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Reliability Basics | |||

Design of Reliability
Tests
In a past issue of the
Weibull++'s Design of Reliability Testing (DRT) utility is shown in Figure 1.
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
From this equation, you
can see that the number of test units, 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.
If Beta is assumed to be equal to 3, then seven units would need to be tested without failure, as shown in Figure 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 n. But R
is no longer 90% since this is the value at 100 hours. So _{Test}R
is the last unknown to be determined in order for _{Test}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:
Therefore, η
= 211.7. Now that η
has been determined along with β
and R can be estimated
using the Weibull reliability equation.
_{Test}
Once
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 -
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: http://reliawiki.org/index.php/Reliability_Test_Design.
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