Function optimization with some conditions conditions

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Jon Bilbao
Jon Bilbao 2023-6-22
评论： Torsten 2023-6-22
Ran in:
I want to find the h values that meet the following conditions, h(j)<Hmax, h(j+1)-h(j)<lvasc(j) and h(j+1)-h(j)>lsa(j), the values of lvasc and lsa are calculated in teh function. About the input values ht is a array with a size ht(n,m), Lc and s are scalar and vmax size is vmax (1,m), i have done the function below but it returns me the error:
Error using vertcat
Dimensions of arrays being concatenated are not consistent.
Error in hp2 (line 38)
b = [b1;b2;b3];
I dont know how to solve this, thanks in davance
Mathematically, I need to find the values of h that minimize the function I, which is the summation of the subtraction between ht(n,m) and h(m), and h has to satisfy several conditions. The first condition is that h(end) < 2.5 * Lc. The next condition is that h(j+1) - h(j) < (1/36 * 0.5 / vmax(j+1)) or (0.75 * s / 9.81) / (Lc * 2), whichever is more restrictive. Finally, h(j+1) - h(j) > (0.5 * s * 0.4 * (ht(j+1) - ht(j))) / (vmax(j+1) * 9.81).
n = 10;
m = 50;
ht = rand(n,m);
Lc = 2;
s = 15;
vmax = rand(1,m);
[h,fval,exitflag] = hp2(ht,Lc,s,vmax)
ans = 1×2
50 1
ans = 1×2
1 49
ans = 1×2
1 49
Error using vertcat
Dimensions of arrays being concatenated are not consistent.

Error in solution>hp2 (line 49)
b = [b1;b2;b3];
function [h,fval,exitflag] = hp2(ht,Lc,s,vmax)
s1=s/1000;
Hmax=2.5*Lc;
Imax=0.75*s1/9.81;
lvasc=1/36*0.5./vmax;
[n,m]=size(ht);
for j=1:m-1
lvasc(j)=1/36*0.5./vmax(j+1);
lv(j)=min(Imax,lvasc(j));
lsa(j) = ((ht(j+1)-ht(j)) * (0.5 * s1 * 0.4)) / (vmax(j+1) * 9.81);
end
I=@(h) sum(sum((ht-h).^2));
h0 = zeros(1,m);
v1 = ones(m,1);
w1 = -ones(m-1,1);
A1 = diag(v1) + diag(w1,1);
b1 = [zeros(m-1,1);Hmax];
v2 = -ones(m,1);
w2 = ones(m-1,1);
A2 = diag(v2) + diag(w2,1);
A2(end,:) = [];
b2 = lv;
v3 = -ones(m-1,1);
w3 = ones(m,1);
A3 = diag(v3,1) + diag(w3);
A3(end,:) = [];
b3 = -lsa;
size(b1)
size(b2)
size(b3)
A = [A1;A2;A3];
b = [b1;b2;b3];
[h,fval,exitflag] = fmincon(I,h0,A,b);
end

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采纳的回答

Torsten
Torsten 2023-6-22
移动：Torsten 2023-6-22
The dimensions would be correct if you use
b = [b1;b2.';b3.'];
instead of
b = [b1;b2;b3];
but you'd better post the problem you are trying to solve in a mathematical form.
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Jon Bilbao
Jon Bilbao 2023-6-22
编辑：Jon Bilbao 2023-6-22
Yeah, but in the problem that i am trying to solve thats physically imposible so if it can be made i would like to change it to 0.
Torsten
Torsten 2023-6-22
function [h,fval,exitflag] = hp2(ht,Lc,s,vmax)
...
[h,fval,exitflag] = fmincon(I,h0,A,b);
h(1) = 0;
end
Now you've set it to 0.

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