带有自定义 function 的 Gekko 优化约束

Question

I am currently trying to solve a Mixed Integer Non Linear Problem with Gekko using its Branch&Bound implementation coupled with its warm-start method to speed up and improve the convergence process compared to vanilla branch& bound. The algorithm finds a solution after a short amount of time. Nevertheless, it hurts the constraint, which I might have defined wrongly: I have a gekko array-variable X and need another gekko array-variable "indices_open" that saves every index of x where x == 1. This "indices_open" goes into another self-defined function which is expecting "indices_open" as an numpy array and does not accept a list or gekko-array of gekko-intermediate variables. The self-defined function returns a numpy array. This final array shall be used in m.Equations and I therefore cast it to a gekko variable array. Needless to say, something went wrong and the current solution hurts the inequality constraint, while the equality constraint is met. While analyzing the result, I came to the conclusion that "indices_open" seems not to have updated in each iteration.

In the following my try so far:

m = GEKKO()
m.options.SOLVER = 1  # APOPT is an MINLP solver
# optional solver settings with APOPT
m.solver_options = ['minlp_maximum_iterations 500', \
                    # minlp iterations with integer solution
                    'minlp_max_iter_with_int_sol 10', \
                    # treat minlp as nlp
                    'minlp_as_nlp 0', \
                    # nlp sub-problem max iterations
                    'nlp_maximum_iterations 50', \
                    # 1 = depth first, 2 = breadth first
                    'minlp_branch_method 1', \
                    # maximum deviation from whole number
                    'minlp_integer_tol 0.05', \
                    # covergence tolerance
                    'minlp_gap_tol 0.01']

#Declare x
x = m.Array(m.Var,(65),lb=0,ub=1,integer=True)
for i, xi in enumerate(x[0:65]):
    xi.value = np.random.choice(np.arange(0, 2), 1, p=[0.4, 0.6])[0]
#constr
m = ineq_constraint_new(x, m)
m = eq_constraint_new(x, m)

#target
m = objective(x,m)

#STArt
start_time = time.time()
#m.solve(disp=False)
m.solve()
print('Objective: ' + x)
print('Objective: ' + str(m.options.objfcnval))

# save x
m.x = [x[j].value[0] for j in range(65)]


def eq_constraint_new(x, m):

    mask = np.isin(list_unique, specific_value)
    indices_fixed = np.nonzero(mask)[0]
    m.Equations([x[j] == 1 for j in indices_fixed])

    return m

def ineq_constraint_new(x, m):

    indices_open = [j for j in range(65) if x[j].value == 1]
    # DOes not work
    #indices_open_banks = [m.Intermediate(j) for j in range(65) if x[j].value == 1]
    array_perc, _, _,_ = self_defined_f(indices_open, some_value)

    #convert to gekko variables
    gekko_vec_perc_upper_bound = m.Array(m.Var, (65))
    for i, xi in enumerate(gekko_vec_perc_upper_bound[0:65]):
        xi.value = some_array[i]

    gekko_arr_perc = m.Array(m.Var, (65))
    for i, xi in enumerate(gekko_arr_perc[0:65]):
        xi.value = arr_perc[i]

    diff = gekko_vec_perc_upper_bound - gekko_arr_perc

    m.Equations([diff[j] >= 0 for j in range(65)])
    
    return m

def objective(x,m):

    
    indices_open = [j for j in range(65) if x[j].value == 1]

    
    _, arr_2, arr_3, arr_4 = self_defined_f(indices_open,some_value )

    # intermediates for objective
    res_dist = [None] * self.ds.n_banks
    res_wand = [None] * self.ds.n_banks
    res_wand_er = [None] * self.ds.n_banks
    
    x_closed = np.array([1]*len(x)) - x

    for j in range(self.ds.n_banks):
        res_dist[j] = m.Intermediate(arr_2[j] * some_factor  )
        res_wand[j] = m.Intermediate(arr_3[j] * some_factor)
        res_wand_er[j] = m.Intermediate(arr_4[j] * some_factor)

    res_sach = some_factor * (some_vector * x_closed)

    # Will be added together
    m.Minimize(sum(res_dist))
    m.Minimize(sum(res_wand))
    m.Minimize(sum(res_wand_er))
    m.Maximize(sum(res_sach))

    return m

Answer 1

There is an undefined function _, arr_2, arr_3, arr_4 = self_defined_f(indices_open,some_value ) that prevents the code from running. From a quick scan of the code, an expression like:

indices_open = [j for j in range(65) if x[j].value == 1]

is not allowed because gekko requires that the equations are all defined before the m.solve() command. A callback to a function in Python is not allowed. Instead, binary variables should be used to turn something On or Off in the optimization problem. This can be an equation such as binary variable b :

b = m.Var(lb=0,ub=1,integer=True)
m.Equation(x*(1-b)<=0)
m.Equation(x*b>=0)

This makes the value of b equal to 0 if x is less than zero and b equal to 1 if x if greater than zero. There is a tutorial on if3() functions in theAPMonitor documentation that may also be useful.