CODEWITHSUNDEEP
pythonpython-3.xgekko
Roberto
Roberto

Reputation: 55

How to build and print in Python - GEKKO a list of optimized values generated by loop?

I have an Objective function with four parameters to be optimized (n1, n2, n3 and n4). This optimization should be done in a loop "for T in T" as the Objective function is a function of T. As a result I get for each T, a set of parameters. How can I build and print individualized lists containing T, n1, n2, n3 and n4?

I have tried different ways of "print", inside the loop and outside of it. I have also searched for examples here and in the GEKKO's documentation but it is still a challenge to me. Thanks in advance.

import math
import numpy as np
from gekko import GEKKO

T = np.arange(1, 10, 2)
for T in T:
    A = 3/(T**2)
    B = 20-T**2
    C = 3+T
    D = T

    mA = A-T*C
    mB = B-T*D

    # Minimization routine

    from gekko import GEKKO
    m = GEKKO()

    # Variables to be minimized:
    n1, n2, n3, n4 = [m.Var() for i in range(4)]
    var = [n1, n2, n3, n4]

    # Initial values:
    n0 = [3,2,3,3]
    nL = np.ones(len(n0))*10**-10
    nU = np.ones(len(n0))*10**10
    for i,x in enumerate(var):
        x.value = n0[i]
        x.lower = nL[i]
        x.upper = nU[i]

    nt = m.Intermediate(n1 + n2 + n3 + n4)

    # Objective:
    m.Obj(T*A*mA + B*mB + C*D/nt)
    # Set global options
    m.options.IMODE = 3

    # Solve minimization
    m.solve()

    print(n1, n2, n3, n4)
    print(str(n1.value), str(n2.value), str(n3.value), str(n4.value))

Upvotes: 1

Views: 777

Answers (2)

John Hedengren
John Hedengren

Reputation: 14346

There are multiple ways to export data from Python as shown here. You need to get the results into a 2 dimensional list, NumPy array, or Pandas data frame. The following are modifications to your original script if you need to get the values into a form that Excel can open, such as a CSV file. This script uses NumPy to write the file.

import math
import numpy as np
from gekko import GEKKO

T = np.arange(1, 10, 2)
z = [None]*len(T)
for j,T in enumerate(T):
    A = 3/(T**2)
    B = 20-T**2
    C = 3+T
    D = T

    mA = A-T*C
    mB = B-T*D

    # Minimization routine

    from gekko import GEKKO
    m = GEKKO()

    # Variables to be minimized:
    n1, n2, n3, n4 = [m.Var() for i in range(4)]
    var = [n1, n2, n3, n4]

    # Initial values:
    n0 = [3,2,3,3]
    nL = np.ones(len(n0))*10**-10
    nU = np.ones(len(n0))*10**10
    for i,x in enumerate(var):
        x.value = n0[i]
        x.lower = nL[i]
        x.upper = nU[i]

    nt = m.Intermediate(n1 + n2 + n3 + n4)

    # Objective:
    m.Obj(T*A*mA + B*mB + C*D/nt)
    # Set global options
    m.options.IMODE = 3

    # Solve minimization
    m.solve(disp=False)

    result = [T]
    for x in var:
        result.append(x.value[0])
    z[j] = result

zn = np.array(z)
np.savetxt('z.txt',zn,delimiter=',',comments='',header='T,n1,n2,n3,n4')

Result Table in Excel

Upvotes: 1

John Hedengren
John Hedengren

Reputation: 14346

Gekko is built for time-varying systems so the results are returned as a list. Steady state optimization only returns one value so you'll need to access the first element of the value property. Try accessing an individual result with n1.value[0]. The T value is a numpy array so you can access the values with T[0], etc. Try something like the following at the end of your program script:

    # Solve minimization
    m.solve(disp=False)

    result = [T]
    for x in var:
        result.append(x.value[0])
    print(result)

It creates an initial list with the value of T and then appends the other values. I set the option disp=False to not show the solver output. The modified script produces these results:

[1, 3457333826.4, 3457319622.4, 3457333826.4, 3457333826.4]
[3, 2482924773.0, 2482917643.6, 2482924773.0, 2482924773.0]
[5, 2201591127.3, 2201587887.6, 2201591127.3, 2201591127.3]
[7, 2105335286.6, 2105331976.7, 2105335286.6, 2105335286.6]
[9, 1630154729.7, 1630153615.7, 1630154729.7, 1630154729.7]

It appears that all of the parameter values are equal for this problem and they don't have a connection to the other equations except for calculation of nt and the objective term C*D/nt.

Upvotes: 1

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