lp: Linear and Integer Programming

Description

Interface to lp_solve linear/integer programming system

Usage

lp (direction = "min", objective.in, const.mat, const.dir, const.rhs,
	transpose.constraints = TRUE, int.vec, presolve=0, compute.sens=0)

Arguments

direction

Character string giving direction of optimization: "min" (default) or "max."

objective.in

Numeric vector of coefficients of objective function

const.mat

Matrix of numeric constraint coefficients, one row per constraint, one column per variable (unless transpose.constraints = FALSE; see below).

const.dir

Vector of character strings giving the direction of the constraint: each value should be one of "<," "<=","" "=","">," or ">=."

const.rhs

Vector of numeric values for the right-hand sides of the constraints.

transpose.constraints

By default each constraint occupies a row of const.mat, and that matrix needs to be transposed before being passed to the optimizing code. For very large constraint matrices it may be wiser to construct the constraints in a matrix column-by-column. In

int.vec

Numeric vector giving the indices of variables that are required to be integer. The length of this vector will therefore be the number of integer variables.

presolve

Numeric: presolve? Default 0 (no); any non-zero value means "yes." Currently ignored.

compute.sens

Numeric: compute sensitivity? Default 0 (no); any non-zero value means "yes."

Value

An lp object. See link{lp.object} for details.

Details

This function calls the lp_solve 5.0 solver. That system has many options not supported here. The current version is maintained at ftp://ftp.es.ele.tue.nl/pub/lp_solve

Examples

Run this code

#
# Set up problem: maximize
#   x1 + 9 x2 +   x3 subject to
#   x1 + 2 x2 + 3 x3  <= 9
# 3 x1 + 2 x2 + 2 x3 <= 15
#
f.obj <- c(1, 9, 3)
f.con <- matrix (c(1, 2, 3, 3, 2, 2), nrow=2, byrow=TRUE)
f.dir <- c("<=", "<=")
f.rhs <- c(9, 15)
#
# Now run.
#
lp ("max", f.obj, f.con, f.dir, f.rhs)
Success: the objective function is 40.5
lp ("max", f.obj, f.con, f.dir, f.rhs)$solution
[1] 0.0 4.5 0.0
#
# Get sensitivities
#
lp ("max", f.obj, f.con, f.dir, f.rhs, compute.sens=TRUE)$sens.coef.from
[1] -1e+30  2e+00 -1e+30
lp ("max", f.obj, f.con, f.dir, f.rhs, compute.sens=TRUE)$sens.coef.to  
[1] 4.50e+00 1.00e+30 1.35e+01
#
# Right now the dual values for the constraints and the variables are
# combined, constraints coming first. So in this example...
#
lp ("max", f.obj, f.con, f.dir, f.rhs, compute.sens=TRUE)$duals     
[1]   4.5   0.0  -3.5   0.0 -10.5
#
# ...the duals of the constraints are 4.5 and 0, and of the variables,
# -3.5, 0.0, -10.5. Here are the lower and upper limits on these:
#
lp ("max", f.obj, f.con, f.dir, f.rhs, compute.sens=TRUE)$duals.from
[1]  0e+00 -1e+30 -1e+30 -1e+30 -6e+00
lp ("max", f.obj, f.con, f.dir, f.rhs, compute.sens=TRUE)$duals.to  
[1] 1.5e+01 1.0e+30 3.0e+00 1.0e+30 3.0e+00
#
# Run again, this time requiring that all three variables be integer
#
lp ("max", f.obj, f.con, f.dir, f.rhs, int.vec=1:3)
Success: the objective function is 37
lp ("max", f.obj, f.con, f.dir, f.rhs, int.vec=1:3)$solution
[1] 1 4 0
#
# You can get sensitivities in the integer case, but they're harder to
# interpret.
#
lp ("max", f.obj, f.con, f.dir, f.rhs, int.vec=1:3, compute.sens=TRUE)$duals
[1] 1 0 0 7 0