Adding available capacity at the hourly time increment
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Hello,
Could you please help me how to cumulate available capacity that is second column of G matrix (10,10,15,15,15) at the hourly time increment. When random number is larger or equal to number from third column of G matrix, these capacities should be cumulated for every time iteration of 1 to 500.
For example, if i=1 => Sum=10+0+15+15+0=40; i=2 => Sum=0+10+15+0+0=25; i=3 => Sum=10+10+0+15+15=50 and so on until 500 iteration. I hope it is clear enough what I want to do.
G=[1,10,0.05;2,10,0.05;3,15,0.03;4,15,0.03;5,15,0.03];
for i=1:500
for j=1:length(G)
X=rand(1);
if X>=G(j,3)
K=G(j,2);
else
K=0;
end
end
end
3 Comments
Guillaume
on 15 Nov 2018
What is X? What is capacity? What is a generator? Where is time so we understand what an hourly iteration is?
We have absolutely no idea what your data represent and you don't explain it. Your basically doesn't give us the basics I'm afraid.
Answers (2)
per isakson
on 23 Dec 2019
This answer is a variation of my answer to your recent question, "Storing and passing all iterations to an array outside the nested for loops"
>> clearvars
>> cssm
>> h = plot(sum(K,2),'.');
>> h.Parent.YLim = [33,67];
Produces 500 realizations of whatever your code represents. The results range from 35 to 65.
where the script, cssm, is your script with a few modifications
%%
G=[1,10,0.05;2,10,0.05;3,15,0.03;4,15,0.03;5,15,0.03];
K = nan( 500, length(G) ); % pre-allocate
for ii=1:500
for jj=1:length(G)
X=rand(1);
if X>=G(jj,3)
K(ii,jj)=G(jj,2);
else
K(ii,jj)=0;
end
end
end
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