Adding available capacity at the hourly time increment

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Banjo
Banjo 2018 年 11 月 15 日
回答済み: per isakson 2019 年 12 月 23 日
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 件のコメント
Banjo
Banjo 2018 年 11 月 15 日
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 2018 年 11 月 15 日
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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回答 (2 件)

madhan ravi
madhan ravi 2018 年 11 月 15 日
have you seen cumsum?
  1 件のコメント
Banjo
Banjo 2018 年 11 月 15 日
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 2019 年 12 月 23 日
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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