R(t₀) = reliability at time "t₀"
t₀ = design life or the life of the part the reliability to be demonstrated at
n = sample size
T = total test time of each part
β = Weibull shape parameter or Weibull slope
C = statistical confidence level specified in percentage

Assumptions behind this equation:

• An appropriate testing is available to correlate “to” test time in the lab (or lab design life) to the real world surviving time (or field design life). The testing must be in some accelerated fashion or typically called accelerated testing. The failure mechanism(s) that the accelerated test will bring out is of the most importance. No one test can surface all potential failure mechanisms of the part. Certain failure mechanisms dominate throughout the useful life time of the part and some may never soccur.
• The Weibull slope is known before the testing. This assumption can be (and will be) misused by people to achieve one goal – fewer samples for the testing. The known Weibull slope should be estimated from experience, engineering knowledge, similar failure mechanisms, etc.
• All parts on test must complete the planned test time without failure to successfully demonstrate the reliability target. If failures occur, the demonstration is no longer valid. However, if failures do occur, you can use Weibull plot to assess the status.

Example 1: Supposed we would like to demonstrate reliability to be 95% at design life “to = 1000 hours” at 90% statistical confidence level. We can test each part to 1000 hours. We also know that Weibull slope is about 1.0. How many parts do we need without failure to demonstrate this target?

Solution: Use the equation, we obtain

Solving for n, we obtain n = 45. That means we need to test 45 parts without failure to 1000 hours each to demonstrate 95% reliability at the design life of 1000 hours of lab testing at 90% statistical confidence level.

Example 2: Continuing on example 1, what if I can test each part to 1500 hours, then how many samples do I need?

Solution: Solving for n, we obtain n = 30.

From example 1 and example 2, we know if we test longer, we can cut down the sample size.

Example 3: Continuing on example 2, suppose we have a different Weibull slope, say the β is 3.0, not 1.0. What is the required sample size?

Solution: Apply the equation with β of 3.0, we obtain

Solving for n, we have n = 14.

From example 2 and example 3, we learned that if Weibull slope is larger, we can cut down the sample size.

This equation can be generalized into the following:

where Ti represents censored times of individual part and n is the total number of sample size.

This equation suggests that each part can be stopped at different times. The usefulness of this equation is for trading off on number of parts and test times when certain parts are no longer available during the testing. This may be due to the fact that certain testing fixtures are broken, certain parts may be required for other purposes and being pulled out of the testing, etc.

Example 4: Continuing on example 3, the plan is to test 14 parts without failure to 1500 hours to demonstrate 95% reliability at 1000 hours (design life in lab testing) at 90% statistical confidence level. The Weibull slope is assumed to be 3.0. Right in the middle of the testing, two parts were stopped at 450 hours due to a broken connector. At this time, another department requires these two parts for other use so you lose these two parts. How long should the other 12 parts be tested to maintain the reliability demonstration requirement?

Solution: Use the above equation, we obtain

Solving for T, we obtain T = 1550 hours. Therefore, the remaining 12 parts will be tested to a total of 1550 hours, rather than 1500 hours as originally planned.

Dr. Bob’s Comments: This is an excellent example of using the Weibayes equations in Chapter 6 to analyze substantiation test data but there is one caution: The test designs are based on a priori declaration of test design. Therefore, if a failure occurs you have flunked the test. There will be people who will claim the failure was different in some sense and would not fail that way in-service. Be skeptical! Please send any questions to me via email.