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EMGT 501 HW #2 Solutions Chapter 6 - SELF TEST 21 Chapter 6 - SELF TEST 22

EMGT 501 HW #2 Solutions Chapter 6 - SELF TEST 21 Chapter 6 - SELF TEST 22. Ch. 6 – 21 a. b. 15. 30. 20. 0. 0. 0. Basis. 15. 1. 0. 1. 1. 0. 0. 4. 30. 0. 1. 1/4. -1/4. 1/2. 0. 1/2. 0. 0. 0. 3/4. -3/4. -1/2. 1. 3/2. 15. 30. 45/2. 15/2. 15. 0. 75. 0. 0.

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EMGT 501 HW #2 Solutions Chapter 6 - SELF TEST 21 Chapter 6 - SELF TEST 22

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  1. EMGT 501 HW #2 Solutions Chapter 6 - SELF TEST 21 Chapter 6 - SELF TEST 22

  2. Ch. 6 – 21 a. b. 15 30 20 0 0 0 Basis 15 1 0 1 1 0 0 4 30 0 1 1/4 -1/4 1/2 0 1/2 0 0 0 3/4 -3/4 -1/2 1 3/2 15 30 45/2 15/2 15 0 75 0 0 5/2 15/2 15 0

  3. c. From the zj values for the surplus variables we see that the optimal primal solution is x1=15/2, x2=15, and x3=0. d. The optimal value for the dual is shown in part b to equal 75. Substituting x1=15/2 and x2=15 into the primal objective function, we find that it gives the same value. 4(15/2)+3(15)=75

  4. Ch. 6 – 22 a. b. The optimal solution to this problem is given by: u1=0, u2=0, u3=0, u4=5/3, and u5=10/3

  5. c. The optimal number of calls is given by the negative of the dual prices for the dual: x1=25 and x2=100. Commission=$750. d. u4=5/3: $1.67 commission increase for an additional call for product 2. u5=10/3: $3.33 commission increase for an additional hour of selling time per month.

  6. Project Scheduling: PERT-CPM

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

  8. Activity Description Estimated Time Immediate Predecessors Activity A B C D E F G H I J K L M N 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 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 - A B C C E D E,G C F,I J J H K,L

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

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

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

  18. LS: Latest Start time for a particular activity LF: Latest Finish time for a particular activity 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.

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

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

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

  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. p o o m 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 M PE Mean Variance 1 2 3 2 A B C 2 1 4 8 4 6 9 18 10 OE: Optimistic Estimate M : Most Likely Estimate PE: Pessimistic Estimate

  26. Activities on Mean Critical Path Mean Variance A B C E F J L N 2 4 10 4 5 8 5 6 1 4 1 1 1 Project Time

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

  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. Activity cost Crash Crash cost Normal Normal cost Crash time Normal time 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. Maximum Reduction in Time (week) Crash Cost per Week Saved Time (week) Cost ($1,000) Activity N C N C 1 2 2 $180 $320 $430 A B J 2 4 8 1 2 6 $280 $420 $490 $100 $ 50 $ 30 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. The Complete linear programming model Minimize

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

  38. EMGT 501 HW #3 Chapter 10 - SELF TEST 7 Chapter 10 - SELF TEST 18 Due Day: Oct 3

  39. Ch. 10 – 7 A project involving the installation of a computer system comprises eight activities. The following table lists immediate predecessors and activity times (in weeks). Immediate Predecessor Activity Time A B C D E F G H - - A B,C D E B,C F,G 3 6 2 5 4 3 9 3 • Draw a project network. • What are the critical activities? • What is the expected project completion time?

  40. Ch. 10 – 18 The manager of the Oak Hills Swimming Club is planning the club’s swimming team program. The first team practice is scheduled for May 1. The activities, their immediate predecessors, and the activity time estimates (in weeks) are as follows. Immediate Predecessor Time (weeks) Most Probable Activity Description Pessimistic Optimistic A B C D E F G H I - A A B,C B A D G E,H,F 1 6 4 2 3 2 2 2 1 2 8 6 3 4 3 3 3 1 Meet with board Hire coaches Reserve pool Announce program Meet with coaches Order team suits Register swimmers Collect fees Plan first practice 1 4 2 1 2 1 1 1 1

  41. Draw a project network. • Develop an activity schedule. • What are the critical activities, and what is the expected project completion time? • If the club manager plans to start the project on February1, what is the probability the swimming program will be ready by the scheduled May 1 date (13 weeks)? Should the manager begin planning the swimming program before February 1?

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