有几种方法可以加快 Gekko 代码的速度：
- 在本地解决而不是在公共服务器上解决。选项是
m=GEKKO(remote=False)。公共服务器可能会因许多作业而变慢。
- Use
sum()代替m.sum()。这可以更快地编译模型。否则，使用m.integral(x)如果你需要积分x.
- 许多方程在每个时间范围步骤中都会重复。 Gekko 使用单个方程定义更加高效
IMODE=2（对于代数方程模型）或IMODE=6（对于微分/代数方程模型），然后在时间范围内创建方程。您可能需要使用m.vsum()代替m.sum().
如需其他诊断，请尝试设置m.options.DIAGLEVEL=1获取编译模型和执行每个函数、一阶导数和二阶导数计算所需时间的详细时序报告。它还提供了求解阶段求解器与模型时间的详细视图。
通过数据文件测试进行更新
感谢您发送数据文件。运行目录显示模型文件长度为 58,682 行。编译这么大的模型需要一段时间。以下是您发送的文件中的解决方案：
--------- APM Model Size ------------
Each time step contains
Objects : 193
Constants : 5
Variables : 20641
Intermediates: 578
Connections : 18721
Equations : 20259
Residuals : 19681
Number of state variables: 20641
Number of total equations: - 19873
Number of slack variables: - 1152
---------------------------------------
Degrees of freedom : -384
* Warning: DOF <= 0
----------------------------------------------
Steady State Optimization with APOPT Solver
----------------------------------------------
Iter Objective Convergence
0 3.37044E+01 5.00000E+00
1 2.81987E+01 1.00000E-10
2 2.81811E+01 5.22529E-12
3 2.81811E+01 2.10942E-15
4 2.81811E+01 2.10942E-15
Successful solution
---------------------------------------------------
Solver : APOPT (v1.0)
Solution time : 10.5119999999879 sec
Objective : 28.1811214884047
Successful solution
---------------------------------------------------
这是一个使用的版本IMODE=6反而。您定义一次变量和方程，然后让 Gekko 处理时间离散化。它建立了一个更有效的模型，因为没有不必要的重复方程。
from gekko import GEKKO
import numpy as np
import pandas as pd
import time
import math
# ------------------------ Import and read input data ------------------------
file = r'Data Sim 1.xlsx'
data = pd.read_excel(file, sheet_name='Input', na_values='NaN')
dataRead = pd.DataFrame(data, columns= ['Timestep','Verbruik woning (kWh)','Prijs afname (€/kWh)',
'Capaciteit batterij (kW)','Capaciteit batterij (kWh)',
'Rendement (%)','Verbruikersprofiel'])
timestep = dataRead['Timestep'].to_numpy()
usage_home = dataRead['Verbruik woning (kWh)'].to_numpy()
price = dataRead['Prijs afname (€/kWh)'].to_numpy()
cap_batt_kW = dataRead['Capaciteit batterij (kW)'].iloc[0]
cap_batt_kWh = dataRead['Capaciteit batterij (kWh)'].iloc[0]
efficiency = dataRead['Rendement (%)'].iloc[0]
usersprofile = dataRead['Verbruikersprofiel'].iloc[0]
# ---------------------------- Optimization model ----------------------------
# Initialise model
m = GEKKO()
m.open_folder()
# Global options
m.options.SOLVER = 1
m.options.IMODE = 6
# Constants
snelheid_laden = cap_batt_kW/4
m.time = timestep
loss_charging = m.Const(value = (1-efficiency)/2)
max_cap_batt = m.Const(value = cap_batt_kWh)
min_cap_batt = m.Const(value = 0)
max_charge = m.Const(value = snelheid_laden) # max battery can charge in 15min
max_decharge = m.Const(value = -snelheid_laden) # max battery can decharge in 15min
# Parameters
usage_home = m.Param(usage_home)
price = m.Param(price)
# Variables
e_batt = m.Var(value=0, lb = min_cap_batt, ub = max_cap_batt) # energy in battery
price_paid = m.Var() # price paid each 15min
charging = m.Var(lb = max_decharge, ub = max_charge) # amount charge/decharge each 15min
usage_net = m.Var(lb=min_cap_batt)
# Equations
m.Equation(e_batt==m.integral(charging*(1-loss_charging)))
m.Equation(usage_net==usage_home + charging)
price_paid = m.Intermediate(usage_net * price / 100)
m.Equation(-charging <= (1-loss_charging)*e_batt)
# Objective
m.Minimize(price_paid)
# Solve problem
m.solve()
import matplotlib.pyplot as plt
plt.plot(m.time,e_batt.value,label='Battery Charge')
plt.plot(m.time,charging.value,label='Charging')
plt.plot(m.time,price_paid.value,label='Price')
plt.plot(m.time,usage_net.value,label='Net Usage')
plt.xlabel('Time'); plt.grid(); plt.legend(); plt.show()
该模型只有 31 行长（参见gk0_model.apm）并且解决速度更快（总共几秒钟）。
--------- APM Model Size ------------
Each time step contains
Objects : 0
Constants : 5
Variables : 8
Intermediates: 1
Connections : 0
Equations : 6
Residuals : 5
Number of state variables: 1337
Number of total equations: - 955
Number of slack variables: - 191
---------------------------------------
Degrees of freedom : 191
----------------------------------------------
Dynamic Control with APOPT Solver
----------------------------------------------
Iter Objective Convergence
0 3.46205E+01 3.00000E-01
1 3.30649E+01 4.41141E-10
2 3.12774E+01 1.98558E-11
3 3.03148E+01 1.77636E-15
4 2.96824E+01 3.99680E-15
5 2.82700E+01 8.88178E-16
6 2.82039E+01 1.77636E-15
7 2.81334E+01 8.88178E-16
8 2.81085E+01 1.33227E-15
9 2.81039E+01 8.88178E-16
Iter Objective Convergence
10 2.81005E+01 8.88178E-16
11 2.80999E+01 1.77636E-15
12 2.80996E+01 8.88178E-16
13 2.80996E+01 8.88178E-16
14 2.80996E+01 8.88178E-16
Successful solution
---------------------------------------------------
Solver : APOPT (v1.0)
Solution time : 0.527499999996508 sec
Objective : 28.0995878585948
Successful solution
---------------------------------------------------
编译时间不长。此外，求解器时间从 10 秒减少到 0.5 秒。目标函数几乎相同（28.18 与 28.10）。
这是一个完整的版本，没有数据文件依赖性（以防数据文件将来不可用）。
from gekko import GEKKO
import numpy as np
import pandas as pd
import time
import math
# ------------------------ Import and read input data ------------------------
timestep = np.arange(1,193)
usage_home = np.array([0.05,0.07,0.09,0.07,0.05,0.07,0.07,0.07,0.06,
0.05,0.07,0.07,0.09,0.07,0.06,0.07,0.07,
0.07,0.16,0.12,0.17,0.08,0.10,0.11,0.06,
0.06,0.06,0.06,0.06,0.07,0.07,0.07,0.08,
0.08,0.06,0.07,0.07,0.07,0.07,0.05,0.07,
0.07,0.07,0.07,0.21,0.08,0.07,0.08,0.27,
0.12,0.09,0.10,0.11,0.09,0.09,0.08,0.08,
0.12,0.15,0.08,0.10,0.08,0.10,0.09,0.10,
0.09,0.08,0.10,0.12,0.10,0.10,0.10,0.11,
0.10,0.10,0.11,0.13,0.21,0.12,0.10,0.10,
0.11,0.10,0.11,0.12,0.12,0.10,0.11,0.10,
0.10,0.10,0.11,0.10,0.10,0.09,0.08,0.12,
0.10,0.11,0.11,0.10,0.06,0.05,0.06,0.06,
0.06,0.07,0.06,0.06,0.05,0.06,0.05,0.06,
0.05,0.06,0.05,0.06,0.07,0.06,0.09,0.10,
0.10,0.22,0.08,0.06,0.05,0.06,0.08,0.08,
0.07,0.08,0.07,0.07,0.16,0.21,0.08,0.08,
0.09,0.09,0.10,0.09,0.09,0.08,0.12,0.24,
0.09,0.08,0.09,0.08,0.10,0.24,0.08,0.09,
0.09,0.08,0.08,0.07,0.06,0.05,0.06,0.07,
0.07,0.05,0.05,0.06,0.05,0.28,0.11,0.20,
0.10,0.09,0.28,0.10,0.15,0.09,0.10,0.18,
0.12,0.13,0.30,0.10,0.11,0.10,0.10,0.11,
0.10,0.21,0.10,0.10,0.12,0.10,0.08])
price = np.array([209.40,209.40,209.40,209.40,193.00,193.00,193.00,
193.00,182.75,182.75,182.75,182.75,161.60,161.60,
161.60,161.60,154.25,154.25,154.25,154.25,150.70,
150.70,150.70,150.70,150.85,150.85,150.85,150.85,
150.00,150.00,150.00,150.00,153.25,153.25,153.25,
153.25,153.25,153.25,153.25,153.25,151.35,151.35,
151.35,151.35,151.70,151.70,151.70,151.70,154.95,
154.95,154.95,154.95,150.20,150.20,150.20,150.20,
153.75,153.75,153.75,153.75,160.55,160.55,160.55,
160.55,179.90,179.90,179.90,179.90,202.00,202.00,
202.00,202.00,220.25,220.25,220.25,220.25,245.75,
245.75,245.75,245.75,222.90,222.90,222.90,222.90,
203.40,203.40,203.40,203.40,205.30,205.30,205.30,
205.30,192.80,192.80,192.80,192.80,177.00,177.00,
177.00,177.00,159.90,159.90,159.90,159.90,152.50,
152.50,152.50,152.50,143.95,143.95,143.95,143.95,
142.10,142.10,142.10,142.10,143.75,143.75,143.75,
143.75,170.80,170.80,170.80,170.80,210.35,210.35,
210.35,210.35,224.45,224.45,224.45,224.45,226.30,
226.30,226.30,226.30,227.85,227.85,227.85,227.85,
225.45,225.45,225.45,225.45,225.80,225.80,225.80,
225.80,224.50,224.50,224.50,224.50,220.30,220.30,
220.30,220.30,220.00,220.00,220.00,220.00,221.90,
221.90,221.90,221.90,230.25,230.25,230.25,230.25,
233.60,233.60,233.60,233.60,225.20,225.20,225.20,
225.20,179.85,179.85,179.85,179.85,171.85,171.85,
171.85,171.85,162.90,162.90,162.90,162.90,158.85,
158.85,158.85,158.85])
cap_batt_kW = 3.00
cap_batt_kWh = 5.00
efficiency = 0.95
usersprofile = 1
# ---------------------------- Optimization model ----------------------------
# Initialise model
m = GEKKO()
#m.open_folder()
# Global options
m.options.SOLVER = 1
m.options.IMODE = 6
# Constants
snelheid_laden = cap_batt_kW/4
m.time = timestep
loss_charging = m.Const(value = (1-efficiency)/2)
max_cap_batt = m.Const(value = cap_batt_kWh)
min_cap_batt = m.Const(value = 0)
max_charge = m.Const(value = snelheid_laden) # max battery can charge in 15min
max_decharge = m.Const(value = -snelheid_laden) # max battery can decharge in 15min
# Parameters
usage_home = m.Param(usage_home)
price = m.Param(price)
# Variables
e_batt = m.Var(value=0, lb = min_cap_batt, ub = max_cap_batt) # energy in battery
price_paid = m.Var() # price paid each 15min
charging = m.Var(lb = max_decharge, ub = max_charge) # amount charge/decharge each 15min
usage_net = m.Var(lb=min_cap_batt)
# Equations
m.Equation(e_batt==m.integral(charging*(1-loss_charging)))
m.Equation(usage_net==usage_home + charging)
price_paid = m.Intermediate(usage_net * price / 100)
m.Equation(-charging <= (1-loss_charging)*e_batt)
# Objective
m.Minimize(price_paid)
# Solve problem
m.solve()
import matplotlib.pyplot as plt
plt.plot(m.time,e_batt.value,label='Battery Charge')
plt.plot(m.time,charging.value,label='Charging')
plt.plot(m.time,price_paid.value,label='Price')
plt.plot(m.time,usage_net.value,label='Net Usage')
plt.xlabel('Time'); plt.grid(); plt.legend(); plt.show()
