问题描述
以下与该问题有关：MPC with ARX Model Using Gekko。
我正在尝试使用15分钟的数据来识别我的系统。而且我正在尝试每天在一天中每小时更新一次MPC MV。这会影响我的控制器吗？
我从上一个问题中运行了更正后的代码，但它似乎并不能保持约束或在一天内更改MV。
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 for 10 min
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
u = 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)
#T = m.CV()
# 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 costs.
m.Obj(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
#Set Up MPC
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()
#Calculate the costs.
c= 0
cost_list = []
for i in range(0,len(u)):
c = c + TOU_v[i]*u[i]
cost_list.append(c)
print('The daily energy cost is',c/100,'Euro')
plt.subplot(5,1,1)
plt.plot(m.time,temp_low,'k--',label='Lower limit')
plt.plot(m.time,temp_upper,label='Upper limit')
plt.plot(m.time,T.value,'r-')
plt.ylabel('Temperature')
plt.legend()
plt.subplot(5,2)
plt.step(m.time,u.value,'b:')
plt.ylabel('Fridge State')
plt.legend()
plt.subplot(5,3)
plt.plot(m.time,eH.value,label='Upper Tempratue Limit Error')
plt.plot(m.time,eL.value,'b--',label='Lower Temprature Limit Error')
plt.ylabel('Cumulative Linar Error')
plt.legend()
plt.subplot(5,4)
plt.plot(m.time,cost_list,'r-')
plt.ylabel('Costs in cent')
plt.show()
结果如下：
我将不胜感激：）
解决方法
在调用u = m.MV()模型之前，需要定义T=m.CV()和m.arx()，以便将这些值用作输入和输出。我还增加了WSPHI值，以使成本目标不会导致温度限制被忽略。当前的制冷系统似乎不足以冷却到该水平。它需要一个功能强大约3倍的系统来维持温度极限。我将制冷系统的上限设置为4，以便它可以将温度保持在极限范围内。最后，它放弃了温度控制，因为它发现节能比在短时间内达到温度极限更有价值。您可以通过增加WSPHI和WSPLO或以TH.UPPER = 0作为硬约束来强制执行限制。如果制冷系统不能满足该约束条件，那么严格的约束条件可能会导致不可行的解决方案。
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 for 10 min
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:
# Controlled variable:
T = m.CV()
# Manipulated variable:
u = m.MV(value=0,lb=0,ub=4,integer=True)
# Create ARX Model
m.arx(p,T,u)
###########################################
#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])
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()
# 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=100000 #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=100000
eL.STATUS = 1
#Linear error (Deviation from the limits)
m.Equations([eH==T-TH,eL==T-TL])
#Objective: minimize 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
#Set Up MPC
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()
m.solve()
#Calculate the costs.
c= 0
cost_list = []
for i in range(0,len(u)):
c = c + TOU_v[i]*u[i]
cost_list.append(c)
print('The daily energy cost is',c/100,'Euro')
plt.subplot(4,1,1)
plt.plot(m.time,temp_low,'k--',label='Lower limit')
plt.plot(m.time,temp_upper,label='Upper limit')
plt.plot(m.time,T.value,'r-')
plt.ylabel('Temperature')
plt.legend()
plt.subplot(4,2)
plt.step(m.time,u.value,'b:',label='u')
plt.ylabel('Fridge State')
#plt.grid()
plt.legend()
plt.subplot(4,3)
plt.plot(m.time,eH.value,label='Upper Temperatue Limit Error')
plt.plot(m.time,eL.value,'b--',label='Lower Temperature Limit Error')
plt.ylabel('Cumulative Linear Error')
plt.legend()
plt.subplot(4,4)
plt.plot(m.time,cost_list,'r-')
plt.ylabel('Costs in cent')
plt.show()
设置时间 控制面板
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