Linear Programming Solver
Solves simple linear programming problems, allowing for inequality and equality constraints as well as lower and upper bounds.
linprog(cc, A = NULL, b = NULL, Aeq = NULL, beq = NULL, lb = NULL, ub = NULL, x0 = NULL, I0 = NULL, bigM = 100, maxiter = 20, maximize = FALSE)
cc |
defines the linear objective function. |
A |
matrix representing the inequality constraints |
b |
vector, right hand side of the inequalities. |
Aeq |
matrix representing the equality constraints |
beq |
vector, right hand side of the inequalities. |
lb |
lower bounds, if not |
ub |
upper bounds, if not |
x0 |
feasible base vector, will not be used at the moment. |
I0 |
index set of |
bigM |
big-M constant, will be used for finding a base vector. |
maxiter |
maximum number of iterations. |
maximize |
logical; shall the objective be minimized or maximized? |
Solves linear programming problems of the form min cc' * x such that
A * x ≤ b
A_{eq} * x = b_{eq}
lb ≤ x ≤ ub
List with
x
the solution vector.
fval
the value at the optimal solution.
errno
, mesage
the error number and message.
This is a first version that will be unstable at times. For real linear
programming problems use package lpSolve
.
HwB <hwborchers@googlemail.com>
Vanderbei, R. J. (2001). Linear Programming: Foundations and Extensions. Princeton University Press.
Eiselt, H. A., and C.-L. Sandblom (2012). Operations Research: A Model-based Approach. Springer-Verlag, Berlin Heidelberg.
linprog::solveLP
, lpSolve::lp
## Examples from the book "Operations research - A Model-based Approach" #-- production planning cc <- c(5, 3.5, 4.5) Ain <- matrix(c(3, 5, 4, 6, 1, 3), 2, 3, byrow=TRUE) bin <- c(540, 480) linprog(cc, A = Ain, b = bin, maximize = TRUE) # $x 20 0 120 # $fval 640 #-- diet problem cc <- c(1.59, 2.19, 2.99) Ain <- matrix(c(-250, -380, -257, 250, 380, 257, 13, 31, 28), 3, 3, byrow = TRUE) bin <- c(-1800, 2200, 100) linprog(cc, A = Ain, b = bin) #-- employee scheduling cc <- c(1, 1, 1, 1, 1, 1) A <- (-1)*matrix(c(1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1), 6, 6, byrow = TRUE) b <- -c(17, 9, 19, 12, 5, 8) linprog(cc, A, b) #-- inventory models cc <- c(1, 1.1, 1.2, 1.25, 0.05, 0.15, 0.15) Aeq <- matrix(c(1, 0, 0, 0, -1, 0, 0, 0, 1, 0, 0, 1, -1, 0, 0, 0, 1, 0, 0, 1, -1, 0, 0, 0, 1, 0, 0, 1), 4, 7, byrow = TRUE) beq <- c(60, 70, 130, 150) ub <- c(120, 140, 150, 140, Inf, Inf, Inf) linprog(cc, Aeq = Aeq, beq = beq, ub = ub) #-- allocation problem cc <- c(1, 1, 1, 1, 1) A <- matrix(c(-5, 0, 0, 0, 0, 0, -4.5, 0, 0, 0, 0, 0, -5.5, 0, 0, 0, 0, 0, -3.5, 0, 0, 0, 0, 0, -5.5, 5, 0, 0, 0, 0, 0, 4.5, 0, 0, 0, 0, 0, 5.5, 0, 0, 0, 0, 0, 3.5, 0, 0, 0, 0, 0, 5.5, -5, -4.5, -5.5, -3.5, -5.5, 10, 10.0, 10.0, 10.0, 10.0, 0.2, 0.2, 0.2, -1.0, 0.2), 13, 5, byrow = TRUE) b <- c(-50, -55, -60, -50, -50, rep(100, 5), -5*64, 700, 0) # linprog(cc, A = A, b = b) lb <- b[1:5] / diag(A[1:5, ]) ub <- b[6:10] / diag(A[6:10, ]) A1 <- A[11:13, ] b1 <- b[11:13] linprog(cc, A1, b1, lb = lb, ub = ub) #-- transportation problem cc <- c(1, 7, 4, 2, 3, 5) Aeq <- matrix(c(1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1), 5, 6, byrow = TRUE) beq <- c(30, 20, 15, 25, 10) linprog(cc, Aeq = Aeq, beq = beq)
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