Solutions to a system of inequalities in R - r

Question

Suppose I have the following system of inequalities:
-2x + y <= -3
1.25x + y <= 2.5
y >= -3
I want to find multiple tuples of (x, y) that satisfy the above inequalities.
library(Rglpk)
obj <- numeric(2)
mat <- matrix(c(-2, 1, 1.25, 1, 0, 1), nrow = 3)
dir <- c("<=", "<=", ">=")
rhs <- c(-3, 2.5, -3)
Rglpk_solve_LP(obj = obj, mat = mat, dir = dir, rhs = rhs)
Using the above code only seems to return 1 possible solution tuple (1.5, 0). Is possible to return other solution tuples?
Edit: Based on the comments, I would be interested to learn if there are any functions that could help me find the corner points.

Answer 1

Actually to understand the possible answers for the given question we can try to solve the system of inequalities graphically.
There was a nice answer concerning plotting of inequations in R at stackowerflow. Using the given aproach we can plot the following graph:
library(ggplot2)
fun1 <- function(x) 2*x - 3 # this is the same as -2x + y <= -3
fun2 <- function(x) -1.25*x + 2.5 # 1.25x + y <= 2.5
fun3 <- function(x) -3 # y >= -3
x1 = seq(-1,5, by = 1/16)
mydf = data.frame(x1, y1=fun1(x1), y2=fun2(x1),y3= fun3(x1))
mydf <- transform(mydf, z = pmax(y3,pmin(y1,y2)))
ggplot(mydf, aes(x = x1)) +
geom_line(aes(y = y1), colour = 'blue') +
geom_line(aes(y = y2), colour = 'green') +
geom_line(aes(y = y3), colour = 'red') +
geom_ribbon(aes(ymin=y3,ymax = z), fill = 'gray60')
All the possible (infinite by number) tuples lie inside the gray triangle.
The vertexes can be found using the following code.
obj <- numeric(2)
mat <- matrix(c(-2, 1.25, 1, 1), nrow = 2)
rhs <- matrix(c(-3, 2.5), nrow = 2)
aPoint <- solve(mat, rhs)
mat <- matrix(c(-2, 0, 1, 1), nrow = 2)
rhs <- matrix(c(-3, -3), nrow = 2)
bPoint <- solve(mat, rhs)
mat <- matrix(c(1.25, 0, 1, 1), nrow = 2)
rhs <- matrix(c(2.5, -3), nrow = 2)
cPoint <- solve(mat, rhs)
Note the order of arguments of matrices.
And you get the coordinates:
> aPoint
[,1]
[1,] 1.6923077
[2,] 0.3846154
> bPoint
[,1]
[1,] 0
[2,] -3
> cPoint
[,1]
[1,] 4.4
[2,] -3.0

Answer 2

All the codes below are with base R only (no need library(Rglpk))
1. Corner Points
If you want to get all the corner points, here is one option
A <- matrix(c(-2, 1.25, 0, 1, 1, -1), nrow = 3)
b <- c(-3, 2.5, 3)
# we use `det` to check if the coefficient matrix is singular. If so, we return `Inf`.
xh <-
combn(nrow(A), 2, function(k) {
if (det(A[k, ]) == 0) {
rep(NA, length(k))
} else {
solve(A[k, ], b[k])
}
})
# We filter out the points that satisfy the constraint
corner_points <- t(xh[, colSums(A %*% xh <= b, na.rm = TRUE) == length(b)])
such that
> corner_points
[,1] [,2]
[1,] 1.692308 0.3846154
[2,] 0.000000 -3.0000000
[3,] 4.400000 -3.0000000
2. Possible Tuples
If you want to have multiple tuples, e.g., n=10, we can use Monte Carlo simulation (based on the obtained corner_points in the previous step) to select the tuples under the constraints:
xrange <- range(corner_points[, 1])
yrange <- range(corner_points[, 2])
n <- 10
res <- list()
while (length(res) < n) {
px <- runif(1, xrange[1], xrange[2])
py <- runif(1, yrange[1], yrange[2])
if (all(A %*% c(px, py) <= b)) {
res[length(res) + 1] <- list(c(px, py))
}
}
and you will see n possible tuples in a list like below
> res
[[1]]
[1] 3.643167 -2.425809
[[2]]
[1] 2.039007 -2.174171
[[3]]
[1] 0.4990635 -2.3363637
[[4]]
[1] 0.6168402 -2.6736421
[[5]]
[1] 3.687389 -2.661733
[[6]]
[1] 3.852258 -2.704395
[[7]]
[1] 1.7571062 0.1067597
[[8]]
[1] 3.668024 -2.771307
[[9]]
[1] 2.108187 -1.365349
[[10]]
[1] 2.106528 -2.134310

Answer 3

First of all, the matrix representing the three equations needs a small correction, because R fills matrices column by column :
-2x + y <= -3
1.25x + y <= 2.5
y >= -3
mat <- matrix(c(-2, 1.25, 0, 1, 1, 1), nrow = 3
# and not : mat <- matrix(c(-2, 1, 1.25, 1, 0, 1), nrow = 3)
To get different tuples, you could modify the objective function :
obj <- numeric(2) results in an objective function 0 * x + 0 * y which is always equal to 0 and can't be maximized : the first valid x,y will be selected.
Optimization on x is achieved by using obj <- c(1,0), resulting in maximization / minimization of 1 * x + 0 * y.
Optimization on y is achieved by using obj <- c(0,1).
#setting the bounds is necessary, otherwise optimization occurs only for x>=0 and y>=0
bounds <- list(lower = list(ind = c(1L, 2L), val = c(-Inf, -Inf)),
upper = list(ind = c(1L, 2L), val = c(Inf, Inf)))
# finding maximum x: obj = c(1,0), max = T
Rglpk_solve_LP(obj = c(10,0), mat = mat, dir = dir, rhs = rhs,bound=bounds, max = T)$solution
# [1] 4.4 -3.0
# finding minimum x: obj = c(1,0), max = F
Rglpk_solve_LP(obj = c(10,0), mat = mat, dir = dir, rhs = rhs,bound=bounds, max = F)$solution
#[1] 0 -3
# finding maximum y: obj = c(0,1), max = T
Rglpk_solve_LP(obj = c(0,1), mat = mat, dir = dir, rhs = rhs,bound=bounds, max = T)$solution
#[1] 1.6923077 0.3846154