Unique, one-time operations designed to accomplish a specific set of objectives in a limited time frame.

How is it different?

Limited time frame

Narrow focus, specific objectives

Less bureaucratic

Why is it used?

Special needs

Pressures for new or improves products or services

What are the Key Metrics Time

Cost

Performance objectives

What are the Key Success Factors? Top-down commitment

Having a capable project manager

Having time to plan

Careful tracking and control

Good communications

What are the Major Administrative Issues?

Executive responsibilities

Project selection

Project manager selection

Organizational structure

Organizational alternatives

Manage within functional unit

Assign a coordinator

Use a matrix organization with a project leader

What are the tools?

Work breakdown structure

Network diagram

Gantt charts

Risk management

Project X

Level 1

Level 2

Level 3

Level 4

MAR APR MAY JUN JUL AUG SEP OCT NOV DEC

Locate new

facilities

Interview staff

Hire and train staff

Select and order

furniture

Remodel and install

phones

Move in/startup

Gantt Chart

PERT: Program Evaluation and Review Technique

CPM: Critical Path Method

Graphically displays project activities Estimates how long the project will take Indicates most critical activities Show where delays will not affect project

Network (precedence) diagram

Diagram of project activities that shows a sequential relationships by use of arrows and nodes

Activity-on-arrow (AOA)

Network diagram convention in which arrows designate activities.

Activity-on-node (AON)

Network diagram convention in which nodes designate activities.

1

2

3

4

5 6

Locate facilities

Order furniture

Furniture setup

Interview Hire and train

Remodel

Move in

AOA

1

2

3

5

6

Locate facilities

Order furniture

Furniture setup

Interview

Remodel Move in

4

Hire and train

7 S

AON

Activities

Project steps that consume resources and/or time.

Events

The starting and finishing of activities, designated by nodes AOA convention.

Path A sequence of activities that leads from the starting node to the

finishing node

Critical path The longest path; determines expected project duration

Critical activities Activities on the critical path

Slack Allowable slippage for path; the difference the length of path and the

length of critical path

Deterministic

Time estimates that are fairly certain

Probabilistic

Estimates of times that allow for variation

1 6 5

4

3

2

1 WEEKS

MOVE IN

Determine the ff:

a. The length of each path c. The expected length of time

b. The critical path d. The amount of slack time for each path

1 6 5

4

3

2

1 WEEKS

MOVE IN

PATH LENGTH(WEEKS) SLACK

1-2-4-5-6 8+6+3+1=18 20-18=2

1-2-5-6 8+11+1=20 20-20=0

1-3-5-6 4+9+1=14 20-14=6

Network activities ES: early start

EF: early finish

LS: late start

LF: late finish

Used to determine Expected project duration

Slack time

Critical path

FOUR PIECES OF INFORMATION ABOUT THE NETWORK ACTIVITIES:

ES-the earliest time activity can start

EF-the earliest time the activity can finish

LS-the latest time the activity can start and not delay the project

LF-the latest time the activity can finish and not delay the project

Computation of earliest starting and finishing time is aided by two simple rules:

1. The earliest finish time for any activity is equal to its earliest start time plus its expected duration, t:

EF = ES + t

2. ES for activities at nodes with one entering arrow is equal to EF of the entering arrow.

ES for activities leaving nodes with multiple entering arrows is equal to the largest EF of the entering arrow.

Computation of latest starting and finishing time is aided by the use two rules:

1. The latest starting time for each activity is equal to its latest finishing time minus its expected duration, t:

LS = LF - t

2. For nodes with one leaving arrow, LF for arrows entering that node equals the LS of the leaving arrow. For nodes with multiple leaving arrows, LF for arrows entering that node equals the smallest LS of leaving arrows.

1 6 5

4

3

2

1 WEEKS

MOVE IN

Determine the ff:

a. ES c. EF

b. LS d. LF

1 6 5

4

3

2

1 WEEKS

MOVE IN

19 20

19 20

Slack = LS – ES

Or

Slack = LF- EF

ACTIVITY LS ES SLACK

1-2 0 0 0

1-3 6 0 6

2-4 10 8 2

2-5 8 8 0

3-5 10 4 6

4-5 16 14 2

5-6 19 19 0

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 2

Write and submit proposal B A 1

Obtain approval C B 1

Develop service vision and goals D C 2

Train employees E C 5

Quality improvement pilot groups F D, E 5

Write assessment report G F 1

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

ES=9

EF=14

ES=14

EF=15

ES=0

EF=2

ES=2

EF=3

ES=3

EF=4

ES=4

EF=9

ES=4

EF=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.

ES=9

EF=14

ES=14

EF=15

ES=0

EF=2

ES=2

EF=3

ES=3

EF=4

ES=4

EF=9

ES=4

EF=6

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

D(2)

E(5)

F(5) G(1)

LS=14

LF=15

LS=9

LF=14

LS=4

LF=9

LS=7

LF=9

LS=3

LF=4

LS=2

LF=3

LS=0

LF=2

Hint: Start with LF=15 or

the total time of the project

and go backward in the

network from G to A.

ES=9

EF=14

ES=14

EF=15

ES=0

EF=2

ES=2

EF=3

ES=3

EF=4

ES=4

EF=9

ES=4

EF=6

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

D(2)

E(5)

F(5) G(1)

LS=14

LF=15

LS=9

LF=14

LS=4

LF=9

LS=7

LF=9

LS=3

LF=4

LS=2

LF=3

LS=0

LF=2

Duration = 15 weeks

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

Optimistic time

Time required under optimal conditions

Pessimistic time

Time required under worst conditions

Most likely time

Most probable length of time that will be required

Activity start

Optimistic time

Most likely time (mode)

Pessimistic time

to tp tm te

te = to + 4tm +tp

6

te = expected time

to = optimistic time

tm = most likely time

tp = pessimistic time

2 = (tp – to)

2

36

2 = variance

to = optimistic time

tp = pessimistic time

The network diagram for a project is shown in the

accompanying figure, with three time estimates for each

activity. Activity times are in weeks. Do the following:

a. Compute the expected time for each activity and the

expected duration for each path

b. Identify the critical path

c. Compute the variance of each activity and the

variance and standard deviation of each path

3-4-5

d

3-5-7

e

5-7-9

f

2-4-6

b

4-6-8

h

Optimistic

time

Most likely

time

Pessimistic

time

a.

b. The path that has the longest expected duration is the

critical path. Because path d-e-f has the largest path total, it

is the critical path.

c.

Z = Specified time – Path mean

Path standard deviation

Z indicates how many standard deviations

of the path distribution the specified tine

is beyond the expected path duration.

Using information given from the previous example answer the ff:

a. What is the probability that the project can be completed within 17

weeks of its start?

b. What is the probability that the project will be completed within 15

weeks of its start?

c. What is the probability that the project will not be completed within

15 weeks of its start?

17

Weeks

Weeks

Weeks

Weeks

10.0

16.0

13.5

1.00

1.00

a-b-c

d-e-f

g-h-i

a.

Path Probability of Completion in 17 weeks

a-b-c + 7.22 1.00

d-e-f +1.00 .8413

g-h-I +3.27 1.00

P(Finish by week 17) = 1.00 x .8413 x 1.00 = .8413

Z = 17– Path Duration

Path standard deviation

b.

Path Probability of Completion in 17 weeks

a-b-c + 5.15 1.00

d-e-f - 1.00 .1587

g-h-I + 1.40 .9192

P(Finish by week 15) = 1.00 x .1587 x .9192 = .1459

c. 1 - .1459 = .8541

Z = 15– Path Duration

Path standard deviation

Task

Immediate

Predecesors Optimistic Most Likely Pessimistic

A None 3 6 15

B None 2 4 14

C A 6 12 30

D A 2 5 8

E C 5 11 17

F D 3 6 15

G B 3 9 27

H E,F 1 4 7

I G,H 4 19 28

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

6

ET(A)=42/6=7 Task

Immediate

Predecesors

Expected

Time

A None 7

B None 5.333

C A 14

D A 5

E C 11

F D 7

G B 11

H E,F 4

I G,H 18

Task

Immediate

Predecesors Optimistic Most Likely Pessimistic

A None 3 6 15

B None 2 4 14

C A 6 12 30

D A 2 5 8

E C 5 11 17

F D 3 6 15

G B 3 9 27

H E,F 1 4 7

I G,H 4 19 28

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

6

Example 2. Network

A(7)

B (5.333)

C(14)

D(5)

E(11)

F(7)

H(4)

G(11)

I(18)

Duration = 54 Days

Example 2. Probability Exercise

What is the probability of finishing this project in

less than 53 days?

p(t < D)

TE = 54

Z = D - TE

cp

2

t D=53

Activity variance, = (Pessim. - Optim.

6)2 2

Task Optimistic Most Likely Pessimistic Variance

A 3 6 15 4

B 2 4 14

C 6 12 30 16

D 2 5 8

E 5 11 17 4

F 3 6 15

G 3 9 27

H 1 4 7 1

I 4 19 28 16

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

There is a 43.8% probability that this project will be

completed in less than 53 weeks.

p(Z < -.156) = .438, or 43.8 % (NORMSDIST(-.156)

Z = D - T

=53- 54

41= -.156E

cp

2

TE = 54

p(t < D)

t

D=53

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

t TE = 54

p(t < D)

D=56

Z = D - T

=56 - 54

41= .312E

cp

2

p(Z > .312) = .378, or 37.8 % (1-NORMSDIST(.312))