1
Chapter 3 part 2
Project Management
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2
CPM With 3 Time
Estimates (PERT)


In past, time estimate is firm
Now, uncertain as to task duration
– Natural variance

Use 3 time estimates to deal with
uncertainty
– a = optimistic estimate
– m = most likely
– b = pessimistic
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3
Critical Path


Path length is sum of expected times of
activities on path (not sum of most likely
times)
Longest expected path length is critical
– ET = expected time of project


Path length is uncertain, and so is project
duration
Path has variance equal to sum of variances
of individual activities on path
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4
Formulas

ET (expected time) = (a + 4m + b)/6

σ 2 (variance) = [(b - a)/6]2

σp (st dev of project) =
sqrt(S (variances on CP))

Z = (D - ET)/σ
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© 2006 The McGraw-Hill Companies, Inc., All Rights Reserved.
PERT Probability Example
5
You’re a project planner for
Apple. A new Ipod project
has an expected
completion time of 40
weeks, with a standard
deviation of 5 weeks.
What is the probability of
finishing the sub in 50
weeks or less?
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6
Converting to
Standardized Variable
X - ET 50 - 40
=
= 2 .0
Z =
s
5
Normal
Distribution
Standardized Normal
Distribution
sZ = 1
s =5
T = 40 50
McGraw-Hill/Irwin
X
mz = 0 2.0
Z
© 2006 The McGraw-Hill Companies, Inc., All Rights Reserved.
Obtaining the Probability
7
Standardized Normal Probability
Table (Portion)
Z
.00
.01
.02
sZ =1
0.0 .50000 .50399 .50798
:
:
:
:
2.0 .97725 .97784 .97831
.97725
2.1 .98214 .98257 .98300
mz = 0 2.0
Z
Probabilities in body
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8
Example 1. CPM with Three
Activity Time Estimates
Immediate
Task 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
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9
Example 1. Expected Time
Calculations ET(A)= 3+4(6)+15
Task
A
B
C
D
E
F
G
H
I
Immediate Expected
Predecesors
Time
None
7
None
5.333
A
14
A
5
C
11
D
7
B
11
E,F
4
G,H
18
6
ET(A)=42/6=7
Immediate
Task 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
Opt. Time + 4(Most Likely Time) + Pess. Time
Expected Time =
6
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10
Ex. 1. Expected Time Calculations
Task
A
B
C
D
E
F
G
H
I
Immediate Expected
Predecesors
Time
None
7
None
5.333
A
14
A
5
C
11
D
7
B
11
E,F
4
G,H
18
ET(B)= 2+4(4)+14
6
ET(B)=32/6=5.333
Immediate
Task 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
Opt. Time + 4(Most Likely Time) + Pess. Time
Expected Time =
6
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11
Ex 1. Expected Time Calculations
Task
A
B
C
D
E
F
G
H
I
Immediate Expected
Predecesors
Time
None
7
None
5.333
A
14
A
5
C
11
D
7
B
11
E,F
4
G,H
18
ET(C)= 6+4(12)+30
6
ET(C)=84/6=14
Immediate
Task 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
Opt. Time + 4(Most Likely Time) + Pess. Time
Expected Time =
6
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12
Example 1. Network
Duration = 54 Days
C(14)
E(11)
H(4)
A(7)
D(5)
F(7)
I(18)
B
(5.333)
McGraw-Hill/Irwin
G(11)
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13
Example 1. Probability Exercise
What is the probability of finishing this project in
less than 53 days?
p(t < D)
D=53
t
TE = 54
Z =
McGraw-Hill/Irwin
D - TE
2

 cp
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14
A ctivity variance, 
Task
A
B
C
D
E
F
G
H
I
2
Pessim . - O ptim . 2
= (
)
6
Optimistic Most Likely Pessimistic Variance
3
6
15
4
2
4
14
6
12
30
16
2
5
8
5
11
17
4
3
6
15
3
9
27
1
4
7
1
4
19
28
16
(Sum the variance along the critical
path.)
McGraw-Hill/Irwin
2

 = 41
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15
p(t < D)
t
TE = 54
D=53
Z =
D - TE

2
cp
53 - 54
=
= -.156
41
p(Z < -.156) = .438, or 43.8 %
There is a 43.8% probability that this project will be
completed in less than 53 weeks. (from the z table)
McGraw-Hill/Irwin
© 2006 The McGraw-Hill Companies, Inc., All Rights Reserved.
Ex 1. Additional Probability
Exercise

16
What is the probability that the
project duration will exceed 56
weeks?
McGraw-Hill/Irwin
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Example 1. Additional Exercise
Solution
17
p(t < D)
TE = 54
Z =
D - TE
2

 cp
t
D=56
56 - 54
=
= .312
41
p(Z > .312) = .378, or 37.8 %
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18
Activity List for Example
Problem
McGraw-Hill/Irwin
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19
Example 2
Activity
A
B
C
D
E
F
G
H
I
J
a
1
3
2
2
4
1
2.5
1
4
1.5
m
3
4.5
3
4
7
1.5
3.5
2
5
3
b
5
9
4
6
16
5
7.5
3
6
4.5
ET
2
ET = (a + 4m + b)/6
σ 2 = (b - a)2/36 = [(b - a)/6]2
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20
Example 2
Activity
A
B
C
D
E
F
G
H
I
J
a
1
3
2
2
4
1
2.5
1
4
1.5
m
3
4.5
3
4
7
1.5
3.5
2
5
3
b
5
9
4
6
16
5
7.5
3
6
4.5
ET
3
5
3
4
8
2
4
2
5
3
2
0.44
1
0.11
0.44
4
0.44
0.69
0.11
0.11
0.25
T = (a + 4m + b)/6
σ 2 = (b - a)2/36 = [(b - a)/6]2
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21
Example 2
Activity
A
B
C
D
E
F
G
H
I
J
a
1
3
2
2
4
1
2.5
1
4
1.5
m
3
4.5
3
4
7
1.5
3.5
2
5
3
b
5
9
4
6
16
5
7.5
3
6
4.5
ET
3
5
3
4
8
2
4
2
5
3
2
0.44
1
0.11
0.44
4
0.44
0.69
0.11
0.11
0.25
T = (a + 4m + b)/6
σ 2 = (b - a)2/36 = [(b - a)/6]2
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22
Calculations
•ET = 23 weeks
•Σσ2CP = σ 2B + σ 2C + σ 2D + σ 2E + σ 2J
= 5.8056
•Z = (D - ET)/σ CP = (22-23)/2.4095
= -0.42
Take .42 to the Z table => 1 – Z[.42] =
.337
•When negative – subtract from 1
•Beware - never add standard deviations!
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23
22 is 0.42 SD below Mean
0.4
16% chance between 22 and 23
(from table)
0.3
0.2
0.1
0
34% chance
< 22
22
McGraw-Hill/Irwin
50% chance
> 23
ET = 23
© 2006 The McGraw-Hill Companies, Inc., All Rights Reserved.
Variability of Completion
Time for Noncritical Paths
24
Variability of times for activities on
noncritical paths must be considered
when finding the probability of finishing
in a specified time.
 Variation in noncritical activity may
cause change in critical path.

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