5/18/2015 L. K. Gaafar PROJECT MANAGEMENT Time Management* Dr. L. K. Gaafar The American University...

04/18/23L. K. Gaafar

PROJECT MANAGEMENT Time Management*

Dr. L. K. GaafarThe American University in Cairo

* This Presentation is Based on information from the PMBOK Guide 2000


Critical Path Method (CPM)

CPM is a project network analysis technique used to predict total project duration

A critical path for a project is the series of activities that determines the earliest time by which the project can be completed

The critical path is the longest path through the network diagram and has the least amount of float


Finding the Critical Path

Develop a network diagram Add the durations of all activities to the project

network diagram Calculate the total duration of every possible

path from the beginning to the end of the project The longest path is the critical path Activities on the critical path have zero float


Activity IPA Duration (days)

A --- 2 B A 5 C B 2 D B 7 E C 1 F D 2

Consider the following project network diagram. Assume all times are in days.

Simple Example


Simple Example

2 3

4

5

A=2 B=5C=2

D=7

1 6

F=2

E=1

start finish

a. 2 paths on this network: A-B-C-E, A-B-D-F.

b. Paths have lengths of 10, 16

c. The critical path is A-B-D-F

d. The shortest duration needed to complete this project is 16 days

Activity-on-arrow network


0 2

0 2

0 2A

2 7

2 7

0 5B

7 14

7 14

0 7D

7 9

13 15

6 2C

14 16

14 16

0 2F

9 10

15 16

6 1E

Dummy

Time ManagementES EF

LS LF

Slack Dur.Act

Key

Activity-on-node network


Activity Days Cost ($) Cost/dayA 2 200 100B 5 500 100C 2 200 100D 7 500 71.4E 1 100 100F 2 100 50

Cash Flow


Cash FlowDaily Expenses

0

20

40

60

80

100

120

140

160

180

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Day

Co

st (

$)

Cumulative Expenses

0

200

400

600

800

1000

1200

1400

1600

1800

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Day

Co

st (

$)

Activity Days Cost ($) Cost/dayA 2 200 100B 5 500 100C 2 200 100D 7 500 71.4E 1 100 100F 2 100 50

Day Activity Cost of day Total cost1 A 100 1002 A 100 2003 B 100 3004 B 100 4005 B 100 5006 B 100 6007 B 100 7008 C,D 171.4 8719 C,D 171.4 104310 D,E 171.4 121411 D 71.4 128612 D 71.4 135713 D 71.4 142814 D 71.4 150015 F 50 155016 F 50 1600


Determining the Critical Path for Project X

a. How many paths are on this network diagram?

b. How long is each path?

c. Which is the critical path?

d. What is the shortest duration needed to complete this project?

04/18/23

Stochastic (non-deterministic) Activity Durations

Project Evaluation and Review Technique (PERT)


Stochastic Times

UniformTriangular

Beta


Important Distributions


Stochastic TimesThe Central Limit Theorem

The sum of n mutually independent random variables is well-approximated by a normal distribution if n is large enough.


PERT: Finding the Critical Path

(Stochastic Times)

Develop a network diagram Calculate the mean duration and variance of each activity Calculate the total mean duration and the variance of every

possible path from the beginning to the end of the project by summing the mean duration and variances of all activities on the path.

The path with the longest mean duration is the critical path If more than one path have the longest mean duration, the

critical path is the one with the largest variance. Calculate possible project durations using the normal

distribution


Example I

Duration (wks) Activity IPA a m b

A --- 4 6 14 B A 3 4 8 C -- 4 5 6 D A,C 7 7 7 E B,D 3 3 6 F A,C 6 8 14 G -- 13 18 20

Assuming that all activities are beta distributed, what is the probability that the project duration will exceed 19 weeks?


DurationActivity IPA a m b

A --- 4 6 14 7.00 2.78B A 3 4 8 4.50 0.69C -- 4 5 6 5.00 0.11D A,C 7 7 7 7.00 0.00E B,D 3 3 6 3.50 0.25F A,C 6 8 14 8.67 1.78G -- 13 18 20 17.50 1.36

A 7, 2.8

C 5,0.1

B 4.5, 0.7

D 7, 0

E 3.5, 0.25

F8.7, 1.8

G 17.5, 1.36

7

7

14 17.5


A 7, 2.8

C 5,0.1

B 4.5, 0.7

D 7, 0

E 3.5, 0.25

F8.7, 1.8

G 17.5, 1.36

7

7

14 17.5

Path

ADE 17.50 3.03 ABE 15.00 3.72 AF 15.67 4.56

CDE 15.50 0.36 CF 13.67 1.89 G 17.50 1.36


Example IIDuration

Activity IPA Distribution a m bA --- Uniform 4 NA 8B --- Triangular 3 4 5C --- Beta 4 5 6D C Beta 5 7 12E A Triangular 3 3 6F A, B Triangular 5 8 8G E, D Uniform 9 NA 9

Construct an activity-on-arrow network for the project above.Provide a 95% confidence interval on the completion time of the project.


Example IIDuration

Activity IPA Distribution a m bA --- Uniform 4 NA 8B --- Triangular 3 4 5C --- Beta 4 5 6D C Beta 5 7 12E A Triangular 3 3 6F A, B Triangular 5 8 8G E, D Uniform 9 NA 9

EC G

FB

Start FinishA

D


Example II

Start Finish

B (4, 0.17)

C (5, 0.11)

A (6, 1.33)

E (4, 0.5) G (9, 0.0)

F (7, 0.5)

D (7.5, 1.36)

Path

BF 11 0.67 AF 13 1.83

AEG 19 1.83 CDG 21.5 1.47


Duration (days) Total cost ($) Activity IPA Normal Crash Normal Crash

A --- 2 2 200 200 B A 5 3 500 700 C B 2 1 200 250 D B 7 4 500 650 E C 1 1 100 100 F D 2 1 100 350

Consider the following project network diagram. Assume all times are in days.

Time Management: Crashing


Time Management

A(2,2,0) B(5,3,100)

C(2,1,50)

D(7,4,50)

E(1,1,0)

F(2,1,250)

Action Critical Path Duration (days) Total Cost ($)No Crashing A-B-D-F 16 1600Crash D by 1 A-B-D-F 15 1650Crash D by 1 A-B-D-F 14 1700Crash D by 1 A-B-D-F 13 1750Crash B by 1 A-B-D-F 12 1850Crash B by 1 A-B-D-F 11 1950Crash F by 1 A-B-D-F 10 2200


Duration/Cost Decision Support Curve

1500

1600

1700

1800

1900

2000

2100

2200

2300

10 11 12 13 14 15 16

Duration (days)

To

tal

Co

st

($)


Time Management

A(2,2,0) B(3,3,100)

C(2,1,50)

D(4,4,50)

E(1,1,0)

F(1,1,250)

Shortest Possible duration with crashing is 10 days.Critical path is not changed.


Duration (days) Total cost ($)Activity IPA Normal Crash Normal Crash

A --- 6 4 100 120B A 4 3 80 93C -- 5 4 95 110D A,C 7 7 115 115E B,D 4 2 64 106F A,C 8 6 75 99G -- 18 13 228 318

Example Problem


Project Network

A 6

C 5

B 4

D 7

E 4

F 8

G 18

Path DurationA-B-E 14A-D-E 17A-F 14C-D-E 16C-F 13G 18

Shortest possible normal duration is 18 at a cost of $757


6 14

10 18

4 8F

0 18

0

0 18G

0 5

2 7

2 5C

0 6

1 7

1 6A

6 13

7 14

1 7D

6 10

10 14

4 4B

Dummy

Time Management

13 17

14 18

1 4E

Dummy

18


A(4,4,10)

C(4,4,15)

B(4,3,13)

D(7,7,0)

E(2,2,21)

F(8,6,12)

G(13,13,18)

Duration Extra TotalAction ABE ADE AF CDE CF G Cost Cost

No Crashing 14 17 14 16 13 18 -- 757Crash G 14 17 14 16 13 17 18 775

Crash G&A 13 16 13 16 13 16 28 803Crash G&E 12 15 13 15 13 15 39 842Crash G&E 11 14 13 14 13 14 39 881

Crash G,A&C 10 13 12 13 12 13 43 924

Crashing


Final Crashed Network

A(4,4,10)

C(4,4,15)

B(4,3,13)

D(7,7,0)

E(2,2,21)

F(8,6,12)

G(13,13,18)

The shortest crashed project duration is 13 days at a minimum total cost of $924.

Further crashing of B or F is useless


Using Critical Path Analysis to Make Schedule Trade-offs Knowing the critical path helps you make

schedule trade-offs Free slack or free float is the amount of time an

activity can be delayed without delaying the early start of any immediately following activities

Total slack or total float is the amount of time an activity may be delayed from its early start without delaying the planned project finish date

This part is from a presentation by Kathy Schwalbe, [email protected]

http://www.augsburg.edu/depts/infotech/

mailto:[email protected]


Techniques for Shortening a Project Schedule

Shortening durations of critical tasks by adding more resources or changing their scope

Crashing tasks by obtaining the greatest amount of schedule compression for the least incremental cost

Fast tracking tasks by doing them in parallel or overlapping them




Shortening Project Schedules

Overlappedtasks

Shortenedduration

Original schedule




Activity Definition

Activity Sequencing


Duration Estimation

Schedule Development


Schedule Control

5/18/2015 L. K. Gaafar PROJECT MANAGEMENT Time Management* Dr. L. K. Gaafar The American University...

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