Hi All
Apologies if this is not the correct list for this question.
The Rglpk package offers the following example in its documentation
library(Rglpk)
## Simple mixed integer linear program.
## maximize: 3 x_1 + 1 x_2 + 3 x_3
## subject to: -1 x_1 + 2 x_2 + x_3 <= 4
## 4 x_2 - 3 x_3 <= 2
## x_1 - 3 x_2 + 2 x_3 <= 3
## x_1, x_3 are non-negative integers
## x_2 is a non-negative real number
obj <- c(3, 1, 3)
mat <- matrix(c(-1, 0, 1, 2, 4, -3, 1, -3, 2), nrow = 3)
dir <- c("<=", "<=", "<=")
rhs <- c(4, 2, 3)
types <- c("I", "C", "I")
max <- TRUE
Rglpk_solve_LP(obj, mat, dir, rhs, types, max)
## Same as before but with bounds replaced by
## -Inf < x_1 <= 4
## 0 <= x_2 <= 100
## 2 <= x_3 < Inf
bounds <- list(lower = list(ind = c(1L, 3L), val = c(-Inf, 2)),
upper = list(ind = c(1L, 2L), val = c(4, 100)))
Rglpk_solve_LP(obj, mat, dir, rhs, types, max, bounds)
I have 2 questions
1. What is the purpose of the L in the bounds statement (e.g. 1L, 3L
etc)?
2. Is it possible to further constrain a variable such that in the
optimal solution to the objective function it will be a specific integer
or an integer multiple of that integer. For example, x_3 must be 2 or
4,6,8,10 etc
Thanks
Pete
This e-mail may contain confidential or proprietary information
belonging to the BP group and is intended only for the use of the
recipients named above. If you are not the intended recipient, please
immediately notify the sender and either delete this email or return to
the sender immediately. You may not review, copy or distribute this
email. Within the bounds of law, this part of BP retains all emails and
IMs and may monitor them to ensure compliance with BP's internal
policies and for other legitimate business purposes.
[[alternative HTML version deleted]]
______________________________________________
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.
