
Fractional Failure Analysis in Weibull++
If a corrective action for a particular failure mode in a product is 100% effective (backed up with strong engineering verification data), then the reliability of the product in the absence of that failure mode can be estimated using available data. First, the failures due to the mode in question are discounted by changing their state from failure to suspension. Then, the data are reanalyzed to predict what the reliability of the product will be after the corrective action is implemented. However, if the corrective action is not 100% effective, then the method above cannot be used to estimate reliability. In Synthesis Version 10, Weibull++ standard folios can perform fractional failure analysis, where one can account for the effectiveness of an assumed corrective action or product redesign in the analysis.
If the corrective action is not 100% effective, then instead of changing the failure to a suspension as in the previous case, one can take credit for a certain percentage of the failure mode being removed by changing it to a partial failure and a partial suspension. Specifically, if the efficiency of the corrective action is E%, where E% ranges from 0% to 100%, then (100%  E%) of the failure remains a failure and E% of the failure becomes a suspension. For example, if the effectiveness of the fix implemented for a failure mode is known or estimated to be 20% effective, a failure at 10 hours would become an 80% failure at 10 hours and a 20% suspension at 10 hours. In Weibull++, you change the Number in State column entry for the original failure from 1 to 0.8 and the software creates the 0.2 suspension in the background.
The same logic can be applied to a redesign that is implemented to improve a product’s reliability upon collection of data from a reliability test or field failures. Assuming that the effectiveness of the redesign is known or estimated, the failure count can be adjusted and reanalyzed for the data set. Determining the theoretical improvement of reliability due to a product redesign is necessary for making decisions such as how to design a reliability demonstration test or how to estimate the improvement in field reliability for the redesigned product.
Example
To better understand how fractional failure analysis works, we will look at a simple example.
Assume that you work at a car manufacturing company and your task is to test the reliability of the heating, ventilation and air conditioning (HVAC) system that is going to be used in the latest model. The HVAC unit has a newly designed compressor. You focus your attention on the compressor, since it is the very heart of the air conditioning system, and try to improve the overall reliability of the HVAC unit by improving the design of the compressor. To meet the overall reliability requirement for the HVAC unit, the reliability requirement for the new compressor should be 90% at 60,000 miles (i.e., the company would like to give a warranty of 60,000 miles on the compressors and does not want to have more than 10% of the units returned by the end of the warranty). You run a reliability test on 20 compressors and collect the data provided in the table below.
MilestoFailure Data  
199,742  114,277  225,487  144,249 
264,677  104,917  282,435  134,718 
183,481  149,342  210,678  178,686 
130,612  39,190  160,521  61,276 
26,029  32,732  44,168  53,055 
Using Weibull++, you analyze the data by using rank regression and assuming a 2PWeibull distribution. The resulting parameters are shown below.
The reliability is estimated to be 81% at 60,000 miles.
You realize that this reliability value is lower than the requirement. You talk to the technical team and learn that you can actually improve the compressor design by 70%. So you reanalyze your data by using fractional failure analysis to see if the compressor can meet the requirement, assuming that your technical team can actually improve the design by 70%. In a new data sheet, you set the Number in State to 0.3 for all data points. Note that you use MLE for running this analysis because you have 14 (i.e., 70% * 20) suspensions and 6 (i.e., 30% * 20) failures.
The increase in the eta (characteristic life) is almost twofold. With the new hypotheticallyimproved design of the compressor, the reliability of the compressor increases to 95% at 60,000 miles.
The probability plot for the original data (black line) and the improved design data (blue line) shows the similar beta values obtained from each analysis, meaning that the failure mechanism stayed the same while the eta value (characteristic life) is increased significantly as the improved design's probability plot shifted to the right.
Now that the compressor surpasses the reliability requirement, you start wondering if there is a way to save money while still achieving the desired 90% reliability. So you talk with your technical team once again and they tell you that the compressor actually fails due to two main failure modes. The first is thrust bearing failures (denoted as TB) and the second is shaft seal failures (denoted as S). In addition, they tell you that they can achieve the same design improvement (70%) for each failure mode. With this new information available, you can check to see if you can meet the overall reliability requirement for the compressor unit by focusing on a single failure mode, either TB or S. You will repeat the analysis for each scenario, where the design is improved by 70% for each failure mode separately. For this, you use the Competing Failure Modes analysis in Weibull++ because you now have two separate failure modes that can cause the compressor to fail.
In the first scenario, you test the effect of the design improvement on the shaft seal failures. Therefore, you enter 0.3 failures for each of the previously observed shaft seal failures (i.e., efficiency of corrective action = 70%). You choose CFMWeibull as your distribution model and use the MLE method for parameter estimation due to the high number of suspensions. Mode 1 is assigned as TB and Mode 2 is assigned as S.
The new reliability for the compressor if only shaft seal failures are addressed is 87% at 60,000 miles. This reliability does not meet your requirement of 90%.
Hence, you use the same methods to evaluate the second scenario where you test the effect of the design improvement on the thrust bearing failures.
The new reliability for the compressor, if only thrust bearing failures are addressed, is 91% at 60,000 miles. Since this reliability meets the requirement of 90%, you decide to go with just the design improvement on the thrust bearings rather than improving the overall design of the compressor.
The following plot shows the reliability requirement and the reliability curves for each given scenario in this example. If the design of the overall compressor or just the thrust bearings is improved by 70%, then the reliability requirement of 90% is met.
Conclusion
In this article, we presented a fractional failure analysis with a basic example where one can account for the effectiveness of an assumed corrective action or product redesign. This approach may be used to estimate the metrics of interest for the improved product until new failure data become available.
Important Note on Selecting the Parameter Estimation Method While Using Fractional Failure Analysis
One thing that should be noted is the importance of selecting the parameter estimation method while using fractional failure analysis. Paying attention to the size of the failure to suspension ratio is essential. This means that both the numbers of failed and suspended data points, and the efficiency % will determine the number of failures and suspensions in fractional failure analysis. For example, if you have a small set of complete data with 5 failures, then rank regression is the most appropriate method to choose. However, if you find out that you are going to apply an 80% design improvement, then the fractional failure data set will have 1 failure and 4 suspensions. In this case, the MLE method would be more appropriate to use. For large data sets and heavily censored data, MLE should be preferred.