CODEWITHSUNDEEP
pythonnonlinear-optimizationgekko
Rojer Vairs
Rojer Vairs

Reputation: 97

Gekko returning a clearly non-optimal solution for a simple quadratic problem?

Given the program:

from gekko import GEKKO
m = GEKKO()  

tlX = m.Var()
tlY = m.Var()

m.Equation(m.sqrt(tlX*tlX + tlY*tlY)  <= 31 )

m.Maximize(tlX * tlY)

m.solve() 

print("solution: ", tlX.value, tlY.value )

Returns:

EXIT: Optimal Solution Found.
 The final value of the objective function is   0.000000000000000E+000
[0.0] [0.0]

Which is clearly not optimal. As tlX=1, tlY=1 is closer to optimal. What am I missing?

Upvotes: 2

Views: 150

Answers (6)

John Hedengren
John Hedengren

Reputation: 14376

The others that responded are correct that any non-zero initial guess will solve the problem. Even setting a small non-zero value for one of the variables resolves the issue.

tlX = m.Var(value=1e-5)

The problem is that the local optimizers find that the initial guess of [0,0] satisfies the Karush-Kuhn Tucker conditions. Because they are local optimizers, they don't check that the Hessian (matrix of partial 2nd derivatives) is positive definite for a minimum or negative definite for a maximum.

KKT conditions

A couple things to consider for global solutions are listed on this page. Here is an example with your problem.

from gekko import GEKKO
from hyperopt import fmin, tpe, hp
from hyperopt import STATUS_OK, STATUS_FAIL

# Define the search space for the hyperparameters
space = {'tlX': hp.quniform('tlX', 0, 1, 0.3),
         'tlY': hp.quniform('tlY', 0, 1, 0.3)}

def objective(params):
    m = GEKKO(remote=False)
    x = m.Array(m.Var,2,lb=0)
    tlX,tlY = x
    tlX.value = params['tlX']
    tlY.value = params['tlY']
    m.Equation(m.sqrt(tlX*tlX + tlY*tlY)  <= 31 )
    m.Maximize(tlX * tlY)
    m.solve(disp=False,debug=False)
    obj = m.options.objfcnval
    if m.options.APPSTATUS==1:
        s=STATUS_OK
    else:
        s=STATUS_FAIL
    m.cleanup()
    return {'loss':-obj, 'status': s, 'x':x}

best = fmin(objective, space, algo=tpe.suggest, max_evals=10)
sol = objective(best)
print(f"Solution Status: {sol['status']}")
print(f"Objective: {sol['loss']:.2f}")
print(f"Solution: {sol['x']}")

Here is the solution:

Solution Status: ok
Objective: 480.50
Solution: [[21.920310221] [21.920310221]]

Upvotes: 2

Motors
Motors

Reputation: 21

You are missing a starting point away from (0,0), which has the unfortunate condition of not having any particularly good search direction

from gekko import GEKKO
m = GEKKO()  


tlX = m.Var(value=1e-5)
tlY = m.Var()

m.Equation(m.sqrt(tlX*tlX + tlY*tlY)  <= 31 )


m.Maximize(tlX * tlY)

m.solve() 


print("solution: ", tlX.value, tlY.value )

Upvotes: 2

Konnor Mueller
Konnor Mueller

Reputation: 11

GEKKO is unable to move from the initial guess variable of 0. In this case set an arbitrary none-zero value this is within the bounds of your problem. Example code for a guess value of 5 for each variable is below.

from gekko import GEKKO
m = GEKKO()  


tlX = m.Var(value=5)
tlY = m.Var(value=5)

m.Equation(m.sqrt(tlX*tlX + tlY*tlY)  <= 31 )


m.Maximize(tlX * tlY)

m.solve(disp=False) 


print("solution: ", tlX.value, tlY.value ) ```

Upvotes: 1

Erik Hobson
Erik Hobson

Reputation: 11

t1X = m.Var(1)

just give an initial guess.

Upvotes: 1

Tyler J Crop
Tyler J Crop

Reputation: 11

Initialize a guess value for the variables tlX and tlY. x = 0, y = 0 is a minimum and the solver gets stuck at the minimum.

tlX = m.Var(value = 0.5)
tlY = m.Var(value = 0.5)

It outputs the answer: x=21.920310224, y= 21.920310224

Upvotes: 1

Keane Stewart
Keane Stewart

Reputation: 21

I got a solution by including bounds on the variables

from gekko import GEKKO
m = GEKKO()  


tlX = m.Var(lb=0,ub=31)
tlY = m.Var(lb=0,ub=31)

m.Equation(m.sqrt(tlX*tlX + tlY*tlY)  <= 31 )


m.Maximize(tlX * tlY)

m.solve() 


print("solution: ", tlX.value, tlY.value )

Upvotes: 1

Related Questions