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How can I do a Monte Carlo simulation for 1000 runs on this operation?

3 vues (au cours des 30 derniers jours)
Uche
Uche le 8 Mar 2023
Commenté : Uche le 11 Mar 2023
Please can someone help to verify that this is the proper way to run a monte carlo simulation on this code?
% A = T(:,1); %Column vector
A = [71.213; 74.499; 79.175; 54.163; 83.008; 52.615];
B = length(A); %size of column (number of elements)
t = 0; %Variable to store harvest time
number_of_runs = 1000;
for n = 1:number_of_runs
for i = 1:B
H = A(i);
while H > 0
H = H - randi([5 10]);
t = t + randi([2 4]);
end
end
end
t_mean = t/number_of_runs;
disp(t_mean);
  2 commentaires
John D'Errico
John D'Errico le 8 Mar 2023
Modifié(e) : John D'Errico le 9 Mar 2023
Hard to say. That is the proper way to simulate what you simulated. Is it the proper way to simulate what you think you were simulating? That is impossible to tell, since we see only the code you wrote. So perhaps you need to explain what is the goal.

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Torsten
Torsten le 9 Mar 2023
Maybe you mean:
% A = T(:,1); %Column vector
A = [71.213; 74.499; 79.175; 54.163; 83.008; 52.615];
B = length(A); %size of column (number of elements)
number_of_runs = 1000;
T = zeros(number_of_runs,1); %Variable to store harvest time
for n = 1:number_of_runs
t = 0;
for i = 1:B
H = A(i);
while H > 0
H = H - randi([5 10]);
t = t + randi([2 4]);
end
end
T(n) = t;
end
plot(1:number_of_runs,T)
t_mean = sum(T)/number_of_runs;
disp(t_mean);
175.9570

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