Simulation Practical

PAPER III SECTION II

SIMULATION AND MODELING

INDEX

No.TopicPage No.DateSign

1Single-Channel Queuing Model2-6

2Multi-Channel Queuing Model7-10

3Inventory System11-14

4Discrete Distribution15-25

5Continuous Distribution26-42

6Generation of Random Numbers43-48

7Random Numbers Tests49-64

8Acceptance-Rejection Technique65-69

Practical-01Single-Channel Queuing ModelWrite a C/C++/Excel Program to simulate a Single-Channel Queuing Model.

For C/C++ Programming use the library functions to generate the n-digited random numbers.

For Excel use the in-built function to generate the n-digited random numbers.

Distribution of Time between Arrivals

Accept probability values for the Inter-arrival time between (1 minute 6 minutes). Generate the Random-digit assignment.

Distribution of Service Time

Accept probability values for the Service Time between (1 minute 6 minutes). Generate the Random-digit assignment.

Total No. of Customers N=20.

Prepare a simulation table and answer the following queries such as find

i) Avg. waiting time (minutes)

ii) Probability that a customer has to wait

iii) Probability of idle server

iv) Avg. service time (minutes)

v) Avg. time between arrivals (minutes)

vi) Avg. waiting time of those who wait (minutes)

vii) Avg. time customer spends in the systems (minutes)

Theory: Simulation of queuing system: A queuing system is described by its calling population, the nature of the arrivals

the service mechanism, the system capacity, and the queuing discipline. Thus for simulation of queuing system, we need to generate arrivals and service times. The

time for arrivals and service times should be random( as time represents real world situation). These random times are generated with the help of random nos.

Example: Single Channel Queue

In the single queue, the system works an (First-In-First-Out) FIFO by a single server or channel.

In a single channel queuing system, inter-arrival times and service times are generated from the distributions of random numbers

Given Data:

Distribution of Service Time:Accept probability values for the Service Time between (1 minute 6 minutes). Generate the Random-digit assignment.

Service-Time Distribution:Service Time Distribution

Service Time (Minutes)ProbabilityCumulative Probability

10.100.10

20.200.30

30.300.60

40.250.85

50.100.95

60.051.00

This is service-time distribution in which service time varies from 1 to 6 minutes with the probabilities.

Distribution of Time between Arrivals:Accept probability values for the Inter-arrival time between (1 minute 6 minutes). Generate the Random-digit assignment.This is the simulation study for Single-Channel Queuing System

Distribution of Time between Arrivals

Time between Arrivals (Minutes)ProbabilityCumulative Probability

10.1250.125

20.1250.250

30.1250.375

40.1250.500

50.1250.625

60.1250.750

70.1250.875

80.1251.000

Explanation of Simulation Table for Single Channel Queuing System ColumnDescription

Customerthe customers in order of their arrival.

Inter-arrival Timethe inter-arrival time of the customer at the random fashion.

Arrival Timethe arrival time of the previous customer and the present customer.

Service Timethe estimated time taken by the system to complete the service.

Time Servicethe starting time of the service.

Waiting Time in Queuethe time taken by the customers wait in queue.

Time Service Endsthe time at which the service to customers ends.

Time Customer spendsthe total time taken by the customer to complete its service in the system.

Idle Time of Serverthe time at which server was in idle state.

1.

2.

3.

4.

5.

6.

7.

Simulation Table for Queuing Example:ClockClockClock

CustomerInterarrivalTime(Minutes)Arrival TimeServiceTime(Minutes)TimeServiceBeginsWaiting Timein Queue(Minutes)Time ServiceEndsTime CustomerSpends in System(Minutes)Idle Timeof Server(Minutes)

100400440

2884801244

31921231450

411021441660

551521611830

662132102433

762762703363

812843353790

953323743960

1063953904450

1134244424860

1274934905231

1325115215320

1415245315750

1535545726160

1645926126340

1746346306740

1846716706810

1947137107433

2047537507831

7577463799258628815

VBA Code:

To insert the Visual Basic Code Tools Macro Visual Basic Editor

Module:

Option Explicit

Function Discrete(Prob As Range, Value As Range)

Dim lRowLast, row As Long

Dim UpperLimit As Double, U As Single

U = Rnd()

lRowLast = Prob.Rows.Count

For row = 1 To lRowLast

UpperLimit = Prob.Cells(row, 1).Value

If U 0 has a normal distribution if has a pdf

Algorithm:

Begin

1) Read the randon variable x,mean,variance.

2) if variance > 0

sigma=sqrt(variance)

pdf=(1/(sigma*sqrt(2*3.142)))*exp((-1/2)*(pow((x-

mean)/sigma,2)));

Note : Its not possible to evaluate cdf in close form

End5.5 Weibull DistributionWrite a C/C++/Excel Program to find Probability Distribution Function (pdf) and cumulative Distribution Function (cdf) for Weibull Distribution Accept for a random variable X with the location parameter (, scale parameter ( > 0, and the shape parameter ( > 0 from the user.

Theory: The random variable X has a weibull distribution if its pdf has the form

The three parameters of the weibull distribution are

1. v () ( the location parameter;

2. ( (( > 0) ( the scale parameter.

3. ( (( > 0) -( the shape parameter.

When the v=0, Weibull distribution becomes

When v = 0 and ( = 1. Letting ( = 1 the Weibull Distribution are given by the following

expression.

The mean & variance of the Weibull are given by the following expression :

Where

Algorithm:

Begin

1) Read the random variable x.

2) Read the location parameter v.

3) Read the scale parameter alpha

4) Read the shape parameter beta.

5) if alpha >0 && beta >0

if x > v || v==0

pdf=(beta/alpha)*(pow((x-v)/alpha,beta-1)*exp(pow(-((x-v)/alpha),beta)));

else

if v == 0 && alpha == 1 && x >= 0

pdf=(1/alpha)*exp(-(x/alpha));

else

pdf=0

if x >= v

cdf=1-exp(-(pow((x-v)/alpha,beta)));

else

cdf=0

End

5.6 Triangular Distribution

Write a C/C++/Excel Program to find Probability Distribution Function (pdf) and Cumulative Distribution Function (cdf) for Triangular Distribution Accept parameters a,b,c where a ( b ( c from the user.

Theory: A random variable X has a triangular distribution if its pdf is given by

The mode occurs at x = b. The parameters ( a, b, c) can be related. The cdf is given by

The mean

5.7 Lognormal Distribution

Write a C/C++/Excel Program to find Probability Distribution Function (pdf) and Cumulative Distribution Function (cdf) for Lognormal Distribution Accept parameters mean ( and variance (2 from the user.

Theory: A random variable X has a Lognormal Distribution if its pdf is given by

Where > 0. The mean and variance of lognormal random variable are

The parameters come from the fact that when Y has N () distribution, then X = e Y has a lognormal distribution with parametes. If the mean & variance of the lognormal are know to be , respectively, then the parameters are given by

Code:#include

#include

#include

#include

void uniform_distribution();

void exponential_distribution();

void weibull_distribution();

void erlang_distribution();

void normal_distribution();

void triangular_distribution();

void lognormal_distribution();

long fact(int);

void main()

{

int choice,flag=1;

clrscr();

while(flag==1)

{

printf("\n\t\t\t Continuous Distribution.");

printf("\n\n1. Uniform Distribution.");

printf("\n2. Exponential Distribution.");

printf("\n3. Erlang Distribution.");

printf("\n4. Normal Distribution.");

printf("\n5. Weibull Distribution.");

printf("\n6. Triangular Distribution.");

printf("\n7. Lognormal Distribution.");

printf("\n\n\nSelect Any One Of The Options : ");

scanf("%d",&choice);

flag=0;

}

switch(choice)

{

case 1:

clrscr();

uniform_distribution();

break;

case 2:

clrscr();

exponential_distribution();

break;

case 3:

clrscr();

erlang_distribution();

break;

case 4:

clrscr();

normal_distribution();

break;

case 5:

clrscr();

weibull_distribution();

break;

case 6:

clrscr();

triangular_distribution();

break;

case 7:

clrscr();

lognormal_distribution();

break;

default:

break;

}

getch();

}

void uniform_distribution()

{

float a,b,x,pdf,cdf,mean,variance;

float f[10];

printf("\t\t\t CONTINIOUS DISTRIBUTION");

printf("\n\n\t\t\t Uniform Distribution ");

printf("\nEnter a = ");

scanf("%f",&a);

printf("\nEnter b = ");

scanf("%f",&b);

printf("\nEnter Random Variable(x) = ");

scanf("%f",&x);

if(a=v || v==0)

{

ans1=(x-v)/alpha;

pdf=(beta/alpha)*pow(ans1,beta-1)*exp(pow(-ans1,beta));

printf("\n\nResults : ");

printf("\nPDF is %.3f",pdf);

}

else

if(v==0 && alpha==1 && x>=0)

{

pdf=(1/alpha)*exp(-(x/alpha));

printf("\n\nResults :");

printf("\nPDF is %.3f",pdf);

}

else

printf("\nPDF is 0.");

if(x>=v)

{

cdf=1-exp(-(pow( ((x-v)/alpha),beta)));

printf("\nCDF is %.3f",cdf);

}

else

{

printf("\nCDF is 0");

}

mean=(float)v+( alpha * fact(1/beta));

ans=fact(2/beta)-pow(fact(1/beta),2);

variance=(float)pow(alpha,2)*ans;

printf("\n\nMean E(X) = %.0f",mean);

printf("\n\nThe variance V(X) = %.0f",variance);

getch();

}

void triangular_distribution()

{

float a,b,c,x,cdf,pdf,r1,mean,var,mode;

int flag=1;

printf("\t\t\t CONTINIOUS DISTRIBUTION");

printf("\n\n\t\t\t Triangular Distribution");

printf("\n\nEnter the random variable x :");

scanf("%f",&x);

printf("\nEnter the parameter values such that a 0 has pmf

N can be interpreted as the number of arrivals from the Poisson arrival process in one unit of time.

Algorithim:

Step1 : n=0, P=1Step2 : Generate a random number R

EMBED Equation.3 and replace P by P. R

Step3 : if p < e- , then accept N = n. Otherwise, reject the surrent n, increase n

by one, and retuen to step 2.

8.2 Gamma Distribution :

Algorithm :Step1 : Compute a =

Step2 : Generate R1 and R2.

Step3 : Compute X =

Step4a : if X use X as the desired variate. The generated variates

from step 4a will have mean & variance both equal to .

Step4b : if X >reject & return to step 2.

Code :

//Acceptance-rejection Technique

#include

#include

#include

#include

void poisson_distribution();

void gamma_distribution();

void main()

{

int choice;

clrscr();

x:

printf("\n\n Acceptance-Rejection Techinque");

printf("\n 1: Poisson Distribution");

printf("\n 2: Gamma Distribution");

printf("\n 3: Exit");

printf("\n Enter Choice : ");

scanf("%d",&choice);

switch(choice)

{

case 1:

clrscr();

poisson_distribution();

goto x;

case 2:

clrscr();

gamma_distribution();

goto x;

case 3:

exit(0);

}

getch();

}

void poisson_distribution()

{

float n=0,P=1,R[20],alpha,q;

int x,count=0,N=0;

printf("\n\n\t\t\t Poisson Distribution");

printf("\n\n Enter the number of Poisson variates : ");

scanf("%d",&x);

printf("\n\n Enter the mean alpha : ");

scanf("%f",&alpha);

//calculating e exp -alpha

q=exp(-(alpha));

randomize();

//printing column values

printf("\n n \t R[n+1] \t P \t Accept/Reject \t Result");

printf("\n--------------------------------------------------------------");

while(count