Gekko：在 Gekko Python 中不能按顺序使用逻辑条件吗？

如何解决Gekko：在 Gekko Python 中不能按顺序使用逻辑条件吗？

我在 Gekko 中编写了一个模型并遇到了错误。于是，我开始逐行检查，发现了一个我无法理解的问题。下面的代码（显示部分模型代码，输入文件link）可以执行，结果也正确。但是，当我取消注释第 31 行时，发生错误。是不是因为我按顺序使用了逻辑条件 (ab3,max3)？有什么办法可以解决这个问题吗？

import numpy as np
from gekko import GEKKO

# Read weather data
weatherData = np.load("Atlanta_TMY3_climate.npy")

# create GEKKO model
m = GEKKO(remote=False)
m.open_folder()
print(m.path)

weatherData = np.insert(weatherData,np.zeros((1,15)),0)

days_to_consider = 1
m.time = np.linspace(0,24*days_to_consider,24*days_to_consider+1)

# Declare parameters
focc = m.Param(value = [0,0.05,0.1,\
                        0.1,0.1],name="focc")

setpoint = [16.00,16.00,\
            16.00,16.38,17.43,17.88,18.08,18.35,\
            17.85,17.13,16.27,16.00]
T_air_set  = m.Param(value=setpoint,name="T_air_set")
Te         = m.Param(value = weatherData[0:24*days_to_consider+1,0],name="Te")
qv_infil_wind  = m.Param(value = 0.0769*0.6*(0.75*0.8* \
                         weatherData[0:24*days_to_consider+1,1]**2)**0.667,\
                         name="qv_infil_wind")

# 1) Airflow
qv_mech_sup     = m.Intermediate(2.519244226731981*focc*1,name="qv_mech_sup")

qv_infil_stack = m.Intermediate(0.0146*0.6*(0.7*8.5* \
                                m.abs3(Te-T_air_set))**0.667,\
                                name="qv_infil_stack")

# When below line code is uncommented,an error occurs.
##qv_infil_sw = m.Intermediate(m.max3(qv_infil_stack,qv_infil_wind) + \
##                             0.14*qv_infil_stack*qv_infil_wind/0.6,\
##                             name="qv_infil_sw")

m.options.IMODE = 6
m.options.DIAGLEVEL = 4
m.options.soLVER = 3
m.options.MAX_ITER = 1000
m.solve(disp=True,GUI=False)

解决方法

我改用 m.options.SOLVER=1 是因为 m.abs3() 和 m.max3() 函数需要混合整数非线性规划 (MINLP) 求解器。然而，这也没有解决问题。似乎每个二进制变量都有难以解决的组合。我不确定为什么，但这是一个求解器问题，而不是 Gekko 公式问题。这是一个简单的问题，它表明 m.abs3() 和 m.max3() 在同一模型中并作为依赖项不是问题。

from gekko import GEKKO
m = GEKKO()
x = m.Param(2)
y = m.Param(3)
z = m.abs3(x-y)
w = m.max3(z,x)
m.solve()
print(z.value[0],w.value[0])

切换到 m.abs() 有一个成功的解决方案。这避免了求解器收敛问题。我通常不建议使用 m.abs()，但所有因变量都是参数，因此非连续梯度不存在问题。

import numpy as np
from gekko import GEKKO
import matplotlib.pyplot as plt

# Read weather data
weatherData = np.load("Atlanta_TMY3_climate.npy")

# create GEKKO model
m = GEKKO(remote=False)
print(m.path)

weatherData = np.insert(weatherData,np.zeros((1,15)),0)

days_to_consider = 1
m.time = np.linspace(0,24*days_to_consider,24*days_to_consider+1)

# Declare parameters
focc = m.Param(value = [0,0.05,0.1,\
                        0.1,0.1],name="focc")

setpoint = [16.00,16.00,\
            16.00,16.38,17.43,17.88,18.08,18.35,\
            17.85,17.13,16.27,16.00]
T_air_set  = m.Param(value=setpoint,name="T_air_set")
Te         = m.Param(value = weatherData[0:24*days_to_consider+1,0],name="Te")
qv_infil_wind = m.Param(value = 0.0769*0.6*(0.75*0.8* \
                         weatherData[0:24*days_to_consider+1,1]**2)**0.667,\
                         name="qv_infil_wind")

# 1) Airflow
qv_mech_sup    = m.Intermediate(2.519244226731981*focc*1,name="qv_mech_sup")

qv_infil_stack = m.Intermediate(0.0146*0.6*(0.7*8.5* \
                                m.abs(Te-T_air_set))**0.667,\
                                name="qv_infil_stack")

mx = m.max3(qv_infil_stack,qv_infil_wind)    
qv_infil_sw = m.Intermediate(mx+0.14*qv_infil_stack*qv_infil_wind/0.6,\
                             name="qv_infil_sw")

m.options.IMODE = 6
m.options.DIAGLEVEL = 0
m.options.SOLVER = 1
m.options.MAX_ITER = 1000
m.solve(disp=True,GUI=False)

plt.plot(m.time,mx.value,color='orange',lw=2)
plt.plot(m.time,qv_infil_stack.value,'r:')
plt.plot(m.time,qv_infil_wind.value,'k:')
plt.show()

我知道这是问题的简化版本，但如果所有这些变量都基于常量 m.Param() 类型，您可以将所有这些变量预先计算为 Numpy 数组。这将避免隐式解决方案中额外的不等式约束和二元变量。