I'm trying to solve a non-linear optimization problem. I've duplicated my issue by creating the code below. Python returns TypeError: object of type 'int' has no len(). How can I include an IF statement in my constraint functions?

Console prints the following:

  File "<ipython-input-196-8d29d410dcea>", line 1, in <module>runfile('C:/Users/***/Documents/***/Project/untitled.py', wdir='C:/Users/***/Documents/***/***/Project')File "C:\Users\***\Anaconda3\lib\site-packages\spyder_kernels\customize\spydercustomize.py", line 704, in runfileexecfile(filename, namespace)File "C:\Users\***\Anaconda3\lib\site-packages\spyder_kernels\customize\spydercustomize.py", line 108, in execfileexec(compile(f.read(), filename, 'exec'), namespace)File "C:/Users/***/Documents/***/***/Project/untitled.py", line 27, in <module>m.Equation(Cx(x1,x2,x3,x4) < 0)File "C:/Users/***/Documents/***/***/Project/untitled.py", line 17, in Cxif K > 15:File "C:\Users\***\Anaconda3\lib\site-packages\gekko\gk_operators.py", line 25, in __len__return len(self.value)File "C:\Users\***\Anaconda3\lib\site-packages\gekko\gk_operators.py", line 134, in __len__return len(self.value)TypeError: object of type 'int' has no len()

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from gekko import GEKKO
m = GEKKO()def Cr(x1,x2,x3,x4):return (x1*x4*(x1+x2+x3)+x3**2)def Cw(x1,x2,x3,x4):return x1*x2*x3*x4def Ck(x1,x2,x3,x4):return x1*x2*x3*x4+1def Cx(x1,x2,x3,x4):K = Ck(x1,x2,x3,x4)if K > 15:  #Issue hereK = 15return x1**2+x2**2+x3**2+x4**2 - Kx1 = m.Var(value=1,lb=-5000,ub=5000)
x2 = m.Var(value=1,lb=-5000,ub=5000)
x3 = m.Var(value=1,lb=-5000,ub=5000)
x4 = m.Var(value=1,lb=-5000,ub=5000)m.Equation(Cw(x1,x2,x3,x4) >= 14)
m.Equation(Cx(x1,x2,x3,x4) < 0)m.Obj(Cr(x1,x2,x3,x4))m.solve(disp=False)
print(x1.value)
print(x2.value)
print(x3.value)
print(x4.value)

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I am looking to have GEKKO run with the IF statement in the constraint, I'm not concerned if the optimization problem in the code has a solution. Thank you in advance.

(Disclaimer: i don't know this lib or what it will do for you)

The if-statement makes this problen non-differentiable which invalidates the assumption of NLP solvers (like Ipopt).

In terms of MINLP solvers (Bonmin, Couenne), this could be achieved by reformulation (and the resulting problem is differentiable when the needed auxiliary binary variables have been relaxed). It's hardly questionable to expect the lib to do it for you.

So it seems you need to play by the rules of some MINLP model like for example described by Bonmin here. There is no concept for "if-based branching".

Either introduce an indicator-variable like it's common in the MIP world, see here. Ignoring the overhead the idea would be something like:

K_ = Ck(x1,x2,x3,x4)
I = K_ > 15 (binary variable; see link for formulation idea)return x1**2+x2**2+x3**2+x4**2 - I*15 - (1-I) * K_

This is an MINLP then.

You might get away without using an additional binary variable (and touching MINLPs) when interpreting your equation like:

return x1**2+x2**2+x3**2+x4**2 - min(Ck(x1,x2,x3,x4), 15)

This is also non-differentiable, but can be easily reformulated (with a quirk) like:

return x1**2+x2**2+x3**2+x4**2 - A# extra constraints
A <= Ck(x1,x2,x3,x4)
A <= 15

if we could enforce moving towards the biggest A possible. This means, it must be part of the objective:

m.Obj(Cr(x1,x2,x3,x4) + c * A) (if it's a maximization problem)

This would be a NLP then, but the value of c needs some care (it must be big enough).