GEKKO：不响应约束，也不求解obj函数

Question

以下与这个问题有关：MPC with ARX Model Using Gekko

我试图用15分钟的数据来识别我的系统。我正试图在一天内每小时更新我的MPC MV。这会影响我的控制器吗

我运行了前一个问题中更正的代码，但它似乎无法维持约束条件，也无法在一天内更改MV

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

m = GEKKO(remote = True)


#initialize variables

#Room Temprature:
T_external = [23,23,23,23,23.5,23.5,23.4,23.5,23.9,23.7,\
              23,23.9,23.9,23.4,23.9,24,23.6,23.7,23.8,\
              23,23,23,23,23]

# Temprature Lower Limit:
temp_low = 10*np.ones(24)

# Temprature Upper Limit:
temp_upper = 12*np.ones(24)

#Hourly Energy prices:
TOU_v = [39.09,34.93,38.39,40.46,40.57,43.93,25,11,9,24,51.28,45.22,45.72,\
            36,35.03,10,12,13,32.81,42.55,8,29.58,29.52,29.52]

###########################################
#System Identification:

#Time 
t = np.linspace(0,10,117)
#State of the Fridge
ud = np.append(np.zeros(78) ,np.ones(39),0)
#Temprature Data for 10 min 
y = [14.600000000000001,14.600000000000001,14.700000000000001,14.700000000000001,14.700000000000001,\
     14.700000000000001,14.700000000000001,14.700000000000001,14.700000000000001,14.700000000000001,\
     14.700000000000001,14.700000000000001,14.700000000000001,14.8,14.8,14.8,14.8,14.8,14.8,14.8,14.8,\
    14.8,14.8,14.9,14.9,14.9,14.9,14.9,14.9,14.9,15,15,15,15,15,15,15,15,15,15,15,15,15.100000000000001,\
    15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,\
    15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,\
    15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,\
    15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,\
    15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,\
    15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,\
    15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,
    15,15,15,15,15,15,15,15,15,15,14.9,14.9,14.9,14.9,14.8,14.9,14.8,14.8,14.8,14.8,14.8,14.8,\
    14.8,14.700000000000001,14.8,14.700000000000001,14.700000000000001,14.700000000000001,\
    14.700000000000001,14.700000000000001,14.700000000000001,14.700000000000001,\
    14.700000000000001,14.600000000000001,14.600000000000001,14.600000000000001,\
    14.600000000000001,14.600000000000001,14.60]

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,y,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(0,23,24)

#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,'k--',label='Upper limit')
plt.plot(m.time,T.value,'r-')
plt.ylabel('Temperature')
plt.legend()
plt.subplot(5,1,2)
plt.step(m.time,u.value,'b:')
plt.ylabel('Fridge State')
plt.legend()
plt.subplot(5,1,3)
plt.plot(m.time, eH.value, 'k--', 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(5,1,4)
plt.plot(m.time, cost_list, 'r-')
plt.ylabel('Costs in cent')

plt.show()

结果如下所示：

我将感谢任何形式的帮助：）

Answer 1

在调用m.arx()模型之前，需要定义u = m.MV()和T=m.CV()，以便将这些值用作输入和输出。我还增加了WSPHI值，以便成本目标不会导致忽略温度限制。目前的制冷系统似乎不足以冷却到这个水平。它需要一个大约3倍更强大的系统来维持温度极限。我将制冷系统的上限设置为4，这样可以将温度保持在限制范围内。它最终放弃了温度控制，因为它发现节能比在很短的时间内达到温度限制更有价值。您可以通过增加WSPHI和WSPLO或以TH.UPPER = 0作为硬约束来强制限制。如果制冷系统不能满足硬约束，硬约束可能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 = [23,23,23,23,23.5,23.5,23.4,23.5,23.9,23.7,\
              23,23.9,23.9,23.4,23.9,24,23.6,23.7,23.8,\
              23,23,23,23,23]

# Temprature Lower Limit:
temp_low = 10*np.ones(24)

# Temprature Upper Limit:
temp_upper = 12*np.ones(24)

#Hourly Energy prices:
TOU_v = [39.09,34.93,38.39,40.46,40.57,43.93,25,11,9,24,51.28,45.22,45.72,\
            36,35.03,10,12,13,32.81,42.55,8,29.58,29.52,29.52]

###########################################
#System Identification:

#Time 
t = np.linspace(0,10,117)
#State of the Fridge
ud = np.append(np.zeros(78) ,np.ones(39),0)
#Temprature Data for 10 min 
y = [14.600000000000001,14.600000000000001,14.700000000000001,14.700000000000001,14.700000000000001,\
     14.700000000000001,14.700000000000001,14.700000000000001,14.700000000000001,14.700000000000001,\
     14.700000000000001,14.700000000000001,14.700000000000001,14.8,14.8,14.8,14.8,14.8,14.8,14.8,14.8,\
    14.8,14.8,14.9,14.9,14.9,14.9,14.9,14.9,14.9,15,15,15,15,15,15,15,15,15,15,15,15,15.100000000000001,\
    15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,\
    15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,\
    15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,\
    15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,\
    15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,\
    15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,\
    15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,15.100000000000001,
    15,15,15,15,15,15,15,15,15,15,14.9,14.9,14.9,14.9,14.8,14.9,14.8,14.8,14.8,14.8,14.8,14.8,\
    14.8,14.700000000000001,14.8,14.700000000000001,14.700000000000001,14.700000000000001,\
    14.700000000000001,14.700000000000001,14.700000000000001,14.700000000000001,\
    14.700000000000001,14.600000000000001,14.600000000000001,14.600000000000001,\
    14.600000000000001,14.600000000000001,14.60]

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(0,23,24)

#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,'k ',label='Upper limit')
plt.plot(m.time,T.value,'r-')
plt.ylabel('Temperature')
plt.legend()
plt.subplot(4,1,2)
plt.step(m.time,u.value,'b:',label='u')
plt.ylabel('Fridge State')
#plt.grid()
plt.legend()
plt.subplot(4,1,3)
plt.plot(m.time, eH.value, 'k ', 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(4,1,4)
plt.plot(m.time, cost_list, 'r-')
plt.ylabel('Costs in cent')

plt.show()