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

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

c. From the z j values for the surplus variables we see that the optimal primal solution is x 1 =15/2, x 2 =15, and x 3 =0. d. The optimal value for the dual is shown in part b to equal 75. Substituting x 1 =15/2 and x 2 =15 into the primal objective function, we find that it gives the same value. 4(15/2)+3(15)=75

Ch. 6 – 22 a. b. The optimal solution to this problem is given by: u 1 =0, u 2 =0, u 3 =0, u 4 =5/3, and u 5 =10/3

c. The optimal number of calls is given by the negative of the dual prices for the dual: x 1 =25 and x 2 =100. Commission=$750. d. u 4 =5/3: $1.67 commission increase for an additional call for product 2. u 5 =10/3: $3.33 commission increase for an additional hour of selling time per month.

Project Scheduling: PERT-CPM

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.

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

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.

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

A B C E M N START FINISH H G D J I F LK nodearc (Estimated) Time 2

START A B C D G H M FINISH = 40 weeks START A B C E F J K N FINISH = 43 weeks START A B C E F J L N FINISH = 44 weeks Path and Length Critical Path

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.

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

A B C E M N START FINISH H G D J I F LK 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

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.

A B C E M N START FINISH H G D J I F L K 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

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

A B C E M N START FINISH H G D J I F L K 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

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

S=(20,20) F=(25,25) A B C E M N START FINISH H G D J I L K 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

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.

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.

o p mo Beta distribution Mean : Variance:

Mean critical path: A path through the project network becomes the critical path if each activity time equals its mean. Activity OE MPEMeanVariance ABCABC OE: Optimistic Estimate M : Most Likely Estimate PE: Pessimistic Estimate

Activities on Mean Critical Path MeanVariance ABCEFJLNABCEFJLN Project Time

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.

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 ) =

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

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

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

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.

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

= 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:

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

Minimize The Complete linear programming model

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

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

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). Activity Immediate Predecessor Time ABCDEFGHABCDEFGH - A B,C D E B,C F,G a.Draw a project network. b.What are the critical activities? c.What is the expected project completion time?

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. Activity Immediate Predecessor ABCDEFGHIABCDEFGHI - A B,C B A D G E,H,F Description Meet with board Hire coaches Reserve pool Announce program Meet with coaches Order team suits Register swimmers Collect fees Plan first practice Optimistic Time (weeks) Most Probable Pessimistic

a.Draw a project network. b.Develop an activity schedule. c.What are the critical activities, and what is the expected project completion time? d.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?