EMGT 501

HW #2

SolutionsChapter 6 - SELF TEST 21

Chapter 6 - SELF TEST 22

s.t.

u20u30u51Max 321

6u2uu

3uu2u0.5

4uu

321

321

31

0u ,u ,u 321

Ch. 6 – 21a.

0u ,5.0u ,4u

75z*3

*2

*1

*

1u

15

10

0

15

0

2u

30

01

0

30

0

3u

20

11/4

3/4

45/2

5/2

1s

0

1-1/4

-3/4

15/2

15/2

2s

0

01/2

-1/2

15

15

41/2

3/2

1530

0

2s

0

00

1

0

0

BCBasis

1u

2u

3s

75jz

jj cz

b.

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

Ch. 6 – 22a.

s.t.

x5x01Max 21

175xx3

100x

100x

20x

20x

21

2

1

2

1

0x ,x 21

s.t.

u175u100u100u20u20-Min 54321

5uuu-

10u3uu-

542

531

0u ,u ,u ,u ,u 54321

b.

The optimal solution to this problem is given by: u1=0, u2=0, u3=0, u4=5/3, and u5=10/3

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.

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.

ActivityActivity

DescriptionImmediate

PredecessorsEstimated

TimeABCDEFGHIJKLMN

-ABCCED

E,GCF,IJJH

K,L

2 weeks4 weeks

10 weeks6 weeks4 weeks5 weeks7 weeks9 weeks7 weeks8 weeks4 weeks5 weeks2 weeks6 weeks

ExcavateLay the foundationPut up the rough wallPut up the roofInstall the exterior plumbingInstall the interior plumbingPut up the exterior sidingDo the exterior paintingDo the electrical workPut up the wallboardInstall the flooringDo the interior paintingInstall the exterior fixturesInstall 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.

AB

C

E

M N

START

FINISH

H

G

D

J

I

F

LK

410

4 76

7

9

8

54

62

nodearc

5

0

0(Estimated)Time

2

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

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

AB

C

E

MN

START

FINISH

H

G

D

J

I

F

LK

4

10

4 76

7

98

54

62

5

0

0

2ES=0EF=2

ES=2EF=6ES=6

EF=16

ES=16EF=20

ES=16EF=23

ES=16EF=22ES=22EF=29

ES=20EF=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.

AB

C

E

MN

START

FINISH

H

G

D

J

I

F

LK

4

10

4 76

7

98

54

62

5

0

0

2ES=0EF=2

ES=2EF=6

ES=6EF=1

ES=16EF=20

ES=16EF=23

ES=16EF=22ES=22EF=29

ES=20EF=25 ES=25

EF=33ES=33EF=37

ES=33EF=38

ES=38EF=44

ES=29EF=38

ES=38EF=40

ES=44EF=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

AB

C

E

MN

START

FINISH

H

G

D

J

I

F

LK

4

10

4 76

7

98

54

62

5

0

0

2

LS=0LF=0

LS=0LF=2

LS=2LF=6

LS=6LF=16

LS=16LF=20

LS=20LF=25

LS=25LF=33

LS=18LF=25

LS=34LF=38

LS=33LF=38

LS=38LF=44

LS=33LF=42

LS=42LF=44

LS=26LF=33

LS=20LF=26

LS=44LF=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)

AB

C

E

MN

START

FINISH

H

G

D

J

I

LK

4

10

4 76

7

9

8

54

62

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

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.

22

6

op

6

pm4o

o pmo

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 M PE Mean Variance

A

B

C

2

13

2

9

1

2

6

3

8

18

2

4

10

9

1

4

1

OE: Optimistic EstimateM : Most Likely EstimatePE: Pessimistic Estimate

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

91

Project Time 44p 92 p

94

94

Approximating Probability of Meeting Deadline

T = a project time has a normal distribution

with mean and ,

d = a deadline for the project = 47 weeks.

44p 92 p

13

4447

p

pdK

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 )

= 1 - 0.1587

0.84.

K

K

Time - Cost Trade - OffsCrashing an activity refers to taking special costly measures to reduce the time of an activity below its normal value.

Crash

Normal

Crashtime

Normaltime

Crash cost

Normal cost

Activitycost

Activitytime

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.

2

000,430$000,490$

Cost($1,000)

Crash Costper Week

Saved

A

B

J

$100

$ 50

$ 30

Activity

Time(week)

MaximumReduction

in Time(week)N NC C

1

2

2

$180

$320

$430

$280

$420

$490

2

4

8

1

2

6

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

jx

.000,60000,50000,100 NBA xxxZ

40FINISHy

FINISHy

= 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:

iy

ix

Ex4

.4 EEF xyy

Immediate predecessor of activity J:

Activity F, which has time =

Activity I, which has time =

Relationship between these activities:

Fx5

Ix7

IIJ

FFJ

xyy

xyy

7

5

.000,60000,50000,100 NBA xxxZ

Minimize

.3,,2,1 NBA xxx

.0,0,,0,0

0,,0,0

FINISHNCB

NBA

yyyy

xxx

The Complete linear programming model

.40FINISHy

CCD

BBC

AB

xyy

xyy

xy

10

4

20

HHM xyy 9

One Immediate

Predecessor

Two ImmediatePredecessors

EEH

GGH

xyy

xyy

4

7

NNFINISH

MMFINISH

xyy

xyy

6

2

Finish Time = 40

Total Cost = $4,690,000

EMGT 501

HW #3Chapter 10 - SELF TEST 7

Chapter 10 - SELF TEST 18

Due Day: Oct 3

Ch. 10 – 7A project involving the installation of a computer system comprises eight activities. The following table lists immediate predecessors and activity times (in weeks).

ActivityImmediate

Predecessor Time

ABCDEFGH

--A

B,CDE

B,CF,G

36254393

a. Draw a project network.b. What are the critical activities?c. What is the expected project completion time?

Ch. 10 – 18The 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.

ActivityImmediatePredecessor

ABCDEFGHI

-AA

B,CBADG

E,H,F

Description

Meet with boardHire coachesReserve poolAnnounce programMeet with coachesOrder team suitsRegister swimmersCollect feesPlan first practice

Optimistic

142121111

Time (weeks)Most Probable

164232221

Pessimistic

286343331

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?