R help - Jul 2009 - lpSolve: how to allow variables to become negative

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 <= 7
#   z1, z2, z3   BINARY {0,1}
#  -5<= b1 <=5 (i.e.  b1 <= 5; -b1 <= 5 )

Using the lpSolve package of R, I wrote:

library (lpSolve)
f.obj <- c(1, 1, 1, 0)
f.con <- matrix (c(7, 0, 0, -1, 0, 8, 0, 1, 0, 0, 6, -1, 0, 0, 0, 1 ), 
nrow=4, byrow=TRUE)
f.dir <- c("<=", "<=", "<=",
"<=")
f.rhs <- c(5, 5, 7, 5)

lp ("max", f.obj, f.con, f.dir, f.rhs, binary.vec=1:3)
lp ("max", f.obj, f.con, f.dir, f.rhs, binary.vec=1:3)$solution

The problem is that BY DEFAULT,  "b1" is constrained to be
">=0",  thus,
the constraint "-b1 <= 5" does not apply.

Given that there is not any global variable in the lp() command such 
that I can change the default value of 0 (to -5 or -inf), I do not get 
the correct solution to the problem, which is indeed b1= -2.

Thank you very much in advance for your help,
Alicia

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_________________________________________________________
Alicia P?rez-Alonso
Departament of Economics
European University Institute
via della Piazzuola 43
50133 Firenze (FI)
ITALY

Room: 40 (Second Floor)
Phone: (+39) 0554685955
Fax: (+39) 0554685902
E-mail: alicia.perez-alonso at eui.eu
URL: http://www.eui.eu/Personal/Fellows/Perez-Alonso/

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, 2009 at 12:37 PM, Alicia
Perez-alonso<alicia.perez-alonso at eui.eu> wrote:
> 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 <= 7
> # ? z1, z2, z3 ? BINARY {0,1}
> # ?-5<= b1 <=5 (i.e. ?b1 <= 5; -b1 <= 5 )
>
> Using the lpSolve package of R, I wrote:
>
> library (lpSolve)
> f.obj <- c(1, 1, 1, 0)
> f.con <- matrix (c(7, 0, 0, -1, 0, 8, 0, 1, 0, 0, 6, -1, 0, 0, 0, 1 ),
> nrow=4, byrow=TRUE)
> f.dir <- c("<=", "<=", "<=",
"<=")
> f.rhs <- c(5, 5, 7, 5)
>
> lp ("max", f.obj, f.con, f.dir, f.rhs, binary.vec=1:3)
> lp ("max", f.obj, f.con, f.dir, f.rhs, binary.vec=1:3)$solution
>
> The problem is that BY DEFAULT, ?"b1" is constrained to be
">=0", ?thus, the
> constraint "-b1 <= 5" does not apply.
>
> Given that there is not any global variable in the lp() command such that I
> can change the default value of 0 (to -5 or -inf), I do not get the correct
> solution to the problem, which is indeed b1= -2.
>
> Thank you very much in advance for your help,
> Alicia
>
> --
> Before printing this message, make sure you really need to.
> Taking care of our environment is everyone's responsibility.
>
> _________________________________________________________
> Alicia P?rez-Alonso
> Departament of Economics
> European University Institute
> via della Piazzuola 43
> 50133 Firenze (FI)
> ITALY
>
> Room: 40 (Second Floor)
> Phone: (+39) 0554685955
> Fax: (+39) 0554685902
> E-mail: alicia.perez-alonso at eui.eu
> URL: http://www.eui.eu/Personal/Fellows/Perez-Alonso/
>
> ______________________________________________
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