X solution is: $$ FIGURE HERE $$

=Published Examples=

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Example 15: 2-Parameter with RRY From Dimitri Kececioglu, ‘‘Reliability & Life Testing Handbook‘‘, Page 418 [20]. Sample of 10 units, all tested to failure. The times-to-failure were recorded at 16; 34; 53; 75; 93; 120; 150; 191; 240; and 339 hours.

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Published Results for Example 15 Published Results (using Rank Regression on Y):

$$ \hat{\beta }=1.20 $$ $$ \hat{\eta }=146.2 $$ $$ \hat{\rho }=0.9998703 $$

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Computed Results for Example 15 This same data set can be entered into Weibull++\ 7 by selecting the ‘‘‘ Times to Failure\ ‘‘‘type. Use RRY for the estimation method. Weibull++\ computed parameters for RRY are: $$ \hat{\beta }=1.1973 $$ $$ \hat{\eta }=146.2545 $$ $$ \hat{\rho }=0.9999 $$

The small difference between the published results and the ones obtained from Weibull++ is due to the difference in the median rank values between the two (in the publication, median ranks are obtained from tables to 3 decimal places, whereas in Weibull++ they are calculated and carried out up to the 15$$ ^{th} $$ decimal point). You will also notice that in the examples that follow, a small difference may exist between the published results and the ones obtained from Weibull++. This can be attributed to the difference between the computer numerical precision employed by Weibull++ and the lower number of significant digits used by the original authors. In most of these publications, no information was given as to the numerical precision used.

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Example 16: Two-parameter MLE with Interval Data From Wayne Nelson, ‘‘ Applied Life Data Analysis‘‘, Page 415 [30]. One hundred and sixty-seven (167) identical parts were inspected for cracks. The following is a table of their last inspection times and times-to-failure:

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Published Results for Example 16 Published results (using MLE):

$$ \hat{\beta }=1.486 $$ $$ \hat{\eta }=71.687 $$

Published 95% FM confidence limits on the parameters: $$ \hat{\beta }=\left\{ 1.224,1.802\right\} $$ $$ \hat{\eta }=\left\{ 61.962,82.938\right\} $$

Published variance/covariance matrix: $$ \left[ \begin{array}{ccc} \hat{Var}\left( \hat{\beta }\right) =0.0215 & \hat{Cov}( \hat{\beta },\hat{\eta })=-0.2791 \\

\hat{Cov}(\hat{\beta },\hat{\eta })=-0.2791 & \hat{Var} \left( \hat{\eta }\right) =28.432 \end{array} \right] $$

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Computed Results for Example 16 This same data set can be entered into Weibull++\ 7 by selecting the data sheet ‘‘‘Times to Failure‘‘‘, ‘‘‘with Right Censored Data (Suspensions)‘‘‘, ‘‘‘with Interval and Left Censored Data‘‘‘ and ‘‘‘ with Grouped Observations‘‘‘ options, and using MLE. Weibull++\ computed parameters for maximum likelihood are: $$ \hat{\beta }=1.485 $$ $$ \hat{\eta }=71.690 $$

Weibull++ computed 95\\ FM\ confidence limits on the parameters: $$ \hat{\beta }=\left\{ 1.224,1.802\right\} $$ $$ \hat{\eta }=\left\{ 61.961,82.947\right\} $$

Weibull++ computed/variance covariance matrix: $$ \left[ \begin{array}{ccc} \hat{Var}\left( \hat{\beta }\right) =0.0215 & \hat{Cov}( \hat{\beta },\hat{\eta })=-0.2792 \\

\hat{Cov}(\hat{\beta },\hat{\eta })=-0.2792 & \hat{Var} \left( \hat{\eta }\right) =28.4461 \end{array} \right] $$

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Example 17: 3-Parameter MLE with Right Censored Data From Dallas R. Wingo, ‘‘ IEEE\ Transactions on Reliability‘‘ Vol. R-22, No 2, June 1973, Pages 96-100 [37]: Wingo uses the following times-to-failure: 37, 55, 64, 72, 74, 87, 88, 89, 91, 92, 94, 95, 97, 98, 100, 101, 102, 102, 105, 105, 107, 113, 117, 120, 120, 120, 122, 124, 126, 130, 135, 138, 182. In addition, the following suspensions are used: 4 at 70, 5 at 80, 4 at 99, 3 at 121 and 1 at 150.

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Published Results for Example 17, using MLE $$ \hat{\beta }=3.7596935 $$ $$ \hat{\eta }=106.49758 $$ $$ \hat{\gamma }=14.451684 $$

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Computed Results for Example 17, using MLE $$ \hat{\beta }=3.7596935 $$ $$ \hat{\eta }=106.49758 $$ $$ \hat{\gamma }=14.451684 $$

Note that you must have the ‘‘‘Use‘‘‘ ‘‘‘True 3-P MLE on Weibull‘‘‘\ option in the Weibull++\ ‘‘‘User Setup‘‘‘ selected to replicate these results.

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Example 18: 2-Parameter MLE with Right Censored Data From Wayne Nelson, Fan Example, ‘‘Applied Life Data Analysis‘‘, page 317 [30]. Seventy diesel engine fans accumulated 344,440 hours in service and twelve of them failed. A table of their life data is shown next (+ denotes non-failed units or suspensions, using Dr. Nelson's nomenclature). Evaluate the parameters with their two-sided 95\ confidence bounds, using MLE\ for the two-parameter Weibull distribution. $$ \text{Table 6.9 - Nelson's Fan Failure Data (hr), for Example 17 [30]} $$ $$  \ $$

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Published Results for Example 18\ Weibull parameters (two-parameter Weibull, MLE): $$ \hat{\beta }=1.0584 $$ $$ \hat{\eta }=26,297 $$

Published 95\\ FM confidence limits on the parameters: $$ \hat{\beta }=\left\{ 0.64408,\text{ }1.7394\right\} $$ $$ \hat{\eta }=\left\{ 10,552,\text{ }65,535\right\} $$

Published variance/covariance matrix: $$ \left[ \begin{array}{ccc} \hat{Var}\left( \hat{\beta }\right) =0.071958 & \hat{Cov}( \hat{\beta },\hat{\eta })=-2,664.5 \\ \hat{Cov}(\hat{\beta },\hat{\eta })=-2,664.5 & \hat{Var} \left( \hat{\eta }\right) =15.010E+7 \end{array} \right] $$ Note that Nelson expresses the results as multiples of $$ 1000 $$ (or $$ \hat{ \eta }=26.297, $$ etc.). The published results were adjusted by this factor to correlate with Weibull++ results.

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Computed Results for Example 18 This same data set can be entered into Weibull++\ 7 by selecting the data sheet ‘‘‘Times to Failure‘‘‘, ‘‘‘with Right Censored Data (Suspensions) ‘‘‘and ‘‘‘I\ want to enter data in groups (‘‘‘in order to group identical values’‘‘)‘‘‘ options, and using two-parameter Weibull and MLE to calculate the parameter estimates. You can also enter the data as given in Table 6.9 without grouping them by opening a Data Sheet with ‘‘‘Times to Failure‘‘‘ and the ‘‘‘with Right Censored Data (Suspensions)‘‘‘ options. Then click the ‘‘‘Group Data\ ‘‘‘icon and chose ‘‘‘Group exactly identical values‘‘‘. $$ FIGURE HERE $$ $$ FIGURE HERE $$ The data will be automatically grouped and put into a new grouped data ‘‘‘\ ‘‘‘sheet. Weibull++\ computed parameters for maximum likelihood are: $$ \hat{\beta }=1.0584 $$ $$ \hat{\eta }=26,296 $$

Weibull++ computed 95\\ FM\ confidence limits on the parameters:

$$ \hat{\beta }=\left\{ 0.6441,\text{ }1.7394\right\} $$ $$ \hat{\eta }=\left\{ 10,552,\text{ }65,532\right\} $$

Weibull++ computed/variance covariance matrix: $$ \left[ \begin{array}{ccc} \hat{Var}\left( \hat{\beta }\right) =0.0720 & \hat{Cov}( \hat{\beta },\hat{\eta })=-2,664.40 \\

\hat{Cov}(\hat{\beta },\hat{\eta })=-2,664.40 & \hat{Var} \left( \hat{\eta }\right) =15.009E+7 \end{array} \right] $$ The two-sided 95\ bounds on the parameters can be determined from the ‘‘‘QCP‘‘‘, in the ‘‘‘Parameter Bounds‘‘‘ tab. $$ FIGURE HERE $$

Example 19
3-Parameter Probability Plotting From Dimitri Kececioglu, ‘‘Reliability & Life Testing Handbook‘‘, Page 406 [20]. Estimate the parameters for three-parameter Weibull, for a sample of ten units all tested to failure. The times-to-failure were recorded at 200; 370; 500; 620; 730; 840; 950; 1,050; 1,160; and 1,400 hours.

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Published Results for Example 19 Published results (using probability plotting):\ $$ \hat{\beta }=3.0,\hat{\eta }=1,220,\hat{ \gamma }=-300 $$

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Computed Results for Example 19 Weibull++\ computed parameters for rank regression on X are: $$ \hat{\beta }=2.9013, \hat{\eta }=1195.5009, \hat{\gamma }=-279.9000 $$ The small difference between the published results and the ones obtained from Weibull++ are due to the difference in the estimation method. In the publication the parameters were estimated using probability plotting ( ‘‘i.e. ‘‘the fitted line was ``eye-balled''). In Weibull++, the parameters were estimated using non-linear regression (a more accurate, mathematically fitted line). Note that $$ \gamma $$ in this example is negative. This means that the unadjusted for $$ \gamma $$ line is concave up, as shown on the next page. FIGURE HERE