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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: Kim Lae Won (laewonee@rotem.co.kr )
Organization:ROTEM company
Date posted: Thu Oct 24 3:54:28 US/Eastern 2002
Subject: MTTR Allocation
Message:
Hi~ I want to know whether the following MTTR allocation is possible and if possible, want to know the appropriate methods. Thanks for your kind help previously. -------------------------------------------------------- Allocation requirements -------------------------------------------------------- The Contractor shall allocate MTTR requirements for the three active CM time intervals for each subsystem. (sample) *system : railway rolling stock *subsystem -Couplers & Draft Gear : MTTR=2.2 & Mmax90%=4.86 -here active CM time=fault isolation time + remove and replace/repair tiem + functional checkout time --------------------------------------------------------


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Subject: maintainability allocation
Reply Posted by: bwd (bdudley@alionscience.com )
Organization: rac
Date Posted: Mon Oct 28 14:59:22 US/Eastern 2002
Message:
In response to your question on allocations of requirements for maintainability, I consulted with the RAC "Maintainability Toolkit", which every engineer should have on their desk, to find the proper process for allocation. The toolkit shows the failure rate complexity method, the statistically-based allocation method and the equal distribution method. I chose the failure rate complexity method and produced the following results based on the facts of two items, a gear and a coupler, and requirements of 2.2 hours MTTR and 4.86 hours M Max: unit quantity failure rate failure rate x qty fr(h) / fr(i) x Ma failure rate x Mi (10^-6) gear 1 1.7 1.7 1.53 Ma 2.6 Ma coupler 1 2.6 2.6 Ma 2.6 Ma total 4.4 4.4 5.2 Ma MTTR = sum of failure rate Mi / sum of failure rate = 5.2 Ma / 4.4 = 2.2, so Ma = 1.86 hours MTTR Max = sum of failure rate MI / sum of failure rate = 5.2 Ma / 4.4 == 4.86, Ma = 4.11 hours Therefore, the coupler has an allocated MTTR of 1.86 hours and a max MTTR of 4.11 hours For the gear, MTTR = 1.53 x Ma or 1.53 x 1.86 = 2.85 hours The M Max for the gear is M Max = 1.53 x Ma = 1.53 x 4.11 = 6.29 From these figures of merit one can allocate to the sub levels as follows: Coupler -- 1.86 hours [ isolation- 0.2 hours(engineering estimate), repair- 1.56 hours, checkout- 0.1 hours(estimate)] -- 4.11 hours [isolation- 0.4 hours (estimate x 2), repair- 3.51 hours, checkout- 0.2 hours (estimate x 2)] Gear -- 2.85 hours [isolation- 0.2 hours (estimate), repair - 2.55 hours, checkout - 0.1 hours (estimate)] -- 6.29 hours [isolation- 0.2 hours (estimation x 2), repair - 5.69 hours, checkout - 0.2 hours (estimate x 2)]


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