具有线性约束的Scipy.optimize.minimize SLSQP失败

Question

考虑以下(凸)优化问题:

minimize 0.5 * y.T * y
s.t.     A*x - b == y

其中优化(矢量)变量x和y和A,b是适当维度的矩阵和向量,分别.

下面的代码使用SLSQPScipy中的方法轻松找到解决方案:

import numpy as np
from scipy.optimize import minimize 

# problem dimensions:
n = 10 # arbitrary integer set by user
m = 2 * n

# generate parameters A, b:
np.random.seed(123) # for reproducibility of results
A = np.random.randn(m,n)
b = np.random.randn(m)

# objective function:
def obj(z):
    vy = z[n:]
    return 0.5 * vy.dot(vy)

# constraint function:
def cons(z):
    vx = z[:n]
    vy = z[n:]
    return A.dot(vx) - b - vy

# constraints input for SLSQP:
cons = ({'type': 'eq','fun': cons})

# generate a random initial estimate:
z0 = np.random.randn(n+m)

sol = minimize(obj, x0 = z0, constraints = cons, method = 'SLSQP',  options={'disp': True})

Optimization terminated successfully.    (Exit mode 0)
            Current function value: 2.12236220865
            Iterations: 6
            Function evaluations: 192
            Gradient evaluations: 6

请注意,约束函数是一个方便的"数组输出"函数.

现在,代替约束的数组输出函数,原则上可以使用一组等效的'标量输出'约束函数(实际上,scipy.optimize文档仅讨论这种类型的约束函数作为输入minimize).

下面是对应的约束集,随后的输出minimize(同样A,b和初始值与上述列表):

# this is the i-th element of cons(z):
def cons_i(z, i):
    vx = z[:n]
    vy = z[n:]
    return A[i].dot(vx) - b[i] - vy[i]

# listable of scalar-output constraints input for SLSQP:
cons_per_i = [{'type':'eq', 'fun': lambda z: cons_i(z, i)} for i in np.arange(m)]

sol2 = minimize(obj, x0 = z0, constraints = cons_per_i, method = 'SLSQP', options={'disp': True})

Singular matrix C in LSQ subproblem    (Exit mode 6)
            Current function value: 6.87999270692
            Iterations: 1
            Function evaluations: 32
            Gradient evaluations: 1

显然,算法失败(返回的目标值实际上是给定初始化的目标值),我觉得有点奇怪.请注意,运行[cons_per_i[i]['fun'](sol.x) for i in np.arange(m)]显示sol.x使用数组输出约束公式获得,满足所有标量输出约束cons_per_i(在数值公差范围内).

如果有人对此问题有一些解释,我将不胜感激.

Answer 1

你已经遇到了"后期绑定关闭" 问题.cons_i使用第二个参数进行的所有调用都等于19.

修复是使用args字典中的字典元素来定义约束而不是lambda函数闭包:

cons_per_i = [{'type':'eq', 'fun': cons_i, 'args': (i,)} for i in np.arange(m)]

有了这个,最小化工作:

In [417]: sol2 = minimize(obj, x0 = z0, constraints = cons_per_i, method = 'SLSQP', options={'disp': True})
Optimization terminated successfully.    (Exit mode 0)
            Current function value: 2.1223622086
            Iterations: 6
            Function evaluations: 192
            Gradient evaluations: 6

您还可以使用链接文章中提出的建议,即使用具有所需默认值的第二个参数的lambda表达式:

cons_per_i = [{'type':'eq', 'fun': lambda z, i=i: cons_i(z, i)} for i in np.arange(m)]