Or Simulation Lecture

8/6/2019 Or Simulation Lecture

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Chapter-Ten

Simulation


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Simulation Cont d

Simulation is an alternative form of analysisSimulation is an alternative form of analysis

when thewhen the problem situations are too complex to beproblem situations are too complex to be

represented by the concise mathematical techniques.represented by the concise mathematical techniques.


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Simulation Cont d

Fig. 10.1 Model Types


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Types of Simulation

Systems may have discrete or continuous state

Based on the use of either a continuous or discrete time

representation.

1.1. Continuous simulation:Continuous simulation: The state changes all the time, not

just at the time of some discrete events .For example: the water level in a reservoir with given in andout flow may change all the time.


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Types of Simulation Cont dDiscrete event simulation has applications in a widerange of sectors including manufacturing and servicesectors.

automotiveautomotivehealthcarehealthcareelectronicselectronicspharmaceuticalspharmaceuticalsfood and beveragefood and beverage

packagingpackaginglogistics, etc.logistics, etc.


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Types of Simulation Cont d

BasedBased onon thethe representationrepresentation of of thethe modelsmodels

1. Analogue simulation (AS):- In this type of simulation, anoriginal physical system is replaced by an analogous physicalmodels that is easier to manipulate.

Examples:M anned space flight

Treadmills that simulate automobile tire wear inlaboratory.


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Steps of Simulation Process

The process of simulating a system consists of following steps:

1. Identify the problem

2. Identify the decision variables, and decide theperformance criterion (objective) and decision rules

3. Construct a numerical model

4. Validate the model

5. Design the experiments

6. Run the simulation model

7. Examine the results


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Identify the problem

M odify the model by changingthe input data, i.e. values of

decision variables

Identify decision variables, performancecriterion and decision rules

Construct simulation model

Validate the model

Design experiments (specify values of decision variables to be tested )

Examine the results and selectthe best course of action

Run or conduct the simulation

Issimulationcompleted?

F ig.10. 2 Steps of the simulation process

No

Yes


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M onte Carlo Simulation

One characteristic of some systems that makes them difficult tosolve analytically is that they consist of random variables

represented by probability distributions.

Thus, a large proportion of the applications of simulations are

for probabilistic models.

The term M onte Carlo has become synonymous with probabilistic

simulation in recent years.

M onte Carlo is a technique for selecting numbers randomly from

a probability distribution .


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M onte Carlo Cont d

InIn casecase itit isis notnot possiblepossible toto describedescribe aa systemsystem inin termsterms of of standardstandard probabilityprobability distributiondistribution suchsuch asas normal,normal, Poisson,Poisson,

exponential,exponential, gamma,gamma, etcetc..,, anan empiricalempirical probabilityprobability distributiondistribution

cancan bebe constructedconstructed. .TheThe M onteM onte CarloCarlo processprocess isis analogousanalogous toto gamblinggambling devicesdevices.

ItIt maymay bebe usedused whenwhen thethe modelmodel containscontains elementselements thatthat exhibitexhibit

chancechance inin theirtheir behavior,behavior,


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M onte Carlo Cont d

TheThe M onteM onte CarloCarlo techniquetechnique consistsconsists of of followingfollowingstepssteps::

1. Set up probability distributions for important variables

2. Build a cumulative probability distribution for each variable3. Establish an interval of random numbers for each variable

4.4. GenerateGenerate randomrandom numbersnumbers

5. Simulate a series of trials by means of random sampling

6. Repeat step 5until the required number of simulation runshas been generated.

7. Design and implement a courses of action and maintaincontrol


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Random Number (RN) Generation

1. Arithmetic ComputationThe nth random number r n consisting of k-digitsgenerated by using multiplicative congruentialmethod given by

W here p and m are positive integers, P


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RN Generation Cont dTo start the process of generating random numbers, the firstnumber r0 is specified by the user.For illustration, let p= 35, m= 100 and arbitrary start with r 0=57.Since m-1 = 99 is the 2=digit number, therefore, it will generate2-digit random numbers:

r1 = pr0 (modulo m)= 35 x 57 (modulo 100)

1,995/100 = 9 5, reminder

r2 = pr1 (modulo m)= 35 x 57 (modulo 100)3,325/100 = 2 5, reminderr3 = pr2 (modulo m)= 35 x 57 (modulo 100)

875/100 = 75, reminder


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RN Generation Cont d2

. Computer GeneratorThe random numbers that are generated by using computer

software are uniformly distributed decimal fractions

between 0 and 1.The software works on the concept of cumulative

distribution function for the random variables for which we

are seeking to generate random numbers.


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RN Generation Cont dFor example, for the negative exponential function with

density function,

The cumulative distribution function is given by,

Taking logarithm on both sides, we have

,0,)( g! xe x f xPP

)(1,

1)(0

x F eor

edxe x F

x

x x x

!

!!

)](1log[)/1(,

)](1log[

x F xor

x F x!!


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RN Generation Cont d

If r = F (x) is a uniformly distributed randomdecimal fraction between 0 and 1, then theexponential variables associated with r is givenby

This is an exponential process generator since 1-ris a random number and can be replaced by r.

.log)/1()1log()/1( r r xn PP !!


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RemarkI. W e can pick up random numbers from

random table, orII. Use built-in Excel formula to generate

random numbers


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W hile picking up random numbers from the random

number table

The starting point could be randomly chosen

Start with any number in any column or row, and

proceed in the same column or row to the nextnumber, but a consistent, unvaried pattern should be

followed in drawing random numbers..

W e should not jump from one number to another

indiscriminately


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Example1: Demand and supply

Using random numbers from the given table;

a) Simulate the demand for the next 10days

b) Also estimate the daily average demand for tires on

the bases of simulated data

c) Compare the results with the expected daily demand .


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Solution

D aily D emand for Tires Frequency

Probability of Occurrence

CumulativeProbability

0 10 10/ 200 = . 05 .051 2 0 20/ 200 = .1 0 .1 5

2 4 0 40/ 200 = .2 0 .35

3 60 60/ 200 = . 30 .65

4 4 0 40/ 200 = .2 0 .8 5

5 30 30/ 200 = .1 5 1. 00

200 days 2 00/ 200 = 1. 00

Table 10.3. Probability of D emand


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Solution Cont d

D aily D emand Probability

CumulativeProbability

Interval of RandomNumbers

0 .05 .05 0 1 through 05

1 .1 0 .1 5 06 through 15

2 .2 0 .35 16 through 35

3 .30 .65 36 through 654 .2 0 .8 5 66 through 85

5 .1 5 1. 00 86 through 99, 00

Table 10.4 Assignment of Random Numbers Assignment of Random Numbers


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Simulation Example 1Simulation Example 1D ay D ay

Number Number

RandomRandom

Number (r)Number (r)

Simulated Simulated D aily D emand D aily D emand

11 3 93 9 33

22 7373 4433 7272 44

44 7575 4455 3 73 7 3366 0 20 2 00

77 8787 55

88 9898 5599 1010 11

1010 4747 33Total= 3 2Total= 3 2

Average= 3 .2Average= 3 .2


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Example 2 : Simulation of a Queuing System

ConsiderConsider thethe CaseCase of of drivedrive--inin marketmarket whichwhich consistsconsists of of oneonecashcash registrarregistrar (the(the serviceservice facility)facility) andand aa singlesingle queuequeue of of

customerscustomers. . TheThe interinter arrivalarrival timetime andand serviceservice timetime isis asas inin tabletable

aa andand bb..

AssumeAssume thatthat thethe timetime intervalsintervals betweenbetween customercustomer arrivalsarrivals areare

discretediscrete randomrandom variablesvariables. .


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Queuing System Cont d

Arrival interval(min), x

ProbabilityP(x)

1.0 . 2 02 .0 .40

3.0 .30

4.0 .10

Service time(min), y

ProbabilityP(y)

0.5 . 2 0

1.0 .50

2 .0 .30

Table a. D istr ibut ion of arr ivalTable a. D istr ibut ion of arr ivalinter val t imeinter val t ime

Table b. D istr ibut ion of

Table b. D istr ibut ion of ser vice t imeser vice t ime

For 10 customers arr ivals to the cash reg istrar, Calculatea ) Average wa iting t imeb) Average queue l inec) Average t ime in the system


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SolutionF irst we have to develop the cumulative probabilityF irst we have to develop the cumulative probabilitydistribution, to determine random number rangesdistribution, to determine random number ranges ..

Arrival

interval(min), x

Probability

P(x)

Cumulative

probability

Random number

range, r 1

1.0 . 2 0 .2 0 01- 2 02 .0 .40 .60 2 1 60

3.0 .30 . 9 0 61 9 04.0 .10 1.00 9 1- 99 , 00

Table a1. Range of random numbers for arrival interval timeTable a1. Range of random numbers for arrival interval time


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Solution Cont dSolution Cont d

Custom-er

r1

Arrivalinterv-al

,x

Arrivalclock

Enterfacilityclock

Waiti-ng

T ime

Lengthof

QueueAfterEnter

r2 Servi-ce

T ime, y

Depar-ture

Clock

T ime insystem

1 - - 0.0 0.0 0.0 0.0 65 1.0 1.0 1.0

2 71 3.0 3.0 3.0 0.0 0.0 18 .5 3.5 .5

3 12 1.0 4.0 4.0 0.0 0.0 17 .5 4.5 .5

4 48 2.0 6.0 6.0 0.0 0.0 89 2.0 8.0 2.0

5 18 1.0 7.0 8.0 1.0 1 83 2.0 10.0 3.0

6 08 1.0 8.0 10.0 2.0 1 90 2.0 12.0 4.0

7 05 1.0 9.0 12.0 3.0 2 89 2.0 14.0 5.08 18 1.0 10.0 14.0 4.0 2 08 .5 14.5 4.5

9 26 2.0 12.0 14.5 2.5 2 47 1.0 15.5 3.5

10 94 4.0 16.0 16.0 0.0 0 06 .5 16.5 .5TotalTotal 1 2 .512 .5 88 2 4.52 4.5

Table c. Simulation of the queuing system for 1 0 customersTable c. Simulation of the queuing system for 1 0 customers


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Solution Cont dSolution Cont dOnce the simulation is complete, we can compute operatingOnce the simulation is complete, we can compute operatingcharacteristics from the simulation results as follows.characteristics from the simulation results as follows.

Average waiting time=Average waiting time=

Average queue length=Average queue length=

Average timeAverage timein the system=in the system=

r percustomecustomers

min25.110

min5.12 !

customer customers

customers80.

108 !

r percustomecustomers

min45.210

min5.24 !


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Example 3: Simulation of a machine breakdown andExample 3: Simulation of a machine breakdown andMaintenance SystemMaintenance System

A continuous probability distribution of the time betweenmachine breakdowns is given by;

W hen a machine breaks down, it must be repaired; and it takes

either one, two, or three days for the repair to be completed,

according to the discrete probability distribution shown intable (I)

w eeks x x

x f 40,8

)( ee!

W here X = W eeks between mach ine breakdowns


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Machine breakdown Cont dMachine breakdown Cont d

The company would like to know if it should implement a

machine maintenance program at a cost of $20,000 per year

that would reduce the frequency of breakdowns and thus the

time for repair.

The maintenance program would result in the following

continuous probability function for time between breakdowns

w eeks x x x f 60,18

)( ee!

W here x= W eeks between mach ine breakdowns


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Machine breakdown Cont dMachine breakdown Cont d

The reduced repair time resulting from the maintenanceprogram is defined by the discrete probability distributionshown in table (II)

M achine repairtime, Y (days)

Probability of repairtime, p(y)

1 .402 .503 .10

Table II. Revised probability distribution of m/c repair timewith the maintenance program


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SolutionSolution

r1 T ime betweenB/D, Xwks

r2 Repair time,ydys

Cost, $ 2 ,000y Cumulative time,xwks

.45.45 2.682.68 .19.19 22 4,0004,000 2.682.68

.90.90 3.803.80 .65.65 22 4,0004,000 6.486.48

.84.84 3.673.67 .51.51 22 4,0004,000 10.1510.15

.17.17 1.651.65 .17.17 22 4,0004,000 11.8011.80

.74.74 3.443.44 .63.63 22 4,0004,000 15.2415.24

.94.94 3.883.88 .85.85 33 6,0006,000 19.1219.12

.07.07 1.061.06 .37.37 22 4,0004,000 20.1820.18

.15.15 1.551.55 .89.89 33 6,0006,000 21.7321.73

.04.04 0.800.80 .76.76 33 6,0006,000 22.5322.53

.31.31 2.232.23 .71.71 33 6,0006,000 24.7624.76

.07.07 1.061.06 .34.34 22 4,0004,000 25.8225.82

.99.99 3.983.98 .11.11 11 2,0002,000 29.8029.80

.97.97 3.943.94 .27.27 22 4,0004,000 33.7433.74

.73.73 3.423.42 .10.10 11 2,0002,000 37.1637.16

.13.13 1.441.44 .59.59 22 4,0004,000 38.6038.60.03.03 0.700.70 .87.87 33 6,0006,000 39.3039.30

.62.62 3.153.15 .08.08 11 2,0002,000 42.4542.45

.47.47 2.742.74 .08.08 11 2,0002,000 45.1945.19

.99.99 3.983.98 .89.89 33 6,0006,000 49.1749.17

.75.75 3.463.46 .42.42 22 4,0004,000 52 .6352 .63$ 84,000

Table Simulation of the m/c breakdown and repair without the maint. Prog.


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SolutionSolution

r1 T ime betweenB/D, Xwks r2 Repair time,ydys Cost,$2 ,000y Cumulative time,xwks

.45.45 4.034.03 .19.19 11 2,0002,000 4.034.03

.90.90 5.695.69 .65.65 22 4,0004,000 9.729.72

.84.84 5.505.50 .51.51 22 4,0004,000 15.2215.22

.17.17 2.472.47 .17.17 11 2,0002,000 17.6917.69

.74.74 5.165.16 .63.63 22 4,0004,000 22.8522.85

.94.94 5.825.82 .85.85 22 4,0004,000 28.6728.67

.07.07 1.591.59 .37.37 11 2,0002,000 30.2930.29

.15.15 2.322.32 .89.89 22 4,0004,000 32.5832.58

.04.04 1.201.20 .76.76 22 4,0004,000 33.7833.78

.31.31 3.343.34 .71.71 22 4,0004,000 37.1237.12

.07.07 1.591.59 .34.34 11 2,0002,000 38.7138.71

.99.99 5.975.97 .11.11 11 2,0002,000 44.6844.68

.97.97 5.915.91 .27.27 11 2,0002,000 50.5950.59

.73.73 5.125.12 .10.10 11 2,0002,000 55.7155.71$ 42 ,000

Table Simulation of the m/c breakdown and repair with the maint. Prog.


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Summary of the resultSummary of the result

Option1:Option1:W

ith out the maintenance programCost = $84,000

Option 2Option 2 : W ith maintenance programCost = $20,000+ $42,000

= $62,000

Therefore Option 2 is better for the organization withProfit = $84,000 - $62,000

= $22 ,000= $22 ,000


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Role of Computers in Simulation

Computers are critical in simulating complex tasksComputers are critical in simulating complex tasks

Computers are used to:Computers are used to:

Generate random numbers;Generate random numbers;

Simulate the given problem with varying values of variablesSimulate the given problem with varying values of variables

in few minutes; andin few minutes; and

help the decisionhelp the decision- -maker to prepare reports which enablemaker to prepare reports which enable

him to make decisions quickly as well as draw validhim to make decisions quickly as well as draw valid

conclusions.conclusions.


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Computers in Simulation Cont d

Com puter languages a va ilableCom puter languages a va ilable1.1. GeneralGeneral- -pur pose languagespur pose languages FOR T R AN,FOR T R AN,

B ASIC, P AS C AL, COBO L, C++ etc.B ASIC, P AS C AL, COBO L, C++ etc.

2.2. Sp ec ialSp ec ial--pur pose s imulat ion languagespur pose s imulat ion languages -- GP SS ,GP SS ,SIMS CR IP T, DYN AMO, SIMULA , AREN A etc.SIMS CR IP T, DYN AMO, SIMULA , AREN A etc.1.1. Requ ire less programm ing t ime for large s imulat ionsRequ ire less programm ing t ime for large s imulat ions

2.2. Usually more eff icient and eas ier to check for errorsUsually more eff icient and eas ier to check for errors

3.3. RandomRandom- -number generators are bu ilt innumber generators are bu ilt in

Spreadsheets such as Excel can be used to developSpreadsheets such as Excel can be used to developsome simulationssome simulations


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Using Software in SimulationUsing Software in Simulation


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Simulation ApplicationsSurvey c onducte d dur ing the 1980s indi cate that a lar ge major ity of

major corporat ions use simulation in such f unctional area s as pr oduct ion, corporate p lannin g, engineer ing, f inancial analysis,

research a nd d eve lo pment, market ing, infor mation systems, and

per sonn el.

Follow ing are descr i ptions of so me of the more comm on

app licat ions of Simulation:

Queu ing

Inventory contr olPr oduct ion and manuf actur ing

Fin ance

Market ing


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Applications Cont d Assembly-l ine balanc ing

Park ing lot and harbor des ign

Distr ibut ion system des ign

S chedul ing a ircraft

Labor-h ir ing dec is ions

Personnel schedul ingT raff ic-light t iming

Bus schedul ing

Ta xi, truck, and ra ilroaddis patch ing

Product ion fac ilityschedul ing

Plant layout

Product ion schedul ing

S ales forecast ing

Inventory plann ing andcontrol


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Advantages of SimulationAdvantages of Simulation1.1. R elatively stra ightfor war d and flex ib leR elatively stra ightfor war d and flex ib le

2.2. Can be used to analyze lar ge and com plex rea lCan be used to analyze lar ge and com plex rea l--wo r ld situat ionswo r ld situat ions

that cann ot be solved by conventional modelsthat cann ot be solved by conventional models

3.3. R ealR eal--wo r ld com plicat ions can be included that most wo r ld com plicat ions can be included that most

mathe matical models cann ot per mi tmathe matical models cann ot per mi t

4.4. Tim e com pression is poss ib leTim e com pression is poss ib le

5.5. A llows whatA llows what --if type s of q uestionsif type s of q uestions

6.6. Does not in ter f ere with rea lDoes not in ter f ere with rea l--wo r ld systemswo r ld systems

7.7. Can study the in teract ive eff ect s of indi vidual com ponents or Can study the in teract ive eff ect s of indi vidual com ponents or

var ia bles in or der to deter min e which ones are im portantvar ia bles in or der to deter min e which ones are im portant


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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. Users must generate all of the conditions and constraints for

solutions they want to examine

4. Each simulation model is unique

Or Simulation Lecture

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