Session 8a

Session 8a

Decision Models -- Prof. Juran

2

OverviewOperations Simulation Models• Reliability Analysis

– RANK, VLOOKUP, MIN• Inventory Order Quantities

– Single Product (MAX)– Multiple Products with Correlated Demand

• Project management– PERT Analysis


3

Example 1: ReliabilityConsider 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?


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1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19

A B C D E F G H I J Start Rank Fail Rank Battery Life Start Time Fail Time

Distribution of battery lifetimes (lognormal) 1 2 Battery A 28.23 0.0 28.2 Mean 20 2 1 Battery B 26.41 0.0 26.4 Stdev 5 3 4 Battery C 25.67 26.4 52.1

4 3 Battery D 22.52 28.2 50.7 5 5 Battery E 28.99 50.7 79.7

Simulation Failure# Insert Battery Current time In Position 1 In Position 2

0 Battery A Battery B 1 Battery C 26.407 Battery A Battery C 26.41 2 Battery D 28.229 Battery D Battery C 28.23 3 Battery E 50.750 Battery E Battery C 50.75 4 (none) 52.077 Battery E (none) 52.08 <---- Device Fails

=RANK(I2,$I$2:$I$6,1)

=VLOOKUP(F4,$B$10:$C$12,2,0) =H4+G4 =F2 =F3

=F10 =IF(MIN(I2:I3)=I2,$F$4,$F$2) =IF(MIN(I2:I3)=I3,$F$4,$F$3)

=(D10=F2)*(I3)+(E10=E9)*(I2)

=IF(((VLOOKUP(E12,$F$2:$I$6,4,0))<(VLOOKUP(D12,$F$2:$I$6,4,0))),(D12),(B13))

=IF(((VLOOKUP(E12,$F$2:$I$6,4,0))<(VLOOKUP(D12,$F$2:$I$6,4,0))),(B13),(E12))

=(E13=B13)*(VLOOKUP(E12,$F$2:$I$6,4,0))+(E13=E12)*(VLOOKUP(D12,$F$2:$I$6,4,0))


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A B C D E F G H IStart Rank Fail Rank Battery Life Start Time Fail Time

Distribution of battery lifetimes (lognormal) 1 1 Battery A 13.40 0.0 13.4Mean 20 2 2 Battery B 18.22 0.0 18.2Stdev 5 3 3 Battery C 15.22 13.4 28.6

4 4 Battery D 31.55 18.2 49.85 5 Battery E 26.09 28.6 54.7

SimulationFailure# Insert Battery Current time In Position 1 In Position 2

0 Battery A Battery B1 Battery C 13.405 Battery C Battery B 13.402 Battery D 18.219 Battery C Battery D 18.223 Battery E 28.627 Battery E Battery D 28.634 (none) 49.767 Battery E (none) 49.77 <---- Device Fails

Battery D

=LARGE(I2:I6,2)

=VLOOKUP(4,E2:F6,2)


6

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.


7

We calculate the fail time of a battery by using the formula:

Fail time = Start time + Lifetime


8

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.


9


10


11

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10111213

A B C D E F G H IStart Rank Fail Rank Battery Life Start Time Fail Time

Distribution of battery lifetimes (lognormal) 1 2 Battery A 25.28 0.0 25.3Mean 20 2 1 Battery B 12.02 0.0 12.0Stdev 5 3 3 Battery C 17.00 12.0 29.0

4 5 Battery D 33.15 25.3 58.45 4 Battery E 23.53 29.0 52.5

SimulationFailure# Insert Battery Current time In Position 1 In Position 2

0 Battery A Battery B1 Battery C 12.017 Battery A Battery C 12.022 Battery D 25.275 Battery D Battery C 25.283 Battery E 29.016 Battery D Battery E 29.024 (none) 52.542 Battery D (none) 52.54 <---- Device Fails


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13


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A confidence interval for the mean time to failure can be calculated as follows:

X nsz 2

00.43 100041.596.1

1711.096.1 3353.0

Or (42.66, 43.34) Note that the Crystal Ball output gives both the standard deviation and the standard error.


15

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.


16

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.


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A B C D E F G H I J K LNumber to print 0 1000 2000 3000 4000 5000 6000 7000Number on hand after printing 2000 3000 4000 5000 6000 7000 8000 9000

Demand 6000 6000 6000 6000 6000 6000 6000 6000Sold at 35 2000 3000 4000 5000 6000 6000 6000 6000Sold at 5 0 0 0 0 0 1000 2000 3000

Fixed cost $50,000 $50,000 $50,000 $50,000 $50,000 $50,000 $50,000 $50,000Variable cost $0 $15,000 $30,000 $45,000 $60,000 $75,000 $90,000 $105,000

Revenue $70,000 $105,000 $140,000 $175,000 $210,000 $215,000 $220,000 $225,000Profit $20,000 $40,000 $60,000 $80,000 $100,000 $90,000 $80,000 $70,000

Probability distribution of demand (triangular) Current inventory 2000Min 4000Most likely 6000 Selling price per book $35Max 9000 Variable cost per book $15

Fixed cost of a printing $50,000Reduced selling price $5

=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)


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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.


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20


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We run for 1000 trials, and then click on the extract data button.

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1011121314

B C D E F G H I0 Profit 1000 Profit 2000 Profit 3000 Profit 4000 Profit 5000 Profit 6000 Profit 7000 Profit

1000 1000 1000 1000 1000 1000 1000 1000$20,000 $40,000 $60,000 $79,179 $92,260 $94,734 $89,364 $80,134$20,000 $40,000 $60,000 $80,000 $100,000 $97,222 $87,222 $77,222$20,000 $40,000 $60,000 $80,000 $100,000 $120,000 $140,000 ---

$0 $0 $0 $3,382 $12,823 $22,914 $29,083 $30,596$0 $0 $0 $11,435,323 $164,429,426 $525,055,988 $845,815,561 $936,106,979

0.00 0.00 0.00 -4.81 -1.65 -0.53 0.07 0.27+Infinity +Infinity +Infinity 26.83 4.78 2.20 2.15 2.47

0.00 0.00 0.00 0.04 0.14 0.24 0.33 0.38$20,000 $40,000 $60,000 $54,823 $44,823 $34,823 $24,823 $14,823$20,000 $40,000 $60,000 $80,000 $100,000 $120,000 $140,000 $158,923

$0 $0 $0 $25,177 $55,177 $85,177 $115,177 $144,100$0.00 $0.00 $0.00 $106.94 $405.50 $724.61 $919.68 $967.53


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Efficient Frontier

$0

$10,000

$20,000

$30,000

$40,000

$50,000

$60,000

$70,000

$80,000

$90,000

$100,000

$0 $5,000 $10,000 $15,000 $20,000 $25,000 $30,000 $35,000

Risk

Expe

cted

Pro

fit

Order 0

Order 1000

Order 2000

Order 3000

Order 4000 Order 5000

Order 6000

Order 7000


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Does your answer change if 4000 copies are currently in stock?

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A B C D E F G H INumber to print 0 1000 2000 3000 4000 5000 6000 7000Number on hand after printing 4000 5000 6000 7000 8000 9000 10000 11000

Demand 6000 6000 6000 6000 6000 6000 6000 6000Sold at 35 4000 5000 6000 6000 6000 6000 6000 6000Sold at 5 0 0 0 1000 2000 3000 4000 5000

Fixed cost $50,000 $50,000 $50,000 $50,000 $50,000 $50,000 $50,000 $50,000Variable cost $0 $15,000 $30,000 $45,000 $60,000 $75,000 $90,000 $105,000

Revenue $140,000 $175,000 $210,000 $215,000 $220,000 $225,000 $230,000 $235,000Profit $90,000 $110,000 $130,000 $120,000 $110,000 $100,000 $90,000 $80,000

Probability distribution of demand (triangular) Current inventory 4000Min 4000Most likely 6000 Selling price per book $35Max 9000 Variable cost per book $15

Fixed cost of a printing $50,000Reduced selling price $5


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Efficient Frontier

$0

$20,000

$40,000

$60,000

$80,000

$100,000

$120,000

$140,000

$0 $5,000 $10,000 $15,000 $20,000 $25,000 $30,000 $35,000

Risk

Expe

cted

Pro

fit

Order 0

Order 1000Order 2000

Order 3000 Order 4000

Order 5000

Order 6000

Order 7000


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Example 3: Walton BookstoreConsider 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.


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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)?


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A B C D E F G H I J K L M N O P QStrategy 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16Product 1 ordered 700 700 700 700 800 800 800 800 900 900 900 900 1000 1000 1000 1000Product 2 ordered 900 1000 1100 1200 900 1000 1100 1200 900 1000 1100 1200 900 1000 1100 1200Demand 1 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000Demand 2 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 12001 sold full price 700 700 700 700 800 800 800 800 900 900 900 900 1000 1000 1000 10002 sold full price 900 1000 1100 1200 900 1000 1100 1200 900 1000 1100 1200 900 1000 1100 12001 sold at refund price 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 02 sold at refund price 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0Full-price revenue 16,000$ 17,000$ 18,000$ 19,000$ 17,000$ 18,000$ 19,000$ 20,000$ 18,000$ 19,000$ 20,000$ 21,000$ 19,000$ 20,000$ 21,000$ 22,000$ Refund revenue -$ -$ -$ -$ -$ -$ -$ -$ -$ -$ -$ -$ -$ -$ -$ -$ Order cost 12,000$ 12,750$ 13,500$ 14,250$ 12,750$ 13,500$ 14,250$ 15,000$ 13,500$ 14,250$ 15,000$ 15,750$ 14,250$ 15,000$ 15,750$ 16,500$ Profit 4,000$ 4,250$ 4,500$ 4,750$ 4,250$ 4,500$ 4,750$ 5,000$ 4,500$ 4,750$ 5,000$ 5,250$ 4,750$ 5,000$ 5,250$ 5,500$

Product 1 Product 2Unit price $10.00 $10.00Unit cost $7.50 $7.50 Means 1000 1200 CorrUnit refund value $2.50 $2.50 Stdevs 250 350 -0.3

Price Cost RefundProduct 1 $10.00 $7.50 $2.50Product 2 $10.00 $7.50 $2.50

=MIN(B2,B4)=MIN(B3,B5)=MAX(0,B2-B4)=MAX(0,B3-B5)=SUMPRODUCT(B6:B7,$B21:$B22)=SUMPRODUCT(B8:B9,$D21:$D22)=SUMPRODUCT(B2:B3,$C21:$C22)=B10+B11-B12

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.


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A B C D E F GStrategy 1 2 3 4 5 6Product 1 ordered 700 700 700 700 800 800Product 2 ordered 900 1000 1100 1200 900 1000Demand 1 1000 1000 1000 1000 1000 1000Demand 2 1200 1200 1200 1200 1200 12001 sold full price 700 700 700 700 800 8002 sold full price 900 1000 1100 1200 900 10001 sold at refund price 0 0 0 0 0 02 sold at refund price 0 0 0 0 0 0Full-price revenue 16,000$ 17,000$ 18,000$ 19,000$ 17,000$ 18,000$ Refund revenue -$ -$ -$ -$ -$ -$ Order cost 12,000$ 12,750$ 13,500$ 14,250$ 12,750$ 13,500$ Profit 4,000$ 4,250$ 4,500$ 4,750$ 4,250$ 4,500$

Product 1 Product 2Unit price $10.00 $10.00Unit cost $7.50 $7.50 Means 1000 1200Unit refund value $2.50 $2.50 Stdevs 250 350


=MIN(B2,B4)=MIN(B3,B5)=MAX(0,B2-B4)=MAX(0,B3-B5)=SUMPRODUCT(B6:B7,$B21:$B22)=SUMPRODUCT(B8:B9,$D21:$D22)=SUMPRODUCT(B2:B3,$C21:$C22)=B10+B11-B12


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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:


30

For more than a few correlated green cells, it’s more efficient to use the matrix view.

You can specify bivariate correlations in the Define Assumption window.


31

Click on “correlate” to enter the correlation information.


32

Switch to “list” view (not the default matrix view).

Enter coefficient by typing, cell reference, or the slider.


33


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A B C D E F G H I J K L M N OStrategy 1 2 3 4 5 6 7 8 9 10 11 12 13 14Product 1 ordered 700 700 700 700 800 800 800 800 900 900 900 900 1000 1000Product 2 ordered 900 1000 1100 1200 900 1000 1100 1200 900 1000 1100 1200 900 1000Demand 1 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000Demand 2 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 12001 sold full price 700 700 700 700 800 800 800 800 900 900 900 900 1000 10002 sold full price 900 1000 1100 1200 900 1000 1100 1200 900 1000 1100 1200 900 10001 sold at refund price 0 0 0 0 0 0 0 0 0 0 0 0 0 02 sold at refund price 0 0 0 0 0 0 0 0 0 0 0 0 0 0Full-price revenue 16,000$ 17,000$ 18,000$ 19,000$ 17,000$ 18,000$ 19,000$ 20,000$ 18,000$ 19,000$ 20,000$ 21,000$ 19,000$ 20,000$ Refund revenue -$ -$ -$ -$ -$ -$ -$ -$ -$ -$ -$ -$ -$ -$ Order cost 12,000$ 12,750$ 13,500$ 14,250$ 12,750$ 13,500$ 14,250$ 15,000$ 13,500$ 14,250$ 15,000$ 15,750$ 14,250$ 15,000$ Profit 4,000$ 4,250$ 4,500$ 4,750$ 4,250$ 4,500$ 4,750$ 5,000$ 4,500$ 4,750$ 5,000$ 5,250$ 4,750$ 5,000$

Product 1 Product 2Unit price $10.00 $10.00Unit cost $7.50 $7.50 Means 1000 1200 CorrUnit refund value $2.50 $2.50 Stdevs 250 350 -0.3



35

7 0 0

8 0 0

9 0 01 0 0 0

9 0 0

1 0 0 0

1 1 0 0

1 2 0 0

$ 3 , 4 0 0

$ 3 , 4 5 0

$ 3 , 5 0 0

$ 3 , 5 5 0

$ 3 , 6 0 0

$ 3 , 6 5 0

$ 3 , 7 0 0

$ 3 , 7 5 0

$ 3 , 8 0 0

$ 3 , 8 5 0

$ 3 , 9 0 0

E x p e c t e dP r o f i t

P r o d u c t 1O r d e r e d


E x p e c t e d P r o f i t( C o r r e la t i o n = - 0 .3 ) 1 0

69

5

1

7 0 0

8 0 0

9 0 01 0 0 0

9 0 0

1 0 0 0

1 1 0 0

1 2 0 0

$ 8 0 0

$ 9 0 0

$ 1 , 0 0 0

$ 1 ,1 0 0

$ 1 , 2 0 0

$ 1 , 3 0 0

$ 1 , 4 0 0

$ 1 , 5 0 0

$ 1 ,6 0 0

$ 1 ,7 0 0

S t d D e vo f P r o f i t



S t d D e v o f P r o f i t( C o r r e l a t i o n = - 0 . 3 )

1

1 0

6

9

5

7 0 0

8 0 0

9 0 01 0 0 0

9 0 0

1 0 0 0

1 1 0 0

1 2 0 0

$ 3 , 4 0 0

$ 3 , 4 5 0

$ 3 , 5 0 0

$ 3 , 5 5 0

$ 3 , 6 0 0

$ 3 , 6 5 0

$ 3 , 7 0 0

$ 3 , 7 5 0

$ 3 , 8 0 0

$ 3 , 8 5 0

$ 3 , 9 0 0




E x p e c t e d P r o f i t( C o r r e la t i o n = - 0 .5 )

1 0

69

5

1

7 0 0

8 0 0

9 0 01 0 0 0

9 0 0

1 0 0 0

1 1 0 0

1 2 0 0

$ 8 0 0

$ 9 0 0

$ 1 , 0 0 0

$ 1 ,1 0 0

$ 1 , 2 0 0

$ 1 , 3 0 0

$ 1 , 4 0 0

$ 1 , 5 0 0

$ 1 ,6 0 0

$ 1 ,7 0 0





1 0

69

5

1

7 0 0

8 0 0

9 0 01 0 0 0

9 0 0

1 0 0 0

1 1 0 0

1 2 0 0

$ 3 , 4 0 0

$ 3 , 4 5 0

$ 3 , 5 0 0

$ 3 , 5 5 0

$ 3 , 6 0 0

$ 3 , 6 5 0

$ 3 , 7 0 0

$ 3 , 7 5 0

$ 3 , 8 0 0

$ 3 , 8 5 0

$ 3 , 9 0 0




E x p e c t e d P r o f i t( C o r r e la t i o n = - 0 .7 )

1 0

6 9

5

1

7 0 0

8 0 0

9 0 01 0 0 0

9 0 0

1 0 0 0

1 1 0 0

1 2 0 0

$ 8 0 0

$ 9 0 0

$ 1 , 0 0 0

$ 1 ,1 0 0

$ 1 , 2 0 0

$ 1 , 3 0 0

$ 1 , 4 0 0

$ 1 , 5 0 0

$ 1 ,6 0 0

$ 1 ,7 0 0





1 0

6

9

5

1


36

Correlation = -0.3

$3,400

$3,500

$3,600

$3,700

$3,800

$3,900

$800 $900 $1,000 $1,100 $1,200 $1,300 $1,400 $1,500 $1,600 $1,700

Std Deviation

Expe

cted

Val

ue

Strategy 1

Strategy 4

Strategy 5

Strategy 6

Strategy 16

Strategy 10

Strategy 9

Strategy 8


37

Correlation = -0.5

$3,400

$3,500

$3,600

$3,700

$3,800

$3,900

$800 $900 $1,000 $1,100 $1,200 $1,300 $1,400 $1,500 $1,600 $1,700

Std Deviation

Expe

cted

Val

ue

Strategy 1 Strategy 4

Strategy 5

Strategy 6

Strategy 16

Strategy 10

Strategy 9

Strategy 8


38

Correlation = -0.7

$3,400

$3,500

$3,600

$3,700

$3,800

$3,900

$800 $900 $1,000 $1,100 $1,200 $1,300 $1,400 $1,500 $1,600 $1,700

Std Deviation

Expe

cted

Val

ue

Strategy 1

Strategy 4

Strategy 5

Strategy 6

Strategy 16

Strategy 10

Strategy 9

Strategy 8


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Beta Distribution

Parameter Description Characteristics Min Minimum Value Any number -∞ to ∞ Max Maximum Value Any number -∞ to ∞ Alpha (α) Shape Factor Must be > 0 Beta (β) Shape Factor Must be > 0

The Beta distribution is a continuous probability distribution defined by four parameters:


40

Here are sixteen different Beta distributions, all with a minimum of 0 and a maximum of 100.

β 0.5 1.0 2.0 4.0

0.5 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75

1.0 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0 25 50 75

2.0 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75

α

4.0 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75


41

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.


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Mean: min - max min

(i)

Standard Deviation:

22 min - max1

(ii)


43

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 mathematically-correct 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.

PERT Approximations


44

3-step Procedure1. 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):

6

max mode 4min (iii)

6min - max

(iv) 3. Use equations (v) and (vi) to calculate shape factors that are consistent with the

mean and standard deviation:

1mean - maxmin -mean

min - maxmin -mean

2 (v)

min -mean

mean - max (vi)


45

Beta Distributions in Crystal Ball


46

Beta Distributions in Crystal Ball


47

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.


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Activity Description Optimistic Pessimistic Most-likely Predecessors A Survey Site 2 4 3 None B Excavation 9 15 12 A C Prepare Drawings 4 18 9 None D Soil Study 1 1 1 B E Prelim. Report 1 3 2 C, D F Approve Plans 1 1 1 E G Concrete Forms 5 9 6 F H Procure Steel 2 10 5 F I Order Cement 1 1 1 F J Deliver Gravel 2 5 3 G K Pour Concrete 8 14 10 H, I, J L Cure Concrete 2 2 2 K M Strength Test 2 2 2 L


49

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.


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Here’s an activity-on-arc diagram of the problem:


51

To use a beta distribution, we would start a spreadsheet model like this, calculating the mean and standard deviation using the PERT formulas:

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101112131415

A B C D E F G H I J K LActivity Description Predecessors Start Node End Node Min Mode Max Mean StDev

A Survey Site None 0 1 2 3 4 3 0.333B Excavation A 1 2 9 12 15 12 1.000C Prepare Drawings None 0 3 4 9 18 9.667 2.333D Soil Study B 2 3 1 1 1 1 0.000E Prelim. Report C, D 3 4 1 2 3 2 0.333F Approve Plans E 4 5 1 1 1 1 0.000G Concrete Forms F 5 7 5 6 9 6.333 0.667H Procure Steel F 5 6 2 5 10 5.333 1.333I Order Cement F 5 8 1 1 1 1 0.000

Dummy H 6 8 0 0 0 0 0.000J Deliver Gravel G 7 8 2 3 5 3.167 0.500K Pour Concrete H, I, J 8 9 8 10 14 10.33 1.000L Cure Concrete K 9 10 2 2 2 2 0.000M Strength Test L 10 11 2 2 2 2 0.000

=(F4+4*G4+H4)/6=(H5-F5)/6


52

Calculating shape and scale parameters:

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101112131415

A B C D E F G H I J K L M N O PActivity Description Predecessors Start Node End Node Min Mode Max Mean StDev Alpha Beta

A Survey Site None 0 1 2 3 4 3 0.333 4.000 4.000B Excavation A 1 2 9 12 15 12 1.000 4.000 4.000C Prepare Drawings None 0 3 4 9 18 9.667 2.333 3.106 4.568D Soil Study B 2 3 1 1 1 1 0.000E Prelim. Report C, D 3 4 1 2 3 2 0.333 4.000 4.000F Approve Plans E 4 5 1 1 1 1 0.000G Concrete Forms F 5 7 5 6 9 6.333 0.667 2.333 4.667H Procure Steel F 5 6 2 5 10 5.333 1.333 3.229 4.521I Order Cement F 5 8 1 1 1 1 0.000

Dummy H 6 8 0 0 0 0 0.000J Deliver Gravel G 7 8 2 3 5 3.167 0.500 2.938 4.617K Pour Concrete H, I, J 8 9 8 10 14 10.33 1.000 2.938 4.617L Cure Concrete K 9 10 2 2 2 2 0.000M Strength Test L 10 11 2 2 2 2 0.000

=((I3-F3)/(H3-F3))*((((I3-F3)*(H3-I3))/(J3^2))-1)

=((H6-I6)/(I6-F6))*K6

123456789101112131415161718192021222324252627282930

A B C D E F G H I J K LActivity Description Predecessors Start Node End Node Min Mode Max Simulated Time Start Time End Time Critical?

A Survey Site None 0 1 2 3 4 3 0.00 3.00 1B Excavation A 1 2 9 12 15 12 3.00 15.00 1C Prepare Drawings None 0 3 4 9 18 9 0.00 9.00 0D Soil Study B 2 3 1 1 1 1 15.00 16.00 1E Prelim. Report C, D 3 4 1 2 3 2 16.00 18.00 1F Approve Plans E 4 5 1 1 1 1 18.00 19.00 1G Concrete Forms F 5 7 5 6 9 6 19.00 25.00 1H Procure Steel F 5 6 2 5 10 5 19.00 24.00 0I Order Cement F 5 8 1 1 1 1 19.00 20.00 0

Dummy H 6 8 0 0 0 0 24.00 24.00J Deliver Gravel G 7 8 2 3 5 3 25.00 28.00 1K Pour Concrete H, I, J 8 9 8 10 14 10 28.00 38.00 1L Cure Concrete K 9 10 2 2 2 2 38.00 40.00 1M Strength Test L 10 11 2 2 2 2 40.00 42.00 1

Node Time Path Total Critical?0 0 A-B-D-E-F-H-K-L-M 38.00 01 3.00 C-E-F-H-K-L-M 31.00 02 15.00 A-B-D-E-F-I-K-L-M 34.00 03 16.00 C-E-F-I-K-L-M 27.00 04 18.00 A-B-D-E-F-G-J-K-L-M 42.00 15 19.00 C-E-F-G-J-K-L-M 35.00 06 24.00 Max Path 42.007 25.008 28.009 38.00

10 40.00 <= 43? Penalty11 42.00 1 -$


53A section to keep track of each node and when it occurs

A section to keep track of each path through the network, to identify the critical path in each simulated project completion

A section for simulating the times of the activities

Model Overview


54

Here’s the section keeping track of the activity times. The numbers in column I will be CB assumption cells.

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A B C D E F G H IActivity Description Predecessors Start Node End Node Min Mode Max Simulated Time

A Survey Site None 0 1 2 3 4 3.00B Excavation A 1 2 9 12 15 12.00C Prepare Drawings None 0 3 4 9 18 9.67D Soil Study B 2 3 1 1 1 1.00E Prelim. Report C, D 3 4 1 2 3 2.00F Approve Plans E 4 5 1 1 1 1.00G Concrete Forms F 5 7 5 6 9 6.33H Procure Steel F 5 6 2 5 10 5.33I Order Cement F 5 8 1 1 1 1.00

Dummy H 6 8 0 0 0 0.00J Deliver Gravel G 7 8 2 3 5 3.17K Pour Concrete H, I, J 8 9 8 10 14 10.33L Cure Concrete K 9 10 2 2 2 2.00M Strength Test L 10 11 2 2 2 2.00


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For each of the random activities, we create an assumption cell, as shown here for Activity A:


56


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Now we set up an area in the spreadsheet to keep track of the nodes and their times:

18192021222324252627282930

A BNode Time

0 0123456789

1011

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.


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Example: Activity C123456789

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A B C D E F G H I J K L M NActivity Description Predecessors Start Node End Node Min Mode Max Simulated Time Start Time End Time Critical?

A Survey Site None 0 1 2 3 4 2.87 0.00 2.87 1B Excavation A 1 2 9 12 15 10.52 2.87 13.39 1C Prepare Drawings None 0 3 4 9 18 6.19 0.00 6.19 0D Soil Study B 2 3 1 1 1 1.00 13.39 14.39 1E Prelim. Report C, D 3 4 1 2 3 2.04 14.39 16.43 1F Approve Plans E 4 5 1 1 1 1.00 16.43 17.43 1G Concrete Forms F 5 7 5 6 9 6.00 17.43 23.43 1H Procure Steel F 5 6 2 5 10 3.61 17.43 21.05 0I Order Cement F 5 8 1 1 1 1.00 17.43 18.43 0

Dummy H 6 8 0 0 0 0.00 21.05 21.05J Deliver Gravel G 7 8 2 3 5 2.71 23.43 26.14 1K Pour Concrete H, I, J 8 9 8 10 14 10.07 26.14 36.20 1L Cure Concrete K 9 10 2 2 2 2.00 36.20 38.20 1M Strength Test L 10 11 2 2 2 2.00 38.20 40.20 1


10 38.20 <= 43? Penalty11 40.20 1 -$

=VLOOKUP(D4,$A$19:$B$30,2,0)

=J4+I4

=MAX(K4,K5)


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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:

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D E F G H I J KStart Node End Node Min Mode Max Simulated Time Start Time End Time

0 1 2 3 4 3.00 0.00 3.001 2 9 12 15 12.00 3.00 15.000 3 4 9 18 9.00 0.00 9.002 3 1 1 1 1.00 15.00 16.003 4 1 2 3 2.00 16.00 18.004 5 1 1 1 1.00 18.00 19.005 7 5 6 9 6.00 19.00 25.005 6 2 5 10 5.00 19.00 24.005 8 1 1 1 1.00 19.00 20.006 8 0 0 0 0.00 24.00 24.007 8 2 3 5 3.00 25.00 28.008 9 8 10 14 10.00 28.00 38.009 10 2 2 2 2.00 38.00 40.0010 11 2 2 2 2.00 40.00 42.00

Path Total Critical?A-B-D-E-F-H-K-L-M 38.00 0C-E-F-H-K-L-M 31.00 0A-B-D-E-F-I-K-L-M 34.00 0C-E-F-I-K-L-M 27.00 0A-B-D-E-F-G-J-K-L-M 42.00 1C-E-F-G-J-K-L-M 35.00 0Max Path 42.00

=SUM(I4,I6,I7,I9,I13,I14,I15)

=IF(G22=$G$25,1,0)

=MAX(G19:G24)


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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:


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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.

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D E F G H I J K L M NStart Node End Node Min Mode Max Simulated Time Start Time End Time Critical?

0 1 2 3 4 3.00 0.00 3.00 11 2 9 12 15 12.00 3.00 15.00 10 3 4 9 18 9.00 0.00 9.00 02 3 1 1 1 1.00 15.00 16.00 13 4 1 2 3 2.00 16.00 18.00 14 5 1 1 1 1.00 18.00 19.00 15 7 5 6 9 6.00 19.00 25.00 15 6 2 5 10 5.00 19.00 24.00 05 8 1 1 1 1.00 19.00 20.00 06 8 0 0 0 0.00 24.00 24.007 8 2 3 5 3.00 25.00 28.00 18 9 8 10 14 10.00 28.00 38.00 19 10 2 2 2 2.00 38.00 40.00 1

10 11 2 2 2 2.00 40.00 42.00 1

Path Total Critical?A-B-D-E-F-H-K-L-M 38.00 0C-E-F-H-K-L-M 31.00 0A-B-D-E-F-I-K-L-M 34.00 0C-E-F-I-K-L-M 27.00 0A-B-D-E-F-G-J-K-L-M 42.00 1C-E-F-G-J-K-L-M 35.00 0Max Path 42.00

=SUM(H20,H22,H24)


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27282930

A B C D E8 47.509 65.83

10 67.83 <= 43?11 69.83 0

=IF(B30<43,1,0)

282930

A B C D E F G H I9 65.8310 67.83 <= 43? Penalty11 69.83 0 268,333$

=IF(B30>43,10000*(B30-43),0)

Here’s a cell to tell whether the project was completed by week 43:

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.


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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).

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A B C D E F G H I J K LActivity Description Predecessors Start Node End Node Min Mode Max Simulated Time Start Time End Time Critical?

A Survey Site None 0 1 2 3 4 3.00 0.00 3.00 1B Excavation A 1 2 9 12 15 12.00 3.00 15.00 1C Prepare Drawings None 0 3 4 9 18 9.00 0.00 9.00 0D Soil Study B 2 3 1 1 1 1.00 15.00 16.00 1E Prelim. Report C, D 3 4 1 2 3 2.00 16.00 18.00 1F Approve Plans E 4 5 1 1 1 1.00 18.00 19.00 1G Concrete Forms F 5 7 5 6 9 6.00 19.00 25.00 1H Procure Steel F 5 6 2 5 10 5.00 19.00 24.00 0I Order Cement F 5 8 1 1 1 1.00 19.00 20.00 0

Dummy H 6 8 0 0 0 0.00 24.00 24.00J Deliver Gravel G 7 8 2 3 5 3.00 25.00 28.00 1K Pour Concrete H, I, J 8 9 8 10 14 10.00 28.00 38.00 1L Cure Concrete K 9 10 2 2 2 2.00 38.00 40.00 1M Strength Test L 10 11 2 2 2 2.00 40.00 42.00 1


10 40.00 <= 43? Penalty11 42.00 1 -$

=K15 =IF(B30<43,1,0)

=IF(B30>43,10000*(B30-43),0)


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A B C D E F GStatistics <= 43? A Critical? B Critical? C Critical? Completion Time Critical? ABDEFGJKLM

Trials 10000 10000 10000 10000 10000 10000Base Case 1.00 1.00 1.00 0.00 42.00 1.00Mean 0.53 0.99 0.99 0.01 42.87 0.98Median 1.00 1.00 1.00 0.00 42.82 1.00Mode 1.00 1.00 1.00 0.00 --- 1.00Standard Deviation 0.50 0.11 0.11 0.11 1.89 0.14Variance 0.25 0.01 0.01 0.01 3.59 0.02Skewness -0.1395 -8.63 -8.63 8.63 0.1271 -6.82Kurtosis 1.02 75.52 75.52 75.52 2.90 47.52Coeff. of Variation 0.9327 0.1143 0.1143 8.75 0.0442 0.1436Minimum 0.00 0.00 0.00 0.00 36.58 0.00Maximum 1.00 1.00 1.00 1.00 49.94 1.00Range Width 1.00 1.00 1.00 1.00 13.36 1.00Mean Std. Error 0.00 0.00 0.00 0.00 0.02 0.00


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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:


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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:


68

Another way would be to use the grabber on the frequency chart for the completion time:


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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:


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Criticality indices for the paths:

Critical? ABDEFGJKLM 0.980Critical? ABDEFHKLM 0.007Critical? ABDEFIKLM 0.000Critical? CEFGJKLM 0.013Critical? CEFHKLM 0.000Critical? CEFIKLM 0.000


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Criticality indices for the activities:

A Critical? 0.987B Critical? 0.987C Critical? 0.013D Critical? 0.987E Critical? 1.000F Critical? 1.000G Critical? 0.993H Critical? 0.008I Critical? 0.000J Critical? 0.993K Critical? 1.000L Critical? 1.000M Critical? 1.000


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SummaryOperations Simulation Models• Reliability Analysis

– RANK, VLOOKUP, MIN• Inventory Order Quantities

– Single Product (MAX)– Multiple Products with Correlated Demand

• Project management– PERT Analysis

Documents

Session 8a