| linp {limSolve} | R Documentation |
Solves a linear programming problem,
min(sum(Cost_i x_i))
subject to
Ex=f
Gx>=h
x_i>=0
(optional)
This function provides a wrapper around lp (see note)
from package lpSolve, written to be consistent with the functions
lsei, and ldei.
It allows for the x's to be negative (not standard in lp).
linp(E = NULL, F = NULL, G = NULL, H = NULL, Cost,
ispos = TRUE, int.vec = NULL, verbose = TRUE, ...)
E |
numeric matrix containing the coefficients of the equality
constraints Ex=F; if the columns of |
F |
numeric vector containing the right-hand side of the equality constraints. |
G |
numeric matrix containing the coefficients of the inequality
constraints Gx>=H; if the columns of |
H |
numeric vector containing the right-hand side of the inequality constraints. |
Cost |
numeric vector containing the coefficients of the cost function;
if |
ispos |
logical, when |
int.vec |
when not |
verbose |
logical to print error messages. |
... |
extra arguments passed to R-function |
a list containing:
X |
vector containing the solution of the linear programming problem. |
residualNorm |
scalar, the sum of absolute values of residuals of equalities and violated inequalities. Should be very small or zero for a feasible linear programming problem. |
solutionNorm |
scalar, the value of the minimised |
IsError |
logical, |
type |
the string "linp", such that how the solution was obtained can be traced. |
If the requirement of nonnegativity are relaxed, then strictly speaking the problem is not a linear programming problem.
The function lp may fail and terminate R for very small problems that
are repeated frequently...
Also note that sometimes multiple solutions exist for the same problem.
Karline Soetaert <karline.soetaert@nioz.nl>
Michel Berkelaar and others (2007). lpSolve: Interface to Lpsolve v. 5.5 to solve linear or integer programs. R package version 5.5.8.
lp the original function from package lpSolve
Blending, a linear programming problem.
#-------------------------------------------------------------------------------
# Linear programming problem 1, not feasible
#-------------------------------------------------------------------------------
# maximise x1 + 3*x2
# subject to
#-x1 -x2 < -3
#-x1 + x2 <-1
# x1 + 2*x2 < 2
# xi > 0
G <- matrix(nrow = 3, data = c(-1, -1, 1, -1, 1, 2))
H <- c(3, -1, 2)
Cost <- c(-1, -3)
(L <- linp(E = NULL, F = NULL, Cost = Cost, G = G, H = H))
L$residualNorm
#-------------------------------------------------------------------------------
# Linear programming problem 2, feasible
#-------------------------------------------------------------------------------
# minimise x1 + 8*x2 + 9*x3 + 2*x4 + 7*x5 + 3*x6
# subject to:
#-x1 + x4 + x5 = 0
# - x2 - x4 + x6 = 0
# x1 + x2 + x3 > 1
# x3 + x5 + x6 < 1
# xi > 0
E <- matrix(nrow = 2, byrow = TRUE, data = c(-1, 0, 0, 1, 1, 0,
0,-1, 0, -1, 0, 1))
F <- c(0, 0)
G <- matrix(nrow = 2, byrow = TRUE, data = c(1, 1, 1, 0, 0, 0,
0, 0, -1, 0, -1, -1))
H <- c(1, -1)
Cost <- c(1, 8, 9, 2, 7, 3)
(L <- linp(E = E, F = F, Cost = Cost, G = G, H = H))
L$residualNorm
#-------------------------------------------------------------------------------
# Linear programming problem 3, no positivity
#-------------------------------------------------------------------------------
# minimise x1 + 2x2 -x3 +4 x4
# subject to:
# 3x1 + 2x2 + x3 + x4 = 2
# x1 + x2 + x3 + x4 = 2
# 2x1 + x2 + x3 + x4 >=-1
# -x1 + 3x2 +2x3 + x4 >= 2
# -x1 + x3 >= 1
E <- matrix(ncol = 4, byrow = TRUE,
data =c(3, 2, 1, 4, 1, 1, 1, 1))
F <- c(2, 2)
G <- matrix(ncol = 4, byrow = TRUE,
data = c(2, 1, 1, 1, -1, 3, 2, 1, -1, 0, 1, 0))
H <- c(-1, 2, 1)
Cost <- c(1, 2, -1, 4)
linp(E = E, F = F, G = G, H = H, Cost, ispos = FALSE)