2WB05 SimulationLecture 1: Introduction andMonte Carlo simulation

Marko Boonhttp://www.win.tue.nl/courses/2WB05

November 12, 2012

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• studeerwijzer available (with notes, slides, programs, assignments)

• examination consists of 4 or 5 take home assignments. Assignments are given on Mondays. Some assign-ments take two weeks to complete, while others have to be handed in one week later.

• assignments done in groups of 2

• register your group (on the homepage of 2WB05) and register yourself in OASE.

Organisation

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Week Dates Monday Thursday1 12-18 Nov Course 12 19-25 Nov Course 2 Course 33 26 Nov - 2 Dec Course 44 3-9 Dec Course 55 10-16 Dec Course 6 Extra course ???6 17-23 Dec Course 77 7-13 Jan Course 8

Warning: this is a preliminary schedule. Check the course website for the definite schedule.

Schedule

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• Monte Carlo simulation

• Modelling of discrete-event systems

• Programming

• Random number generation

• Output analysis

Topics

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State changes at (random) discrete points in time

Examples:

• Manufacturing systemsCompletion times of jobs at machines, machine break-downs

• Inventory systemsArrival times of customer demand, replenishments

• Communication systemsArrival times of messages at communication links

Discrete-event systems

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Programming tools:

• General purpose languages (C, C++, Java, Mathematica, . . .);

• Simulation language χ developed by the Systems Engineering group

• Simulation system Arena

Discrete-event systems

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The law of large numbersZ1, Z2, . . . Zn are i.i.d. random variables with mean z := E[Z1] and finite variance.The probability of the sample mean being close to z is large. In fact, for every ε > 0,

limn→∞

P(∣∣∣∣ Z1 + Z2 + . . .+ Zn

n− z

∣∣∣∣ < ε

)= 1.

Estimator for z:

Z ≡ Zn :=Z1 + Z2 + · · · + Zn

nEstimator for P(Zi > t):

E[1{Zi>t}]

Monte Carlo simulation

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Simulation is a perfect tool to develop and sharpen your intuitionfor probabilistic models (see also Tijms’ book Spelen met kansen)

In no time probabilistic properties can be illustrated by a simulationexperiment and the results can be shown in graphs or tables!

Problems

• coin tossing

• (nearly) birth day problem

• lottery

Simple probabilistic problems

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Coin tossing

Two players A and B throw a fair coin N times.If Head, then A gets 1 point; otherwise B.

• What happens to the absolute difference in points as N increases?

• What is the probability that one of the players is leading between 50% and 55% of the time? Or more than95% of the time?

• In case of 20 trials, say, what is the probability of 5 Heads in a row?

Simple probabilistic problems

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Birthday problem

Consider a group of N randomly chosen persons.What is the probability that at least 2 persons have the same birthday?

Nearly birthday problem

What is the probability that at least 2 persons have their birthday within r days of each other?

Simple probabilistic problems

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Lottery

Each week a very popular lottery in Andorra prints 104 tickets. Each tickets has two 4-digit numbers on it, onevisible and the other covered. The numbers are randomly distributed over the lots. If someone, after uncoveringthe hidden number, finds two identical numbers, he wins a large amount of money.

• What is the average number of winners per week?

• What is the probability of at least one winner?

The same lottery prints 107 tickets in Spain. What about the answers to the questions above?

Simple probabilistic problems

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Monte Carlo simulation of Coin tossing

int n = 0;int pointsA = 0;int pointsB = 0;

while (n < N) {if (rand.nextDouble() < p)pointsA++;

elsepointsB++;

n++;System.out.println(pointsA - pointsB);

}

Note: rand is instance of java.util.Random

Coin Tossing

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Complete Java-code of simulating coin tossing

import java.util.Random;

public class CoinToss {

/* total number of trials */protected int N;/* probability on head */protected double p;/* random number generator */protected Random rand;

/*** Constructs a CoinToss object.* The probability on head is taken 0.5.* @param N the number of realisations*/

public CoinToss(int N) {this(N, 0.5);

}

/*** Constructs a CoinToss object.* The probability on head is 0.5.* @param N the number of realisations* @param seed the random seed for the random number generator*/

public CoinToss(int N, long seed) {this(N, 0.5, seed);

}

Coin Tossing

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/*** Constructs a CoinToss object.* The probability on head is 0.5.* @param N the number of realisations* @param p the probability on head.*/

public CoinToss(int N, double p) {this.N = N;this.p = p;this.rand = new Random();

}

/*** Constructs a CoinToss object.* @param N the number of realisations* @param p the probability on head.* @param seed the random seed for the random number generator*/

public CoinToss(int N, double p, long seed) {this.N = N;this.p = p;this.rand = new Random(seed);

}

Coin Tossing

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/*** Tosses a coin <i>N</i> times, and prints* <i>number of heads - number of tails</i>*/

public void printRealisation() {int n = 0; /* initialization */int points_A = 0;int points_B = 0;

while (n < N) {if (rand.nextDouble() < p) /* coin tossing */

points_A = points_A + 1; /* it is Head */else

points_B = points_B + 1;n = n + 1;System.out.println("" + (points_A - points_B));

}

}

public static void main(String[] arg) {CoinToss c = new CoinToss(1000, 0.5);c.printRealisation();

}

}

Coin Tossing

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Mathematica

realisation = RandomChoice[{-1, 1}, 10000]Count[realisation, 1]

cumulative = Accumulate[realisation];ListLinePlot[cumulative]

Coin Tossing

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2000 4000 6000 8000 10 000

50

100

Oscillations become bigger and bigger (grow as√

N )

Coin Tossing

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Fraction of time one of the players is leading

Let P(α, β) be the probability that one of the players is leading between 100α% and 100β% of the time

To determine P(α, β) do the experiment “Throw N times with a coin” many times; an experiment is successfulif one of the players is leading between 100α% and 100β% of the time

Then

P(α, β) ≈number of successful experiments

total number of experiments

Coin Tossing

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Simulation of M experiments

int success = 0

for (int run = 0; run < M; run++) {coinTossing();if ((alpha < 1.0 * time_A / N &&

1.0 * time_A / N < beta) ||(alpha < 1.0 * time_B / N &&1.0 * time_B / N < beta))

success++;}System.out.println(1.0 * success / M);

Coin Tossing

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Simulation of coin tossing

int n = 0;int points_A = 0;int points_B = 0;int time_A = 0

while (n < N) {if (rand.nextDouble() < 0.5) points_A++;else points_B++;if (points_A - points_B >= 0)time_A++;

n++;}

time_B = N - time_A;

Coin Tossing

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Results for M = 104, N = 104:

(α, β) P(α, β)(0.50, 0.55) 0.0648(0.50, 0.60) 0.1294(0.90, 1.00) 0.3972(0.95, 1.00) 0.2743(0.98, 1.00) 0.1692

Coin Tossing

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Histogram of time_A/N

0.2 0.4 0.6 0.8 1

100

200

300

400

500

600

Coin Tossing

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Successive Heads

Let P(k) be the probability of at least k successive Heads in case of 20 trials

To determine P(k) do the experiment “Throw 20 times with a coin” many times; an experiment is successful isa row of at least k Heads appears

Then

P(k) ≈number of successful experiments

total number of experiments

Coin Tossing

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Simulation of M experiments

int success = 0;

for (int run = 0; run < M; run++) {coinTossing();if (k_row) success++;

}

System.out.println(1.0 * success / M);

Coin Tossing

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Simulation of coin tossing

int n = 0;int nr_Heads = 0;boolean k_row = false;

while (n < 20 && !k_row) {if (rand.nextDouble() < 0.5)nr_Heads++;

else nr_Heads = 0;if (nr_Heads >= k) k_row = true;n++;

}

Coin Tossing

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Results for M = 104:

k P(k)1 1.000002 0.983173 0.786434 0.476675 0.250786 0.120817 0.058178 0.027169 0.0125510 0.00582 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

0.0

0.2

0.4

0.6

0.8

1.0

Coin Tossing

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Histogram of the maximum number of successive heads

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 170

5000

10 000

15 000

20 000

25 000

30 000

Coin Tossing

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Birthday problem

Let P(N ) be the probability that at least two persons have the same birthday in a group of size N

To determine P(N ) do the experiment “Take a group of N randomly chosen persons and get their birthdays”many times; an experiment is successful is at least two persons have the same birthday

Then

P(N ) ≈number of successful experiments

total number of experiments

Birthday problem

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Simulation of birthday problem

int success = 0;

for (int run = 0; run < M; run++) {"take group of size N"if (same_birthday)success++;

}

System.out.println(1.0 * success / M);

Birthday problem

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Simulation of taking random group

int n = 0;boolean[] birthday = new boolean[365];boolean same_birthday = false;

while (n < N && !same_birthday) {int newBirthday = rand.nextInt(birthday.length);if (birthday[newBirthday]) same_birthday = true;else birthday[newBirthday] = true;n++;

}

Birthday problem

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Results for M = 103 and seed = 1

N P(N )10 0.12615 0.26920 0.42225 0.57230 0.69340 0.89350 0.974

Birthday problem

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Simulation of the nearly birthday problem

int n = 0;boolean[] birthday = new boolean[365];boolean same_birthday = false;

while (n < N && !same_birthday) {int newBirthday = rand.nextInt(365);for (int i = -r; i < r; i++) {if (birthday[(newBirthday + i) % 365])

same_birthday = true;birthday[newBirthday] = true;

}n++;

}

Birthday problem

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Results for M = 103 and seed = 1

N r P(N )10 0 0.126

1 0.3242 0.4767 0.878

20 0 0.4221 0.811

30 0 0.6931 0.971

Birthday problem

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Lottery problem

Tickets are numbered 1, . . . , N

To print covered numbers, first generate a random permutation of 1, . . . , N and then print first number on firstticket, second one on second ticket, and so on; hence:

Number of winners is equal to number of numbers that stay on their position after this permutation

Average number of winners in a lottery

≈total number of winners in all experiments

total number of experiments

Lottery problem

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Generating a random permutation of 1, . . . , N

1. Initialise t = N and A[i] = i for i = 1, . . . , N ;

2. Generate a random number u between 0 and 1;

3. Set k = 1+ btuc; swap values of A[k] and A[t];

4. Set t = t − 1;If t > 1, then return to step 2;otherwise stop and A[1], . . . , A[N ] yields a permutation.

Complexity is O(N )

Lottery problem

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Results for M = 103, N = 104 and seed = 1

Average numbers of winners is 1.005

k P(k winners)0 0.3511 0.3852 0.1853 0.0664 0.013

Lottery problem

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Results for M = 103, N = 105 and seed = 1

Average numbers of winners is 1.047

k P(k winners)0 0.3501 0.3802 0.1723 0.0744 0.020

Lottery problem

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Observe that number of winners is equal to number of loops of length 1

Further, consider 1, . . . , s;then length of loop generated by 1 is uniform on 1, . . . , s

Loops of a random permutation of 1, . . . , N

1. Initialise t = N and W = 0;

2. Generate a random number u between 0 and 1;

3. Set l = 1+ btuc; if l = 1, then W = W + 1;

4. Set t = t − l; if t ≥ 1, return to step 2;otherwise stop and W yields number of loops of length 1.

Complexity is O(log N )

Lottery problem

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Results for M = 105, N = 107 and seed = 1

Average numbers of winners is 1.001

k P(k winners)0 0.3671 0.3702 0.1833 0.0614 0.016

Lottery problem