Question

我正在尝试解决cvxpy中的LMI约束问题。对于某些参数，该程序可以运行，但是我得到了变量值（NoneType）gamma2.value。将None作为变量值的原因是什么，这是因为问题不可行吗？我觉得在文档上找不到任何有用的信息。另外，我在Matlab中也遇到了同样的问题，并且得到了这组参数的结果。

破坏程序的参数是q和r。可以重现此问题的最小代码：

import numpy as np
from cvxpy import *

q = 0.1
r = 0.1

Ad = np.array([[0.8217916 , 0.        , 0.        ],
        [0.        , 0.8217916 , 0.        ],
        [0.        , 0.        , 0.38560449]])

Bd = np.array([[ 0.        ,  0.03995538, -0.03995538],
        [-0.0461365 ,  0.02306825,  0.02306825],
        [ 0.81953625,  0.81953625,  0.81953625]])

Cd = np.eye(3)

Dd = np.zeros((3, 3))

Z = np.zeros((3,3))
I = np.eye(3);
A = np.block([[Ad, Z], [I, I]])
B1 = np.vstack((I, Z))
B2 = np.vstack((Bd, Z))
C1 = np.block([[I, Z], [Z, Z]])
D11 = np.block([[Z], [Z]])
D12 = np.block([[Z], [I]])
C2 = np.block([I, Z])
D21 = Z
D22 = Z

tol = 1e-15
n = A.shape[0]

L = Variable((B2.shape[1], n))
P = Variable((n, n), PSD=True)
gamma2 = Variable()

LMI1 = bmat([
        [P, A*P + B2*L, B1, np.zeros((B1.shape[0], D11.shape[0]))],
        [P*A.T + L.T * B2.T, P, np.zeros((P.shape[0], B1.shape[1])), P*C1.T + L.T*D12.T],
        [B1.T, np.zeros((B1.shape[1], P.shape[1])), np.eye(B1.shape[1]), D11.T],
        [np.zeros((C1.shape[0], B1.shape[0])), C1*P + D12*L, D11, gamma2*np.eye(D11.shape[0])]
          ])

cons1 = LMI1 >> tol

LMI2 = bmat([
        [-r*P, -q*P + A*P + B2*L],
        [-q*P + P*A.T + L.T*B2.T, -r*P]
        ])

cons2 = LMI2 << tol

cons3 = gamma2 >= tol

cons4 = P == P.T

cons5 = P >> tol

optprob = Problem(Minimize(gamma2), constraints=[cons1, cons2, cons3, cons4, cons5])
optprob.solve()

Hinf_norm = np.sqrt(gamma2.value)
Pop = P.value
Lop = L.value

K = Lop.dot(np.linalg.inv(Pop))

无类型作为cvxpy中的variable.value

0 个答案: