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3rd year Univeristy question on critical path watch

    • Thread Starter


    I am stuck on this question

    Q) We assume that we can reduce the duration of any task, one day at a time, up to the maximum reduction, and that the relationship between reduction and cost is linear. Using the information in the table below, find the crash gradients for the activities.

    Activity = A
    Original duration(days) = 30
    Max duration(days) = 2
    Total crash cost ($) = 3000

    My answer, which I am not sure is

    Cost gradient = (crash cost-normal time cost)/(normal time duration-crash duration)

    Crash gradient = 3000/2=1500

    The question is only worth 2 marks, for working out the crash gradient for 10 activities. I only shown 1 activity here.

    • Study Helper

    Study Helper
    Whilst I have some familiarity with CPA, I've not come across "crash gradient" in that context or otherwise, and googling the two together gives nothing.

    I can't really make sense of what you've done; can you post a link/scan for the actual question and I might be able to help.

    As ghostwalker said, no idea what crash gradients are and I've done ALOT of Decision Maths.

    As before! Well, I've done D1 and D2 - did critical paths in D1... but never even so much as mentioned crash gradients.
    • Thread Starter

    What I posted is the actual question. It's university 3rd year maths. I edited the actual title.
    • Study Helper

    Study Helper
    (Original post by GuyUK)
    What I posted is the actual question. It's university 3rd year maths. I edited the actual title.
    My volumes on OR are somewhat dated, and whilst there is some coverage on crash points and normal points and the linear relationship between the two regarding costs, it doesn't mention crash gradients. Your interpretation however seems perfectly reasonable, but that's as far as I can go.

    Edit: Just found the section on cost-time slopes, which looks to be what you are calling the "crash gradient" and that uses the formula you've got.
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Updated: September 28, 2010
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