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2.43 kB
| # 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) | |