问题描述
我正在对MPC进行建模，以控制冰箱并将温度保持在给定的时间间隔内，同时将成本降至最低。我正在使用GEKKO为我的算法建模。
我编写了以下代码。首先，我使用系统中的传感器数据来确定模型（我使用了GEKKO的sysif函数）。然后，我建立了一个ARX模型（使用GEKKO中的arx函数），该模型成为sysid（）的结果作为输入。
我试图在将其实施到Pi之前编写一种“虚拟”算法以在本地进行测试。
我收到以下错误：
KeyError Traceback (most recent call last)
<ipython-input-13-108148376700> in <module>
107 #Solve the optimization problem.
108
--> 109 m.solve()
~/opt/anaconda3/lib/python3.8/site-packages/gekko/gekko.py in solve(self,disp,debug,GUI,**kwargs)
2214 if timing == True:
2215 t = time.time()
-> 2216 self.load_JSON()
2217 if timing == True:
2218 print('load JSON',time.time() - t)
~/opt/anaconda3/lib/python3.8/site-packages/gekko/gk_post_solve.py in load_JSON(self)
48 vp.__dict__[o] = dpred
49 else: #everything besides value,dpred and pred
---> 50 vp.__dict__[o] = data[vp.name][o]
51 for vp in self._variables:
52 if vp.type != None: #(FV/MV/SV/CV) not Param or Var
KeyError: 'int_p6'
这是我的代码
from gekko import GEKKO
import numpy as np
import matplotlib.pyplot as plt
m = GEKKO(remote = True)
#initialize variables
#Room Temprature:
T_external = [23,23,23.5,23.4,23.9,23.7,\
23,24,23.6,23.8,23]
# Temprature Lower Limit:
temp_low = 10*np.ones(24)
# Temprature Upper Limit:
temp_upper = 12*np.ones(24)
#Hourly Energy prices:
TOU_v = [39.09,34.93,38.39,40.46,40.57,43.93,25,11,9,51.28,45.22,45.72,\
36,35.03,10,12,13,32.81,42.55,8,29.58,29.52,29.52]
###########################################
#System Identification:
#Time
t = np.linspace(0,117)
#State of the Fridge
ud = np.append(np.zeros(78),np.ones(39),0)
#Temprature Data
y = [14.600000000000001,14.600000000000001,14.700000000000001,\
14.700000000000001,14.8,\
14.8,14.9,15,15.100000000000001,\
15.100000000000001,\
14.700000000000001,\
14.600000000000001,14.60]
na = 1 # output coefficients
nb = 1 # input coefficients
print('Identification')
yp,p,K = m.sysid(t,ud,y,na,nb,objf=10000,scale=False,diaglevel=1)
#create control ARX model:
y = m.Array(m.CV,1)
uc = m.Array(m.MV,1)
m.arx(p,uc)
# rename CVs
T= y[0]
# rename MVs
uc = uc[0]
# steady state initialization
m.options.IMODE = 1
m.solve(disp=True)
###########################################
#Parameter
P = m.Param(value =100) #power
TL = m.Param(value=temp_low)
TH = m.Param(value=temp_upper)
c = m.Param(value=TOU_v)
# Manipilated variable:
u = m.MV(lb=0,ub=1,integer=True)
u.STATUS = 1 # allow optimizer to change the variable to attein the optimum.
# Controlled Variable (Affected with changes in the manipulated variable)
T = m.CV(value=11) # Temprature will start at 11.
# Soft constraints on temprature.
eH = m.CV(value=0)
eL = m.CV(value=0)
eH.SPHI=0 #Set point high for linear error model.
eH.WSPHI=100 #Objective function weight on upper set point for linear error model.
eH.WSPLO=0 # Objective function weight on lower set point for linear error model
eH.STATUS =1 # eH : Error is considered in the objective function.
eL.SPLO=0
eL.WSPHI=0
eL.WSPLO=100
eL.STATUS = 1
#Linear error (Deviation from the limits)
m.Equations([eH==T-TH,eL==T-TL])
#Objective : minimize the costs.
m.Minimize(c*P*u)
#Optimizer Options.
m.options.IMODE = 6 # MPC mode in Gekko.
m.options.NODES = 2 # Collocation nodes.
m.options.SOLVER = 1 # APOT solver for mixed integer linear programming.
m.time = np.linspace(0,24)
#Solve the optimization problem.
m.solve()
解决方法
问题在于：
T = m.CV(value=11) # Temperature will start at 11.
您正在重新定义T变量，但变量在内部都存储。如果您需要重新初始化为11，请使用T.value=11。另外，我在稳态初始化之前添加了eH和eL变量。这是一个可以成功运行的完整脚本。
from gekko import GEKKO
import numpy as np
import matplotlib.pyplot as plt
m = GEKKO(remote = True)
#initialize variables
#Room Temprature:
T_external = [23,23,23.5,23.4,23.9,23.7,\
23,24,23.6,23.8,23]
# Temprature Lower Limit:
temp_low = 10*np.ones(24)
# Temprature Upper Limit:
temp_upper = 12*np.ones(24)
#Hourly Energy prices:
TOU_v = [39.09,34.93,38.39,40.46,40.57,43.93,25,11,9,51.28,45.22,45.72,\
36,35.03,10,12,13,32.81,42.55,8,29.58,29.52,29.52]
###########################################
#System Identification:
#Time
t = np.linspace(0,117)
#State of the Fridge
ud = np.append(np.zeros(78),np.ones(39),0)
#Temprature Data
y = [14.600000000000001,14.600000000000001,14.700000000000001,\
14.700000000000001,14.8,\
14.8,14.9,15,15.100000000000001,\
15.100000000000001,\
14.700000000000001,\
14.600000000000001,14.60]
na = 1 # output coefficients
nb = 1 # input coefficients
print('Identification')
yp,p,K = m.sysid(t,ud,y,na,nb,objf=10000,scale=False,diaglevel=1)
#create control ARX model:
y = m.Array(m.CV,1)
uc = m.Array(m.MV,1)
m.arx(p,uc)
# rename CVs
T= y[0]
# rename MVs
uc = uc[0]
###########################################
#Parameter
P = m.Param(value =100) #power
TL = m.Param(value=temp_low[0])
TH = m.Param(value=temp_upper[0])
c = m.Param(value=TOU_v[0])
# Manipilated variable:
u = m.MV(lb=0,ub=1,integer=True)
u.STATUS = 1 # allow optimizer to change the variable to attein the optimum.
# Controlled Variable (Affected with changes in the manipulated variable)
# Soft constraints on temprature.
eH = m.CV(value=0)
eL = m.CV(value=0)
eH.SPHI=0 #Set point high for linear error model.
eH.WSPHI=100 #Objective function weight on upper set point for linear error model.
eH.WSPLO=0 # Objective function weight on lower set point for linear error model
eH.STATUS =1 # eH : Error is considered in the objective function.
eL.SPLO=0
eL.WSPHI=0
eL.WSPLO=100
eL.STATUS = 1
#Linear error (Deviation from the limits)
m.Equations([eH==T-TH,eL==T-TL])
#Objective : minimize the costs.
m.Minimize(c*P*u)
#Optimizer Options.
# steady state initialization
m.options.IMODE = 1
m.solve(disp=True)
TL.value = temp_low
TH.value = temp_upper
c.value = TOU_v
T.value = 11 # Temprature starts at 11
m.options.IMODE = 6 # MPC mode in Gekko.
m.options.NODES = 2 # Collocation nodes.
m.options.SOLVER = 1 # APOT solver for mixed integer linear programming.
m.time = np.linspace(0,24)
#Solve the optimization problem.
m.solve()
这是控制器的输出：
--------- APM Model Size ------------
Each time step contains
Objects : 1
Constants : 0
Variables : 9
Intermediates: 0
Connections : 2
Equations : 3
Residuals : 3
Number of state variables: 1035
Number of total equations: - 1012
Number of slack variables: - 0
---------------------------------------
Degrees of freedom : 23
----------------------------------------------
Dynamic Control with APOPT Solver
----------------------------------------------
Iter: 1 I: 0 Tm: 0.07 NLPi: 3 Dpth: 0 Lvs: 0 Obj: 6.76E+03 Gap: 0.00E+00
Successful solution
---------------------------------------------------
Solver : APOPT (v1.0)
Solution time : 8.319999999366701E-002 sec
Objective : 6763.77971670735
Successful solution
---------------------------------------------------
设置时间 控制面板
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