SRC Forum - Message Replies
Forum: Reliability & Maintainability Questions and Answers
Topic: Reliability & Maintainability Questions and Answers
Topic Posted by: Reliability & Maintainability Forum
Organization: System Reliability Center
Date Posted: Mon Aug 31 12:47:36 US/Eastern 1998
Posted by: B.G.Hebalkar
Organization:Cummins India Limited
Date posted: Mon Apr 29 1:55:04 US/Eastern 2002
Subject: Reliability predictions
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.
Subject: Weibull Analysis
Reply Posted by: Patrick Hetherington
Organization: Reliability Analysis Center
Date Posted: Mon Apr 29 7:04:13 US/Eastern 2002
To accurately assess the reliability distribution for your component, you must include the data points for items that have not failed. A simplistic approach for plotting manually is to consider that you have 10015 components in your sample, 15 of, which are failures and 10000 which have not failed and all have time greater than 5000 hours. Calculated your Y plotting position (Median Rank) using Bernardís Approximation where Median Rank = (i-.3)/(N+.4). If your Weibull paper has Y-axis in percent, multiply the median rank by 100. i is the rank order of the failures, in your case 1 through 15 and N is the sample size which in your case is 10015. Replot your fifteen points. In reality you probably have many components that have not failed but have not reached 5000 hours. Any item that has not failed (a suspension) will influence your results. Right suspensions are units that all have more time than the highest failure time will tend to increase the scale parameter and have little impact on the shape parameter. Left suspensions (items that have not failed but have less time than highest failure time can influence both parameters. There are probably many components that have failed outside the warranty period. You would be very well served to get information on about 30 of these and include in your analysis because you have a pretty good idea of what percentage fail in the warranty, but can not make an accurate judgement of the results of extending the warranty. An excellent resource for handling suspensions grouped data and interval data in the Weibull Handbook by Dr. R. Abernethy. Software by Reliasoft and Fulton Findings is also very useful.
Subject: Nonparametric Analysis
Reply Posted by: Larry George
Organization: Problem Solving Tools
Date Posted: Sat May 4 15:54:04 US/Eastern 2002
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.