添加约束以将电池循环限制为每 24 小时 1 次充电和 1 次放电

Question

此查询是对以下问题的扩展：'@Error: Solution not found' being returned when using gekko for optimization。 “ind_1”和“ind_2”是长度为 8760 的列表，包含 0s/1s。一年中的某些时间可能会获得额外的收入，因此这些指标列表用于区分这些时间（进一步用于最大化功能 我试图通过将电池循环限制为每 24 小时最多 1 次充电和放电来构建此模型。作为最初的简单方法，我试图总结每个 24 小时段的电池命令信号，并将其限制为最多 8000 kWh。您可以在下面找到我的方法：

m = Gekko(remote=False)
#variables
e_battery = m.Var(lb=0, ub=4000, value=2000) #energy in battery at time t, battery size 4 MWh, initial value is 2MWh
command = m.Var(lb=-1000, ub=1000) #command power -1 to 1 (in MW)
e_price = m.Param(value = price) #price is a list of 8760 values
ind_1 = m.Param(value = ind_1) 
ind_2 = m.Param(value = ind_2)
peak_list = m.Param(value = peak_load_list) #list of the monthly peaks (an array of length 8760)
load_list = m.Param(value = load) #hourly electric load
m.time = np.linspace(0,8759, 8760)
m.Equation(e_battery.dt() == command) 

#The next 2 constraints are to ensure that the new load (original load + battery operation) is greater than 0, but less than the peak load for that month
m.Equation(load_list + command >= 0)
m.Equation(load_list + command <= peak_list)

#Here is the code to limit the cycling. "abs(command)" is used since "command" can be negative (discharge) or positive (charge), and a full charge and full discharge will equate to 8000 kWh. 
daily_sum=0
for i in range(8760):
    daily_sum += abs(command)
    if i%24==0 and i!=0: #when i=0, it's the beginning of the first day so we can skip it
        m.Equation(daily_sum <= 8000)
        daily_sum = 0 #reset to 0 in preparation for the first hour of the next day
m.Maximize((-command)*(e_price + ind_1*ind1_price + ind_2*ind2_price))
m.options.IMODE = 6
m.solve()

添加循环约束时，返回如下输出：

 --------- APM Model Size ------------
 Each time step contains
   Objects      :  0
   Constants    :  0
   Variables    :  373
   Intermediates:  0
   Connections  :  0
   Equations    :  368
   Residuals    :  368
Error: At line 1545 of file apm.f90
Traceback: not available, compile with -ftrace=frame or -ftrace=full
Fortran runtime error: Out of memory

这个特定的实现是否使用 gekko 的框架工作？我是否必须为“命令”初始化不同类型的变量？此外，我还没有找到许多将 for 循环用于方程式的相关示例，因此我非常清楚我的实现可能会很顺利。很想听听任何人的想法和/或建议，谢谢。

Answer 1

二进制变量指示何时到达目的地（e_battery>3999 或 e_battery

以下是两个具有软约束和硬约束的示例。用于测试的时间点数量从 8760 个减少到 120 个（5 天）。

from gekko import Gekko
import numpy as np
m = Gekko(remote=False)

n = 120 # hours
price=np.ones(n)
e_battery = m.Var(lb=0, ub=4000, value=2000) #energy in battery at time t
# battery size 4 MWh, initial value is 2MWh
command = m.Var(lb=-1000, ub=1000) #command power -1 to 1 (in MW)
e_price = m.Param(value = price) #price is a list of 8760 values
ind_1=1; ind_2=1
ind1_price=1; ind2_price=1
ind_1 = m.Param(value = ind_1) 
ind_2 = m.Param(value = ind_2)
m.time = np.linspace(0,n-1,n)
m.Equation(e_battery.dt() == command)

day = 24
discharge = m.Intermediate(m.integral(m.if3(e_battery+1,1,0)))
charge    = m.Intermediate(m.integral(m.if3(e_battery-3999,0,1)))
x         = np.ones_like(m.time)
for i in range(1,n):
    if i%day==0:
        x[i] = x[i-1] + 1
    else:
        x[i] = x[i-1]
limit = m.Param(x)

soft_constraints = True
if soft_constraints:
    derr = m.CV(value=0)
    m.Equation(derr==limit-discharge)
    derr.STATUS = 1
    derr.SPHI = 1; derr.WSPHI = 1000
    derr.SPLO = 0; derr.WSPLO = 1000

    cerr = m.CV(value=0)
    m.Equation(cerr==limit-charge)
    cerr.STATUS = 1
    cerr.SPHI = 1; cerr.WSPHI = 1000
    cerr.SPLO = 0; cerr.WSPLO = 1000
else:
    # Hard Constraints
    m.Equation(charge<=limit)
    m.Equation(charge>=limit-1)

    m.Equation(discharge<=limit)
    m.Equation(discharge>=limit-1)

m.Minimize(command*(e_price + ind_1*ind1_price + ind_2*ind2_price))
m.options.IMODE = 6
m.solve()

import matplotlib.pyplot as plt
plt.figure(figsize=(10,6))
plt.subplot(3,1,1)
plt.plot(m.time,limit.value,'g-.',label='Limit')
plt.plot(m.time,discharge.value,'b:',label='Discharge')
plt.plot(m.time,charge.value,'r--',label='Charge')
plt.legend(); plt.xlabel('Time'); plt.ylabel('Cycles'); plt.grid()
plt.subplot(3,1,2)
plt.plot(m.time,command.value,'k-',label='command')
plt.legend(); plt.xlabel('Time'); plt.ylabel('Command'); plt.grid()
plt.subplot(3,1,3)
plt.plot(m.time,e_battery.value,'g-.',label='Battery Charge')
plt.legend(); plt.xlabel('Time'); plt.ylabel('Battery'); plt.grid()
plt.show()

原始问题中的应用程序内存不足，因为 8760 个方程分别在 8760 个时间步上同时积分。尝试提出写过一次但在整个框架内有效的方程。当前的目标函数是最小化用电量。您可能需要包含约束或目标函数以满足需求。否则，解决方案是永远不要使用电力，因为它被最小化（例如最大化（-command））。这里有类似的Grid Energy Benchmark problems，可能会有所帮助。