The facility location problem

This tutorial was originally contributed by Mathieu Tanneau (@mtanneau) and Alexis Montoison (@amontoison). It requires the following packages:

using JuMP
import HiGHS
import LinearAlgebra
import Plots
import Random

Uncapacitated facility location

Problem description

We are given

• A set $M=\{1, \dots, m\}$ of clients
• A set $N=\{ 1, \dots, n\}$ of sites where a facility can be built

Decision variables Decision variables are split into two categories:

• Binary variable $y_{j}$ indicates whether facility $j$ is built or not
• Binary variable $x_{i, j}$ indicates whether client $i$ is assigned to facility $j$

Objective The objective is to minimize the total cost of serving all clients. This costs breaks down into two components:

• Fixed cost of building a facility.

In this example, this cost is $f_{j} = 1, \ \forall j$.

• Cost of serving clients from the assigned facility.

In this example, the cost $c_{i, j}$ of serving client $i$ from facility $j$ is the Euclidean distance between the two.

Constraints

• Each customer must be served by exactly one facility
• A facility cannot serve any client unless it is open

MILP formulation

The problem can be formulated as the following MILP:

\[\begin{aligned} \min_{x, y} \ \ \ & \sum_{i, j} c_{i, j} x_{i, j} + \sum_{j} f_{j} y_{j} \\ s.t. & \sum_{j} x_{i, j} = 1, && \forall i \in M \\ & x_{i, j} \leq y_{j}, && \forall i \in M, j \in N \\ & x_{i, j}, y_{j} \in \{0, 1\}, && \forall i \in M, j \in N \end{aligned}\]

where the first set of constraints ensures that each client is served exactly once, and the second set of constraints ensures that no client is served from an unopened facility.

Problem data

To ensure reproducibility, we set the random number seed:

Random.seed!(314)

MersenneTwister(314)

Here's the data we need:

# Number of clients
m = 12
# Number of facility locations
n = 5

# Clients' locations
x_c, y_c = rand(m), rand(m)

# Facilities' potential locations
x_f, y_f = rand(n), rand(n)

# Fixed costs
f = ones(n);

# Distance
c = zeros(m, n)
for i in 1:m
    for j in 1:n
        c[i, j] = LinearAlgebra.norm([x_c[i] - x_f[j], y_c[i] - y_f[j]], 2)
    end
end

Display the data

Plots.scatter(
    x_c,
    y_c;
    label = "Clients",
    markershape = :circle,
    markercolor = :blue,
)
Plots.scatter!(
    x_f,
    y_f;
    label = "Facility",
    markershape = :square,
    markercolor = :white,
    markersize = 6,
    markerstrokecolor = :red,
    markerstrokewidth = 2,
)

JuMP implementation

Create a JuMP model

model = Model(HiGHS.Optimizer)
set_silent(model)
@variable(model, y[1:n], Bin);
@variable(model, x[1:m, 1:n], Bin);
# Each client is served exactly once
@constraint(model, client_service[i in 1:m], sum(x[i, j] for j in 1:n) == 1);
# A facility must be open to serve a client
@constraint(model, open_facility[i in 1:m, j in 1:n], x[i, j] <= y[j]);
@objective(model, Min, f' * y + sum(c .* x));

Solve the uncapacitated facility location problem with HiGHS

optimize!(model)
println("Optimal value: ", objective_value(model))

Optimal value: 5.226417970467934

Visualizing the solution

The threshold 1e-5 ensure that edges between clients and facilities are drawn when x[i, j] ≈ 1.

x_is_selected = isapprox.(value.(x), 1; atol = 1e-5);
y_is_selected = isapprox.(value.(y), 1; atol = 1e-5);

p = Plots.scatter(
    x_c,
    y_c;
    markershape = :circle,
    markercolor = :blue,
    label = nothing,
)

Plots.scatter!(
    x_f,
    y_f;
    markershape = :square,
    markercolor = [(y_is_selected[j] ? :red : :white) for j in 1:n],
    markersize = 6,
    markerstrokecolor = :red,
    markerstrokewidth = 2,
    label = nothing,
)

for i in 1:m, j in 1:n
    if x_is_selected[i, j]
        Plots.plot!(
            [x_c[i], x_f[j]],
            [y_c[i], y_f[j]];
            color = :black,
            label = nothing,
        )
    end
end

p

Capacitated facility location

Problem formulation

The capacitated variant introduces a capacity constraint on each facility, i.e., clients have a certain level of demand to be served, while each facility only has finite capacity which cannot be exceeded.

Specifically,

• The demand of client $i$ is denoted by $a_{i} \geq 0$
• The capacity of facility $j$ is denoted by $q_{j} \geq 0$

The capacity constraints then write

\[\begin{aligned} \sum_{i} a_{i} x_{i, j} &\leq q_{j} y_{j} && \forall j \in N \end{aligned}\]

Note that, if $y_{j}$ is set to $0$, the capacity constraint above automatically forces $x_{i, j}$ to $0$.

Thus, the capacitated facility location can be formulated as follows

\[\begin{aligned} \min_{x, y} \ \ \ & \sum_{i, j} c_{i, j} x_{i, j} + \sum_{j} f_{j} y_{j} \\ s.t. & \sum_{j} x_{i, j} = 1, && \forall i \in M \\ & \sum_{i} a_{i} x_{i, j} \leq q_{j} y_{j}, && \forall j \in N \\ & x_{i, j}, y_{j} \in \{0, 1\}, && \forall i \in M, j \in N \end{aligned}\]

For simplicity, we will assume that there is enough capacity to serve the demand, that is, there exists at least one feasible solution.

We need some new data:

# Demands
a = rand(1:3, m);

# Capacities
q = rand(5:10, n);

Display the data

Plots.scatter(
    x_c,
    y_c;
    label = nothing,
    markershape = :circle,
    markercolor = :blue,
    markersize = 2 .* (2 .+ a),
)

Plots.scatter!(
    x_f,
    y_f;
    label = nothing,
    markershape = :rect,
    markercolor = :white,
    markersize = q,
    markerstrokecolor = :red,
    markerstrokewidth = 2,
)

JuMP implementation

Create a JuMP model

model = Model(HiGHS.Optimizer)
set_silent(model)
@variable(model, y[1:n], Bin);
@variable(model, x[1:m, 1:n], Bin);
# Each client is served exactly once
@constraint(model, client_service[i in 1:m], sum(x[i, :]) == 1);
# Capacity constraint
@constraint(model, capacity, x' * a .<= (q .* y));
# Objective
@objective(model, Min, f' * y + sum(c .* x));

Solve the problem

optimize!(model)
println("Optimal value: ", objective_value(model))

Optimal value: 6.17371506253207

Visualizing the solution

The threshold 1e-5 ensure that edges between clients and facilities are drawn when x[i, j] ≈ 1.

x_is_selected = isapprox.(value.(x), 1; atol = 1e-5);
y_is_selected = isapprox.(value.(y), 1; atol = 1e-5);

Display the solution

p = Plots.scatter(
    x_c,
    y_c;
    label = nothing,
    markershape = :circle,
    markercolor = :blue,
    markersize = 2 .* (2 .+ a),
)

Plots.scatter!(
    x_f,
    y_f;
    label = nothing,
    markershape = :rect,
    markercolor = [(y_is_selected[j] ? :red : :white) for j in 1:n],
    markersize = q,
    markerstrokecolor = :red,
    markerstrokewidth = 2,
)

for i in 1:m, j in 1:n
    if x_is_selected[i, j]
        Plots.plot!(
            [x_c[i], x_f[j]],
            [y_c[i], y_f[j]];
            color = :black,
            label = nothing,
        )
    end
end

p