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Kelly Devilbiss - Problem 13-281 Problem 13-28 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-91011 B-41016 CA91011 DB58

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Kelly Devilbiss - Problem 13-282 (a)Calculate the expected time and variance for each activity. Activityamb Expected time t= [(a+4m+b)/6] Variance [(b-a)/6] 2 A91011104/36 B4101610144/36 C91011104/36 D5811836/36

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Kelly Devilbiss - Problem 13-283 (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 13-284 (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 = 0.222 Critical Path BD: σ 2 CP = σ 2 B + σ 2 D = 144/36 + 36/36 = 180/36 = 5

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Kelly Devilbiss - Problem 13-285 (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 ) = 0.222 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 13-286 (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 = 0.96327

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Kelly Devilbiss - Problem 13-287 (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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