Sara505I've got a question in this, for part b of this question, you did 0.8-0.5 because the completion time should be beyond the mean which is 50% right? As in it should be somewhere less than 0.8 but greater than 0.5 right? Another this, where did the 0.84 value come from? And thank you!
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The optimistic, most probable and pessimistic time (in week)
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Question (a)1• R.C Coleman’s top management established a required 40-week completion time for the project. Can this completion time be achieved?
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D I ASTART C E F K FINISH H J B G Project Network
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Expected Time•Optimistic time (a)The minimum activity time if everything progresses ideally•Most probable time (m)The most probable activity time under normal conditions•Pessimistic time (b)The maximum activity time if significant delays areencountered•Expected time (t)The average time for the activity a 4m b t 6
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ACTIVITY EXPECTED TIME (WEEKS) A 6 B 9 C 4 D 12 E 10 F 6 G 8 H 6 I 7 J 4 K 4 TOTAL 76 Expected Time In Weeks
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D 13 25 12 17 29 A 0 6 6 3 9 I 29 36 7 32 39 C 9 13 E 13 23 K 39 43START 4 39 43 4 9 13 10 13 23 F 23 29 H 29 35 B 0 9 6 23 29 6 29 35 FINISH 9 0 9 J 35 39 4 35 39 G 13 21 8 21 29 Project Network (Earliest Start, Finish Times, Latest Start and Finish Times)
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ACTIVITY ES EF LS LF SLACK A 0 6 3 9 3 B 0 9 0 9 0 (critical) C 9 13 9 13 0 (critical) D 13 25 17 29 4 E 13 23 13 23 0 (critical) F 23 29 23 29 0 (critical) G 13 21 21 29 8 H 29 35 29 35 0 (critical) I 29 36 32 39 3 J 35 39 35 39 0 (critical) K 39 43 39 43 0 (critical) Activity Slack Time
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Critical Path:B-C-E-F-H-J-KProject Completion Time:43 weeksR.C Coleman’s required 40-week completion time. This completion time can’t be achieved
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Question (a)2Include probability information in yourdiscussion. What recommendations do youhave if the 40-week completion time isrequired?
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Variance ( 2)• used to describe the dispersion or variation in the activity time values 2 b a 2 6
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ACTIVITY VARIANCE A 0.44 B 2.78 C 0.44 D 7.11 E 1.00 F 0.44 G 7.11 H 0.44 I 2.78 J 0.11 K 0.44 Variance of Each Activity
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Variance in the project completion time ( 2) 2 = 2 + 2 + 2 + 2 + 2 2 2 B C E F H+ J + K = 2.78 + 0.44 + 1.00 + 0.44 + 0.44 + 0.11 + 0.44 = 5.65Standard deviation ( ) = √5.65 = 2.38
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Normal Distribution of Project Completion Time = 2.38 Expected Completion Time 43 Time (weeks)
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Probability the project will meet the 40-week completion time = 2.38 40 43 Time (weeks)
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At 40 ≤ T ≤ 43z = (40 – 43) / 2.38 = 1.26using table,Pr (40 ≤ T ≤ 43) = 0.3962Pr (T ≤ 40) = 0.5000 – 0.3962 = 0.103 = 10.38% We find that the probability of the project meeting the 40-weeks deadline is 10.38%. Thus, R.C Collman have a low chance if they want the project completed at 40-weeks
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Question (b)• Suppose that management request that activity times be shortened to provide an 80 percent chance of meeting the 40-week completion time. If the variance in the project completion time is the same as you found in (a), how much should the expected project completion time be shortened to achieved the goal of an 80 percent chance of completion within 40 weeks?
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Expected completion time to achieve 80% chance of completion within 40 weeks = 2.38 X 40 Time (weeks)
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Pr (T ≤ 40) = 0.8000Pr (x ≤ T ≤ 40) = 0.8000 – 0.5000 = 0.3000using table,z (x ≤ T ≤ 40) = 0.84(40 – x) / 2.38 = 0.84x = 38 It shows the project has to be shortened to 38 weeks to achieve the goal of an 80 % chance of completion within 40 weeks.
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Question (c)• Using the expected completion times as the normal times and the following crashing information, determine the activity crashing decisions and revised activity schedule for the warehouse expansion project
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Maximum possible reduction in time, Mi M i i i (3)with i = expected time for activity I ’i = time for activity i under maximum crashingCrash cost per unit of time, Ki C C K i i i M iwith Ci = cost for activity I C’i = cost for activity i under maximum crashing
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TIME COSTACTIVITY Mj Kj NORMAL CRASHED NORMAL CRASHED A 6 4 1000 1900 2 450 B 9 7 1000 1800 2 400 C 4 2 1500 2700 2 600 D 12 8 2000 3200 4 300 E 10 7 5000 8000 3 1000 F 6 4 3000 4100 2 550 G 8 5 8000 10259 3 753 H 6 4 5000 6400 2 700 I 7 4 10000 12400 3 800 J 4 3 4000 4400 1 400 K 4 3 5000 5500 1 500
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CRASHINGACTIVITY TIME (WEEKS) (WEEKS) A 6 0 B 7 2 C 4 0 D 12 0 E 10 0 F 5 1 G 8 0 H 6 0 I 7 0 J 3 1 K 3 1 Total Crashing Time 5 weeks Total Crashing Cost $ 2,250.00 Activity Crashing Decision
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