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About Michael Drew

Michael Drew has over 30 year’s experience in reliability engineering, maintenance management, reliability centered maintenance and root cause analysis and is CEO of ARMS Reliability

12 Thoughts on “5 Minute Brain Teaser – Post your answer in comments below

  1. Michael Drew on January 19, 2015 at 1:20 am said:

    Lets try another. Lets say all tanks are required in service but if any tank fails it can be bypassed and there is enough volume in the other three to maintain operations for 10 hours. How would you model that?

    • Mike Spence on January 19, 2015 at 2:32 am said:

      Hmmm… I have no experience with modelling buffers in RBD’s, so I would treat it like this:
      Depends on the acceptable mission duration. If <10 hours is acceptable, then I'd model it as a parallel 3oo4 block. You'll complete the mission before the buffer runs out, so you have 3oo4 redundancy.
      If 10 hours isn't acceptable, then I'd model it as a series A-B-C-D. No redundancy here, single failure means ultimate system failure, the buffer buys you time but doesn't get you out of the hole.

      Cheers,
      Mike

  2. David Robson on January 11, 2015 at 6:54 pm said:

    AB series, in parallel with CD series.

    • Michael Drew on January 19, 2015 at 1:11 am said:

      So if C and D fails, the path through AB would pass. Would this pass the test that two tanks side by side can’t fail?

      • David Robson on January 19, 2015 at 8:18 am said:

        No. Stochastic failure is always possible at any point in time. All one can do is try to ensure that components have a very high mean-time-to fail.

        • Michael Drew on January 19, 2015 at 9:11 am said:

          Well yes a random failure can occur at any time. The purpose of a Reliability Block Diagram is to represent the success path that defines the up state of the system.

          So if both C and D fail then two tanks side by side would be in a failed state, which breaches the rule of logic required. (same goes for A and B). Whereas if the logic diagram has A and C in series and parallel to B and D in series, the logic of this diagram dictates if A or C AND B or D fail, the system is in a failed state. If A and C fail and B and D do not, then system is in an Available state. Similarly is B and D fail and A and C do not, then system is available.

          So the solution is A and C in series and parallel to B and D in series. With a success vote of 1 at the node.

  3. I suggest that BD in series,AC in series, and these two serious structure in parallel.

    • Michael Drew on January 19, 2015 at 1:16 am said:

      So in this case if any or both of A and C can fail and the rule is met. Similarly any or both of B and D can fail and rule is met. If A and D or A and B fail the system fails. Similar logic for other combos as represented in your logic.
      Well done.

  4. I just saw this question, it is interesting. I suggest that ABD in series,ADC in series, CBD in series, BAD in series, and these four serious structure in parallel.

  5. Mike Spence on December 30, 2014 at 12:35 am said:

    (A series C) in parallel with (B series D)

    • Michael Drew on January 19, 2015 at 1:17 am said:

      So in this case if any or both of A and C can fail and the rule is met. Similarly any or both of B and D can fail and rule is met. If A and D or A and B fail the system fails. Similar logic for other combos as represented in your logic.
      Well done.

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