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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: Clay Davis
Date posted: Tue Oct 5 17:12:16 US/Eastern 1999
Subject: Reliability vs Availability Calculations
Message:
I am trying to determine the availability for a new design for a computer system. Since availability is highly dependent of the software and the MTTR, I have decided to focus only on the hardware at this point. My question is, say for a system that has an MTBF of 10 yrs, is the availability (from a hardware standpoint) 0.9048 at the end of the first year? I calculated this number using the standard R=e(-lambda*t). Lambda = 1 / 87600 hours and t = 8760 hrs. This would lead me to believe the computer system is going to be un-available for more than 1 month during the first year, which cannot be correct. Any thought on what I am doing wrong? Thanks for your help. Clay Davis


Reply:

Subject: Availability vs. Reliability
Reply Posted by: Patrick Hetherington (phetherington@alionscience.com )
Organization: RAC
Date Posted: Wed Oct 6 6:06:55 US/Eastern 1999
Message:
Reliability for a complex system having a random failure pattern (such a computer) equals R=e (-t/MTBF) where MTBF is 1/Lambda (system) as you point out. However, Reliability is the probability of system success (no failures) for the period of time you are interested in (1 year in your case). You could interpret this for your situation as, if you have ten identical computers operating continuously each for one year, you can expect one to fail. Availability in the general sense is equal to MTBF/(MTBF+MTTR). If repairs on your computers were instantaneous, Availability is 100% (best case). If the average time to repair is 1 month (maybe still hopeful) the availability is 87600/(87600+720) or 99.2%


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