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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%


Reply:

Subject: Reliability vs Availability Calculations
Reply Posted by: Clay Davis
Date Posted: Wed Oct 6 12:26:14 US/Eastern 1999
Message:
So how does one determine, from looking at a design and a preliminary BOM, what the availability (approximately) will be? In other words, we are trying to design a computer system that will have an availability of 99.5%. For the 1st year, that equals 43.8 minutes of downtime (24*7*.005). If the avg repair time is 15 minutes (a guess), what does the MTBF have to be? From the standard availability formula it looks like the MTBF would need to be 2,985 hrs - which does not seem correct. Also, say we do not want the system to fail more than twice a year (with the est 15 min repair time for each failure) - does that affect the calculations? Thanks again. --Clay


Reply:

Subject: Availability Calculations
Reply Posted by: Patrick Hetherington
Date Posted: Mon Oct 11 7:15:19 US/Eastern 1999
Message:
Downtime during the first year assuming continual operation and 99.5% availability is 2628 minutes (24*365*60*.005). To determine MTBF solve Availabilty=MTBF/(MTBF+MTTR) for MTBF. In your case with a 15 minute (.25 hour) MTBF=A*MTTR/(1-A). MTBF = 49.75 hours (2985 minutes). By saying you want no more than 2 failures per year, you are implying a failure rate. Generally you would like to put confidence levels on this, but using the averages you are saying you would like a failure rate of 2/8760 or 0.00022 failures per hour. This equates to a MTBF of 4380 hours. Given you have the same MTTR (.25 hours) your availability is now 99.99%.


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