CPM PERT

CPM/PERTCPM/PERT

S. Ajit

ProjectProject

“A project is a series of activities directed to the accomplishment of a desired objective.”

HistoryHistory

PERT was developed by the US Navy for the planning and control of the Polaris missile program and the emphasis was on completing the program in the shortest possible time. In addition PERT had the ability to cope with uncertain activity completion times (e.g. for a particular activity the most likely completion time is 4 weeks but it could be anywhere between 3 weeks and 8 weeks).

CPM was developed by Du Pont and the emphasis was on the trade-off between the cost of the project and its overall completion time (e.g. for certain activities it may be possible to decrease their completion times by spending more money - how does this affect the overall completion time of the project?)

Developed in the 1950s

CPM - Critical Path MethodCPM - Critical Path Method

Definition: In CPM activities are shown as a network of precedence relationships using activity-on-node network construction– Single estimate of activity time– Deterministic activity times

USED IN : Production management - for the jobs of repetitive in nature where the activity time estimates can be predicted with considerable certainty due to the existence of past experience.

PERT - PERT - Project Evaluation & Review Project Evaluation & Review

TechniquesTechniques Definition: In PERT activities are shown as a network of

precedence relationships using activity-on-arrow network construction– Multiple time estimates – Probabilistic activity times

USED IN : Project management - for non-repetitive jobs (research and development work), where the time and cost estimates tend to be quite uncertain. This technique uses probabilistic time estimates.

The Project NetworkThe Project Network

Use of nodes and arrows

Arrows An arrow leads from tail to head directionally– Indicate ACTIVITY, a time consuming effort that is

required to perform a part of the work.

Nodes A node is represented by a circle- Indicate EVENT, a point in time where one or more

activities start and/or finish.

– An activity is a time-consuming effort that is required to complete part of a project. Shown as an arrow on the diagram

– An event is denoted by a circle and defines the end of one activity and beginning of the next. An event may be a decision point.

Activity Event

The Project NetworkThe Project Network

Activity SlackActivity Slack

Each event has two important times associated with it :

- Earliest time , Te , which is a calendar time when an event can occur when all the predecessor events completed at the earliest possible times

- Latest time , TL , which is the latest time the event can occur with out delaying the subsequent events and completion of project.

Difference between the latest time and the earliest time of an event is the slack time for that event

Positive slack : Slack is the amount of time an event can be delayed without delaying the project completion

Critical PathCritical Path

Is that the sequence of activities and events where there is no “slack” i.e.. Zero slack

Longest path through a network

minimum project completion time

Questions Answered by CPM & PERTQuestions Answered by CPM & PERT

Completion date?

On Schedule?

Within Budget?

Critical Activities?

How can the project be finished early at the least cost?

ExampleExample1. Construct the CPM Network using the details below and determine the critical path

Activity Immediate Predecessor

Duration

A - 1

B A 4

C A 2

D A 2

E D 3

F D 3

G E 2

H F,G 1

I C,H 3

J B 2

1 2

CPM NETWORK

A

1 7

4

2

3

CPM NETWORK

A

B

D

C

1 7

6

5

4

2

3

CPM NETWORK

A

B

DF

E

C

1 7

6

5

4

2

3

CPM NETWORK

A

B

DF

GE

C

1 7

6

5

4

2

3

CPM NETWORK

A

B

DF

GE

C

H

1 7

6

5

4

2 8

3

CPM NETWORK

A

B

DF

GE

C

IH

1 7

6

5

4

2 8

3

CPM NETWORK

A

J

B

DF

GE

C

IH

1 7

6

5

4

2 8

3

CPM NETWORK with duration of each activity

A(1)

J(2)

B(4)

D(2)

F(3)G(2)

E(3)

C(2)

I(3)

H(1)

1 7

6

5

4

2 8

3

To find the Critical Path

A(1)

J(2)

B(4)

D(2)F(3)

G(2)E(3)

C(2)

I(3)H(1) 0

0

LFT EST

1 7

6

5

4

2 8

3

A(1)

J(2)

B(4)

D(2)F(3)

G(2)E(3)

C(2)

I(3)H(1) 0

0 0 1

LFT EST

To find the Critical Path

1 7

6

5

4

2 8

3

A(1)

J(2)

B(4)

D(2)F(3)

G(2)E(3)

C(2)

I(3)H(1) 0

0 0 1

0 3

To find the Critical Path LFT EST

1 7

6

5

4

2 8

3

A(1)

J(2)

B(4)

D(2)F(3)

G(2)E(3)

C(2)

I(3)H(1) 0

0

0 6

0 1

0 3


1 7

6

5

4

2 8

3

A(1)

J(2)

B(4)

D(2)F(3)

G(2)E(3)

C(2)

I(3)H(1) 0

0

0 6

0 1

0 3

0 8


1 7

6

5

4

2 8

3

A(1)

J(2)

B(4)

D(2)F(3)

G(2)E(3)

C(2)

I(3)H(1) 0

00 9

0 6

0 1

0 3

0 8


1 7

6

5

4

2 8

3

A(1)

J(2)

B(4)

D(2)F(3)

G(2)E(3)

C(2)

I(3)H(1) 0

0 0 9

0 6

0 5

0 1

0 3

0 8


1 7

6

5

4

2 8

3

A(1)

J(2)

B(4)

D(2)F(3)

G(2)E(3)

C(2)

I(3)H(1) 0

00 9

0 12

0 6

0 5

0 1

0 3

0 8


1 7

6

5

4

2 8

3

A(1)

J(2)

B(4)

D(2)F(3)

G(2)E(3)

C(2)

I(3)H(1) 0

00 9

12 12

0 6

0 5

0 1

0 3

0 8


1 7

6

5

4

2 8

3

A(1)

J(2)

B(4)

D(2)F(3)

G(2)E(3)

C(2)

I(3)H(1) 0

09 9

12 12

0 6

0 5

0 1

0 3

0 8


1 7

6

5

4

2 8

3

A(1)

J(2)

B(4)

D(2)F(3)

G(2)E(3)

C(2)

I(3)H(1) 0

09 9

12 12

0 6

0 5

0 1

0 3

8 8


1 7

6

5

4

2 8

3

A(1)

J(2)

B(4)

D(2)F(3)

G(2)E(3)

C(2)

I(3)H(1) 0

09 9

12 12

6 6

0 5

0 1

0 3

8 8


1 7

6

5

4

2 8

3

A(1)

J(2)

B(4)

D(2)F(3)

G(2)E(3)

C(2)

I(3)H(1) 0

09 9

12 12

6 6

0 5

0 1

3 3

8 8


1 7

6

5

4

2 8

3

A(1)

J(2)

B(4)

D(2)F(3)

G(2)E(3)

C(2)

I(3)H(1) 0

09 9

12 12

6 6

0 5

1 1

3 3

8 8


1 7

6

5

4

2 8

3

A(1)

J(2)

B(4)

D(2)F(3)

G(2)E(3)

C(2)

I(3)H(1) 0

09 9

12 12

6 6

10 5

1 1

3 3

8 8


1 7

6

5

4

2 8

3

A(1)

J(2)

B(4)

D(2)F(3)

G(2)E(3)

C(2)

I(3)H(1) 0

09 9

12 12

6 6

10 5

1 1

3 3

8 8

Critical Path

Critical Path

A – B – E – G – H - I

Critical Path

1 - 2 - 4 – 5 – 6 – 7 - 8

CPM - ExampleCPM - Example

Critical Path

1 - 2 - 4 – 5 – 6 – 7 – 8

Critical Path

A – B – E – G – H - IProject Completion Time = 12 months

PERTPERTProbabilisticThree estimates

– to (a) – optimistic time

– tp (b) – pessimistic time

– tl (m) – most likely time

Mean µ = (to + 4* tl + tp)/6

Variance σ2 = [(tp – to)/6]2

Activity Predecessor

a m b

A - 6 7 8

B - 1 2 9

C - 1 4 7

D A 1 2 3

E A,B 1 2 9

F C 1 5 9

G C 2 2 8

H E,F 4 4 4

I E,F 4 4 10

J D,H 2 5 14

K G,I 2 2 8

PERT – Example: PERT – Example: Construct Project network, find the Construct Project network, find the expected duration and variance for each activity & find the expected duration and variance for each activity & find the

critical path and expected project completion timecritical path and expected project completion time

Activity

Predecessor

a m b µte

A - 6 7 8 7

B - 1 2 9

C - 1 4 7

D A 1 2 3

E A,B 1 2 9

F C 1 5 9

G C 2 2 8

H E,F 4 4 4

I E,F 4 4 10

J D,H 2 5 14

K G,I 2 2 8

PERT – Example: PERT – Example: Mean µ = (tMean µ = (too + 4* t + 4* tll + t + tpp)/6)/6 Variance Variance σσ22 = [(t = [(tpp – t – too)/6])/6]22

Activity

Predecessor

a m b µte

A - 6 7 8 7

B - 1 2 9 3

C - 1 4 7

D A 1 2 3

E A,B 1 2 9

F C 1 5 9

G C 2 2 8

H E,F 4 4 4

I E,F 4 4 10

J D,H 2 5 14

K G,I 2 2 8


Activity

Predecessor

a m b µte

σ2

A - 6 7 8 7

B - 1 2 9 3

C - 1 4 7 4

D A 1 2 3 2

E A,B 1 2 9 3

F C 1 5 9 5

G C 2 2 8 3

H E,F 4 4 4 4

I E,F 4 4 10 5

J D,H 2 5 14 6

K G,I 2 2 8 3


Activity

Predecessor

a m b µte

σ2

A - 6 7 8 7 0.11

B - 1 2 9 3

C - 1 4 7 4

D A 1 2 3 2

E A,B 1 2 9 3

F C 1 5 9 5

G C 2 2 8 3

H E,F 4 4 4 4

I E,F 4 4 10 5

J D,H 2 5 14 6

K G,I 2 2 8 3


Activity

Predecessor

a m b µte

σ2

A - 6 7 8 7 0.11

B - 1 2 9 3 1.78

C - 1 4 7 4

D A 1 2 3 2

E A,B 1 2 9 3

F C 1 5 9 5

G C 2 2 8 3

H E,F 4 4 4 4

I E,F 4 4 10 5

J D,H 2 5 14 6

K G,I 2 2 8 3


Activity

Predecessor

a m b µte

σ2

A - 6 7 8 7 0.11

B - 1 2 9 3 1.78

C - 1 4 7 4 1.00

D A 1 2 3 2 0.11

E A,B 1 2 9 3 1.78

F C 1 5 9 5 1.78

G C 2 2 8 3 1.00

H E,F 4 4 4 4 0

I E,F 4 4 10 5 1.00

J D,H 2 5 14 6 4.00

K G,I 2 2 8 3 1.00


1

4

2

3

A(7)

J(6)

B(3)

D(2)

F(5)

G(3)

E(3)

C(4) I(5)

H(4)

6 6


K(3)

1

4

26

3

A(7)

J(6)

B(3)

D(2)

F(5)

G(3)

E(3)

C(4) I(5)

H(4)

6 6


K(3)

1

6

5

4

2

3

A(7)

J(6)

B(3)

D(2)

F(5)

G(3)E(3)

C(4) I(5)

H(4)

6 6


K(3)

1

7

6

5

4

2

3

A(7)

J(6)

B(3)

D(2)

F(5)

G(3)

E(3)

C(4) I(5)

H(4)

6 6


K(3)

1

7

6

5

4

2

83

A(7) J(6)

B(3)

D(2)

F(5)

G(3)

E(3)

C(4) I(5)

H(4)

6 6


K(3)

1

7

6

5

4

2

83

A(7)J(6)

B(3)

D(2)

F(5)

G(3)

E(3)

C(4) I(5)

H(4)

6 6


K(3)

1

7

6

5

4

2

83

A(7) J(6)

B(3)

D(2)

F(5)

G(3)

E(3)

C(4) I(5)

H(4) 0 0


K(3)

1414

4 5

7 7

77

10 10

15 17

20 20

1

7

6

5

4

2

83

A(7) J(6)

B(3)

D(2)

F(5)

G(3)

E(3)

C(4) I(5)

H(4) 0 0

K(3)

1414

4 5

7 7

77

10 10

15 17

20 20

Critical path 1 – 2 – 3 – 5 – 6 – 8

Critical path A – Dummy – E – H – J

PERT - ExamplePERT - Example What is the probability of completing the project within

25 weeks?

Σσ2 = 5.89 σ = √(Σσ2) = 2.43 Ts (Scheduled time) = 25 weeks Z = (Ts – Te) / σ = (25 – 20) / 2.43 = 2.06 From z table (normal table) z(2.06) = 97.68 %

Activities

Variance

A 0.11

Dummy 0

E 1.78

H 0

J 4

PERT - ExamplePERT - Example What is the probability of completing the project within

15 weeks?

Σσ2 = 5.89 σ = √(Σσ2) = 2.43 Ts (Scheduled time) = 15 weeks Z = (Ts – Te) / σ = (15 – 20) / 2.43 = -2.06 From z table (normal table) z(-2.06) = 100 - 97.68 %

• = 2.02 %

Activities

Variance

A 0.11

Dummy 0

E 1.78

H 0

J 4

Limitations of CPM/PERTLimitations of CPM/PERT

Clearly defined, independent and stable activities

Specified precedence relationships Over emphasis on critical paths

CPM PERT

Documents

Transcript of CPM PERT