Why does GEKKO return a decimal value for an integer variable in my optimization code?

Why does GEKKO optimisation return a decimal value for an Integer variable?

Hello everyone,
With my GEKKO code I want to optimise a cutting pattern, with the maximum surface area of rectangles in the area of a circle.

All my variables are predefined and are integer:

• diam: 1 diameter from the list
• ep and larg: 1 thickness and 1 width linked together by dictionary
• pcs: number of rectangles pieces, between 1 and 14 and also integer.

I could optimise my problem without Gekko, by testing all the possibilities with nested loops, but I want to add new variables afterwards, which would lead to too many possibilities for the loops.

I therefore use GEKKO with the principle of binary decision variables for almost all my variables.

from gekko import GEKKO 
m = GEKKO(remote=False)

#define diameter of circle
diam = [205,225,237,255,269,286,295,305]
xdiam = [m.Var(lb=0, ub=1, integer=True) for _ in diam]
m.Equation(sum(xdiam) == 1)

#define numbers of pieces
pcs = m.Var(lb=1, ub=14, integer=True, value=10)
#define thickness et width
dic = {
    12:[75,95,100],
    15:[95,100,150],
    10:[100,135,120]}
pair = []
for cle in dic:
    for val in dic[cle]:
        xpair = m.Var(lb=0,ub=1, integer=True)
        pair.append((cle,val, xpair))
m.Equation(sum(xl[2] for xl in pair) == 1)

#constraint : diagonal of all rectangles (hypotenuse) 
#  must not exceed the diameter
pyth = m.Intermediate((((sum((e*xe) for e,_,xe in pair))*pcs \
  +((pcs-1)*3.5))**2+(sum((l*xl) for _,l,xl in pair))**2)**0.5)
m.Equation(pyth <= sum((d*xd) for d, xd in zip(diam, xdiam)))

#Maximise area of rectangles/area of circle
rend = m.Intermediate(((sum((l*xl) for _,l,xl in pair))*(sum((e*xe) \
  for e,_,xe in pair))*pcs)/((((sum((d*xd) \
  for d, xd in zip(diam, xdiam)))*0.5)**2)*3.14))
m.Maximize(rend)

#Solver
m.options.SOLVER = 3
m.options.MAX_ITER = 1000 
m.solve()

#Display results
for ep,larg,bin in pair:
    if float(bin.value[0]) > 0.90:
        print("thi",ep)
        print("wid",larg)
for diam, xdiam in zip(diam,xdiam):
    if xdiam.VALUE[0] > 0.9:
        print("diam",diam)
print("pcs",pcs.VALUE[0])
print("yield",rend.VALUE[0])
print("pyth",pyth.VALUE[0])

And I don't know why the optimal solution proposed for the number of pieces is not an integer value?

In this example the pcs result is pretty close to an integer value, but by modifying the sections in the dictionary I can come up with values like : pcs = 5.820173.

 The solution was found.
The final value of the objective function is  -0.5289469114205082
---------------------------------------------------
Solver         :  IPOPT (v3.12)
Solution time  :  0.0211 sec
Objective      :  -0.5289469114205082
Successful solution
---------------------------------------------------
thi 15
wid 150
diam 205
pcs 7.9775445419
yield 0.52894691142
pyth 205.00005502

Thank you in advance

The correct solver needs to be selected. It is currently IPOPT with m.options.SOLVER = 3 as a Nonlinear Programming (NLP) solver that does not attempt to find an integer solution. However, the IPOPT solver is very good at finding an initial guess for the APOPT solver as a Mixed Integer Nonlinear Programming (MINLP) solver. By default, the minlp_integer_tol (integer tolerance) is 0.01. This means that 0.99 and 1.01 are both considered integer solutions. This tolerance can be adjusted to a more stringent requirement such as 1e-6. Try this sequence of solutions to find a more exact integer solution:

# Initialize with IPOPT (NLP Solver)
m.options.SOLVER = 3
m.solve()

# Find Integer solution with APOPT (MINLP Solver)
m.options.SOLVER = 1
m.solver_options = ['minlp_gap_tol 1.0e-4',\
                    'minlp_integer_tol 1.0e-6',\
                    'minlp_max_iter_with_int_sol 500']
m.solve()

The solution is:

 Successful solution
---------------------------------------------------
Solver         :  APOPT (v1.0)
Solution time  :   3.590000000258442E-002 sec
Objective      :  -0.510288956224352     
Successful solution
---------------------------------------------------
thi 15
wid 150
diam 237
pcs 10.0
yield 0.51028895622
pyth 235.46178034

Here is the complete script:

from gekko import GEKKO 
m = GEKKO(remote=True)

#define diameter of circle
diam = [205,225,237,255,269,286,295,305]
xdiam = [m.Var(lb=0, ub=1, integer=True) for _ in diam]
m.Equation(sum(xdiam) == 1)

#define numbers of pieces
pcs = m.Var(lb=1, ub=14, integer=True, value=10)
#define thickness et width
dic = {
    12:[75,95,100],
    15:[95,100,150],
    10:[100,135,120]}
pair = []
for cle in dic:
    for val in dic[cle]:
        xpair = m.Var(lb=0,ub=1, integer=True)
        pair.append((cle,val, xpair))
m.Equation(sum(xl[2] for xl in pair) == 1)

#constraint : diagonal of all rectangles (hypotenuse) 
#  must not exceed the diameter
pyth = m.Intermediate((((sum((e*xe) for e,_,xe in pair))*pcs \
  +((pcs-1)*3.5))**2+(sum((l*xl) for _,l,xl in pair))**2)**0.5)
m.Equation(pyth <= sum((d*xd) for d, xd in zip(diam, xdiam)))


#Maximise area of rectangles/area of circle
rend = m.Intermediate(((sum((l*xl) for _,l,xl in pair))*(sum((e*xe) \
  for e,_,xe in pair))*pcs)/((((sum((d*xd) \
  for d, xd in zip(diam, xdiam)))*0.5)**2)*3.14))
m.Maximize(rend)

m.options.MAX_ITER = 1000 

# Initialize with IPOPT (NLP Solver)
m.options.SOLVER = 3
m.solve()

# Find Integer solution with APOPT (MINLP Solver)
m.options.SOLVER = 1
m.solver_options = ['minlp_gap_tol 1.0e-4',\
                    'minlp_integer_tol 1.0e-6',\
                    'minlp_max_iter_with_int_sol 500']
m.solve()

#Display results
for ep,larg,bin in pair:
    if float(bin.value[0]) > 0.90:
        print("thi",ep)
        print("wid",larg)
for diam, xdiam in zip(diam,xdiam):
    if xdiam.VALUE[0] > 0.9:
        print("diam",diam)
print("pcs",pcs.VALUE[0])
print("yield",rend.VALUE[0])
print("pyth",pyth.VALUE[0])