# The facility location problem

*This tutorial was generated using Literate.jl.* *Download the source as a .jl file*.

**This tutorial was originally contributed by Mathieu Tanneau and Alexis Montoison.**

## Required packages

This tutorial 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)`

`Random.TaskLocalRNG()`

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: 5.7018394545724185`

### 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, that is, 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: 6.1980444155009975`

### 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
```