# A Warehouse Allocation Example using Julia

This post has been upgraded to use Julia 1.1 and JuMP 0.19.

## Location of Warehouses

This is based on *Example 5.1 (Location of Warehouses)* from Applied Linear Programming

A firm has 5 distribution centers and we want to determine which subset of these should serve as a site for a warehouse. The goal is the build a minimum number of warehouses that can cover all distribution centers so that every warehouse is within 10 miles of each distribution center.

Per the problem statement, is the number of distribution centers.

We are given a table of distances between distribution centers, :

```
m = 5
max_miles = 10
D = [0 10 15 20 18;
10 0 20 15 10;
15 20 0 8 17;
20 15 8 0 5;
18 10 17 5 0
]
```

```
5×5 Array{Int64,2}:
0 10 15 20 18
10 0 20 15 10
15 20 0 8 17
20 15 8 0 5
18 10 17 5 0
```

For example, it is *18* miles between distribution centers *1* (column 1) and *5* (row 5).

To convert this to a binary coverage vector , we convert each distance into a binary variable indicating whether the distribution centers are 10 or fewer miles from one another:

```
A = [Int(D[i, j] <= max_miles) for i=1:m, j=1:m]
```

```
5×5 Array{Int64,2}:
1 1 0 0 0
1 1 0 0 1
0 0 1 1 0
0 0 1 1 1
0 1 0 1 1
```

Now we can model this problem using the JuMP package and the (open source) `Cbc`

solver:

(First we import the relevant packages)

```
using JuMP, Cbc
```

```
model = Model(with_optimizer(Cbc.Optimizer))
# decision variable (binary): whether to build warehouse near distribution center i
@variable(model, y[1:m], Bin)
# Objective: minimize number of warehouses
@objective(model, Min, sum(y))
# Constraint: has to cover all warehouses
# (.>= is the element-wise dot comparison operator)
@constraint(model, A*y .>= 1)
model
```

We have an additional constraint that at least 1 warehouse should be within 10 miles of distribution center 1, but our activity matrix already covers that, so technically we do not need this explicit constraint.

```
@constraint(model, y[1] + y[2] >= 1)
```

```
# Solve problem using MIP solver
optimize!(model)
```

```
println("Total # of warehouses: ", objective_value(model))
println("Build warehouses at distribution center(s):")
for i=1:m
if value(y[i]) == 1
println("Warehouse $i")
end
end
```

```
Total # of warehouses: 2.0
Build warehouses at distribution center(s):
Warehouse 2
Warehouse 3
```

We should build warehouses at distribution centers 2 and 3.