PMedianProblemWithDistances

Fields

• distancematrix::Matrix: n x n matrix of distances
• ncenters::Int: Number of centers

See also PMedianProblemWithCoordinates, solve(::PMedianProblemWithDistances)

PMedianProblemWithCoordinates

Fields

• coordinates::Matrix: n x d matrix of coordinates
• ncenters::Int: Number of centers

See also PMedianProblemWithDistances, solve(::PMedianProblemWithCoordinates)

PMedianResult

Fields

• centers::Vector: Indices of selected centers.
• model::JuMP.Model: JuMP model.
• objective::Float64: Objective value.
• z::Matrix: Binary matrix of assignments. If z[i, j] = 1 then ith point is connected to the jth point.
• y::Vector: Binary vector. If y[i] is 1 then ith point is a depot.

See also solve(::PMedianProblemWithDistances), solve(::PMedianProblemWithCoordinates)

solve(p::PMedianProblemWithDistances)

Description

The function selects ncenters locations that minimize the total distances to the nearest rows based on the provided distance matrix.

Arguments

• p::PMedianProblemWithDistances: PMedian problem with distances.

Output

• PMedianResult: PMedianResult object.

See also solve(::PMedianProblemWithCoordinates)

solve(p::PMedianProblemWithCoordinates)

Description

The function calculates Euclidean distances between all possible rows of the matrix data. ncenters locations are then selected that minimizes the total distances to the nearest rows.

Arguments

• p::PMedianProblemWithCoordinates: PMedian problem with coordinates.

Output

• PMedianResult: PMedianResult object.

Example

data1 = rand(10, 2)

data2 = rand(10, 2) .+ 50

data3 = rand(10, 2) .+ 100

data = vcat(data1, data2, data3)

problem = PMedianProblemWithCoordinates(data, 3)

result = solve(problem)

println(result.centers)

# 3-element Vector{Int64}:
#  1
# 16
# 21

println(result.objective)
# 11.531012240599605