如何解决投资组合优化的方程和约束系统？

Question

我有一个DataFrame如下：

Name  Volatility    Return
a      0.0243        0.212
b      0.0321        0.431
c      0.0323        0.443
d      0.0391        0.2123
e      0.0433        0.3123

我想要一个Volatility{}的Volatility和该波动性的最大Return

也就是说，我想，在一个新的Df中Name和在我的portfolio中的资产百分比，它给出了Volatility的最大Return等于0.035

因此，我需要求解一个具有多个条件的方程组，以获得固定结果（Volatility == 0.035）的最佳解（最高Return）

条件是：

• 每个资源的权重介于0和1之间
• 权重之和为1
• 权重之和乘以各资产的波动率即为“期望波动率”
• 权重之和乘以每项资产的收益即为“总收益”。这应该最大化

Answer 1

下面是一种使用Z3Py的方法，这是一种开源的SAT/SMT解算器。 在SAT/SMT解算器中，您可以将代码编写为条件列表，程序会找到最佳解决方案（或者在使用Z3作为解算器时，仅找到满足所有条件的解决方案）

最初SAT解算器只处理纯布尔表达式，但现代SAT/SMT解算器也允许将固定位和无限整数、分数、实数甚至函数作为中心变量

为了将给定的方程写入Z3，它们被逐字转换为Z3表达式。下面的代码注释了每个步骤

import pandas as pd
from z3 import *

DesiredVolatility = 0.035
df = pd.DataFrame(columns=['Name', 'Volatility', 'Return'],
                  data=[['a', 0.0243, 0.212],
                        ['b', 0.0321, 0.431],
                        ['c', 0.0323, 0.443],
                        ['d', 0.0391, 0.2123],
                        ['e', 0.0433, 0.3123]])

# create a Z3 instance to optimize something
s = Optimize()
# the weight of each asset, as a Z3 variable
W = [Real(row.Name) for row in df.itertuples()]
# the total volatility
TotVol = Real('TotVol')
# the total return, to be maximized
TotReturn = Real('TotReturn')

# weights between 0 and 1, and sum to 1
s.add(And([And(w >= 0, w <= 1) for w in W]))
s.add(Sum([w for w in W]) == 1)
# the total return is calculated as the weighted sum of the asset returns
s.add(TotReturn == Sum([w * row.Return for w, row in zip(W, df.itertuples())]))
# the volatility is calculated as the weighted sum of the asset volatility
s.add(TotVol == Sum([w * row.Volatility for w, row in zip(W, df.itertuples())]))
# the volatility should be equal to the desired volatility
s.add(TotVol == DesiredVolatility)
# we're maximizing the total return
h1 = s.maximize(TotReturn)
# we ask Z3 to do its magick
res = s.check()
# we check the result, hoping for 'sat': all conditions satisfied, a maximum is found
if res == sat:
    s.upper(h1)
    m = s.model()
    #for w in W:
    #    print(f'asset {w}): {m[w]} = {m[w].numerator_as_long() / m[w] .denominator_as_long() : .6f}')
    # output the total return
    print(f'Total Return: {m[TotReturn]} = {m[TotReturn].numerator_as_long() / m[TotReturn] .denominator_as_long() :.6f}')
    # get the proportions out of the Z3 model
    proportions = [m[w].numerator_as_long() / m[w] .denominator_as_long() for w in W]
    # create a dataframe with the result
    df_result = pd.DataFrame({'Name': df.Name, 'Proportion': proportions})
    print(df_result)
else:
    print("No satisfiable solution found")

结果:

Total Return: 452011/1100000 = 0.410919
  Name  Proportion
0    a    0.000000
1    b    0.000000
2    c    0.754545
3    d    0.000000
4    e    0.245455

您可以轻松添加其他约束，例如“任何资产都不能超过总资产的30%”：

# change 
s.add(And([And(w >= 0, w <= 1) for w in W]))`
# to
s.add(And([And(w >= 0, w <= 0.3) for w in W]))`

这将导致：

Total Return: 558101/1480000 = 0.377095
  Name  Proportion
0    a    0.082432
1    b    0.300000
2    c    0.300000
3    d    0.017568
4    e    0.300000