08a - NYU Stern

Report
Session 8a
Overview
Operations Simulation Models
• Reliability Analysis
– RANK, VLOOKUP, MIN
• Inventory Order Quantities
– Single Product (MAX)
– Multiple Products with Correlated Demand
• Project management
– PERT Analysis
Decision Models -- Prof. Juran
2
Example 1: Reliability
Consider a device that requires two batteries to function. If either
of these batteries dies, the device will not work. Currently there
are two brand new batteries in the device, and there are three
extra brand new batteries. Each battery, once it is placed in the
device, lasts a random amount of time that is lognormally
distributed with mean 20 hours and standard deviation 5 hours.
When any of the batteries in the device dies, it is immediately
replaced by an extra (if an extra is still available).
Simulate the time the device can last with the batteries currently
available. What is the estimated probability that the device will
last longer than 50 hours?
Decision Models -- Prof. Juran
3
A
B
C
D
Start Rank
1
2
3
4
5
E
Fail Rank
2
1
4
3
5
F
G
H
Battery
Life
Start Time
=RANK(I2,$I$2:$I$6,1)
Battery A 28.23
0.0
Battery B 26.41
0.0
Battery C 25.67
26.4
Battery D 22.52
28.2
Battery E 28.99
50.7
I
Fail Time
28.2
26.4
52.1
50.7
79.7
J
1
istribution 2of battery lifetimes (lognormal)
3
Mean
20
4
Stdev
5
5
6
=F2
=F3
7
Simulation
=VLOOKUP(F4,$B$10:$C$12,2,0)
=H4+G4
8
Failure#
Insert Battery Current time In Position 1 In Position 2
9
0
Battery A
Battery B
=F10
=(D10=F2)*(I3)+(E10=E9)*(I2)
10
1
Battery C
26.407
Battery A
Battery C
26.41
=IF(MIN(I2:I3)=I3,$F$4,$F$3)
11
2
Battery D
28.229 =IF(MIN(I2:I3)=I2,$F$4,$F$2)
Battery D
Battery C
28.23
12
3
Battery E
50.750
Battery E
Battery C
50.75
13
4
(none)
52.077
Battery E
(none)
52.08
<---- Device Fails
14
=IF(((VLOOKUP(E12,$F$2:$I$6,4,0))<(VLOOKUP(D12,$F$2:$I$6,4,0))),(D12),(B13))
15
16
=IF(((VLOOKUP(E12,$F$2:$I$6,4,0))<(VLOOKUP(D12,$F$2:$I$6,4,0))),(B13),(E12))
17
18
=(E13=B13)*(VLOOKUP(E12,$F$2:$I$6,4,0))+(E13=E12)*(VLOOKUP(D12,$F$2:$I$6,4,0))
19
Decision Models -- Prof. Juran
4
A
B
C
D
Start Rank
1
2
3
4
5
E
Fail Rank
1
2
3
4
5
F
Battery
Battery A
Battery B
Battery C
Battery D
Battery E
1
istribution2 of battery lifetimes (lognormal)
3
Mean
20
4
Stdev
5
5
6
7
Simulation
8
Failure#
Insert Battery Current time In Position 1 In Position 2
9
0
Battery A
Battery B
10
1
Battery C
13.405
Battery C
Battery B
13.40
11
2
Battery D
18.219
Battery C
Battery D
18.22
12
3
Battery E
28.627
Battery E
Battery D
28.63
13
4
(none)
49.767
Battery E
(none)
49.77
14
15
Battery D
Decision Models -- Prof. Juran
G
Life
13.40
18.22
15.22
31.55
26.09
H
Start Time
0.0
0.0
13.4
18.2
28.6
I
Fail Time
13.4
18.2
28.6
49.8
54.7
=LARGE(I2:I6,2)
<---- Device Fails
=VLOOKUP(4,E2:F6,2)
5
B3:B4 contain the parameters for the random lifetimes
of batteries.
D1:I6 keep track of all five batteries, including when
they start (and in what order) and when they fail (and
in what order). Note that the random variables are
G2:G6, which will be simulated by Crystal Ball. These
are the “lifetimes” of the batteries.
Decision Models -- Prof. Juran
6
We calculate the fail time of a battery by using the
formula:
Fail time = Start time + Lifetime
Decision Models -- Prof. Juran
7
For example, in this realization, Batteries A and B both start at time 0.
Battery B fails at time 26.41, and is replaced by Battery C (see row 10).
Battery A fails at time 28.23 and is replaced by Battery D (row 11).
Battery D fails at time 50.75 and is replaced by E (D started at time
28.23 and lasted 22.52; see row 12).
Finally, Battery C fails at time 52.08 (it started at time 26.41 and lasted
25.67; see row 13). At this point the device fails, having lasted 52.08
hours.
Decision Models -- Prof. Juran
8
Decision Models -- Prof. Juran
9
Decision Models -- Prof. Juran
10
A
B
C
D
E
1
Start Rank
Fail Rank
stribution2of battery lifetimes (lognormal)
1
2
3
Mean
20
2
1
4
Stdev
5
3
3
5
4
5
6
5
4
7 Simulation
8
Failure#
Insert Battery Current time In Position 1 In Position 2
9
0
Battery A
Battery B
10
1
Battery C
12.017
Battery A
Battery C
11
2
Battery D
25.275
Battery D
Battery C
12
3
Battery E
29.016
Battery D
Battery E
13
4
(none)
52.542
Battery D
(none)
Decision Models -- Prof. Juran
F
Battery
Battery A
Battery B
Battery C
Battery D
Battery E
G
Life
25.28
12.02
17.00
33.15
23.53
H
Start Time
0.0
0.0
12.0
25.3
29.0
12.02
25.28
29.02
52.54
<----
Device Fails
I
Fail Time
25.3
12.0
29.0
58.4
52.5
11
Decision Models -- Prof. Juran
12
Decision Models -- Prof. Juran
13
A confidence interval for the mean time to failure can be calculated as follows:
X
43.00
 z 2
s
n
 1.96
5.41
1000
 1.960.1711
 0.3353
Or (42.66, 43.34)
Note that the Crystal Ball output gives both the standard deviation and the
standard error.
Decision Models -- Prof. Juran
14
Using the frequency chart’s “certainty” feature, we can drag the lower
“grabber” to a point where the left window reads 50 (or Crystal Ball will
allow you simply to type 50 into the left window).
The certainty window indicates the estimated area under the curve in the
blue shaded region. In this case, we estimate a 89.89% probability that the
device will fail before 50 hours.
Decision Models -- Prof. Juran
15
Example 2: Blockbuster Publishers
At present, 2000 copies of the book are in stock, and Blockbuster
must determine how many copies of the book to print for the next
year.
Demand has a triangular distribution with parameters 4000, 6000,
and 9000.
Each copy sold during brings the publisher a revenue of $35.
Any copies left at the end of the year can be sold for $5.
The cost of printing the book is $50,000 plus $15 per book printed.
Decision Models -- Prof. Juran
16
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
A
Number to print
Number on hand after printing
B
Demand
Sold at 35
Sold at 5
Fixed cost
Variable cost
Revenue
Profit
Probability distribution of demand (triangular)
Min
Most likely
Max
0
2000
6000
2000
0
$50,000
$0
$70,000
$20,000
C
D
E
F
G
H
I
1000
2000
3000
4000
5000
6000
7000
3000
4000
5000
6000
7000
8000
9000
6000
6000
6000
6000
6000
6000
6000
3000
4000
5000
6000
6000
6000
6000
0
0
0
0
1000
2000
3000
$50,000 $50,000 $50,000 $50,000 $50,000 $50,000 $50,000
$15,000 $30,000 $45,000 $60,000 $75,000 $90,000 $105,000
$105,000 $140,000 $175,000 $210,000 $215,000 $220,000 $225,000
$40,000 $60,000 $80,000 $100,000 $90,000 $80,000 $70,000
J
K
L
=I1+$E$11
=$B$3
=MIN(I2,I3)
=MAX(0,I2-I3)
=$E$15
=$E$14*I1
=I4*$E$13+I5*$E$16
=I8-(I6+I7)
Current inventory
2000
4000
6000 Selling price per book
$35
9000 Variable cost per book
$15
Fixed cost of a printing $50,000
Reduced selling price
$5
Decision Models -- Prof. Juran
17
Somewhat arbitrarily, we have decided to consider
possible order quantities of 0, 1000, 2000, 3000, 4000, 5000,
6000, and 7000 (B1:I1).
For each possible order quantity, we have a fairly simple
income statement-like analysis in rows 2 through 9.
All columns get their demand number from B3. (Note that
the demand level in B3 will be a random variable later.)
The lower part of the spreadsheet (rows 11 through 16)
contains the basic parameters of the problem.
Decision Models -- Prof. Juran
18
Decision Models -- Prof. Juran
19
Decision Models -- Prof. Juran
20
We run for 1000 trials, and then click on the extract data button.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
B
C
D
E
F
G
H
I
0 Profit
1 0 0 0 Profit
2 0 0 0 Profit
3 0 0 0 Profit
4 0 0 0 Profit
5 0 0 0 Profit
6 0 0 0 Profit
7 0 0 0 Profit
1000
$20,000
$20,000
$20,000
$0
$0
0.00
+Infinity
0.00
$20,000
$20,000
$0
$0.00
1000
$40,000
$40,000
$40,000
$0
$0
0.00
+Infinity
0.00
$40,000
$40,000
$0
$0.00
1000
$60,000
$60,000
$60,000
$0
$0
0.00
+Infinity
0.00
$60,000
$60,000
$0
$0.00
Decision Models -- Prof. Juran
1000
$79,179
$80,000
$80,000
$3,382
$11,435,323
-4.81
26.83
0.04
$54,823
$80,000
$25,177
$106.94
1000
$92,260
$100,000
$100,000
$12,823
$164,429,426
-1.65
4.78
0.14
$44,823
$100,000
$55,177
$405.50
1000
$94,734
$97,222
$120,000
$22,914
$525,055,988
-0.53
2.20
0.24
$34,823
$120,000
$85,177
$724.61
1000
$89,364
$87,222
$140,000
$29,083
$845,815,561
0.07
2.15
0.33
$24,823
$140,000
$115,177
$919.68
1000
$80,134
$77,222
--$30,596
$936,106,979
0.27
2.47
0.38
$14,823
$158,923
$144,100
$967.53
21
Efficient Frontier
$100,000
$90,000
Order 4000
$80,000
Order 5000
Order 6000
Expected Profit
$70,000
Order 3000
Order 7000
Order 2000
$60,000
$50,000
Order 1000
$40,000
$30,000
Order 0
$20,000
$10,000
$0
$0
$5,000
$10,000
$15,000
$20,000
$25,000
$30,000
$35,000
Risk
Decision Models -- Prof. Juran
22
Does your answer change if 4000 copies are currently in stock?
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
A
Number to print
Number on hand after printing
B
Demand
Sold at 35
Sold at 5
Fixed cost
Variable cost
Revenue
Profit
Probability distribution of demand (triangular)
Min
Most likely
Max
Decision Models -- Prof. Juran
0
4000
6000
4000
0
$50,000
$0
$140,000
$90,000
C
D
E
F
G
H
I
1000
2000
3000
4000
5000
6000
7000
5000
6000
7000
8000
9000
10000
11000
6000
6000
6000
6000
6000
6000
6000
5000
6000
6000
6000
6000
6000
6000
0
0
1000
2000
3000
4000
5000
$50,000 $50,000 $50,000 $50,000 $50,000 $50,000 $50,000
$15,000 $30,000 $45,000 $60,000 $75,000 $90,000 $105,000
$175,000 $210,000 $215,000 $220,000 $225,000 $230,000 $235,000
$110,000 $130,000 $120,000 $110,000 $100,000 $90,000 $80,000
Current inventory
4000
4000
6000 Selling price per book
$35
9000 Variable cost per book
$15
Fixed cost of a printing $50,000
Reduced selling price
$5
23
Efficient Frontier
$140,000
$120,000
Order 3000
Order 2000
$100,000
Expected Profit
Order 4000
Order 1000
Order 5000
Order 6000
$80,000
Order
0
Order 7000
$60,000
$40,000
$20,000
$0
$0
$5,000
$10,000
$15,000
$20,000
$25,000
$30,000
$35,000
Risk
Decision Models -- Prof. Juran
24
Example 3: Walton Bookstore
Consider two competing products sold by a company. Sales of
either product tend to take away sales from the other product.
That is, demands for the two products are negatively correlated.
The company first places an order for each product. Then during a
period of time, there is demand D1 for product 1 and demand D2
for product 2.
These demands are normally distributed with means 1000 and
1200 and standard deviations 250 and 350. The correlation
between D1 and D2 is , where  is a negative number between –1
and 0.
Decision Models -- Prof. Juran
25
The unit cost of each product is $7.50, the unit price for each
product is $10, and the unit refund for any unit of either product
not sold is $2.50.
The company must decide how many units of each product to
order.
Use simulation to help the company by experimenting with
different order quantities. Try this for  = -0.3,  = -0.5, and  = 0.7. What recommendations can you give about the “best” order
quantities as the demands become more highly correlated (in a
negative direction)?
Decision Models -- Prof. Juran
26
We have laid out (somewhat arbitrarily) 16 different combinations of order
quantities for the two products (B2:Q3). Each of the columns from B to Q
represents a model of one order quantity strategy.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
A
Strategy
Product 1 ordered
Product 2 ordered
Demand 1
Demand 2
1 sold full price
2 sold full price
1 sold at refund price
2 sold at refund price
Full-price revenue
Refund revenue
Order cost
Profit
$
$
$
$
Unit price
Unit cost
Unit refund value
Product 1 Product 2
$10.00
$10.00
$7.50
$7.50
$2.50
$2.50
Product 1
Product 2
B
1
700
900
1000
1200
700
900
0
0
16,000
12,000
4,000
C
$
$
$
$
D
E
F
2
3
4
5
700
700
700
800
1000
1100
1200
900
1000
1000
1000
1000
1200
1200
1200
1200
=MIN(B2,B4)
700
700
700
800
=MIN(B3,B5)
1000
1100
1200
900
=MAX(0,B2-B4)
0
0
0
0
=MAX(0,B3-B5)
0
0
0
0
=SUMPRODUCT(B6:B7,$B21:$B22)
17,000
$ 18,000 $ 19,000 $ 17,000
=SUMPRODUCT(B8:B9,$D21:$D22)
$
$
$
=SUMPRODUCT(B2:B3,$C21:$C22)
12,750
$ 13,500 $ 14,250 $ 12,750
=B10+B11-B12
4,250 $ 4,500 $ 4,750 $ 4,250
Means
Stdevs
1000
250
G
H
I
J
K
L
M
N
O
P
Q
6
800
1000
1000
1200
800
1000
0
0
$ 18,000
$
$ 13,500
$ 4,500
7
800
1100
1000
1200
800
1100
0
0
$ 19,000
$
$ 14,250
$ 4,750
8
800
1200
1000
1200
800
1200
0
0
$ 20,000
$
$ 15,000
$ 5,000
9
900
900
1000
1200
900
900
0
0
$ 18,000
$
$ 13,500
$ 4,500
10
900
1000
1000
1200
900
1000
0
0
$ 19,000
$
$ 14,250
$ 4,750
11
900
1100
1000
1200
900
1100
0
0
$ 20,000
$
$ 15,000
$ 5,000
12
900
1200
1000
1200
900
1200
0
0
$ 21,000
$
$ 15,750
$ 5,250
13
1000
900
1000
1200
1000
900
0
0
$ 19,000
$
$ 14,250
$ 4,750
14
1000
1000
1000
1200
1000
1000
0
0
$ 20,000
$
$ 15,000
$ 5,000
15
1000
1100
1000
1200
1000
1100
0
0
$ 21,000
$
$ 15,750
$ 5,250
16
1000
1200
1000
1200
1000
1200
0
0
$ 22,000
$
$ 16,500
$ 5,500
1200
350
Corr
-0.3
Price
Cost
Refund
$10.00
$7.50
$2.50
$10.00
$7.50
$2.50
Decision Models -- Prof. Juran
27
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
B
1
700
900
1000
1200
700
900
0
0
16,000
12,000
4,000
C
Strategy
Product 1 ordered
Product 2 ordered
Demand 1
Demand 2
1 sold full price
2 sold full price
1 sold at refund price
2 sold at refund price
Full-price revenue
Refund revenue
Order cost
Profit
$
$
$
$
Unit price
Unit cost
Unit refund value
Product 1 Product 2
$10.00
$10.00
$7.50
$7.50
$2.50
$2.50
$
$
$
$
D
E
F
2
3
4
5
700
700
700
800
1000
1100
1200
900
1000
1000
1000
1000
1200
1200
1200
1200
=MIN(B2,B4)
700
700
700
800
=MIN(B3,B5)
1000
1100
1200
900
=MAX(0,B2-B4)
0
0
0
0
=MAX(0,B3-B5)
0
0
0
0
=SUMPRODUCT(B6:B7,$B21:$B22)
17,000
$ 18,000 $ 19,000 $ 17,000
=SUMPRODUCT(B8:B9,$D21:$D22)
$
$
$
=SUMPRODUCT(B2:B3,$C21:$C22)
12,750
$ 13,500 $ 14,250 $ 12,750
=B10+B11-B12
4,250 $ 4,500 $ 4,750 $ 4,250
Means
Stdevs
1000
250
G
6
800
1000
1000
1200
800
1000
0
0
$ 18,000
$
$ 13,500
$ 4,500
1200
350
Price
Cost
Refund
Product 1
$10.00
$7.50
$2.50
Product 2
$10.00
$7.50
$2.50
Decision Models -- Prof. Juran
28
The random variables (demand D1 for product 1 and demand D2 for product 2)
appear in rows 3 and 4. Note that the demand stream for all strategies is the
same, because each value in rows 3 and 4 refers to B4:B5.
The correlation between D1 and D2 () appears in I18.
Here are the two assumption cells:
Decision Models -- Prof. Juran
29
You can specify bivariate correlations in
the Define Assumption window.
For more than a few correlated green cells,
it’s more efficient to use the matrix view.
Decision Models -- Prof. Juran
30
Click on “correlate” to enter the correlation information.
Decision Models -- Prof. Juran
31
Switch to “list” view (not the default matrix
view).
Enter coefficient by typing, cell reference, or the slider.
Decision Models -- Prof. Juran
32
Decision Models -- Prof. Juran
33
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
A
Strategy
Product 1 ordered
Product 2 ordered
Demand 1
Demand 2
1 sold full price
2 sold full price
1 sold at refund price
2 sold at refund price
Full-price revenue
Refund revenue
Order cost
Profit
Unit price
Unit cost
Unit refund value
Product 1
Product 2
B
$
$
$
$
1
700
900
1000
1200
700
900
0
0
16,000
12,000
4,000
Product 1
$10.00
$7.50
$2.50
C
$
$
$
$
2
700
1000
1000
1200
700
1000
0
0
17,000
12,750
4,250
D
$
$
$
$
3
700
1100
1000
1200
700
1100
0
0
18,000
13,500
4,500
Product 2
$10.00
$7.50
$2.50
E
$
$
$
$
4
700
1200
1000
1200
700
1200
0
0
19,000
14,250
4,750
Means
Stdevs
F
$
$
$
$
G
5
800
900
1000
1200
800
900
0
0
17,000
12,750
4,250
1000
250
$
$
$
$
6
800
1000
1000
1200
800
1000
0
0
18,000
13,500
4,500
1200
350
H
$
$
$
$
7
800
1100
1000
1200
800
1100
0
0
19,000
14,250
4,750
I
$
$
$
$
J
8
800
1200
1000
1200
800
1200
0
0
20,000
15,000
5,000
$
$
$
$
9
900
900
1000
1200
900
900
0
0
18,000
13,500
4,500
K
$
$
$
$
10
900
1000
1000
1200
900
1000
0
0
19,000
14,250
4,750
L
$
$
$
$
11
900
1100
1000
1200
900
1100
0
0
20,000
15,000
5,000
M
$
$
$
$
12
900
1200
1000
1200
900
1200
0
0
21,000
15,750
5,250
N
$
$
$
$
O
13
1000
900
1000
1200
1000
900
0
0
19,000
14,250
4,750
Corr
-0.3
Price
Cost
Refund
$10.00
$7.50
$2.50
$10.00
$7.50
$2.50
Decision Models -- Prof. Juran
34
$
$
$
$
14
1000
1000
1000
1200
1000
1000
0
0
20,000
15,000
5,000
$3,900
Expected Profit
(Correlation = -0.3)
$3,850
6
$1,700
Std Dev of Profit
(Correlation = -0.3)
10
$1,600
$3,800
9
$1,500
$3,750
$1,400
5
$3,700
$3,650
$1,300
Expected
Profit
10
$1,200
$3,600
$1,100
6
$3,550
1
9
$3,500
$1,000
$3,450
$900
5
$3,400
1000
1200
Product 2
Ordered
Product 1
Ordered
800
1000
700
Product 2
Ordered
$800
1000
1200
900
1100
Std Dev
of Profit
1
900
1100
Product 1
Ordered
800
1000
700
900
900
$3,900
Expected Profit
(Correlation = -0.5)
$1,700
Std Dev of Profit
(Correlation = -0.5)
$3,850
$1,600
10
$3,800
$1,500
6
$3,750
9
$1,400
$3,700
5
$3,650
$1,300
Expected
Profit
10
$1,200
$3,600
6
$3,550
$1,100
9
$1,000
$3,500
1
$3,450
$900
5
$3,400
Product 2
Ordered
$800
1
1000
1200
Product 1
Ordered
800
1000
700
1000
1200
900
1100
Std Dev
of Profit
Product 2
Ordered
900
900
1100
Product 1
Ordered
800
1000
700
900
$3,900
Expected Profit
(Correlation = -0.7)
$1,700
Std Dev of Profit
(Correlation = -0.7)
$3,850
$1,600
$3,800
$1,500
10
$3,750
$1,400
6
$3,700
9
$3,650
5
$1,300
Expected
Profit
10
$1,200
$3,600
6
Std Dev
of Profit
$1,100
$3,550
9
$1,000
$3,500
1
$3,450
$900
5
$3,400
1000
1200
Product 2
Ordered
900
1100
800
1000
700
Product 1
Ordered
900
Decision Models -- Prof. Juran
$800
1
1000
1200
Product 2
Ordered
900
1100
800
1000
700
Product 1
Ordered
900
35
Correlation = -0.3
$3,900
Strategy 10
$3,800
Strategy 6
Strategy 9
Expected Value
Strategy 5
$3,700
Strategy 8
Strategy 16
Strategy 1
$3,600
Strategy 4
$3,500
$3,400
$800
$900
$1,000
$1,100
$1,200
$1,300
$1,400
$1,500
$1,600
$1,700
Std Deviation
Decision Models -- Prof. Juran
36
Correlation = -0.5
$3,900
Strategy 10
$3,800
Strategy 6
Expected Value
Strategy 9
$3,700
Strategy 8
Strategy 5
Strategy 16
$3,600
Strategy 1
Strategy 4
$3,500
$3,400
$800
$900
$1,000
$1,100
$1,200
$1,300
$1,400
$1,500
$1,600
$1,700
Std Deviation
Decision Models -- Prof. Juran
37
Correlation = -0.7
$3,900
$3,800
Strategy 10
Expected Value
$3,700
Strategy 9
Strategy 6
$3,600
Strategy 5
Strategy 8
Strategy 16
$3,500
Strategy 1
Strategy 4
$3,400
$800
$900
$1,000
$1,100
$1,200
$1,300
$1,400
$1,500
$1,600
$1,700
Std Deviation
Decision Models -- Prof. Juran
38
Beta Distribution
The Beta distribution is a continuous probability
distribution defined by four parameters:
Parameter
Min
Max
Alpha ( α)
Beta (β)
Decision Models -- Prof. Juran
Description
Minimum Value
Maximum Value
Shape Factor
Shape Factor
Characteristics
Any number -∞ to ∞
Any number -∞ to ∞
Must be > 0
Must be > 0
39
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
0.0100
0.0050
0.0050
0.0050
0.0000
0.0000
0.0000
0
25
50
75
0
25
50
75
0.0050
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
0.0100
0.0050
0.0000
0
Decision Models -- Prof. Juran
25
50
75
0.0000
0
25
50
75
40
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.
Decision Models -- Prof. Juran
41
Mean:
  min 

 max - min 
   
   
2




max
min

Standard Deviation:
   2      1
Decision Models -- Prof. Juran
(i)
(ii)
42
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.
Decision Models -- Prof. Juran
43
3-step Procedure
1. Get estimates for the optimistic (minimum), most-likely (mode), and pessimistic
(maximum) completion times 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 
Decision Models -- Prof. Juran
(v)
(vi)
44
Beta Distributions in Crystal Ball
Decision Models -- Prof. Juran
45
Beta Distributions in Crystal Ball
Decision Models -- Prof. Juran
46
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.
Decision Models -- Prof. Juran
47
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
Decision Models -- 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
48
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, (b) the “by-hand” PERT method from
the textbook, and (c) HOM.
Decision Models -- Prof. Juran
49
Here’s an activity-on-arc diagram of the problem:
Decision Models -- Prof. Juran
50
To use a beta distribution, we would start a spreadsheet model like this,
calculating the mean and standard deviation using the PERT formulas:
A
B
C
1 Activity
Description
Predecessors
2
A
Survey Site
None
3
B
Excavation
A
4
C
Prepare Drawings
None
5
D
Soil Study
B
6
E
Prelim. Report
C, D
7
F
Approve Plans
E
8
G
Concrete Forms
F
9
H
Procure Steel
F
10
I
Order Cement
F
11 Dummy
H
12
J
Deliver Gravel
G
13
K
Pour Concrete
H, I, J
14
L
Cure Concrete
K
15
M
Strength Test
L
Decision Models -- Prof. Juran
D
Start Node
0
1
0
2
3
4
5
5
5
6
7
8
9
10
E
F
G
End Node Min Mode
1
2
3
2
9
12
3
4
9
3
1
1
4
1
2
5
1
1
7
5
6
6
2
5
8
1
1
8
0
0
8
2
3
9
8
10
10
2
2
11
2
2
H
Max
4
15
18
1
3
1
9
10
1
0
5
14
2
2
I
Mean
3
12
9.667
1
2
1
6.333
5.333
1
0
3.167
10.33
2
2
J
K
L
StDev
0.333
1.000
=(F4+4*G4+H4)/6
2.333
=(H5-F5)/6
0.000
0.333
0.000
0.667
1.333
0.000
0.000
0.500
1.000
0.000
0.000
51
Calculating shape and scale parameters:
A
B
C
1 Activity
Description
Predecessors
2
A
Survey Site
None
3
B
Excavation
A
4
C
Prepare Drawings
None
5
D
Soil Study
B
6
E
Prelim. Report
C, D
7
F
Approve Plans
E
8
G
Concrete Forms
F
9
H
Procure Steel
F
10
I
Order Cement
F
11 Dummy
H
12
J
Deliver Gravel
G
13
K
Pour Concrete
H, I, J
14
L
Cure Concrete
K
15
M
Strength Test
L
D
Start Node
0
1
0
2
3
4
5
5
5
6
7
8
9
10
Decision Models -- Prof. Juran
E
F
G
End Node Min Mode
1
2
3
2
9
12
3
4
9
3
1
1
4
1
2
5
1
1
7
5
6
6
2
5
8
1
1
8
0
0
8
2
3
9
8
10
10
2
2
11
2
2
H
Max
4
15
18
1
3
1
9
10
1
0
5
14
2
2
I
Mean
3
12
9.667
1
2
1
6.333
5.333
1
0
3.167
10.33
2
2
J
StDev
0.333
1.000
2.333
0.000
0.333
0.000
0.667
1.333
0.000
0.000
0.500
1.000
0.000
0.000
K
Alpha
4.000
4.000
3.106
L
M
N
O
P
Beta
4.000
=((I3-F3)/(H3-F3))*((((I3-F3)*(H3-I3))/(J3^2))-1)
4.000
4.568
4.000
4.000
2.333
3.229
4.667
4.521
2.938
2.938
4.617
4.617
=((H6-I6)/(I6-F6))*K6
52
Model Overview
A section for simulating the times of the activities
A
B
1 Activity
Description
2
A
Survey Site
3
B
Excavation
4
C
Prepare Drawings
5
D
Soil Study
6
E
Prelim. Report
7
F
Approve Plans
8
G
Concrete Forms
9
H
Procure Steel
10
I
Order Cement
11 Dummy
12
J
Deliver Gravel
13
K
Pour Concrete
14
L
Cure Concrete
15
M
Strength Test
16
17
18 Node
Time
19
0
0
20
1
3.00
21
2
15.00
22
3
16.00
23
4
18.00
24
5
19.00
25
6
24.00
26
7
25.00
27
8
28.00
28
9
38.00
29
10
40.00
30
11
42.00
C
Predecessors
None
A
None
B
C, D
E
F
F
F
H
G
H, I, J
K
L
D
E
F
G
Start Node End Node Min Mode
0
1
2
3
1
2
9
12
0
3
4
9
2
3
1
1
3
4
1
2
4
5
1
1
5
7
5
6
5
6
2
5
5
8
1
1
6
8
0
0
7
8
2
3
8
9
8
10
9
10
2
2
10
11
2
2
Path
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
Max Path
<= 43?
Penalty
1 $
-
A section
to keep
track of
and when it occurs
Decision
Models
-- each
Prof.node
Juran
H
Max
4
15
18
1
3
1
9
10
1
0
5
14
2
2
I
Simulated Time
3
12
9
1
2
1
6
5
1
0
3
10
2
2
J
K
L
Start Time End Time Critical?
0.00
3.00
1
3.00
15.00
1
0.00
9.00
0
15.00
16.00
1
16.00
18.00
1
18.00
19.00
1
19.00
25.00
1
19.00
24.00
0
19.00
20.00
0
24.00
24.00
25.00
28.00
1
28.00
38.00
1
38.00
40.00
1
40.00
42.00
1
Total Critical?
38.00
0
31.00
0
34.00
0
27.00
0
42.00
1
35.00
0
42.00
A section to keep track of each
path through the network, to
identify the critical path in each
simulated project completion 53
Here’s the section keeping track of the activity times. The numbers in
column I will be CB assumption cells.
A
B
C
1 Activity
Description
Predecessors
2
A
Survey Site
None
3
B
Excavation
A
4
C
Prepare Drawings
None
5
D
Soil Study
B
6
E
Prelim. Report
C, D
7
F
Approve Plans
E
8
G
Concrete Forms
F
9
H
Procure Steel
F
10
I
Order Cement
F
11 Dummy
H
12
J
Deliver Gravel
G
13
K
Pour Concrete
H, I, J
14
L
Cure Concrete
K
15
M
Strength Test
L
Decision Models -- Prof. Juran
D
Start Node
0
1
0
2
3
4
5
5
5
6
7
8
9
10
E
F
G
End Node Min Mode
1
2
3
2
9
12
3
4
9
3
1
1
4
1
2
5
1
1
7
5
6
6
2
5
8
1
1
8
0
0
8
2
3
9
8
10
10
2
2
11
2
2
H
Max
4
15
18
1
3
1
9
10
1
0
5
14
2
2
I
Simulated Time
3.00
12.00
9.67
1.00
2.00
1.00
6.33
5.33
1.00
0.00
3.17
10.33
2.00
2.00
54
For each of the random activities, we create an assumption cell, as shown
here for Activity A:
Decision Models -- Prof. Juran
55
Decision Models -- Prof. Juran
56
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.
Decision Models -- Prof. Juran
57
Example: Activity C
A
B
1 Activity
Description
2
A
Survey Site
3
B
Excavation
4
C
Prepare Drawings
5
D
Soil Study
6
E
Prelim. Report
7
F
Approve Plans
8
G
Concrete Forms
9
H
Procure Steel
10
I
Order Cement
11 Dummy
12
J
Deliver Gravel
13
K
Pour Concrete
14
L
Cure Concrete
15
M
Strength Test
16
17
18 Node
Time
19
0
0
20
1
2.87
21
2
13.39
22
3
14.39
23
4
16.43
24
5
17.43
25
6
21.05
26
7
23.43
27
8
26.14
28
9
36.20
29
10
38.20
30
11
40.20
C
Predecessors
None
A
None
B
C, D
E
F
F
F
H
G
H, I, J
K
L
D
E
F
G
Start Node End Node Min Mode
0
1
2
3
1
2
9
12
0
3
4
9
2
3
1
1
3
4
1
2
4
5
1
1
5
7
5
6
5
6
2
5
5
8
1
1
6
8
0
0
7
8
2
3
8
9
8
10
9
10
2
2
10
11
2
2
Path
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
=MAX(K4,K5) 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
Max Path
<= 43?
H
Max
4
15
18
1
3
1
9
10
1
0
5
14
2
2
I
Simulated Time
2.87
10.52
6.19
1.00
2.04
1.00
6.00
3.61
1.00
0.00
2.71
10.07
2.00
2.00
J
K
L
M
N
Start Time End Time Critical?
0.00
2.87
1
=J4+I4
2.87
13.39
1
0.00
6.19
0
13.39
14.39
1
14.39
16.43
1
=VLOOKUP(D4,$A$19:$B$30,2,0)
16.43
17.43
1
17.43
23.43
1
17.43
21.05
0
17.43
18.43
0
21.05
21.05
23.43
26.14
1
26.14
36.20
1
36.20
38.20
1
38.20
40.20
1
Total Critical?
35.11
0
26.92
0
32.50
0
24.30
0
40.20
1
32.01
0
40.20
Penalty
1 $
-
Decision Models -- Prof. Juran
58
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
D
E
F
G
H
I
J
K
paths:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Start Node End Node Min Mode
0
1
2
3
1
2
9
12
0
3
4
9
2
3
1
1
3
4
1
2
4
5
1
1
5
7
5
6
5
6
2
5
5
8
1
1
6
8
0
0
7
8
2
3
8
9
8
10
9
10
2
2
10
11
2
2
Path
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
Max Path
Decision Models -- Prof. Juran
Max
4
15
18
1
3
1
9
10
1
0
5
14
2
2
Simulated Time
3.00
12.00
9.00
1.00
2.00
1.00
6.00
5.00
1.00
0.00
3.00
10.00
2.00
2.00
Start Time End Time
0.00
3.00
3.00
15.00
0.00
9.00
15.00
16.00
16.00
18.00
18.00
19.00
19.00
25.00
19.00
24.00
19.00
20.00
24.00
24.00
25.00
28.00
28.00
38.00
38.00
40.00
40.00
42.00
Total Critical?
38.00
0
31.00 =SUM(I4,I6,I7,I9,I13,I14,I15)
0
34.00
0
27.00
0 =IF(G22=$G$25,1,0)
42.00
1
35.00
0
=MAX(G19:G24)
42.00
59
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:
Decision Models -- Prof. Juran
60
Now, for each activity, we can set up an IF statement to say whether the activity
was critical for any particular realization of the model. Note that Activity H
(Procure Steel, in row 9) is part of two paths (A-B-D-E-F-H-K-L-M, in row 19, and
C-E-F-H-K-L-M, in row 20). In this example, neither of those was the critical path,
so Activity H is non-critical.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
D
E
F
G
Start Node End Node Min Mode
0
1
2
3
1
2
9
12
0
3
4
9
2
3
1
1
3
4
1
2
4
5
1
1
5
7
5
6
5
6
2
5
5
8
1
1
6
8
0
0
7
8
2
3
8
9
8
10
9
10
2
2
10
11
2
2
Path
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
Max Path
H
Max
4
15
18
1
3
1
9
10
1
0
5
14
2
2
I
Simulated Time
3.00
12.00
9.00
1.00
2.00
1.00
6.00
5.00
1.00
0.00
3.00
10.00
2.00
2.00
J
K
L
Start Time End Time Critical?
0.00
3.00
1
3.00
15.00
1
0.00
9.00
0
15.00
16.00
1
16.00
18.00
1
18.00
19.00
1
19.00
25.00
1
19.00
24.00
0
19.00
20.00
0
24.00
24.00
25.00
28.00
1
28.00
38.00
1
38.00
40.00
1
40.00
42.00
1
M
N
=SUM(H20,H22,H24)
Total Critical?
38.00
0
31.00
0
34.00
0
27.00
0
42.00
1
35.00
0
42.00
Decision Models -- Prof. Juran
61
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
Decision Models -- Prof. Juran
C
<= 43?
D
Penalty
0 $ 268,333
E
F
G
H
I
=IF(B30>43,10000*(B30-43),0)
62
Now we create output cells to track: (a) the completion time of the whole project (B30), (b)
the criticalities of the various paths (H19:H24) and (c) activities (N2:N15), and (d) whether
the project took longer than 43 weeks, and what the penalty was (C30:D30).
A
B
1 Activity
Description
2
A
Survey Site
3
B
Excavation
4
C
Prepare Drawings
5
D
Soil Study
6
E
Prelim. Report
7
F
Approve Plans
8
G
Concrete Forms
9
H
Procure Steel
10
I
Order Cement
11 Dummy
12
J
Deliver Gravel
13
K
Pour Concrete
14
L
Cure Concrete
15
M
Strength Test
16
17
18 Node
Time
19
0
0
20
1
3.00
21
2
15.00
22
3
16.00
23
4
18.00
24
5
19.00
25
6
24.00
26
7
25.00
27
8
28.00
28
9
38.00
29
10
40.00
30
11
42.00
C
Predecessors
None
A
None
B
C, D
E
F
F
F
H
G
H, I, J
K
L
D
E
F
G
Start Node End Node Min Mode
0
1
2
3
1
2
9
12
0
3
4
9
2
3
1
1
3
4
1
2
4
5
1
1
5
7
5
6
5
6
2
5
5
8
1
1
6
8
0
0
7
8
2
3
8
9
8
10
9
10
2
2
10
11
2
2
Path
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
Max Path
=K15
<= 43?
H
Max
4
15
18
1
3
1
9
10
1
0
5
14
2
2
I
Simulated Time
3.00
12.00
9.00
1.00
2.00
1.00
6.00
5.00
1.00
0.00
3.00
10.00
2.00
2.00
J
K
L
Start Time End Time Critical?
0.00
3.00
1
3.00
15.00
1
0.00
9.00
0
15.00
16.00
1
16.00
18.00
1
18.00
19.00
1
19.00
25.00
1
19.00
24.00
0
19.00
20.00
0
24.00
24.00
25.00
28.00
1
28.00
38.00
1
38.00
40.00
1
40.00
42.00
1
Total Critical?
38.00
0
31.00
0
34.00
0
27.00
0
42.00
1
35.00
0
42.00
=IF(B30<43,1,0)
Penalty
1 $
-
Decision Models -- Prof. Juran
=IF(B30>43,10000*(B30-43),0)
63
Decision Models -- Prof. Juran
64
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
A
B
C
D
E
F
G
Statistics
<= 43? A Critical? B Critical? C Critical? Completion Time Critical? ABDEFGJKLM
Trials
10000
10000
10000
10000
10000
10000
Base Case
1.00
1.00
1.00
0.00
42.00
1.00
Mean
0.53
0.99
0.99
0.01
42.87
0.98
Median
1.00
1.00
1.00
0.00
42.82
1.00
Mode
1.00
1.00
1.00
0.00
--1.00
Standard Deviation
0.50
0.11
0.11
0.11
1.89
0.14
Variance
0.25
0.01
0.01
0.01
3.59
0.02
Skewness
-0.1395
-8.63
-8.63
8.63
0.1271
-6.82
Kurtosis
1.02
75.52
75.52
75.52
2.90
47.52
Coeff. of Variation 0.9327
0.1143
0.1143
8.75
0.0442
0.1436
Minimum
0.00
0.00
0.00
0.00
36.58
0.00
Maximum
1.00
1.00
1.00
1.00
49.94
1.00
Range Width
1.00
1.00
1.00
1.00
13.36
1.00
Mean Std. Error
0.00
0.00
0.00
0.00
0.02
0.00
Decision Models -- Prof. Juran
65
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:
Decision Models -- Prof. Juran
66
Question 3: What is the probability of completion by week 43?
We can answer that using the statistics from the “<= 43?”
forecast cell, or by using the mean line on the frequency chart:
Decision Models -- Prof. Juran
67
Another way would be to use the grabber on the frequency
chart for the completion time:
Decision Models -- Prof. Juran
68
Give an estimated probability distribution for the amount of penalties
Sharon will have to pay. What is the expected value of the penalty?
Here’s the frequency chart and the summary statistics:
Decision Models -- Prof. Juran
69
Criticality indices for the paths:
Critical? ABDEFGJKLM
Critical? ABDEFHKLM
Critical? ABDEFIKLM
Critical? CEFGJKLM
Critical? CEFHKLM
Critical? CEFIKLM
Decision Models -- Prof. Juran
0.980
0.007
0.000
0.013
0.000
0.000
70
Criticality indices for the activities:
A Critical?
B Critical?
C Critical?
D Critical?
E Critical?
F Critical?
G Critical?
H Critical?
I Critical?
J Critical?
K Critical?
L Critical?
M Critical?
Decision Models -- Prof. Juran
0.987
0.987
0.013
0.987
1.000
1.000
0.993
0.008
0.000
0.993
1.000
1.000
1.000
71
Summary
Operations Simulation Models
• Reliability Analysis
– RANK, VLOOKUP, MIN
• Inventory Order Quantities
– Single Product (MAX)
– Multiple Products with Correlated Demand
• Project management
– PERT Analysis
Decision Models -- Prof. Juran
72

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