How to force intlinprog to give integer solution ?

The following command

 x=intlinprog([0; 0; 0],[1 2 3],-[0 2 0;0 1 2;3 1 0],-[6;13;11],[],[],[0;0;0],[])

produced x=[2.6667 3 5]' which is not integer.

Answer 1

[Matt J]

I think it's a bug. If your goal is simply to find some feasible solution to the constraints, I think you can workaround the bug with any nonzero f vector, e.g.,

>> x=intlinprog([1e-10,0,0],[1 2 3],-[0 2 0;0 1 2;3 1 0],-[6;13;11],[],[],[0;0;0],[]);

x =

     0
    11
     1

[Jan]

Thanks for the prompt answer, but try the following real optimization:

x=intlinprog([0; 1; 0],[1 2 3],-[0 2 0;0 1 2;3 1 0],-[6;13;11],[],[],[0;0;0],[]);

This gives x=[2.6667 3 5]'. Moreover, I have no idea how to find a proper f in a bigger problem.

[Walter Roberson]

What is the exitflag being returned? You do not appear to be recording the exitflag so you do not know what the output represents.

[Jan]

Walter,

[x,y,exitflag,output]=intlinprog([0;1;0],[1 2 3],-[0 2 0;0 1 2;3 1 0],-[6;13;11],[],[],[0;0;0],[])

gives:

Optimal solution found.
Intlinprog stopped at the root node because the objective value is within a gap tolerance of the
optimal value, options.TolGapAbs = 0 (the default value). The intcon variables are
integer within tolerance, options.TolInteger = 1e-05 (the default value).

x =

      2.6667
      3.0000
      5.0000
  y =
       3
  exitflag =
       1
  output = 
          relativegap: 0
          absolutegap: 0
        numfeaspoints: 1
             numnodes: 0
      constrviolation: 0
              message: 'Optimal solution found.…'

Answer 2

[Alan Weiss]

I think that this must be a bug in your Optimization Toolbox™ version. Please report the issue to technical support. There might be a workaround.

Alan Weiss

MATLAB mathematical toolbox documentation