Help with fmincon, non-linear constraints and binary variables

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Hello, I'm working on a problem that requires me to find the solution of a system subject to several inequalities so I decided to use fmincon to solve it.
One of the kinds of constraint is
x = min{c1, c2}
x is a variable part of a set of variables and c1 and c2 are dependent on the same set.
To implement this constraint I added the additional constraints:
x <= c1
x <= c2
x >= c1 + M*y1
x >= c2 + M*y2
y1 + y2 = 1
Where M is an adequately big number and y1 and y2 are additional binary variables (I know i could have used one, it seems to be working better with 2).
To impose y1 and y2 to be binary i used as lower bound 0 and as upper bound 1 in addition to the non linear condition mod([y1 y2],1) = [0 0].
Trying to figure out why it isn't working I tested it with the following code:
A = [1;1];
B = [5;1.6];
M = 10e5;
A_en = [A, [0 0;0 0];
-A,[M 0;0 M]];
B_en = [B;-B];
Aeq = [0 1 1];
Beq = 1;
obj = @(x) [-1 0 0]*x';
nf = [0 1 1];
nonlcon = @con;
OPTIONS = optimoptions('fmincon','Algorithm','interior-point');
fmincon(obj,[0 0 0],A_en,B_en,Aeq,Beq,[0 0 0],[Inf 1 1],nonlcon,OPTIONS)
where con is
function [c, ceq] = con(x)
c = [];
ceq = mod([x(2) x(3)],1);
end
I tried removing the non linear constraints too but the result never satisfies all the constraints.
Why shouldn't it work?

采纳的回答

Torsten
Torsten 2023-2-20
编辑:Torsten 2023-2-20
It doesn't work because fmincon only handles continuous variables and functions.
Use "intlinprog" instead. Here you can define variables to be binary.
Another option is to run your code twice: Once with x = c1 and once with x = c2 and see which run gives the better value for the objective function. This is most probably the way to go because you seem to have nonlinear constraints in a function "con" that intlinprog cannot handle.
As a last option, use "ga".
  1 个评论
Giosuè Basso
Giosuè Basso 2023-2-21
The nonlinear constraints are just there to try and make y1 and y2 integers, but I'll try what you said, thanks a lot!

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