Introduction
In this article, I will briefly explain how the gamma value is solved when you are fitting a dataset to a 3-Parameter Weibull (3P-Weibull), so that you have enough knowledge to decide whether you should avoid this analysis.
I will also provide an example that demonstrates an appropriate use case of the 3-Parameter Weibull distribution.
3-Parameter Weibull
The 3-Parameter Weibull pdf is given by:
Beta (β), also known as shape parameter, is a positive value that governs the rate at which the failure rate changes over time. If β < 1, the failure rate decreases over time, indicating the population is getting more stable (in terms of failure) as it ages. If β = 1, the failure rate is constant, and if β > 1, the failure rate increases over time.
Eta (η), also known as characteristic life, represents the time at which 63.2% of the population fails.
Gamma (γ) is a time-shift term used to modify the original 2-Parameter Weibull distribution. If γ = 0, the 3-Parameter Weibull reduces to a 2-Parameter Weibull distribution.
A positive γ means that the population cannot fail before t = γ. From reliability perspective, this is a desirable attribute.
What if the γ is negative? The item from this population can start to fail from a negative time! Note that in reliability engineering, we are only interested in the positive time domain.
If you have no knowledge of the gamma value, stick to the 2P-Weibull
Fitting failure data to a 3-Parameter Weibull
As a reliability engineer, the main reason to use 3-P Weibull for your dataset is that you believe that a “no failure time”, (positive gamma), exist in your population, and you hope that the analysis can tell you this value.
Consider the following two time-to-failure (hours) datasets (generated using Monte-Carlos simulation from 2P-Weibull with the same parameter β =9 and η =3600):
Dataset 1: {3000, 3086, 3322, 3650, 4001} and
Dataset 2: {2850, 3503, 3606, 3800, 4000}
Both the datasets are fitted with 2P-Weibull using RRX, and followings are the results:
From the contour plots, we can observe that there is no statistical evidence that the two sets are different (i.e., there is a significant overlap between the two contours).
You want to know if γ exist, and so, you proceed with 3P-Weibull analysis. Perhaps you are convinced that it is not possible to fail within the first 1000 hours.
Note that you are expecting a “positive” γ.
Now, let me explain how γ is calculated with Dataset 1: {3000, 3086, 3322, 3650, 4001}
The points are plotted on the Probability-Weibull graph. A 2P-Weibull line regresses along these points. Notice that the points form a curvature facing “downward” as indicated in Figure 2.
By reducing all the values (times to failure) in the dataset by an appropriate constant (i.e., γ), we can remove the curvature. Let’s subtract 2965 from all values in Dataset 1.
Now we plot these “adjusted” points on the Probability-Weibull graph.
The γ value is 2965 hours- the software search for a gamma value such that the adjusted dataset does not form a curvature.
The β and η are now 0.7611 and 485.8 (hours) respectively (based on the new straight line formed). We have just fitted the original dataset with a 3-Parameter Weibull:
Now, let us proceed to analyze Dataset 2: {2850, 3503, 3606, 3800, 4000}. We would expect that the gamma value to be close to that of Dataset 1 (2965). Since smallest failure point for Dataset 2 is 2850, it should be somewhere below 2850.
Dataset 2 is fitted to a 2-parameter Weibull:
To “straighten” the data points, a negative γ value is required. To form the best possible straight line, the γ value is -5852112 hours!
The corresponding β and η are 1,364 hours and 585,875 respectively!
We have negative γ! What is the explanation for that? And, we have a β value that is way beyond 50!
Side node: A large β means that the failure rate increases very shapely when the age is approaching η value. Personally, I would consider other distributions, like Normal or Lognormal if the β is greater than 20.
Comments
These 2 datasets were generated using Monte-Carlo simulation with 2-parameter Weibull. There was no γ value, but we can always come up with a γ, by fitting a 3-parameter Weibull!
When you obtain data from your lab test (or collecting data from the field, or generate random data through simulation), there is a 50% chance that your dataset forms a “downwards” curvature (and the other 50% “upwards”). In which case, you get some positive γ. Otherwise, you get an absurd negative γ.
Given that you know the math (or Madness) behind this analysis, would you trust the γ value even if the result looks reasonable (a γ value between 0 and the smallest value in your dataset)?
What if we use MLE solution instead of Regression method? Well, I will not go into details about applying MLE for 3P-Weibull, but I will just state that MLE provides even more absurd results most of the time!
Let me end my commentary with an example of an appropriate use case of 3P-Weibull.
The following dataset contains arrival time of a spare part from a supplier. You need a distribution to describe the lead-time for simulation purpose.
Historical lead-time data (day): {30, 30, 30, 30, 31, 31, 35, 36, 42}
You are aware that the admin plus shipment time is 27 days i.e., lead-time cannot be short than 27 days. You can set the γ =27 days, and the new dataset become {3, 3, 3, 3, 4, 4, 8, 9, 15}. Fitting it to 2P-Weibull to get β =1.64 and η =6.5 days. So, the distribution can be described using a 3P-Weibull with β=1.64, η=6.5 days and γ=27 days.
This example illustrates the case where gamma is a value you need to provide, and then transform the data for fitting a 2P-Weibull. It is very unlikely to determine via statistical analysis/fitting that the admin plus shipment time is 27 days. The analyst knows this because they understand the situation in which the data is collected.
Conclusion
We cannot expect that one could extract a gamma value through fitting the data to a 3P-Weibull (i.e., mathematically). Gamma should be a quantity that is known to the analyst. If you have no knowledge of the gamma value, stick to the 2P-Weibull.
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