1. EMGT 501 HW #2 Answer

2. 020/3 X 3 05/601-1/62/3050/3 X 6 0-5/300-2/3-1/3180/3 (c).3/230with )3/80,0,0,3/50,3/20,0(*)*, ( solution Optimal 654321   Z xxxxxx

3. 6.1-4 (a) Dual formulation becomes Min s.t. # of constraints of Dual = 3 # of constraints of Primal = 5 So, Dual is better than Primal because the size of B -1 in Dual is smaller than that of Primal.

4. (b) Dual formulation becomes Min s.t. # of constraints of Dual = 5 # of constraints of Primal = 3 So, Primal is better than Dual because the size of B -1 in Primal is smaller than that of Dual.

5. 6.1-5 (a) Min s.t. (b) It is clear that Z*=0, y 1 *=0, y 2 *=0.

6. Project Scheduling: PERT-CPM

7. PERT (Program evaluation and review technique) and CPM (Critical Path Method) makes a managerial technique to help planning and displaying the coordination of all the activities.

8. Activity Activity Description Immediate Predecessors Estimated Time ABCDEFGHIJKLMNABCDEFGHIJKLMN - A B C E D E,G C F,I J H K,L 2 weeks 4 weeks 10 weeks 6 weeks 4 weeks 5 weeks 7 weeks 9 weeks 7 weeks 8 weeks 4 weeks 5 weeks 2 weeks 6 weeks Excavate Lay the foundation Put up the rough wall Put up the roof Install the exterior plumbing Install the interior plumbing Put up the exterior siding Do the exterior painting Do the electrical work Put up the wallboard Install the flooring Do the interior painting Install the exterior fixtures Install the interior fixtures

9. Immediate predecessors: For any given activity, its immediate predecessors are the activities that must be completed by no later than the starting time of the given activity.

10. AON (Activity-on-Arc): Each activity is represented by a node. The arcs are used to show the precedence relationships between the activities.

11. A B C E M N START FINISH H G D J I F LK 4 10 4 7 6 7 9 8 5 4 6 2 nodearc 5 0 0 (Estimated) Time 2

12. START A B C D G H M FINISH 2 + 4 + 10 + 6 + 7 + 9 + 2 = 40 weeks START A B C E F J K N FINISH 2 + 4 + 10 + 4 + 5 + 8 + 4 + 6 = 43 weeks START A B C E F J L N FINISH 2 + 4 + 10 + 4 + 5 + 8 + 5 + 6 = 44 weeks Path and Length Critical Path

13. Critical Path: A project time equals the length of the longest path through a project network. The longest path is called “critical path”. Activities on a critical path are the critical bottleneck activities where any delay in their completion must be avoided to prevent delaying project completion.

14. ES : Earliest Start time for a particular activity EF : Earliest Finish time for a particular activity

15. A B C E M N START FINISH H G D J I F LK 4 10 4 7 6 7 9 8 5 4 6 2 5 0 0 2 ES=0 EF=2 ES=2 EF=6 ES=6 EF=16 ES=16 EF=20 ES=16 EF=23 ES=16 EF=22 ES=22 EF=29 ES=20 EF=25

16. If an activity has only a single immediate predecessor, then ES = EF for the immediate predecessor. Earliest Start Time Rule: The earliest start time of an activity is equal to the largest of the earliest finish times of its immediate predecessors. ES = largest EF of the immediate predecessors.

17. A B C E M N START FINISH H G D J I F L K 4 10 4 7 6 7 9 8 5 4 6 2 5 0 0 2 ES=0 EF=2 ES=2 EF=6 ES=6 EF=1 ES=16 EF=20 ES=16 EF=23 ES=16 EF=22 ES=22 EF=29 ES=20 EF=25 ES=25 EF=33 ES=33 EF=37 ES=33 EF=38 ES=38 EF=44 ES=29 EF=38 ES=38 EF=40 ES=44 EF=44

18. Latest Finish Time Rule: The latest finish time of an activity is equal to the smallest of the latest finish times of its immediate successors. LF = the smallest LS of immediate successors. LS: Latest Start time for a particular activity LF: Latest Finish time for a particular activity

19. A B C E M N START FINISH H G D J I F L K 4 10 4 7 6 7 9 8 5 4 6 2 5 0 0 2 LS=0 LF=0 LS=0 LF=2 LS=2 LF=6 LS=6 LF=16 LS=16 LF=20 LS=20 LF=25 LS=25 LF=33 LS=18 LF=25 LS=34 LF=38 LS=33 LF=38 LS=38 LF=44 LS=33 LF=42 LS=42 LF=44 LS=26 LF=33 LS=20 LF=26 LS=44 LF=44

20. Latest Start Time Earliest Start Time S=( 2, 2 ) F=( 6, 6 ) Latest Finish Time Earliest Finish Time

21. S=(20,20) F=(25,25) A B C E M N START FINISH H G D J I L K 4 10 4 7 6 7 9 8 5 4 6 2 5 0 0 2 S=(0,0) F=(0,0) S=(0,0) F=(2,2) S=(2,2) F=(6,6) S=(16,16) F=(20,20) S=(25,25) F=(33,33) S=(16,18) F=(23,25) S=(33,34) F=(37,38) S=(33,33) F=(38,38) S=(38,38) F=(44,44) S=(29,33) F=(38,42) S=(38,42) F=(40,44) S=(22,26) F=(29,33) S=(16,20) F=(22,26) S=(44,44) F=(44,44) F S=(6,6) F=(16,16) Critical Path

22. Slack: A difference between the latest finish time and the earliest finish time. Slack = LF - EF Each activity with zero slack is on a critical path. Any delay along this path delays a whole project completion.

23. Three-Estimates Most likely Estimate (m) = an estimate of the most likely value of time. Optimistic Estimate (o) = an estimate of time under the most favorable conditions. Pessimistic Estimate (p) = an estimate of time under the most unfavorable conditions.

24. o p mo Beta distribution Mean : Variance:

25. Mean critical path: A path through the project network becomes the critical path if each activity time equals its mean. Activity OE MPEMeanVariance ABCABC 2 9 1 2 6 3 8 18 2 4 10 4 1 OE: Optimistic Estimate M : Most Likely Estimate PE: Pessimistic Estimate

26. Activities on Mean Critical Path MeanVariance ABCEFJLNABCEFJLN 2 4 10 4 5 8 5 6 1411114111 Project Time

27. Approximating Probability of Meeting Deadline T = a project time has a normal distribution with mean and, d = a deadline for the project = 47 weeks. Assumption: A probability distribution of project time is a normal distribution.

28. Using a table for a standard normal distribution, the probability of meeting the deadline is P ( T d ) = P ( standard normal ) = 1 - P( standard normal ) = 1 - 0.1587 0.84.

29. Time - Cost Trade - Offs Crashing an activity refers to taking special costly measures to reduce the time of an activity below its normal value. Crash Normal Crash time Normal time Crash cost Normal cost Activity cost Activity time

30. Activity J: Normal point: time = 8 weeks, cost = $430,000. Crash point: time = 6 weeks, cost = $490,000. Maximum reduction in time = 8 - 6 = 2 weeks. Crash cost per week saved = = $30,000.

31. Cost ($1,000) Crash Cost per Week Saved ABJABJ $100 $ 50 $ 30 Activity Time (week) Maximum Reduction in Time (week) NNCC 122122 $180 $320 $430 $280 $420 $490 248248 126126 N: Normal C: Crash

32. Using LP to Make Crashing Decisions Let Z be the total cost of crashing activities. A problem is to minimize Z, subject to the constraint that its project duration must be less than or equal to the time desired by a project manager.

33. = the reduction in the time of activity j by crashing it = the project time for the FINISH node

34. = the start time of activity j Duration of activity j = its normal time Immediate predecessor of activity F: Activity E, which has duration = Relationship between these activities:

35. Immediate predecessor of activity J: Activity F, which has time = Activity I, which has time = Relationship between these activities:

36. Minimize The Complete linear programming model

37. One Immediate Predecessor Two Immediate Predecessors Finish Time = 40 Total Cost = $4,690,000

38. EMGT 501 HW #3 10.3-5 10.4-5 Due Day: Sep 27

39. 10.3-5. You are given the following information about a project consisting of six activities: (a) Construct the project network for this project. (b) Find the earliest times, latest times, and slack for each activity. Which of the paths is a critical path? (c) If all other activities take the estimated amount of time, what is the maximum duration of activity D without delaying the completion of the project?

40. 10.4-5 Sharon Lowe, vice president for marketing for the Electronic Toys Company, is about to begin a project to design an advertising campaign for a new line of toys. She wants the project completed within 57 days in time to launch the advertising campaign at the beginning of the Christmas season. Sharon has identified the six activities (labeled A, B, …, F) needed to execute this project. Considering the order in which these activities need to occur, she also has constructed the following project network. STARTFINISH ACEF BD

41. Using the PERT three-estimate approach, Sharon has obtained the following estimates of the duration of each activity. Optimistic Most Likely Pessimistic Activity Estimate Estimate Estimate A 12 days 12 days 12 days B 15 days 21 days 39 days C 12 days 15 days 18 days D 18 days 27 days 36 days E 12 days 18 days 24 days F 2 days 5 days 14 days (a) Find the estimate of the mean and variance of the duration of each activity. (b) Find the mean critical path.

42. (c) Use the mean critical path to find the approximate probability that the advertising campaign will be ready to launch within 57 days. (d) Now consider the other path through the project network. Find the approximate probability that this path will be completed within 57 days. (e) Since these paths do not overlap, a better estimate of the probability that the project will finish within 57 days can be obtained as follows. The project will finish within 57 days if both paths are completed within 57 days. Therefore, the approximate probability that the project will finish within 57 days is the product of the probabilities found in parts (c) and (d). Perform this calculation. What does this answer say about the accuracy of the standard procedure used in part (c)?