GEKKO RTO vs MPC模式 [英] GEKKO RTO vs MPC MODES
问题描述
这是从一个派生的问题.发布我的问题后，我找到了解决方案(更像是迫使优化器进行优化的补丁程序).有什么让我感到困惑. John Hedengren 正确指出，ODE中的b=1.0导致使用IMODE=6的不可行解决方案.但是，在我对IMODE=3进行的零星工作中，我确实找到了解决方案.
我正在尝试阅读 GEKKO的文档以了解此处发生的情况IMODE=3和6，但我不清楚
IMODE=3
RTO实时优化(RTO)是一种稳态模式，它允许 决策变量(STATUS = 1的FV或MV类型)或其他 超出方程数量的变量.目标函数 指导选择其他变量以选择最佳变量 可行的解决方案. RTO是Gekko的默认模式，如果 未指定m.options.IMODE.
IMODE=6
MPC模型预测控制(MPC)通过IMODE = 6实现 同时解决方案或以IMODE = 9作为顺序拍摄方法.
为什么b = 1.可以在一种模式下工作，而不能在另一种模式下工作?
这是我对IMODE=3和b=1.0的零星工作:
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:
不可行的IMODE=6和可行的IMODE=3之间的区别在于，IMODE=3情况允许优化器调整温度的初始条件.优化器认识到可以更改初始条件，因此将其修改为75，以保持可行并最小化未来的能源消耗.
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()
如果再待一天(48小时)，您可能会发现该问题最终将不可行，因为较小的加热器b=1无法满足较低的温度限制.
使用IMODE=6的优点之一是您可以编写微分方程，而不用自己进行离散化.对于IMODE=3，您可以对微分方程使用Euler方法. IMODE>=4的默认离散化为NODES=2，等效于您的Euler有限差分法.设置NODES=3-6可以通过有限元素上的正交搭配来提高准确性.. >
This is a question derived from this one. 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 correctly points out that a b=1.0 in the ODE leads to an infeasible solution with IMODE=6. However in my patchy work around with IMODE=3 I do get a solution.
I'm trying to understand what's happening here reading GEKKO's documentation for IMODE=3 and 6 but it is not clear to me
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.
Why b=1. works in one mode but not in the other?
This is my patchy work around with IMODE=3 and b=1.0:
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()
Results:
The difference between the infeasible IMODE=6 and the feasible IMODE=3 is that the IMODE=3 case allows the temperature initial condition to be adjusted by the optimizer. The optimizer recognizes that the initial condition can be changed and so it modifies it to 75 to both stay feasible and also minimize the future energy consumption.
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()
If you went for another day (48 hours), you'd probably see that the problem would eventually be infeasible because the smaller heater b=1 wouldn't be able to meet the temperature lower constraint.
One of the advantages of using IMODE=6 is that you can write the differential equation instead of doing the discretization yourself. With IMODE=3, you use an Euler's method for the differential equation. The default discretization for IMODE>=4 is NODES=2, equivalent to your Euler finite difference method. Setting NODES=3-6 increases the accuracy with orthogonal collocation on finite elements.
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