How to get postive dt without changing x(without the commented while loop)
1 view (last 30 days)
Show older comments
I have written this code
%parameters
Max=1;
Min=0.01;
nvar=8;
I = [414 414 414 414 414 414 414 414];
Ikg = [8532 9342 9342 10435 10435 9031 9031 8532];
pgr = [1 6; 2 4; 3 1; 4 7; 8 3; 7 5; 6 8; 5 2]';
Ikr = [8532 9342 9342 10435 10435 9031 9031 8532];
C = 0.3;
gvp = [];
gp = [];
rvp = [];
rp=[];
gk = [];
rk = [];
dt = [];
vp = size (pgr);
bp = vp(2);
%calculation
a = (Max-Min)*rand(1,nvar)+Min;
x = a;
dim = size(a);
r = dim(1);
s = dim(2);
x = a;
for i = 1:r
for j = 1:bp
gk(i,j)= Ikg(i,pgr(1,j));
gp(i,j)=I(i,pgr(1,j));
gvp(i,j)=0.14*x(i,pgr(1,j))/((gk(i,j)/gp(i,j))^0.02 -1);
rk(i,j)=Ikr(j);
rp(i,j)=I(i,pgr(2,j));
rvp(i,j)=(0.14*(x(i,pgr(2,j))))/((rk(i,j)/rp(i,j))^0.02 -1);
dt(i,j)=rvp(i,j)-gvp(i,j)-C;
% while dt(i,j)<0
% x = (Max-Min)*rand(1,nvar)+Min;
% gvp(i,j)=0.14*x(i,pgr(1,j))/((gk(i,j)/gp(i,j))^0.02 -1);
% rvp(i,j)=(0.14*(x(i,pgr(2,j))))/((rk(i,j)/rp(i,j))^0.02 -1);
% dt(i,j)=rvp(i,j)-gvp(i,j)-C;
% end
end
end
Can someone give me an idea how to get dt positive without using commented section of code written above. If I use that commented section I get all positive values for dt, but x then changed and I need x to be one randomly generated vector [1 nvar] always.
If commmented section is not used, I get correct values since I generated x outside the for loop and it didn't changed in the proces. But then I have negative values of dt and that doesn't fit for my project.
4 Comments
Torsten
on 23 Mar 2022
Try this:
%parameters
Max=1;
Min=0.01;
nvar=8;
I = [414 414 414 414 414 414 414 414];
Ikg = [8532 9342 9342 10435 10435 9031 9031 8532];
pgr = [1 6; 2 4; 3 1; 4 7; 8 3; 7 5; 6 8; 5 2]';
Ikr = [8532 9342 9342 10435 10435 9031 9031 8532];
C = 0.3;
gvp = [];
gp = [];
rvp = [];
rp=[];
gk = [];
rk = [];
dt = [];
vp = size (pgr);
bp = vp(2);
%calculation
a = (Max-Min)*rand(1,nvar)+Min;
x = a;
dim = size(a);
r = dim(1);
s = dim(2);
%x = a;
X = 0.001:0.001:1;
fail = zeros(size(X));
for k = 1:numel(X)
x = X(k);
for i = 1:r
for j = 1:bp
gk(i,j)= Ikg(i,pgr(1,j));
gp(i,j)=I(i,pgr(1,j));
gvp(i,j)=0.14*x/((gk(i,j)/gp(i,j))^0.02 -1);
rk(i,j)=Ikr(j);
rp(i,j)=I(i,pgr(2,j));
rvp(i,j)=(0.14*x)/((rk(i,j)/rp(i,j))^0.02 -1);
dt(i,j)=rvp(i,j)-gvp(i,j)-C;
end
end
dt = dt(:);
dt = dt(dt<=0);
if ~isempty(dt)
fail(k) = 1;
else
fail(k) = 0;
end
end
plot(X,fail)
As you can see, your dt is <=0 for all values for x in the respective interval.
The initial code should work if you let
dt(i,j)=-(rvp(i,j)-gvp(i,j)-C);
Answers (0)
See Also
Categories
Find more on Optimization in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!