table of Contents
Maximum Flows
"""From Taha 'Introduction to Operations Research', example 6.4-2."""
from __future__ import print_function
from ortools.graph import pywrapgraph
def main():
"""MaxFlow simple interface example."""
# Define three parallel arrays: start_nodes, end_nodes, and the capacities
# between each pair. For instance, the arc from node 0 to node 1 has a
# capacity of 20.
start_nodes = [0, 0, 0, 1, 1, 2, 2, 3, 3]
end_nodes = [1, 2, 3, 2, 4, 3, 4, 2, 4]
capacities = [20, 30, 10, 40, 30, 10, 20, 5, 20]
# Instantiate a SimpleMaxFlow solver.
max_flow = pywrapgraph.SimpleMaxFlow()
# Add each arc.
for i in range(0, len(start_nodes)):
max_flow.AddArcWithCapacity(start_nodes[i], end_nodes[i], capacities[i])
# Find the maximum flow between node 0 and node 4.
if max_flow.Solve(0, 4) == max_flow.OPTIMAL:
print('Max flow:', max_flow.OptimalFlow())
print('')
print(' Arc Flow / Capacity')
for i in range(max_flow.NumArcs()):
print('%1s -> %1s %3s / %3s' % (
max_flow.Tail(i),
max_flow.Head(i),
max_flow.Flow(i),
max_flow.Capacity(i)))
print('Source side min-cut:', max_flow.GetSourceSideMinCut())
print('Sink side min-cut:', max_flow.GetSinkSideMinCut())
else:
print('There was an issue with the max flow input.')
if __name__ == '__main__':
main()Minimum Cost Flows
task
To transform configuration data center traffic simulation, the input two-dimensional matrix, in accordance with the algorithm given in the paper a number of defined conditions, solved for the new topology.
Minimum Cost Flows simple count of Demo
Compared with the maximum flows problems, and more unit_costs supply, and the supply inflow outflow equal to the sum of the sum, i.e. to 0 and supply elements.
Negative elements which represents a supply of demand.
or-tools in AddArcWithCapacityAndUnitCost support directed graph node index and capacity (Capacity) must be non-negative, cost unit cost may be any integer, and the cycle is repeated from the support arcs.
Adds a directed arc from tail to head to the underlying graph with a given capacity and cost per unit of flow. * Node indices and the capacity must be non-negative (>= 0). * The unit cost can take any integer value (even negative). * Self-looping and duplicate arcs are supported. * After the method finishes, NumArcs() == the returned ArcIndex + 1.
# """From Bradley, Hax, and Magnanti, 'Applied Mathematical Programming', figure 8.1."""
from __future__ import print_function
from ortools.graph import pywrapgraph
def main():
"""MinCostFlow simple interface example."""
# Define four parallel arrays: start_nodes, end_nodes, capacities, and unit costs
# between each pair. For instance, the arc from node 0 to node 1 has a
# capacity of 15 and a unit cost of 4.
start_nodes = [ 0, 0, 1, 1, 1, 2, 2, 3, 4]
end_nodes = [ 1, 2, 2, 3, 4, 3, 4, 4, 2]
capacities = [15, 8, 20, 4, 10, 15, 4, 20, 5]
unit_costs = [ 4, 4, 2, 2, 6, 1, 3, 2, 3]
# Define an array of supplies at each node.
supplies = [20, 0, 0, -5, -15]
# Instantiate a SimpleMinCostFlow solver.
min_cost_flow = pywrapgraph.SimpleMinCostFlow()
# Add each arc.
for i in range(0, len(start_nodes)):
min_cost_flow.AddArcWithCapacityAndUnitCost(start_nodes[i], end_nodes[i],
capacities[i], unit_costs[i])
# Add node supplies.
for i in range(0, len(supplies)):
min_cost_flow.SetNodeSupply(i, supplies[i])
# Find the minimum cost flow between node 0 and node 4.
if min_cost_flow.Solve() == min_cost_flow.OPTIMAL:
print('Minimum cost:', min_cost_flow.OptimalCost())
print('')
print(' Arc Flow / Capacity Cost')
for i in range(min_cost_flow.NumArcs()):
cost = min_cost_flow.Flow(i) * min_cost_flow.UnitCost(i)
print('%1s -> %1s %3s / %3s %3s' % (
min_cost_flow.Tail(i),
min_cost_flow.Head(i),
min_cost_flow.Flow(i),
min_cost_flow.Capacity(i),
cost))
else:
print('There was an issue with the min cost flow input.')
if __name__ == '__main__':
main()