Sorry. Of course, I meant the DIFFERENCE of two non-negative
variables. So, for example, write b1 = b1p - b1n, where both b1p and
b1n are non-negative.
On Thu, Jul 2, 2009 at 4:18 PM, Rob Helpert wrote:
> Hi Alicia,
>
> I think the trick may be to split b1 into the sum of two non-negative
> varia
Hi Alicia,
I think the trick may be to split b1 into the sum of two non-negative
variables. You will then also have to alter your constraints and
objective to include the two new variables with negative values in
appropriate places, but I believe that this will solve the problem.
On Thu, Jul 2,
Dear all,
I am interested in solving a MIP problem with binary outcomes and
continuous variables, which ARE NOT RESTRICTED TO BE NEGATIVE. In
particular,
Max {z1,z2,z3,b1} z1 + z2 + z3
(s.t.)
# 7 z1 + 0 z2 + 0 z3 + b1 <= 5
# 0 z1 + 8 z2 + 0 z3 - b1 <= 5
# 0 z1 + 0 z2 + 6 z3 + b1 <=
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