Skip to content

Instantly share code, notes, and snippets.

@bogdan-kulynych
Last active May 10, 2024 15:27
Show Gist options
  • Star 11 You must be signed in to star a gist
  • Fork 5 You must be signed in to fork a gist
  • Save bogdan-kulynych/7984367 to your computer and use it in GitHub Desktop.
Save bogdan-kulynych/7984367 to your computer and use it in GitHub Desktop.
Transportation problem solver in Python
import numpy as np
from collections import Counter
def transport(supply, demand, costs):
# Only solves balanced problem
assert sum(supply) == sum(demand)
s = np.copy(supply)
d = np.copy(demand)
C = np.copy(costs)
n, m = C.shape
# Finding initial solution
X = np.zeros((n, m))
indices = [(i, j) for i in range(n) for j in range(m)]
xs = sorted(zip(indices, C.flatten()), key=lambda (a, b): b)
# Iterating C elements in increasing order
for (i, j), _ in xs:
if d[j] == 0:
continue
else:
# Reserving supplies in a greedy way
remains = s[i] - d[j] if s[i] >= d[j] else 0
grabbed = s[i] - remains
X[i, j] = grabbed
s[i] = remains
d[j] -= grabbed
# Finding optimal solution
while True:
u = np.array([np.nan]*n)
v = np.array([np.nan]*m)
S = np.zeros((n, m))
_x, _y = np.where(X > 0)
nonzero = zip(_x, _y)
f = nonzero[0][0]
u[f] = 0
# Finding u, v potentials
while any(np.isnan(u)) or any(np.isnan(v)):
for i, j in nonzero:
if np.isnan(u[i]) and not np.isnan(v[j]):
u[i] = C[i, j] - v[j]
elif not np.isnan(u[i]) and np.isnan(v[j]):
v[j] = C[i, j] - u[i]
else:
continue
# Finding S-matrix
for i in range(n):
for j in range(m):
S[i, j] = C[i, j] - u[i] - v[j]
# Stop condition
s = np.min(S)
if s >= 0:
break
i, j = np.argwhere(S == s)[0]
start = (i, j)
# Finding cycle elements
T = np.copy(X)
T[start] = 1
while True:
_xs, _ys = np.nonzero(T)
xcount, ycount = Counter(_xs), Counter(_ys)
for x, count in xcount.items():
if count <= 1:
T[x,:] = 0
for y, count in ycount.items():
if count <= 1:
T[:,y] = 0
if all(x > 1 for x in xcount.values()) \
and all(y > 1 for y in ycount.values()):
break
# Finding cycle chain order
dist = lambda (x1, y1), (x2, y2): abs(x1-x2) + abs(y1-y2)
fringe = set(tuple(p) for p in np.argwhere(T > 0))
size = len(fringe)
path = [start]
while len(path) < size:
last = path[-1]
if last in fringe:
fringe.remove(last)
next = min(fringe, key=lambda (x, y): dist(last, (x, y)))
path.append(next)
# Improving solution on cycle elements
neg = path[1::2]
pos = path[::2]
q = min(X[zip(*neg)])
X[zip(*neg)] -= q
X[zip(*pos)] += q
return X, np.sum(X*C)
if __name__ == '__main__':
supply = np.array([200, 350, 300])
demand = np.array([270, 130, 190, 150, 110])
costs = np.array([[24., 50., 55., 27., 16.],
[50., 40., 23., 17., 21.],
[35., 59., 55., 27., 41.]])
routes, z = transport(supply, demand, costs)
assert z == 23540
print routes
@shamzos
Copy link

shamzos commented Apr 12, 2015

Hello, Your algorithm doesn't work with my parameters

@shamzos
Copy link

shamzos commented Apr 12, 2015

Here my parameters please try it
aCost = [[ 1, 8, 1, 5, 4]
,[ 5, 5, 3, 6, 7]
,[ 2, 9, 5, 9, 8]]

aDemand = [ 120, 130, 145, 125, 140]
aSupply = [ 240, 160, 260]

@dzhuang
Copy link

dzhuang commented Jun 11, 2016

This won't work it there is a degenerate solution, because there will be less than m+n-1 non-zero numbers in the route and some cycle chain order won't be found and result in endless loops.

@dzhuang
Copy link

dzhuang commented Aug 21, 2016

The generated cycle chain will be wrong for more complex chains because the dist function only considered distance between two node while ignoring that the two neighboring nodes should have exactly one common coordinate. My revised version is here.

@koty-koty
Copy link

bonjour,j'utilise actuellement votre code et ça me donne comme erreur ceci:

next = min(fringe, key=lambda x, y: dist(last, (x, y)))
TypeError: () missing 1 required positional argument: 'y'

@redspirit
Copy link

Line while any(np.isnan(u)) or any(np.isnan(v)):
endless loop on my data

Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment