Reliability: 3 of 4

[TomTidal]

Hello,
I'm trying to calculate the reliability of three out of four generators working when in parallel.

Does anyone know how to do this?

Thanks in advance, Tom

 

[Wrenchbender]

I would use the general relation that reliability, R = exp(-t/MTBF) where t is mission time. (How long are you going to run the generator system?)

There are ‘n’ generators: 1,2,3 … n, which are subsystems.

I think you need to know (or assume) the mean time between failure (MTBF) for each generator.

http://en.wikipedia.org/wiki/Mean_time_between_failures

(That’s the difficult part.) e.g. MTBF_1, MTBF_2, MTBF_3 … MTBF_n.
Then calculate individual R_1, R_2, R_3… R_n.

The block diagram would show 'n' subsystems connected in parallel. Overall reli of this type of system is computed by R = 1 – (1 – R_1)(1 – R_2)(1 – R_3)…(1 – R_n).

You should be able to consider portions within the total system in a similar manner.

 

[unclesyd]

Are you talking about having at least three of the four generators come online at the same time on demand?

If it's keeping 3 of 4 online all time then the previous post by BestWrench covers it

 

[TomTidal]

Thank you Bestwrench and Unlcesyd.

Unlcesyd, it is calculating three out of four. i.e. one is broken.

Bestwrench,
Does: R = 1 – (1 – R_1)(1 – R_2)(1 – R_3)(1 – R_4). calculate the probiblity of any one of the four working? Can you reconfiure that to show three working one broken?

Tom

 

[zekeman]

You take the function
(p+q)^4 and do a binomial expansion to get
p^4+4p^3q+6p^2q^2+4pq^3+q4=1 (1)
where
assuming equal probabilities for the 4 generators
q= probability of one failing
p=probability of one surviving
therefore p+q=1
and the terms developed binomial expansion have the meaning
p^4 probability of all surviving
4p^3q probability of exactly 3 surviving
6q^2p^2 probability of exactly 2 surviving
4pq^3 probability of exactly 1 surviving
q^4 probability of exactly none surviving
So the probability of at least 3 surviving is
p^4+4p^3q (2)
and using the MTBF= T proposed by Bestwrench (There are numerous other reliability curves depending on the application)
we can write p=e^(-t/T); q=1-e^(-t/T)substituted in eq (2)
where t is the time in service.

 

[Artisi]

Don't forget to factor in scheduled maintenance or lack of into your analysis

 

[zekeman]

Yes, and that is factored into the reliability curves which is more of an art than a science; also the simple analysis is flawed since it is predicated on t being the same for all units.
To do it accurately you would have to track the individual t's and do the permutations for each probability, the sum of zero failure probability and exactly one failure.