CODEWITHSUNDEEP
pythongekko
Chet
Chet

Reputation: 451

Accessing specific values in GEKKO optimal solution

Following up on a question posted here earlier:

GEKKO - optimization in matrix form

I notice that I can get specific values for q (optimal solution) when I reference them individually:

q[0,1] = [2.5432017412] q[0,2] = [3.7228765674]

I can also add these values individually:

q[0,1][0] + q[0,2][0] = 6.2660783086

I am not able to use the sum function to get the values of q however.E.g.(code: Output):

q[0].sum(): (((v1+v2)+v3)+v4)
sum(q[0,:]) : ((((0+v1)+v2)+v3)+v4)
q[0,:].sum(axis=0) : (((v1+v2)+v3)+v4)

I'm not sure why sum is not able to access the variables inside q. Do I need to perform any intermediate operation on the output (i.e. q)? Thanks,

Upvotes: 3

Views: 331

Answers (1)

John Hedengren
John Hedengren

Reputation: 14346

The numpy array stores gekko objects at each row and column location. To access the gekko value for variable x, you need to get x.value[0]. You can convert the q matrix to a numpy matrix with

# convert to matrix form
qr = np.array([[q[i,j].value[0] for j in range(4)] for i in range(4)])

Here is a complete version of your code that checks that the row and column constraints are satisfied.

import numpy as np
import scipy.optimize as opt
from gekko import GEKKO
p= np.array([4, 5, 6.65, 12]) #p = prices
pmx = np.triu(p - p[:, np.newaxis]) #pmx = price matrix, upper triangular
m = GEKKO(remote=False)
q = m.Array(m.Var,(4,4),lb=0,ub=10)
# only upper triangular can change
for i in range(4):
    for j in range(4):
        if j<=i:
            q[i,j].upper=0 # set upper bound = 0
def profit(q):
    profit = np.sum(q.flatten() * pmx.flatten())
    return profit
for i in range(4):
    m.Equation(np.sum(q[i,:])<=10)
    m.Equation(np.sum(q[:,i])<=8)
m.Maximize(profit(q))
m.solve()
print(q)
# convert to matrix form
qr = np.array([[q[i,j].value[0] for j in range(4)] for i in range(4)])
for i in range(4):
    rs = qr[i,:].sum()
    print('Row sum ' + str(i) + ' = ' + str(rs))
    cs = qr[:,i].sum()
    print('Col sum ' + str(i) + ' = ' + str(cs))

The results of this script are:

[[[0.0] [2.5432017412] [3.7228765674] [3.7339217013]]
 [[0.0] [0.0] [4.2771234426] [4.2660783187]]
 [[0.0] [0.0] [0.0] [0.0]]
 [[0.0] [0.0] [0.0] [0.0]]]
Row sum 0 = 10.000000009899999
Col sum 0 = 0.0
Row sum 1 = 8.5432017613
Col sum 1 = 2.5432017412
Row sum 2 = 0.0
Col sum 2 = 8.00000001
Row sum 3 = 0.0
Col sum 3 = 8.00000002

There is a slight violation of the constraints because of the solver tolerance. You can adjust the solver tolerance with m.options.ATOL=1e-6 to something that is more like 1e-8 if you'd like to have a more precise answer.

Upvotes: 1

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