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Formulate a nonlinear optimization problem in Python (gekko) and with m.Equations

Tags:

python

gekko

I would like to solve an optimization problem with Gekko. The non-linear constraints are generated in a function and passed on as a vector. Since there are vectors, I use m.Equations and m.Array. To make it as simple as possible, I have simplified my problem so that the code remains short and simple. I think my mistake is in how I use m.Equations. The error message is:

Exception: @error: Equation Definition An equation without equality (=) or inequality (>,<) false STOPPING...

This is my code:

import numpy as np
from gekko import GEKKO

def powerf(x):
    U = np.array([[x[2]], [x[3] * m.exp(x[4] * 1j)]])
    G = np.array([[x[0] + x[1] * 1j], [0]])
    L = np.array([[0],[1]])

    y = np.array([[-18.8301*1j, 18.8566*1j], [18.8566*1j, -18.8301*1j]])
    I = np.matmul(y, U)
    Iconj = np.real(I) - np.imag(I) * 1j

    S = np.matmul(np.diagflat(Iconj), U)
    P = np.real(S) - np.real(G) + np.real(L)
    Q = np.imag(S) - np.imag(G) + np.imag(L)
    return np.concatenate((P,Q)).reshape(4,)

m = GEKKO()
x =  m.Array(m.Var,5)
ig = [0,  0,    1,    1,      0]
lb = [0, -2, 0.95, 0.95, -np.pi]
ub = [5,  2, 1.10, 1.10,  np.pi]

for i in range(len(x)):
    x[i].value = ig[i]
    x[i].lower = lb[i]
    x[i].upper = ub[i]

m.Equations([powerf(x) == np.zeros(4,)])
m.Obj(x[0])

m.solve()
like image 245
FragezeichenPython Avatar asked Mar 20 '26 05:03

FragezeichenPython


1 Answers

The m.Equations() function needs a list of equations. Here is one way to pose those:

y = powerf(x)
m.Equations([y[i]==0 for i in range(4)])

The resulting model can be viewed with m.open_folder().

Model
Variables
    v1 = 0, <= 5, >= 0
    v2 = 0, <= 2, >= -2
    v3 = 1, <= 1.1, >= 0.95
    v4 = 1, <= 1.1, >= 0.95
    v5 = 0, <= 3.141592653589793, >= -3.141592653589793
End Variables
Equations
    ((((((((((-0-18.8301j))*(v3))+((18.8566j)*(((v4)*(exp(((v5)*(1j))))))))-0j))*(v3))+((0)*(((v4)*(exp(((v5)*(1j))))))))-(v1+((v2)*(1j))))+0)=0
    (((((0)*(v3))+((((((18.8566j)*(v3))+(((-0-18.8301j))*(((v4)*(exp(((v5)*(1j))))))))-0j))*(((v4)*(exp(((v5)*(1j))))))))-0)+1)=0
    True
    True
    minimize v1
End Equations

End Model

Equations 3 and 4 from the list do not appear to have Gekko variables so they return True. One other thing to consider is that Gekko doesn't currently support imaginary numbers natively so the problem needs to reformed into separate variables for imaginary value and real values such as x_real[i] and x_im[i]. A more readable form of the objective is m.Minimize(x[0]).

like image 98
John Hedengren Avatar answered Mar 22 '26 19:03

John Hedengren



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