MISDP/MISOCP在cvxpy

0

的问题

我试图解决以下问题在CVXPY.

问题是一个混合整数的SDP是由于私营部门司的矩阵,我们的解决。 然而,根据 该列表, 它看起来好像没有解决可以处理这样的问题。

我们可以使用的事实, A 是2x2矩阵以某种方式将这一混合整数SOCP问题?

import cvxpy as cp
import matplotlib.pyplot as plt
import numpy as np

np.random.seed(271828) 
m = 2; n = 50
x = np.random.randn(m,n)

off = cp.Variable(boolean=True)

A = cp.Variable((2,2), PSD=True)
b = cp.Variable(2)
obj = cp.Maximize(cp.log_det(A))
constraints = [ cp.norm(A@x[:,i] + b) <= 1 + 20*off for i in range(n) ]
constraints += [cp.sum(off) <= 20]

prob = cp.Problem(obj, constraints)
optval = prob.solve(solver='XPRESS', verbose=False) # seems to work, although it's not super accurate

print(f"Optimum value: {optval}")

# plot the ellipse and data
angles = np.linspace(0, 2*np.pi, 200)
rhs = np.row_stack((np.cos(angles) - b.value[0], np.sin(angles) - b.value[1]))
ellipse = np.linalg.solve(A.value, rhs)

plt.scatter(x[0,:], x[1,:])
plt.plot(ellipse[0,:].T, ellipse[1,:].T)
plt.xlabel('Dimension 1'); plt.ylabel('Dimension 2')
plt.title('Minimum Volume Ellipsoid')
plt.show()
1

最好的答案

1

让我们说 A=[[x,z], [z,y]]然后你就可以最大限度地提高 sqrt(det(A)) (这相当于你的目标). 注意,

det(A) = xy-z^2

所以最大化 sqrt(det(A)) 是相同的最大化 u

xy - z^2 >= u^2

等价

xy >= z^2 + u^2

这是(几乎)一个旋转的第二阶锥体意义上的 https://docs.mosek.com/modeling-cookbook/cqo.html#rotated-quadratic-cones

我觉得喜欢的东西

x >= quad_over_lin([z,u], y)

(没有测试过的语法)可以最方便的方式来表达它在cvxpy.

注意定义的锥的quad_over_lin还规定 x,y>=0 所以你不需要分开,并且圆锥约束的自动保证PSDness的 A.

2021-11-24 09:08:37

这里是一些光线娱乐的相关问题: twitter.com/mosektw/status/1190259248250445824
Michal Adamaszek

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