GEKKO RTO 与 MPC 模式

Question

这是从这个 . After posting my question I found a solution (more like a patch to force the optimizer to optimize). There is something that baffles me. John Hedengren 派生的问题，它正确地指出 ODE 中的 b=1.0 导致 IMODE=6 的不可行解。然而，在我对 IMODE=3 的零碎工作中，我确实找到了解决方案。

我正在阅读 GEKKO's IMODE=3 和 6 的文档，但我并不清楚

IMODE=3

RTO Real-Time Optimization (RTO) is a steady-state mode that allows decision variables (FV or MV types with STATUS=1) or additional variables in excess of the number of equations. An objective function guides the selection of the additional variables to select the optimal feasible solution. RTO is the default mode for Gekko if m.options.IMODE is not specified.

IMODE=6

MPC Model Predictive Control (MPC) is implemented with IMODE=6 as a simultaneous solution or with IMODE=9 as a sequential shooting method.

为什么 b=1。在一种模式下工作但在另一种模式下不工作？

这是我对 IMODE=3 和 b=1.0:

的零碎处理

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

m = GEKKO(remote=False)
m.time = np.linspace(0,23,24)

#initialize variables
T_e = [50.,50.,50.,50.,45.,45.,45.,60.,60.,63.,\
       64.,45.,45.,50.,52.,53.,53.,54.,54.,53.,52.,51.,50.,45.]
temp_low = [55.,55.,55.,55.,55.,55.,55.,68.,68.,68.,68.,55.,55.,68.,\
            68.,68.,68.,55.,55.,55.,55.,55.,55.,55.]
temp_upper = [75.,75.,75.,75.,75.,75.,75.,70.,70.,70.,70.,75.,75.,\
              70.,70.,70.,70.,75.,75.,75.,75.,75.,75.,75.]
TOU = [0.05,0.05,0.05,0.05,0.05,0.05,0.05,200.,200.,\
       200.,200.,200.,200.,200.,200.,200.,200.,200.,\
       200.,200.,200.,0.05,0.05,0.05]

b = m.Param(value=1.)
k = m.Param(value=0.05)

u = [m.MV(0.,lb=0.,ub=1.) for i in range(24)]

# Controlled Variable
T = [m.SV(60.,lb=temp_low[i],ub=temp_upper[i]) for i in range(24)]

for i in range(24):
    u[i].STATUS = 1

for i in range(23):
    m.Equation( T[i+1]-T[i]-k*(T_e[i]-T[i])-b*u[i]==0.0 )

m.Obj(np.dot(TOU,u))

m.options.IMODE = 3
m.solve(debug=True)
myu =[u[0:][i][0] for i in range(24)]
print myu
myt =[T[0:][i][0] for i in range(24)]
plt.plot(myt)
plt.plot(temp_low)
plt.plot(temp_upper)
plt.show()
fig, ax1 = plt.subplots()
ax2 = ax1.twinx()
ax1.plot(myu,color='b')
ax2.plot(TOU,color='k')
plt.show()

结果：

Answer 1

不可行IMODE=6和可行IMODE=3的区别在于IMODE=3情况允许优化器调整温度初始条件。优化器认识到初始条件可以更改，因此将其修改为 75 以保持可行并最大限度地减少未来的能源消耗。

from gekko import GEKKO
import numpy as np

m = GEKKO(remote=False)
m.time = np.linspace(0,23,24)

#initialize variables
T_external = [50.,50.,50.,50.,45.,45.,45.,60.,60.,63.,\
              64.,45.,45.,50.,52.,53.,53.,54.,54.,\
              53.,52.,51.,50.,45.]
temp_low = [55.,55.,55.,55.,55.,55.,55.,68.,68.,68.,68.,\
            55.,55.,68.,68.,68.,68.,55.,55.,55.,55.,55.,55.,55.]
temp_upper = [75.,75.,75.,75.,75.,75.,75.,70.,70.,70.,70.,75.,\
              75.,70.,70.,70.,70.,75.,75.,75.,75.,75.,75.,75.]
TOU_v = [0.05,0.05,0.05,0.05,0.05,0.05,0.05,200.,200.,200.,200.,\
         200.,200.,200.,200.,200.,200.,200.,200.,200.,200.,0.05,\
         0.05,0.05]

b = m.Param(value=1.)
k = m.Param(value=0.05)
T_e = m.Param(value=T_external)
TL = m.Param(value=temp_low)
TH = m.Param(value=temp_upper)
TOU = m.Param(value=TOU_v)

u = m.MV(lb=0, ub=1)
u.STATUS = 1  # allow optimizer to change

# Controlled Variable
T = m.SV(value=75)

m.Equations([T>=TL,T<=TH])
m.Equation(T.dt() == k*(T_e-T) + b*u)

m.Minimize(TOU*u)

m.options.IMODE = 6
m.solve(disp=True,debug=True)

import matplotlib.pyplot as plt
plt.subplot(2,1,1)
plt.plot(m.time,temp_low,'k--')
plt.plot(m.time,temp_upper,'k--')
plt.plot(m.time,T.value,'r-')
plt.ylabel('Temperature')
plt.subplot(2,1,2)
plt.step(m.time,u.value,'b:')
plt.ylabel('Heater')
plt.xlabel('Time (hr)')
plt.show()

如果再试一天（48 小时），您可能会发现问题最终将无法解决，因为较小的加热器 b=1 无法满足温度下限限制。

使用IMODE=6的优点之一是您可以编写微分方程而不是自己进行离散化。使用 IMODE=3，您可以对微分方程使用欧拉方法。 IMODE>=4 的默认离散化为 NODES=2，相当于您的欧拉有限差分法。设置 NODES=3-6 可提高 orthogonal collocation on finite elements 的准确性。

GEKKO RTO 与 MPC 模式

GEKKO RTO vs MPC MODES

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