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Basics of Monte Carlo Simulations: practical exercises

Francesco DI MAIO

EXERCISE 1

Francesco DI MAIO

Exercise 1

Consider the Weibull distribution:

with

1. Sample N=400 values from

2. Show a Figure reporting the distribution of the

obtained samples

3. Provide an estimate of with N=400

and N=6400

4. Estimate the variance of with N=400 and N=6400

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x

X exF 1

Sample R from UR(r) and find X:

RFX X

1

1

1 r 0

rUrR RPr xFX

R X x

x

X exFR 1

RRFX X 1ln11

Example: Exponential distribution

Sampling Random Numbers from FX(x)

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Example: Weibull Distribution

• Time-dependent hazard rate

cdf:

pdf:

• Sampling a failure time T (by the inverse transform)

1 t

TF t P T t e

1 t

Tf t dt P t T t dt t e dt

1 t

R TR F r F t e

1

1 1ln 1TT F R R

1 tt

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% Exercise 1

clear all; close all

% Weibull parameters

beta = 1;

alpha = 1.5;

% Sample N values from the Weibull distribution

N = 400; % number of samples

r = rand(N,1);

t = (-log(1-r)/beta).^(1/alpha); % inverse transform method

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% Verify the distribution

delta_channel = 0.1;

channel_centers = [delta_channel/2:delta_channel:5-

delta_channel/2];

num_samples = hist(t,channel_centers);

pdf_est = num_samples/(N*delta_channel); %

Normalization

% Compute the analytic pdf

analytic_weibull = alpha*beta*channel_centers.^(alpha-

1).*exp(-beta*channel_centers.^alpha);

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figure % pdf

plot(channel_centers,analytic_weibull); hold on;

plot(channel_centers,pdf_est,'-sr'); grid

legend('Analytic pdf','Sampled values distribution pdf',0)

figure % cdf

cdf_est = cumsum(pdf_est)*delta_channel;

plot(channel_centers,

cumsum(analytic_weibull)*delta_channel);hold on;

plot(channel_centers, cdf_est,'*r'); grid

legend('Analytic cdf','Sampled values distribution cdf',0)

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0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Analytic pdf

Sampled values distribution pdf

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0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.2

0.4

0.6

0.8

1

1.2

1.4

Analytic cdf

Sampled values distribution cdf

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MC Evaluation of Definite Integrals (1D)

Analog Case

MC analog dart game: sample x from f(x)

• the probability that a shot hits x dx is f(x)dx

• the award is g(x)

Consider N trials with result {x1, x2, …, xn} and define the estimator:

N

i

iN xgN

G1

1

1 ; 0 pdf

b

a

b

a

dxxfxfxf

dxxfxgG

iN

N

xgVarN

GVar

GGE

1

2

2

1 11

1 2 11 1

E T

Var T

Real values (from theory)

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% Estimating the mean of the distribution (definite integral)

GN = mean(t); % Sample mean

disp(['G_N = ',num2str(GN)])

S2GN = var(t)/N;

disp(['G_N variance = ', num2str(S2GN)])

disp(['Estimate = ',num2str(GN),' +-

',num2str(sqrt(S2GN))]);

disp([' '])

disp(['True Value = ', num2str(gamma(5/3))])

% mean_weibull=1/beta(gamma(1+1/alfa))

disp([' '])

disp(['Error = ', num2str(abs(gamma(5/3)-GN))])

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Interpretation of the Variance of GN (1)

GN

1st run of the code:

400 samples from f

2nd run of the code:

400 different samples

100th run of the code:

400 different samples

GN is a random variable:

GN distribution is approximated by a Normal

iN

N

xgVarN

GVar

GGE

1

N

i

iN xgN

G1

1

GN

0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.70

500

1000

1500

2000

2500

3000

3500

G

h(GN)

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Interpretation of the Variance of GN (2)

GN

G

h(GN)

NGG NGG

0.68… ; 0.68

N NN G GP G G G

unknown

; 0.68N NN G N G

P G G G

GN

1st run of the code:

400 samples from f

2nd run of the code:

400 different sample

G

100th run of the code:

400 different samples

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% Validation

clc; clear all;

Nval = 1000; % Number of estimates of the integral

inside = 0;

% Weibull parameters

beta = 1;

alpha = 1.5;

N = 400; % number of samples

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for i=1:Nval

% Sampling N values from the Weibull distribution

r = rand(N,1);

t = (-log(r)/beta).^(1/alpha); % inverse transform method

% Verifying the distribution

GN(i) = mean(t);

S2GN(i) = var(t)/N;

SGN(i) = S2GN(i)^0.5;

ext_inf(i) = GN(i)-(S2GN(i))^0.5;

ext_sup(i) = GN(i)+S2GN(i)^0.5;

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if((gamma(5/3)>ext_inf(i)) & (gamma(5/3)<ext_sup(i)))

inside = inside+1;

end

end

figure

errorbar(GN(1:30),SGN(1:30),'.');

hold on; plot(gamma(5/3)*ones(1,30),'r');

axis([0 31 0.5 1]);

%plot(ext_inf(1:100),'.b');hold on;

plot(ext_sup(1:100),'.k'),hold on;

plot(gamma(5/3)*ones(1,100),'r')

disp(['Validation = ',num2str(inside/Nval)]);

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0 5 10 15 20 25 300.5

0.55

0.6

0.65

0.7

0.75

0.8

0.85

0.9

0.95

1

Validation = 0.692

Francesco DI MAIO

EXERCISE 2

Francesco DI MAIO

Exercise 2

• Consider the following system

• Transition rates (in arbitrary time-units):

Failure: λA = 0.001; λB = 0.002; λC = 0.005;

Repair: μA = 0.1; μB = 0.15; μC = 0.05;

• Estimate the reliability of the system at Tmiss = 500

A

CB

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Components’ times of transition between states are exponentially distributed

Arrival

Initial1 2 3

1(nominal) 0 ,C)(BA )(31BA

2 (failed) ,C)(BA 0 )(32

BA

3 (failed) )(13BA

)(23

BA 0

Indirect Monte Carlo

A

CB

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• The rate of transition of the system out of its current configuration (1,1, 1) is:

• We are now in the position of sampling the first system transitiontime t1, by applying the inverse transform method:

where Rt ~ U[0,1)

CBA 1,1,1

)1ln(1

1,1,101 tRtt

Analog Monte Carlo Trial

Sampling the time of transition

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• Assuming that t1 < TM (otherwise we would proceed to the

successive trial), we now need to determine which component

has undergone the transition

• The probabilities of components A, B, C undergoing a transition

out of their initial nominal states 1, given that a transition occurs

at time t1, are:

• Thus, we can apply the inverse transform method to the discrete

distribution

1,1,11,1,11,1,1 , ,

CBA

RC U[0,1)

Analog Monte Carlo Trial

Sampling the kind of Transition (1)

100

Rc

C 1,1,1

A

1,1,1

B

1,1,1

C

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• As a result of this first transition, at t1 the system is

operating in configuration (1,2,1).

• The simulation now proceeds to sampling the next

transition time t2 with the updated transition rate

CBA 1,2,1

Analog Monte Carlo Trial

Sampling the kind of Transition (2)

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% Reliability of Simple Systems (A\\(B+C))

clc

choice = input('1 --> Simulation mode 2 --> Step by step

mode');

% System parameters

lambda_A = 0.001; mu_A = 0.1;

lambda_B = 0.002; mu_B = 0.15;

lambda_C = 0.005; mu_C = 0.05;

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Trans_A = [0 lambda_A; mu_A 0];

Trans_B = [0 lambda_B; mu_B 0];

Trans_C = [0 lambda_C; mu_C 0];

% 2 = Failure; 1 = Working

failed_states = [2 1 2; 2 2 1; 2 2 2]; % cutset - the last one is

useless

initial_state = [1 1 1];

% Mission time

Tmiss = 500;

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% MC simulation parameters

N = 1e3;

% MC cycle

unrel_counter = zeros(1,N);

for n = 1:N % Main Monte Carlo cycle

if choice == 2

disp([' '])

disp(['Cycle = ', num2str(n)])

pause

end

unrel_flag = 0; % 0 if no failures before Tmiss, 1 otherwise

t = 0;

current_state = initial_state;

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while (t < Tmiss)

for jj = 1:3 % 3 is the number of cutset

if sum(current_state == failed_states(jj,1:3)) == 3

unrel_flag = 1;

end

end

if unrel_flag ~= 1

lambda_A_out = sum(Trans_A(current_state(1),:));

lambda_B_out = sum(Trans_B(current_state(2),:));

lambda_C_out = sum(Trans_C(current_state(3),:));

lambda_sys = lambda_A_out + lambda_B_out +

lambda_C_out;

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t_trans = -1/lambda_sys*log(rand); % Sample transition time

t = t+t_trans;

if choice == 2

disp(['System transition time = ', num2str(t)])

pause

end

if t < Tmiss % if the transition time falls within Tmiss, then

sample the kind of transition

r = rand;

sum_l = cumsum([lambda_A_out lambda_B_out

lambda_C_out])/lambda_sys;

comp = min(find(sum_l>r)); % Component that makes

the transition

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if choice == 2

disp(['Component = ', int2str(comp)])

pause

end

old_st = current_state(comp);

current_state(comp) = 3-current_state(comp);

if choice == 2

disp(['Transition = ', int2str(old_st), ' --> ',

int2str(current_state(comp))])

pause

end

end

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else % if unrel_flag = 1 (failure before Tmiss)

break % break the while loop

end

end

unrel_counter(n) = unrel_flag;

end

% Compute the analytic reliability

rel_analytic = 1-(1-exp(-lambda_A*Tmiss))*(1-exp(-

lambda_B*Tmiss)*exp(-lambda_C*Tmiss));

% Estimate the reliability by the MC samples

rel_MC = 1-mean(unrel_counter);

var_MC = var(unrel_counter)/N;