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 Pérez-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-alo...@eui.eu
URL: http://www.eui.eu/Personal/Fellows/Perez-Alonso/
______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
