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Time between process problems (Probability distributions)

  1. Feb 7, 2012 #1
    1. The problem statement, all variables and given/known data

    The time between process problems in a manufacturing line is exponentially distributed with a mean of 30 days.

    (a) Let T be the waiting time (in days) for four problems. What is the distribution of T?

    (b) What is the expected waiting time for four problems?

    (c) What is the probability that the time until the fourth problem exceeds 120 days?

    (d) Suppose that the problems can be classified into three mutually exclusive classes: I, II and III. The probability that a problem is of type I, II, and III, are respectively, 0.75, 0.2, 0.05.

    (i) Give the distribution of the waiting time for a process problem of type I.

    (ii) Suppose that the problems I, II and III, cost (per problem): 1,000, 2,500 and 6,000 $, respectively. Give the mean total cost for process problems for a period of 90 days.

    (e) There are no process problems for 30 days, what is the probability that there will be at least one process problem in the next 30 days?

    2. Relevant equations

    3. The attempt at a solution

    For (a) (b) and (c) I have managed to answer the following:

    (a) The waiting time distribution for T is Erlang since this is a sum of exponential distributions.

    (b) E[X] = r.1/lambda = 4/lambda , where lambda = 1/30 ==> 4/0.033

    (c) This would be a Poisson distribution P(X>120)

    However for (d) and (e) I am kind of stuck. Any help would be greatly appreciated!

    Thank you,

  2. jcsd
  3. Feb 7, 2012 #2

    Ray Vickson

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    (a) An Erlang distribution has two parameters in it. What are those parameters in this case?
    (b) The answer is not 4/0.033 = 121.212121... ; it is equal to 3/(1/30) = 120 exactly. There is no need to round off; just use the fact that 1/(1/30) = 30. If you are going to round off, keep more digits during the computation. For example, 3/0.03333 = 120.0120012..., which is much closer to the true answer.
    (c) Your answer is meaningless unless you specify what Poisson distribution X has (in other words, the mean of X or some equivalent information). Anyway, you seem to be asking for the probability that more than 120 events have occurred; after all, the Poisson distribution is the distribution of the number of events!
    (d) Read the textbook, or if it does not have the required material, look on-line. For example, the document http://www.netlab.tkk.fi/opetus/s38143/luennot/E_poisson.pdf will have everything you need to start (d) and get (d.i). Once you understand (d.i), (d.ii) will not be difficult.
    (e) Are you sure you don't know how to do (e)? Surely you must know about some basic, very important properties of exponential distributions! I am going to assume that you really do know the answer but are just forgetting something or are afraid to commit yourself. However, if I am wrong, I suggest you read your textbook or look on line for properties of exponential distributions.

  4. Feb 7, 2012 #3
    Actually I am ok with (e), correct me if I am wrong but we should be dealing with the "memorylessness" property of the exponential distribution right?

    For (a) I said distribution Erlang since we are dealing with a sum of exponential distributions: the cdf of an erlang is: 1 - exp^-(mean.x) * sum(i=0 to r-1) * (mean.x)^i / i! (hopefully this formula is clear enough). Isn't this right?

    For c) we want the fourth problem to exceed 120 days. So I think here we want P(N<=3) with parameter 120*(1/30)?

    I will look at d and see how I can solve it.

    Am I right for the others though?
  5. Feb 7, 2012 #4

    Ray Vickson

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    For (a): your answer is OK; I just wanted you to specify the parameters, so you could either give the formula for the density or cdf, or you could say E4(30) instead of just "erlang". Actually, the density is a lot easier than the cumulative: the density of the sum of n independent expl(a) variables is
    [tex] f(t) = \frac{a^n t^{n-1} e^{-at}}{(n-1)!}, \mbox{ for } t \geq 0. [/tex]

    You have (e) OK.

    For (c) the only issue is whether or not you need {N <= 3} or {N = 3}; a lawyer could argue that the question is ambiguous, so {4th later than 120} could mean {1st later than 120} or {1st before 120, 2nd later} or {first two before 120 and third later} or {first 3 before 120 and 4th later}. Which one do you think is wanted, and which of {N <= 3} or {N = 3} should you use?

  6. Feb 7, 2012 #5
    I completely understand what you mean for (c) as in just one problem might be before 120 days but the remaining 3 could happen after 120 days. But I believe, in this case, it is just asking for the exact fourth problem to happen after 120 days.
  7. Feb 7, 2012 #6
    I'm having a bit of trouble with d.(ii)

    For d.(i) the distribution for waiting time for Problem of Type 1 is exponential with a mean of 30/0.75.

    How do I proceed with d.(ii)?
  8. Feb 7, 2012 #7

    Ray Vickson

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    You have a fixed span of time (90 days in this case). On average, how many problems of types I, II and III occur in that time span? What is the cost?

  9. Feb 7, 2012 #8
    Would this be the product of the poisson distributions of each problem type?
  10. Feb 8, 2012 #9

    Ray Vickson

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    Sorry: the Forum rules/guidelines prevent me from offering any more help on this question. You just have to do the best you can now.

  11. Feb 8, 2012 #10
    Figured it out. Thanks RGV.
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