具有自定义函数的Gekko优化约束

Question

我目前正试图用Gekko的分支来解决一个混合整数非线性问题

以下是我迄今为止的尝试：

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

有一个未定义的函数，阻止代码运行。从代码的快速扫描中，可以看到如下表达式：

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

不允许，因为gekko要求在m.solve（）命令之前定义所有方程。不允许在Python中回调函数。相反，应该使用二进制变量来打开或关闭优化问题中的某些内容。这可以是一个等式，例如二进制变量b：

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

如果x小于零，则b的值等于0，如果x大于零，则b的值等于1。APMonitor留档中有一个关于if3（）函数的教程也可能很有用。