Python3中函数的约束最大化和不成功的尝试

Question

首先，我将提出问题。

我的尝试

我试着用解决这个问题科学优化包装。我必须传递初始向量（初始猜测）。算法给出的值依赖于这个初始猜测。所以我尝试了随机的初始猜测，试图在越来越多的初始猜测中找到最大值

import numpy as np
from scipy.optimize import minimize

def J(data):
    l = len(data)
    if (l%2==0 | l<2):
        raise Exception('Wrong length or arguments!')

    phi = data[0:l//2]
    p = data[l//2:l-1]
    theta = data[l-1]

    phi = np.array(phi)    
    p = np.array(p)

    w = (1 + np.cos(phi)*np.cos(theta))*p
    lam = (1+np.cos(phi-theta))/(2*(1+np.cos(phi)*np.cos(theta)))

    return -(1 - sum(w * ((-lam*np.log2(lam)) - (1-lam)*np.log2(1-lam))))

def probability_constraint(x):
    return sum(x[len(x)//2:len(x)-1])-1

def lambda_w_constraint(data):
    l = len(data)
    phi = data[0:l//2]
    p = data[l//2:-1]
    theta = data[l-1]

    phi = np.array(phi)    
    p = np.array(p)

    w = (1 + np.cos(phi)*np.cos(theta))*p
    lam = (1+np.cos(phi-theta))/(2*(1+np.cos(phi)*np.cos(theta)))    
    return sum(w*lam)-1/2

def lambda_greater_than_zero(data):
    l = len(data)
    phi = data[0:l//2]
    p = data[l//2:-1]
    theta = data[l-1]

    lam = (1+np.cos(phi-theta))/(2*(1+np.cos(phi)*np.cos(theta)))    
    return lam

def probability_positive(x):
    return x

cons = ({'type': 'ineq',
'fun' : lambda x: lambda_greater_than_zero(x)},
        {'type': 'eq',
'fun' : lambda x: probability_constraint(x)},
        {'type': 'eq',
'fun' : lambda x: lambda_w_constraint(x)},
        {'type':'ineq', 
'fun' : lambda x: probability_positive(x)}

)

def maximize(numiter):
    maxJ = -1000
    maxx = []

    for i in range(0, numiter):
        x0 = np.random.rand(5)*10
        res = minimize(J, x0, constraints=cons)
        r = J(res.x)
        if (r>maxJ):
            maxJ = r
            maxx = res.x

    return maxJ, maxx

maximize(10)

maximize(100)

我知道，这种方法是非常幼稚和不正确的。你知道吗

所以我的问题是：你知道怎么解决这个问题吗？它可以是python3，但其他语言（Matlab？）对我也有好处。你知道吗