Network flow problems

Originally Contributed by: Arpit Bhatia

In graph theory, a flow network (also known as a transportation network) is a directed graph where each edge has a capacity and each edge receives a flow. The amount of flow on an edge cannot exceed the capacity of the edge.

Often in operations research, a directed graph is called a network, the vertices are called nodes and the edges are called arcs.

A flow must satisfy the restriction that the amount of flow into a node equals the amount of flow out of it, unless it is a source, which has only outgoing flow, or sink, which has only incoming flow.

A network can be used to model traffic in a computer network, circulation with demands, fluids in pipes, currents in an electrical circuit, or anything similar in which something travels through a network of nodes.

using JuMP
import GLPK
import LinearAlgebra

The shortest path problem

Suppose that each arc $(i, j)$ of a graph is assigned a scalar cost $a_{i,j}$, and suppose that we define the cost of a forward path to be the sum of the costs of its arcs.

Given a pair of nodes, the shortest path problem is to find a forward path that connects these nodes and has minimum cost.

\[\begin{aligned} \min && \sum_{\forall e(i,j) \in E} a_{i,j} \times x_{i,j} \\ s.t. && b(i) = \sum_j x_{ij} - \sum_k x_{ki} = \begin{cases} 1 &\mbox{if $i$ is the starting node,} \\ -1 &\mbox{if $i$ is the ending node,} \\ 0 &\mbox{otherwise.} \end{cases} \\ && x_{e} \in \{0,1\} && \forall e \in E \end{aligned}\]

G = [
    0 100 30 0 0
    0 0 20 0 0
    0 0 0 10 60
    0 15 0 0 50
    0 0 0 0 0
]

n = size(G)[1]

shortest_path = Model(GLPK.Optimizer)

@variable(shortest_path, x[1:n, 1:n], Bin)
5×5 Matrix{VariableRef}:
 x[1,1]  x[1,2]  x[1,3]  x[1,4]  x[1,5]
 x[2,1]  x[2,2]  x[2,3]  x[2,4]  x[2,5]
 x[3,1]  x[3,2]  x[3,3]  x[3,4]  x[3,5]
 x[4,1]  x[4,2]  x[4,3]  x[4,4]  x[4,5]
 x[5,1]  x[5,2]  x[5,3]  x[5,4]  x[5,5]

Arcs with zero cost are not a part of the path as they do no exist

@constraint(shortest_path, [i = 1:n, j = 1:n; G[i, j] == 0], x[i, j] == 0)
JuMP.Containers.SparseAxisArray{ConstraintRef{Model, MathOptInterface.ConstraintIndex{MathOptInterface.ScalarAffineFunction{Float64}, MathOptInterface.EqualTo{Float64}}, ScalarShape}, 2, Tuple{Int64, Int64}} with 18 entries:
  [1, 1]  =  x[1,1] = 0.0
  [1, 4]  =  x[1,4] = 0.0
  [2, 2]  =  x[2,2] = 0.0
  [2, 4]  =  x[2,4] = 0.0
  [2, 5]  =  x[2,5] = 0.0
  [3, 1]  =  x[3,1] = 0.0
  [3, 2]  =  x[3,2] = 0.0
  [3, 3]  =  x[3,3] = 0.0
          ⋮
  [4, 1]  =  x[4,1] = 0.0
  [4, 3]  =  x[4,3] = 0.0
  [4, 4]  =  x[4,4] = 0.0
  [5, 2]  =  x[5,2] = 0.0
  [5, 3]  =  x[5,3] = 0.0
  [5, 4]  =  x[5,4] = 0.0
  [5, 5]  =  x[5,5] = 0.0

Flow conservation constraint

@constraint(
    shortest_path,
    [i = 1:n; i != 1 && i != 2],
    sum(x[i, :]) == sum(x[:, i])
)
JuMP.Containers.SparseAxisArray{ConstraintRef{Model, MathOptInterface.ConstraintIndex{MathOptInterface.ScalarAffineFunction{Float64}, MathOptInterface.EqualTo{Float64}}, ScalarShape}, 1, Tuple{Int64}} with 3 entries:
  [3]  =  x[3,1] + x[3,2] - x[1,3] - x[2,3] - x[4,3] - x[5,3] + x[3,4] + x[3,5] = 0.0
  [4]  =  x[4,1] + x[4,2] + x[4,3] - x[1,4] - x[2,4] - x[3,4] - x[5,4] + x[4,5] = 0.0
  [5]  =  x[5,1] + x[5,2] + x[5,3] + x[5,4] - x[1,5] - x[2,5] - x[3,5] - x[4,5] = 0.0

Flow coming out of source = 1

@constraint(shortest_path, sum(x[1, :]) - sum(x[:, 1]) == 1)

\[ -x_{2,1} - x_{3,1} - x_{4,1} - x_{5,1} + x_{1,2} + x_{1,3} + x_{1,4} + x_{1,5} = 1.0 \]

Flowing coming out of destination = -1 i.e. Flow entering destination = 1

@constraint(shortest_path, sum(x[2, :]) - sum(x[:, 2]) == -1)
@objective(shortest_path, Min, LinearAlgebra.dot(G, x))

optimize!(shortest_path)
objective_value(shortest_path)
55.0
value.(x)
5×5 Matrix{Float64}:
 0.0  0.0  1.0  0.0  0.0
 0.0  0.0  0.0  0.0  0.0
 0.0  0.0  0.0  1.0  0.0
 0.0  1.0  0.0  0.0  0.0
 0.0  0.0  0.0  0.0  0.0

The assignment problem

Suppose that there are $n$ persons and $n$ objects that we have to match on a one-to-one basis. There is a benefit or value $a_{i,j}$ for matching person $i$ with object $j$, and we want to assign persons to objects so as to maximize the total benefit.

There is also a restriction that person $i$ can be assigned to object $j$ only if $(i, j)$ belongs to a given set of pairs $A$.

Mathematically, we want to find a set of person-object pairs $(1, j_{1}),..., (n, j_{n})$ from $A$ such that the objects $j_{1},...,j_{n}$ are all distinct, and the total benefit $\sum_{i=1}^{y} a_{ij_{i}}$ is maximized.

\[\begin{aligned} \max && \sum_{(i,j) \in A} a_{i,j} \times y_{i,j} \\ s.t. && \sum_{\{j|(i,j) \in A\}} y_{i,j} = 1 && \forall i = \{1,2....n\} \\ && \sum_{\{i|(i,j) \in A\}} y_{i,j} = 1 && \forall j = \{1,2....n\} \\ && y_{i,j} \in \{0,1\} && \forall (i,j) \in \{1,2...k\} \end{aligned}\]

G = [
    6 4 5 0
    0 3 6 0
    5 0 4 3
    7 5 5 5
]

n = size(G)[1]

assignment = Model(GLPK.Optimizer)
@variable(assignment, y[1:n, 1:n], Bin)
4×4 Matrix{VariableRef}:
 y[1,1]  y[1,2]  y[1,3]  y[1,4]
 y[2,1]  y[2,2]  y[2,3]  y[2,4]
 y[3,1]  y[3,2]  y[3,3]  y[3,4]
 y[4,1]  y[4,2]  y[4,3]  y[4,4]

One person can only be assigned to one object

@constraint(assignment, [i = 1:n], sum(y[:, i]) == 1)
4-element Vector{ConstraintRef{Model, MathOptInterface.ConstraintIndex{MathOptInterface.ScalarAffineFunction{Float64}, MathOptInterface.EqualTo{Float64}}, ScalarShape}}:
 y[1,1] + y[2,1] + y[3,1] + y[4,1] = 1.0
 y[1,2] + y[2,2] + y[3,2] + y[4,2] = 1.0
 y[1,3] + y[2,3] + y[3,3] + y[4,3] = 1.0
 y[1,4] + y[2,4] + y[3,4] + y[4,4] = 1.0

One object can only be assigned to one person

@constraint(assignment, [j = 1:n], sum(y[j, :]) == 1)
@objective(assignment, Max, LinearAlgebra.dot(G, y))

optimize!(assignment)
objective_value(assignment)
20.0
value.(y)
4×4 Matrix{Float64}:
 0.0  1.0  0.0  0.0
 0.0  0.0  1.0  0.0
 1.0  0.0  0.0  0.0
 0.0  0.0  0.0  1.0

The max-flow problem

In the max-flow problem, we have a graph with two special nodes: the $source$, denoted by $s$, and the $sink$, denoted by $t$.

The objective is to move as much flow as possible from $s$ into $t$ while observing the capacity constraints.

\[\begin{aligned} \max && \sum_{v:(s,v) \in E} f(s,v) \\ s.t. && \sum_{u:(u,v) \in E} f(u,v) = \sum_{w:(v,w) \in E} f(v,w) && \forall v \in V - \{s,t\} \\ && f(u,v) \leq c(u,v) && \forall (u,v) \in E \\ && f(u,v) \geq 0 && \forall (u,v) \in E \end{aligned}\]

G = [
    0 3 2 2 0 0 0 0
    0 0 0 0 5 1 0 0
    0 0 0 0 1 3 1 0
    0 0 0 0 0 1 0 0
    0 0 0 0 0 0 0 4
    0 0 0 0 0 0 0 2
    0 0 0 0 0 0 0 4
    0 0 0 0 0 0 0 0
]

n = size(G)[1]

max_flow = Model(GLPK.Optimizer)

@variable(max_flow, f[1:n, 1:n] >= 0)
8×8 Matrix{VariableRef}:
 f[1,1]  f[1,2]  f[1,3]  f[1,4]  f[1,5]  f[1,6]  f[1,7]  f[1,8]
 f[2,1]  f[2,2]  f[2,3]  f[2,4]  f[2,5]  f[2,6]  f[2,7]  f[2,8]
 f[3,1]  f[3,2]  f[3,3]  f[3,4]  f[3,5]  f[3,6]  f[3,7]  f[3,8]
 f[4,1]  f[4,2]  f[4,3]  f[4,4]  f[4,5]  f[4,6]  f[4,7]  f[4,8]
 f[5,1]  f[5,2]  f[5,3]  f[5,4]  f[5,5]  f[5,6]  f[5,7]  f[5,8]
 f[6,1]  f[6,2]  f[6,3]  f[6,4]  f[6,5]  f[6,6]  f[6,7]  f[6,8]
 f[7,1]  f[7,2]  f[7,3]  f[7,4]  f[7,5]  f[7,6]  f[7,7]  f[7,8]
 f[8,1]  f[8,2]  f[8,3]  f[8,4]  f[8,5]  f[8,6]  f[8,7]  f[8,8]

Capacity constraints

@constraint(max_flow, [i = 1:n, j = 1:n], f[i, j] <= G[i, j])
8×8 Matrix{ConstraintRef{Model, MathOptInterface.ConstraintIndex{MathOptInterface.ScalarAffineFunction{Float64}, MathOptInterface.LessThan{Float64}}, ScalarShape}}:
 f[1,1] ≤ 0.0  f[1,2] ≤ 3.0  f[1,3] ≤ 2.0  …  f[1,7] ≤ 0.0  f[1,8] ≤ 0.0
 f[2,1] ≤ 0.0  f[2,2] ≤ 0.0  f[2,3] ≤ 0.0     f[2,7] ≤ 0.0  f[2,8] ≤ 0.0
 f[3,1] ≤ 0.0  f[3,2] ≤ 0.0  f[3,3] ≤ 0.0     f[3,7] ≤ 1.0  f[3,8] ≤ 0.0
 f[4,1] ≤ 0.0  f[4,2] ≤ 0.0  f[4,3] ≤ 0.0     f[4,7] ≤ 0.0  f[4,8] ≤ 0.0
 f[5,1] ≤ 0.0  f[5,2] ≤ 0.0  f[5,3] ≤ 0.0     f[5,7] ≤ 0.0  f[5,8] ≤ 4.0
 f[6,1] ≤ 0.0  f[6,2] ≤ 0.0  f[6,3] ≤ 0.0  …  f[6,7] ≤ 0.0  f[6,8] ≤ 2.0
 f[7,1] ≤ 0.0  f[7,2] ≤ 0.0  f[7,3] ≤ 0.0     f[7,7] ≤ 0.0  f[7,8] ≤ 4.0
 f[8,1] ≤ 0.0  f[8,2] ≤ 0.0  f[8,3] ≤ 0.0     f[8,7] ≤ 0.0  f[8,8] ≤ 0.0

Flow conservation constraints

@constraint(max_flow, [i = 1:n; i != 1 && i != 8], sum(f[i, :]) == sum(f[:, i]))
@objective(max_flow, Max, sum(f[1, :]))

optimize!(max_flow)
objective_value(max_flow)
6.0
value.(f)
8×8 Matrix{Float64}:
 0.0  3.0  2.0  1.0  0.0  0.0  0.0  0.0
 0.0  0.0  0.0  0.0  3.0  0.0  0.0  0.0
 0.0  0.0  0.0  0.0  1.0  1.0  0.0  0.0
 0.0  0.0  0.0  0.0  0.0  1.0  0.0  0.0
 0.0  0.0  0.0  0.0  0.0  0.0  0.0  4.0
 0.0  0.0  0.0  0.0  0.0  0.0  0.0  2.0
 0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0
 0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0

Tip

This tutorial was generated using Literate.jl. View the source .jl file on GitHub.