CODEWITHSUNDEEP
pythonoptimizationgekko
Mohit Tuteja
Mohit Tuteja

Reputation: 95

Trying to maximize this simple non linear problem using #gekko but getting this error

(@error: No solution found)

positions = ["AAPL", "NVDA", "MS","CI", "HON"]
cov = df_ret.cov()
ret = df_ret.mean().values
weights = np.array(np.random.random(len(positions)))

def maximize(weights):
    std = np.sqrt(np.dot(np.dot(weights.T,cov),weights))
    p_ret = np.dot(ret.T,weights)
    sharpe = p_ret/std
    return sharpe

a = GEKKO()
w1 = a.Var(value=0.2, lb=0, ub=1)
w2 = a.Var(value=0.2, lb=0, ub=1)
w3 = a.Var(value=0.2, lb=0, ub=1)
w4 = a.Var(value=0.2, lb=0, ub=1)
w5 = a.Var(value=0.2, lb=0, ub=1)

a.Equation(w1+w2+w3+w4+w5<=1)
weight = np.array([w1,w2,w3,w4,w5])

a.Obj(-maximize(weight))
a.solve(disp=False)

**** trying to figure out why is it giving no solution as an error

# df_ret is a data frame with returns (for stocks in position)

Df_ret looks like this

# trying to maximize the sharpe ratio

# w(1 to n) are the weights with sum less than or equal to 1****

Upvotes: 5

Views: 362

Answers (2)

John Hedengren
John Hedengren

Reputation: 14376

Here is a solution with gekko:

from gekko import GEKKO
import numpy as np
import pandas as pd
a = GEKKO()

positions = ["AAPL", "NVDA", "MS","CI", "HON"]

df_ret = pd.DataFrame(np.array([[.001729, .014603, .036558, .016772, .001983],
[-0.015906, .006396, .012796, -.002163, 0],
[-0.001849, -.019598, .014484, .036856, .019292],
[.006648, .002161, -.020352, -.007580, 0.022083],
[-.008821, -.014016, -.006512, -.015802, .012583]]))
cov = df_ret.cov().values
ret = df_ret.mean().values

def obj(weights):
    std = a.sqrt(np.dot(np.dot(weights.T,cov),weights))
    p_ret = np.dot(ret.T,weights)
    sharpe = p_ret/std
    return sharpe

a = GEKKO()
w = a.Array(a.Var,len(positions),value=0.2,lb=1e-5, ub=1)
a.Equation(a.sum(w)<=1)
a.Maximize(obj(w))
a.solve(disp=False)

print(w)

A couple things that I've done to problem is to use the Array function to create the variable weights w. I also switched to using the gekko sqrt so that it does the automatic differentiation for the objective function. I also added a lower bound of 1e-5 to avoid sqrt(0) and divide by zero. The Obj() function minimizes so I removed the negative sign and use the Maximize() function to make it more readable. It produces this solution for w:

[[1e-05] [0.15810629919] [0.19423029287] [1e-05] [0.6476428726]]

Many are more familiar with scipy. Here is a benchmark problem where the same problem is solved with scipy.minimize.optimize and gekko. There is also a link for that same solution with MATLAB fmincon or gekko with MATLAB.

Upvotes: 3

Anna Nevison
Anna Nevison

Reputation: 2759

I am not familiar with GEKKO so I can't really help with that package, but incase someone doesn't answer how to do it using GEKKO, here's a potential solution with scipy.optimize.minimize:

from scipy.optimize import minimize
import numpy as np
import pandas as pd



def OF(weights, cov, ret, sign = 1.0):
  std = np.sqrt(np.dot(np.dot(weights.T,cov),weights))
  p_ret = np.dot(ret.T,weights)
  sharpe = p_ret/std
  return sign*sharpe


if __name__ == '__main__':

  x0 = np.array([0.2,0.2,0.2,0.2,0.2])
  df_ret = pd.DataFrame(np.array([[.001729, .014603, .036558, .016772, .001983],
[-0.015906, .006396, .012796, -.002163, 0],
[-0.001849, -.019598, .014484, .036856, .019292],
[.006648, .002161, -.020352, -.007580, 0.022083],
[-.008821, -.014016, -.006512, -.015802, .012583]]))
  cov = df_ret.cov()
  ret = df_ret.mean().values


  minx0 = np.repeat(0, [len(x0)] , axis = 0)
  maxx0 = np.repeat(1, [len(x0)] , axis = 0)
  bounds = tuple(zip(minx0, maxx0))

  cons = {'type':'ineq', 
  'fun':lambda weights: 1 - sum(weights)}
  res_cons = minimize(OF, x0, (cov, ret, -1), bounds = bounds, constraints=cons, method='SLSQP')



  print(res_cons)
  print('Current value of objective function: ' + str(res_cons['fun']))
  print('Current value of controls:')
  print(res_cons['x'])

which outputs:

     fun: -2.1048843911794486
     jac: array([ 5.17067784e+00, -2.36839056e-04, -6.24716282e-04,  6.56819057e+00,
        2.45392323e-04])
 message: 'Optimization terminated successfully.'
    nfev: 69
     nit: 9
    njev: 9
  status: 0
 success: True
       x: array([5.47832097e-14, 1.52927443e-01, 1.87864415e-01, 5.32258098e-14,
       6.26433468e-01])
Current value of objective function: -2.1048843911794486
Current value of controls:
[5.47832097e-14 1.52927443e-01 1.87864415e-01 5.32258098e-14
 6.26433468e-01]

The sign parameter is added here because in order to maximize the objective function you just minimize OF*(-1). I set the default to 1 (minimize), but I pass -1 in args to change it.

Upvotes: 2

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