Bayesian Testing

Overview

Bogey Testing

Bayesian Testing

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Effect of the Shape Parameter
The shape of the distribution has a tremendous effect on the amount of testing required to demonstrate reliability. Consider the histogram of time to fail for a population with a Weibull time to fail distribution with a shape parameter of 3.6 shown in the figure below. Ninety-five percent of the items survived for more than 1 bogey, thus the reliability at 1 bogey is 95%. This histogram appears to show a normally distributed population because the Weibull distribution is nearly normal when the shape parameter is 3.6. Also, notice that when the shape parameter is 3.6, to demonstrate 95% reliability at 1 bogey the mean must be at least 2.06 bogeys.

Weibull distributed population with 95% reliability at 1 bogey and a shape parameter of 3.6.

The figure below shows a population having a Weibull time to fail distribution with a shape parameter of 1. This figure shows a reliability of 95% at 1 bogey, and a mean of 19.5 bogeys; nearly 10 times greater than the Weibull distribution with a shape parameter of 3.6. The variance of the distribution increases as the shape parameter decreases. Variance is the equivalent of uncertainty, and the amount of testing is required to demonstrate reliability is dependent on the amount of variance in the population.

Weibull distributed population with 95% reliability at 1 bogey and a shape parameter of 1.0.

The figure below shows a population having a Weibull time to fail distribution with a shape parameter of 1.8. This figure shows a reliability of 95% at 1 bogey, and a mean of 5.2 bogeys.

Weibull distributed population with 95% reliability at 1 bogey and a shape parameter of 1.8.

The figure below shows a population having a Weibull time to fail distribution with a shape parameter of 8.0. This figure shows a reliability of 95% at 1 bogey, and a mean of 1.37 bogeys.

Weibull distributed population with 95% reliability at 1 bogey and a shape parameter of 8.0.

Estimating the Shape Parameter
From the figures above it can be seen that assumed shape greatly affects the testing requirements. How is the shape parameter estimated? The shape parameter is governed by the physics of failure. Some phenomena are skewed right while others are skewed left or have no skewness. In general, the more the failure mode is a result of mechanical wear the larger the shape parameter will be. The shape parameter is usually stable and tends to remain constant. For example, if the shape parameter for master cylinders will be similar for master cylinders for small cars, large trucks, and for different designs because the physics of how the master cylinder fails is similar.

The best way to estimate the shape parameter is through prior testing. Many automotive companies require some testing to failure to allow the shape parameter to be determined. Keep detailed records of all tests and build a database of shape parameters. It is recommended to use the lower 90% confidence limit for the shape parameter because to the magnitude the shape parameter has on test requirements. In lack of any prior knowledge there are data sources available on the internet, or the shape parameter can be estimated based on the knowledge of the physics of failure.

Be careful when estimating the shape parameter for electronics. Many sources state the shape parameter for electronics is 1.0 because there are is no mechanical wear in electronics. For electronic modules located in environmentally harsh conditions, such as under the hood of an automobile of in an airplane, fail as a result of mechanical ear. The extreme vibration, temperature cycling and in some cases contaminants cause mechanical failures. It is not uncommon to have shape parameters greater than 8.0 for electronic components.

Determining Test Parameters
The statistical properties for bayesian testing with the Weibull distribution are based on the exponential distribution. If the parameter t follows a Weibull distribution, then the parameter t^b is exponentially distributed. The lower (1-a) confidence limit for reliability is

where

n is the number of units tested, both failed and surviving, and c²_a,d is the critical value of the chi-square distribution with significance of a (0.05 for a confidence level of 95%) and d degrees of freedom. For failure truncated testing d is equal to 2r where r is the number of failed units. For time truncated testing d is equal to 2r+2.

Example
Fourteen units are placed on a test stand; the first unit fails after 324 hours, the second unit fails after 487 hours of testing, a third unit failed after 528 hours of testing, and the remaining 11 units were removed from testing. Given a Weibull time to fail distribution with a shape parameter of 2.2, what is the lower 90% confidence limit for reliability at 400 hours?

Solution
Since testing was suspended at the time of the last failure, this is failure truncated testing. With failure truncated testing and 3 failures, the degrees-of-freedom for the chi-square distribution is 2(3) = 6. The critical value of the chi-square distribution given 6 degrees-of-freedom and a significance of 10% is 10.64. This value can be found using the following expression in Microsoft Excel.

=CHIINV(1-0.9,6)

The lower 90% confidence limit for the mean of the transformed data is

The lower 90% confidence limit for the reliability at 400 hours is

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Example
Fourteen units are placed on a test stand; the first unit fails after 324 hours, the second unit fails after 487 hours of testing, a third unit failed after 528 hours of testing, and the remaining 11 units were removed from testing after 550 hours without failing. Given a Weibull time to fail distribution with a shape parameter of 2.2, what is the lower 90% confidence limit for reliability at 400 hours?

Solution
Since testing was not suspended at the time of the last failure, this is time truncated testing. With time truncated testing and 3 failures, the degrees-of-freedom for the chi-square distribution is 2(3)+2 = 8. The critical value of the chi-square distribution given 8 degrees-of-freedom and a significance of 10% is 13.36. This value can be found using the following expression in Microsoft Excel.

=CHIINV(1-0.9,8)

The lower 90% confidence limit for the mean of the transformed data is

The lower 90% confidence limit for the reliability at 400 hours is

Click Here to download this example in a spreadsheet.

Reliability tests are often designed in a 2 step procedure; 1) how many test stands are available, and 2) what is the test duration given the number of test fixtures. For a sample of n units the required test duration to demonstrate a reliability of R at time t with a confidence level of 1-a assuming no failures is

Example
How long must 20 units be tested to demonstrate 99% reliability with 80% confidence at 200 hours of operation given a Weibull time to fail with a shape parameter of 3.5 and assuming no failures?

Solution
The required test duration is

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For a test duration of T the required sample size to demonstrate a reliability of R at time t with a confidence level of 1-a assuming no failures is

Example
How many units must be tested for 300 hours to demonstrate 99% reliability with 80% confidence at 200 hours of operation given a Weibull time to fail with a shape parameter of 3.5 and assuming no failures?

Solution
The required sample size is

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