When I use fmincon, the optimised result does not satisfy my non liner constraints

clear variables
close all
clc
fun = @(x)4*x(1)+x(2);
x0=[0.4,0.28]

x0 = 1×2

0.4000 0.2800

lb = [0.01,0.01];
ub = [5,0.8];
A = [];
b = [];
Aeq = [];
beq = [];
options = optimoptions('fmincon','Algorithm','sqp');
c = @(x)(4+4*x(1)+9*x(1)^2)*(1+x(1))^2-(2*x(1)^3*x(2)+2+3*x(1)+3*x(1)^2)^2+0.001;
nonlcon = @(x)deal(c(x),[]);
[x,fval,exitflag,output] = fmincon(fun,x0,A,b,Aeq,beq,lb,ub,nonlcon,options);

Feasible point with lower objective function value found.


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.

exitflag

exitflag = -2

checkinitialpoint=(4+4*x0(1)+9*x0(1)^2)*(1+x0(1))^2-(2*x0(1)^3*x0(2)+2+3*x0(1)+3*x0(1)^2)^2+0.001;
checkconstraits=(4+4*x(1)+9*x(1)^2)*(1+x(1))^2-(2*x(1)^3*x(2)+2+3*x(1)+3*x(1)^2)^2+0.001;

The above is my matlab code, I input the nonlear constraints, but the results give to me is obviously not satisfy the constraints (you can see that checkconstraits is positive), my initial point is within the range.

Can anyone help me? Many thanks.

[Torsten]

The solver converged to an infeasible point (see above). Your observation is the same as the exitflag from "fmincon" indicates.

Answer 1

[Alan Weiss]

You would do better to use the default 'interior-point' algorithm, which arrives at a feasible solution.

fun = @(x)4*x(1)+x(2);
x0=[0.4,0.28];
lb = [0.01,0.01];
ub = [5,0.8];
A = [];
b = [];
Aeq = [];
beq = [];
% options = optimoptions('fmincon','Algorithm','sqp');
c = @(x)(4+4*x(1)+9*x(1)^2)*(1+x(1))^2-(2*x(1)^3*x(2)+2+3*x(1)+3*x(1)^2)^2+0.001;
nonlcon = @(x)deal(c(x),[]);
[x,fval,exitflag,output] = fmincon(fun,x0,A,b,Aeq,beq,lb,ub,nonlcon)

Feasible point with lower objective function value found.


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.

x = 1×2

0.1004 0.5349

fval = 0.9363

exitflag = 1

output = struct with fields:
         iterations: 51
          funcCount: 199
    constrviolation: 0
           stepsize: 2.4326e-07
          algorithm: 'interior-point'
      firstorderopt: 5.5041e-07
       cgiterations: 4
            message: '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.↵↵<stopping criteria details>↵↵Optimization completed: The relative first-order optimality measure, 1.376030e-07,↵is less than options.OptimalityTolerance = 1.000000e-06, and the relative maximum constraint↵violation, 0.000000e+00, is less than options.ConstraintTolerance = 1.000000e-06.'
       bestfeasible: [1×1 struct]

For more information about dealing with infeasible solutions, see Solver Lost Feasibility.

Alan Weiss

MATLAB mathematical toolbox documentation

[Tianshu Gao]

Thank you

Answer 2

[Matt J]

Since it is a 2D problem, it practically begs you to pre-sweep for a good initial guess:

fun = @(x)4*x(1)+x(2);

lb = [0.01,0.01];
ub = [5,0.8];
A = [];
b = [];
Aeq = [];
beq = [];
c = @(x)(4+4*x(1)+9*x(1)^2)*(1+x(1))^2-(2*x(1)^3*x(2)+2+3*x(1)+3*x(1)^2)^2+0.001;
nonlcon = @(x)deal(c(x),[]);

[x0,fval0]=sweep(fun,c,lb,ub)

x0 = 1×2

0.1821 0.4186

fval0 = 1.1469

options = optimoptions('fmincon','Algorithm','sqp');

[x,fval,exitflag,output] = fmincon(fun,x0,A,b,Aeq,beq,lb,ub,nonlcon,options);

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.

x,fval

x = 1×2

0.1004 0.5349

fval = 0.9363

function [x0,fval]=sweep(fun,c,lb,ub)

 F=@(x1,x2) fun([x1,x2])+eps./(c([x1,x2])<=0);

 [X1,X2]=ndgrid(linspace(lb(1),ub(1),30), linspace(lb(2),ub(2),30));

 v=arrayfun(F,X1,X2);

  [fval,i]=min(v(:));

  if ~isfinite(fval)
  
      disp 'No feasible point found'; x0=[];

  else
      x0=[X1(i),X2(i)];
  end

end

[Tianshu Gao]

Thank you.

[Matt J]

You're welcome, but please Accept-click whichever answer solved your issue.