Python Linprog可以解LP原矩阵,但不能解对偶矩阵?
我能够用矩阵A_ub用scipy linprog解决以下最小化问题:Python Linprog可以解LP原矩阵,但不能解对偶矩阵?,python,optimization,scipy,Python,Optimization,Scipy,我能够用矩阵A_ub用scipy linprog解决以下最小化问题: A_ub = [[ 1 10 0 3] [ 6 2 3 6] [ 3 5 4 2] [ 4 9 2 2]] 及 最小化问题是c=[-1,-1,-1](即范数1的负数)。 从scipy调用linprog将得到以下结果(如预期的那样): 然而,我还需要找到问题的双重解决方案 根据我对极大极小定理的理解,上述问题等价于: scipy.optimize.linpr
A_ub = [[ 1 10 0 3]
[ 6 2 3 6]
[ 3 5 4 2]
[ 4 9 2 2]]
及
最小化问题是c=[-1,-1,-1]
(即范数1的负数)。
从scipy调用linprog将得到以下结果(如预期的那样):
然而,我还需要找到问题的双重解决方案
根据我对极大极小定理的理解,上述问题等价于:
scipy.optimize.linprog(-b_ub, A_ub=A_ub.T, b_ub=c)
但是,运行此命令会导致错误:
con: array([], dtype=float64)
fun: 0.0
message: "Phase 1 of the simplex method failed to find a feasible solution. The pseudo-objective function evaluates to 4.0e+00 which exceeds the required tolerance of 1e-12 for a solution to be considered 'close enough' to zero to be a basic solution. Consider increasing the tolerance to be greater than 4.0e+00. If this tolerance is unacceptably large the problem may be infeasible."
nit: 0
slack: array([-1., -1., -1., -1.])
status: 2
success: False
x: array([0., 0., 0., 0.])
如果我将公差增加到一个大值(10),那么它确实会以一个解决方案终止,但我认为这是不正确的,因为函数值与原始值不同。
我真的很感谢任何关于这个问题的帮助和提示,以及如何找到双重问题的解决方案
最好的,
你好。我打电话给linprog是个错误 问题的双重性质应该是:
minimizing b_ub
s.t
-A_transpose *x <= c
con: array([], dtype=float64)
fun: 0.0
message: "Phase 1 of the simplex method failed to find a feasible solution. The pseudo-objective function evaluates to 4.0e+00 which exceeds the required tolerance of 1e-12 for a solution to be considered 'close enough' to zero to be a basic solution. Consider increasing the tolerance to be greater than 4.0e+00. If this tolerance is unacceptably large the problem may be infeasible."
nit: 0
slack: array([-1., -1., -1., -1.])
status: 2
success: False
x: array([0., 0., 0., 0.])
minimizing b_ub
s.t
-A_transpose *x <= c
linprog(b_ub, -A_transpose, c)