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我正在研究一个整数优化问题，并想尝试 GEKKO。

问题描述：x1,x2,x3,...,x9,x10,x11（它们是 1-16 范围内的整数）是目标函数的十一个整数参数。我想找到一组 x 值来最小化目标函数的输出。然而，目标值是通过运行另一个 C/C++ 程序获得的，因为该问题不能用数学表示。像你的例子一样的公式。

我怀疑目标函数的返回值 j 是否与规范不匹配。在这种情况下 j python float 可以吗？但从日志和复查来看，init 状态的客观值是 0.06，这是正确的。我不知道为什么它只运行一次迭代然后停止。

def f_per_particle(x1,x2,x3,x4,x5,x6,x7,x8,x9,x10,x11):
    inputs_list = [x1.value,x2.value,x3.value,x4.value,x5.value,x6.value,x7.value,x8.value,x9.value,x10.value,x11.value]
    #Use the inputs_list as input to run another SW program 
    #The SW program will return an objective value j which is float value
    #The goal is to find min. objective value with corresponding x1-x11 value
    return j



m = GEKKO() # Initialize 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']

# Initialize variables
# Integer constraints for x1 to x11
x1 = m.Var(value=8,lb=1,ub=16,integer=True)
x2 = m.Var(value=8,lb=1,ub=16,integer=True)
x3 = m.Var(value=8,lb=1,ub=16,integer=True)
x4 = m.Var(value=8,lb=1,ub=16,integer=True)
x5 = m.Var(value=8,lb=1,ub=16,integer=True)
x6 = m.Var(value=8,lb=1,ub=16,integer=True)
x7 = m.Var(value=8,lb=1,ub=16,integer=True)
x8 = m.Var(value=8,lb=1,ub=16,integer=True)
x9 = m.Var(value=8,lb=1,ub=16,integer=True)
x10 = m.Var(value=8,lb=1,ub=16,integer=True)
x11 = m.Var(value=8,lb=1,ub=16,integer=True)

m.Obj(f_per_particle(x1,x2,x3,x4,x5,x6,x7,x8,x9,x10,x11)) # Objective
m.solve() # Solve

Execution result as follows :
apm 130.226.87.11_gk_model0 <br><pre> ----------------------------------------------------------------
 APMonitor, Version 0.9.1
 APMonitor Optimization Suite
 ----------------------------------------------------------------


 --------- APM Model Size ------------
 Each time step contains
   Objects      :            0
   Constants    :            0
   Variables    :           11
   Intermediates:            0
   Connections  :            0
   Equations    :            1
   Residuals    :            1

 ________________________________________________
 WARNING: objective equation           1 has no variables
 ss.Eqn(1)
 0 = 0.06
 ________________________________________________
 Number of state variables:             11
 Number of total equations: -            0
 Number of slack variables: -            0
 ---------------------------------------
 Degrees of freedom       :             11

 ----------------------------------------------
 Steady State Optimization with APOPT Solver
 ----------------------------------------------
Iter:     1 I:  0 Tm:      0.00 NLPi:    1 Dpth:    0 Lvs:    0 Obj:  6.00E-02 Gap:  0.00E+00
 Successful solution

 ---------------------------------------------------
 Solver         :  APOPT (v1.0)
 Solution time  :   1.410000000032596E-002 sec
 Objective      :   6.000000000000000E-002
 Successful solution
 ---------------------------------------------------

Results
x1: [8.0]
x2: [8.0]
x3: [8.0]
x4: [8.0]
x5: [8.0]
x6: [8.0]
x7: [8.0]
x8: [8.0]
x9: [8.0]
x10: [8.0]
x11: [8.0]
Objective: 0.06
(virtual-env) nan@TEK-CB-NAN-01:~/Swarm_Sandbox/GEKKO/mlp_fi_fwl/run$
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1 回答 1

1

这是您的问题的完整但简化的版本：

from gekko import GEKKO
def f(x1,x2):
    return (x1.value-3.3)**2 + (x2.value-5.7)**2
m = GEKKO() # Initialize gekko
m.options.SOLVER=1  # APOPT is an MINLP solver
x1 = m.Var(value=8,lb=1,ub=16,integer=True)
x2 = m.Var(value=8,lb=1,ub=16,integer=True)
m.Obj(f(x1,x2)) # Objective
m.solve() # Solve
print('Objective: ' + str((8-3.3)**2 + (8-5.7)**2))
print(x1.value[0],x2.value[0])

它会产生一个类似于您的解决方案，其中求解器不会从初始猜测中移动值。这是因为您的目标函数返回的是标量，而不是 Gekko 表达式。Gekko 需要一个表达式，以便它可以执行运算符重载，以便为求解器提供自动微分，并以稀疏形式提供精确的一阶和二阶导数。您需要修改代码以使用cspline或bspline 之类的东西来提供目标函数的函数逼近，基于对一系列变量的目标函数进行采样，或者使用 Gekko 表达式。这是使用 Gekko 表达式的修改版本：

from gekko import GEKKO
def f(x1,x2):
    return (x1-3.3)**2 + (x2-5.7)**2
m = GEKKO() # Initialize gekko
m.options.SOLVER=1  # APOPT is an MINLP solver
x1 = m.Var(value=8,lb=1,ub=16,integer=True)
x2 = m.Var(value=8,lb=1,ub=16,integer=True)
m.Obj(f(x1,x2)) # Objective
m.solve() # Solve
print(x1.value[0],x2.value[0])

它产生正确的解决方案：x1=3.0和x2=6.0。

 ----------------------------------------------
 Steady State Optimization with APOPT Solver
 ----------------------------------------------
Iter:     1 I:  0 Tm:      0.00 NLPi:    4 Dpth:    0 Lvs:    3 Obj:  2.05E-27 Gap:       NaN
--Integer Solution:   1.80E-01 Lowest Leaf:   2.05E-27 Gap:   1.80E-01
Iter:     2 I:  0 Tm:      0.00 NLPi:    1 Dpth:    1 Lvs:    2 Obj:  1.80E-01 Gap:  1.80E-01
Iter:     3 I:  0 Tm:      0.00 NLPi:    1 Dpth:    1 Lvs:    2 Obj:  9.00E-02 Gap:  1.80E-01
Iter:     4 I:  0 Tm:      0.00 NLPi:    1 Dpth:    1 Lvs:    1 Obj:  4.90E-01 Gap:  1.80E-01
--Integer Solution:   1.80E-01 Lowest Leaf:   5.80E-01 Gap:  -4.00E-01
Iter:     5 I:  0 Tm:      0.00 NLPi:    1 Dpth:    2 Lvs:    1 Obj:  5.80E-01 Gap: -4.00E-01
 Successful solution

 ---------------------------------------------------
 Solver         :  APOPT (v1.0)
 Solution time  :   1.390000000174041E-002 sec
 Objective      :   0.180000000000000     
 Successful solution
 ---------------------------------------------------
于 2019-11-01T13:59:46.693 回答