EMGT 501

HW #2

Answer

4.4-6 (b)Basis Z X1 X2 X3 X4 X5 X6 RHS

Z 1 -2 -4 -3 0 0 0 0X4 0 3 4 2 1 0 0 60X5 0 2 1 2 0 1 0 40X6 0 1 3 2 0 0 1 80Z 1 1 0 -1 1 0 0 60X2 0 3/4 1 1/2 1/4 0 0 15X5 0 5/4 0 3/2 -1/4 1 0 25X6 0 -5/4 0 1/2 -3/4 0 1 35Z 1 11/6 0 0 5/6 2/3 0 230/3X2 0 1/3 1 0 1/3 -1/3 0 20/3X3 0 5/6 0 1 -1/6 2/3 0 50/3X6 0 -5/3 0 0 -2/3 -1/3 1 80/3

(c)

.3/230with

)3/80,0,0,3/50,3/20,0(*)*,*,*,*,*,( solution Optimal 654321

Z

xxxxxx

6.1-4 (a) Dual formulation becomes

Min

s.t.

711232

411121

1021513

54321

54321

54321

yyyyy

yyyyy

yyyyy

54321 2090402525 yyyyy

.0,,,, 54321 yyyyy

# 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.

(b) Dual formulation becomes

Mins.t.

21 156 yy

1

473

352

563

24

21

21

21

21

21

yy

yy

yy

yy

yy

# 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.

6.1-5 (a)

Mins.t.

21 112 yy

12

2

1

21

21

21

yy

yy

yy

0,0 21 yy

(b)

It is clear that Z*=0, y1*=0, y2*=0.

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

4

op

pmo

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 #3

10.3-5

10.4-5

Due Day: Sep 27

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?

ActivityImmediate

PredecessorsEstimatedDuration

A - 5 monthsB - 1 monthsC B 2 monthsD A, C 4 monthsE A 6 monthsF D, E 3 months

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.

START FINISH

A C E F

B D

Using the PERT three-estimate approach, Sharon has obtained the following estimates of the duration of each activity.

Optimistic Most Likely PessimisticActivity 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.

(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)?