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I'm trying to calculate the reliability of a system that has proven to be a bit more complicated than I thought. Here is the block diagram for the system: Block Diagram

Every single component in this diagram is a Line Replaceable Unit with a mean downtime of 2 hours. If any of A-F fail, that's a failure for the "subsystem" indicated by the dotted line. The entire system will successfully work as long as only 1 of the dotted line, A-F, "subsystems" has failed. Because of the mean downtime of 2 hours, if two of the dotted line subsystems fail within 2 hours of each other, the system fails. This subsystem highlighted in green is not particularly difficult. What's throwing me off is the inclusion of components G1 and G2, routers that fail when either A (UPS) or B (DC power) fails. This makes up the subsystem highlighted in red. The entire system fails if both routers (components G1 and G2) fail, within 2 hours of each other of course.

Basically, I thought that the reliability of the whole system can be calculated with Bayes Theorem which would give the reliability of the system as:

Reliability of green subsystem * probability that G1 and G2 do not fail

PLUS

Reliability of red system * probability that another of the dotted line subsystems does not fail

PLUS

reliability of red system * probability that the other red subsystem does not fail

Unfortunately, I think I'm vastly misunderstanding Bayes Theorem and maybe the system as a whole. Can someone lead me in the right direction to how the inclusion of these G components affects the system reliability? Maybe I just need to re-read up on Bayes Theorem?

Thanks!

****EDIT**** I've simplified the diagram a little bit and I think it moves things along a little bit.

Simplified Block Diagram

With this diagram (note modified lettering) I was able to determine all possible failure cases, 11 to be exact. The logical expression for whether it passes or not is: ~[(~F & ~G) v (~D & ~C) v (~D & ~E) v (~C & ~E) v (~B & ~C) v (~B & ~D) v (~B & ~A) v (~B & ~F) v (~A & ~C) v (~A & ~E) v (~A & ~G)] which gives the possible failure scenarios: (X, Y)=pair of failed components resulting in a failure.

(F, G) (D, E) (C, E) (C, D) (B, F) (B, D) (B, C) (A, G) (A, E) (A, C) (A, B)

So now my issue is definitely just a matter of Bayes and making sure not to double count.

trevinator
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    Interesting problem! Your description of the system doesn't say anything about the probabilities of failure of the components A-G, only the downtimes. – jonathanjo Feb 18 '20 at 01:01
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    You might benefit from reading this paper. – relayman357 Feb 18 '20 at 05:00
  • @jonathanjo I didn't include that because I wasn't really looking for someone to do the actual calculation for me, but I do have the failure rates – trevinator Feb 18 '20 at 17:50
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    You only really have to deal with making sure you don't double-count the situation where CDEF block 1 or 2 fails at the same time as matching G block. – jonathanjo Feb 18 '20 at 17:52
  • @jonathanjo Yeah that's definitely what I'm running into now, don't want to double count anything. I've edited with a bit more thought into that, but I feel like I have tunnel vision on these three scenarios from the original post. – trevinator Feb 18 '20 at 18:48

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