Simulation Technique

8/3/2019 Simulation Technique

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SIMULATION


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SIMULATION vs. OPTIMIZATION

In an optimization model, the values of the decision

variables are outputs.

In a simulation model, the values of the decisionvariables are inputs. The model evaluates theobjective function for a particular set of values.

The result of the model is a set of values for thedecision variables that will maximize (or minimize)

the value of the objective function.

The result of the model is a measure of the qualityof a suggested solution and the variability in variousperformance measures due to randomness in the

inputs.


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When should simulation be used?

Simulation is one of the most frequently used tools of

quantitative analysis today because:

1. Analytical models may be difficult or impossible toobtain, depending on complicating factors.

2. Analytical models typically predict only average orsteady-state (long-run) behavior.

3. Simulation can be performed with a variety of

software on a PC or workstation. The level ofcomputing and mathematical skill required todesign and run a simulator has been substantiallyreduced.


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8. Experimental Design

9. Model runs and analysis

10. More runs

NoYes

3. Model conceptualization 4. Data Collection

5. Model Translation

6. Verified

7. Validated

Yes

No

No No

Yes

Phase 3

Experimentation

1. Problem formulation

2. Set objectives and overall project plan

Phase 1

Problem Definition

Phase 2

Model Building

11. Documentation, reporting and

implementation

Phase 4

Implementation

Steps in simulation


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Verification (efficiency)

Is the model correctly built/programmed? Is it doing what it is intended to do?

Validation (effectiveness) Is the right model built? Does the model adequately describe the reality

you want to model? Does the involved decision makers trust the

model?

Two of the most important and most challenging

issues in performing a simulation study

Model Verification and Validation


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Find alternative ways of describing/evaluating the system

and compare the results

Simplification enables testing of special cases with

predictable outcomes

Removing variability to make the model deterministic Removing multiple job types, running the model with one job type

at a time

Reducing labor pool sizes to one worker

Build the model in stages/modules and incrementally test

each module

Uncouple interacting sub-processes and run them separately

Test the model after each new feature that is added

Simple animation is often a good first step to see if things are working

as intended

Model Verification Methods


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Simulation and Random Variables

Simulation models are often used to analyze a decisionunder risk. Under risk, the behavior of one or morefactors is not known with certainty. For example:

demand for a product during the next month

the return on an investment

the number of trucks that will arrive to be unloaded

The factor that is not known with certainty is called therandom variable.

The behavior of the random variable can be describedby a probability distribution.

MONTE CARLO METHOD:


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Design of Docking Facilities. In the following model,trucks of different sizes carrying different types of loads,arrive at a warehouse to be unloaded.

Exit Entrance

Truck

Truck

Truck

Dock3

Dock2

Dock1

Truck waiting Truck waiting


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The uncertainties are:

When will a truck arrive?

What kind and size of load will it be carrying?How long will it take to unload the trucks?

Each uncertain quantity would be a random variable

characterized by a probability distribution.The planners must address a variety of designquestions:

How many docks should be built?What type and quantity of material-handlingequipment are required?

How many workers are required over what

periods of time?


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The design of the unloading dock will affect its cost ofconstruction and operation. Management must balancethe cost of acquiring and using the various resourcesagainst the cost of having trucks wait to be unloaded.


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Determination of Inventory Control Policies. Simulationcan be used to study inventory control models.

In this model, the factory produces goods that are sentto the warehouses to satisfy customer demand.

The random variables are: daily demand at eachwarehouse and shipping times from factory to

warehouse.

Factory

Warehouse 1 Warehouse 2 Warehouse 3

Demand Demand Demand


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Simulation can be used to study inventory controlmodels.

Some of the operational questions are:

The main costs are:

When should a warehouse reorder from thefactory and how much?

How much stock should the factory maintain tosatisfy the orders of the warehouses?

Cost of holding the inventory

Cost of shipping goods from a factory to a

warehouseCost of not being able to satisfy customerdemand at the warehouse

The objective is to find a stocking and ordering policythat keeps the total cost low while meeting demand.


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To generate a random variable, draw a random sample

from a given probability distribution.

Generating Random Variables

Two broad categories of random variables:

Can assume only certain specific values(e.g., integers)

Can take on any fractional value (an infinitenumber of values)

Continuous

Discrete

Using a Random Number Generator in a Spreadsheet


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To generate a discrete random variable with the RAND()function in a spreadsheet, two things are needed:

A GENERALIZED METHOD:

1. The ability to generate discrete uniform randomvariables

2. The distribution of the discrete random variable

to be generatedTo generate a continuous random variable, two thingsare needed:

1. The ability to generate continuous uniformrandom variables on the interval 0 to 1

2. The distribution (in the form of the cumulativedistribution function) of the random variable to

be generated


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Conversion of a Random No. to a Uniform Distribution


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200 Random Numbers Generated Between 0 and 100


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200 Uniform Discrete Random Numbers GeneratedBetween 20 and 100

Th C l i Di ib i F i (CDF) C id


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The Cumulative Distribution Function (CDF). Consider arandom variable, D, the demand. The CDF for D[calledF(x)] is then defined as the probability that Dtakes on avalue < x.

F(x) = Prob{D < x}

Knowing the probability distribution for D, the CDF for

key values of Dis:

X 8 9 10 11 12 13F(x) 0.1 0.3 0.6 0.8 0.9 1.0

H i h f h CDF T di


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0

0.2

0.4

0.6

0.8

1

1.2

7 8 9 10 11 12 13 14

Probability

x

F(x)

Here is a graph of the CDF. To generate a discretedemand using the graph:

Step 2: Read theparticular valueof the randomquantity, d, on

this axis

Step 1: Locatethe particularvalue of Uon

this axis

u

d

S t t d l di t if


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Suppose you want to model a discrete uniformdistribution of demand where the values of 8 through 12all have the same probability of occurring (uniform,equally likely).

The spreadsheet has a function, =RAND(), that returns arandom number between 0 and 1. However, this willresult in a continuous uniform distribution.

To create a discrete uniform distribution, use the INT()function. For example:

In general, if you want a discrete, uniform distribution ofinteger values between xand y, use the formula:

INT(x+ (yx+ 1)*RAND() )

G ti f th E ti l Di t ib ti


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Generating from the Exponential Distribution.The exponential distribution is often used to model thetime between arrivals in a queuing model. Its CDF isgiven by:

F(x) = Prob{t>T} = e-lT

Where 1/l is the mean of the random variable T.Therefore, we want to solve the following equation for w:

u = e-lT

The solution is: T = -1/l ln(u)where we can get u from Random number which

represents cumulative distribution function

G ti f th N l Di t ib ti


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Generating from the Normal Distribution.The normal distribution plays an important role in manysimulation and analytic models.

Consider drawing a random demand from a normaldistribution with a mean (m) of 1000 and a standarddeviation (s) of 100.If Z is a unit normal random variable (normallydistributed with a mean of 0 and a standard deviation of1) then m + Zs is a normal random variable with mean mand standard deviation s.So, we can draw from a unit normal distribution. Excelhas a built-in function that can do this:

= NORMINV( RAND() , 1000, 100)

Excel will automatically return a normally distributedrandom number with mean 1000 and std. dev. 100.


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CAPITAL BUDGETING

PROBLEM

A CAPITAL BUDGETING EXAMPLE


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A CAPITAL BUDGETING EXAMPLE:ADDING A NEW PRODUCT LINE

Airbus Industry is considering adding a new jet airplane

(model A3XX) to its product line. The following financialinformation is available:

Startup Costs $150,000Sales Price $ 35,000Fixed Costs (per year) $ 15,000Variable Costs (per year) 75% of revenues

Tax depreciation on the new equipment would be

$10,000 per year over the 4 year expected product life.

Salvage value of the equipment at the end of the 4 yearsis estimated to be 0.

Airbus cost of capital is 10% and tax rate is 34%.

If demand is known then a spreadsheet can be used to


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If demand is known, then a spreadsheet can be used tocalculate the net present value(NPV). For example,assume that the demand for A3XXs is 10 units for eachof the next 4 years:

=C16 + C13

=-$B$2

=NPV($D$3,C17:F17)+B17

THE MODEL WITH RANDOM DEMAND
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It is unlikely that demand will be the same every year. Amore realistic model would be one in which demandeach year is a sequence of random variables.


This model of demand is appropriate when there is aconstant base level of demand that is subject to randomfluctuations from year to year.

Sampling Demand with a Spreadsheet: Assume initiallythat the demand in a year will be either 8, 9, 10, 11, or 12units with each value being equally likely to occur.

This is an example of a discrete uniform distribution.

Now, use the formula =INT(8+ 5*RAND() ) to samplefrom a discrete uniform distribution on the integers 8, 9,10, 11, 12 .

Multiple trials can be performed by pressing the

recalculation key for the spreadsheet (e.g., F9).

Using this formula results in random demands
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=INT(8+5*RAND() )

Using this formula results in random demands.

Hitting the F9 key would result in a different sample ofdemands, and possibly a different NPV.

The demands are random variables, therefore, the NPVis also a random variable.

EVALUATING THE PROPOSAL
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Two questions need to be answered about the NPVdistribution:

EVALUATING THE PROPOSAL

1. What is the meanor expected value of the NPV?

2. What is the probability that the NPV assumes anegative value (making the proposal to add theA3XX less attractive)?

To answer these questions, a simulation model must bebuilt. To run the simulation automatically and capture

the resulting NPV in a separate spreadsheet, use theData Tablecommand.

The resulting analysis gives the estimated mean NPV


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The resulting analysis gives the estimated mean NPVand standard deviation.

Downside Risk and Upside Risk: To get a better ideaabout the range of possible NPVs that could occur, look

at the minimum and maximum NPVs.

In the resulting analysis the Frequency (column B)
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In the resulting analysis, the Frequency(column B)indicates the number of trials that fell into the bins(categories) defined by column A.

The cumulative % column indicates the cumulativepercentage of observations that fall into each categoryor bin.

The histogram gives a visual representation of the


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The histogram gives a visual representation of thedistribution of NPVs. Note that it is somewhat bellshaped.

How Reliable is the Simulation? Now the two questions
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The next questions to ask are:

How Reliable is the Simulation? Now the two questionsabout the distribution can be answered:

1. What is the meanor expected value of the NPV?

2. What is the probability that the NPV assumes anegative value (making the proposal to add the

A3XX less attractive)?

In this trial, the mean is $12,100.

In this trial, the probability is


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For a 95% confidence interval, the formula is:

estimated mean + 1.96(standard deviation)

In this case, the standard deviation is the standard error

(the standard deviation divided by the square root of thenumber of trials).

Based on this trial, the upper and lower confidence

limits are: =$E$4-1.96*$E$8/SQRT($E$16)

=$E$4+1.96*$E$8/SQRT($E$16)

So, we have 95% confidence that the true mean NPV is

somewhere between $9,679 and $14,521.



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It is unlikely that demand will be the same every year. Amore realistic model would be one in which demandeach year is a sequence of random variables.


This model of demand is appropriate when there is aconstant base level of demand that is subject to randomfluctuations from year to year.

Sampling Demand with a Spreadsheet: Assume initiallythat the demand in a year will be either 8, 9, 10, 11, or 12units with each value being equally likely to occur.

This is an example of a discrete

uniform distribution.

OTHER DISTRIBUTIONS OF DEMAND
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Originally, we started with equal mean demands of 10 foreach period (year). Then, we allowed for randomvariation in mean demand (between 8 and 12 units) anddiscrete distribution.

OTHER DISTRIBUTIONS OF DEMAND

Now, assume the mean demand will stay the same over

the next four years, somewhere between 6 and 14 units ayear, with all values being equally likely.

This scenario can be modeled as a continuous, uniformdistribution between 6 and 14.

In addition, we can explore the impact of differentdemand distributions on the NPV.


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MAINTAINANCE PROBLEM

Corrective Maintenance versus


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Corrective Maintenance versusPreventive Maintenance

The Heavy Duty Company has just purchased a large machine for anew production process.

The machine is powered by a motor that occasionally breaks down and

requires a major overhaul. Therefore, a second standby motor is kept,

and the two motors are rotated in use.

The breakdowns always occur on the fourth, fifth, or sixth day that themotor is in use. Fortunately, it takes fewer than three days to overhaul

a motor, so a replacement is always ready.

Cost of Replacement Cycle that Begins with Breakdown

Replace a Motor $2,000

Lost production during replacement 5,000

Overhaul a motor 4,000

Total $11,000


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Probability Distribution of Breakdowns

Day

Probability of

a Breakdown

Corresponding

Random Numbers

1, 2, 3 0

4 0.25 0.0000 to 0.2499

5 0.5 0.2500 to 0.7499

6 0.25 0.7500 to 0.99997 or more

Computer Simulation of Corrective


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Computer Simulation of CorrectiveMaintenance

Random Time Since Last Cumulative Cumulative Distribution of

Breakdown Number Breakdown Day Cost Cost Time Between Breakdowns

1 0.1343 4 4 $11,000 $11,000 Number

2 0.1523 4 8 $11,000 $22,000 Prob Cumu. Prob. of Days

3 0.9091 6 14 $11,000 $33,000 0.25 0 4

4 0.6161 5 19 $11,000 $44,000 0.5 0.25 5

5 0.6223 5 24 $11,000 $55,000 0.25 0.75 6

6 0.0026 4 28 $11,000 $66,000

7 0.6920 5 33 $11,000 $77,000 $11,000

8 0.6155 5 38 $11,000 $88,000

9 0.7805 6 44 $11,000 $99,000

10 0.9189 6 50 $11,000 $110,000

26 0.6869 5 128 $11,000 $286,00027 0.6415 5 133 $11,000 $297,000

28 0.3599 5 138 $11,000 $308,000

29 0.1493 4 142 $11,000 $319,000 Average Cost per Day

30 0.4700 5 147 $11,000 $330,000 $2,245

Breakd

own

Cost


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Preventive Maintenance Options

Preventive maintenance would involve scheduling the motor to beremoved (and replaced) for an overhaul at a certain time, even if a

breakdown has not occurred.

The goal is to provide maintenance early enough to prevent a

breakdown.

Scheduling the overhaul enables removing and replacing themotor at a convenient time when the machine is not in use, so no

production is lost.

Cost of Replacement Cycle that Begins without a BreakdownReplace a motor on overtime $3,000

Lost production during replacement 0

Overhaul a motor before a breakdown 3,000

Total $6,000

R l M t 4 D


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Replace Motor on 4 DaysRandom Time Until Scheduled Time Event That umulative Cumulative Distribution of

Cycle Number Breakdown ntil Replacemen Concludes Cycle Day Cost Cost Time Between Breakdowns

1 0.3926 5 4 Replacement 4 $6,000 $6,000 Number

2 0.4904 5 4 Replacement 8 $6,000 $12,000 Probability Cumulative of Days

3 0.8207 6 4 Replacement 12 $6,000 $18,000 0.25 0 44 0.6811 5 4 Replacement 16 $6,000 $24,000 0.5 0.25 5

5 0.1084 4 4 Breakdown 20 $11,000 $35,000 0.25 0.75 6

6 0.1032 4 4 Breakdown 24 $11,000 $46,000

7 0.5737 5 4 Replacement 28 $6,000 $52,000 Breakdown Cost $11,000

8 0.0718 4 4 Breakdown 32 $11,000 $63,000 Replacement Cost $6,000

9 0.5408 5 4 Replacement 36 $6,000 $69,000

10 0.9918 6 4 Replacement 40 $6,000 $75,000 Replace After 4 days

11 0.9948 6 4 Replacement 44 $6,000 $81,000

12 0.1657 4 4 Breakdown 48 $11,000 $92,00013 0.1035 4 4 Breakdown 52 $11,000 $103,000

14 0.3365 5 4 Replacement 56 $6,000 $109,000

15 0.1650 4 4 Breakdown 60 $11,000 $120,000

16 0.4908 5 4 Replacement 64 $6,000 $126,000

17 0.0267 4 4 Breakdown 68 $11,000 $137,000

18 0.1388 4 4 Breakdown 72 $11,000 $148,000

19 0.9297 6 4 Replacement 76 $6,000 $154,000

20 0.5908 5 4 Replacement 80 $6,000 $160,00021 0.1035 4 4 Breakdown 84 $11,000 $171,000

22 0.6132 5 4 Replacement 88 $6,000 $177,000

23 0.5361 5 4 Replacement 92 $6,000 $183,000

24 0.0726 4 4 Breakdown 96 $11,000 $194,000

25 0.0593 4 4 Breakdown 100 $11,000 $205,000

26 0.7241 5 4 Replacement 104 $6,000 $211,000

27 0.3092 5 4 Replacement 108 $6,000 $217,000

28 0.4092 5 4 Replacement 112 $6,000 $223,000

29 0.2356 4 4 Breakdown 116 $11,000 $234,000 Average Cost per Day

30 0.6483 5 4 Re lacement 120 $6 000 $240 000 $2 000

Replace Motor After 5 Days


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Replace Motor After 5 DaysRandom Time Unt il Scheduled Time Event That Cumulat ive Cumulat ive Distribution of

Cycle Number Breakdown Until Replacement Conc ludes Cycle Day Cost Cost Time Between Breakdowns

1 0.3543 5 5 Breakdown 5 $11,000 $11,000 Number

2 0.2204 4 5 Breakdown 9 $11,000 $22,000 Probability Cumulative of Days

3 0.0583 4 5 Breakdown 13 $11,000 $33,000 0.25 0 44 0.8282 6 5 Replacement 18 $6,000 $39,000 0.5 0.25 5

5 0.9815 6 5 Replacement 23 $6,000 $45,000 0.25 0.75 6

6 0.4620 5 5 Breakdown 28 $11,000 $56,000

7 0.7658 6 5 Replacement 33 $6,000 $62,000 Breakdown Cost $11,000

8 0.4318 5 5 Breakdown 38 $11,000 $73,000 Replacement Cost $6,000

9 0.8745 6 5 Replacement 43 $6,000 $79,000

10 0.9448 6 5 Replacement 48 $6,000 $85,000 Replace After 5 days

11 0.0987 4 5 Breakdown 52 $11,000 $96,000

12 0.5796 5 5 Breakdown 57 $11,000 $107,000

13 0.7489 5 5 Breakdown 62 $11,000 $118,000

14 0.2480 4 5 Breakdown 66 $11,000 $129,000

15 0.5809 5 5 Breakdown 71 $11,000 $140,000

16 0.9055 6 5 Replacement 76 $6,000 $146,000

17 0.5844 5 5 Breakdown 81 $11,000 $157,000

18 0.0157 4 5 Breakdown 85 $11,000 $168,00019 0.0949 4 5 Breakdown 89 $11,000 $179,000

20 0.1892 4 5 Breakdown 93 $11,000 $190,000

21 0.9239 6 5 Replacement 98 $6,000 $196,000

22 0.1051 4 5 Breakdown 102 $11,000 $207,000

23 0.8739 6 5 Replacement 107 $6,000 $213,000

24 0.5229 5 5 Breakdown 112 $11,000 $224,000

25 0.6667 5 5 Breakdown 117 $11,000 $235,000

26 0.3945 5 5 Breakdown 122 $11,000 $246,000

27 0.7721 6 5 Replacement 127 $6,000 $252,000

28 0.2253 4 5 Breakdown 131 $11,000 $263,000

29 0.6972 5 5 Breakdown 136 $11,000 $274,000 Average Cost per Day

30 0.5078 5 5 Breakdown 141 $11,000 $285,000 $2,021

Inventory Problem


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Inventory Problem

The daily demand of an item has the following distribution:

Demand per day(units)

0 1 2 3 4 5

No. of days on whichdemand occurred

4 10 15 39 22 10

When an order is placed to replenish inventory there is a delivery time

lag, which follows the following distribution:

Lead time (days) 2 3 4

No. of times ofoccurrence 24 12 4

The management is of the opinion that proportion of stockouts should

not exceed 5%.


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Queuing Problem


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Single server queue

A small grocery store with just one

checkout counter

Interarrival time uniform between 1-8 mins

Service time varies between 1-6 mins.

Probability distribution given.

The problem is to analyze the system bysimulating the arrival and service of 20

customers.


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Given distributions

Minimum 1Maximum 8

Interarrival Times

(minutes) 1 0.10 0.102 0.20 0.30

3 0.30 0.60

4 0.25 0.85

5 0.10 0.95

6 0.05 1.00

Service

Times

(Minutes)

ProbabilityCumulative

Probability

Uniform

Simulating of a M/M/1 Queue


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Assume a small branch office of a local bank with only

one teller. Empirical data gathering indicates that inter-arrival

and service times are exponentially distributed.

The average arrival rate = l = 5 customers per hour The average service rate = m = 6 customers per hour

Using our knowledge of queuing theory we obtain

= the server utilization = 5/6 0.83 Lq = the average number of people waiting in line

Wq = the average time spent waiting in line

Lq = 0.832/(1-0.83) 4.2 Wq = Lq/l 4.2/5 0.83 How do we go about simulating this system?

How do the simulation results match the analytical ones?

Simulating of a M/M/1 Queue


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NEWSPAPER BOY PROBLEM

bl


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Newspaper Boy Problem

Newsvendor sells newspapers on the street Buys forc = $0.55 each, sells forr = $1.00 each

Each morning, buys q copies

q is a fixed number, same every day

Demand during a day: D = max (X, 0) X~ normal (m = 135.7, s = 27.1), from historical data X roundsX to nearest integer

IfDq, satisfy all demand, and qD 0 left over, sell forscrap ats = $0.03 each

IfD > q, sells out (sells all q copies), no scrap

But missed out onDq > 0 sales

What should q be?

N B P bl F l i


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Newspaper Boy Problem Formulation

Choose q to maximize expected profit per day

q too smallsell out, miss $0.45 profit per paper

q too bighave left over, scrap at a loss of $0.52 per paper

Classic operations-research problem

Many versions, variants, extensions, applications

Much research on exact solution in certain cases

But easy to simulate, even in a spreadsheet

Profit in a day, as a function ofq:

W(q) =r min (D, q) +s max (q

D, 0)

cq

W(q) is a random variableprofit varies from day to day

Maximize E(W(q)) over nonnegative integers q

Sales revenue Scrap revenue Cost

Steps: Newspaper boy Problem


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Steps: Newspaper boy Problem

Simulation

Set trial value ofq, generate demandD, compute profit for

that day

Then repeat this for many days independently, average

to estimate E(W(q))

Also get confidence interval, estimate of P(loss),

histogram ofW(q)

Try for a range of values ofq

Need to generate demandD = max (X, 0) So need to generateX~ normal (m = 135.7, s = 27.1)

Demands = MAX(ROUND(NORMINV(RAND(),, ), 0)

Advantage of Simulation


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1. It is particularly well-suited for problems that are difficult or

impossible to solve mathematically.

2. It allows an analyst or decision maker to experiment with

system behavior in a controlled environment instead of in a

real-life setting that has inherent risks.

3. It enables a decision maker to compress time in order to

evaluate the long-term effects of various alternatives.

4. It can serve as a mode for training decision makers by

enabling them to observe the behavior of a system under

different conditions.

Advantage of Simulation

Limitations of Simulation


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1. Probabilistic simulation results are approximations, rather

than optimal solutions.

2. Good simulations can be costly and time-consuming to

develop properly; they also can be time-consuming to run,

especially in cases in which a large number of trials are

indicated.

3. A certain amount of expertise is required in order to design a

good simulation, and this may not be readily available.

4. Analytical techniques may be available that can provide better

answers to problems.

Limitations of Simulation


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THANK YOU

Documents

Simulation Technique