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Forum: Reliability & Maintainability Questions and AnswersTopic: Reliability & Maintainability Questions and Answers
Topic Posted by: Reliability & Maintainability Forum
(src_forum@alionscience.com
)
Organization: System Reliability Center
Date Posted: Mon Aug 31 12:47:36 US/Eastern 1998
Original Message:
Posted by: Wayne
Date posted: Mon Feb 24 8:05:38 US/Eastern 2003
Subject: AMSAA model fail
Message: I have a query that is i have only 2 failures during my Reliability growth testing. I fit in into the AMSAA model and it fails the goodness of fit test. In this case can i represent the system reliability by it's Observed MTBF which is = total test duration/ total failures? If not are there a better alternative since i have only 2 data points.
Reply:
Subject: AMSAA data
Reply Posted by: Jorge L. Romeu
(jromeu@alionscience.com
)
Organization: RAC
Date Posted: Mon Mar 3 15:47:18 US/Eastern 2003
Message: For starters, n=2 is the bare minimum of failures that one can have to calculate the AMSAA model parameters (or any estimate of variation in a sample). There are also differences in the analysis, regarding to whether the test ended at the time of the second failure, or if testing is continued beyond this failure event for some additional time.
Then, two points determine a line. It is risky to establish that reliability growth is linear (as opposed to following a nonlinear, concavedown, functional form such as a square root, quadratic, etc.). My personal experience is that reliability growth flattens with time.
Also, in a sample there are always sampling errors. Hence, the true slope of the resulting line (assuming the line is the true representation of the reliability growth process) may be very different from the slope we obtain from this twopoint sample.
Finally, assume the test was stopped at the time of the second failure. A rough estimate of the MTBF is test time divided by two. But if the test was allowed to continue beyond the second failure, for a significant period of time, a bias may have been introduced in the estimation of the MTBF. We illustrate the problem with the figure below, where the Xi represent the failure times and T is the total test time:
 X1 X2 T…. X3
The test observes two failures (at times X1 and X2) up to time T. Hence, an estimate of MTBF is T/2. Assume that (T X2) is large (with respect to an unobserved failure X3 that would have occurred if the test had not been curtailed at T). Then, the estimate of MTBF based on T/2 would be significantly larger than a MTBF estimated by X3/3. This is now especially so, because there are so few failures (only two) in the present case. Had there been, say a dozen or more, this bias effect would be decreased.
We suggest that the user takes T/2 as a very rough estimate of MTBF. For, at this time there is no other alternative. Then, if the test was stopped at the second failure or shortly thereafter, this rough estimate will point to the true MTBF. Otherwise, it may well be very biased. In any case, an estimate based on two failures is rather weak.
In general it is better to either stop at the time of the (second) failure or continue until the next (third) one occurs. The derivation of several design of experiment DOE performance measures before the testing starts would be extremely helpful. For example, it would help to know how many failures to expect in the allotted test time or what minimum test time should we have, in order to obtain meaningful results. Or we would need to know how many units should we place on test, to obtain a meaningful result in the allotted test time.
Knowledge of such experimental design parameters helps prevent that situations like the present one reoccur and save lots of time and resources. Organizations like the RAC are prepared to help with these types of planning.
Reply:
Subject: Reply on the above
Reply Posted by: Wayne
Date Posted: Tue Mar 11 21:24:15 US/Eastern 2003
Message: Thanks for your long reply. As far as planning is concern, the total test duration is derived based on the Duane Model reliability planning.We currently have only 1 prototype so the number of units is constraint in our case. We only realise that there are only 2 chargeable failures after completing the whole test duration.
In our case we could not stop at the 2nd or even 3rd failure, we have to complete the whole test duration. I thought it is suppose to be a time terminate test? Please advise!! Thanks
