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Chapter 6Simulation

Advantages and Disadvantages of Using Simulation

Modeling Random Variables and Pseudo-Random

Numbers Time Increments Simulation Languages Validation and Statistical Considerations Examples

• Risk Analysis• Waiting Line Simulation

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What is Simulation?

An attempt to duplicate the features, appearance, and characteristics of a real system

1. To imitate a real-world situation mathematically

2. To study its properties and operating characteristics

3. To draw conclusions and make action decisions based on the results of the simulation

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Simulation Applications

Bus scheduling

Design of library operations

Taxi, truck, and railroad dispatching

Production facility scheduling

Plant layout

Capital investments

Production scheduling

Sales forecasting

Inventory planning and control

Ambulance location and dispatching

Assembly-line balancing

Parking lot and harbor design

Distribution system design

Scheduling aircraft

Labor-hiring decisions

Personnel scheduling

Traffic-light timing

Voting pattern prediction

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Select best course

Examine results

Conduct simulation

Specify valuesof variables

Construct model

Introduce variables

The Process of Simulation

Define problem

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Advantages of Simulation

1. Relatively straightforward and flexible

2. Can be used to analyze large and complex real-world situations that cannot be solved by conventional models

3. Real-world complications can be included that most OR models cannot permit

4. “Time compression” is possible

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Advantages of Simulation

5. Allows “what-if” types of questions

6. Does not interfere with real-world systems

7. Can study the interactive effects of individual components or variables in order to determine which ones are important

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Disadvantages of Simulation

1. Can be very expensive and may take months to develop

2. It is a trial-and-error approach that may produce different solutions in repeated runs

3. Managers must generate all of the conditions and constraints for solutions they want to examine

4. Each simulation model is unique

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Monte Carlo Simulation

Select numbers randomly from a probability distribution

Use these values to observe how a model performs over time

Random numbers each have an equal likelihood of being selected at random

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Distribution of Demand

LAPTOPS DEMANDED FREQUENCY OF PROBABILITY OFPER WEEK, DEMAND DEMAND, P(x) CUMULATIVE

0 20 0.20 0 1 40 0.40 0.20 2 20 0.20 0.60 3 10 0.10 0.80 4 10 0.10 0.90

100 1.00

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Roulette Wheel of Demand

90

80

60

20

0

x = 2

x = 0x = 4

x = 3

x = 1

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Generating Demand from Random Numbers

DEMAND, RANGES OF RANDOM NUMBERS,x r

0 0-191 20-59 r = 392 60-793 80-894 90-99

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Random Number Table

39 65 76 45 45 19 90 6964 6173 71 23 70 90 65 97 6012 1172 18 47 33 84 51 67 4797 1975 12 25 69 17 17 95 2178 5837 17 79 88 74 63 52 0634 30

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15 Weeks of Demand

Average demand = 31/15

= 2.07 laptops/week

WEEK r DEMAND (x) REVENUE (S)

1 39 1 4,3002 73 2 8,6003 72 2 8,6004 75 2 8,6005 37 1 4,3006 02 0 07 87 3 12,9008 98 4 17,2009 10 0 0

10 47 1 4,30011 93 4 17,20012 21 1 4,30013 95 4 17,20014 97 4 17,20015 69 2 8,600

= 31 $133,300

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Computing Expected Demand

E(x) = (0.20)(0) + (0.40)(1) + (0.20)(2) + (0.10)(3) + (0.10)(4)= 1.5 laptops per week

Not particularly close to simulated result of 2.07 laptops

Difference is due to small number of periods analyzed

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Random Numbers in Excel

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Simulation in Excel

Enter this formula in G6 and copy to

G7:G20

Enter “=4300*G6” in H6 and copy to

H7:H20

Generate random numbers for cells F6:F20 with the

formula “=RAND()” in F6 and copying to

F7:F20

= AVERAGE (G6:G20)

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Simulation in Excel

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Example of Risk AnalysisPortaCom Project

PortCom’s product design group has developed a prototype for a new high-quality portable printer. The new printer has an innovative design and the potential to capture a significant share of the portable printer market. Preliminary marketing and financial analysis have provided the following information. Selling price = $249 per unitAdministrative cost = $400,000Advertising cost = $600,000

PortaCom believes that the costs and the demand range as follows:Unit direct labor cost = $43~$47Unit parts cost = $80~$100First-year demand = 1500~28,500 units

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Simulation The advantage of simulation is that it allows

us to assess the probability of a profit and the probability of a loss.

Procedure of simulation 1. Check parameters2. Check controllable inputs3. Check probabilistic inputs * Generate random numbers

4. Formulate a model5. Draw a flowchart

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Simulation

1. Check parametersSelling price = $249 per unitAdministrative cost = $400,000Advertising cost = $600,000

2. Check controllable inputsWhether or not introduce the product

3. Check probabilistic inputsUnit direct labor cost range = $43~$47Unit parts cost range = $80~$100First-year demand range = 1500~28,500 units

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Simulation4. Formulate a model

Profit=(249-c1-c2)X-1,000,0005. Draw a flowchart

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Probability Distribution of the Direct Labor Cost

Direct labor cost Probability $43 0.1 $44 0.2 $45 0.4 $46 0.2 $47 0.1

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Probability Distribution of the Parts Costs

The probability distribution for the parts cost per unit is the uniform distribution as follows:

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Probability Distribution of the First-year Demand

The first-year demand is described by the normal probability distribution with mean 15,000 units and the standard deviation 45000 units as follows:

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How to Generate Random Numbers

Computer-generated random numbers

* Assign ranges of random numbers to to corresponding values of probabilistic inputs. The prob. of any input value is identical to the prob. of its occurrence in the real system. * Placing =RAND() in a cell of an Excel worksheet will result in a random number.

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Generate Random Value for Direct Labor Cost

Interval ofDirect labor cost Probability random

numbers $43 0.1 0.0~0.1 $44 0.2 0.1~0.3 $45 0.4 0.3~0.7

$46 0.2 0.7~0.9 $47 0.1 0.9~1.0

*Excel Statement =Vlookup(Rand(),range, Col_index)

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Generate Random Numbers for Parts Cost

With a uniform probability distribution, the following relationship between the random number and the associated value of the parts cost is used.

Parts cost=a+r(b-a) where r=random number a=smallest value for parts cost b=largest value for parts costParts cost=80+r(100-80)=80+r20

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Generate Random Numbers for First-year Demand

Because first-year demand is normally distributed, we need a procedure for generating random values from a normal distribution.

We use the following formula of Excell

=NORMINV(RAND(),mean,standard deviation)

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Waiting Line Simulation

HKSB Savings Bank will open several new branch bank during the coming year. Each new branch is designed to have one automated teller machine (ATM). A concern is that during busy periods several customer may have to wait to use the ATM. This concern prompted the bank to undertake a study of the waiting line system. The bank’s vise president want to determine whether one ATM will be sufficient. The bank established service guidelines for its ATM system stating that the average customer waiting time for an ATM should be one minute or less

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Waiting Line Simulation

Customer Arrival Times

Interval Time = a + r (b-a) r = random number between o and 1

a = minimum interarrival time b = maximum interarrival time

For the HKSB ATM System, the minimum interarrival time is a = 0 minutes, and the maximum interarrival time is b = 5 minutes Interval Time = 0 + r (5 - 0)= 5r

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Customer Arrival Times

For the HKSB ATM System, the minimum interarrival time is a = 0 minutes, and the maximum interarrival time is b = 5 minutes Interval Time = 0 + r (5 - 0)= 5rAssume that the simulation run begins at time = 0, A random number of r= 0.2804 generates an interval time of 5(0.2804) = 1.4 minutes for customer 1. A second random number of r=0.2598 generates an interarrival time of 5(0.2598) = 1.3 minutes, indicating that customer 2 arrive 1.3 minutes after customer 1. Thus customer 2 arrives 1.4 + 1.3 = 2.7 minutes after the simulation begin. Continuing, a third random number of r = 0.9802 indicates that customer 3 arrives 4.9 minutes after customer 2, which 7,6 minutes after the simulation begin.

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Waiting Line Simulation

HKSB ATM Simulation Model

ModelNumber of ATMs

Interarrival Time

Operating

Characteristic

Service Time

Interarrival Times (Uniform Distribution)Smallest Value 0Largest Value 5  

Service Times (Normal Distribution)Mean 2Std Deviation 0.5  

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Waiting Line Simulation

Customer

InterarrivalTime

Arrival time

Service Start Time

Waiting Time

Service Time

Completion Time

Time in System

1 1.4 1.4 1.4 0.0 2.3 3.7 2.3

2 1.3 2.7 3.7 1.0 1.5 5.2 2.5

3 4.9 7.6 7.6 0.0 2.2 9.8 2.2

4 3.5 11.1 11.1 0.0 2.5 13.6 2.5

5 0.7 11.8 13.6 1.8 1.8 15.4 3.6

6 2.8 14.6 15.4 0.8 2.4 17.8 3.2

7 2.1 16.7 17.8 1.1 2.1 19.9 3.2

8 0.6 17.3 19.9 2.6 1.8 21.7 4.4

9 2.5 19.8 21.7 1.9 2.0 23.7 3.9

10 1.9 21.7 23.7 2.0 2.3 26.0 4.3

Total 21.7 11.2 20.9

Averages 2.17 1.12 2.09

• Simulation results for 10 ATM Customer