Pert Cpm (Unit 6) Copy

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Project Management

PERT & CPM

Presented by:

Anubhav Geetey



Contents

Project Management Basics

CPM

PERT

Crashing



Characteristics of a project:

A project is a tem porary endeavor involving a connected sequenceof activities and a range of resources, which is designed toachieve a s pecific and unique outcome and which o perateswithin time, cost and quality constraints and which is oftenused to introduce change.

A unique, one-time o perational activity or effort,

Requires the com pletion of a large number of interrelated activities,

Established to achieve specific objective,

Resources (such as time and/or money) are limited,

Ty pically has its own management structure.

What is a Project?



What is Project Management?

The application of a collection of tools and techniques todirect the use of diverse resources towards the accom plishmentof a unique, com plex, one-time task within time, cost andquality constraints.

Its origins lie in World War II, when the military authoritiesused the techniques of o perational research to plan theoptimum use of resources.

One of these techniques was the use of networks to re present asystem of related activities.



Project Management Process:

Project planning Project scheduling

Project control

Project team -

made u p of individuals from various areas and de partments within a com pany

Matrix organization -

team structure with members from functional areas, de pending on skills required

Project Manager -

most im portant member of project team

Scope statement -

document providing an understanding, justification & ex pected result of project

Statement of work -written descri ption of objectives of a project

Organizational Breakdown Structure -

a chart that shows which organizational units are res ponsible for work items

Responsibility Assignment Matrix -

shows who is res ponsible for which work in a project

}



What is Work Breakdown Structure?

A method of breaking down a project into individual elements(com ponents, subcom ponents, activities and tasks) in ahierarchical structure which can be scheduled and cost.

It is foundation of project planning.

It is develo ped before identification of de pendencies andestimation of activity durations.

It can be used to identify the tasks in the CPM and PERT.



Project Planning

Resource Availability and/or Limits: (Time & Money)

Due date, late penalties, early com pletion incentives

Budget

Activity Information:

Identify all required activities

Estimate the resources (such as time and/or money)

required to com plete each activity

Immediate predecessor(s) to each activity to create

interrelationshi ps



Gantt Chart

Critical Path Method (CPM)

Program Evaluation and Review Technique (PERT)

Project Scheduling & Control techniques



Gra ph or bar chart with a bar for each project activity that shows passage of time. It provides a visual dis play of the project schedule.

Gantt Chart



History of CPM & PERT

Critical Path Method (CPM) DuPont Co. (1957) for construction and maintenance of

new chemical plant,

Deterministic task times,

Activity-on-node network construction,

Re petitive nature of jobs.

Project Evaluation and Review Technique (PERT)

US Navy (1958) for the POLARIS missile program,

Multi ple task time estimates ( probabilistic nature),

Activity-on-arrow network construction,

Non-re petitive jobs (R&D work).



Project Network

Network shows the sequential relationshi ps among various

activities. A network is made u p of:

Arrows

(An arrow leads from tail to head directionally. It indicates ACTIVITY, atime consuming effort that is required to perform a part of the work.)

Nodes

(A node is re presented by a circle. It indicates EVENT, a point in timewhere one or more activities start and/or finish.)

Dummy Activity

(A dummy activity is re presented by a dashed arrow. It indicates onlyPRECEDENCE relationshi ps & does not require any time or effort.)



Situations in network diagram

AB

CA must finish before either of B or C canstart

A

B

C both A and B must finish before C can start

D

C

B

A both A and C must finish before either of B

or D can start

A

C

B

D

DummyA must finish before B can start &

both A and C must finish before D can start



Activity-on-arrow (AOA):

arrows re present activities and nodes are events for points in time.

3

2 0

13

1 1

11 2 4 6 7

3

5

Lay

foundation

Design house

and obtain

financing

Order and

receive

materials

Dummy

Finish

work

Select

carpet

Select

paint

Build

house

(AOA) Project Network forH

ouse



Concurrent Activities:

2 3

Lay foundationLay foundation

Order materialOrder material

(a)(a) Incorrect precedenceIncorrect precedence

relationshiprelationship

(b)(b) Correct precedenceCorrect precedence

relationshiprelationship

3

42

DummyDummyLayLay

foundationfoundation

Order materialOrder material

11

22 00



(AON) Project Network forH

ouse

13

22

4

3

31 5

1

61

7

1Start

Design house and

obtain financing

Order and receive

materialsSelect paint

Select carpet

Lay foundations Build house

Finish work

Activity-on-node (AON):

nodes re present activities, and arrows show precedence

relationshi ps



?

?

?

?

?



Activity ImmediatePreceding

Activities

Duration(in weeks)

a - 6

b - 8

c - 5

d b 13

e c 9

f a 15

g a 17

h f 9

i g 6

j d, e 12

CPM Example:



Network

1 8

2

7

4

3

5

a, 6a, 6

f, 15f, 15f, 15f, 15

b, 8b, 8b, 8b, 8

c, 5c, 5c, 5c, 5

e, 9e, 9e, 9e, 9

d, 13d, 13d, 13d, 13

g, 17g, 17g, 17g, 17 h, 9h, 9h, 9h, 9

i, 6i, 6i, 6i, 6

j, 12 j, 12 j, 12 j, 12

6



CPM calculations

Path

A connected sequence of activities leading from thestarting event to the ending event.

Critical Path

The longest path (in terms of time); determines the projectduration i.e. minimum project com pletion time.

Critical ActivitiesAll the activities that make u p the critical path.



Paths (1-2-6-8), (a-f-h), 30 (1-2-7-8), (a-g-i), 29

(1-3-5-8), (b-d-j), 33 (1-4-5-8), (c-e-j), 26

1 8

2

7

4

3

5

a, 6a, 6

f, 15f, 15f, 15f, 15

b, 8b, 8b, 8b, 8

c, 5c, 5c, 5c, 5

e, 9e, 9e, 9e, 9

d, 13d, 13d, 13d, 13

g, 17g, 17g, 17g, 17 h, 9h, 9h, 9h, 9

i, 6i, 6i, 6i, 6

j, 12 j, 12 j, 12 j, 12

6



1 8

2

7

4

3

5

a, 6a, 6a, 6a, 6

f, 15f, 15f, 15f, 15

b, 8b, 8b, 8b, 8

c, 5c, 5c, 5c, 5

e, 9e, 9e, 9e, 9

d, 13d, 13d, 13d, 13

g, 17g, 17g, 17g, 17 h, 9h, 9h, 9h, 9

i, 6i, 6i, 6i, 6

j, 12 j, 12 j, 12 j, 12

6

Critical Path & Critical Activities



Forward Pass

Earliest Start Time (ES)

Earliest time an activity can start.

ES = maximum EF of immediate predecessors

Earliest finish time (EF)

Earliest time an activity can finish.

EF= ES + t (where; t = activity time)



ES and EF Times

1 8

2

7

4

3

5

a, 6a, 6a, 6a, 6

f, 15f, 15f, 15f, 15

b, 8b, 8b, 8b, 8

c, 5c, 5c, 5c, 5

e, 9e, 9e, 9e, 9

d, 13d, 13d, 13d, 13

g, 17g, 17g, 17g, 17 h, 9h, 9h, 9h, 9

i, 6i, 6i, 6i, 6

j, 12 j, 12 j, 12 j, 12

6

0 6

0 8

0 5



ES and EF Times

1 8

2

7

4

3

5

a, 6a, 6a, 6a, 6

f, 15f, 15f, 15f, 15

b, 8b, 8b, 8b, 8

c, 5c, 5c, 5c, 5

e, 9e, 9e, 9e, 9

d, 13d, 13d, 13d, 13

g, 17g, 17g, 17g, 17 h, 9h, 9h, 9h, 9

i, 6i, 6i, 6i, 6

j, 12 j, 12 j, 12 j, 12

6

0 6

0 8

0 5

5 14

8 21

6 23

6 21



1 8

2

7

4

3

5

a, 6a, 6a, 6a, 6

f, 15f, 15f, 15f, 15

b, 8b, 8b, 8b, 8

c, 5c, 5c, 5c, 5

e, 9e, 9e, 9e, 9

d, 13d, 13d, 13d, 13

g, 17g, 17g, 17g, 17 h, 9h, 9h, 9h, 9

i, 6i, 6i, 6i, 6

j, 12 j, 12 j, 12 j, 12

6

0 6

0 8

0 5

5 14

8 21 21 33

6 2321 30

23 29

6 21

Project¶s EF = 33

ES and EF Times



Backward Pass

Latest Start Time (LS)

Latest time an activity can start without delaying critical path time.

LS= LF - t

Latest finish time (LF)

Latest time an activity can be com pleted without delayingcritical path time.

LF = minimum LS of immediate predecessors



1 8

2

7

4

3

5

a, 6a, 6a, 6a, 6

f, 15f, 15f, 15f, 15

b, 8b, 8b, 8b, 8

c, 5c, 5c, 5c, 5

e, 9e, 9e, 9e, 9

d, 13d, 13d, 13d, 13

g, 17g, 17g, 17g, 17

h, 9h, 9h, 9h, 9

i, 6i, 6i, 6i, 6

j, 12 j, 12 j, 12 j, 12

6

0 6

0 8

0 5

5 14

8 2121 33

6 23

21 30

23 29

6 21

21 33

27 33

24 33

LS and LF Times







Event Slack (by activity)

Slack is the maximum amount of time that this activity can

be delayed in its com pletion before it becomes a critical

activity.

Slack = (LS ± ES) = (LF ± EF)

On the critical path, slack = 0.

Activity Floats (by events i,j)

Total float : (LS j ± ESi - t) most favorable

Free float : (ES j ± ESi - t) moderate

Inde pendent float : (ES j ± LSi - t) adverse





1 8

2

7

4

3

5

a, 6a, 6a, 6a, 6

f, 15f, 15f, 15f, 15

b, 8b, 8b, 8b, 8

c, 5c, 5c, 5c, 5

e, 9e, 9e, 9e, 9

d, 13d, 13d, 13d, 13

g, 17g, 17g, 17g, 17 h, 9h, 9h, 9h, 9

i, 6i, 6i, 6i, 6

j, 12 j, 12 j, 12 j, 12

6

Critical Path



CPM analysis

D

raw the CPM network.

Analyze the paths through the network.

Determine the slack for each activity.

Find the critical path that is the sequence of activities andevents where there is no ³slack´ i.e., Zero slack.

Find the project duration i.e., minimum project com pletiontime.

Find out floats.



Find out floats?

Activity Immediate

Preceding

Activities

Duration

(in weeks)

a - 13

b - 12

c a 2

d b 8

e a 15

f c, d 2



15

2

3 4

a, 13a, 13e, 15e, 15e, 15e, 15

b, 12b, 12b, 12b, 12

c, 2c, 2c, 2c, 2

d, 8d, 8d, 8d, 8

f, 2f, 2f, 2f, 2

Network



15

2

3 4

a, 13a, 13e, 15e, 15e, 15e, 15

b, 12b, 12b, 12b, 12

c, 2c, 2c, 2c, 2

d, 8d, 8d, 8d, 8

f, 2f, 2f, 2f, 2

Paths

(1-2-5), (a-e), 28

(1-2-4-5), (a-c-f), 17

(1-3-4-5), (b-d-f), 22



15

2

3 4

a, 13a, 13e, 15e, 15e, 15e, 15

b, 12b, 12b, 12b, 12

c, 2c, 2c, 2c, 2

d, 8d, 8d, 8d, 8

f, 2f, 2f, 2f, 2




15

2

3 4

a, 13a, 13e, 15e, 15e, 15e, 15

b, 12b, 12b, 12b, 12

c, 2c, 2c, 2c, 2

d, 8d, 8d, 8d, 8

f, 2f, 2f, 2f, 2

130

120

ES and EF Times a.

b.

c.

d.

e.

f.



15

2

3 4

a, 13a, 13e, 15e, 15e, 15e, 15

b, 12b, 12b, 12b, 12

c, 2c, 2c, 2c, 2

d, 8d, 8d, 8d, 8

f, 2f, 2f, 2f, 2

130

120

1513

2012

2813

ES and EF Times a.

b.

c.

d.

e.

f.



15

2

3 4

a, 13a, 13e, 15e, 15e, 15e, 15

b, 12b, 12b, 12b, 12

c, 2c, 2c, 2c, 2

d, 8d, 8d, 8d, 8

f, 2f, 2f, 2f, 2

130

120

1513

2012

2813

2220

ES and EF Times a.

b.

c.

d.

e.

f.

Project¶s EF

= 28



15

2

3 4

a, 13a, 13e, 15e, 15e, 15e, 15

b, 12b, 12b, 12b, 12

c, 2c, 2c, 2c, 2

d, 8d, 8d, 8d, 8

f, 2f, 2f, 2f, 2

130

120

1513

2012

281313 28

2220

26 28

LS and LF Times a.

b.

c.

d.

e.

f.



15

2

3 4

a, 13a, 13e, 15e, 15e, 15e, 15

b, 12b, 12b, 12b, 12

c, 2c, 2c, 2c, 2

d, 8d, 8d, 8d, 8

f, 2f, 2f, 2f, 2

130

120

1513

24 26

2012

18 26

281313 28

2220

26 28

LS and LF Times a.

b.

c.

d.

e.

f.



15

2

3 4

a, 13a, 13e, 15e, 15e, 15e, 15

b, 12b, 12b, 12b, 12

c, 2c, 2c, 2c, 2

d, 8d, 8d, 8d, 8

f, 2f, 2f, 2f, 2

130

0 13120

6 18

1513

24 26

2012

18 26

281313 28

2220

26 28

LS and LF Times a.

b.

c.

d.

e.

f.



15

2

3 4

a, 13a, 13e, 15e, 15e, 15e, 15

b, 12b, 12b, 12b, 12

c, 2c, 2c, 2c, 2

d, 8d, 8d, 8d, 8

f, 2f, 2f, 2f, 2

130

0 13120

6 18

1513

24 26

2012

18 26

281313 28

2220

26 28

Slack

0

6

11

6

0

6

a.

b.

c.

d.

e.

f.



15

2

3 4

a, 13a, 13e, 15e, 15e, 15e, 15

b, 12b, 12b, 12b, 12

c, 2c, 2c, 2c, 2

d, 8d, 8d, 8d, 8

f, 2f, 2f, 2f, 2




Activity

(i-j)

Duration

(in weeks)ES EF LS LF Total

Float

Free

Float

Indp.

Float

a (1-2) 13 0 13 0 13 0 0 0

b (1-3) 12 0 12 6 18 6 0 0

c (2-4) 2 13 15 24 26 11 5 5

d (3-4) 8 12 20 18 26 6 0 (6)

e (2-5) 15 13 28 13 28 0 0 0

f (4-5) 2 20 22 26 28 6 6 0

Floats



?

?

?

?

?



PERT

PERT is based on the assum ption that an activity¶s duration

follows a probability distribution instead of being a single value.

Three time estimates are required to com pute the parameters of an activity¶s duration distribution:

Pessimistic time (tp ) - the time the activity would take if

things did not go well.

Most likely time (tm ) - the consensus best estimate of theactivity¶s duration.

Optimistic time (to ) - the time the activity would take if

things did go well.



Mean (expected time):

(by beta distribution)

Variance:

Probability:

that the project will be com pleted within a s pecified time

Z =x - QWc p( )

te = ( )t p + 4tm + to

6

V = ² = ( )t p ± to ²

6

Where; Z = Std. N.D. scale value,

x = S pecified time,

µ = Mean project EF,

cp = Std. dev. of critical path.

Vc p = Sum of Variances of critical activities.



Draw the network.

Com pute the Mean (ex pected time) to analyze the paths throughthe network and find the critical path.

The length of the critical path is the mean of the project duration probability distribution which is assumed to be normal.

The standard deviation of the project duration probabilitydistribution is com puted by adding the variances of the criticalactivities (all activities that make u p the critical path) and taking

the square root of that sum.

Probability com putations can now be made using the normaldistribution table.

PERT analysis



PERT Exam ple:Activity Immediate

PrecedingActivities

To Tm Tp

a - 4 6 8

b - 1 4.5 5

c a 3 3 3d a 4 5 6

e a 0.5 1 1.5

f b, c 3 4 5

g b, c 1 1.5 5h e, f 5 6 7

i e, f 2 5 8

j d, h 2.5 2.75 4.5

k g, i 3 5 7

(in days)



aa

dd

cc

bb

f f

ee

gg

ii

hh

kk

j j

Network without Time estimates



Activity Mean

(exp. time)

Variance

a 6 4/9

b 4 4/9

c 3 0

d 5 1/9e 1 1/36

f 4 1/9

g 2 4/9

h 6 1/9

i 5 1

j 3 1/9

k 5 4/9



a, 6a, 6

d, 5d, 5

c, 3c, 3

b, 4b, 4 f, 4f, 4

e, 1e, 1

g, 2g, 2

i, 5i, 5

h, 6h, 6

k, 5k, 5

j, 3 j, 3

Network with Mean (expected time)



Activity ES EF LS LF Slack

a 0 6 0 6 0* b 0 4 5 9 5

c 6 9 6 9 0*

d 6 11 15 20 9

e 6 7 12 13 6

f 9 13 9 13 0*

g 9 11 16 18 7

h 13 19 14 20 1

i 13 18 13 18 0*

j 19 22 20 23 1

k 18 23 18 23 0*

Critical



a, 6a, 6

d, 5d, 5

c, 3c, 3

b, 4b, 4 f, 4f, 4

e, 1e, 1

g, 2g, 2

i, 5i, 5

h, 6h, 6

k, 5k, 5

j, 3 j, 3




Vcp = (Va + Vc + Vf + Vi + Vk )= (4/9 + 0 + 1/9 + 1 + 4/9)

= 2

cp = 1.414

Z = (24-23) / 1.414 = 0.71

From the Standard Normal Distribution table:

P(z < 0.71) = 0.5 + 0.2612 = 0.7612

Com puting the probability to com plete the project in 24 days:

(?>24 days)



Normal Distribution of Project Time

Q Time x

Z W

@ Z =0

Probability = 0.50



?

?

?

?

?



Project managers may have the o ption or requirement to crash the project, or accelerate the com pletion of the project.

This is accom plished by reducing the length of the critical path(s).

The length of the critical path is reduced by reducing the duration of the activities on the critical path.

If each activity requires the ex penditure of an amount of money to reduce itsduration by one unit of time, then the project manager selects the least costcritical activity, reduces it by one time unit, and traces that change throughthe remainder of the network.

As a result of a reduction in an activity¶s time, a new critical path may

be created.

When there is more than one critical path, each of the critical paths must bereduced.

If the length of the project needs to be reduced further, the process is

re peated.

Cost consideration in project



Project Crashing

Crashing

reducing project time by ex pending additional resources.

Crash time

an amount of time to which an activity is reduced.

Crash cost

total cost after reducing the activity time.

Goal

reduce project duration at minimum cost.



Activity crashing

Activity time

Crashing activity

Crash

time

Crash

cost

Normal Activity

Normal

time

Normal

cost

Slo pe = crash cost per unit time



Time-Cost Relationship Crashing costs increase as project duration decreases.

Indirect costs increase as project duration increases. Reduce project length as long as crashing cost is less than indirect cost.

Time-Cost Tradeoff

time

Direct cost

Indirect

cost

Total project costMin total cost =o ptimal project

time



Project Crashing Example:

Activity Normal

time

(in weeks)

Normal

cost

(in Rs.)

Crash

time

(in weeks)

Crash

cost

(in Rs.)

Slope

(in Rs.)

1-2 8 3,000 4 6,000 750

1-3 5 4,000 3 8,000 2,000

2-4 9 4,000 6 5,500 500

3-5 7 2,000 5 3,200 600

2-5 5 8,000 1 12,000 1,000

4-6 3 10,000 2½ 11,200 2,400

5-6 6 4,000 2 6,800 700

6-7 10 6,000 7 8,700 900

5-7 9 4,200 5 9,000 1,200

45,200 70,400

The indirect cost is Rs. 2,000 per week.



Network

1 7

2 6

3 5

88

9999

5555

7777

5555

3333

10101010

9999

4

6666

C.P. = (1-2-4-6-7)

D. Cost = 45,200 Rs.

I. Cost = 60,000 Rs.

Total = 1,05,200 Rs.

Time = 30 weeks



Activity Slope

(in Rs.)

1-2 750

1-3 2,000

2-4 500

3-5 600

2-5 1,000

4-6 2,400

5-6 700

6-7 900

5-7 1,200

5,50064,0009

Crash

cost

Crash

time

Normal

cost

Normal

time

2-4

Activity



Crashing

1 7

2 6

3 5

88

6666

5555

7777

5555

3333

10101010

9999

4

6666

C.P. = (1-2-5-6-7) Time = 29 weeks

D. Cost = 46,700 Rs.

I. Cost = 58,000 Rs.

Total = 1,04,700 Rs.



Activity Slope

(in Rs.)

1-2 750

1-3 2,000

2-4 500

3-5 600

2-5 1,000

4-6 2,400

5-6 700

6-7 900

5-7 1,200

6,80024,0006

Crash

cost

Crash

time

Normal

cost

Normal

time

5-6

Activity



Crashing

1 7

2 6

3 5

88

6666

5555

7777

5555

3333

10101010

9999

4

2222

C.P. = (1-2-4-6-7) Time = 27 weeks

D. Cost = 49,500 Rs.

I. Cost = 54,000 Rs.

Total = 1,03,500 Rs.



Activity Slope

(in Rs.)

1-2 750

1-3 2,000

2-4 500

3-5 600

2-5 1,000

4-6 2,400

5-6 700

6-7 900

5-7 1,200

6,00043,0008

Crash

cost

Crash

time

Normal

cost

Normal

time

1-2

Activity



Crashing

1 7

2 6

3 5

44

6666

5555

7777

5555

3333

10101010

9999

4

2222

C.P. = (1-3-5-6-7) Time = 24 weeks

D. Cost = 52,500 Rs.

I. Cost = 48,000 Rs.

Total = 1,00,500 Rs.



Activity Slope

(in Rs.)

1-2 750

1-3 2,000

2-4 500

3-5 600

2-5 1,000

4-6 2,400

5-6 700

6-7 900

5-7 1,200

3,20052,0007

Crash

cost

Crash

time

Normal

cost

Normal

time

3-5

Activity



Crashing

1 7

2 6

3 5

44

6666

5555

5555

5555

3333

10101010

9999

4

2222

C.P. = (1-2-4-6-7) Time = 23 weeks

D. Cost = 53,700 Rs.

I. Cost = 46,000 Rs.

Total = 99,700 Rs.



Activity Slope

(in Rs.)

1-2 750

1-3 2,000

2-4 500

3-5 600

2-5 1,000

4-6 2,400

5-6 700

6-7 900

5-7 1,200

8,70076,00010

Crash

cost

Crash

time

Normal

cost

Normal

time

6-7

Activity



Crashing

1 7

2 6

3 5

44

6666

5555

5555

5555

3333

7777

9999

4

2222

C.P. = (1-2-4-6-7) Time = 20 weeks

D. Cost = 56,400 Rs.

I. Cost = 40,000 Rs.

Total = 96,400 Rs.



Activity Slope

(in Rs.)

1-2 750

1-3 2,000

2-4 500

3-5 600

2-5 1,000

4-6 2,400

5-6 700

6-7 900

5-7 1,200

11,2002½10,0003

Crash

cost

Crash

time

Normal

cost

Normal

time

4-6

Activity



Crashing

1 7

2 6

3 5

44

6666

5555

5555

5555

2½2½2½2½

7777

9999

4

2222

C.P. = (1-2-4-6-7) Time = 19½ weeks

D. Cost = 57,600 Rs.

I. Cost = 39,000 Rs.

Total = 96,600 Rs.



Crashing

stage

Completion

time

(in weeks)

Direct

cost

(in Rs.)

Indirect

cost

(in Rs.)

Total

cost

(in Rs.)

0 30 45,200 60,000 1,05,200

1 29 46,700 58,000 1,04,700

2 27 49,500 54,000 1,03,500

3 24 52,500 48,000 1,00,500

4 23 53,700 46,000 99,700

5 20 56,400 40,000 96,400

6 19½ 57,600 39,000 96,600



If the objective is to minimize the com pletion time of the

project then the critical

path, time & total cost will be:

C.P. = (1-2-4-6-7)

Time = 19½ weeks

D. Cost = 57,600 Rs.

I. Cost = 39,000 Rs.

Total = 96,600 Rs.

If the objective is to minimize the total cost of the project

then the critical path, time & total cost will be:

C.P. = (1-2-4-6-7)

Time = 20 weeks

D. Cost = 56,400 Rs.I. Cost = 40,000 Rs.

Total = 96,400 Rs.



How long will the entire project take to be com pleted? What are

the risks involved?

Which are the critical activities or tasks in the project whichcould delay the entire project if they were not com pleted on

time?

Is the project on schedule, behind schedule or ahead of

schedule?

If the project has to be finished earlier than planned, what is the

best way to do this at the least cost?

CPM / PERT can answer the following

important questions:



Useful at many stages of project management.

Mathematically sim ple.

Give critical path and slack time.

Provide project documentation.

Useful in monitoring costs.

Benefits of CPM / PERT:



Limitations to CPM / PERT:

Clearly defined, inde pendent and stable activities.

S pecified precedence relationshi ps.

Over em phasis on critical paths.

Deterministic CPM model.

Activity time estimates are subjective and de pend on judgment.

PERT assumes a beta distribution for these time estimates, butthe actual distribution may be different.

To overcome these limitations, Monte Carlo simulations can be

performed on the network to eliminate the o

ptimistic bias.



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Documents

Pert Cpm (Unit 6) Copy