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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: Beth Chappel (elizabeth.chappel@goodrich.com )
Organization:Goodrich
Date posted: Tue Aug 7 18:46:41 US/Eastern 2001
Subject: Weight Averaging
Message:
Question- can you weight average reliability over time? For instance, if you have a system that operates at one reliability value R for a third of the time, and another reliability R2 for two-thirds of the time, would the reliability for one hour be = 1/3 R + 2/3 R2?


Reply:

Subject: Weight Averaging
Reply Posted by: Gary Sunada (gsunada@alionscience.com )
Organization: Reliability Analysis Center
Date Posted: Thu Sep 13 9:54:19 US/Eastern 2001
Message:

Unfortunately, no: you cannot weight average reliability over time.  Assuming that the system has an exponential distribution, R(t) = e^-[(lambda)t], then for t=T, the failure rate lambda = -(1/T) (ln R).  So, to calculate the new R value, R = R1 * R2 = e^[(1/3)(ln R1)+(2/3)(ln R2)].  If I understand your question correctly, you postulate R = (1/3)R1 + (2/3)R2.  By plugging in a series of values for R1 and R2 using a spreadsheet and comparing the answers, it shows that both equations will roughly agree as long as the difference between R1 and R2 is small (e.g., R1=0.9, R2=0.8), but the error increases substantially when those values diverge even more, which is logical.  The weights provide some variance, but most of the disparity is caused by the R1 and R2 values.

 An example: R1=0.9, R2=0.8; weights are 1/3 and 2/3.  The correct formula produces an R of 0.833333, and the proposed formula’s R = 0.832034.  When R1=0.2 and R2=0.6, the results are 0.466667 and 0.416017, respectively.
 

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