How to add the iteration count to the code

Hi,

I started using Matlab to run Monte Carlo Simulation just a few weeks ago. I have a question about the iteration count:

My code looks like this:

% Toll Road Brownian Motion Problem

% Set the initial parameters
mu=0.2;
sigma=0.1;
price=3;
n=35;

% Generate the initial traffic amount

R(1,1)=normrnd(10000,3000);

% Brownian Motion Iteration (Traffic)

for i=2:36

      R=R.*exp((mu-sigma-0.5*sigma^2)*1+sigma*normrnd(0,1)*1);

end;

% Revenue Calculation (Revenue)

Revenue=price*R*365;

This code runs good except that since I generated random numbers, I would like to generate them for 1000 times, and get the average value out of all the 1000 simulations.

I wanted to add

for k=1:1000

But I don't know how my R matrix can interact with this extra k count.

I'd really appreciate it someone could offer a bit of advice.

Thanks a lot.

Ying

[Rick Rosson]

Please format your code.

Answer 1

[Sargondjani]

you could make a R into a row vector (this will be faster than looping over k)

N=1000; %number of simulations

R=normrnd(10000,3000,[1,N]);
for i=2:36

      R=R.*exp((mu-sigma-0.5*sigma^2)*1+sigma*normrnd(0,1,[1,N])*1);
  end

This should give you a row vector containing the R's for N simulations

[Ying]

Hi Sargondjani,

Thanks so much for your answer. I checked the size of R after using your above code, and it gave me a 1 by 10000 as opposed to 36 10000. I guess the i from 1 to 36 was not counted in the loop. So I went from this matrix representation and used the following code:

% Generate the traffic amount per day

N=10000; %number of simulations

for j=1:N

for i=1;

R(i,j)=norminv(rand(1,1),10000,3000); % Random Initial Traffic

end;

for i=2:36

R(i,j)=R(i-1,j).*exp((mu-sigma-0.5*sigma^2)*1+sigma*z*1);

end;

end;

Basically, I'm using the rows to represent different years (36 rows in total), and using the columns to represent different iterations (10000 in total), and this code works.

And you are right, this is much efficient than using the loop. Thanks a lot for providing a starting point and a really efficient way for simulation iterations.

Ying

Answer 2

[Sargondjani]

With a loop, you could it something like this:

for ik=1:N;
 R(1,ik)=normrnd(10000,3000);
 for it=2:36

      R(1,ik)=R(1,ik).*exp((mu-sigma-0.5*sigma^2)*1+sigma*normrnd(0,1)*1);
 end;
end;

And if you want to keep track of the changes in R then you could do:

R(it,ik)=R(it-1,ik).*exp((mu-sigma-0.5*sigma^2)*1+sigma*normrnd(0,1)*1);