maximization problem , fmincon optimization

Hello,

I'm trying to solve this maximization problem with matlab.

I use the following code

objective = @(x) - ( -.858*x(1)-.04182*x(2)-1.645*sqrt((.5882*x(1) -.055*x(2)).^2 + (-.0558*x(1) +.352*x(2)).^2 ));
Aeq = [1,1];
beq = 1;
lb = zeros*(2);
ub = ones*(2);
x0 = [0,0];
x = fmincon(objective,x0,[],[],Aeq,beq,lb,ub)

once I run the code I received the following :

Warning: Length of lower bounds is < length(x); filling in missing lower bounds
with -Inf. 
> In checkbounds (line 33)
  In fmincon (line 324) 
Warning: Length of upper bounds is < length(x); filling in missing upper bounds
with +Inf. 
> In checkbounds (line 47)
  In fmincon (line 324) 

Any advice to improve my code.

Thank you

Answer 1

[John D'Errico]

syms x y
objxy = ( -.858*x-.04182*y-1.645*sqrt((.5882*x -.055*y).^2 + (-.0558*x +.352*y).^2 ));

If x+y == 1 then we can replace y with 1-x.

objx = subs(objxy,y,1-x)

objx = 

Clearly, the solution must be between 0 and 1 for both x and y given the constraint.

fplot(objx,[0,1])

At a glance, that tends to suggest the function is monotone decreasing, with the maximum at x ==0. But in fact, if we expand the axis down near zero, we do find a non-trivial maximum.

fplot(objx,[0,0.02])

It looks like a max exists around x= 0.06.

Now, how will fmincon do it. First, this is meaningless:

lb = zeros*(2);
ub = ones*(2);

That is NOT how you create vectors of length 2 in MATLAB. READ THE HELP!!!!

Now, fmincon is a MINIMIZER. You state you want to MAXIMIZE. So negate the objective. You actually got that part right.

objective = @(x) - ( -.858*x(1)-.04182*x(2)-1.645*sqrt((.5882*x(1) -.055*x(2)).^2 + (-.0558*x(1) +.352*x(2)).^2 ));
Aeq = [1,1];
beq = 1;
lb = zeros(1,2);
ub = ones(1,2);
x0 = [.5 .5];
fmincon(@(x) objective(x),x0,[],[],Aeq,beq,lb,ub)

Local minimum found that satisfies the constraints.

Optimization completed because the objective function is non-decreasing in 
feasible directions, to within the value of the optimality tolerance,
and constraints are satisfied to within the value of the constraint tolerance.

ans = 1×2

0.0066 0.9934

That is the solution we found graphically.

[Az.Sa]

Thank you very much