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Project Scheduling
CPM/PERT

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 Network techniques
 Developed in 1950’s
 CPM by DuPont for chemical plants (1957)
 PERT by Booz, Allen & Hamilton with the U.S.
Navy, for Polaris missile (1958)
 Consider precedence relationships and
interdependencies
 Each uses a different estimate of activity times
PERT and CPM

why networks/CPM/PERT?
 Gantt charts don’t explicitly show
task relationships
 don’t show impact of delays or
shifting resources well
 network models clearly show
interdependencies

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Is the project on schedule, ahead of schedule,
or behind schedule?
Is the project over or under cost budget?
Are there enough resources available to finish
the project on time?
If the project must be finished in less than the
scheduled amount of time, what is the way to
accomplish this at least cost?
Questions Which May Be
Addressed by PERT & CPM

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The Six Steps Common to PERT & CPM
 Define the project and prepare the work breakdown
structure,
 Develop relationships among the activities. (Decide
which activities must precede and which must follow
others.)
 Draw the network connecting all of the activities
 Assign time and/or cost estimates to each activity
 Compute the longest time path through the network.
This is called the critical path
 Use the network to help plan, schedule, monitor, and
control the project

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Terminology
 Activity: A specific or set of tasks required by
the project
 Event: Outcome of one or more activities
 Network: Combination of all activities and
events
 Path: Series of connected activities or
between any two events
 Critical path: Longest - Any delay would delay
the project
 Slack/float: Allowable slippage for a path

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Activity Relationships
Predecessor – an activity that is required to
start or finish before the next activity(s) can
proceed
Successor – an activity that must start or
finish after the previous activity can finish
Types of relationships are defined from the
predecessor to the successor

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1
A
B
A & B can occur
concurrently
2
3
Activity Relationships

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1 4
2
3
A
B
C
A must be done
before C & D can
begin
D
Activity Relationships

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1 4
2
3
A
B E
C
B & C must be done
before E can begin
D
Activity Relationships

AOA Project Network for
a House
3
2 0
1
3
1 1
1
1 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

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 Activities are defined often by beginning &
ending events
 Every activity must have unique pair of
beginning & ending events
 Otherwise, computer programs get confused
 Dummy activities maintain precedence
 Consume no time or resources
Dummy Activities

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Job on Arc Network
 Not allowed: no two
jobs can have the
same starting and
ending node!
 Need to introduce a
dummy job.
A
B
D
C
A
B
D
C

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1 4
3
1-2
2-3
Incorrect
1 4
2
3
5
2
2-3
3-4
1-2
2-3
2-4 4-5
3-4: Dummy
activity
Correct
Dummy Activity Example

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Critical Path
The longest continuous path of activities
through a project, which determines the
project end date

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A General Hospital’s Activities and
Predecessors
Activity Description Immediate
Predecessors
A -
B -
C A
D A, B
E C
F C
G D, E
H F, G

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AON Network for General Hospital
Start
A
B
C
D
F
F
G
H

Program Evaluation and
Review Technique (PERT)
PERT is based on the assumption that an
activity’s duration follows a probability
distribution instead of being a single value.
The probabilistic information about the
activities is translated into probabilistic
information about the project.

PERT
 reflects PROBABILISTIC nature of durations
 assumes BETA distribution
 same as CPM except THREE duration
estimates
optimistic
most likely
pessimistic

PERT Calculation
a = optimistic duration estimate
m = most likely duration estimate
b = pessimistic duration estimate
expected duration:
variance:
Te
a + 4m + b
6
V =
b - a
6







2

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 3 time estimates
 Optimistic times (a)
 Most-likely time (m)
 Pessimistic time (b)
 Follow beta distribution
 Expected time: t = (a + 4m + b)/6
 Variance of times: v = (b - a)2/6
PERT Activity Times

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Variability of Completion Time for
Noncritical Paths
Variability of times for activities on non-critical
paths must be considered when finding the
probability of finishing in a specified time.
Variation in non-critical activity may cause
change in critical path.

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Advantages of PERT/CPM
 Especially useful when scheduling and controlling
large projects.
 Straightforward concept and not mathematically
complex.
 Graphical networks aid perception of relationships
among project activities.
 Critical path & slack time analyses help pinpoint
activities that need to be closely watched.
 Project documentation and graphics point out who is
responsible for various activities.
 Applicable to a wide variety of projects.
 Useful in monitoring schedules and costs.

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Questions Answered by CPM & PERT
Completion date?
On Schedule?
Within Budget?
Critical Activities?
How can the project be finished early at the
least cost?

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 Assumes clearly defined, independent, &
stable activities
 Specified precedence relationships
 Activity times (PERT) follow beta distribution
 Subjective time estimates
 Over-emphasis on critical path
Limitations of PERT/CPM

Example 2. CPM with Three Activity Time
Estimates
Ta s k
Im m e d ia t e
P re d e c e s o rs O p t im is t ic M o s t L ik e ly P e s s im is t ic
A N o n e 3 6 1 5
B N o n e 2 4 1 4
C A 6 1 2 3 0
D A 2 5 8
E C 5 1 1 1 7
F D 3 6 1 5
G B 3 9 2 7
H E , F 1 4 7
I G , H 4 1 9 2 8

Example 2. Expected Time
Calculations
T a s k
Im m e d i a t e
P re d e c e s o rs
E x p e c t e d
T i m e
A N o n e 7
B N o n e 5 . 3 3 3
C A 1 4
D A 5
E C 1 1
F D 7
G B 1 1
H E , F 4
I G , H 1 8

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.
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)
2 2

Ta s k O p tim is tic M o s t L ik e ly P e s s im is tic V a ria n c e
A 3 6 1 5 4
B 2 4 1 4
C 6 1 2 3 0 1 6
D 2 5 8
E 5 1 1 1 7 4
F 3 6 1 5
G 3 9 2 7
H 1 4 7 1
I 4 1 9 2 8 1 6
(Sum the variance along the critical path.) 2
 = 41

p(Z < -0.156) = 0.5 - 0.0636 = 0.436, or 43.6 %
Z =
D - T
=
53- 54
41
= -.156
E
cp
2


TE = 54
p(t < D)
t
D=53

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A Comparison of AON and AOA
Network Conventions