如何在 Gekko 中模拟时间相关的约束？

Question

我是 Python 中的 Gekko 库的新手，想知道我是否可以在 Gekko 中按照 LP 公式建模。

LP 公式的意思是我想找到最优的电器调度转移，从而使总电力成本最小化。但是，如约束所示，每天的总耗电量 (f_app*P_app) 必须保持一致。

我的代码如下所示，但我收到以下错误“x 必须是 GEKKO 参数、变量或表达式的 python 列表”，我不知道如何解决。

代码：

import numpy as np
from gekko import GEKKO

m = GEKKO()
m.time = np.linspace(1, 24, 24)

TOU_list = [0.074646,0.074646,0.074646,0.074646,0.074646,0.074646,\
            0.074646,0.099206,0.099206,0.099206,0.099206,\
            0.169230,0.169230,0.169230,0.169230,0.169230,0.099206,\
            0.099206,0.099206,0.099206,0.074646,0.074646,\
            0.074646,0.074646] # Electricity Cost
        
TOU = m.Param(value= TOU_list)
P_app = 10.5 # Appliance power (kW)
f_app = m.MV(lb=0.0, ub=1.0, integer=False) # electric appliance schedule (0-1)

m.Equation(m.sum(f_app) == 15) # Summation of the appliance schedule for 24 hour time horizon         

m.Minimize(TOU*f_app*P_app)

m.options.IMODE = 6
m.options.SOLVER = 3
m.solve(disp=True, GUI=False)

Answer 1

m.sum() 是该特定时间点的值的总和。尝试使用m.integral() 来生成时间范围内的总和。使用m.fix_final(c,15) 将积分的最终值固定为15。将m.options.SOLVER=1 与integer=True 一起使用，否则设备可以打开小数值。

import numpy as np
from gekko import GEKKO

m = GEKKO()
m.time = np.linspace(0, 24, 25)

# Electricity Cost
TOU_list = [0.074646,0.074646,0.074646,0.074646,0.074646,0.074646,\
            0.074646,0.074646,0.099206,0.099206,0.099206,0.099206,\
            0.169230,0.169230,0.169230,0.169230,0.169230,0.099206,\
            0.099206,0.099206,0.099206,0.074646,0.074646,\
            0.074646,0.074646] 
        
TOU = m.Param(value= TOU_list)
P_app = 10.5 # Appliance power (kW)
# electric appliance schedule (0-1)
f_app = m.MV(value=0,lb=0.0, ub=1.0, integer=True)
f_app.STATUS = 1

# Summation of the appliance schedule for 24 hour time horizon
c = m.integral(f_app)
m.fix_final(c,15)

m.Minimize(TOU*f_app*P_app)

m.options.IMODE = 6
m.options.SOLVER = 1
m.solve(disp=True, GUI=False)

import matplotlib.pyplot as plt
plt.plot(m.time,f_app.value,label='On / Off')
plt.plot(m.time,TOU_list,label='Electricity Cost')
plt.legend()
plt.show()

我在开始时添加了一个额外的时间点，因为它是 24 小时，总共 25 个时间点。如果有 1 小时，那么优化问题需要两个时间点来定义开始和结束。 24小时，需要25分。

您可以通过使用m.Array() 对 24 个时间段进行编程来避免动态控制模式的一些复杂性。这样您就可以使用m.sum()，因为f_app 是一个包含24 个值的数组。

import numpy as np
from gekko import GEKKO

m = GEKKO()

# Electricity Cost
TOU_list = [0.074646,0.074646,0.074646,0.074646,0.074646,0.074646,\
            0.074646,0.099206,0.099206,0.099206,0.099206,\
            0.169230,0.169230,0.169230,0.169230,0.169230,0.099206,\
            0.099206,0.099206,0.099206,0.074646,0.074646,\
            0.074646,0.074646] 
n = len(TOU_list)
        
P_app = 10.5 # Appliance power (kW)
# electric appliance schedule (0-1)
f_app = m.Array(m.Var,n,value=0,lb=0.0, ub=1.0, integer=True)

# Summation of the appliance schedule for 24 hour time horizon
m.Equation(m.sum(f_app)==15)

for i in range(n):
    m.Minimize(TOU_list[i]*f_app[i]*P_app)

m.options.IMODE = 3
m.options.SOLVER = 1
m.solve(disp=True)

f = []
t = []
for i in range(n):
    t.append(i+1)
    f.append(f_app[i].value[0])

import matplotlib.pyplot as plt
plt.bar(x=t,height=f,label='On / Off')
plt.bar(x=t,height=TOU_list,label='Electricity Cost')
plt.legend()
plt.show()

因为问题没有微分方程，我推荐第二种方法。