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Multicommodity flow problem

Description
Create a random directed graph using NumPy's random functions and then model and solve a minimum-cost multicommodity flow problem constructed on that graph.

Further explanation of this example: 'Xpress Python Reference Manual'

multicommodity_flow_python.zip[download all files]

Source Files





example_multicommodity_flow.py

import xpress as xp
import numpy as np
import math

# random network generation

n = 3 + math.ceil(30 * np.random.random())  # number of nodes

thres = 0.4     # density of network
thresdem = 0.8  # density of demand mesh

# generate random forward stars for each node

fwstars = {}

for i in range(n):
    fwstar = []
    for j in range(n):
        if j != i:
            if np.random.random() < thres:
                fwstar.append(j)
    fwstars[i] = fwstar

# backward stars are generated based on the forward stars

bwstars = {i: [] for i in range(n)}

for j in fwstars.keys():
    for i in fwstars[j]:
        bwstars[i].append(j)

# Create arc array

arcs = []
for i in range(n):
    for j in fwstars[i]:
        arcs.append((i, j))

# Create random demand between node pairs

dem = []

for i in range(n):
    for j in range(n):
        if i != j and np.random.random() < thresdem:
            dem.append((i, j, math.ceil(200*np.random.random())))

# U is the unit capacity of each edge
U = 1000
# edge cost
c = {(i, j): math.ceil(10 * np.random.random()) for (i, j) in arcs}

# flow variables
f = {(i, j, d): xp.var(name='f_{0}_{1}_{2}_{3}'.format(i, j, dem[d][0],
                                                       dem[d][1]))
     for (i, j) in arcs for d in range(len(dem))}

# capacity variables
x = {(i, j): xp.var(vartype=xp.integer, name='cap_{0}_{1}'.format(i, j))
     for (i, j) in arcs}

p = xp.problem()
p.addVariable(f, x)


def demand(i, d):
    if dem[d][0] == i:  # source
        return 1
    elif dem[d][1] == i:  # destination
        return -1
    else:
        return 0


# Flow conservation constraints: total flow balance at node i for each demand d
# must be 0 if i is an intermediate node, 1 if i is the source of demand d, and
# -1 if i is the destination.

flow = {(i, d):
        xp.constraint(constraint=xp.Sum(f[i, j, d]
                                        for j in range(n) if (i, j) in arcs) -
                      xp.Sum(f[j, i, d] for j in range(n) if (j, i) in arcs)
                      == demand(i, d),
                      name='cons_{0}_{1}_{2}'.format(i, dem[d][0], dem[d][1]))
        for d in range(len(dem)) for i in range(n)}

# Capacity constraints: weighted sum of flow variables must be contained in the
# total capacity installed on the arc (i, j)
capacity = {(i, j):
            xp.constraint(constraint=xp.Sum(dem[d][2] * f[i, j, d]
                                            for d in range(len(dem)))
                          <= U * x[i, j],
                          name='capacity_{0}_{1}'.format(i, j))
            for (i, j) in arcs}

p.addConstraint(flow, capacity)

p.setObjective(xp.Sum(c[i, j] * x[i, j] for (i, j) in arcs))

p.solve()

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