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我正在尝试解决以下 MINLP，基本上是试图最大化某个投资组合达到“天花板”性能的可能性。我对代码的第一次尝试如下。

编辑：数学说最大化，应该说最小化

基本 MINLP

from pyscipopt import Model, quicksum
import numpy as np
import pandas as pd
from random import uniform, normalvariate
model=Model()
t=20000
stocks_portfolio = {}
stocks_df = pd.DataFrame(np.zeros((150,4)),columns = {'ids','Mean','cost','stdev'})
noptions = len(stocks_df)
stocks_df['ids'] = [i for i in range(noptions)]
stocks_df['Mean'] = [uniform(500,2500) for i in range(noptions)]
stocks_df['cost'] = [stocks_df.loc[i,'Mean']*uniform(50,250) for i in range(noptions)]
stocks_df['stdev'] = [stocks_df.loc[i,'Mean']*uniform(0.2,0.5) for i in range(noptions)]
cov_mat = np.array([[normalvariate(0,0.3) for i in range(noptions)] for j in range(noptions)])
for i in range(len(stocks_df)):
    stocks_portfolio[i] = model.addVar(vtype='B')
model.addCons(quicksum(stocks_portfolio[i] for i in range(noptions))==15)
model.addCons(quicksum(stocks_df.loc[i, 'cost']*stocks_portfolio[i] for i in range(noptions)) <= 600000)
stand_in = model.addVar(vtype='C')
model.addCons(stand_in>=(t-quicksum(stocks_df.loc[i,'Mean']*stocks_portfolio[i] for i in range(noptions)))/((quicksum(stocks_portfolio[i]*stocks_df.loc[i,'stdev']**2 for i in range(noptions))+quicksum(2*stocks_portfolio[i]*stocks_portfolio[j]*cov_mat[i,j] for i in range(noptions) for j in range(noptions)))**0.5))
model.setObjective(stand_in,'minimize')
model.optimize()
model.getCondition()
portfolios = []
for i in range(noptions):
    if model.getVal(stocks_portfolio[i]) > 0.9:
        portfolios.append(i)

这里的表现缓慢而笨拙，我想知道我是否在考虑这个问题完全错误。

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