| Title: | Interface to 'Lp_solve' v. 5.5 to Solve Linear/Integer Programs |
|---|---|
| Description: | Lp_solve is freely available (under LGPL 2) software for solving linear, integer and mixed integer programs. In this implementation we supply a "wrapper" function in C and some R functions that solve general linear/integer problems, assignment problems, and transportation problems. This version calls lp_solve version 5.5. |
| Authors: | Gábor Csárdi [cre], Michel Berkelaar [aut] |
| Maintainer: | Gábor Csárdi <[email protected]> |
| License: | LGPL-2 |
| Version: | 5.6.21.9000 |
| Built: | 2024-10-07 12:22:52 UTC |
| Source: | https://github.com/gaborcsardi/lpsolve |
Interface to lp_solve linear/integer programming system
lp (direction = "min", objective.in, const.mat, const.dir, const.rhs, transpose.constraints = TRUE, int.vec, presolve=0, compute.sens=0, binary.vec, all.int=FALSE, all.bin=FALSE, scale = 196, dense.const, num.bin.solns=1, use.rw=FALSE, timeout = 0L)lp (direction = "min", objective.in, const.mat, const.dir, const.rhs, transpose.constraints = TRUE, int.vec, presolve=0, compute.sens=0, binary.vec, all.int=FALSE, all.bin=FALSE, scale = 196, dense.const, num.bin.solns=1, use.rw=FALSE, timeout = 0L)
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 ">=". (In each pair the two values are identical.) |
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 that case set transpose.constraints to FALSE. |
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." |
binary.vec |
Numeric vector like int.vec giving the indices of variables that are required to be binary. |
all.int |
Logical: should all variables be integer? Default: FALSE. |
all.bin |
Logical: should all variables be binary? Default: FALSE. |
scale |
Integer: value for lpSolve scaling. Details can be found in the lpSolve documentation. Set to 0 for no scaling. Default: 196 |
dense.const |
Three column dense constraint array. This is ignored if const.mat is supplied. Otherwise the columns are constraint number, column number, and value; there should be one row for each non-zero entry in the constraint matrix. |
num.bin.solns |
Integer: if all.bin=TRUE, the user can request up to num.bin.solns optimal solutions to be returned. |
use.rw |
Logical: if TRUE and num.bin.solns > 1, write the lp out to a file and read it back in for each solution after the first. This is just to defeat a bug somewhere. Although the default is FALSE, we recommend you set this to TRUE if you need num.bin.solns > 1, until the bug is found. |
timeout |
Integer: timeout variable in seconds, defaults to 0L which means no limit is set. |
This function calls the lp_solve 5.5 solver. That system has many options not
supported here. The current version is maintained at
https://lpsolve.sourceforge.net/5.5/
Note that every variable is assumed to be >= 0!
An lp object. See lp.object for details.
Sam Buttrey, [email protected]
# # 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, 1) 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) ## Not run: Success: the objective function is 40.5 lp ("max", f.obj, f.con, f.dir, f.rhs)$solution ## Not run: [1] 0.0 4.5 0.0 # # The same problem using the dense constraint approach: # f.con.d <- matrix (c(rep (1:2,each=3), rep (1:3, 2), t(f.con)), ncol=3) lp ("max", f.obj, , f.dir, f.rhs, dense.const=f.con.d) ## Not run: Success: the objective function is 40.5 # # Get sensitivities # lp ("max", f.obj, f.con, f.dir, f.rhs, compute.sens=TRUE)$sens.coef.from ## Not run: [1] -1e+30 2e+00 -1e+30 lp ("max", f.obj, f.con, f.dir, f.rhs, compute.sens=TRUE)$sens.coef.to ## Not run: [1] 4.50e+00 1.00e+30 1.35e+01 # # Right now the dual values for the constraints and the variables are # combined, constraints coming first. So in this example... # lp ("max", f.obj, f.con, f.dir, f.rhs, compute.sens=TRUE)$duals ## Not run: [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)$duals.from ## Not run: [1] 0e+00 -1e+30 -1e+30 -1e+30 -6e+00 lp ("max", f.obj, f.con, f.dir, f.rhs, compute.sens=TRUE)$duals.to ## Not run: [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) ## Not run: Success: the objective function is 37 lp ("max", f.obj, f.con, f.dir, f.rhs, int.vec=1:3)$solution ## Not run: [1] 1 4 0 # # You can get sensitivities in the integer case, but they're harder to # interpret. # lp ("max", f.obj, f.con, f.dir, f.rhs, int.vec=1:3, compute.sens=TRUE)$duals ## Not run: [1] 1 0 0 7 0 # # Here's an example in which we want more than one solution to a problem # in which all variables are binary: the 8-queens problem, # with dense constraints. # chess.obj <- rep (1, 64) q8 <- make.q8 () chess.dir <- rep (c("=", "<"), c(16, 26)) chess.rhs <- rep (1, 42) lp ('max', chess.obj, , chess.dir, chess.rhs, dense.const = q8, all.bin=TRUE, num.bin.solns=3)# # 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, 1) 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) ## Not run: Success: the objective function is 40.5 lp ("max", f.obj, f.con, f.dir, f.rhs)$solution ## Not run: [1] 0.0 4.5 0.0 # # The same problem using the dense constraint approach: # f.con.d <- matrix (c(rep (1:2,each=3), rep (1:3, 2), t(f.con)), ncol=3) lp ("max", f.obj, , f.dir, f.rhs, dense.const=f.con.d) ## Not run: Success: the objective function is 40.5 # # Get sensitivities # lp ("max", f.obj, f.con, f.dir, f.rhs, compute.sens=TRUE)$sens.coef.from ## Not run: [1] -1e+30 2e+00 -1e+30 lp ("max", f.obj, f.con, f.dir, f.rhs, compute.sens=TRUE)$sens.coef.to ## Not run: [1] 4.50e+00 1.00e+30 1.35e+01 # # Right now the dual values for the constraints and the variables are # combined, constraints coming first. So in this example... # lp ("max", f.obj, f.con, f.dir, f.rhs, compute.sens=TRUE)$duals ## Not run: [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)$duals.from ## Not run: [1] 0e+00 -1e+30 -1e+30 -1e+30 -6e+00 lp ("max", f.obj, f.con, f.dir, f.rhs, compute.sens=TRUE)$duals.to ## Not run: [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) ## Not run: Success: the objective function is 37 lp ("max", f.obj, f.con, f.dir, f.rhs, int.vec=1:3)$solution ## Not run: [1] 1 4 0 # # You can get sensitivities in the integer case, but they're harder to # interpret. # lp ("max", f.obj, f.con, f.dir, f.rhs, int.vec=1:3, compute.sens=TRUE)$duals ## Not run: [1] 1 0 0 7 0 # # Here's an example in which we want more than one solution to a problem # in which all variables are binary: the 8-queens problem, # with dense constraints. # chess.obj <- rep (1, 64) q8 <- make.q8 () chess.dir <- rep (c("=", "<"), c(16, 26)) chess.rhs <- rep (1, 42) lp ('max', chess.obj, , chess.dir, chess.rhs, dense.const = q8, all.bin=TRUE, num.bin.solns=3)
Interface to lp_solve linear/integer programming
system specifically for solving assignment problems
lp.assign (cost.mat, direction = "min", presolve = 0, compute.sens = 0)lp.assign (cost.mat, direction = "min", presolve = 0, compute.sens = 0)
cost.mat |
Matrix of costs: the ij-th element is the cost of assigning source i to destination j. |
direction |
Character vector, length 1, containing either "min" (the default) or "max" |
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." In that case presolving is attempted. |
This is a particular integer programming problem. All the decision variables are assumed to be integers; each row has the constraint that its entries must add up to 1 (so that there is one 1 and the remaining entries are 0) and each column has the same constraint. This is assumed to be a minimization problem.
An lp object. See documentation for details. The constraints
are assumed (each row adds to 1, each column adds to 1, and no others) and
are not returned.
Sam Buttrey, [email protected]
assign.costs <- matrix (c(2, 7, 7, 2, 7, 7, 3, 2, 7, 2, 8, 10, 1, 9, 8, 2), 4, 4) ## Not run: > assign.costs [,1] [,2] [,3] [,4] [1,] 2 7 7 1 [2,] 7 7 2 9 [3,] 7 3 8 8 [4,] 2 2 10 2 ## End(Not run) lp.assign (assign.costs) ## Not run: Success: the objective function is 8 lp.assign (assign.costs)$solution ## Not run: [,1] [,2] [,3] [,4] [1,] 0 0 0 1 [2,] 0 0 1 0 [3,] 0 1 0 0 [4,] 1 0 0 0 ## End(Not run)assign.costs <- matrix (c(2, 7, 7, 2, 7, 7, 3, 2, 7, 2, 8, 10, 1, 9, 8, 2), 4, 4) ## Not run: > assign.costs [,1] [,2] [,3] [,4] [1,] 2 7 7 1 [2,] 7 7 2 9 [3,] 7 3 8 8 [4,] 2 2 10 2 ## End(Not run) lp.assign (assign.costs) ## Not run: Success: the objective function is 8 lp.assign (assign.costs)$solution ## Not run: [,1] [,2] [,3] [,4] [1,] 0 0 0 1 [2,] 0 0 1 0 [3,] 0 1 0 0 [4,] 1 0 0 0 ## End(Not run)
Structure of lp object
An lp.object is a list containing the following elements:
direction |
Optimization direction, as entered |
x.count |
Number of variables in objective function |
objective |
Vector of objective function coefficients, as entered |
const.count |
Number of constraints entered |
constraints |
Constraint matrix, as entered (not returned
by |
int.count |
Number of integer variables |
int.vec |
Vector of integer variables' indices, as entered |
objval |
Value of objective function at optimum |
solution |
Vector of optimal coefficients |
num.bin.solns |
Numeric indicator of number of solutions returned |
status |
Numeric indicator: 0 = success, 2 = no feasible solution |
Sam Buttrey, [email protected]
Interface to lp_solve linear/integer programming
system specifically for solving transportation problems
lp.transport (cost.mat, direction="min", row.signs, row.rhs, col.signs, col.rhs, presolve=0, compute.sens=0, integers = 1:(nc*nr) )lp.transport (cost.mat, direction="min", row.signs, row.rhs, col.signs, col.rhs, presolve=0, compute.sens=0, integers = 1:(nc*nr) )
cost.mat |
Matrix of costs; ij-th element is the cost of transporting one item from source i to destination j. |
direction |
Character, length 1: "min" or "max" |
row.signs |
Vector of character strings giving the direction of the row constraints: each value should be one of "<," "<=," "=," "==," ">," or ">=." (In each pair the two values are identical.) |
row.rhs |
Vector of numeric values for the right-hand sides of the row constraints. |
col.signs |
Vector of character strings giving the direction of the column constraints: each value should be one of "<," "<=," "=," "==," ">," or ">=." |
col.rhs |
Vector of numeric values for the right-hand sides of the column constraints. |
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." |
integers |
Vector of integers whose ith element gives the index of the ith integer variable. Its length will be the number of integer variables. Default: all variables are integer. Set to NULL to have no variables be integer. |
This is a particular integer programming problem. All the decision variables are assumed to be integers, and there is one constraint per row and one per column (and no others). This is assumed to be a minimization problem.
An lp object. Constraints are implicit and not returned.
See documentation for details.
Sam Buttrey, [email protected]
Example problem from Bronson (1981), Operations Research, Scahum's Outline Series, McGraw-Hill.
# # Transportation problem, Bronson, problem 9.1, p. 86 # # Set up cost matrix # costs <- matrix (10000, 8, 5); costs[4,1] <- costs[-4,5] <- 0 costs[1,2] <- costs[2,3] <- costs[3,4] <- 7; costs[1,3] <- costs[2,4] <- 7.7 costs[5,1] <- costs[7,3] <- 8; costs[1,4] <- 8.4; costs[6,2] <- 9 costs[8,4] <- 10; costs[4,2:4] <- c(.7, 1.4, 2.1) # # Set up constraint signs and right-hand sides. # row.signs <- rep ("<", 8) row.rhs <- c(200, 300, 350, 200, 100, 50, 100, 150) col.signs <- rep (">", 5) col.rhs <- c(250, 100, 400, 500, 200) # # Run # lp.transport (costs, "min", row.signs, row.rhs, col.signs, col.rhs) ## Not run: Success: the objective function is 7790 lp.transport (costs, "min", row.signs, row.rhs, col.signs, col.rhs)$solution ## Not run: [,1] [,2] [,3] [,4] [,5] [1,] 0 100 0 100 0 [2,] 0 0 300 0 0 [3,] 0 0 0 350 0 [4,] 200 0 0 0 0 [5,] 50 0 0 0 50 [6,] 0 0 0 0 50 [7,] 0 0 100 0 0 [8,] 0 0 0 50 100 ## End(Not run)# # Transportation problem, Bronson, problem 9.1, p. 86 # # Set up cost matrix # costs <- matrix (10000, 8, 5); costs[4,1] <- costs[-4,5] <- 0 costs[1,2] <- costs[2,3] <- costs[3,4] <- 7; costs[1,3] <- costs[2,4] <- 7.7 costs[5,1] <- costs[7,3] <- 8; costs[1,4] <- 8.4; costs[6,2] <- 9 costs[8,4] <- 10; costs[4,2:4] <- c(.7, 1.4, 2.1) # # Set up constraint signs and right-hand sides. # row.signs <- rep ("<", 8) row.rhs <- c(200, 300, 350, 200, 100, 50, 100, 150) col.signs <- rep (">", 5) col.rhs <- c(250, 100, 400, 500, 200) # # Run # lp.transport (costs, "min", row.signs, row.rhs, col.signs, col.rhs) ## Not run: Success: the objective function is 7790 lp.transport (costs, "min", row.signs, row.rhs, col.signs, col.rhs)$solution ## Not run: [,1] [,2] [,3] [,4] [,5] [1,] 0 100 0 100 0 [2,] 0 0 300 0 0 [3,] 0 0 0 350 0 [4,] 200 0 0 0 0 [5,] 50 0 0 0 50 [6,] 0 0 0 0 50 [7,] 0 0 100 0 0 [8,] 0 0 0 50 100 ## End(Not run)
Generate sparse constraint matrix for 8-queens problem
make.q8 ()make.q8 ()
None.
Sparse constraints come in a three-column matrix or data frame. Each row gives the row number, column number, and value of a particular non-zero entry in the constraint matrix. This function produces the sparse constraint matrix for the 8-queens problem (in which the object is to place eight queens on a chessboard with no two sharing a row, column or diagonal). The resulting sparse representation is 252 x 3, compared to 42 x 64 for the usual representation.
A 252 x 3 numeric matrix. See lp for the complete example.
Sam Buttrey, [email protected]
Print method for lp objects
## S3 method for class 'lp' print(x, ...)## S3 method for class 'lp' print(x, ...)
x |
List with items named |
... |
Other arguments, all currently ignored |
This function prints the objective function value, together with the word "Success" if the operation is successful, or an indication of the error if not. If multiple solutions have been produced (because this was an all-binary problem and lp was called with num.bin.solns > 1) the number of solutions is also displayed.
None
Sam Buttrey, [email protected]