Reliability HotWire: eMagazine for the Reliability Professional
Reliability HotWire

Issue 24, February 2003

Reliability Basics
Design of Reliability Tests

In a past issue of the Reliability Edge (see Cumulative Binomial for Test Design and Analysis), an article was presented on the cumulative binomial distribution and how it can be applied towards test design. Given a specified reliability at a certain time along with a desired confidence level, the required test units (given test time) can be determined to meet the required goal. Calculating the required number of test units is fairly straightforward if the test time is equal to the time at the reliability goal. If this is not the case, then a failure distribution needs to be assumed so that the calculations can be completed. This article presents how this failure distribution can be applied to generate the results using Weibull++ and illustrates the concept of allowable failures.

Design of Reliability Testing Utility

Weibull++'s Design of Reliability Testing (DRT) utility is shown in Figure 1.

The Weibull++ 6 DRT (Design of Reliability Tests) interface

Figure 1: The Weibull++ 6 DRT utility

Given the specified goal of a 90% reliability at 100 hours and a desired confidence level of 90%, 22 test units must be put under test for 100 hours with no failures. But is this result only valid for an assumed Weibull failure distribution with Beta = 2? And if Beta is changed to 3, then will the required number of test units change? To answer these questions, consider the cumulative binomial equation shown next.

Given f = 0, and CL = 0.90, the cumulative binomial equation can be reduced to the following:


From this equation, you can see that the number of test units, n, can easily be calculated given RTest. RTest is simply R(100) and is given under Requirements Input. In this case, it is not necessary to assume a failure distribution since the reliability at the test time has already been given based on the goal requirements. Therefore, the previous result of 22 units is independent of the distribution. This is because the test time is equal to the required time. Now, what if the units were to be tested for 150 hours instead of 100 using the same goal? Assuming a Weibull distribution with Beta = 2 and 0 failures, ten units would need to be tested to meet the required goal, as shown in Figure 2.

Calculate number of test units with Beta = 2

Figure 2: Calculate number of test units with Beta = 2

If Beta is assumed to be equal to 3, then seven units would need to be tested without failure, as shown in Figure 3.

Calculate number of test units with Beta = 3

Figure 3: Calculate number of test units with Beta = 3

In this case, the required number of test units is dependent upon the assumed value of Beta because the entered test time is not equal to the demonstration time. As shown in Eqn. (1), the cumulative binomial equation can be reduced to a function of RTest and n. But RTest is no longer 90% since this is the value at 100 hours. So RTest is the last unknown to be determined in order for n to be calculated. But what is R(150)? The first part of the process in determining this value is to determine the value of η for the Weibull distribution. This can be accomplished using the Weibull reliability equation.

This can be rewritten as:

η can now be determined by simply plugging in the values on the right side of the equation where:

RRequired = 0.9

tRequired = 100

β = 3

Therefore, η = 211.7. Now that η has been determined along with β and tTest = 150, RTest can be estimated using the Weibull reliability equation.

Once RTest has been determined, then the number of test units can be calculated from Eqn. (1). The same process can be applied for any specified value of Beta or for the other available distributions. A graphical representation of this process is shown in Figure 4.

Graphical representation

Figure 4: Graphical representation

When time equals 100 hours, the reliability is given to be 90%. This corresponds to an unreliability of 10% and is represented on the plot by the blue arrows. The value that needs to be determined is the reliability at 150 hours. This value can be estimated by traversing along the plot line (given β = 3, η = 211.7), represented by the red arrow, to the point where time is equal to 150 hours. This point is equal to 1 - RTest.

Allowable Failures

Allowing failures during the test is a very simple concept. In allowing failures to occur during the test, you are simply specifying the maximum number of failures that you can observe and still pass the test. Of course, it is possible to not allow any failures (which is what has been done up to this point). The more failures you allow, the greater the number of units that you must test to demonstrate the specified goal. But what if you specify that you do not want to allow any failures and one is observed before the test has been completed? Can you just add more units to the test to account for the observed failure? You could. But think about this: you originally designed the test to not allow any failures but you observed a failure, which indicates that you have failed the test. So what do you hope to gain by adding more units to a test that you have already failed? Adding units to a test that you know you have failed to prove a specified goal does not make much sense. In this case, it may be more beneficial to test the remaining units or a portion of the units to failure and then analyze the data as life data in Weibull++.

Additional information on the Design of Reliability Testing utility in Weibull++ can be found at:


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