Here is a simple example of solving the system of linear equations and the example of using for loop for many equations.
import numpy as np
from gekko import GEKKO
m = GEKKO(remote=False)
# Random 3x3
A = np.random.rand(3,3)
# Random 3x1
b = np.random.rand(3)
# Gekko array 3x1
x = m.Array(m.Var,(3))
# solve Ax = b
eqn = np.dot(A,x)
for i in range(3):
m.Equation(eqn[i]==b[i])
m.solve(disp=False)
X = [x[i].value for i in range(3)]
print(X)
print(b)
print(np.dot(A,X))
with the correct output. With the result X (np.dot(A,X)==b) - correct!
[[-0.45756768428], [1.0562541773], [0.10058435163]]
[0.64342498 0.34894335 0.5375324 ]
[[0.64342498]
[0.34894335]
[0.5375324 ]]
In the recent Gekko 0.2rc6 there is also introduced axb() function for linear programing. This might be the same problem solved with this function, but I am not sure how to get the correct result.
m = GEKKO(remote=False)
# Random 3x3
A = np.random.rand(3,3)
# Random 3x1
b = np.random.rand(3)
# Gekko array 3x1
x = m.Array(m.Var,(3))
# solve Ax = b
m.axb(A,b,x=x)
m.solve(disp=False)
X = [x[i].value for i in range(3)]
print(X)
print(b)
print(np.dot(A,X))
but it seems I missed something because the output is not the solution??? With the result X (np.dot(A,X)==b) - is not correct!
[[0.2560342704], [0.7543346092], [-0.084190799732]]
[0.27262652 0.61028723 0.74616952]
[[0.4201021 ]
[0.5206979 ]
[0.39195592]]
