Function optimization meeting conditions

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How can i optimize the I function, i want to find the values of h(j) that minimize I, meetentig the conditions h(j+1)>h(j), h(end)<120 and h(j+1)-h(j)<1.25 ?
ht is a array beiing its size ht(lt,lc) or the same ht(i,j) and it is calculated in another function. The formula of Ins is Ins=ht(i,j)-h(j).
Thanks for the help
function [h] = hp(ht, Lc, Lt)
lt = 0:0.5:Lt;
lc = 0:0.5:Lc;
Ins = cell(length(lt), length(lc));
h= cell(length(Lc));
for i = 1:length(lt)
for j = 1:length(lc)
Ins{i,j} = @(h) (ht(i,j) - h);
end
end
h0 = zeros(size(lc));
A = [];
b = [];
Aeq = [];
beq = [];
lb = [];
ub = [];
nonlcon = @constraints;
h = fmincon(@(h) Ins, h0, A, b, Aeq, beq, lb, ub, nonlcon);
end
function [c] = constraints(h)
c>0;
c = h(2:end) - h(1:end-1);
end
  10 个评论
Torsten
Torsten 2023-6-8
编辑:Torsten 2023-6-8
If ht is nxm, the linear constraints can be defined by A and b as in the code below.
A and b are then used in the call to the optimizer, e.g.
Now it's your turn to define the objective function and the call to "fmincon" (or some similar optimizer).
(And incidentally the .^2 appears for the summands in the objective :-) )
m = 4;
v1 = ones(m,1);
w1 = -ones(m-1,1);
A1 = diag(v1) + diag(w1,1)
A1 = 4×4
1 -1 0 0 0 1 -1 0 0 0 1 -1 0 0 0 1
b1 = [zeros(m-1,1);120];
v2 = -ones(m,1);
w2 = ones(m-1,1);
A2 = diag(v2) + diag(w2,1);
A2(end,:) = []
A2 = 3×4
-1 1 0 0 0 -1 1 0 0 0 -1 1
b2 = 1.25*ones(m-1,1);
A = [A1;A2]
A = 7×4
1 -1 0 0 0 1 -1 0 0 0 1 -1 0 0 0 1 -1 1 0 0 0 -1 1 0 0 0 -1 1
b = [b1;b2]
b = 7×1
0 0 0 120.0000 1.2500 1.2500 1.2500
Jon Bilbao
Jon Bilbao 2023-6-8
it cant be solved like thah because ht is too big, ht is a 401x51 matrix
Error using fmincon
A must have 2601 column(s).
Error in hpp (line 22)
h = fmincon(I,h0,A,b);
function [h] = hpp(ht)
[n,m]=size(ht);
I=@(h) (ht-h);
h0=zeros(m);
v1 = ones(m,1);
w1 = -ones(m-1,1);
A1 = diag(v1) + diag(w1,1);
b1 = [zeros(m-1,1);120];
v2 = -ones(m,1);
w2 = ones(m-1,1);
A2 = diag(v2) + diag(w2,1);
A2(end,:) = [];
b2 = 1.25*ones(m-1,1);
A = [A1;A2];
b = [b1;b2];
h = fmincon(I,h0,A,b);
end

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采纳的回答

Torsten
Torsten 2023-6-8
移动:Torsten 2023-6-8
ht = rand(401,51);
[n,m]=size(ht);
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);120];
v2 = -ones(m,1);
w2 = ones(m-1,1);
A2 = diag(v2) + diag(w2,1);
A2(end,:) = [];
b2 = 1.25*ones(m-1,1);
A = [A1;A2];
b = [b1;b2];
[h,fval,exitflag] = fmincon(I,h0,A,b)
Local minimum possible. Constraints satisfied. fmincon stopped because the size of the current step is less than the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance.
h = 1×51
0.4889 0.5003 0.5003 0.5003 0.5003 0.5042 0.5042 0.5042 0.5042 0.5042 0.5042 0.5042 0.5042 0.5042 0.5042 0.5042 0.5042 0.5042 0.5042 0.5042 0.5042 0.5042 0.5042 0.5042 0.5042 0.5042 0.5042 0.5042 0.5042 0.5042
fval = 1.7157e+03
exitflag = 2

更多回答(1 个)

rakshit gupta
rakshit gupta 2023-6-7
You can consider following changes to the code to optimize the function while meeting the condition h(j+1)>h(j).
  1. Modify the Ins cell array to a function handle that takes in the h array.
Ins = @(h) ht - h;
2. Modify the h array to a vector instead of a cell array.
h = zeros(size(lc));
3. Add the upper bound constraint to ensure h(j+1) > h(j).
ub = inf(size(h));
ub(end) = h(end);
4. Modify the constraints function to return the inequality constraint.
function [c, ceq] = constraints(h)
c = h(2:end) - h(1:end-1);
ceq = [];
end
5. Call the fmincon function with the changes made above.
h = fmincon(Ins, h, A, b, Aeq, beq, lb, ub, @constraints);
These changes could help in optimizing the function.
  6 个评论
rakshit gupta
rakshit gupta 2023-6-8
Yes, you can try modifying 'h' vector by changing the creation of the 'h' vector to use the same size and data type as 'ht',
h = zeros(size(ht), 'like', ht);
This may help in making Ins scalar.

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