Advanced Simulation Methods
Overview
Advanced Simulation Applications
• Beta Distribution
• Operations
– Project Management (PERT)
• “Textbook” method
• Crystal Ball
• Marketing
– New Product Development decision
Operations -- Prof. Juran
2
©The McGraw-Hill Companies, Inc., 2004
Beta Distribution
The Beta distribution is a continuous probability distribution defined
by four parameters:
Parameter
Min
Max
Alpha ( α)
Beta (β)
Operations -- Prof. Juran
Description
Minimum Value
Maximum Value
Shape Factor
Shape Factor
Characteristics
Any number -∞ to ∞
Any number -∞ to ∞
Must be > 0
Must be > 0
3
©The McGraw-Hill Companies, Inc., 2004
Here are sixteen different Beta distributions, all with a minimum of 0 and a maximum
of 100.
β
0.5
0.5
1.0
2.0
4.0
0.0500
0.0500
0.0500
0.0500
0.0450
0.0450
0.0450
0.0450
0.0400
0.0400
0.0400
0.0400
0.0350
0.0350
0.0350
0.0350
0.0300
0.0300
0.0300
0.0300
0.0250
0.0250
0.0250
0.0250
0.0200
0.0200
0.0200
0.0200
0.0150
0.0150
0.0150
0.0150
0.0100
0.0100
0.0100
0.0050
0.0050
0.0050
0.0000
0.0000
0
25
50
75
0.0100
0.0050
0.0000
0
25
50
75
0.0000
0
0.0500
0.0500
0.0500
0.0450
0.0450
0.0450
0.0400
0.0400
0.0400
0.0350
0.0350
0.0350
0.0300
0.0300
0.0300
0.0250
0.0250
0.0250
0.0200
0.0200
0.0200
0.0150
0.0150
0.0150
0.0100
0.0100
0.0100
0.0050
0.0050
0.0050
0.0000
0.0000
0.0000
25
50
75
0
25
50
75
0
25
50
75
0
25
50
75
0
25
50
75
0.0400
0.0350
0.0300
1.0
0.0250
0.0200
0.0150
0.0100
α
0
2.0
25
50
75
25
50
75
0.0050
0.0000
0
25
50
75
0.0500
0.0500
0.0500
0.0500
0.0450
0.0450
0.0450
0.0450
0.0400
0.0400
0.0400
0.0400
0.0350
0.0350
0.0350
0.0350
0.0300
0.0300
0.0300
0.0300
0.0250
0.0250
0.0250
0.0250
0.0200
0.0200
0.0200
0.0200
0.0150
0.0150
0.0150
0.0150
0.0100
0.0100
0.0100
0.0050
0.0050
0.0050
0.0000
0.0000
0
4.0
0
25
50
75
0.0100
0.0050
0.0000
0
25
50
75
0.0000
0
25
50
75
0.0500
0.0500
0.0500
0.0500
0.0450
0.0450
0.0450
0.0450
0.0400
0.0400
0.0400
0.0400
0.0350
0.0350
0.0350
0.0350
0.0300
0.0300
0.0300
0.0300
0.0250
0.0250
0.0250
0.0250
0.0200
0.0200
0.0200
0.0200
0.0150
0.0150
0.0150
0.0150
0.0100
0.0100
0.0100
0.0050
0.0050
0.0050
0.0000
0.0000
0
25
50
75
Operations -- Prof. Juran
0.0100
0.0050
0.0000
0
25
50
75
0.0000
0
25
50
75
4
©The McGraw-Hill Companies, Inc., 2004
The Beta distribution is popular among simulation
modelers because it can take on a wide variety of
shapes, as shown in the graphs above.
The Beta can look similar to almost any of the
important continuous distributions, including
Triangular, Uniform, Exponential, Normal, Lognormal,
and Gamma.
For this reason, the Beta distribution is used
extensively in PERT, CPM and other project
planning/control systems to describe the time to
completion of a task.
Operations -- Prof. Juran
5
©The McGraw-Hill Companies, Inc., 2004
Mean:
  min 

 max - min 
   
   
2




max
min

Standard Deviation:
   2      1
Operations -- Prof. Juran
(i)
(ii)
6
©The McGraw-Hill Companies, Inc., 2004
PERT Approximations
The project management community has evolved approximations for
the Beta distribution which allow it to be handled with three
parameters, rather than four.
The three parameters are the minimum, mode, and maximum activity
times (usually referred to as the optimistic, most-likely, and
pessimistic activity times).
This doesn’t give exactly the same results as the mathematicallycorrect version, but has important practical advantages.
Most real-life managers are not comfortable talking about things like
probability functions and Greek-letter parameters, but they are
comfortable talking in terms of optimistic, most-likely, and
pessimistic.
Operations -- Prof. Juran
7
©The McGraw-Hill Companies, Inc., 2004
3-step Procedure
1. Get estimates for the minimum optimistic, most-likely, and pessimistic time to
completion for the activity.
2. Estimate the mean and standard deviation using equations (iii) and (iv):
min  4  mode  max
 
(iii)
6
max - min
(iv)
 
6
3. Use equations (v) and (vi) to calculate shape factors that are consistent with the
mean and standard deviation:


 mean - min   mean - min max - mean  
 

 1 
2
 
 max - min  

max - mean 
 
 
 mean - min 
Operations -- Prof. Juran
(v)
(vi)
8
©The McGraw-Hill Companies, Inc., 2004
Beta Distributions in Crystal Ball
The Crystal Ball distribution gallery includes the Beta distribution, but in a form
slightly different from the description above.
Specifically, Crystal Ball assumes the minimum is zero. Instead of “maximum” or
“pessimistic”, it asks for a “Scale” parameter.
Operations -- Prof. Juran
9
©The McGraw-Hill Companies, Inc., 2004
We can still use Crystal Ball to simulate Betas, but it requires following these steps.
1. Make an assumption cell with the alpha and beta calculated with formulas (v)
and (vi) above.
2. For the “Scale” parameter, enter the difference between the maximum and
minimum.
3. In the spreadsheet model, create a separate cell for the activity time. This cell
should be the minimum plus the random number generated by Crystal Ball in
the assumption cell.
Operations -- Prof. Juran
10
©The McGraw-Hill Companies, Inc., 2004
Example
Assume we are given optimistic, most-likely, and pessimistic times of 1, 2, and 3
time units, respectively.
We first use these parameters to calculate the mean (formula (iii)), standard
deviation (formula (iv)), alpha (formula (v)), beta (formula (vi)), and the difference
between the maximum and minimum, as shown here:
A
1
2
3
4
5
6
7
8
9
10
B
C
D
E
Optimistic (o) Most Likely (m)
1
2
Operations -- Prof. Juran
F
G
H
I
J
K
Pessimistic (p) Mean StDev Alpha Beta Max - Min
3
2
0.333 4.000 4.000
2
L
M
=F2-D2
=(D2+4*E2+F2)/6
=((F2-G2)/(G2-D2))*I2
=(F2-D2)/6
=((G2-D2)/(F2-D2))*((((G2-D2)*(F2-G2))/(H2^2))-1)
11
©The McGraw-Hill Companies, Inc., 2004
Next, we create a Crystal Ball assumption cell in A2, using the parameters shown:
We make a cell next to the assumption cell, adding the random number to the minimum.
1
2
A
B
2
3
C
D
E
F
G
H
I
J
K
Optimistic (o) Most Likely (m) Pessimistic (p) Mean StDev Alpha Beta Max - Min
=A2+D2
1
2
3
2
0.333 4.000 4.000
2
Cell B3 will now be a Beta-distributed random variable with the optimistic, most-likely,
and pessimistic activity times we specified.
Operations -- Prof. Juran
12
©The McGraw-Hill Companies, Inc., 2004
Operations Example:
Project Management (PERT)
Sharon Katz is project manager in charge of laying the foundation for the
new Brook Museum of Art in New Haven, Connecticut.
Liya Brook, the benefactor and namesake of the museum, wants to have the
work done within 41 weeks, but Sharon wants to quote a completion time
that she is 90% confident of achieving.
The contract specifies a penalty of $10,000 per week for each week the
completion of the project extends beyond week 43.
Operations -- Prof. Juran
13
©The McGraw-Hill Companies, Inc., 2004
Activity
A
B
C
D
E
F
G
H
I
J
K
L
M
Description
Survey Site
Excavation
Prepare Drawings
Soil Study
Prelim. Report
Approve Plans
Concrete Forms
Procure Steel
Order Cement
Deliver Gravel
Pour Concrete
Cure Concrete
Strength Test
Operations -- Prof. Juran
Optimistic
2
9
4
1
1
1
5
2
1
2
8
2
2
Pessimistic
4
15
18
1
3
1
9
10
1
5
14
2
2
Most-likely
3
12
9
1
2
1
6
5
1
3
10
2
2
Predecessors
None
A
None
B
C, D
E
F
F
F
G
H, I, J
K
L
14
©The McGraw-Hill Companies, Inc., 2004
Create a PERT model of this project and use it to answer these questions:
1. What is the expected completion time of this project?
2. What completion time should Sharon use, if she wants to be 90%
confident?
3. What is the probability of completion by week 43?
4. Give an estimated probability distribution for the amount of penalties
Sharon will have to pay.
5. What is the expected value of the penalty?
6. Which activities are most likely to be on the critical path?
7. Compare the PERT results to those you would have found using (a) basic
CPM using the most-likely times, and (b) the “by-hand” PERT method
from the textbook.
Operations -- Prof. Juran
15
©The McGraw-Hill Companies, Inc., 2004
Here’s an activity-on-arc diagram of the problem:
Operations -- Prof. Juran
16
©The McGraw-Hill Companies, Inc., 2004
We start a spreadsheet model like this, calculating the mean and
standard deviation using the PERT formulas:
A
B
C
D
E
F
G
H
I
1 Activity Description
Optimistic Pessimistic Most-likely Predecessors Start Node End Node Mean
2
A
Survey Site
2
4
3
None
0
1
3.00
3
B
Excavation
9
15
12
A
1
2
12.00
4
C
Prepare Drawings
4
18
9
None
0
3
9.67
5
D
Soil Study
1
1
1
B
2
3
1.00
6
E
Prelim. Report
1
3
2
C, D
3
4
2.00
7
F
Approve Plans
1
1
1
E
4
5
1.00
8
G
Concrete Forms
5
9
6
F
5
7
6.33
9
H
Procure Steel
2
10
5
F
5
6
5.33
10
I
Order Cement
1
1
1
F
5
8
1.00
11 Dummy
0
0
0
H
6
8
0.00
12
J
Deliver Gravel
2
5
3
G
7
8
3.17
13
K
Pour Concrete
8
14
10
H, I, J
8
9
10.33
14
L
Cure Concrete
2
2
2
K
9
10
2.00
15
M
Strength Test
2
2
2
L
10
11
2.00
Operations -- Prof. Juran
J
K
L
StDev
0.33
=(C3+4*E3+D3)/6
1.00
2.33
=(D5-C5)/6
0.00
0.33
0.00
0.67
1.33
0.00
0.00
0.50
1.00
0.00
0.00
M
17
©The McGraw-Hill Companies, Inc., 2004
Now we calculate shape and scale parameters:
A
B
C
D
E
F
G
H
I
1 Activity Description
Optimistic Pessimistic Most-likely Predecessors Start Node End Node Mean
2
A
Survey Site
2
4
3
None
0
1
3.00
3
B
Excavation
9
15
12
A
1
2
12.00
4
C
Prepare Drawings
4
18
9
None
0
3
9.67
5
D
Soil Study
1
1
1
B
2
3
1.00
6
E
Prelim. Report
1
3
2
C, D
3
4
2.00
7
F
Approve Plans
1
1
1
E
4
5
1.00
8
G
Concrete Forms
5
9
6
F
5
7
6.33
9
H
Procure Steel
2
10
5
F
5
6
5.33
10
I
Order Cement
1
1
1
F
5
8
1.00
11 Dummy
0
0
0
H
6
8
0.00
12
J
Deliver Gravel
2
5
3
G
7
8
3.17
13
K
Pour Concrete
8
14
10
H, I, J
8
9
10.33
14
L
Cure Concrete
2
2
2
K
9
10
2.00
15
M
Strength Test
2
2
2
L
10
11
2.00
Operations -- Prof. Juran
J
StDev
0.33
1.00
2.33
0.00
0.33
0.00
0.67
1.33
0.00
0.00
0.50
1.00
0.00
0.00
K
Alpha
4.00
4.00
3.11
4.00
2.33
3.23
2.94
2.94
L
M
N
O
P
Q
R
S
Beta Scale
4.00
2
=((I3-C3)/(D3-C3))*((((I3-C3)*(D3-I3))/(J3^2))-1)
4.00
6
=((D4-I4)/(I4-C4))*K4
4.57 14
=D5-C5
0
4.00
2
0
4.67
4
4.52
8
0
0
4.62
3
4.62
6
0
0
18
©The McGraw-Hill Companies, Inc., 2004
Model Overview
A section for simulating the times of the activities
A
B
C
D
E
F
G
1 Activity Description
Predecessors Start Node End Node Min Alpha
2
A
Survey Site
None
0
1
2 4.00
3
B
Excavation
A
1
2
9 4.00
4
C
Prepare Drawings
None
0
3
4 3.11
5
D
Soil Study
B
2
3
1
6
E
Prelim. Report
C, D
3
4
1 4.00
7
F
Approve Plans
E
4
5
1
8
G
Concrete Forms
F
5
7
5 2.33
9
H
Procure Steel
F
5
6
2 3.23
10
I
Order Cement
F
5
8
1
11 Dummy
H
6
8
0
12
J
Deliver Gravel
G
7
8
2 2.94
13
K
Pour Concrete
H, I, J
8
9
8 2.94
14
L
Cure Concrete
K
9
10
2
15
M
Strength Test
L
10
11
2
16
17
18 Node
Time
Path
Total
19
0
0
A-B-D-E-F-H-K-L-M
60.67
20
1
5.00
A-C-E-F-H-K-L-M
52.33
21
2
26.00
A-B-D-E-F-I-K-L-M
54.33
22
3
27.00
A-C-E-F-I-K-L-M
46.00
23
4
30.00
A-B-D-E-F-G-J-K-L-M
69.83
24
5
31.00
A-C-E-F-G-J-K-L-M
61.50
25
6
38.33
69.83
26
7
42.33
27
8
47.50
28
9
65.83
29
10
67.83
<= 43?
Penalty
30
11
69.83
0 $ 268,333
H
I
J
K
Beta Scale CB Time Simulated Time
4.00
2
3.00
5.00
4.00
6
12.00
21.00
4.57 14
9.67
13.67
0
1.00
1.00
4.00
2
2.00
3.00
0
1.00
1.00
4.67
4
6.33
11.33
4.52
8
5.33
7.33
0
1.00
1.00
0
0.00
0.00
4.62
3
3.17
5.17
4.62
6
10.33
18.33
0
2.00
2.00
0
2.00
2.00
Critical?
0
0
0
0
1
0
L
M
N
Start Time End Time Critical?
0.00
5.00
1
5.00
26.00
1
0.00
13.67
0
26.00
27.00
1
27.00
30.00
1
30.00
31.00
1
31.00
42.33
1
31.00
38.33
0
31.00
32.00
0
38.33
38.33
42.33
47.50
1
47.50
65.83
1
65.83
67.83
1
67.83
69.83
1
A section to keep track of each
path through the network, to
identify the critical path in each
simulated project completion
A section to keep track of each node and when it occurs
Operations -- Prof. Juran
19
©The McGraw-Hill Companies, Inc., 2004
Here’s the section keeping track of the activity times (we’ll fill in the stuff in
column J with Crystal Ball assumption cells later). Because of the way Crystal
Ball handles Beta distributions, we need to add the random numbers (column J)
to the minimums (column F) to get the simulated activity times (column K).
A
B
C
D
E
F
G
H
I
J
K
1 Activity Description
Predecessors Start Node End Node Min Alpha Beta Scale CB Time Simulated Time
2
A
Survey Site
None
0
1
2 4.00 4.00
2
3.00
5.00
3
B
Excavation
A
1
2
9 4.00 4.00
6
12.00
21.00
4
C
Prepare Drawings
None
0
3
4 3.11 4.57 14
9.67
13.67
5
D
Soil Study
B
2
3
1
0
1.00
1.00
6
E
Prelim. Report
C, D
3
4
1 4.00 4.00
2
2.00
3.00
7
F
Approve Plans
E
4
5
1
0
1.00
1.00
8
G
Concrete Forms
F
5
7
5 2.33 4.67
4
6.33
11.33
9
H
Procure Steel
F
5
6
2 3.23 4.52
8
5.33
7.33
10
I
Order Cement
F
5
8
1
0
1.00
1.00
11 Dummy
H
6
8
0
0
0.00
0.00
12
J
Deliver Gravel
G
7
8
2 2.94 4.62
3
3.17
5.17
13
K
Pour Concrete
H, I, J
8
9
8 2.94 4.62
6
10.33
18.33
14
L
Cure Concrete
K
9
10
2
0
2.00
2.00
15
M
Strength Test
L
10
11
2
0
2.00
2.00
Operations -- Prof. Juran
L
Start Time
=F4+J4
=F5
20
©The McGraw-Hill Companies, Inc., 2004
Now we set up an area in the spreadsheet to keep track of the nodes and their
times:
18
19
20
21
22
23
24
25
26
27
28
29
30
A
Node
0
1
2
3
4
5
6
7
8
9
10
11
B
Time
0
We need to link the node times to the starting and ending times for the activities.
The start time for any activity is the time at which its beginning node occurs. The
end time for any activity is the start time plus the activity time.
Operations -- Prof. Juran
21
©The McGraw-Hill Companies, Inc., 2004
Example: Activity C
A
B
C
D
E
F
G
H
I
J
K
L
M
N
1 Activity Description
Predecessors Start Node End Node Min Alpha Beta Scale CB Time Simulated Time Start Time End Time
2
A
Survey Site
None
0
1
2 4.00 4.00
2
3.00
5.00
0
5.00
3
B
Excavation
A
1
2
9 4.00 4.00
6
12.00
21.00
0
21.00
4
C
Prepare Drawings
None
0
3
4 3.11 4.57 14
9.67
13.67
0
13.67
5
D
Soil Study
B
2
3
1
0
1.00
1.00
0
1.00=L4+K4
6
E
Prelim. Report
C, D
3
4
1 4.00 4.00
2
2.00
3.00
0
3.00
7
F
Approve Plans
E
4
5
1
0
1.00 =VLOOKUP(D4,$A$19:$B$30,2,0)
1.00
0
1.00
8
G
Concrete Forms
F
5
7
5 2.33 4.67
4
6.33
11.33
0
11.33
9
H
Procure Steel
F
5
6
2 3.23 4.52
8
5.33
7.33
0
7.33
10
I
Order Cement
F
5
8
1
0
1.00
1.00
0
1.00
11 Dummy
H
6
8
0
0
0.00
0.00
0
0.00
12
J
Deliver Gravel
G
7
8
2 2.94 4.62
3
3.17
5.17
0
5.17
13
K
Pour Concrete
H, I, J
8
9
8 2.94 4.62
6
10.33
18.33
0
18.33
14
L
Cure Concrete
K
9
10
2
0
2.00
2.00
0
2.00
15
M
Strength Test
L
10
11
2
0
2.00
2.00
0
2.00
16
17
18 Node
Time
19
0
0
20
1
21
2
22
3
23
4
24
5
25
6
26
7
27
8
28
9
29
10
30
11
Operations -- Prof. Juran
1
H
5
6
Dummy
I
22
©The McGraw-Hill Companies, Inc., 2004
It’s important to be careful with the nodes that have multiple activities leading into them
(in this model, Nodes 3 and 8). The times for those nodes must be the maximum ending
time for the set of activities leading in.
Nodes with only one preceding activity are easier (see Nodes 4 and 8 below).
A
B
C
D
E
F
G
H
I
J
K
1 Activity Description
Predecessors Start Node End Node Min Alpha Beta Scale CB Time Simulated Time
2
A
Survey Site
None
0
1
2 4.00 4.00
2
3.00
5.00
3
B
Excavation
A
1
2
9 4.00 4.00
6
12.00
21.00
4
C
Prepare Drawings
None
0
3
4 3.11 4.57 14
9.67
13.67
5
D
Soil Study
B
2
3
1
0
1.00
1.00
6
E
Prelim. Report
C, D
3
4
1 4.00 4.00
2
2.00
3.00
7
F
Approve Plans
E
4
5
1
0
1.00
1.00
8
G
Concrete Forms
F
5
7
5 2.33 4.67
4
6.33
11.33
9
H
Procure Steel
F
5
6
2 3.23 4.52
8
5.33
7.33
10
I
Order Cement
F
5
8
1
0
1.00
1.00
11 Dummy
H
6
8
0
0
0.00
0.00
12
J
Deliver Gravel
G
7
8
2 2.94 4.62
3
3.17
5.17
13
K
Pour Concrete
H, I, J
8
9
8 2.94 4.62
6
10.33
18.33
14
L
Cure Concrete
K
9
10
2
0
2.00
2.00
15
M
Strength Test
L
10
11
2
0
2.00
2.00
16
17
18 Node
Time
19
0
0
20
1
5.00
21
2
26.00
22
3
27.00
=M6
23
4
30.00
24
5
31.00
25
6
38.33
26
7
42.33
=MAX(M10:M12)
27
8
47.50
28
9
65.83
29
10
67.83
30
11
69.83
Operations -- Prof. Juran
L
M
Start Time End Time
0.00
5.00
5.00
26.00
0.00
13.67
26.00
27.00
27.00
30.00
30.00
31.00
31.00
42.33
31.00
38.33
31.00
32.00
38.33
38.33
42.33
47.50
47.50
65.83
65.83
67.83
67.83
69.83
23
©The McGraw-Hill Companies, Inc., 2004
Now we set up an area in the spreadsheet to track each of the paths through
the network, to see which one is critical. This network happens to have six
paths, so we set up a cell to add up all of the activity times for each of these
paths:
A
B
C
D
E
F
G
1 Activity Description
Predecessors Start Node End Node Min Alpha
2
A
Survey Site
None
0
1
2 4.00
3
B
Excavation
A
1
2
9 4.00
4
C
Prepare Drawings
None
0
3
4 3.11
5
D
Soil Study
B
2
3
1
6
E
Prelim. Report
C, D
3
4
1 4.00
7
F
Approve Plans
E
4
5
1
8
G
Concrete Forms
F
5
7
5 2.33
9
H
Procure Steel
F
5
6
2 3.23
10
I
Order Cement
F
5
8
1
11 Dummy
H
6
8
0
12
J
Deliver Gravel
G
7
8
2 2.94
13
K
Pour Concrete
H, I, J
8
9
8 2.94
14
L
Cure Concrete
K
9
10
2
15
M
Strength Test
L
10
11
2
16
17
18 Node
Time
Path
Total
19
0
0
A-B-D-E-F-H-K-L-M
60.67
20
1
5.00
A-C-E-F-H-K-L-M
52.33
21
2
26.00
A-B-D-E-F-I-K-L-M
54.33
22
3
27.00
A-C-E-F-I-K-L-M
46.00
23
4
30.00
A-B-D-E-F-G-J-K-L-M
69.83
24
5
31.00
A-C-E-F-G-J-K-L-M
61.50
25
6
38.33
69.83
Operations -- Prof. Juran
H
I
J
K
Beta Scale CB Time Simulated Time
4.00
2
3.00
5.00
4.00
6
12.00
21.00
4.57 14
9.67
13.67
0
1.00
1.00
4.00
2
2.00
3.00
0
1.00
1.00
4.67
4
6.33
11.33
4.52
8
5.33
7.33
0
1.00
1.00
0
0.00
0.00
4.62
3
3.17
5.17
4.62
6
10.33
18.33
0
2.00
2.00
0
2.00
2.00
L
M
Start Time End Time
0.00
5.00
5.00
26.00
0.00
13.67
26.00
27.00
27.00
30.00
30.00
31.00
31.00
42.33
31.00
38.33
31.00
32.00
38.33
38.33
42.33
47.50
47.50
65.83
65.83
67.83
67.83
69.83
=SUM(K2,K3,K5,K6,K7,K9,K13,K14,K15)
=MAX(G19:G24)
24
©The McGraw-Hill Companies, Inc., 2004
Now, for each path, and for each activity, we can set up an IF statement to
say whether the path (or activity) was critical for any particular
realization of the model:
A
B
C
D
E
F
G
1 Activity Description
Predecessors Start Node End Node Min Alpha
2
A
Survey Site
None
0
1
2 4.00
3
B
Excavation
A
1
2
9 4.00
4
C
Prepare Drawings
None
0
3
4 3.11
5
D
Soil Study
B
2
3
1
6
E
Prelim. Report
C, D
3
4
1 4.00
7
F
Approve Plans
E
4
5
1
8
G
Concrete Forms
F
5
7
5 2.33
9
H
Procure Steel
F
5
6
2 3.23
10
I
Order Cement
F
5
8
1
11 Dummy
H
6
8
0
12
J
Deliver Gravel
G
7
8
2 2.94
13
K
Pour Concrete
H, I, J
8
9
8 2.94
14
L
Cure Concrete
K
9
10
2
15
M
Strength Test
L
10
11
2
16
17
18 Node
Time
Path
Total
19
0
0
A-B-D-E-F-H-K-L-M
60.67
20
1
5.00
A-C-E-F-H-K-L-M
52.33
21
2
26.00
A-B-D-E-F-I-K-L-M
54.33
22
3
27.00
A-C-E-F-I-K-L-M
46.00
23
4
30.00
A-B-D-E-F-G-J-K-L-M
69.83
24
5
31.00
A-C-E-F-G-J-K-L-M
61.50
25
6
38.33
69.83
Operations -- Prof. Juran
H
I
J
K
Beta Scale CB Time Simulated Time
4.00
2
3.00
5.00
4.00
6
12.00
21.00
4.57 14
9.67
13.67
0
1.00
1.00
4.00
2
2.00
3.00
0
1.00
1.00
4.67
4
6.33
11.33
4.52
8
5.33
7.33
0
1.00
1.00
0
0.00
0.00
4.62
3
3.17
5.17
4.62
6
10.33
18.33
0
2.00
2.00
0
2.00
2.00
L
M
N
Start Time End Time Critical?
0.00
5.00
1
5.00
26.00
1
0.00
13.67
0
26.00
27.00
1
27.00
30.00
1
30.00
31.00
1
31.00
42.33
1
31.00
38.33
0
31.00
32.00
0
38.33
38.33
42.33
47.50
1
47.50
65.83
1
65.83
67.83
1
67.83
69.83
1
O
P
Q
=SUM(H19,H20)
Critical?
0
0
0
0
=IF(G23=$G$25,1,0)
1
0
25
©The McGraw-Hill Companies, Inc., 2004
Here’s a cell to tell whether the project was completed by week 43:
A
8
9
10
11
27
28
29
30
B
47.50
65.83
67.83
69.83
C
D
<= 43?
E
=IF(B30<43,1,0)
0
Here’s a cell to keep track of the penalty (if any) Sharon will have to
pay. Note that we have assumed that the penalty applies continuously
to any part of a week.
28
29
30
A
9
10
11
B
65.83
67.83
69.83
Operations -- Prof. Juran
C
<= 43?
D
Penalty
0 $ 268,333
E
F
G
H
I
=IF(B30>43,10000*(B30-43),0)
26
©The McGraw-Hill Companies, Inc., 2004
Crystal Ball
For each of the random activities, we create an assumption cell, as shown
here for Activity A:
Operations -- Prof. Juran
27
©The McGraw-Hill Companies, Inc., 2004
Here’s the model after doing this for every random activity time (Activities
D, F, I, L, M, and the Dummy activity have no variability):
A
B
C
D
E
F
G
H
I
J
K
1 Activity Description
Predecessors Start Node End Node Min Alpha Beta Scale CB Time Simulated Time
2
A
Survey Site
None
0
1
2 4.00 4.00
2
1.03
3.03
3
B
Excavation
A
1
2
9 4.00 4.00
6
2.97
11.97
4
C
Prepare Drawings
None
0
3
4 3.11 4.57 14
4.41
8.41
5
D
Soil Study
B
2
3
1
0
1.00
1.00
6
E
Prelim. Report
C, D
3
4
1 4.00 4.00
2
1.46
2.46
7
F
Approve Plans
E
4
5
1
0
1.00
1.00
8
G
Concrete Forms
F
5
7
5 2.33 4.67
4
1.54
6.54
9
H
Procure Steel
F
5
6
2 3.23 4.52
8
6.93
8.93
10
I
Order Cement
F
5
8
1
0
1.00
1.00
11 Dummy
H
6
8
0
0
0.00
0.00
12
J
Deliver Gravel
G
7
8
2 2.94 4.62
3
0.41
2.41
13
K
Pour Concrete
H, I, J
8
9
8 2.94 4.62
6
2.57
10.57
14
L
Cure Concrete
K
9
10
2
0
2.00
2.00
15
M
Strength Test
L
10
11
2
0
2.00
2.00
Operations -- Prof. Juran
L
M
N
Start Time End Time Critical?
0.00
3.03
1
3.03
15.00
1
0.00
8.41
0
15.00
16.00
1
16.00
18.45
1
18.45
19.45
1
19.45
25.99
1
19.45
28.39
0
19.45
20.45
0
28.39
28.39
25.99
28.41
1
28.41
38.97
1
38.97
40.97
1
40.97
42.97
1
28
©The McGraw-Hill Companies, Inc., 2004
Now we create forecast cells to track the completion time of the whole
project (B30) as well as the criticalities of the various paths (H19:H24) and
activities (N2:N15).
We also make forecast cells to track whether the project took longer than 43
weeks, and what the penalty was.
A
B
C
D
E
F
G
1 Activity Description
Predecessors Start Node End Node Min Alpha
2
A
Survey Site
None
0
1
2 4.00
3
B
Excavation
A
1
2
9 4.00
4
C
Prepare Drawings
None
0
3
4 3.11
5
D
Soil Study
B
2
3
1
6
E
Prelim. Report
C, D
3
4
1 4.00
7
F
Approve Plans
E
4
5
1
8
G
Concrete Forms
F
5
7
5 2.33
9
H
Procure Steel
F
5
6
2 3.23
10
I
Order Cement
F
5
8
1
11 Dummy
H
6
8
0
12
J
Deliver Gravel
G
7
8
2 2.94
13
K
Pour Concrete
H, I, J
8
9
8 2.94
14
L
Cure Concrete
K
9
10
2
15
M
Strength Test
L
10
11
2
16
17
18 Node
Time
Path
Total
19
0
0
A-B-D-E-F-H-K-L-M
60.67
20
1
5.00
A-C-E-F-H-K-L-M
52.33
21
2
26.00
A-B-D-E-F-I-K-L-M
54.33
22
3
27.00
A-C-E-F-I-K-L-M
46.00
23
4
30.00
A-B-D-E-F-G-J-K-L-M
69.83
24
5
31.00
A-C-E-F-G-J-K-L-M
61.50
25
6
38.33
69.83
26
7
42.33
27
8
47.50
28
9
65.83
29
10
67.83
<= 43?
Penalty
30
11
69.83
0 $ 268,333
Operations -- Prof. Juran
H
I
J
K
Beta Scale CB Time Simulated Time
4.00
2
3.00
5.00
4.00
6
12.00
21.00
4.57 14
9.67
13.67
0
1.00
1.00
4.00
2
2.00
3.00
0
1.00
1.00
4.67
4
6.33
11.33
4.52
8
5.33
7.33
0
1.00
1.00
0
0.00
0.00
4.62
3
3.17
5.17
4.62
6
10.33
18.33
0
2.00
2.00
0
2.00
2.00
L
M
N
Start Time End Time Critical?
0.00
5.00
1
5.00
26.00
1
0.00
13.67
0
26.00
27.00
1
27.00
30.00
1
30.00
31.00
1
31.00
42.33
1
31.00
38.33
0
31.00
32.00
0
38.33
38.33
42.33
47.50
1
47.50
65.83
1
65.83
67.83
1
67.83
69.83
1
Critical?
0
0
0
0
1
0
29
©The McGraw-Hill Companies, Inc., 2004
Question 2: What completion time should Sharon use, if she wants to be 90%
confident?
The best way to answer that is to look at the percentiles for the Project Time
forecast cell:
Operations -- Prof. Juran
30
©The McGraw-Hill Companies, Inc., 2004
Question 3: What is the probability of completion by week 43?
We can answer that using the statistics from the “<= 43?” forecast cell:
Operations -- Prof. Juran
31
©The McGraw-Hill Companies, Inc., 2004
Question 4: Give an estimated probability distribution for the amount of
penalties Sharon will have to pay.
Question 5: What is the expected value of the penalty?
Here’s the frequency chart and the summary statistics:
Operations -- Prof. Juran
32
©The McGraw-Hill Companies, Inc., 2004
Question 6: Which activities are most likely to be on the critical path?
Here are results for the various paths:
Path
A-B-D-E-F-H-K-L-M
A-C-E-F-H-K-L-M
A-B-D-E-F-I-K-L-M
A-C-E-F-I-K-L-M
A-B-D-E-F-G-J-K-L-M
A-C-E-F-G-J-K-L-M
Operations -- Prof. Juran
Estimated Probability of being Critical
0.0010
0.0000
0.0000
0.0000
0.8940
0.1050
33
©The McGraw-Hill Companies, Inc., 2004
Here are results for the various activities, sorted in descending order of
criticality:
Activity
A
E
F
K
L
M
G
J
B
D
C
H
I
Operations -- Prof. Juran
Estimated Probability of being Critical
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
0.9990
0.9990
0.8950
0.8950
0.1050
0.0010
0.0000
34
©The McGraw-Hill Companies, Inc., 2004
Question 7: Compare the PERT results to those you would have found
using (a) basic CPM using the most-likely times, (b) the “by-hand”
PERT method from the textbook, and (c) HOM.
CPM analysis gives a completion time of 42 weeks. The critical path is
A-B-D-E-F-G-J-K-L-M
Activity
Name
========
A
B
C
D
E
F
G
H
I
J
K
L
M
Early
Start
========
0
3
0
15
16
18
19
19
19
25
28
38
40
Operations -- Prof. Juran
Early
Finish
========
3
15
9
16
18
19
25
24
20
28
38
40
42
Late
Start
========
0
3
7
15
16
18
19
23
27
25
28
38
40
Late
Finish
========
3
15
16
16
18
19
25
28
28
28
38
40
42
Slack
========
0
0
7
0
0
0
0
4
8
0
0
0
0
35
©The McGraw-Hill Companies, Inc., 2004
“Textbook” Method
The textbook method involves (a) finding the means and standard deviations
for each path, (b) determining which path has the longest expected total
time, and (c) summing the variances of the activities on that path to get the
variance of the path.
In our case, the longest path would be A-B-D-E-F-G-J-K-L-M, with a mean of
42.83 weeks and a variance of 2.92 weeks.
A
B
C
D
E
F
G
H
I
J
1 Activity Description
Optimistic Pessimistic Most-likely Predecessors Start Node
End Node Mean
StDev
2
A
Survey Site
2
4
3
None
0
1
3.00
0.33
3
B
Excavation
9
15
12
A
1
2
12.00
1.00
4
C
Prepare Drawings
4
18
9
None
0
3
9.67
2.33
5
D
Soil Study
1
1
1
B
2
3
1.00
0.00
6
E
Prelim. Report
1
3
2
C, D
3
4
2.00
0.33
7
F
Approve Plans
1
1
1
E
4
5
1.00
0.00
8
G
Concrete Forms
5
9
6
F
5
7
6.33
0.67
9
H
Procure Steel
2
10
5
F
5
6
5.33
1.33
10
I
Order Cement
1
1
1
F
5
8
1.00
0.00
11 Dummy
0
0
0
H
6
8
0.00
0.00
12
J
Deliver Gravel
2
5
3
G
7
8
3.17
0.50
13
K
Pour Concrete
8
14
10
H, I, J
8
9
10.33
1.00
14
L
Cure Concrete
2
2
2
K
9
10
2.00
0.00
15
M
Strength Test
2
2
2
L
10
11
2.00
0.00
16
17
18
Path
Sum of Means Sum of Variances
19
A-B-D-E-F-H-K-L-M
38.67
20
A-C-E-F-H-K-L-M
35.33
=SUM(I2,I3,I5,I6,I7,I8,I12:I15)
21
A-B-D-E-F-I-K-L-M
34.33
22
A-C-E-F-I-K-L-M
31.00
23
A-B-D-E-F-G-J-K-L-M
42.83
2.92
24
A-C-E-F-G-J-K-L-M
39.50
25
=SUM(J2^2,J3^2,J5^2,J6^2,J7^2,J8^2,J12^2, J13^2,J14^2,J15^2)
26
Operations -- Prof. Juran
36
©The McGraw-Hill Companies, Inc., 2004
Once we have the mean and variance, we can estimate the probability of
finishing by week 43, assuming that the total time is normally distributed:
D
18
19
20
21
22
23
24
25
26
27
E
Path
A-B-D-E-F-H-K-L-M
A-C-E-F-H-K-L-M
A-B-D-E-F-I-K-L-M
A-C-E-F-I-K-L-M
A-B-D-E-F-G-J-K-L-M
A-C-E-F-G-J-K-L-M
F
G
H
Sum of Means Sum of Variances
38.67
35.33
34.33
31.00
42.83
2.92
39.50
90% completion time
Prob(X < 43)
Operations -- Prof. Juran
I
J
1.71
45.022
0.5389
K
L
M
N
=SQRT(H23)
=NORMINV(0.9,G23,I23)
=NORMDIST(43,G23,I23,1)
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©The McGraw-Hill Companies, Inc., 2004
90% Completion Time:
Crystal Ball
Textbook
45.11
46.57
Discussion: These results are consistent with each other; the estimates are both
within a narrow range. The “Textbook” method is based on the assumption that
the probability distribution of the total project time is normal.
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©The McGraw-Hill Companies, Inc., 2004
Probability of Completion by Week 43
Crystal Ball
Textbook
0.553
0.5389
Discussion: Again, the estimates are consistent with each other. The “Textbook”
method, as before, is based on a normal distribution for the total project time.
Operations -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
Expected Penalty
Crystal Ball
Textbook
$6,257
N/A
Discussion: Crystal Ball has a distinct advantage in answering this question; not
only does it provide a precise estimate of the expected penalty, but it also
provides a standard error for this estimate, which would be necessary if we were
interested in constructing a confidence interval around the estimate.
The “Textbook” method cannot be used to answer this question without
employing some difficult calculus on the normal distribution.
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©The McGraw-Hill Companies, Inc., 2004
Criticality
Paths
A-B-D-E-F-H-K-L-M
C-E-F-H-K-L-M
A-B-D-E-F-I-K-L-M
C-E-F-I-K-L-M
A-B-D-E-F-G-J-K-L-M
C-E-F-G-J-K-L-M
Activities
A
B
C
D
E
F
G
H
I
J
K
L
M
Crystal Ball
0.995
0.995
0.005
0.995
1.000
1.000
0.999
0.001
0.000
0.999
1.000
1.000
1.000
Crystal Ball
0.001
0.000
0.000
0.000
0.994
0.005
Textbook
1.000
1.000
0.000
1.000
1.000
1.000
1.000
0.000
0.000
1.000
1.000
1.000
1.000
Discussion: The textbook method assumes that there is only one path that could
be critical (the one with the longest expect total time). With this model, any
discussion of criticality is not very interesting. To the extent that several paths
have the potential to be critical, the textbook method may underestimate the total
time of the project and/or the variability of the project time.
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©The McGraw-Hill Companies, Inc., 2004
Marketing Example:
New Product Development decision
Cavanaugh Pharmaceutical Company (CPC) has enjoyed a monopoly on
sales of its popular antibiotic product, Cyclinol, for several years.
Unfortunately, the patent on Cyclinol is due to expire. CPC is considering
whether to develop a new version of the product in anticipation that one of
CPC’s competitors will enter the market with their own offering.
The decision as to whether or not to develop the new antibiotic (tentatively
called Minothol) depends on several assumptions about the behavior of
customers and potential competitors. CPC would like to make the decision
that is expected to maximize its profits over a ten-year period, assuming a
15% cost of capital.
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©The McGraw-Hill Companies, Inc., 2004
Costs and Revenues
Cyclinol costs $1.00 per dose to manufacture, and sells for $7.50 per dose. The proposed
Minothol product would cost $0.90 per dose and sell for $6.00, allowing CPC to protect its
market share against lower-priced competition. This would, however, require a one-time
investment of $140 million.
Fixed Cost
Variable Cost
Selling Price
Cyclinol
None
$1.00
$7.50
Minothol
$140 million
$0.90
$6.00
Competition
There is really only one other company with the potential to enter the market, Ahrens
MethLabs, Inc. (AMI). Competitive analysis indicates that AMI is 20% likely to introduce a
competing product if CPC stays with the higher-priced Cyclinol product, but only 5%
likely to enter the market if CPC introduces Minothol.
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©The McGraw-Hill Companies, Inc., 2004
Customer Demand
Analysts estimate that the average annual demand over the next ten years
will be normally distributed with a mean of 40 million doses and a
standard deviation of 10 million doses, as shown below.
This demand is believed to be independent of whether CPC introduces
Minothol or whether Cyclinol/Minothol has a competitor.
Average Antibiotic Demand
0.030
0.025
Probability
0.020
0.015
0.010
0.005
0.000
10
25
40
55
Millions of Doses
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©The McGraw-Hill Companies, Inc., 2004
CPC’s market share is expected to be 100% of demand, as long as there is
no competition from AMI. In the event of competition, CPC will still enjoy
a dominant market position because of its superior brand recognition.
However, AMI is likely to price its product lower than CPC’s in an effort
to gain market share. CPC’s best analysis indicates that its share of total
sales, in the event of competition, will be a function of the price it chooses
to charge per dose, as shown below.
Market Share vs. Price
Market Share
100%
80%
60%
40%
20%
0%
$-
$2
$4
$6
$8
$10
$12
$14
Price
The Cyclinol product at $7.50 would only retain a 38.1% market share,
whereas the Minothol product at $6.00 would have a 55.0% market share.
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©The McGraw-Hill Companies, Inc., 2004
Questions
What is the best decision for CPC, in terms of maximizing the expected
value of its profits over then next ten years?
What is the least risky decision, using the standard deviation of the
ten-year profit as a measure of risk?
What is the probability that introducing Minothol will turn out to be
the best decision?
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©The McGraw-Hill Companies, Inc., 2004
U~(0, 1)
(whether or not AMI
enters market)
N~(40, 10)
(Total market
demand)
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
B
0.70928
Price
Total Demand
P(Competition)
Competition?
Market Share
Cyclinol Units Sold
Revenue
Fixed Cost
Variable Cost
Annual profit
Discount Rate
10-year PV
Results
Cyclinol
Minothol
Minothol Better?
Income statement-like calculations
for each of four scenarios
C
Cyclinol
No Competition
Competition
$
7.50 $
7.50
40.2
20%
No
100.0%
38.1%
40.2
15.3
$
301.28 $
114.86
$
$
$
1.00 $
1.00
$
261.11 $
99.55
15%
$1,310.45
$499.61
$
$
D
E
Minothol
No Competition
Competition
$
6.00 $
6.00
5%
No
$
$
$
$
100.0%
40.2
241.02
140.00
0.90
204.87
$888.20
$
$
$
$
55.0%
22.1
132.56
140.00
0.90
112.68
$425.51
1,310.45
888.20
0
3 Forecasts:
NPV in $millions for each decision
Yes/No New Product Better
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©The McGraw-Hill Companies, Inc., 2004
Summary
Advanced Simulation Applications
• Beta Distribution
• Operations
– Project Management (PERT)
• “Textbook” method
• Crystal Ball
• Marketing
– New Product Development decision
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©The McGraw-Hill Companies, Inc., 2004