Basics of Monte Carlo Simulations: practical exercises

Transcription

1 Basics of Monte Carlo Simulations: practical exercises Piero Baraldi Politecnico di Milano, Energy Department Phone:

2 EXERCISE 1 (first part)

3 Exercise 1 Consider the Weibull distribution: with 1. Sample =400 values from 2. Show a Figure reporting the distribution of the obtained samples 3. Verify whether the obtained distribution provides a good approximation of the Weibull distribution

4 Sampling Random umbers from F X (x)

5 MATLAB COMMAD rand(m,1) provides a column of M random numbers sorted from an uniform distribution in the range (0,1) = hist(y) bins the elements of vector Y into 10 equally spaced containers and returns the number of elements in each container. More options if you write help hist

6 Example: Weibull Distribution Time-dependent hazard rate 1 t t cdf: T 1 F t P T t e t pdf: T 1 t f t dt P t T t dt t e dt Sampling a failure time T (by the inverse transform) R 1 R F r F t e T t 1 1 T FT R ln 1 R 1

7 % Exercise 1 clear all; close all % Weibull parameters beta = 1; alpha = 1.5; % Sample values from the Weibull distribution = 400; % number of samples r = rand(,1); t = (-log(1-r)/beta).^(1/alpha); % inverse transform method

8 % 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/(*delta_channel); % ormalization % Compute the analytic pdf analytic_weibull = alpha*beta*channel_centers.^(alpha- 1).*exp(-beta*channel_centers.^alpha);

9 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)

10 1 0.9 Analytic pdf Sampled values distribution pdf

11 Analytic cdf Sampled values distribution cdf

12 EXERCISE 1 (second part)

13 MC Evaluation of Definite Integrals 13 G b a g x f xdx E[ gx] X MC = 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 trials with results x 1, x 2,, x : the average award is: G 1 i1 g x g i

14 MC Evaluation of Definite Integrals: Why G is a good estimator of G? 14 G 1 g x i i1 G is a random variable with: E 1 G E gx Egx Egx Var i1 1 G Var gx Vargx Vargx Varg( x) i1 i i 1 i1 1 2 i1 i 1 i1 i 1 2 i1 G 1 G is an unbiased estimator of G: G is a consistent estimator of G: EG G lim Var G 0

15 Exercise 1 (Second Part) Consider the Weibull distribution: with 1. Sample =400 values from 2. Show a Figure reporting the distribution of the obtained samples 3. Provide an estimate of with =400 and = Estimate the variance of with =400 and =6400

16 % Estimating the mean of the distribution (definite integral) G = mean(t); % Sample mean disp(['g_ = ',num2str(g)]) S2G = var(t)/; disp(['g_ variance = ', num2str(s2g)]) disp(['estimate = ',num2str(g),' +- ',num2str(sqrt(s2g))]); disp([' ']) disp(['true Value = ', num2str(gamma(5/3))]) disp([' ']) % mean_weibull=1/beta(gamma(1+1/alfa)) disp(['error = ', num2str(abs(gamma(5/3)-g))])

17 Interpretation of the Variance of G (1) G 1 st run of the code: 2 nd run of the code: 400 samples from f 400 different samples 100 th run of the code: 400 different samples G is a random variable: G distribution is approximated by a ormal 2000 E G Var G Vargx G 1 i G 1 g x i i1 h(g3500 ) G G

18 Interpretation of the Variance of G (2) unknown P G ; 0.68 G G G G h(g ) 0.68 P G G ; G 0.68 G G G G G G G G G G 1 st run of the code: 400 samples from f 2 nd run of the code: 400 different sample 100 th run of the code: 400 different samples

19 Validation =

20 EXERCISE 2

21 Exercise 2 Consider the following system A B C Transition rates (in arbitrary time-units): Failure: λ A = 0.001; λ B = 0.002; λ C = 0.005; Repair: μ A = 0.1; μ B = 0.15; μ C = 0.05; Estimate the reliability of the system at T miss = 500

22 Indirect Monte Carlo A B C Components times of transition between states are exponentially distributed Arrival Initial 1(nominal) 0 A( B,C) A B 1 2 (failed) ( B,C) A 0 ( 3 ) A( B) (failed) A( B) 3 1 A( B) 3 2 0

23 Analog Monte Carlo Trial Sampling the time of transition 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 transition time t 1, by applying the inverse transform method: where R t ~ U[0,1) t 1,1,1 t A B C 1 ln(1 R ) 1,1,1 1 0 t

24 Analog Monte Carlo Trial Sampling the kind of Transition (1) Assuming that t 1 < T M (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 t 1, are: A B C,, 1,1,1 1,1,1 1,1,1 Thus, we can apply the inverse transform method to the discrete distribution Rc A B C C 1,1,1 1,1,1 1,1,1 0 R C U[0,1) 1

25 Analog Monte Carlo Trial Sampling the kind of Transition (2) As a result of this first transition, at t 1 the system is operating in configuration (1,2,1). The simulation now proceeds to sampling the next transition time t 2 with the updated transition rate 1,2,1 A B C

26 Unreliability and Unavailability Estimation 26 When the system enters a failed, tallies are appropriately collected for the unreliability estimates; After performing a large number of trials M, we can obtain estimates of the system unreliability by simply dividing by M, the number of MC simulation ended in a failure configuration

27 % Reliability of Simple Systems (A\\(B+C)) % 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;

28 % 2 = Failure; 1 = Working Trans_A = [0 lambda_a; mu_a 0]; Trans_B = [0 lambda_b; mu_b 0]; Trans_C = [0 lambda_c; mu_c 0]; failed_states = [2 1 2; 2 2 1; 2 2 2]; initial_state = [1 1 1]; % Mission time Tmiss = 500;

29 % MC simulation parameters = 1e3; % MC cycle unrel_counter = zeros(1,); for n = 1: % Main Monte Carlo cycle unrel_flag = 0; % 0 if no failures before Tmiss, 1 otherwise t = 0; current_state = initial_state;

30 while (t < Tmiss) % check if the system is in a failure configuration for jj = 1:3 % 3 is the number of system failure states end if sum(current_state == failed_states(jj,1:3)) == 3 end unrel_flag = 1; % find the system transition rate 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;

31 % Sample transition time t_trans = -1/lambda_sys*log(rand); t = t+t_trans; if t < Tmiss % if the transition time falls within Tmiss, end % 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 %Change the state of the component that makes the transition old_st = current_state(comp); current_state(comp) = 3-current_state(comp);

32 else %(unreliability flag is 1) break % break the while loop end end %(of the while loop) unrel_counter(n) = unrel_flag; end %of the simulation % Estimate the reliability by the MC samples rel_mc = 1-mean(unrel_counter); var_mc = var(unrel_counter)/;