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Practical No: 1

Single channel queuing model.

Aim:-

Implement single channel queuing model using an excel

program. Perform the following tasks to implement the

following concept.

Write an excel program to do the following:

1.Use built in function to generate the n-digit randomnumbers.

2.Distribution of time between arrivals.a.Accept probability values for the interval-

arrival time between (1-6 minutes).

Generate the random digit assignment

b.Distribution of service time. Accept probvalues for the service time between(1-6 minutes). Generate the random- digit

assignment.

3.Total No. of customers N=20.4.Prepare a simulation table & answer the following

queries such as find

i) Avg. waiting time (in minutes)ii) Probability that a customer has to

wait.

iii) Probability of idle server.iv) Avg. Service time (in minutes)v) Avg. time between arrivals (in minutes)vi) Avg. waiting time of those who wait (in

minutes)

vii) Avg. time customer spends in thesystem(in minutes)

Concept:-

Its calling population, the nature of the arrivals,

the service mechanism, the system capacity, & the queuing

discipline, describes a queuing system.In the single channel queue, the calling population is

finite. Arrivals for service occur one at a time in a

random fashion; once they join the waiting line, they are

eventually served. In the addition, service times are of

some random length according to a probability

distribution, which does not change over time. The system

capacity has no limits. A single server or channel serves

units in order of their arrivals.

Arrivals & services are defined by the distribution of the

time between arrivals and the distribution of the service

times, respectively. For any single or multi channel queue,the overall effective arrival rate must be less than the

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total service rate, or the waiting line will grow without

bound, they are termed explosive or unstable.Concepts in queuing system:

1.State: The no. of units in the system & the status ofthe server, busy or idle.

2.Event: Set of circumstances that cause an instantaneouschange in the state of the system.3.In a single channel queue there are only 2 possible

events that can change the state of the system. There

are entry of a unit into the system or the completion

of service.

4.The system includes a server, the unit being served &units in the queue.

5.Simulation clock is used to track the simulated time.

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Practical No: 2

Multi-channel queuing model.

Aim:-

Implement Multi-channel queuing model using a C programwith one server getting higher performance. Perform the

following tasks to implement the following concept.

Write a C program to do the following:

1)Use built-in function to generate the n-digited randomnumbers

2)Distribution of inter-arrivals of cars.a) Accept probability values for the inter-arrival time

between (1-4 minutes).

Generate the random-digit assignment.

b) Service distribution of Baker. Accept probability

values for the service time between (3-6 min). Generatethe random-digit assignment.

c) Service distribution of baker. Accept probability

values for the service time between (3-6 min). Generate

the random-digit assignment

3)Total no. of customers N=20.4)Prepare a simulation table & answer the following

queries such as find.

1)Percentage of time Able & Baker individually werebusy over total service time.

2)Percentage of arrivals had to wait.3)Avg. Service time (minutes)4)Avg. time between arrivals (minutes)5)Avg. waiting time for all customers6)Avg. waiting time for all customers who had to

wait

Draw the conclusion regarding the system & cost of

waiting.

Concept:-

Its calling population, the nature of the arrival, the

service mechanism, the system capacity, and the queuing

discipline, describes a queuing system.

In the single channel queue, the calling population is finite.Arrivals of service occur one at a time in a random fashion;

once they join the waiting line, they are eventually served.

In the addition, service times are of some random length

according to, a probability distribution that does not change

over time. The system capacity has no limit. Units are served

in order of their arrivals by a single channel or server.

Arrivals & services are defined by a distribution of the

time between arrivals & the distribution of the service times,

respectively. For any single or multi channel queue, the

overall effective arrival rate must be less than total service

rate, or the waiting line will grow without bound; they aretermed explosive or unstable.

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Concepts in queuing systems:

1.State: The number of units in the system & the status ofthe server busy or idle.

2.Event: Set of circumstances that cause an instantaneouschange in the state of the system.

3.In the single channel queue, there are only two possibleevents that can change the state of the system. There arethe entries of the units into the system or the

completion of the service.

4.The system includes a server, the unit being served &units in the queue.

5.Simulation clock is used to track the simulated time.6.We find how well the current arrangement is working.

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Practical No: 3

Inventory system

Aim:

Simulate Inventory system The Newspaper Sellers

Problem

Perform the following:

1.Use the in-built function to generate the n-digitrandom number.

2.Random digit assignment for type of Newsday.3.Accept the probabilities for the 3 types of news days:

Good, Fair, & Poor.

4.Prepare the simulation table for purchase of 70newspapers & find out the total profit.

5.Total profit = (revenue from sales)-(cost ofnewspaper)- (lost profit from excess demand) + (salvagefrom sale of scrap papers)

6.For C/C++ programming use the library functions togenerate the n-digited random numbers.

Given:

Distribution of Newspapers Demanded

Demand Probability Distribution

Demand Good Fair Poor

40 0.03 0.10 0.4450 0.05 0.18 0.22

60 0.15 0.40 0.16

70 0.20 0.20 0.12

80 0.35 0.08 0.06

90 0.15 0.04 0.00

100 0.07 0.00 0.00

Random digit assignment for Type of Newsday

Accept the probability for the three type of news days:

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Good, Fair and Poor.

Prepare the simulation table for Purchase of 70

newspapers and find out the total profit.

Total Profit= (revenue from sales)(Cost of newspapers)(lost profit from excess demand) + (Salvage from sale of

scrap papers).

Concept:-

An important class of simulation problems involves

inventory systems. A simple Inventory system includes a

periodic view length N, at which time the Inventory level is

checked. An order is make to bring the inventory up to the

level M. At the end of the first review period, an order

quantity, Q1, is placed. In the Inventory system the lead time(i.e. the length of time between the placement & the receipt

of an order) is zero. Since the demands are not usually known

with certainty, the order quantities are probabilistic. Demand

is uniform over the time period. One possibility is that

demands all occur at the beginning of the cycle. Another is

that the lead-time is random of some positive length.

If the amount in Inventory drops below zero, indicating a

shortage. These units are backordered; when the order arrives,

the demand for the backordered items is satisfied first. To

avoid shortage, a buffer, or safety, stock would need to becarried.

Carrying stock in inventory has an associated cost

attributed to the interest paid on the funds borrowed to buy

the items (this also could be considered as the loss from not

having the funds available for other investments purposes).

Other costs can be placed in the carrying or holding cost

column: renting of storage space, hiring guards, & so on. An

alternative to carrying high inventory is to make more

frequent reviews, & consequently, more frequent purchases or

replenishments. This has an associated cost: the orderingcost. Also, there is a cost in being short. Customers may get

angry, at subsequent loss of good will. Larger inventories

decrease the possibilities of shortages. These costs must be

traded off in order to minimize the total cost of an inventory

system.

The total cost (pr total profit) of an inventory system

is the measure of performance. This can be affected by the

policy alternatives; the decision maker can control the

maximum inventory level M, & length of cycle N.

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Practical No: 4

Discrete Distributions

Aim:

Implement discrete distribution for statistical models with

discrete random variable.Perform the following task to show the concept.

a)Define discrete distribution.b)Describe the purpose of discrete distribution.c)Write C/C++/Excel code for

i) Bernoulli distribution.

1.Accept probability of success p, the probability offailure, no. of trials from the user.

2.Calculate probability mass function (pmf) P(X),mean E(X)& Variance V(X).

3.Display calculated terms.ii) Binomial distribution.

1. Accept probability of success p, the probability of

failure, number of trails from the user.

2.Calculate probability mass function (pmf) P(X),mean

E(X) & Variance V(X)

3.Display calculated terms.

iii) Geometric Distribution:

1.Accept probability of success p, the probability of failure,

number of trials x to achieve first success from the user

2.Calculate probability mass function (pmf) P(X), mean E(X) &Variance V(X)

3.Display calculated terms.

iv) Poisson Distribution:-

1.Accept the mean & variance from the user.

2.Calculate probability mass function (pmf)

3.Display calculated terms.

Concept:-

In modeling real world systems there are few situations wherethe actions of the entities within the system under study cannot

be completely predicted in advance. There may be many causes of

variations. Some statistical models can describe these

variations.

Sampling the appropriate parameters can develop an appropriate

model. The model builder selects a known distribution form; make

an estimate of the parameters of the distribution.

Discrete random variable:

Let X is a random variable. If the number of possible values

is X is finite, or countable infinite, X is called a discreterandom variable. The possible values of X may be listed as x1,

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x2 In the finite case the list terminates. In the countableinfinite case the list continues indefinitely.

Discrete random variables are used to describe random

phenomenon in which only integer values can occur.

1)Bernoulli trials & Bernoulli distribution.Consider an experiment with n trials, each of which can

be a success or failure. These are called Bernoulli process if

the trials are independent; each trial has only two possible

outcomes (Success or Failure), & the probability of a success

remains constant from trial to trial.

Then p(x1, x2, x3 xn)= p1(x1). p2(x2). p3(x3).. pn(xn).& pj(xj)= p(xj) =p, xj=1=success, j=1,2,..,n.

=1-p=q xj=1=failure, j=1,2,.,n.=0 otherwise.

The mean = E(Xj) = 0 . q+1 . p = p.

The variance = V(Xj) = [(02

.q)+ (12

.p)]- p2

= p(1-p)

2) Binomial Distribution:-

The random variable X that denotes the number of

successes in n Bernoulli trials has a Binomial Distribution

given by p(x), where

p(x) = nCxpxqn-x , x=0,1,2,.....,n

= 0,

E(X) = np

V(X) = npq

3) Geometric distribution:-The Geometric distribution is related to the sequence of

Bernoulli trials; the random variable of interest, X, is

defined to be the number of trials to achieve the first

success. The distribution of X is given by:

p(x) = qx-1 p, x=1,2,.= 0 , otherwise

E(X) = 1/p

V(X) = q/p2

Geometric distribution is memoryless.

4) Poisson distribution:-The Poisson distribution describes many random processes.

The Poisson pmf is given by:

p(x) = e-@ @x

x!

E(X) = @ =V(X)

F(x) = e-@ @i/i!

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Practical no 5

Continuous Distribution

Aim:-

Implement continuous distribution for statistical models

with discrete random variable.

Perform the following task to show the concept.

a) Define continuous distribution.b) Describe the purpose of continuous distributionc) Write Excel code for

i) Uniform distribution:

a) Accept parameters a,b, from the userb)Calculate Probability distribution function (pdf) f(X),

Cumulative distribution function (cdf) F(X), where X isdistributed uniformly between a & b.

c)Display calculated terms.

ii) Exponential distribution:

a) Accept parameters lamda from the user

b)Calculate Probability distribution function (pdf) f(X),

Cumulative distribution function (cdf) F(X)

c)Display calculated terms.

iii) Gamma distribution:

a) Accept parameters & from the userb)

Calculate Probability distribution function (pdf) f(X),Cumulative distribution function (cdf) F(X)

c) Display calculated terms.iv) Erlang distribution:

a) Accept parameters k & from the userb) Calculate Probability distribution function (pdf) f(X),

Cumulative distribution function (cdf) F(X)

c) Display calculated terms.v) Normal distribution:

a) Accept parameters mean & variance 2 & transformationvariable

b) z=t-/ from the user.c) Calculate Probability distribution function (pdf) f(X),

Cumulative distribution function (cdf) F(X)

d) Display calculated terms.vi) Weibull Distribution:

a) Accept parameters of shape >0, scale >0 & scale vfrom the user.

b) Calculate Probability distribution function (pdf)f(X), Cumulative distribution function (cdf) F(X)

c) Display calculated terms.

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vii) Triangular Distribution:

a)Accept parameters a, b, c from the user where a

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= 0, otherwise.

The cdf is given by

F(x) =0 x0 if its pdf is given by, f(x)=1* e

-1*

It is used to model interarrival times when arrivals are

completely random & to model service time that are highly

variable. In these instances, is the rate.F(x) = 0, x0

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Practical No: 6

Generation of Random Numbers:

Aim: -

Implement Generation of Random Numbers using followingmethods:

a.Define Linear Congruential Method.b.Define Combined Linear Congruential Methodc.Write C/C++/Excel program for:

i) Linear Congruential Method.a)Accept constant multiplier a, increment c,

modulus m, and seed X0 from the user.b)Display random numbers.

ii) Combined Linear Congruential MethodConcept:-

Linear Congruential Method: -

The Linear Congruential Method, initially proposed

by Lehmer(1951), produces a sequence of integer X1, X2between zero and m-1 according to the following recursive

relationship:

Xi+1 = (aXi+ c) mod m, i = 0, 1, 2, . ------(1)The initial value X0 is called the seed, a is called

the constant multiplier, c is the incremental and m is the

modulus. If c!=0 in equation (1), the form is called the

mixed Congruential Method. When c=0, the form is known as

the multiplicative Congruential Method. The selection of

the value for a, c, m and X0 drastically affects the

statistical properties and cycle length. Variation of

equation (1), are quite common in the computer generation

of random numbers.

Combined Linear Congruential Method: -

As computing power has increased, the complexity of

the system that we are able to simulate gas also

increased. Random number generators will period 2

31

-1 =2*109.

One fruitful approach is to combine two or more

multiplicative congruential generators in such a way that

the combined generator has a good statistical property

and a longer period. The following result from

LEcuyer[1998] suggest how this can be done.

If Wi,1, Wi,2, ., Wi,k, are any independent, discretevalued random variables but one of them say Wi,1, is

uniformly distributed on the integer 0 to mi-2, then

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k

Wi = Wi,j mod mi-1j=1

is uniformly distributed on the integer 0 to mi-2.To see how this result can be used to form combined

generator, let Xj,1, Xj,2, .., Xj, k be the ith output

from k different multiplicative congruential

generator, where the jth generator has prime modulus

mj, and the multiplier aj is chosen so that the

period is mj-1. Then the jth generator is producing

integer Xi,j , that are approximately uniformly

distributed on 1 mj-1 and Wi,j-1 is approximatelyuniformly distributed on 0 to mj-2 LEcuyer[1998]thus suggested combined generator of the form:

k

Xi = (-1)j-1Xi,j mod mi-1j=1

with

Xi/m1 Xi>0

Ri=

m1-1/m1 Xi=0

Notice that the (-1)j-1 Xi,j-1; for eg: if k=2, then

(-10 ) (Xi,1-1) (-11 ) (Xi,2-1) = 2 (-1j-1) (Xi,j)

The maximum possible period for such a generator is

(m1-1)(m2-1).(mk-1)P= 2k-1

which is achieved by the following generator.

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Practical No: 7

Test for Random Number

Aim: -

Implement a menu driven program to test the Random

Number using following test:

1)Frequency Test:a)Chi-Square Test: Accept n random numbers, level of

significance number of intervals n, set ofobserved values & set of estimated values

b)Kolmogrov Test: Accept n random numbers, number ofobservations N, the set of Observations (Ri) where

1

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Sampling distribution D is known and is tabulated as a

function of N, for testing against a uniform cdf, the

test procedure follows these steps:

Step1.: Rank the data from smallest to largest. Let

R(i) denote the ith smallest observation, so thatR(1)

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Inverse Transform Technique

Aim: -

Implement Inverse Transform technique to find the randomvariants:

Perform the following task to show the concept.

a)Define Inverse Transformb)Write Algorithm for continuous distributions.Write C code for implementing Inverse Transform

technique.

i) Exponential distribution:a.Accept parameters & from the user.b.Calculate Probability distribution function

(pdf) f(X), Cumulative distribution function(cdf) F(X).

c.Display calculated terms.ii) Weibull distribution:

a.Accept parameters of shape >0, scale >0 &scale v from the user.

b.Calculate Probability distribution function(pdf) f(X), Cumulative distribution function

(cdf) F(X).

c.Display calculated terms.iii) Gamma distribution:

a.Accept parameters lamda from the user.b.Calculate Probability distribution function

(pdf) f(X), Cumulative distribution function

(cdf) F(X).

c.Display calculated terms.iv) Triangular distribution:

a.Accept parameters a, b, & c from the user,where abc.

b.Calculate Probability distribution function(pdf) f(X), Cumulative distribution function

(cdf) F(X), where X is distributed uniformly

between a, b.

c.Display calculated terms

Concept:-

The inverse transform technique can be used to sample

from the exponential, the uniform, the Weibull, and the

Triangular distributions and the empirical distributions

Additionally, it is the underlying principle for sampling

from a wide variety of discrete distributions. The technique

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will be explained in detail for the exponential distribution

and then applied to other distributions. It is most straight

forward, but not always more efficient, technique

computationally.

(1) Exponential distribution:Exponential distribution, has probability density

function (pdf) given by

f(x)= e- x, x>= 00, x< 0

And cumulative distribution function (cdf) given by

F (x)= -xf (t)dt = 1- e-x, x>= 0

0, x< 0

The parameter can be interpreted as the mean number ofoccurrences per time unit. For e.g. if inter arrival times

X1, X2, X3 had an exponential distribution with rate , then could be interpreted as the mean number of arrivals pertime unit, or the arrival rate. Notice that for any i

E (Xi) = 1/So that 1/ is the mean inter arrival time. The goal is

to develop a procedure for generating values X1, X2, X3 that

have an Exponential distribution. The inverse transformtechnique can be utilized, at least in principle, for any

distribution, but it is most useful when the cdf, F(x), is of

such simple form that its inverse, F-1, can be easily

computed. A step-by-step procedure for the

Inverse transform technique, illustrated by the exponential

distributions, is as follows:

Step1: Compute the cdf of the desired random variable X. For

the exponential distribution, the cdf is F(x)= 1-e- x

,x>=0.

Step2: Set F(x)=R on the range of X. For the exponential

distribution, it becomes 1-e- x = R on the range x>=0.

Since X is a random variable, it follows that 1-e- x is also

a random variable, here called R. As will be showed later, R

has a uniform distribution over the interval (0,1).

Step3: Solve the equation F (x)=R on the range of X. For the

exponential distribution, the solution proceeds as follows:

1-e- x = R

-e- x =1- R

- x= In (1-R)

X = -1/ In (1-R) (I)

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Equation I is called a random variate generator for the

exponential distribution. In general, equation I is written

as X = F-1(R). Generating a sequence of values is accomplished

through step 4.

Step4: Generate (as needed) uniform random numbers R1, R2,

R3, and compute the desired random variates by:Xi=F-1 (Ri)

For the exponential case, F-1 (R) = (-1/)In (1-R) byequation(I), so that

Xi = -1/1 In (1-R) (II)

For I = 1, 2, 3, one simplification that is usually employedin equation(II), is to be replace 1-Ri by Ri to yield

Xi = -1/1 In (1-R) (III)

which is justified since both Ri and 1- Ri are uniformly

distributed on (0,1)

(2)Weibull distribution: -

Step 1 : The cdf is given by

F (x)= 1-e-(x/) x>=0

Step 2 : F (x)= 1-e-(x/

) = R

Step 3 : Solving for X in terms of yields

X=[-In(1-R)]1-

(3)Gamma distribution

If X is following exponention with parameter

nthen X is gamma(n, ). To generate one observation or

i=1

One random variable from gamma, we have to first

generate n observations from exponential say X1, X2,...,

Xn then Y= X is random observation from gamma.n

Therefore Y=-1/*log(1-R)i=1

(4)Triangular distribution:

Consider a random variable X that has pdf.

This distribution is called Triangular distribution with

endpoints (0,2) & mode at 1. Its cdf is given by:

f(x) = 2(x-a) a x b(b-a)(c-a)

= 2(c-x) b x c(c-b)(c-a)

= 0 otherwise (I)

F(x) = (x-a)2 a x b(b-a)(c-a)

= (c-x)2 b x c(c-b)(c-a)

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= 0 x>c (II)

By equation (I), 0

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printf("| %d\t | %lf\t | %f\t

|\n",i+1,R[i],x[i]);

}

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

getch();

}

Output

Random Variate Generation for Exponential Distribution

------------------------------------------------------

Pdf of Exponential Distribution is lamda*e^(-lamda*x)

Cdf of Uniform Distribution is 1-e^(-lamda*x)

Enter the value of lamda(mean): 2

Enter the number of random numbers you are going to provide :

5

Random Numbers Generated are:

0.98 0.89 0.06 0.91 0.77

-----------------------------------------------

| i | R[i] |X[i]=-1/a*log(1-R[i]|

-----------------------------------------------

| 1 | 0.980000 | 1.956012 |

| 2 | 0.890000 | 1.103637 |

| 3 | 0.060000 | 0.030938 |

| 4 | 0.910000 | 1.203973 |

| 5 | 0.770000 | 0.734838 |

----------------------------------------------

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(2) Weibull Distribution#include

#include

#include

#include

#include

int n,i;

double x[50];

float a,b,p;

float R[10],r1;

void main()

{clrscr();

printf("\nRandom Variate Generation for Weibull Distribution

\n");

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

printf("\nPdf of Weibull Distribution is

(beta/(alpha)^beta)*(x^beta-1)(e^(-x/alpha)^beta)\n");

printf("\nCdf of Weibull Distribution is 1-e^(-x/alpha)^beta\n");

printf("\nEnter the value of alpha(float): ");

scanf("%f",&a);

printf("\nEnter the value of beta(float): ");

scanf("%f",&b);p=a/b;

printf("\nEnter the number of random numbers you are going to

provide : ");

scanf("%d",&n);

randomize();

printf("\n\n Random Numbers Generated are:\n\n");

for (i=0;i

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

getch();

}

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Output

Random Variate Generation for Weibull Distribution

--------------------------------------------------

Pdf of Weibull Distribution is (beta/(alpha)^beta)*(x^beta-

1)(e^(-x/alpha)^beta)

Cdf of Weibull Distribution is 1-e^(-x/alpha)^beta

Enter the value of alpha(float): 2.5

Enter the value of beta(float): 3.4

Enter the number of random numbers you are going to provide : 10

Random Numbers Generated are:

0.0400 0.6300 0.4200 0.5700 0.2400 0.4200 0.1600 0.5900

0.0500 0.6100

----------------------------------------------------------

| i | R[i] | X[i]=(-alpha/beta)*log(1-R[i])|

----------------------------------------------------------

| 1 | 0.040000 | 0.030016 |

| 2 | 0.630000 | 0.731068 |

| 3 | 0.420000 | 0.400535 |

| 4 | 0.570000 | 0.620566 |

| 5 | 0.240000 | 0.201792 |

| 6 | 0.420000 | 0.400535 |

| 7 | 0.160000 | 0.128201 |

| 8 | 0.590000 | 0.655587 |

| 9 | 0.050000 | 0.037716 |

| 10 | 0.610000 | 0.692359 |

----------------------------------------------------------

(3) Gamma Distribution

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#include

#include

#include

#include

#includevoid main()

{

int n,i,j,n1;

float r[100],a,r1,r2,y,R[10];

clrscr();

printf("\nRandom variate generation using Gamma

Distribution\n");

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

printf("\n\tConvolution Method");

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

printf("\nHow many gamma variates should be generated: ");scanf("%d",&n1);

for(j=0;j

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Output

Random variate generation using Gamma Distribution

--------------------------------------------------

Convolution Method

------------------

How many gamma variates should be generated: 2

Enter the number of random numbers required to generate 1

gamma variate:5

Random Numbers Generated are:

0.3800 0.9200 0.4500 0.4600 0.2600

Enter the value of lamda: 0.5

Gamma variate is : 9.037786

Enter the number of random numbers required to generate 2

gamma variate:4

Random Numbers Generated are:

0.5300 0.8300 0.0800 0.2300

Enter the value of lamda: 0.3

Gamma variate is : 9.572418

(4)Triangular Distribution

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#include

#include

#include

#include

#includeint a,b,n,i;

float x[50],r1;

double R[100];

void main()

{

clrscr();

printf("\nRandom Variate Generation for Triangular

Distribution \n");

printf("\n---------------------------------------------------

--\n");

printf("\nEnter the number of random numbers you are going toprovide : ");

scanf("%d",&n);

randomize();

printf("\n\n Random Numbers Generated are:\n\n");

for (i=0;i

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Output

Random Variate Generation for Triangular Distribution

Enter the value of lamda(mean): 1

Enter the number of random numbers you are going to provide :

20

Random Numbers Generated are:

0.92 0.71 0.62 0.09 0.32 0.76 0.72 0.77

0.63 0.07

0.70 0.61 0.98 0.12 0.64 0.35 0.63 0.36

0.33 0.02

-----------------------------------------------

| i | R[i] | X[i]=-1/alog(1-R[i]|

-----------------------------------------------

| 0 | 0.920000 | 2.525729 |

| 1 | 0.710000 | 1.237874 |

| 2 | 0.620000 | 0.967584 |

| 3 | 0.090000 | 0.094311 |

| 4 | 0.320000 | 0.385662 |

| 5 | 0.760000 | 1.427116 |

| 6 | 0.720000 | 1.272966 || 7 | 0.770000 | 1.469676 |

| 8 | 0.630000 | 0.994252 |

| 9 | 0.070000 | 0.072571 |

| 10 | 0.700000 | 1.203973 |

| 11 | 0.610000 | 0.941609 |

| 12 | 0.980000 | 3.912023 |

| 13 | 0.120000 | 0.127833 |

| 14 | 0.640000 | 1.021651 |

| 15 | 0.350000 | 0.430783 |

| 16 | 0.630000 | 0.994252 |

| 17 | 0.360000 | 0.446287 || 18 | 0.330000 | 0.400478 |

| 19 | 0.020000 | 0.020203 |

---------------------------------------------

7/29/2019 Simulation prac

28/28

Practical No:09

Acceptance-Rejection Technique:

Aim: -

Implement methods for generation of random variates, X for a

given Distribution.

Perform the following tasks to show the concept:

a.Define Acceptance Rejection Technique.b.Write algorithm for Acceptance Rejection Technique.c.Write C/C++/excel code for implementing:

1)Poisson distributiona)Accept value of mean , arrivals n and

probability of random variate p.

b)Display the calculated Poisson variates.Concept:-

Method for generating random variates, X, uniformly

distributed between and one way to proceed would be to

follow these steps:

Step1 : Generate a random numbers R.

Step2 : a) If R>=1/4, accept X=R then goto step 3.

b) If R0 has pmfp(n)= P(N=n)=e-

n/n!, n=0,1,2,3,.N can be interpreted as the number of arrivals from

Poisson arrival process in one

unit of time.

Step1 : Set n=0, P=1

Step2 :.Generate the random numbers Rn+1 and replace P by P*Rn+1.Step3 : If P