Probability assessment

  1. Dear folks,

    Please help me in assessing the following scenario. I have a cooling circuit arranged in series: (1) Heat exchanger (probability = 10^(-4)/requirement), (2) Pump (probability = 10^(-3)/requirement), (3) Heat exchanger (probability = 10^(-3)/requirement) and connection of pipes (probability = 10^(-4)/requirement) (name the pipes as 4th component). In addition there is 5th component supplying energy to the pump (probability = 10^(-3)/requirement).

    What is the probability that the cooling circuit corresponding to the scenario arranged above fails?

    Thank you in advance
     
  2. jcsd
  3. Unlike choosing two winning numbers in a row, where the probabilities are simply multiplied, in your case, a failure of any one of the services makes failure more likely. You add them all up somehow, but the result must not exceed 1.0, however many are the ingredients, or how likely they are.
     
    Last edited: Nov 10, 2007
  4. OK - I do have the answer. It comes from my son, who is is in his Master's degree Math finals.
    We put the problem to him slightly differently, but you will understand. Your example is likely the simplest case, where the probabilities of failure you set out are independent.

    Please be aware that if you are calculating for a real situation, then the values for any component may well be modified by an fail event in one of the others. For example, a pump breakdown might be a minor inconvenience, but the collateral effect on the other probabilities might modify them dramatically to swiftly alter the overall failure probability. His answer allows for this.
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    We hope this helps. :smile:
     
  5. "What is the probability that the cooling circuit corresponding to the scenario arranged above fails?"

    Let us assume component (1) Heat exchanger (probability = 10^(-4)/requirement), (2) Pump (probability = 10^(-3)/requirement) and energy to the pump (probability = 10^(-3)/requirement) are available to fulfil the function. How would you determine the probability above?

    Since the three components above are available to fulfil the function then the system can only fail due to component 3 and 4. That is to say: If one of them fails: system fails (probability = (1-P3/4)*P3/4) or if both are not functioning then system fails (Probability = P3*P4).

    Will this be correct?
     
    Last edited: Nov 21, 2007
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