# The facility location problem

It was originally contributed by Mathieu Tanneau (@mtanneau) and Alexis Montoison (@amontoison).

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

Random.seed!(314)

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

# Clients' locations
Xc, Yc = rand(m), rand(m)

# Facilities' potential locations
Xf, Yf = 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([Xc[i] - Xf[j], Yc[i] - Yf[j]], 2)
end
end

Display the data

Plots.scatter(
Xc,
Yc;
label = "Clients",
markershape = :circle,
markercolor = :blue,
)
Plots.scatter!(
Xf,
Yf;
label = "Facility",
markershape = :square,
markercolor = :white,
markersize = 6,
markerstrokecolor = :red,
markerstrokewidth = 2,
)

### JuMP implementation

Create a JuMP model

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

Solve the uncapacitated facility location problem with HiGHS

optimize!(ufl)
println("Optimal value: ", objective_value(ufl))
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_ = value.(x) .> 1 - 1e-5
y_ = value.(y) .> 1 - 1e-5

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

Plots.scatter!(
Xf,
Yf;
markershape = :square,
markercolor = [(y_[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_[i, j] == 1
Plots.plot!(
[Xc[i], Xf[j]],
[Yc[i], Yf[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, i.e., there exists at least one feasible solution.

Demands

a = rand(1:3, m);

Capacities

q = rand(5:10, n);

Display the data

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

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

### JuMP implementation

Create a JuMP model

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

Solve the problem

optimize!(cfl)
println("Optimal value: ", objective_value(cfl))
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_ = value.(x) .> 1 - 1e-5;
y_ = value.(y) .> 1 - 1e-5;

Display the solution

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

Plots.scatter!(
Xf,
Yf;
label = nothing,
markershape = :rect,
markercolor = [(y_[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_[i, j] == 1
Plots.plot!(
[Xc[i], Xf[j]],
[Yc[i], Yf[j]];
color = :black,
label = nothing,
)
end
end

p

Tip