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Forum: Reliability & Maintainability Questions and Answers

Topic: 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

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Original Message:

Posted by: B.G.Hebalkar (Bhalchandra.G.Hebalkar@cummins.com )
Organization:Cummins India Limited
Date posted: Mon Apr 29 1:55:04 US/Eastern 2002
Subject: Reliability predictions
Message:
I have a failure data for a component as follows(failed hours) 1,574,611,887,1250,1445,2480,2750,3234,3540,3900,4001,4124, 4180 and 4189 hours I calculated the relibility of this part using this failed data after plotting manually on weibull plot as 0.112 considering 5000 hours Warranty life. But how can I derive the actual reliability of this component as already some 10000 components are running in field and have already crossed 5000 hours which is our warranty life. Please advice me the appropriate formula to get the actual reliability of this component. Thanks,


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Reply:

Subject: Nonparametric Analysis
Reply Posted by: Larry George (pstlarry@attbi.com )
Organization: Problem Solving Tools
Date Posted: Sat May 4 15:54:04 US/Eastern 2002
Message:
Weibull gives a lousy fit to the nonparametric, Kaplan-Meier reliability function estimate. Weibull hides potentially actionable information. The Weibull fit hides the infant mortality and wearout apparent in the nonparametric failure rate function estimate. The number of failures is not very great, so the variance of the Kaplan-Meier estimator is pretty large, but at least you can see suspicious behavior. I will send you the analysis, if I can figure out your email address.


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