Trouble with very basic usage of glpk.ilp

[Rajhans Samdani]

Hi!

I am using cvxopt 1.1 and glpk version 4.4.2. I tried running a very
simple and vacuous (effectively unconstrained) ILP as a sanity-check
but I am getting the wrong answer. I am solving
max c'x
s.t.
Gx <=h
where x is 3 dimensional binary vector, G=1,1,1 and h = 5 (so the
problem is effectively unconstrained as x1+x2+x3 is always less than
5.) The answer irrespective of the value of c is always 0,0,0!!

Following shows the python command line operations. Am I doing
something obviously wrong? Please help!

>>> G=matrix([1.0,1.0,1.0],(1,3))
>>> h=matrix([5.0])
>>> c=matrix([-10,-10,-15],(3,1))
>>> b=set([0,1,2])
>>> (s,y)=glpk.ilp(G=G,c=c,h=h,B=b)
GLPK Integer Optimizer, v4.42
1 row, 3 columns, 3 non-zeros
3 integer variables, all of which are binary
Preprocessing...
Objective value = nan
INTEGER OPTIMAL SOLUTION FOUND BY MIP PREPROCESSOR
>>> print y
[ 0.00e+00]
[ 0.00e+00]
[ 0.00e+00]

>>> c=matrix([10,10,15],(3,1))
>>> (s,y)=glpk.ilp(G=G,c=c,h=h,B=b)
GLPK Integer Optimizer, v4.42
1 row, 3 columns, 3 non-zeros
3 integer variables, all of which are binary
Preprocessing...
Objective value = 0.000000000e+00
INTEGER OPTIMAL SOLUTION FOUND BY MIP PREPROCESSOR
>>> print y
[ 0.00e+00]
[ 0.00e+00]
[ 0.00e+00]


Thanks!
Rajhans

[alexander...@gmail.com]

Hi Rajhans!

Did you manage to solve the problem? I am also facing it now and can not find solution.

Thanks in advance,
Alexander