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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