Question

有没有办法对一个变量建模，该变量是前一个时间步（t-1）的值和当前时间步（t）的值（例如，eq1) x(t) = (some equation), eq2) a(t) = (x(t-1) + x(t))/2）的平均值IMODE=6 在壁虎？

下面的简化代码显示了我在上面提出的问题（代码本身（在物理上）没有意义，但代码从原始代码简化为以更简洁的方式提出问题）。

import numpy as np
from gekko import GEKKO

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

m.time = np.linspace(0,5,6)

# Variables
T_air_set_prev = m.Var(value=10, name="T_air_set_prev") # previous -> previous time step (t-1)
T_m0 = m.Var(value=15, name="T_m0")
T_m_ac_t = m.Var(value=25, name="T_m_ac_t") # t -> current time step (t)
T_m_ac = m.Var(value=25, name="T_m_ac") 
Q_heating =  m.Var(value=0, name="Q_heating")

actuator = m.MV(value=6, lb=3, ub=9, name="actuator")
actuator.STATUS = 1
actuator.DMAX = 1.5

# Equations
m.Equations([T_m0 == 0.5*T_air_set_prev + 1,\
             T_m_ac_t == 0.7* T_m0 + 10,\
             T_m_ac == (T_air_set_prev + T_m_ac_t)/2,\
             Q_heating == 5*T_m_ac+20])

# This line is meant for saving the previous time step's value, but I don't know the correct way to implement this.
T_air_set_prev = m.Intermediate(T_m_ac_t)

m.Obj(0.5*Q_heating + actuator)

# Solve
m.options.IMODE = 6
m.options.DIAGLEVEL = 4
m.options.MAX_ITER = 1000
m.options.SOLVER = 3
m.solve(disp=True)

Answer 1

尝试将 Gekko 中的 delay 函数用作 shown in the documentation。数字 1 是多少时间步长（不是时间），所以这应该是您所需要的。

m.delay(T_m_ac_t,T_air_set_prev,1)

这是完整的脚本。

import numpy as np
from gekko import GEKKO

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

m.time = np.linspace(0,5,6)

# Variables
# previous -> previous time step (t-1)
T_air_set_prev = m.Var(value=10, name="T_air_set_prev") 
T_m0 = m.Var(value=15, name="T_m0")
# t -> current time step (t)
T_m_ac_t = m.Var(value=25, name="T_m_ac_t") 
T_m_ac = m.Var(value=25, name="T_m_ac") 
Q_heating =  m.Var(value=0, name="Q_heating")

actuator = m.MV(value=6, lb=3, ub=9, name="actuator")
actuator.STATUS = 1
actuator.DMAX = 1.5

# Equations
m.Equations([T_m0 == 0.5*T_air_set_prev + 1,\
             T_m_ac_t == 0.7* T_m0 + 10,\
             T_m_ac == (T_air_set_prev + T_m_ac_t)/2,\
             Q_heating == 5*T_m_ac+20])

# Time delay
m.delay(T_m_ac_t,T_air_set_prev,1)

m.Obj(0.5*Q_heating + actuator)

# Solve
m.options.IMODE = 6
m.options.DIAGLEVEL = 0
m.options.MAX_ITER = 1000
m.options.SOLVER = 3
m.solve(disp=True)

Answer 2

您可能想在 GEKKO 中使用 time delay 函数，m.delay。基本上，添加一个新的 gekko 变量，其中 m.delay() 移动一个时间步长，然后您可以像往常一样在等式中使用它。

T_m_ac_t = m.Var(value=25, name="T_m_ac_t") # t -> current time step (t)
T_air_set_prev = m.Var(value=10, name="T_air_set_prev") # previous -> previous time step (t-1)

m.delay(T_m_ac_t, T_air_set_prev, 1)

查看 apmonitor 页面中的示例：Tiem Delay (Dead-Time)

Gekko：计算一个变量，该变量是前一个和当前时间步长值的函数

2 个答案: