Multi-commodity Network Flows#
Multi-commodity network flows are a mathematical model used to optimize the simultaneous routing of multiple resources through a single network. These models can be applied in various fields, including logistics, telecommunications, and urban planning to improve efficiency and resource allocation. In our example, we will explore the application of this model using a small network consisting of nodes A, B, C, D, and E, connected by directed edges with specific capacities: A->B, A->D, B->C, B->D, C->E, D->C, and D->E. The network includes two commodities, with the first aiming to maximize flow from node A to node C, and the second from node A to node E. Transporting each commodity through the network has a different profit per edge, and the objective is to maximize the total profit while respecting the capacity constraints of each edge.
import numpy as np
import pandas as pd
import corneto as cn
cn.info()
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G = cn.Graph()
# We create the transportation network, and we add attributes to the edges.
# These attributes are the capacity of the edge, and the profit of the edge for the two commodities.
G.add_edge("A", "B", capacity=10, profit_c1=4, profit_c2=1)
G.add_edge("A", "D", capacity=15, profit_c1=3, profit_c2=2)
G.add_edge("B", "C", capacity=12, profit_c1=2, profit_c2=3)
G.add_edge("B", "D", capacity=5, profit_c1=1, profit_c2=4)
G.add_edge("C", "E", capacity=10, profit_c1=5, profit_c2=5)
G.add_edge("D", "C", capacity=4, profit_c1=2, profit_c2=6)
G.add_edge("D", "E", capacity=8, profit_c1=3, profit_c2=4)
G.plot()
Now we will manually create a flow graph by adding incoming edges for flow through the input vertices and outgoing flow edges for the output vertices.
# First commodity, routing from A to C
G.add_edge((), "A", capacity=1000)
G.add_edge("C", (), capacity=1000)
# Second commodity, routing from A to E.
G.add_edge("E", (), capacity=1000)
G.plot()
G.get_attr_from_edges("capacity")
[10, 15, 12, 5, 10, 4, 8, 1000, 1000, 1000]
# P.expr.flow.sum(axis=0).shape
# We create a flow problem with 2 flows, one per commodity
P = cn.opt.Flow(G, ub=G.get_attr_from_edges("capacity"), n_flows=2, shared_bounds=True)
# NOTE: Using shared bounds links shares the capacity of the edges across flows. It is equivalent to adding this constraint here:
# P += P.expressions.flow[:, 0] + P.expressions.flow[:, 1] <= G.get_attr_from_edges('capacity')
P.expressions
{'_flow': _flow: Variable((10, 2), _flow),
'flow': _flow: Variable((10, 2), _flow)}
c1 = np.array(G.get_attr_from_edges("profit_c1", 0))
c2 = np.array(G.get_attr_from_edges("profit_c2", 0))
# Finally, we add the objective function: maximize the amount of flow per commodity, taking into account the profit
# of transporting each commodity through each edge.
P.add_objectives(sum(P.expressions.flow[:, 0].multiply(c1)), weights=-1)
P.add_objectives(sum(P.expressions.flow[:, 1].multiply(c2)), weights=-1)
P.solve(verbosity=1)
===============================================================================
CVXPY
v1.6.6
===============================================================================
-------------------------------------------------------------------------------
Compilation
-------------------------------------------------------------------------------
-------------------------------------------------------------------------------
Numerical solver
-------------------------------------------------------------------------------
presolving:
(0.0s) symmetry computation started: requiring (bin +, int +, cont +), (fixed: bin -, int -, cont -)
(0.0s) no symmetry present (symcode time: 0.00)
presolving (0 rounds: 0 fast, 0 medium, 0 exhaustive):
0 deleted vars, 0 deleted constraints, 0 added constraints, 0 tightened bounds, 0 added holes, 0 changed sides, 0 changed coefficients
0 implications, 0 cliques
presolved problem has 20 variables (0 bin, 0 int, 0 impl, 20 cont) and 70 constraints
70 constraints of type <linear>
Presolving Time: 0.00
time | node | left |LP iter|LP it/n|mem/heur|mdpt |vars |cons |rows |cuts |sepa|confs|strbr| dualbound | primalbound | gap | compl.
* 0.0s| 1 | 0 | 16 | - | LP | 0 | 20 | 70 | 30 | 0 | 0 | 0 | 0 |-1.900000e+02 |-1.900000e+02 | 0.00%| unknown
0.0s| 1 | 0 | 16 | - | 829k | 0 | 20 | 70 | 30 | 0 | 0 | 0 | 0 |-1.900000e+02 |-1.900000e+02 | 0.00%| unknown
SCIP Status : problem is solved [optimal solution found]
Solving Time (sec) : 0.00
Solving Nodes : 1
Primal Bound : -1.90000000000000e+02 (1 solutions)
Dual Bound : -1.90000000000000e+02
Gap : 0.00 %
-------------------------------------------------------------------------------
Summary
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Problem(Minimize(Expression(AFFINE, UNKNOWN, ())), [Inequality(Constant(CONSTANT, ZERO, (10, 2))), Inequality(Variable((10, 2), _flow)), Equality(Expression(AFFINE, UNKNOWN, (5, 2)), Constant(CONSTANT, ZERO, ())), Inequality(Expression(AFFINE, UNKNOWN, (10,))), Inequality(Constant(CONSTANT, ZERO, (10,)))])
P.objectives[0].value, P.objectives[1].value
(np.float64(110.0), np.float64(80.0))
df_result = pd.DataFrame(P.expressions.flow.value, index=G.E, columns=["Commodity 1", "Commodity 2"])
df_result["total"] = df_result.sum(axis=1)
df_result["capacity"] = G.get_attr_from_edges("capacity")
df_result["Profit c1"] = df_result["Commodity 1"] * c1
df_result["Profit c2"] = df_result["Commodity 2"] * c2
df_result["Total profit"] = df_result["Profit c1"] + df_result["Profit c2"]
df_result
| Commodity 1 | Commodity 2 | total | capacity | Profit c1 | Profit c2 | Total profit | ||
|---|---|---|---|---|---|---|---|---|
| (A) | (B) | 10.0 | 0.0 | 10.0 | 10 | 40.0 | 0.0 | 40.0 |
| (D) | 0.0 | 12.0 | 12.0 | 15 | 0.0 | 24.0 | 24.0 | |
| (B) | (C) | 10.0 | 0.0 | 10.0 | 12 | 20.0 | 0.0 | 20.0 |
| (D) | 0.0 | 0.0 | 0.0 | 5 | 0.0 | 0.0 | 0.0 | |
| (C) | (E) | 10.0 | 0.0 | 10.0 | 10 | 50.0 | 0.0 | 50.0 |
| (D) | (C) | 0.0 | 4.0 | 4.0 | 4 | 0.0 | 24.0 | 24.0 |
| (E) | 0.0 | 8.0 | 8.0 | 8 | 0.0 | 32.0 | 32.0 | |
| () | (A) | 10.0 | 12.0 | 22.0 | 1000 | 0.0 | 0.0 | 0.0 |
| (C) | () | 0.0 | 4.0 | 4.0 | 1000 | 0.0 | 0.0 | 0.0 |
| (E) | () | 10.0 | 8.0 | 18.0 | 1000 | 0.0 | 0.0 | 0.0 |
df_result["Total profit"].sum()
np.float64(190.0)
P.objectives[0].value + P.objectives[1].value
np.float64(190.0)