MTBF vs Availability in Parallel systems

[Vajooo]

Hi everyone,
I have a parallel system consisting of two identical blocks with an MTBF equal to 49567 hours and MTTR equal to 0,16. Putting this scenario on BlockSim gives a total MTBF of 74323 and a total Availability of 99,99999999%.
As a proof i tried to calculate analitically the availability with the formula MTBF/(MTBF+MTTR) considering an MTBF of 74323. In this case the total availability is 99,9997848%
Can somebody exlplain me why I obtain this two different values of Availability?
Thanks in advance...
Joseph

[Pantelis]

Well many reasons...
The first issue is what is the MTTF/MTTF+MTTR equation giving you … its giving you the Inherent availability (which is the steady state availability when considering only the corrective downtime of the system). The big word here is steady state (the limit on A as t approaches infinity). Now if you run something long enough it will eventually get there. In turn when you are using BlockSim you are computing availability up to a time (a much more useful metric), not infinity. So that depends on what time you run up to. See http://www.weibull.com/SystemRelWeb/availability.htm (http://www.weibull.com/SystemRelWeb/availability.htm) for details on different definitions of availability.
Secondly the inherent availability equation you provided is valid only at the component level, not at a system level (although If you treat the system as a single block then you could use it). In this case is the MTTR given at the system level or is it for each block? If it is for each block again the equation does not apply.
Hope that helped.

[Vajooo]

I considered a total MTBF of 74323 (obtained as a result of the simulation and verified analitically by inverting failure rate as in http://www.reliability-discussion.net/showthread.php?t=2435).
I considered a system level MTTR obtained analitically (the two blocks are identical so the formula shows that total MTTR remains the same as the single block value)
With these assumptions i suppose the formula MTBF/(MTBF+MTTR) is applicable to this case. Am I wrong with this thought?

[Pantelis]

With regards to your specific question, if those metrics are at the system level, in other words you are assuming that the system itself is a single component, then yes the equation will give you the steady state solution.

Now with regards to the other comments made in your post, and as I said in the aforementioned thread, inverting the failure rate to get a mean for a parallel system configuration is not valid, however in this case, and also why the MTTR seems the same for the system as it does for each block, is due to the fact that the MTTR for each is much much smaller than the MTTF. It has nothing to do with the fact that the two blocks are identical. In other words in the case of two parallel blocks it is possible for one block to fail while the other one is being restored yielding a system failure and a longer repair time. Given how short the repair time is however, this is not likely in this case and that’s why you are observing what you are observing. If you change the MTTR to something much larger you will find out that this is no longer the case.

If you want to and "just for fun" :-)... try the calculations again with an MTTR>=MTTF and see what you get.

Hope this answers your question.

[Vajooo]

The major problem i'm facing with is that internet searching or theory books brought me lots of different versions about these concepts...While for series systems (i think) there's a common "modus operandi" for Reliability, failure rate and MTBF/MTTF (Reliability and Availability of the system as the product of single avalabilities and failure rate as the sum, MTBF as the inverse of FR...) i'm figuring out this is no more the case when considering parallel systems.
Using BlockSim I uncovered also that for series systems, system failure rate is always costant with time, whereas system failure rate for parallel system changes with time. Is this maybe the major difference that explains why we can only calculate rightly series systems parameters instead of parallel ones?
Thanks a lot for your support.
Joseph

P.S. I consider only exponential distributions for the simulations...that's why i say that FR=1/MTBF...I don't care about other distributions

[Pantelis]

Well really there is only one concept and one way to it, not many. Unfortunately the right way to do it becomes mathematically intractable very quickly (for any real system) so many people used approximations (some valid and some not).

To make a long answer short, given the computers we have today and the tools out there there is no reason to be using approximations that are in most cases not accurate.

The same goes with your last statement about an exponential distribution. Why only consider that? It assumes a constant failure rate and not many things out there have a constant failure rate! Another approximation that may not be the most appropriate.