中间变量条件的 Gekko 和

Question

我有一个简单的优化问题，我试图优化潜在客户的数量，每个潜在客户都有一定的成本（这取决于潜在客户的数量，例如 1 个潜在客户可能需要 1 美元，而 2 个潜在客户可能需要花费每个 1.5 美元）。使用以下函数计算每个潜在客户的成本。 cmets 是我为解决不同问题所做的尝试。

# Here x is leads and returns spend
def objective_inverse(x, a, b, c):
     if c == 0:
         return 0
     else:
         return np.real(a + (x/c)**(1./b))

这里是变量reg_feats，其中列表中的每个元组都是前面函数中的(a,b,c)。

reg_feats = [(0.0, 1.0, 0.0008237661018136143), (0.9999999999999811, 0.8536915943397881, 0.004454688129841911), (0.0, 1.0, 0.0004855869694355375), (0.0, 1.0, 0.00038427101866404334), (8.308985001840891e-10, 1.097417548409198, 0.0009170956872015353), (0.999999999998488, 1.9999999897072056, 1.144122017534284e-07), (0.9999999999825329, 0.8097037523302283, 0.20606857088111075), (0.9999999999999963, 1.1649279015402045, 0.009316713936903972), (0.0, 1.0, 0.0034430519212229715), (0.0, 1.0, 0.0007980573950249244), (0.0, 1.0, 0.0009069008844368589)]

并且每个变量的边界都有一个列表。

bounds = [(0, 4), (0, 20), (0, 3), (0, 1), (0, 114), (0, 14), (0, 208), (0, 529), (0, 1), (0, 4), (0, 4)]

这是优化的代码：

m = GEKKO(remote=not bool(i))

m.options.SOLVER=1
# optional solver settings with APOPT
m.solver_options = ['minlp_maximum_iterations 50000', \
                # minlp iterations with integer solution
                'minlp_max_iter_with_int_sol 100', \
                # treat minlp as nlp
                'minlp_as_nlp 0', \
                # nlp sub-problem max iterations
                'nlp_maximum_iterations 500', \
                # 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']

这里的 x 是每个广告系列的潜在客户数量。

x = m.Array(m.Var, (len(bounds)),integer=True)
# Here x is the number of leads and is the decision variable
# x variables are nb of leads, not $ to be allocated
for i, xi in enumerate(x):
    lb, ub = bounds[i]
    xi.lower = lb
    xi.upper = ub
    xi.value = ub

这个中间变量是获取每个潜在客户的成本。这用于下一个约束。

sums = [m.Intermediate(objective_inverse(xi,*y)) for xi,y in zip(x,reg_feats)]

max_budget = 150000
m.Maximize(m.sum(x))
c = m.Const(max_budget, 'Budget')
m.Equation(m.sum(sums) < c)
m.solve(disp=True)

在这里，如果我运行此代码，我会收到一条错误消息，提示找不到解决方案和Warning: no more possible trial points and no integer solution

此外，如果我删除 max_budget 约束，我会得到一个总花费低于 max_budget 的解决方案，这意味着添加约束不应该改变解决方案。

我知道我如何定义约束存在问题，但不知道如何解决。

Answer 1

问题似乎不在于约束，而在于x 的起点。当我运行您的代码时，我得到了解决方案（我没有收到任何错误）。但是当我将max_budget 更改为100000 时开始出现同样的错误。我怀疑这是因为 APPT 需要一个好的初始猜测。由于我没有很好的猜测，所以我使用np.random 来生成起点。我尝试了几个不同的 max_budget 并且它可以工作（尽管尝试了几次才能获得一个好的初始点）。

import numpy as np
from gekko import GEKKO

# Here x is leads and returns spend
def objective_inverse(x, a, b, c):
     if c == 0:
         return 0
     else:
         return np.real(a + (x/c)**(1./b))

reg_feats = [(0.0, 1.0, 0.0008237661018136143), (0.9999999999999811, 0.8536915943397881, 0.004454688129841911), (0.0, 1.0, 0.0004855869694355375), (0.0, 1.0, 0.00038427101866404334), (8.308985001840891e-10, 1.097417548409198, 0.0009170956872015353), (0.999999999998488, 1.9999999897072056, 1.144122017534284e-07), (0.9999999999825329, 0.8097037523302283, 0.20606857088111075), (0.9999999999999963, 1.1649279015402045, 0.009316713936903972), (0.0, 1.0, 0.0034430519212229715), (0.0, 1.0, 0.0007980573950249244), (0.0, 1.0, 0.0009069008844368589)]
bounds = [(0, 4), (0, 20), (0, 3), (0, 1), (0, 114), (0, 14), (0, 208), (0, 529), (0, 1), (0, 4), (0, 4)]
m = GEKKO(remote=False)

m.options.SOLVER=1
# optional solver settings with APOPT
m.solver_options = ['minlp_maximum_iterations 50000', \
                # minlp iterations with integer solution
                'minlp_max_iter_with_int_sol 100', \
                # treat minlp as nlp
                'minlp_as_nlp 0', \
                # nlp sub-problem max iterations
                'nlp_maximum_iterations 500', \
                # 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']

x = m.Array(m.Var, (len(bounds)),integer=True)
# Here x is the number of leads and is the decision variable
# x variables are nb of leads, not $ to be allocated
for i, xi in enumerate(x):
    lb, ub = bounds[i]
    xi.lower = lb
    xi.upper = ub
    xi.value = np.random.randint(lb,ub)

print("Initial Value: ",x)

sums = [m.Intermediate(objective_inverse(xi,*y)) for xi,y in zip(x,reg_feats)]

max_budget = 90000
# max_budget = 100000
# max_budget = 120000
# max_budget = 150000
m.Maximize(m.sum(x))
c = m.Const(max_budget, 'Budget')
m.Equation(m.sum(sums) < c)
m.solve(disp=True)


x_ans =[i[0] for i in x]
print('x : ', x_ans)
print('Constraint: ', np.sum([(objective_inverse(xi,*y)) for xi,y in zip(x_ans,reg_feats)]))

输出：

Initial Value:  [2 7 0 0 113 5 205 151 0 3 2]
 ----------------------------------------------------------------
 APMonitor, Version 0.9.2
 APMonitor Optimization Suite
 ----------------------------------------------------------------


 --------- APM Model Size ------------
 Each time step contains
   Objects      :  1
   Constants    :  1
   Variables    :  24
   Intermediates:  11
   Connections  :  12
   Equations    :  24
   Residuals    :  13

 Number of state variables:    24
 Number of total equations: -  13
 Number of slack variables: -  1
 ---------------------------------------
 Degrees of freedom       :    10

 ----------------------------------------------
 Steady State Optimization with APOPT Solver
 ----------------------------------------------
Iter:     1 I:  0 Tm:      0.00 NLPi:   12 Dpth:    0 Lvs:    2 Obj: -8.85E+02 Gap:       NaN
Iter:     2 I:  0 Tm:      0.00 NLPi:    3 Dpth:    1 Lvs:    3 Obj: -8.85E+02 Gap:       NaN
Iter:     3 I:  0 Tm:      0.00 NLPi:    3 Dpth:    2 Lvs:    4 Obj: -8.85E+02 Gap:       NaN
Iter:     4 I:  0 Tm:      0.00 NLPi:    3 Dpth:    3 Lvs:    5 Obj: -8.85E+02 Gap:       NaN
Iter:     5 I:  0 Tm:      0.00 NLPi:    3 Dpth:    4 Lvs:    6 Obj: -8.84E+02 Gap:       NaN
Iter:     6 I:  0 Tm:      0.00 NLPi:    3 Dpth:    5 Lvs:    7 Obj: -8.84E+02 Gap:       NaN
--Integer Solution:  -8.84E+02 Lowest Leaf:  -8.85E+02 Gap:   8.39E-04
Iter:     7 I:  0 Tm:      0.00 NLPi:    2 Dpth:    6 Lvs:    7 Obj: -8.84E+02 Gap:  8.39E-04
 Successful solution

 ---------------------------------------------------
 Solver         :  APOPT (v1.0)
 Solution time  :  0.0677 sec
 Objective      :  -884.
 Successful solution
 ---------------------------------------------------


x :  [0.0, 18.0, 0.0, 0.0, 114.0, 14.0, 208.0, 529.0, 1.0, 0.0, 0.0]
Constraint:  89187.87455034524

我知道，如果没有良好的初始猜测，对于更大的问题来说，随机生成起点可能是不可行的。但不幸的是，我不知道在这种情况下如何使用 APPT。