First, I'll try my MILP problems in double-precision (probably using 
CPLEX).  I'll re-post my question about arbitrary precision if it turns out 
that I need it.

We definitely needed arbitrary precision when we did some related problems 
using semi-definite programming (over the real numbers -- without the 
integrality constraints).  To use arbitrary precision floating point 
arithmetic, we had to employ SDPA-GMP as our SDP-solver, invoking it from 
Sage via a shell script.

Thanks again for the help.

Daniel

On Tuesday, January 29, 2013 3:12:45 PM UTC, Daniel Friedan wrote:
>
> The following example from Sage Reference v5.6 >> Numerical Optimization 
> >> Mixed integer linear programming
>       http://www.sagemath.org/doc/reference/sage/numerical/mip.html
> gives a wrong answer when solver = 'PPL' is used.  The equality 
> constraints are violated.
>
> Sage 5.6-OSX-64bit-10.6 under OS X 10.6.8
>
> sage: p = MixedIntegerLinearProgram(maximization=False, solver = "PPL")
> sage: print p.base_ring()
> sage: w = p.new_variable()
> sage: p.add_constraint(w[0] + w[1] + w[2] - 14*w[3] == 0)
> sage: p.add_constraint(w[1] + 2*w[2] - 8*w[3] == 0)
> sage: p.add_constraint(2*w[2] - 3*w[3] == 0)
> sage: p.add_constraint(w[0] - w[1] - w[2] >= 0)
> sage: p.add_constraint(w[3] >= 1)
> sage: _ = [ p.set_min(w[i], None) for i in range(1,4) ]
> sage: p.set_objective(w[3])
> sage: p.show()
> sage: print 'Objective Value:', p.solve()
> sage: for i, v in p.get_values(w).iteritems():\
> sage:          print 'w_%s = %s' % (i, int(round(v)))
>
> Rational Field
> Minimization:
>   x_3
> Constraints:
>   constraint_0: 0 <= x_0 + x_1 + x_2 - 14 x_3 <= 0
>   constraint_1: 0 <= x_1 + 2 x_2 - 8 x_3 <= 0
>   constraint_2: 0 <= 2 x_2 - 3 x_3 <= 0
>   constraint_3: - x_0 + x_1 + x_2 <= 0
>   constraint_4: - x_3 <= -1
> Variables:
>   x_0 is a continuous variable (min=0, max=+oo)
>   x_1 is a continuous variable (min=-oo, max=+oo)
>   x_2 is a continuous variable (min=-oo, max=+oo)
>   x_3 is a continuous variable (min=-oo, max=+oo)
> Objective Value: 1
> w_0 = 8
> w_1 = 5
> w_2 = 2
> w_3 = 1
>
>
>

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