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Adding available capacity at the hourly time increment

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Banjo
Banjo on 15 Nov 2018
Answered: per isakson on 23 Dec 2019
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
Guillaume on 15 Nov 2018
" I hope it is clear enough what I want to do."
No. I don't understand why for i = 1 you don't add row 2 and 5, for i = 2 you don't add row 1, 4 and 5 and for i = 3 you don't add row 3. There seems to be no logic to that. You need to explain better.
edit: and your code doesn't help at all. notwithstanding the fact that it will error if G has less than 3 rows, it doesn't make much sense. The whole lot is equivalent to:
G=[1,10,0.05;2,10,0.05;3,15,0.03;4,15,0.03;5,15,0.03];
K = G(end, 2) * (rand >= G(end, 3)); %since K is never indexed, only the last step of the loop matters
Banjo
Banjo on 15 Nov 2018
Basically I just want that every of 500 hourly iterations represent a summation of available capacity for 5 generators. Generator capacity is available if X>=G(j,3). Regards
Guillaume
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.

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Answers (2)

madhan ravi
madhan ravi on 15 Nov 2018
have you seen cumsum?

  1 Comment

Banjo
Banjo on 15 Nov 2018
When I put cumsum(K)
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
cumsum(K)
end
I get for example
K = 10
K = 10
K = 0
K = 15
K = 15
ans = 15
but as an answer I should get
ans = 50

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per isakson
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.
Capture.PNG
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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