【发布时间】:2020-05-27 05:25:14
【问题描述】:
我正在尝试使用 CVXOPT 求解一个简单的二次程序,但我对我可以比求解器提供的最优解更好地猜测可行解这一事实感到困扰。优化的形式为:
我将在最后提供 P、q、G、h、A 和 b 的定义。当我导入并运行时:
from cvxopt import matrix, spmatrix, solvers
# Code that creates matrices goes here
sol = solvers.qp(P, q, G, h, A, b)
结果是:
pcost dcost gap pres dres
0: 0.0000e+00 -5.5000e+00 6e+00 6e-17 4e+00
1: 0.0000e+00 -5.5000e-02 6e-02 1e-16 4e-02
2: 0.0000e+00 -5.5000e-04 6e-04 3e-16 4e-04
3: 0.0000e+00 -5.5000e-06 6e-06 1e-16 4e-06
4: 0.0000e+00 -5.5000e-08 6e-08 1e-16 4e-08
Optimal solution found.
Objective = 0.0
但是我可以定义一个不同的解决方案guessed_solution,它是可行的并进一步最小化目标:
guessed_solution = matrix([0.5,0.5,0.0,0.0,0.0,0.0,0.5,0.5,0.0,0.0,1.0])
# Check Ax = b; want to see zeroes
print(A * guessed_solution - b)
>>>
[ 0.00e+00]
[ 0.00e+00]
[ 2.78e-17]
# Check Gx <= h; want to see non-positive entries
print(G * guessed_solution - h)
>>>
[-5.00e-01]
[-5.00e-01]
[ 0.00e+00]
[ 0.00e+00]
[ 0.00e+00]
[ 0.00e+00]
[-5.00e-01]
[-5.00e-01]
[-1.00e+00]
[-1.00e+00]
[ 0.00e+00]
[ 0.00e+00]
[ 0.00e+00]
[-1.00e+00]
# Check objective
print(guessed_solution.T * P * guessed_solution + q.T * guessed_solution)
>>>[-6.67e-01]
这导致目标为 -2/3,明显小于 0。我认为 Ax=b 测试中的 2.78e-17 错误无关紧要。
任何有关解决此问题的帮助将不胜感激!以下是代码中相关矩阵的定义(最大矩阵为11乘11)。
P = matrix([[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0],[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0], [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0/3.0, 0.0, 2.0/3.0], [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0/3.0, 2.0/3.0], [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, -1.0, 0.0, 0.0],[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, -1.0, 0.0], [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, -1.0/3.0, 0.0, -2.0/3.0], [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, -1.0/3.0, -2.0/3.0], [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0], [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0], [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0]]).T
q = matrix([0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0])
A = matrix([[0.0, 0.0, 0.0, 0.0, 1.0, 1.0, 1.0, 1.0, 0.0, 0.0, 0.0],[1.0, 1.0, 1.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0],[0.0, 1.0, 1.0/3.0, 2.0/3.0, 0.0, -1.0, -1.0/3.0, -2.0/3.0, 0.0, 0.0, 0.0]]).T
b = matrix([1.0, 1.0, 0.0])
G = spmatrix([-1.0, -1.0, -1.0, -1.0, -1.0, -1.0, -1.0, -1.0, 1.0, 1.0, 1.0, -1.0, -1.0, -1.0], [0,1,2,3,4,5,6,7,8,9,10,11,12,13], [0,1,2,3,4,5,6,7,8,9,10,8,9,10])
h = matrix([0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 1.0, 1.0, 0.0, 0.0, 0.0])
【问题讨论】:
标签: python cvxopt quadratic-programming