凸优化-货币兑换问题-cvxpy错误
Convex Optimization - Currency exchange problem - cvxpy error
我正在解决货币兑换问题，我们有一组 10 种不同的货币持有量。我们的 objective 是要对每种货币的持有量设定一个特定的最小值，同时最大限度地减少由于我们将一种货币的某些单位兑换另一种货币的某些单位时出现的买卖价差而造成的损失。
如果有帮助的话，这里是完整的问题：
https://drive.google.com/file/d/1DG8xfM3ayQ0Oj9VQGYSNbHF_rbKHMTLq/view
代码：
import numpy as np
import matplotlib.pyplot as plt
import cvxpy as cp
# Exchange rate data. Obtained from website
tickers = ["USD", "EUR", "GBP", "CAD", "JPY", "CNY", "RUB", "MXN", "INR", "BRL"]
n = len(tickers)
F = np.zeros((n, n))
# USD
data = ([1.0, 0.87, 0.76, 1.31, 108.90, 6.72, 65.45, 19.11, 71.13, 3.69],
# EUR
[1.0, 0.88, 1.51, 125.15, 7.72, 75.23, 21.96, 81.85, 4.24],
# GBP
[1.0, 1.72, 142.94, 8.82, 85.90, 25.08, 93.50, 4.84],
# CAD
[1.0, 82.93, 5.11, 49.82, 14.54, 54.23, 2.81],
# JPY
[1.0, 0.062, 0.60, 0.18, 0.65, 0.034],
# CNY
[1.0, 9.74, 2.85, 10.61, 0.55],
# RUB
[1.0, 0.29, 1.09, 0.056],
# MXN
[1.0, 3.73, 0.19],
# INR
[1.0, 0.052],
# BRL
[1.0])
for i in range(n):
F[i,i:] = data[i]
for j in range(n):
for i in range(j+1,n):
F[i,j] = 1.035/F[j,i]
# Initial and final portfolios.
c_req = np.arange(1,n+1)
c_req = 1e4*c_req/c_req.sum()
c_init = c_req[::-1]
X = cp.Variable((n,n))
c_final = c_init - cp.matmul(cp.transpose(X/F),np.ones(n)) + cp.matmul(X/F,np.ones(n)) # final number of currency holdings
value_per_currency = np.zeros(n) # value per currency unit in USD
for i in range(n):
value_per_currency[i] = pow(F[i][0]/F[0][i] , 0.5) # value per currency unit in USD
Loss = cp.matmul((c_init - c_final),value_per_currency) # Loss in currency valued as the GM of bid and ask price in USD
# Defining the convex optimization problem
objective = cp.Minimize(Loss) # Loss is minimised
constraints = [X>=0, cp.diag(X)==0] # constraints added
constraints += [cp.matmul(cp.transpose(X),np.ones(n))<=c_init]
for i in range(n):
constraints += [c_final[i] >= c_req[i]]
prob = cp.Problem(objective, constraints) # Problem defined
result = prob.solve() # solution
initial_value = np.dot(value_per_currency, c_init)
final_value = np.dot(value_per_currency, c_final)
print("The initial currency holdings is:\n\n", c_init);print("\n")
print("The final currency holdings is:\n\n", c_final);print("\n")
print("The initial value of holdings = ", initial_value, "USD")
print("The final value of holdings = \n", final_value, "USD")
输出：
The initial currency holdings is:
[1818.18181818 1636.36363636 1454.54545455 1272.72727273 1090.90909091
909.09090909 727.27272727 545.45454545 363.63636364 181.81818182]
The final currency holdings is:
[1818.18181818 1636.36363636 1454.54545455 1272.72727273 1090.90909091
909.09090909 727.27272727 545.45454545 363.63636364 181.81818182] + -var562 / [[1.00000000e+00 8.70000000e-01 7.60000000e-01 1.31000000e+00
1.08900000e+02 6.72000000e+00 6.54500000e+01 1.91100000e+01
7.11300000e+01 3.69000000e+00]
[1.18965517e+00 1.00000000e+00 8.80000000e-01 1.51000000e+00
1.25150000e+02 7.72000000e+00 7.52300000e+01 2.19600000e+01
8.18500000e+01 4.24000000e+00]
[1.36184211e+00 1.17613636e+00 1.00000000e+00 1.72000000e+00
1.42940000e+02 8.82000000e+00 8.59000000e+01 2.50800000e+01
9.35000000e+01 4.84000000e+00]
[7.90076336e-01 6.85430464e-01 6.01744186e-01 1.00000000e+00
8.29300000e+01 5.11000000e+00 4.98200000e+01 1.45400000e+01
5.42300000e+01 2.81000000e+00]
[9.50413223e-03 8.27007591e-03 7.24080034e-03 1.24804052e-02
1.00000000e+00 6.20000000e-02 6.00000000e-01 1.80000000e-01
6.50000000e-01 3.40000000e-02]
[1.54017857e-01 1.34067358e-01 1.17346939e-01 2.02544031e-01
1.66935484e+01 1.00000000e+00 9.74000000e+00 2.85000000e+00
1.06100000e+01 5.50000000e-01]
[1.58135982e-02 1.37578094e-02 1.20488941e-02 2.07747892e-02
1.72500000e+00 1.06262834e-01 1.00000000e+00 2.90000000e-01
1.09000000e+00 5.60000000e-02]
[5.41601256e-02 4.71311475e-02 4.12679426e-02 7.11829436e-02
5.75000000e+00 3.63157895e-01 3.56896552e+00 1.00000000e+00
3.73000000e+00 1.90000000e-01]
[1.45508224e-02 1.26450825e-02 1.10695187e-02 1.90853771e-02
1.59230769e+00 9.75494816e-02 9.49541284e-01 2.77479893e-01
1.00000000e+00 5.20000000e-02]
[2.80487805e-01 2.44103774e-01 2.13842975e-01 3.68327402e-01
3.04411765e+01 1.88181818e+00 1.84821429e+01 5.44736842e+00
1.99038462e+01 1.00000000e+00]].T @ [1. 1. 1. 1. 1. 1. 1. 1. 1. 1.] + (var562 / [[1.00000000e+00 8.70000000e-01 7.60000000e-01 1.31000000e+00
1.08900000e+02 6.72000000e+00 6.54500000e+01 1.91100000e+01
7.11300000e+01 3.69000000e+00]
[1.18965517e+00 1.00000000e+00 8.80000000e-01 1.51000000e+00
1.25150000e+02 7.72000000e+00 7.52300000e+01 2.19600000e+01
8.18500000e+01 4.24000000e+00]
[1.36184211e+00 1.17613636e+00 1.00000000e+00 1.72000000e+00
1.42940000e+02 8.82000000e+00 8.59000000e+01 2.50800000e+01
9.35000000e+01 4.84000000e+00]
[7.90076336e-01 6.85430464e-01 6.01744186e-01 1.00000000e+00
8.29300000e+01 5.11000000e+00 4.98200000e+01 1.45400000e+01
5.42300000e+01 2.81000000e+00]
[9.50413223e-03 8.27007591e-03 7.24080034e-03 1.24804052e-02
1.00000000e+00 6.20000000e-02 6.00000000e-01 1.80000000e-01
6.50000000e-01 3.40000000e-02]
[1.54017857e-01 1.34067358e-01 1.17346939e-01 2.02544031e-01
1.66935484e+01 1.00000000e+00 9.74000000e+00 2.85000000e+00
1.06100000e+01 5.50000000e-01]
[1.58135982e-02 1.37578094e-02 1.20488941e-02 2.07747892e-02
1.72500000e+00 1.06262834e-01 1.00000000e+00 2.90000000e-01
1.09000000e+00 5.60000000e-02]
[5.41601256e-02 4.71311475e-02 4.12679426e-02 7.11829436e-02
5.75000000e+00 3.63157895e-01 3.56896552e+00 1.00000000e+00
3.73000000e+00 1.90000000e-01]
[1.45508224e-02 1.26450825e-02 1.10695187e-02 1.90853771e-02
1.59230769e+00 9.75494816e-02 9.49541284e-01 2.77479893e-01
1.00000000e+00 5.20000000e-02]
[2.80487805e-01 2.44103774e-01 2.13842975e-01 3.68327402e-01
3.04411765e+01 1.88181818e+00 1.84821429e+01 5.44736842e+00
1.99038462e+01 1.00000000e+00]]) @ [1. 1. 1. 1. 1. 1. 1. 1. 1. 1.]
The initial value of holdings = 6910.667218038612 USD
The final value of holdings =
[Expression(AFFINE, UNKNOWN, (10,)) Expression(AFFINE, UNKNOWN, (10,))
Expression(AFFINE, UNKNOWN, (10,)) Expression(AFFINE, UNKNOWN, (10,))
Expression(AFFINE, UNKNOWN, (10,)) Expression(AFFINE, UNKNOWN, (10,))
Expression(AFFINE, UNKNOWN, (10,)) Expression(AFFINE, UNKNOWN, (10,))
Expression(AFFINE, UNKNOWN, (10,)) Expression(AFFINE, UNKNOWN, (10,))] USD
查看 c_final，我意识到“var562”未解决，因此我们无法完全解决它（对吗？）
我该如何解决这个问题？
无论您在何处对 cvxpy 对象进行操作，都应该使用 cp. 函数，而不是 np. 函数。所以 cp.dot 等等
我正在解决货币兑换问题，我们有一组 10 种不同的货币持有量。我们的 objective 是要对每种货币的持有量设定一个特定的最小值，同时最大限度地减少由于我们将一种货币的某些单位兑换另一种货币的某些单位时出现的买卖价差而造成的损失。
如果有帮助的话，这里是完整的问题： https://drive.google.com/file/d/1DG8xfM3ayQ0Oj9VQGYSNbHF_rbKHMTLq/view
代码：
import numpy as np
import matplotlib.pyplot as plt
import cvxpy as cp
# Exchange rate data. Obtained from website
tickers = ["USD", "EUR", "GBP", "CAD", "JPY", "CNY", "RUB", "MXN", "INR", "BRL"]
n = len(tickers)
F = np.zeros((n, n))
# USD
data = ([1.0, 0.87, 0.76, 1.31, 108.90, 6.72, 65.45, 19.11, 71.13, 3.69],
# EUR
[1.0, 0.88, 1.51, 125.15, 7.72, 75.23, 21.96, 81.85, 4.24],
# GBP
[1.0, 1.72, 142.94, 8.82, 85.90, 25.08, 93.50, 4.84],
# CAD
[1.0, 82.93, 5.11, 49.82, 14.54, 54.23, 2.81],
# JPY
[1.0, 0.062, 0.60, 0.18, 0.65, 0.034],
# CNY
[1.0, 9.74, 2.85, 10.61, 0.55],
# RUB
[1.0, 0.29, 1.09, 0.056],
# MXN
[1.0, 3.73, 0.19],
# INR
[1.0, 0.052],
# BRL
[1.0])
for i in range(n):
F[i,i:] = data[i]
for j in range(n):
for i in range(j+1,n):
F[i,j] = 1.035/F[j,i]
# Initial and final portfolios.
c_req = np.arange(1,n+1)
c_req = 1e4*c_req/c_req.sum()
c_init = c_req[::-1]
X = cp.Variable((n,n))
c_final = c_init - cp.matmul(cp.transpose(X/F),np.ones(n)) + cp.matmul(X/F,np.ones(n)) # final number of currency holdings
value_per_currency = np.zeros(n) # value per currency unit in USD
for i in range(n):
value_per_currency[i] = pow(F[i][0]/F[0][i] , 0.5) # value per currency unit in USD
Loss = cp.matmul((c_init - c_final),value_per_currency) # Loss in currency valued as the GM of bid and ask price in USD
# Defining the convex optimization problem
objective = cp.Minimize(Loss) # Loss is minimised
constraints = [X>=0, cp.diag(X)==0] # constraints added
constraints += [cp.matmul(cp.transpose(X),np.ones(n))<=c_init]
for i in range(n):
constraints += [c_final[i] >= c_req[i]]
prob = cp.Problem(objective, constraints) # Problem defined
result = prob.solve() # solution
initial_value = np.dot(value_per_currency, c_init)
final_value = np.dot(value_per_currency, c_final)
print("The initial currency holdings is:\n\n", c_init);print("\n")
print("The final currency holdings is:\n\n", c_final);print("\n")
print("The initial value of holdings = ", initial_value, "USD")
print("The final value of holdings = \n", final_value, "USD")
输出：
The initial currency holdings is:
[1818.18181818 1636.36363636 1454.54545455 1272.72727273 1090.90909091
909.09090909 727.27272727 545.45454545 363.63636364 181.81818182]
The final currency holdings is:
[1818.18181818 1636.36363636 1454.54545455 1272.72727273 1090.90909091
909.09090909 727.27272727 545.45454545 363.63636364 181.81818182] + -var562 / [[1.00000000e+00 8.70000000e-01 7.60000000e-01 1.31000000e+00
1.08900000e+02 6.72000000e+00 6.54500000e+01 1.91100000e+01
7.11300000e+01 3.69000000e+00]
[1.18965517e+00 1.00000000e+00 8.80000000e-01 1.51000000e+00
1.25150000e+02 7.72000000e+00 7.52300000e+01 2.19600000e+01
8.18500000e+01 4.24000000e+00]
[1.36184211e+00 1.17613636e+00 1.00000000e+00 1.72000000e+00
1.42940000e+02 8.82000000e+00 8.59000000e+01 2.50800000e+01
9.35000000e+01 4.84000000e+00]
[7.90076336e-01 6.85430464e-01 6.01744186e-01 1.00000000e+00
8.29300000e+01 5.11000000e+00 4.98200000e+01 1.45400000e+01
5.42300000e+01 2.81000000e+00]
[9.50413223e-03 8.27007591e-03 7.24080034e-03 1.24804052e-02
1.00000000e+00 6.20000000e-02 6.00000000e-01 1.80000000e-01
6.50000000e-01 3.40000000e-02]
[1.54017857e-01 1.34067358e-01 1.17346939e-01 2.02544031e-01
1.66935484e+01 1.00000000e+00 9.74000000e+00 2.85000000e+00
1.06100000e+01 5.50000000e-01]
[1.58135982e-02 1.37578094e-02 1.20488941e-02 2.07747892e-02
1.72500000e+00 1.06262834e-01 1.00000000e+00 2.90000000e-01
1.09000000e+00 5.60000000e-02]
[5.41601256e-02 4.71311475e-02 4.12679426e-02 7.11829436e-02
5.75000000e+00 3.63157895e-01 3.56896552e+00 1.00000000e+00
3.73000000e+00 1.90000000e-01]
[1.45508224e-02 1.26450825e-02 1.10695187e-02 1.90853771e-02
1.59230769e+00 9.75494816e-02 9.49541284e-01 2.77479893e-01
1.00000000e+00 5.20000000e-02]
[2.80487805e-01 2.44103774e-01 2.13842975e-01 3.68327402e-01
3.04411765e+01 1.88181818e+00 1.84821429e+01 5.44736842e+00
1.99038462e+01 1.00000000e+00]].T @ [1. 1. 1. 1. 1. 1. 1. 1. 1. 1.] + (var562 / [[1.00000000e+00 8.70000000e-01 7.60000000e-01 1.31000000e+00
1.08900000e+02 6.72000000e+00 6.54500000e+01 1.91100000e+01
7.11300000e+01 3.69000000e+00]
[1.18965517e+00 1.00000000e+00 8.80000000e-01 1.51000000e+00
1.25150000e+02 7.72000000e+00 7.52300000e+01 2.19600000e+01
8.18500000e+01 4.24000000e+00]
[1.36184211e+00 1.17613636e+00 1.00000000e+00 1.72000000e+00
1.42940000e+02 8.82000000e+00 8.59000000e+01 2.50800000e+01
9.35000000e+01 4.84000000e+00]
[7.90076336e-01 6.85430464e-01 6.01744186e-01 1.00000000e+00
8.29300000e+01 5.11000000e+00 4.98200000e+01 1.45400000e+01
5.42300000e+01 2.81000000e+00]
[9.50413223e-03 8.27007591e-03 7.24080034e-03 1.24804052e-02
1.00000000e+00 6.20000000e-02 6.00000000e-01 1.80000000e-01
6.50000000e-01 3.40000000e-02]
[1.54017857e-01 1.34067358e-01 1.17346939e-01 2.02544031e-01
1.66935484e+01 1.00000000e+00 9.74000000e+00 2.85000000e+00
1.06100000e+01 5.50000000e-01]
[1.58135982e-02 1.37578094e-02 1.20488941e-02 2.07747892e-02
1.72500000e+00 1.06262834e-01 1.00000000e+00 2.90000000e-01
1.09000000e+00 5.60000000e-02]
[5.41601256e-02 4.71311475e-02 4.12679426e-02 7.11829436e-02
5.75000000e+00 3.63157895e-01 3.56896552e+00 1.00000000e+00
3.73000000e+00 1.90000000e-01]
[1.45508224e-02 1.26450825e-02 1.10695187e-02 1.90853771e-02
1.59230769e+00 9.75494816e-02 9.49541284e-01 2.77479893e-01
1.00000000e+00 5.20000000e-02]
[2.80487805e-01 2.44103774e-01 2.13842975e-01 3.68327402e-01
3.04411765e+01 1.88181818e+00 1.84821429e+01 5.44736842e+00
1.99038462e+01 1.00000000e+00]]) @ [1. 1. 1. 1. 1. 1. 1. 1. 1. 1.]
The initial value of holdings = 6910.667218038612 USD
The final value of holdings =
[Expression(AFFINE, UNKNOWN, (10,)) Expression(AFFINE, UNKNOWN, (10,))
Expression(AFFINE, UNKNOWN, (10,)) Expression(AFFINE, UNKNOWN, (10,))
Expression(AFFINE, UNKNOWN, (10,)) Expression(AFFINE, UNKNOWN, (10,))
Expression(AFFINE, UNKNOWN, (10,)) Expression(AFFINE, UNKNOWN, (10,))
Expression(AFFINE, UNKNOWN, (10,)) Expression(AFFINE, UNKNOWN, (10,))] USD
查看 c_final，我意识到“var562”未解决，因此我们无法完全解决它（对吗？）
我该如何解决这个问题？
无论您在何处对 cvxpy 对象进行操作，都应该使用 cp. 函数，而不是 np. 函数。所以 cp.dot 等等