使用带有快速傅里叶变换的 GEKKO

Question

我实际上是在尝试使用 GEKKO 上提供的 IPOPT Optimisor 来优化大型非凸非线性问题。为了做到这一点，我需要使用带有 scipy 的快速傅里叶变换。首先，让我们修复我们的样本数据（为简单起见）：

import numpy as np
import pandas as pd
import math
from scipy import *
from scipy.integrate import quad
import scipy.stats as ss
import scipy.optimize as scpo
from scipy import sparse
from scipy.fftpack import ifft
from scipy.interpolate import interp1d
from scipy.optimize import fsolve
from functools import partial

r,prices, strikes, spreads,s0,T  = 0,array([1532.45 , 1507.55 , 1482.65 , 1457.8  , 1432.95 , 1408.1  ,
        1383.25 , 1358.45 , 1333.6  , 1308.8  , 1284.   , 1259.2  ,
        1234.45 , 1209.7  , 1037.15 , 1012.55 ,  988.05 ,  963.35 ,
         938.9  ,  914.3  ,  889.8  ,  865.5  ,  841.   ,  816.65 ,
         792.45 ,  768.1  ,  743.95 ,  719.85 ,  695.85 ,  672.   ,
         648.1  ,  624.5  ,  600.9  ,  577.5  ,  554.2  ,  531.   ,
         508.15 ,  485.35 ,  462.9  ,  440.65 ,  418.65 ,  396.85 ,
         375.55 ,  354.5  ,  333.9  ,  313.65 ,  293.85 ,  255.75 ,
         237.55 ,  219.8  ,  202.8  ,  186.35 ,  170.55 ,  155.4  ,
         141.05 ,  127.4  ,  113.5  ,  101.35 ,   90.1  ,   79.65 ,
          70.   ,   61.3  ,   53.4  ,   46.35 ,   34.5  ,   29.6  ,
          25.35 ,   18.55 ,   15.85 ,   13.55 ,   11.55 ,    9.9  ,
           7.35 ,    5.45 ,    3.075,    2.7  ]),array([12500., 12525., 12550., 12575., 12600., 12625., 12650., 12675.,
        12700., 12725., 12750., 12775., 12800., 12825., 13000., 13025.,
        13050., 13075., 13100., 13125., 13150., 13175., 13200., 13225.,
        13250., 13275., 13300., 13325., 13350., 13375., 13400., 13425.,
        13450., 13475., 13500., 13525., 13550., 13575., 13600., 13625.,
        13650., 13675., 13700., 13725., 13750., 13775., 13800., 13850.,
        13875., 13900., 13925., 13950., 13975., 14000., 14025., 14050.,
        14075., 14100., 14125., 14150., 14175., 14200., 14225., 14250.,
        14300., 14325., 14350., 14400., 14425., 14450., 14475., 14500.,
        14550., 14600., 14700., 14725.]),array([29.7 , 29.7 , 29.7 , 29.8 , 29.7 , 29.8 , 29.7 , 29.7 , 29.8 ,
        29.8 , 29.8 , 29.8 , 29.7 , 29.8 , 10.3 , 10.3 , 10.5 , 10.3 ,
        10.6 , 10.4 , 10.4 , 10.6 , 10.4 , 10.5 , 10.7 , 10.4 , 10.5 ,
        10.5 , 10.5 , 10.8 , 10.6 , 10.8 , 10.6 , 10.8 , 10.6 , 10.6 ,
        10.9 , 10.5 , 10.8 , 10.7 , 10.7 , 10.3 , 10.5 , 10.4 , 10.2 ,
        10.1 ,  9.9 ,  9.5 ,  9.3 ,  9.  ,  8.8 ,  8.5 ,  8.3 ,  8.2 ,
         7.9 ,  7.6 ,  3.8 ,  3.7 ,  3.6 ,  3.5 ,  3.4 ,  3.2 ,  3.2 ,
         3.1 ,  2.8 ,  2.6 ,  2.5 ,  2.3 ,  2.1 ,  2.1 ,  1.9 ,  1.8 ,
         1.7 ,  1.5 ,  1.25,  1.2 ]),14000,0.05

然后是傅里叶函数：

class Heston_pricer():

    def __init__(self, Option_info, Process_info ):
        """
        Process_info: a instance of "Heston_process.", which contains (mu, rho, sigma, theta, kappa)
        Option_info: of type Option_param, which contains (S0,K,T)
        """
        self.r = Process_info.mu              # interest rate
        self.sigma = Process_info.sigma       # Heston parameters
        self.theta = Process_info.theta       
        self.kappa = Process_info.kappa       
        self.rho = Process_info.rho           

        self.S0 = Option_info.S0          # current price
        self.v0 = Option_info.v0          # spot variance
        self.K = Option_info.K            # strike
        self.T = Option_info.T            # maturity(in years)
        self.exercise = Option_info.exercise
        self.payoff = Option_info.payoff

    # payoff function
    def payoff_f(self, S):
        if self.payoff == "call":
            Payoff = np.maximum( S - self.K, 0 )
        elif self.payoff == "put":    
            Payoff = np.maximum( self.K - S, 0 )  
        return Payoff

    # FFT method. It returns a vector of prices.
    def FFT(self, K): # K: strikes
        K = np.array(K)

        # Heston characteristic function (proposed by Schoutens 2004)
        def cf_Heston_good(u, t, v0, mu, kappa, theta, sigma, rho):
            xi = kappa - sigma*rho*u*1j
            d = m.sqrt( xi**2 + sigma**2 * (u**2 + 1j*u) ) 
            g1 = (xi+d)/(xi-d)  
            g2 = 1/g1
            cf = m.exp( 1j*u*mu*t + (kappa*theta)/(sigma**2) * ( (xi-d)*t - 2*m.log( (1-g2*m.exp(-d*t))/(1-g2) ))\
                      + (v0/sigma**2)*(xi-d) * (1-m.exp(-d*t))/(1-g2*m.exp(-d*t))) 
            return cf

        cf_H_b_good = partial(cf_Heston_good, t=self.T, v0=self.v0, mu=self.r, theta=self.theta, 
                                  sigma=self.sigma, kappa=self.kappa, rho=self.rho)
        if self.payoff == "call":
            return fft_(K, self.S0, self.r, self.T, cf_H_b_good)
        elif self.payoff == "put":        # put-call parity
            return fft_(K, self.S0, self.r, self.T, cf_H_b_good) - self.S0 + K*m.exp(-self.r*self.T)

class Heston_process():
  def __init__(self, mu=0, rho=0, sigma=0.00001, theta=0.4, kappa=.00001):
            """
            r: risk free constant rate
            rho: correlation between stock noise and variance noise (|rho| must be <=1)
            theta: long term mean of the variance process(positive)
            sigma: volatility coefficient(positive)
            kappa: mean reversion coefficient for the variance process(positive)
            """
            self.mu, self.rho, self.theta, self.sigma, self.kappa = mu, rho, theta, sigma, kappa

def fft_(K, S0, r, T, cf): # interp support cubic 
    """ 
    K = vector of strike
    S0 = spot price scalar
    cf = characteristic function
    """
    N=2**15                         # FFT more efficient for N power of 2
    B = 500                         # integration limit 

    dx = B/N
    x = np.arange(N) * dx

    weight = 3 + (-1)**(np.arange(N)+1) # Simpson weights
    weight[0] = 1; weight[N-1]=1

    dk = 2*np.pi/B
    b = N * dk /2
    ks = -b + dk * np.arange(N)

    integrand = m.exp(- 1j * b *(np.arange(N))*dx) * cf(x - 0.5j) * 1/(x**2 + 0.25) * weight * dx/3
    integral_value = np.real(ifft(integrand)*N)
    spline_cub = interp1d(ks, integral_value, kind="cubic") # cubic will fit better than linear
    prices = S0 - m.sqrt(S0 * K) * m.exp(-r*T)/np.pi * spline_cub( m.log(S0/K) )
    return prices

# A class that stores option parameters (in order to write BS/Heston class neatly)
class Option_param():  
    def __init__(self, S0=10000, K=10000, T=.1, v0=0.04, payoff="call", exercise="European"):
        """
        S0: current stock price
        K: Strike price
        T: time to maturity
        v0: (optional) spot variance 
        exercise: European or American
        """
        self.S0, self.v0, self.K, self.T, self.exercise, self.payoff = S0, v0, K, T, exercise, payoff

现在让我们使用 GEKKO：

#Initialize Model
m = GEKKO()

#define parameter
eq = m.Param(value=5)

#initialize variables
x1,x2,x3,x4,x5 = [m.Var(lb=-1, ub=1),m.Var(lb=1e-3, ub=1),m.Var(lb=1e-3, ub=1),m.Var(lb=1e-3, ub=20),m.Var(lb=1e-3, ub=1)]

#initial values
x1.value = 0
x2.value = 0.5
x3.value = 0.5
x4.value = 0.5
x5.value = 0.5

X = [x1,x2,x3,x4,x5]

#Equations,Feller Condition
m.Equation(2*x3*x4 - x2*x2 >=0)

def least_sq(x):
    """ Objective function """
    Heston_param = Heston_process(mu=0, rho=X[0], sigma=X[1], theta=X[2], kappa=X[3])
    m = 1/(spreads**2)
    if len(m) == 1:
      l = 1
    else:
      l = (m - np.min(m))/(np.max(m)-np.min(m))
    opt_param = Option_param(S0=s0, K=strikes, T=T, v0=X[4], exercise="European", payoff="call" )
    Hest = Heston_pricer(opt_param, Heston_param)
    prices_calib = Hest.FFT(strikes)
    results = (l * (prices_calib-prices)**2)/len(prices)
    return results

m.Obj(m.sum(least_sq(X)))
#Set global options
m.options.IMODE = 3 #steady state optimization

#Solve simulation
m.solve()

#Results
print('')
print('Results')
print('x1: ' + str(x1.value))
print('x2: ' + str(x2.value))
print('x3: ' + str(x3.value))
print('x4: ' + str(x4.value))
print('x5: ' + str(x5.value))

这里的问题是 ifft scipy 函数由于 GEKKO 给出的变量类型不同而不起作用。问题是我不知道如何替换或避免它。

错误是这样的：

<ipython-input-16-305a2bb0769b> in fft_(K, S0, r, T, cf)
     78 
     79     integrand = m.exp(- 1j * b *m.CV(np.arange(N))*dx) * cf(x - 0.5j) * 1/(x**2 + 0.25) * weight * dx/3
---> 80     integral_value = np.real(ifft(integrand)*N)
     81     spline_cub = interp1d(ks, integral_value, kind="cubic") # cubic will fit better than linear
     82     prices = S0 - m.sqrt(S0 * K) * m.exp(-r*T)/np.pi * spline_cub( m.log(S0/K) )

/usr/local/lib/python3.7/dist-packages/scipy/_lib/deprecation.py in call(*args, **kwargs)
     18             warnings.warn(msg, category=DeprecationWarning,
     19                           stacklevel=stacklevel)
---> 20             return fun(*args, **kwargs)
     21         call.__doc__ = msg
     22         return call

<__array_function__ internals> in ifft(*args, **kwargs)

/usr/local/lib/python3.7/dist-packages/numpy/fft/_pocketfft.py in ifft(a, n, axis, norm)
    274     a = asarray(a)
    275     if n is None:
--> 276         n = a.shape[axis]
    277     if norm is not None and _unitary(norm):
    278         inv_norm = sqrt(max(n, 1))

IndexError: tuple index out of range

谁能帮我调试一下。谢谢

Answer 1

Gekko 要求表达式不是黑盒，而是能够用特殊类型的变量（Gekko 类型）表示，以进行自动微分和稀疏检测。使用 Scipy.optimize.minimize 等求解器可能会更好地解决此问题。这是comparison of the two on a simple problem。

Scipy

import numpy as np
from scipy.optimize import minimize

def objective(x):
    return x[0]*x[3]*(x[0]+x[1]+x[2])+x[2]

def constraint1(x):
    return x[0]*x[1]*x[2]*x[3]-25.0

def constraint2(x):
    sum_eq = 40.0
    for i in range(4):
        sum_eq = sum_eq - x[i]**2
    return sum_eq

# initial guesses
n = 4
x0 = np.zeros(n)
x0[0] = 1.0
x0[1] = 5.0
x0[2] = 5.0
x0[3] = 1.0

# show initial objective
print('Initial Objective: ' + str(objective(x0)))

# optimize
b = (1.0,5.0)
bnds = (b, b, b, b)
con1 = {'type': 'ineq', 'fun': constraint1} 
con2 = {'type': 'eq', 'fun': constraint2}
cons = ([con1,con2])
solution = minimize(objective,x0,method='SLSQP',\
                    bounds=bnds,constraints=cons)
x = solution.x

# show final objective
print('Final Objective: ' + str(objective(x)))

# print solution
print('Solution')
print('x1 = ' + str(x[0]))
print('x2 = ' + str(x[1]))
print('x3 = ' + str(x[2]))
print('x4 = ' + str(x[3]))

壁虎

from gekko import GEKKO    

m = GEKKO()

#initialize variables
x1,x2,x3,x4 = [m.Var(lb=1,ub=5) for i in range(4)]
x1.value = 1; x2.value = 5; x3.value = 5; x4.value = 1

m.Equation(x1*x2*x3*x4>=25)
m.Equation(x1**2+x2**2+x3**2+x4**2==40)

m.Minimize(x1*x4*(x1+x2+x3)+x3)

m.solve()

print('x1: ' + str(x1.value))
print('x2: ' + str(x2.value))
print('x3: ' + str(x3.value))
print('x4: ' + str(x4.value))