How to get postive dt without changing x(without the commented while loop)

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MIch 2022-3-23

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https://ww2.mathworks.cn/matlabcentral/answers/1678579-how-to-get-postive-dt-without-changing-x-without-the-commented-while-loop

评论： MIch 2022-3-24

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.

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Torsten 2022-3-23

在 MATLAB Online 中打开

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);

MIch 2022-3-24

Okay, thanks for the help. I rewrote my code, added some if statements and now it works well.

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How to get postive dt without changing x(without the commented while loop)

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How to get postive dt without changing x(without the commented while loop)

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