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Homework Help: Critical Path PERT - Time to Complete a Task

  1. Sep 3, 2015 #1
    1. The problem statement, all variables and given/known data
    A small purse manufacturer has a single machine that makes the metal parts of a purse. This takes 2 minutes. Another single machine makes the cloth parts in 3 minutes. Then it takes a worker 4 minutes to sew the cloth and metal parts together. How long will it take to make 6 purses?

    2. Relevant equations
    The solution in the back of the book states that it will take 4.5 minutes to make each purse but I have no idea how they got this value. So, it will obviously take 6 x 4.5 = 27 minutes to make 6 purses. How did they get a time of 4.5 minutes?
    3. The attempt at a solution

    Task Time Preceding Task
    A | 2 | None
    B | 3 | None
    C | 4 | A, B

    The following is incorrect but it is my attempt at solving the problem.
    A to C = 2 + 4 = 6
    B to C = 3 + 4 = 7
    So, B to C is the critical path which yields a time of 7 minutes.
    6 x 7 = 42
    So, it would take 42 minutes to make six purses.

  2. jcsd
  3. Sep 3, 2015 #2


    Staff: Mentor

    Your answer seems reasonable to me. It is not unheard of for book answers to be wrong, on occasion, possibly due to a problem being changed from one edition to the next, or even due to sloppy work on the part of the person solving the problem. If I were you, I would ask my instructor what he/she thinks (assuming you are taking this as a class and not self-studying).
  4. Sep 3, 2015 #3
    Hello, and thank you for the response. I have E-mailed my instructor and he hasn't responded yet.
  5. Sep 3, 2015 #4

    Ray Vickson

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    This does not look anything like a small PERT/CPM problem to me, but it IS a large one.

    Say we start producing at time t = 0. Let's measure all time in minutes, so we can dispense with units.

    At time t = 2 machine A is done on part 1 and sends its output to the assembly stage C; but C waits for the first job to complete on machine B, which is at t = 3. So, at t = 3, C starts, and so completes part 1 at t = 3+4 = 7. Meanwhile, from t = 2 to t = 4, A is working on part 2, and from t = 3 to t = 6, B is working on part 2. Therefore, at t = 4, A sends its output to C, where it waits in a "metal" queue. At time t = 6, B sends its output to C where it, too, sits in a "cloth" queue. At t = 7, C withdraws one part from each queue and begins assembly of part 2, which completes at time t = 7 + 4 = 11. Meanwhile, from t = 4 to t = 6, A is working on part 3 and sends its output to C, which goes into the "metal parts" queue; since C is still working on part 1 then, there are two parts in the metal queue from t = 6 to t = 7. At t = 6, B starts working on part 3 and finishes at t = 9, etc, etc.

    You see ... it gets complicated and keeping track of things can be tricky. In problems of this type it is often recommended that you make use of a so-called Gantt Chart; see, eg.,
    https://en.wikipedia.org/wiki/Gantt_chart .

    However, if you prefer you CAN represent the problem as a (large) PERT network, with tasks A1, A2, A3, ..., A6 (metal parts), B1, B2, ..., B6 (cloth) and C1, C2, ...,C6 (assembly). The precedence constraint are A1 ← A2 ← A3 ← ⋅ ⋅ ⋅ ← A6, B1 ← B2 ← ⋅ ⋅ ⋅ ← B6, and (A1, B1) ← C1, (A2, B2) ← C2, ⋅ ⋅ ⋅, (A6, B6) ← C6. If you apply standard CPM methodology, the time you want is the project completion time, when C6 is finished. Applying the standard Critical Path Method should not be too hard, but it will be tedious and possible error-prone. You can do it in Excel or a package such a Matlab, Maple or Mathematica. I did it in Maple and got the book's answer of 27 minutes. Of course, 27/6 = 4.5 is the average time per purse, but I'm not sure that means very much. (For example, the average time per purse for 100 purses would be 4.03, and for 1000 purses would be 4.003, etc.)
    Last edited: Sep 3, 2015
  6. Sep 3, 2015 #5


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    I think the essential point is to look at the start time for each sub-task..
    A1 and B1 can start at zero, but C1 can start at 3 when both A1 and B1 are complete.
    Then A2 can start at 2, B2 can start at 3 and C2 can start at 7 when C1, A2 and B2 are complete.
    You could then continue your PERT table up to C6, but the solution is already obvious (if you've drawn the diagram) and all sub-tasks are complete at 27, giving an average time of 27 / 6 = 4.5

    PS. Looking back at the OP, is the CP not A-C, but B1-C1-C2-C3-C4-C5-C6 for the whole task?
    You can't just take making one purse and multiply it up by the number of purses.
    Last edited: Sep 3, 2015
  7. Sep 4, 2015 #6

    Vanadium 50

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    Merlin is correct, but I think there is a clearer way to look at it.

    Once he gets going, the sewer - that is, the person who sews, not the place where sewage goes - is never starved for parts. The second he finishes, there are more parts waiting for him. So, he will take 6x4 minutes = 24 minutes. When can he start? After 3 minutes, when he has both parts. So the total time is 3+24 minutes or 27.

    It is true that the average time is 27/6 = 4.5 minutes. But that's a consequence of the 27, not its cause. If he did only two purses, the average time would be 5.5 minutes. If he did a million, it would be 4 minutes.
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