SRC Forum - Message Replies
Forum: Reliability & Maintainability Questions and Answers
Topic: Reliability & Maintainability Questions and AnswersTopic Posted by: Reliability & Maintainability Forum (firstname.lastname@example.org )
Organization: System Reliability Center
Date Posted: Mon Aug 31 12:47:36 US/Eastern 1998
Posted by: Beth Chappel
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.