lpsugar

lpsugar is an AML (Algebraic Modeling Language) inspired in R’s vectorized syntax, used to define and solve Mixed Integer Linear Programs (MILP).

This package is in development, and being made by a solo student. For a safer choice, I recommend using ompr.

Installation

You can install the development version of lpsugar from GitHub with:

# install.packages("pak")
pak::pak("benet1one/lpsugar")

Quick Example

Start with a simple problem to show the basic syntax.

$$\begin{array}{rl} \text{max} & x+y \\ \text{st} & 2x + y \le 8 \\ & 2x + 3y \le 12 \end{array}$$

Let’s write the problem in lpsugar and solve it!

library(lpsugar)
library(ROI.plugin.highs) # Use any ROI solver you like!

my_problem <- lp_problem() |> 
    lp_variable(x) |> 
    lp_variable(y) |> 
    lp_maximize(x + y) |> 
    lp_constraint(
        2*x + y <= 8,
        2*x + 3*y <= 12
    )

my_solution <- lp_solve(my_problem)
my_solution$variables
#> $x
#> [1] 3
#> 
#> $y
#> [1] 2
my_solution$objective
#> [1] 5

Production Problem

Given a set of limited resources, we can make products. Each product takes a certain amount of resources and sells at a price. We wish to maximize earnings.

available <- c(Material = 20, Labour = 10)
price <- c(A = 150, B = 220, C = 160)

resources <- names(available)
products <- names(price)

required <- c(
    # A, B, C
      6, 3, 4, # Material
      2, 3, 1  # Labour
      
) |> parameter(resources, products)

# parameter() sets the dimensions and names
required
#>           products
#> resources  A B C
#>   Material 6 3 4
#>   Labour   2 3 1

production_problem <- lp_problem() |> 
    lp_var(units[products], integer = TRUE, lower = 0) |> 
    lp_max(sum(price * units)) |> 
    lp_con(
        for (r in resources) {
            sum_over(p=products, units[p] * required[r,p]) <= available[r]
        }
    )

production_solution <- lp_solve(production_problem)
production_solution$objective
#> [1] 920
production_solution$variables
#> $units
#> products
#> A B C 
#> 0 2 3

Transportation Problem

Here’s an example solving the classic transportation problem. We shall transport u[f, w] units from factory f to warehousse w, subject to a supply[f] and a demand[w]. We will minimize the total cost.

factory <- paste0("fct", 1:3)
warehouse <- c("A", "B", "C", "D")

supply <- c(20, 25, 15) |> parameter(factory)
demand <- c(8, 10, 12, 10) |> parameter(warehouse)

cost <- c(
    # A, B, C, D
      2, 7, 3, 1,
      6, 2, 9, 2,
      4, 6, 2, 8
      
) |> parameter(factory, warehouse)

cost
#>        warehouse
#> factory A B C D
#>    fct1 2 7 3 1
#>    fct2 6 2 9 2
#>    fct3 4 6 2 8

transportation_problem <- lp_problem() |> 
    lp_variable(u[factory, warehouse], integer = TRUE, lower = 0) |> 
    lp_minimize(sum(u * cost)) |> 
    lp_constraint(
        for (f in factory)   sum(u[f, ]) <= supply[f],
        for (w in warehouse) sum(u[, w]) >= demand[w]
        # Alternatively, using vectorization:
        # rowSums(u) <= supply,
        # colSums(u) >= demand
    )

transportation_solution <- lp_solve(transportation_problem)
transportation_solution$objective
#> [1] 70
transportation_solution$variables
#> $u
#>        warehouse
#> factory A  B  C  D
#>    fct1 8  0  0 10
#>    fct2 0 10  0  0
#>    fct3 0  0 12  0