A question regarding a optimization problem by using "fmincon" solver.

Hi, I've faced with an error in my optimization code. When I consider the parameter "options.MaxFunEvals" as a large number (like 10000), the system will be busy. Otherwise, if this parameter be a small number (e.g., 5), the output will be equal to the primary value of parameters. The code is presented as follows.

clear
clc

sc=5;
sp=4;
lo=21;
t=0.5;

A=[-1 -2 -2;...
    1  2  2];
b=[0;72];
x0=10*ones(11073,1);
lb=2*ones(11073,1);
ub=100*ones(11073,1);
options = optimoptions(@fmincon,...
'Display','iter','Algorithm','interior-point');
options.MaxFunEvals=1000000;
[x,fval,exitflag,output]=fmincon(@(x)OFT(x),x0,[],[],[],[],lb,ub,[],options);

Also, the function namely OFT is called via below code:

function f=OFT(x)
% f=-x(1,1)*x(2,1)*x(3,1);
f=0;
for i=1:1:11073;
    f=f+((cos(i*(pi))*x(i,1)^.2)/125);
end

end

Thank you so much for your considerations, in advance.

[Torsten]

Why are you talking about an error in your code ? The behaviour of "fmincon" seems plausible for me.

[Matt J]

You have defined what look like inequality constraint data,

A=[-1 -2 -2;...
  1  2  2];
b=[0;72];

but they are for only a 3-dimensional problem and are never passed to fmincon. Is this deliberate? Are there only supposed to be lb,ub constraints?

[Stephan]

Your code is confusing a little bit - seeing what you wrote you have 11073 decision variables - considering the inequalities Matt has commented above, it could also be three decision variables and you were confused when you wrote the code.

I am looking forward to the solution of the riddle ?

Answer 1

[Matt J]

Use the default MaxFunEvals, but you need to vectorize OFT for faster performance. You also need to provide vectorized computations of the gradient and Hessian, or you will incur a lot of overhead from finite difference calculations.

e=(-1).^(1:11073)/125;
g=2*e; 

OTF= @(x) deal( e*x(:).^2 , g(:).*x(:) ); %provide objective AND gradient

options = optimoptions(@fmincon,'Algorithm','interior-point',...
      'SpecifyObjectiveGradient',true,'SpecifyConstraintGradient',true,...
      'SubproblemAlgorithm','cg','HessianMultiplyFcn',@(x,l,v) g(:).*v(:));

[x,fval,exitflag,output]=fmincon(OTF,x0,[],[],[],[],lb,ub,[],options);

However, your objective is quite non-convex, so I expect it could be difficult for fmincon to minimize. ga() might be a better choice.

[Stephan]

Hi Matt,

i tried to use your solution - with no success

[Matt J]

Worked fine for me.

[Stephan]

You're right - without A & b it works ... Still not sure if there is not a 3-dimensional problem here. Your question about inequality conditions has brought me to it.

Answer 2

[Stephan]

Hi,

is it what you want:

A=[-1 -2 -2;...
    1  2  2];
b=[0;72];
x0=10*ones(1,3);
lb=2*ones(1,3);
ub=100*ones(1,3);
options = optimoptions(@fmincon,...
'Display','iter','Algorithm','interior-point');
[x,fval,exitflag,output]=fmincon(@(x)OFT(x),x0,A,b,[],[],lb,ub,[],options);
function f=OFT(x)
% f=-x(1,1)*x(2,1)*x(3,1);
f=0;
for i=1:11073
    f=f+((cos(i*(pi))*x(:,1)^.2)/125);
end
end

This results in:

                                              First-order      Norm of
   Iter F-count            f(x)  Feasibility   optimality         step
      0       4   -1.267915e-02    0.000e+00    2.530e-04
      1       8   -1.267923e-02    0.000e+00    2.530e-04    3.428e-04
      2      12   -1.267967e-02    0.000e+00    2.530e-04    1.765e-03
      3      16   -1.268120e-02    0.000e+00    2.529e-04    6.190e-03
      4      20   -1.268883e-02    0.000e+00    2.523e-04    3.095e-02
      5      24   -1.272666e-02    0.000e+00    2.493e-04    1.544e-01
      6      28   -1.290714e-02    0.000e+00    2.356e-04    7.622e-01
      7      32   -1.364494e-02    0.000e+00    1.885e-04    3.592e+00
      8      36   -1.502162e-02    0.000e+00    1.271e-04    9.135e+00
      9      48   -1.501503e-02    0.000e+00    1.140e-04    5.251e-02
     10      54   -1.501813e-02    0.000e+00    9.823e-05    2.465e-02
     11      58   -1.501774e-02    0.000e+00    9.824e-05    5.967e-03
     12      62   -1.502349e-02    0.000e+00    9.804e-05    4.631e-02
     13      66   -1.504799e-02    0.000e+00    9.721e-05    2.028e-01
     14      70   -1.516734e-02    0.000e+00    9.323e-05    1.007e+00
     15      74   -1.569717e-02    0.000e+00    7.734e-05    4.869e+00
     16      78   -1.741822e-02    0.000e+00    4.758e-05    2.106e+01
     17      82   -1.835169e-02    0.000e+00    4.540e-05    1.550e+01
     18      87   -1.834569e-02    0.000e+00    2.897e-05    1.495e-01
     19      92   -1.834905e-02    0.000e+00    1.454e-05    5.888e-02
     20      96   -1.834932e-02    0.000e+00    9.991e-06    4.905e-03
     21     100   -1.837306e-02    0.000e+00    2.047e-06    4.230e-01
     22     104   -1.837911e-02    0.000e+00    2.214e-08    1.081e-01

Local minimum found that satisfies the constraints.

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

<stopping criteria details>

with x:

x =

     64.0000    2.0000    2.0000

and

>> fval

fval =

     -32.7680

This x meets the inequality conditions you made. It was very confusing that you made x0, lb and ub a size of 11073 --> So i first thought your solution needs to have 11073 components...

Best regards

Stephan

Answer 3

[Abdollah Louni]

Hi everybody, Thank you for your help and considerations.

Best wishes, A. Louni