Intlinprog does not retun integer solution

I'se setup a problem for intlinprog, solutions are constrained to be binary:

lb =

       0     0     0     0     0     0     0     0     0

ub =

       1     1     1     1     1     1     1     1     1

However this is what I get when I run intlingprog:

LP:                Optimal objective value is 6.000000.                                             

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).

ans =

      0.5000    0.5000         0         0    0.5000    0.5000         0         0         0

Which is clearly a not integer solution within the tollerance.

What am I missing?

[Matt J]

How about we see the code that produced it?

Answer 1

[Bo Li]

Yes, the INTCON parameter is very important to constrain the value of the variables to be integer. It does NOT mean the number of integer, but means the integer constraints. I made the same mistake.

See what it is explained:

intcon — Vector of integer constraints vector of integers Vector of integer constraints, specified as a vector of positive integers. The values in intcon indicate the components of the decision variable x that are integer-valued. intcon has values from 1 through numel(f).

intcon can also be an array. Internally, intlinprog converts an array intcon to the vector intcon(:).

Example: intcon = [1,2,7] means x(1), x(2), and x(7) take only integer values.

[Roberto]

Thanks, should have read the documentation with more attention.

Answer 2

[Alan Weiss]

Without seeing your intlinprog call, I can only guess that you neglected to include an intcon vector of the requisite length.

Alan Weiss

MATLAB mathematical toolbox documentation

[Roberto]

This is the entire problem:

f =

       3     3     3     3     3     3     0     0     3     3     0     0     3     3     3     3     3     3

intcon =

      18

A =

       2     2     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
       0     0     2     2     0     0     0     0     0     0     0     0     0     0     0     0     0     0
       0     0     0     0     2     2     0     0     0     0     0     0     0     0     0     0     0     0
       1     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
       0     0     1     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0
       0     0     0     0     1     1     0     0     0     0     0     0     0     0     0     0     0     0
       0     0     0     0     0     0     2     2     0     0     0     0     0     0     0     0     0     0
       0     0     0     0     0     0     0     0     2     2     0     0     0     0     0     0     0     0
       0     0     0     0     0     0     0     0     0     0     2     2     0     0     0     0     0     0
       0     0     0     0     0     0     0     0     0     0     0     0     2     2     0     0     0     0
       0     0     0     0     0     0     0     0     0     0     0     0     0     0     2     2     0     0
       0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     2     2

b =

      10    10     0     1     1     1     0   100     0   100   100   100

Aeq =

       1     0     1     0     1     0     0     0     0     0     0     0     0     0     0     0     0     0
       0     1     0     1     0     1     0     0     0     0     0     0     0     0     0     0     0     0
       1    -1     0     0     0     0     0     0    -1     0     0     0     0     0     1     0     0     0
      -1     1     0     0     0     0     0     0     0    -1     0     0     0     0     0     1     0     0
       0     0     1    -1     0     0     0     0     0     0     0     0    -1     0     0     0     1     0
       0     0    -1     1     0     0     0     0     0     0     0     0     0    -1     0     0     0     1
       0     0     0     0     1    -1     0     0     1     0     0     0     1     0    -1     0    -1     0
       0     0     0     0    -1     1     0     0     0     1     0     0     0     1     0    -1     0    -1

beq =

       1     1     0     0     0     0     0     0

lb =

       0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0

ub =

       1     1     1     1     1     1     1     1     1     1     1     1     1     1     1     1     1     1

and this is the output of

intlinprog(f,intcon,A,b,Aeq,beq,lb,ub);

LP: Optimal objective value is 6.000000.

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).

ans =

      0.5000    0.5000    0.5000    0.5000         0         0         0         0         0         0         0         0         0         0         0         0         0         0

[Matt J]

As Alan guessed, you are interpreting intcon incorrectly. Setting intcon=18 means that you are requiring only x(18) to be integer-valued. You should really have

   intcon=1:18;