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Kelly Devilbiss - Problem Problem The estimated times (in weeks) and immediate predecessors for the activities in a project are given in the following table. Assume that the activity times are independent. Activity Immediate Predecessoramb A B CA91011 DB58

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Kelly Devilbiss - Problem (a)Calculate the expected time and variance for each activity. Activityamb Expected time t= [(a+4m+b)/6] Variance [(b-a)/6] 2 A /36 B /36 C /36 D /36

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Kelly Devilbiss - Problem (b)What is the expected completion time of the critical path? What is the expected completion time of the other patient in the network? Critical Path AC takes 20 weeks Critical Path BD takes 18 weeks

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Kelly Devilbiss - Problem (c)What is the variance of the critical path? What is the variance of the other path in the network? Critical Path AC: σ 2 CP = σ 2 A + σ 2 C = 4/36 + 4/36 = 8/36 = Critical Path BD: σ 2 CP = σ 2 B + σ 2 D = 144/ /36 = 180/36 = 5

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Kelly Devilbiss - Problem (d)If the time to complete path A-C is normally distributed, what is the probability that this path will be finished in 22 weeks or less? t = 20 variance (σ 2 ) = standard deviation (√0.222) = 0.47 P(x ≤ 22) = P[(22 – 20)/0.47] = 4.26 Using Appendix A:4.26 = 1.00

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Kelly Devilbiss - Problem (e)If the time to complete path B-D is normally distributed, what is the probability that this path will be finished in 22 weeks or less? t = 18 variance (σ 2 ) = 5 standard deviation (√5) = 2.24 P(x ≤ 22) = P[(22 – 18)/2.24] = 1.79 Using Appendix A:1.79 =

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Kelly Devilbiss - Problem (f)Explain why the probability that the critical path will be finished in 22 weeks or less is not necessarily the probability that the project will be finished in 22 weeks or less. Path B-D has more variability and has a higher probability of exceeding 22 weeks.

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