Typical problem and execution method
In the directed graph $ G = (V, E) $, given the capacity and weight of each side and the demand for nodes, the sum of the weights for the flow rate of each side is minimized without exceeding the capacity of each side. Find the flow.
--At each node, "inflow --outflow" is equal to demand. --If the demand is negative, it represents the supply. --The sum of the demands of all nodes must be 0.
usage
Signature: nx.min_cost_flow(G, demand='demand', capacity='capacity', weight='weight')
Docstring:
Return a minimum cost flow satisfying all demands in digraph G.
G is a digraph with edge costs and capacities and in which nodes
have demand, i.e., they want to send or receive some amount of
flow. A negative demand means that the node wants to send flow, a
positive demand means that the node want to receive flow. A flow on
the digraph G satisfies all demand if the net flow into each node
is equal to the demand of that node.
python
#CSV data
import pandas as pd, networkx as nx
from ortoolpy import graph_from_table, networkx_draw
tbn = pd.read_csv('data/node0.csv')
tbe = pd.read_csv('data/edge0.csv')
g = graph_from_table(tbn, tbe, directed=True)[0]
result = nx.min_cost_flow(g)
for i, d in result.items():
for j, f in d.items():
if f: print((i, j), f)
result
(0, 1) 1
(0, 3) 1
(0, 4) 2
(4, 5) 1
python
# pandas.DataFrame
from ortoolpy.optimization import MinCostFlow
MinCostFlow('data/node0.csv','data/edge0.csv')
| node1 | node2 | capacity | weight | flow | |
|---|---|---|---|---|---|
| 0 | 0 | 1 | 2 | 1 | 1 |
| 1 | 0 | 3 | 2 | 2 | 1 |
| 2 | 0 | 4 | 2 | 2 | 2 |
| 3 | 4 | 5 | 2 | 1 | 1 |
python
#Random number data
import networkx as nx
g = nx.fast_gnp_random_graph(8, 0.2, 1, True)
g.nodes[1]['demand'] = -2 #Supply
g.nodes[7]['demand'] = 2 #demand
g.adj[2][7]['capacity'] = 1 #capacity
result = nx.min_cost_flow(g)
for i, d in result.items():
for j, f in d.items():
if f: print((i, j), f)
result
(1, 2) 2
(2, 3) 1
(2, 7) 1
(3, 7) 1
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