Minimize multivariable function with multivariable nonlinear constraints in MATLAB

Hi !

So I'm trying to :

• minimize a function (non linear and multivariable)
• with constraints also non linear, multivariable and with inequations (maybe we can use slack variables) and equations

And for now I've tried this

x=[H, P, L, A, B, M]; %name of my variables
fun=@(x) (x(1)*x(3)*cos(x(4))); %function to minimize
nonlcon = % ??????
x0=[0,0 ,0 ,0 ,0 ,0] %I will have to find a first solution
A=[x(3)*cos(x4), x(6)*cos(x5),x(3)*sin(x4)+ x(6)*sin(x5)]; %constraints with inequality
b=[12.5, 12.5,25]; %boundary for constraints with inequality
Aeq=[x(1)*cos(x(2)), x(3)*cos(x(4))]; %constraints with equality
beq=[x(1)-14,x(6)*cos(x5)]; %boundary for constraints with inequality
x = fmincon(fun,x0,A,b,Aeq,beq); 

but this only works for linear constraint and can't understand how works nonlcon in

x = fmincon(fun,x0,A,b,Aeq,beq,lb,ub,nonlcon,options)

Can you help, thanks :)

[Torsten]

I'm not sure which nonlinear constraints you try to set with your A, b, Aeq and beq.

A and Aeq must have 6 columns and b and beq must have as many rows as A resp. Aeq in order that

A*x <= b or Aeq*x = beq

make sense.

Answer 1

[Bruno Luong]

x0 = [0;0;0;0;0;0];
x = fmincon(@mycost,x0,[],[],[],[], [],[],@mycon)

Converged to an infeasible point.

fmincon stopped because the size of the current step is less than
the value of the step size tolerance but constraints are not
satisfied to within the value of the constraint tolerance.


Consider enabling the interior point method feasibility mode.

x = 6×1

1.0e+13 * -1.4175 0.0000 1.1787 0.0000 0.0000 0.3344

function f = mycost(x)
f = (x(1)*x(3)*cos(x(4)));
end

function [c,ceq] = mycon(x)
A=[x(3)*cos(x(4));
   x(6)*cos(x(5));
   x(3)*sin(x(4))+ x(6)*sin(x(5))];
b = [12.5; 12.5; 25];
c = A-b;
Aeq=[x(1)*cos(x(2));
     x(3)*cos(x(4))]; %constraints with equality
beq=[x(1)-14;
    x(6)*cos(x(5))];
ceq = Aeq-beq;
end

Answer 2

[Matt J]

The problem is unbounded. Consider x=[ 14, pi/2, x3,0,0,x6]. Then the problem reduces to the linear program,

This imposes no lowerbound on , so just send it to to push the objective to as well.