# 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)