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.
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