I am actually trying to use the IPOPT Optimisor available on GEKKO in order to optimise a large non-convex and non-linear problem.In order to do that I need to use the Fast Fourrier Transform with scipy.First,lets fix our sample data(for simplicity):
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
then the fourrier functions:
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
Now let's use 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))
The problem here is the ifft scipy function that doesn't work because of the different type of variable given by GEKKO.The problem is I don't see how I can replace or avoid it.
The error is such :
<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
Could someone help me debugging this.Thank you
Gekko requires that expressions are not black box but are able to be expressed with special types of variables (Gekko type) for automatic differentiation and sparsity detection. This may be better solved with a solver such as Scipy.optimize.minimize. Here is a 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]))
Gekko
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))