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# This little solver works on LPs of the form
# maximize cx
# subject to: Ax <= b
from sympy import *
def solve(A, b, c, B, verbose):
'''requires a feasible basis B for the LP'''
m, n = A.rows, A.cols
itr = 0
# find vertex corresponding to feasible basis B
AB = A.extract(B, range(n))
bB = b.extract(B, [0])
x = AB.LUsolve(bB)
ABi = AB.inv()
while True:
itr += 1
# compute lambda
l = (c.T * ABi).T
# check for optimality
if all(e >= 0 for e in l):
return x, itr
# find leaving index B[r]
r = min(i for i in range(l.rows) if l[i] < 0)
# compute direction to move int
d = -ABi[:, r]
# determine the set K
K = [i for i in range(m) if (A[i, :] * d)[0] > 0]
if not K:
return 'unbounded', itr
# find entering index e
e, v = None, None
for k in K:
w = (b[k] - (A[k, :] * x)[0]) / ((A[k, :] * d)[0])
if v is None or w < v:
v = w
e = k
# update basis
B[r] = e
AB[r, :] = A[e, :]
bB[r, :] = b[e, :]
# update inverse
ABi = AB.inv()
# move to the new vertex
x = x + v*d
def simplex(A, b, c, B, verbose=False):
x, itr = solve(A, b, c, B, verbose)
if x == 'infeasible':
print 'LP is infeasible'
elif x == 'unbounded':
print 'LP is unbounded'
else:
print 'Vertex', x.T, 'is optimal'
print 'Optimal value is', (c.T * x)[0]
print 'Found after', itr, 'simplex iterations'
if __name__ == '__main__':
# example
A = Matrix([[1, 0, 0, 0],
[20, 1, 0, 0],
[200, 20, 1, 0],
[2000, 200, 20, 1],
[-1, 0, 0, 0],
[0, -1, 0, 0],
[0, 0, -1, 0],
[0, 0, 0, -1]])
b = Matrix([1,100,10000,1000000, 0,0,0,0])
c = Matrix([1000, 100, 10, 1])
B = range(4, 8)
simplex(A, b, c, B, verbose=True)