Presentazione di PowerPoint - LASAR€¦ · Francesco DI MAIO FX xx 1 e O Sample R from U R (r) and X find X: F X R 1 1 r X1 0 Pr ^ R d r ` U R r F X x R x x R F X x e 1 O X F X R

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

Francesco DI MAIO

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)

Francesco DI MAIO

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

Francesco DI MAIO

% 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

Francesco DI MAIO

% 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);

Francesco DI MAIO

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)

Francesco DI MAIO

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

Francesco DI MAIO

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

Francesco DI MAIO

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)

Francesco DI MAIO

% 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))])

Francesco DI MAIO

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)

Francesco DI MAIO

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

Francesco DI MAIO

% 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

Francesco DI MAIO

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;

Francesco DI MAIO

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)]);

Francesco DI MAIO

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

Francesco DI MAIO

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

Francesco DI MAIO

• 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

Francesco DI MAIO

• 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

Francesco DI MAIO

• 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)

Francesco DI MAIO

% 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;

Francesco DI MAIO

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;

Francesco DI MAIO

% 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;

Francesco DI MAIO

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;

Francesco DI MAIO

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

Francesco DI MAIO

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

Francesco DI MAIO

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;