### Production and Operations Management: Manufacturing and Services

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Chapter 3 part 2
Project Management
McGraw-Hill/Irwin
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
McGraw-Hill/Irwin
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
McGraw-Hill/Irwin
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
McGraw-Hill/Irwin
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?
McGraw-Hill/Irwin
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
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
McGraw-Hill/Irwin
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
McGraw-Hill/Irwin
9
Example 1. Expected Time
Calculations ET(A)= 3+4(6)+15
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
McGraw-Hill/Irwin
10
Ex. 1. Expected Time Calculations
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
McGraw-Hill/Irwin
11
Ex 1. Expected Time Calculations
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
McGraw-Hill/Irwin
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)
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
14
A ctivity variance, 
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
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
Exercise

16
What is the probability that the
project duration will exceed 56
weeks?
McGraw-Hill/Irwin
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 %
McGraw-Hill/Irwin
18
Activity List for Example
Problem
McGraw-Hill/Irwin
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
McGraw-Hill/Irwin
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
McGraw-Hill/Irwin
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
McGraw-Hill/Irwin
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!
McGraw-Hill/Irwin
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
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.

McGraw-Hill/Irwin