© the mcgraw-hill companies, inc., 2004 1 project management cpm and pert

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©The McGraw-Hill Companies, Inc., 2004 1 Project Management CPM and PERT

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Page 1: © The McGraw-Hill Companies, Inc., 2004 1 Project Management CPM and PERT

©The McGraw-Hill Companies, Inc., 2004

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Project ManagementCPM and PERT

Page 2: © The McGraw-Hill Companies, Inc., 2004 1 Project Management CPM and PERT

©The McGraw-Hill Companies, Inc., 2004

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Project is a series of related jobs usually directed toward some major output and requiring a significant period of time to perform

Project Management are the management activities of planning, directing, and controlling resources (people, equipment, material) to meet the technical, cost, and time constraints of a project

Page 3: © The McGraw-Hill Companies, Inc., 2004 1 Project Management CPM and PERT

©The McGraw-Hill Companies, Inc., 2004

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A project is made up of a sequence of activities that form a network representing a project

The path taking longest time through this

network of activities is called the “critical path”

The critical path provides a wide range of scheduling information useful in managing a project

Critical Path Method (CPM) helps to identify the critical path(s) in the project networks

Page 4: © The McGraw-Hill Companies, Inc., 2004 1 Project Management CPM and PERT

©The McGraw-Hill Companies, Inc., 2004

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A project must have:

well-defined jobs or tasks whose completion marks the end of the project;

independent jobs or tasks;

and tasks that follow a given sequence.

Page 5: © The McGraw-Hill Companies, Inc., 2004 1 Project Management CPM and PERT

©The McGraw-Hill Companies, Inc., 2004

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CPM with a Single Time Estimate› Used when activity times are known with

certainty› Used to determine timing estimates for the

project, each activity in the project, and slack time for activities

CPM with Three Activity Time Estimates› Used when activity times are uncertain › Used to obtain the same information as the Single

Time Estimate model and probability information

Time-Cost Models› Used when cost trade-off information is a major

consideration in planning› Used to determine the least cost in reducing total

project time

Page 6: © The McGraw-Hill Companies, Inc., 2004 1 Project Management CPM and PERT

©The McGraw-Hill Companies, Inc., 2004

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1. Activity Identification

2. Activity Sequencing and Network Construction

3. Determine the critical path

› From the critical path all of the project and activity timing information can be obtained

Page 7: © The McGraw-Hill Companies, Inc., 2004 1 Project Management CPM and PERT

©The McGraw-Hill Companies, Inc., 2004

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Consider the following consulting project:

Develop a critical path diagram and determine the duration of the critical path and slack times for all activities.

Activity Designation Immed. Pred. Time (Weeks)Assess customer's needs A None 2Write and submit proposal B A 1Obtain approval C B 1Develop service vision and goals D C 2Train employees E C 5Quality improvement pilot groups F D, E 5Write assessment report G F 1

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A(2) B(1) C(1)

D(2)

E(5)

F(5) G(1)

A None 2

B A 1

C B 1

D C 2

E C 5

F D,E 5

G F 1

Act. Imed. Pred. Time

Page 9: © The McGraw-Hill Companies, Inc., 2004 1 Project Management CPM and PERT

©The McGraw-Hill Companies, Inc., 2004

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ES=9EF=14

ES=14EF=15

ES=0EF=2

ES=2EF=3 ES=3

EF=4

ES=4EF=9

ES=4EF=6

A(2) B(1) C(1)

D(2)

E(5)

F(5) G(1)

Hint: Start with ES=0 and go forward in the network from A to G.

Hint: Start with ES=0 and go forward in the network from A to G.

Page 10: © The McGraw-Hill Companies, Inc., 2004 1 Project Management CPM and PERT

©The McGraw-Hill Companies, Inc., 2004

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ES=9EF=14

ES=14EF=15

ES=0EF=2

ES=2EF=3

ES=3EF=4

ES=4EF=9

ES=4EF=6

A(2) B(1) C(1)

D(2)

E(5)

F(5) G(1)

LS=14LF=15

LS=9LF=14

LS=4LF=9

LS=7LF=9

LS=3LF=4

LS=2LF=3

LS=0LF=2

Hint: Start with LF=15 or the total time of the project and go backward in the network from G to A.

Hint: Start with LF=15 or the total time of the project and go backward in the network from G to A.

Page 11: © The McGraw-Hill Companies, Inc., 2004 1 Project Management CPM and PERT

©The McGraw-Hill Companies, Inc., 2004

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ES=9EF=14

ES=14EF=15

ES=0EF=2

ES=2EF=3

ES=3EF=4

ES=4EF=9

ES=4EF=6

A(2) B(1) C(1)

D(2)

E(5)

F(5) G(1)

LS=14LF=15

LS=9LF=14

LS=4LF=9

LS=7LF=9

LS=3LF=4

LS=2LF=3

LS=0LF=2

Duration = 15 weeks

Slack=(7-4)=(9-6)= 3 Wks

Page 12: © The McGraw-Hill Companies, Inc., 2004 1 Project Management CPM and PERT

©The McGraw-Hill Companies, Inc., 2004

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TaskImmediate

Predecesors Optimistic Most Likely PessimisticA None 3 6 15B None 2 4 14C A 6 12 30D A 2 5 8E C 5 11 17F D 3 6 15G B 3 9 27H E,F 1 4 7I G,H 4 19 28

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ET(A)= 3+4(6)+15

6

ET(A)= 3+4(6)+15

6

ET(A)=42/6=7ET(A)=42/6=7Task

Immediate Predecesors

Expected Time

A None 7B None 5.333C A 14D A 5E C 11F D 7G B 11H E,F 4I G,H 18

TaskImmediate

Predecesors Optimistic Most Likely PessimisticA None 3 6 15B None 2 4 14C A 6 12 30D A 2 5 8E C 5 11 17F D 3 6 15G B 3 9 27H E,F 1 4 7I G,H 4 19 28

Expected Time = Opt. Time + 4(Most Likely Time) + Pess. Time

6Expected Time =

Opt. Time + 4(Most Likely Time) + Pess. Time

6

Page 14: © The McGraw-Hill Companies, Inc., 2004 1 Project Management CPM and PERT

©The McGraw-Hill Companies, Inc., 2004

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TaskImmediate

PredecesorsExpected

TimeA None 7B None 5.333C A 14D A 5E C 11F D 7G B 11H E,F 4I G,H 18

ET(B)=32/6=5.333ET(B)=32/6=5.333

ET(B)= 2+4(4)+14

6

ET(B)= 2+4(4)+14

6

TaskImmediate

Predecesors Optimistic Most Likely PessimisticA None 3 6 15B None 2 4 14C A 6 12 30D A 2 5 8E C 5 11 17F D 3 6 15G B 3 9 27H E,F 1 4 7I G,H 4 19 28

Expected Time = Opt. Time + 4(Most Likely Time) + Pess. Time

6Expected Time =

Opt. Time + 4(Most Likely Time) + Pess. Time

6

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©The McGraw-Hill Companies, Inc., 2004

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TaskImmediate

PredecesorsExpected

TimeA None 7B None 5.333C A 14D A 5E C 11F D 7G B 11H E,F 4I G,H 18

ET(C)= 6+4(12)+30

6

ET(C)= 6+4(12)+30

6

ET(C)=84/6=14ET(C)=84/6=14

TaskImmediate

Predecesors Optimistic Most Likely PessimisticA None 3 6 15B None 2 4 14C A 6 12 30D A 2 5 8E C 5 11 17F D 3 6 15G B 3 9 27H E,F 1 4 7I G,H 4 19 28

Expected Time = Opt. Time + 4(Most Likely Time) + Pess. Time

6Expected Time =

Opt. Time + 4(Most Likely Time) + Pess. Time

6

Page 16: © The McGraw-Hill Companies, Inc., 2004 1 Project Management CPM and PERT

©The McGraw-Hill Companies, Inc., 2004

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Step 2. Draw the Network DiagramAnd determine the Critical Path

A(7)

B(5.333)

C(14)

D(5)

E(11)

F(7)

H(4)

G(11)

I(18)

Duration = 54 Days

Page 17: © The McGraw-Hill Companies, Inc., 2004 1 Project Management CPM and PERT

©The McGraw-Hill Companies, Inc., 2004

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Step 3 : Probability Exercise

What is the probability of finishing this project in less than 53 days?

What is the probability of finishing this project in less than 53 days?

p(t < D)

TE = 54

Z = D - TE

cp2

Z = D - TE

cp2

tD=53

Page 18: © The McGraw-Hill Companies, Inc., 2004 1 Project Management CPM and PERT

©The McGraw-Hill Companies, Inc., 2004

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Activity variance, = (Pessim. - Optim.

6)2 2Activity variance, = (

Pessim. - Optim.

6)2 2

Task Optimistic Most Likely Pessimistic VarianceA 3 6 15 4B 2 4 14C 6 12 30 16D 2 5 8E 5 11 17 4F 3 6 15G 3 9 27H 1 4 7 1I 4 19 28 16

(Sum the variance along the critical path) 2 = 41 2 = 41

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There is a 43.64% probability that this project will be completed in less than 53 weeks.

There is a 43.64% probability that this project will be completed in less than 53 weeks.

P(Z < -.16) = 0.5- 0.0636=0 .4364, or 43.64% (NORMSDIST(-.16)

P(Z < -.16) = 0.5- 0.0636=0 .4364, or 43.64% (NORMSDIST(-.16)

Z = D - T

=53- 54

41= -.156E

cp2

Z = D - T

=53- 54

41= -.156E

cp2

TE = 54

p(t < D)

t

D=53

Z ( - 0.16) = 0.0636

0.0636 = 6.36 %

Page 20: © The McGraw-Hill Companies, Inc., 2004 1 Project Management CPM and PERT

©The McGraw-Hill Companies, Inc., 2004

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What is the probability that the project duration will exceed 56 weeks?

What is the probability that the project duration will exceed 56 weeks?

Page 21: © The McGraw-Hill Companies, Inc., 2004 1 Project Management CPM and PERT

©The McGraw-Hill Companies, Inc., 2004

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tTE = 54

P(t < D)

D=56

Z = D - T

=56 - 54

41= .312E

cp2

Z = D - T

=56 - 54

41= .312E

cp2

P (Z > .31) = 1- 0.6217= 0.3783, or 37.83 %P (Z > .31) = 1- 0.6217= 0.3783, or 37.83 %

Z( 0.31) = 0.1217So, P( Z= 0.31) = 0.5 + 0.1217 = 0.6217

0.1217 = 12.17 %P= 50% = 0.5

Page 22: © The McGraw-Hill Companies, Inc., 2004 1 Project Management CPM and PERT

©The McGraw-Hill Companies, Inc., 2004

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Basic Assumption: Relationship between activity completion time and project cost

Time Cost Models: Determine the optimum point in time-cost tradeoffs› Activity direct costs› Project indirect costs› Activity completion times

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Project activities can be identified as entities (There is a clear beginning and ending point for each activity.)

Project activity sequence relationships can be specified and networked.

Project control should focus on the critical

path

The activity times follow the beta distribution, with the variance of the project assumed to equal the sum of the variances along the critical path.

Page 24: © The McGraw-Hill Companies, Inc., 2004 1 Project Management CPM and PERT

©The McGraw-Hill Companies, Inc., 2004

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