Every day I receive questions by email and phone. I prefer email. If you have The New Weibull Handbook and/or the SuperSMITH software, there is no charge for the answers so feel free to email your questions to me. Ask Dr. Bob. Over the years I’ve answered thousands of questions. I have selected the ones herein because they are either frequently asked or they are interesting in some sense. I hope you will find the answers helpful in rounding out your Weibull background.

The index is upfront so your search is simple and easy. Each word in the index is hyperlinked to the answer. Another hyperlink returns you to the index. If you want to find every instance of the word in this document, use your search function—remember some of the definitions you need are answered under different issues. All references are to The New Weibull Handbook (NWH) unless otherwise indicated.

This document may be downloaded as a PDF (200 KB) from Paul Barringer's website. Send me your email comments and other questions for inclusion into this document.

Last revision: May 27, 2005
 

Quality, Probability, Sampling [index]

July 2005 From Richard Unkle at GE Transportation:

Hi Bob. I'd butter you up a bit for some help on this next question, but know it's probably not necessary, as I believe you'll have a simple answer.

Here's the situation. There's a particular part on our locomotive that has failed at least 5 times, that we know of, in a given failure mode. Now, all five failures have come from parts made by the same supplier. No other supplier's part has experienced this failure mode. However, it's been 5 out of more than 1 million parts made, versus zero failures out of much more than a million parts made by all other suppliers combined. I thought about doing a weibull on the 5 failures, and then comparing that to a one-parameter weibull for the rest of the data. The problem is we don't have all the data necessary (in terms of accurate suspension and even failed part ages) to do that, and would be too expensive and time consuming to get
the data.

So, barring the above approach, is there some kind of procedure that works as follows? If I have zero failures out of X million parts made by multiple suppliers, what's the likelihood that I would see 5 failures out of Y million parts made by any single supplier? I have some thoughts on how to do that, but thought you might now without having to think about it too hard.

As always, I am in your debt for sharing your knowledge with me.

Best Regards, Rich

Dr. Bob’s reply: Let's assume the bad vendor has produced B parts and the good vendors have produced G parts.

If B=G: the probability is (1/2)^5 = 1/32. This is based on the probability of randomly selecting a B part equals 1/2. Rigorously we should use sampling without replacement for the second and later draws, .but this is a negligible difference for large samples. Conclusion or inference: This is unlikely but it could happen.

If there are 2B=G parts the probability is (1/3)^5 = 1/243. Inference: This probability is so small (0.4%) I would say it is highly unlikely to have happen due to random chance.

If there were 3B=G parts the probability is (1/4)^5 = 1/1024. Inference:This is extremely small, on the order of three sigma for the normal. Statisticians would say it could not happen due to random chance. You have one bad vendor, producing a batch problem, as a result.

Take Care, Bob

Adjusted Rank Algorithm, Plotting Positions with Suspensions [index]

April 2003. I want to know if there is more information about how the adjusted rank algorithm (that adjusts the plotting positions for suspensions) was calculated. You reference a Drew Auth as the engineer that came up with the simplified version of something originally done by Leonard G. Johnson. His book, "The Statistical Treatment of Fatigue Experimentation," has been out of print for some time. Any help you could give in tracking down how this adjusted rank algorithm was derived would be most appreciated.

Dr. Bob’s reply: The derivation in Johnson’s book is not too clear; however Charles Mischke’s ASME paper [pdf] is excellent, very clear. See also Chapter Two for a case study of using the Drew Auth formula.

Aggregate Cumulative Hazard, Batch Problems [index]

A batch problem means that one of the failure modes only applies to a portion, a subset of the full sample. The other units are immune to the batch failure mode. Batch problems are common in production when something is changed. Usually the newer units are affected. Throughout the Weibull Handbook there are clues for identifying a batch problem. They are summarized in Chapter 8. The Aggregated Cumulative Hazard method, (ACH), was developed by Rolls Royce. It detects batch problems using the hazard function and provides a plot of the batch. The size of the batch is approximately indicated by the plot. The hazard function is the probability density function divided by the reliability function which gives the instantaneous failure rate at any age. The cumulative hazard function is summation of the hazard function over time. See Appendix F for a complete presentation with examples.

Average & Standard Deviation [index]

January 2003. Hi Dr. Bob, I was your student last year in the Weibull course at Lohmar, Germany. I see that the average and the standard deviation estimates in the SuperSMITH software are different from the simple average and standard deviation (as calculated by Excel, for example). Please explain why there are differences. Thanks a lot for your help.

Dr. Bob’s reply: You cannot use the usual equations for the mean and standard deviation because they do not account for suspensions. Instead we solve for eta and beta by whatever method you have selected and then use the Weibull equations in Appendix G to estimate the mean, mu, and standard deviations. Bob you may have to life these equations from Paul's CD.

Batch Problem Clues [index]

July 2003. In my research, I have come across an example wherein I get some conflicting results when comparing the standard Weibull 2-parameter regression model to the MLE-RBA model (four failures in the example with lots of suspensions). The MLE and regression solutions agree quite well, but the MLE-RBA does not. When doing risk analysis, the MLE and regression models predict quite closely what the actual number of failures are, but the RBA model would lead one to believe that a batch problem may exist. Without delving deeper into the handbook, perhaps you could provide some guidance on this example, as it could impact the research I am doing when I have to use small failure samples.

Dr. Bob’s reply: I will have to give you the prize for coming up with interesting problems. Although I admit the "now risk" and the ACH batch detection methods show only a small probability of a batch problem, the number one test - lots of right suspensions, shows strong evidence of a batch problem or phony data. The probability of having 47 right suspensions at the B10 life without a failure is significantly less than 1%. The fact that MLE beta is less than the MRR beta is more evidence, (discovered by Geoff Cole of Rolls Royce). For this small sample we expect the MLE beta to be greater than the MRR beta. You either have a batch problem or bad data. Note that if you delete the 47 right suspensions you have almost a perfect fit and a steeper beta. I suspect there is something wrong about the right suspensions.

Confidence Intervals, System Reliability, Component Estimates [index]

June 2003. I need to compute the confidence intervals (upper and lower) for the failure rate of a system with three identical components in series. I have the individual component reliability and of course the individual failure rate with its confidence bounds based upon a particular set of measurements. The distribution is log normal and I have the estimated parameters for the probability density function. This data can give me a median CDF with confidence intervals. What I want is to take this same data and get CDF (probability of failure) for a system with three of these components in place and the associated confidence intervals for this system. The CDF would be for one to three failures. I know I can get probability of failure from 1-(Rel)^3 but I do not know how to get the associated confidence intervals.

Dr. Bob’s reply: You won't like my answer. Confidence interval estimates are a function of data, and you have no data on the system. Further, confidence interval estimates are not invariant under transformation which means you must always calculate the confidence estimates last in the calculation. You cannot calculate confidence intervals from confidence intervals as the result is garbage. Therefore, the answer is there is no way you can make the confidence interval estimate for the system without data on the system. ... Can you play games with this? Of course you can, but not rigorously. You might divide the time on the components tested by three and say "if we had tested them in series" we would have had F failures in T time and go from there...As long as your customer understands the meaning of "if" you could do this. Strictly speaking the answer would not be a confidence interval...but your customer might buy it.

Goodness of Fit, Cramer-Mise and Kolmogorov-Smirnoff [index]

May 2003. My statisticians recommend Cramer-Mise and Kolmogorov-Smirnoff as measures of goodness of fit, but you do not. Why?

Dr. Bob’s reply: Comparing all the known goodness of fit tests the two best goodness of fit tests are the likelihood ratio test and our "p" value for the correlation coefficient squared (Coefficient of Determination), [as shown in Chi Chao Liu's thesis]. The likelihood ratio test described in Chapter 5, is used with MLE or better with MLE Reduced Bias Adjustment. The p value for r squared presented I Chapter 3, goes with median rank regression, X on Y. Of course, either technique may be used with either method. For testing goodness of fit both are excellent and we recommend using both. For distribution analysis, both work well, but you need at least 21 failures to have enough information to make a credible choice with any method. Less than 21 failures always use the Weibull 2p even if you know the data are log normal. If you do not have Chi Chao Liu's thesis you may download in two parts: 1) Abstract (1.2 Meg) 2) Dissertation (15.8 Meg).

Mixtures of Failure Modes, Competing Failure Modes [index]

April 2003. The failures were produced by two competitive, "dueling," failure modes. We cannot separate the data into two failure modes. How can we do a failure forecast?

Dr. Bob’s reply: The cumulative probability of failure considering both modes is [1-(F(t1))x(F(t2))]. The first step is do a mixture analysis with WSW. If the "p" value supports two Weibulls rather than one Weibull, it is now possible to do an Abernethy risk analysis with the WSW, version 4.0V and later. Alternatively, you could use Monte Carlo Simulation. "Raptor" would be a good choice for the simulation software.

Exponential Distribution [index]

May 2003. In Chapter 8 discussing the exponential distribution, you mention the mean time between in-flight shutdowns for commercial engines of 25,000 engine operating hours. As typical commercial flights are 2-4 hours this number seems extraordinary.

Dr. Bob’s reply: At the time I wrote that section, 25,000 was the standard; today it is much higher and it is extraordinary. To put that in context, years ago I had lunch with Dr. Von Ohain with the President of Pratt & Whitney Aircraft. Dr. Von Ohain invented the gas turbine and developed the Jumo engine for Germany . He said the mean time between in-flight shutdowns on the Jumo engines in the ME262 was 25 hours. When I expressed shock, he said that was good enough as the mean life of the ME262 was 7 hours and 10 minutes. In the six decades that have followed World War II we have made progress [with gas turbines]!

Small Samples Sizes, Failures (few data points) [index]

July 2003 When I recently attended your excellent Weibull Workshop you said best practice for 20 failures or less was to always use the two parameter Weibull even if you know that the log normal is the parent distribution. This is very hard for me to believe. Could you give some reasoning for this recommendation?

Dr. Bob’s reply: This recommendation comes from Chi Chao Liu’s thesis. His massive study of thousands of Weibull and log normal data sets with and without all types of suspensions showed that the Weibull 2P is always more conservative in the lower tail than the log normal. It also showed that for betas of one or less the mean square error of the log normal was much greater, even an order of magnitude greater than the Weibull 2P even when the data was log normal. For betas greater than one the log normal MSE is slightly less than the Weibull 2p for log normal data but the log normal B lives from the lower tail are always more optimistic than the Weibull 2P B lives. Therefore, for engineering problems, we recommend the Weibull 2P for all samples with 20 failures or less. For more detail see Liu’s thesis available as a free download [abstract or dissertation].

Interval and Inspection Data, Inspection Option, Data Input [index]

February 2003. With interval and inspection data I use the data shortcut to input points at the same value. For example, 88x9 means there nine failures occurred at time eighty-eight. This data appears as the number 9 on the plot. If I change from median rank regression (MRR) to the Inspection Option the plot position of the 9 changes. Why?

Dr. Bob’s reply: If you input the 9 failures in separately they would appear as a vertical column of points on the Weibull plot. With MRR the 9 is located in the middle of this column which is approximately where we expect the standard Weibull line to appear. With the Inspection Option the 9 is located at the topmost point of the column, where we expect the Inspection Option line to appear.

Mean Time Between Failure (MTBF) and Mean Time To Failure (MTTF) [index]

May 2003. What is the difference between MTTF and MTBF?

Dr. Bob’s reply: Mean Time To Failure (MTTF) is the average life of the part. Mean Time Between Failures (MTBF) is the average time between failures for all the parts in the fleet. To estimate MTBF divide the total operating time on all parts by the number of failures. With a "complete" sample, (no suspensions) the two parameters are identical. MTTF does not change with or without suspensions other than statistical scatter. MTBF is heavily influenced by suspensions. For example for wearout failure modes a young fleet will have enormous MTBF but MTTF does not change. For a very old fleet that has been through many overhauls and parts replacements, MTBF will converge toward MTTF.

These two parameters are different and have different applications. MTTF is the parameter of the exponential distribution and is related to eta, the characteristic life of the Weibull distribution. MTBF is used for fleets of repairable systems and is useful for maintainability analysis.

Unbiased Estimates, Bias, Mean Square Error, MLE VS [index]

February 2003. In the New Weibull Handbook you use median bias for most of the comparison of methods and yet you write about unbiased estimates. I thought an unbiased estimate is one whose expected value equals the true value. I am confused as the expected value is the mean value. Correct?

Dr. Bob’s reply: You are correct that an unbiased estimate is one whose expected value equals the true value and that the mean value is the expected value. However, the mean value is not a good measure, or a typical measure, for skewed distributions and most life distributions are skewed. We recommend the median value instead of the mean value for skewed distributions. Statistical estimates will split 50/50 around the median. For example, we recommend the median rank plotting instead of the mean rank positions for life data analysis. In the case you selected, five failures, B1 and beta are highly skewed. See the simulation results below. For comparing the accuracy of alternative methods like MRR versus MLE, I use the median bias rather than the mean bias. With MLE-RBA Weibull we employ the median bias correction as the standard or default, but we also offer a mean bias correction in our software. These are quite different corrections; the median bias correction is C4^3.5 versus C4^6 for mean bias. The mean bias correction is much larger.

MonteCarlo Simulation N=5, True Values eta=1000, beta=1, B1=10….. Median /Mean

Note that statisticians prefer mean square error to bias as a measure of accuracy. Engineers prefer bias as a measure because they want to know if the estimate is optimistic or pessimistic. If a failure produces a health, or death, or crash risk, engineers want either an unbiased estimate or a conservative estimate. Optimistic bias estimators like small sample MLE failure forecasts and B life are unacceptable. Our conclusion is to recommend methods with the smallest median bias as our best practice. MRR and MLE-RBA for small and moderate sample sizes are examples.

Mixtures of Failure Modes, System Models, Exponential [index]

June 2003. I have read much of your Handbook and have been learning how to apply Weibull analysis to benefit our company. I am writing to see if you could elaborate for me on a statement made on p. 46 of the Handbook: "Weibulls for a system or component with many modes mixed together will tend toward a beta of one. These Weibulls should not be employed if there is any way to categorize the data into separate, more accurate failure modes. Using a Weibull plot with mixtures of many failure modes is the equivalent of assuming the exponential distribution applies. The exponential results are often misleading and yet this is common practice." … I have performed analyses on individual failure modes pertaining to our product, but I anticipate a management will request a Weibull analysis for a product with all failure modes lumped together, for the purposes of making decisions regarding warranty terms. According to your statement above, this would appear to be a risky proposition. However, I do not fully understand why. What would be the harm in lumping all failure modes together and performing Weibull analysis in order to determine the overall failure risk for the product?

Dr. Bob’s reply: Your method will forecast the same number of claims each month for a constant fleet size...This does this make sense? A better answer is use Monte Carlo simulation. There is free software available to do that called "Raptor." Download it from the net at Raptor.com. Or better yet, make a failure-warranty forecast for each failure mode from the individual Weibulls and plot them all in WinSMITH Visual, then sum the Y values to obtain the system forecast of cumulative failures by months. For repairable systems, this method produces accurate forecasts. Lumping them all together assumes no wearout modes, no infant mortality, everything is exponential. Does that make sense? You might consider attending one of our Weibull Workshops. We will teach you everything about warranty forecasting with Kaplan Meier, the inspection option and Crow-AMSAA.

Random or Exponential Data, Confidence Bounds for Beta [index]

May 2003. Hi Dr. Bob, I'm lost. I have a Beta of 0.80 and the BetaU is 1.03 and BetaL is 0.70. How does that tell me if the failure is close enough to be "random"/exponential?

Dr. Bob’s reply: Your data is not significantly different from an exponential at whatever level of confidence you selected to get the bounds on beta because the value one lies within the interval. If your double sided confidence bounds are at 90% confidence, your data is not significantly different from an exponential at 90% confidence.

Statistical Text References [index]

May 2003. I have recently been reassigned and need to use statistics in my new position. Could you recommend some introductory statistical texts?

Dr. Bob’s reply: Some recommendations:

• "Introduction to Statistical Analysis," Dixon and Massey, McGraw Hill
• "Introduction to the Theory of Statistics," Mood & Graybill, Mcgraw Hill
• "Introduction to Statistical Inference," Keeping, Van Nostrand...this one is a little heavier..
• "The Cartoon Guide to Statistics," Gonick and Smith, HarperPerennial, This is very easy to read and really quite good. paperback.

The first three are old standards. Suggest Amazon.com books.

Unbiased Estimate [index]

In the New Weibull Handbook you use median bias for most of the comparison of methods and yet you write about unbiased estimates. I thought an unbiased estimate is one whose expected value equals the true value. I am confused as the expected value is the mean value. Correct?

Dr. Bob’s reply: You are correct that an unbiased estimate is one whose expected value equals the true value and that the mean value is the expected value. However, the mean value is not a good measure, or a typical measure, for skewed distributions and most life distributions are skewed. We recommend the median value instead of the mean value for skewed distributions. Statistical estimates will split 50/50 around the median. For example, we recommend the median rank plotting instead of the mean rank positions for life data analysis. In the case you selected, five failures, the sampling distributions of B1 and beta are highly skewed. See the simulation results below. For comparing the accuracy of alternative methods like MRR versus MLE, I use the median bias rather than the mean bias. With MLE-RBA Weibull we employ the median bias correction as the standard or default, but we also offer a mean bias correction in our software. These are quite different corrections; the median bias correction is C4^3.5 versus C4^6 for mean bias. The mean bias correction is much larger.

MonteCarlo Simulation N=5, True Values eta=1000, beta=1, B1=10….. Median /Mean
 

Note that statisticians prefer mean square error to bias as a measure of accuracy. Engineers prefer bias as a measure because they want to know if the estimate is optimistic or pessimistic. If a failure produces a health, or death, or crash risk, engineers want either an unbiased estimate or a conservative estimate. Optimistic bias estimators like small sample MLE failure forecasts and B life are unacceptable. Our conclusion is to recommend methods with the smallest median bias as our best practice. MRR and MLE-RBA for small and moderate sample sizes are examples.

Mixtures of Two Failure Modes [index]

The failures were produced by two competitive, "dueling," failure modes. We cannot separate the data into two failure modes. How can we do a failure forecast?

Dr. Bob’s reply: The cumulative probability of failure considering both modes is [1-(R(t1))x(R(t2))]. The first step is do a mixture analysis with WSW. If the "p" value supports two Weibulls rather than one Weibull, it is now possible to do an Abernethy risk analysis with the WSW, version 4.0V and later. YBATH is a good alternative to the WSW mixture solution. You could also use Monte Carlo Simulation. "Raptor" would be a good choice for the simulation software.

Warranty Claims Forecast, Kaplan-Meier, Inspection Option [index]

April 2003. For warranty claims forecasting by age you recommend both the Inspection Option and the Kaplan-Meier model. Which should I use?

Dr. Bob’s reply: We have research underway comparing all the interval methods, Inspection Option, Probit, K-M, and Interval MLE but it is not completed. Industry seems to prefer the Inspection Option for warranty claims by age but many use K-M.

There is a problem with KM. When we use the KM with the actuarial correction, some of your suspensions are eliminated from the data set. This prohibits using the Abernethy Risk forecast as the suspension histogram is wrong. However, this does not effect the estimate of per cent claims at the end of the warranty period which is the usual objective. The Inspection Option does not have this problem.

If the probability of repeat warranty claims for the same problem is significant you may want to consider Wayne Nelson’s Graphical Repair method described in Appendix M. May 2005 update. The research mentioned above is almost completed and suggests the MLE-Interval method is most accurate, Probit next, and then the Inspection Option although the differences are not large. K-M was not included in the study. Based on these results from Todd Marquardt, we now recommend the MLE-Interval method for warranty claims predictions by age.

Weibull Analysis [index]

Use of the Weibull distribution provides accurate failure analysis and risk predictions with extremely small samples using a simple and useful graphical plots. Solutions are possible at the earliest stage of a problem without the requirement to "crash a few more". Small samples also allow cost-effective component testing. Weibull analysis is a key discipline for reliability, maintainability, safety, and supportability (RMS) engineering largely because of new, credible, and accurate quantitative methods. See Chapter One.

Weibull 3-Parameter [index]

March 2003. When I use a three parameter Weibull the slope, beta, is less than the two parameter beta. Which beta is correct?

Dr. Bob’s reply: The 3-P Weibull is a much more complex distribution than the two parameter and we have fixed requirements to meet before we adopt the 3P solution. Remember the four hard fixed rules for using 3-parameter:

• you must have 21 or more failures, some experts say 100.
• you must be able to explain why the physics of failure support a guaranteed failure free zone,
• the 2-p plot should show curvature,
• and the distribution analysis must favor the 3-p.

If you meet all these criteria above the 3-p distribution is the best distribution and the 3-P beta is the correct beta. The 2-p beta is irrelevant. For example, when we have some data sets with missing data and use the 3-p and compare it to data sets without missing data from the same source the 2-p with all the data fall on top of the 3-p with missing data, same beta.

Temperature Weibulls, Challenger Space Shuttle, NASA [index]

The Challenger Accident. I was honored by being invited to do a Weibull Workshop at the NASA Goddard Space Flight Center. One of the NASA engineers asked me if he could do a Weibull plot with the data on O-ring damage from the Challenger-Space Shuttle history. He provided the following data and linear plot which are available on the Internet [Dunar]:

This curve fit on the linear plot is obviously not very good. More complex regression modeling is attempted in [Wujek] but it also has poor goodness of fit. These models are better than nothing but inadequate.

My Weibull experience with temperature is based on increased temperatures accelerate failures but here the reverse is true. What to do? I thought perhaps if we took the reciprocal of absolute temperature as the accelerating parameter that might work.

The figure below shows my results. Beta is 83 indicating extremely rapidly increasing O-ring damage with colder temperatures. The fit is excellent with a "p" value for the coefficient of determination of 74.5%. (Any value above 10% is considered an acceptable fit.) The probability of damage at the Challenger’s 31 degrees F is 100%. It is easy to be smart retrospectively and I do not mean to criticize NASA for whom I have great respect. However, it certainly would have been enlightening if they had made this Weibull plot.

References:

1. A. J. Dunar & Stephen P. Waring, "Power To Explore: History of Marshall Space flight Center 1960-1990," Government Printing Office, 033-000-01221-7, Chapter IX, "The Challenger Accident."
2. J. H. Wujek, "Challenger O-Ring Data Analysis," OnLine Ethics Center.

Are two designs significantly different? [index]

A problem from Ronald Shop, DAF Trucks, Netherlands:

The data is extremely small:
1. Old Design: Failures at 5768, 6230, 2394, and 3390.
2. New Design: 1 Failure at 11020; 1 Suspension at 3666.

Using Weibayes the lower 90% bound on the new set is to the right of the Weibull line. Does this mean the new design is significantly better? If I use the likelihood ratio test the p value is 74% using the Fulton Factor. These results seem inconsistent. Which method should I believe? Your comments please.

Dr. Bob’s reply: To the degree that you have a good estimate of beta, Weibayes would be the best test, and more precise than the likelihood ratio (LR) test. Weibayes eliminates the uncertainty in beta which is larger than the uncertainty in eta; the LR test includes both uncertainties. My analysis using MRR estimated beta as 2.287 as you show on your plot. If I use MLE-RBA the estimate is 1.954. Using either of these betas shows the 90% bound to the right of the Weibull plot indicating a significant difference. Further, in another E-mail Ronald told me they had prior experience with this failure mode showing a beta of 2.2 which he used for his Weibayes lower bound. Ronald is commended for an excellent analysis.

MLE vs. Regression [index]

Dr. Bob, I'd appreciate hearing any comments you might have about this article. It is a discussion of MLE vs. Regression in the case of many suspensions. Thanks, James

In previous issues, we have discussed the maximum likelihood estimation (MLE) and rank regression methods for parameter estimation. We have also mentioned that when your data set contains a large number of suspensions, it is suggested that you use MLE for parameter estimation. But why? We will take a closer look at this "rule of thumb" and illustrate how rank regression does not take into account the characteristics of the entered data when there are a large number of suspensions.

Even though the median rank adjustment method is the most widely used method for performing suspended items analysis, there is a shortcoming to the method that must be understood. As you may have noticed when using this analysis method for suspended items, the position where the failure occurred is taken into account but not the exact time-to-suspension. For example, rank regression would yield the exact same results if late suspension times are changed but are still late, that is, beyond the last failure.

This shortfall is significant when the number of failures is small and the number of suspensions is large and not spread uniformly between failures, as with these data sets. In cases like this, it is highly recommended that you use maximum likelihood estimation (MLE) to solve for the parameters instead of least squares, since maximum likelihood does not look at ranks or plotting positions, but rather considers each unique time-to-failure or suspension.

Dr. Bob’s reply: James, they missed the point! Which method is most accurate? MRR or MLE with suspensions? My objective is to use the method that is most accurate and conservative. MRR is almost more accurate than MLE with small samples and suspensions.

To answer the question we would use Monte Carlo simulation with all kinds of suspensions in large and small numbers with small samples of failures. The combinations are huge!! I have of course completed many of these studies and in they favor MRR in general but you can find cases that favor MLE.

Many others have looked at this. Dr. Wenham at GKN International in Wolverhampton was promoted to be the worldwide reliability geru for the corporation. Her predecessor, a mathematical statistician, had made MLE the corporate standard. Questioning this decision Dr. Wenham did a massive study with and without suspensions and concluded MRR was more accurate and changed the corporate standard to MRR. "The Rank Regression (RR) method calculates Weibull parameters with the best combination of accuracy and ease of interpretation over more of the range of conditions thought to be common within GKN than any other method investigated." Further, under Recommendations: "The RR method should be adopted for Weibull analysis throughout GKN."

The largest study ever completed was done by Dr. Chi Chao Liu for his doctoral thesis. He looked at thousands of simulations with all kinds and sample sizes of suspensions. His conclusion was the same as Dr. Wenham, MRR is more accurate than MLE for small samples with and without suspensions. His thesis is available as a download from Barringer's website below if you want to read it. "The results show that the empirical equations fit the data very well if the Median Rank Regression method is used. Points estimated (from the Weibull and lognormal) by MLE are more scattered."

Both studies are referenced in The New Weibull Handbook. Waloddi Weibull's position on the subject: [Weibull 1967] states, "The Maximum Likelihood Method is the most efficient one of the general methods, but it requires complicated equations, and, above all, the estimates from small samples are heavily biased." This was written by Weibull after decades of study and numerous research projects, many for the US Air Force.

Finally, the criticism of MRR that it does not weigh suspensions as much as MLE is a two-sided sword. Most in-service data includes suspension data that is somewhere between inaccurate, wrong and non-existent. The fact that MLE heavily weighs suspensions is bad with bad suspension data, but with MRR it makes little difference, in other words MRR is more robust.

Zero Failure Testing

Subject: n = -(eta/t)^beta ln(1-Confidence)

Dear Dr. Robert Abernethy, This is how I thought about the above equation. Please comment about it. The equation was created by combining n =ln (1-conf) / ln (Reliability) and R(t)= -(eta/t)^beta. They are like Oranges and Apples, [Discrete and continuous, Non-time Dependent and Time Dependent, respectively] He has combined two valid equations, coming from two different disciplines to create a non-valid equation. In one hand, you take beta=1 while allowing to use whole spectrum of beta values. Therefore, the equation is not a valid equation.

Dr. Bob’s reply: The equation relates to equation 6-3 in The New Weibull Handbook and is the basis for the zero failure Weibayes testing. These tests have the minimum test-time requirements for substantiation among all other types of Weibull tests so they are very popular. I developed or designed these tests in the seventies so they have been around a long time. If you have "The New Weibull Handbook" it provides the background.

It is apples and oranges but in this case it is quite proper and rigorous. Start with the zero failure term of the binomial for successes and failures. Probability of zero failures = p^n where p is the reliability and n is the number of trials. This is the apples, (discrete) part of the problem. The null hypothesis for substantiation tests is that the reliability is no better than the value to be demonstrated and the probability of passing the test if that is true is the complement of the confidence. Therefore, (1-C) = R^n. Now we substitute the Weibull reliability function for R and solve for t, the test duration. This is oranges part, the continuous distribution. This is a simple conjugation.

System reliability often requires mixing apples and oranges. More often the conjugation of apples and oranges for systems is complex and may require Monte Carlo. For example, my doctoral thesis (1965) modeled the reliability of large liquid rocket engines, conjugating the binomial requirement of start, accelerate, operate and shutdown without mechanical failure, with the performance requirements of specific impulse, mixture ratio control and thrust, all normally distributed. Although I was able to derive analytical solutions for conjugate Bayesian and likelihood solutions, the method was too complex to ever be widely accepted. Fortunately my thesis was accepted.

Hope I have helped. Let me know if you have questions.

Zero Failure Weibayes [index]

Dr. Bob: We have a real problem with a new component under test. We want to run a zero failure Weibayes test to prove it meets the contractual requirement but we have absolutely no information on what the dominant failure mode may be, let alone the beta. The requirement is 99% reliability at 10,000 cycles with 90% confidence. How long should we test how many components?

Dr. Bob’s reply: Beta will be somewhere between 0.7 and 5.0 for almost all good components. Design your test with each beta, 0.7 and 5.0, and use the test with the longest test duration. The alternative is to use a Binomial test, see Table 8-1, but the sample sizes are too large as the binomial is imprecise. 230 tests for 10,000 cycles without failure would meet your requirement.