python - 更改参数、目标函数和初始条件时使用 Gekko 优化套件出现 "Solution Not Found"错误

标签 python optimization gekko

我正在尝试使用 Gekko 套件解决具有各种初始条件的多个优化问题。分配初始条件，使用 Gekko 运行优化，并收集每个解决方案。当我尝试更改参数、目标函数或初始条件时，Gekko 经常给我“找不到解决方案错误:第 2130 行，解决引发异常(apm_error)”。我在下面展示了一些案例，希望得到建议来解决我遇到的这个问题。我之前发布过类似的问题，但我希望这个问题更简洁明了。谢谢。

案例 1. 运行良好，没有错误。

from gekko import GEKKO
import pandas as pd
import numpy as np

dat = {'A0': [23221, 2198, 4296, 104906, 691], 'h': [0.04, 0.25, 0.07, 0, 12.58],'emax': [23221, 2198, 4296, 104906, 691] }
dftemp = pd.DataFrame(data=dat)
na=len(dftemp)

# time points
n=51
year=50

# constants
Pa0 = 3.061 
Pe0 = 10.603 
C0 = 100 
r = 0.05 
k=50 
shift=10000000 # to make positive inside log function
ll=0.15

  
for i in range(0,na):
    # create GEKKO model
    m = GEKKO(remote=False)

    m.time = np.linspace(0,year,n)
    t=m.time
    
    A0=dftemp.loc[i][0]
    h=dftemp.loc[i][1]
    emax=dftemp.loc[i][2]    
    
    A = m.SV(value=A0, lb=0, ub=A0)
    E = m.SV(value=0, lb=0, ub=A0)

    u = m.MV(value=0, lb=-emax, ub=emax) 
    u.STATUS = 1

    t = m.Param(value=m.time)
    C = m.Var(value=C0)
    d = m.Var(value=1)
    l = m.Param(value=ll)
    
    # Equation
    m.Equation(A.dt()==-u)
    m.Equation(E.dt()==u)
    m.Equation(C.dt()==-C/k)
    m.Equation(d==m.exp(-t*r))
    # Objective (Utility)
    J = m.Var(value=0)
    
    # Final objective
    Jf = m.FV()
    Jf.STATUS = 1
    m.Connection(Jf,J,pos2='end')
    m.Equation(J.dt() == m.log((A+E*(1-l))*h*Pa0-C*u+E*Pe0+shift)*d)
    
    # maximize U
    m.Maximize(Jf)
    
    # options
    m.options.IMODE = 6  # optimal control
    m.options.NODES = 3  # collocation nodes
    m.options.SOLVER = 3 # solver (IPOPT)
    
    # solve optimization problem
    m.solve()
    
    # print profit
    print('Optimal Profit: ' + str(Jf.value[0]))

案例2.将“时间点”的个数从n=51改为n=501出现错误

from gekko import GEKKO
import pandas as pd
import numpy as np

dat = {'A0': [23221, 2198, 4296, 104906, 691], 'h': [0.04, 0.25, 0.07, 0, 12.58],'emax': [23221, 2198, 4296, 104906, 691] }
dftemp = pd.DataFrame(data=dat)
na=len(dftemp)

# time points
n=501 
year=50

# constants
Pa0 = 3.061 
Pe0 = 10.603 
C0 = 100 
r = 0.05 
k=50 
shift=10000000 # to make positive inside log function
ll=0.15

   
for i in range(0,na):
    # create GEKKO model
    m = GEKKO(remote=False)

    m.time = np.linspace(0,year,n)
    t=m.time
    
    A0=dftemp.loc[i][0]
    h=dftemp.loc[i][1]
    emax=dftemp.loc[i][2]    
    
    A = m.SV(value=A0, lb=0, ub=A0)
    E = m.SV(value=0, lb=0, ub=A0)

    u = m.MV(value=0, lb=-emax, ub=emax) 
    u.STATUS = 1

    t = m.Param(value=m.time)
    C = m.Var(value=C0)
    d = m.Var(value=1)
    l = m.Param(value=ll)
    
    # Equation
    m.Equation(A.dt()==-u)
    m.Equation(E.dt()==u)
    m.Equation(C.dt()==-C/k)
    m.Equation(d==m.exp(-t*r))
    # Objective (Utility)
    J = m.Var(value=0)
    
    # Final objective
    Jf = m.FV()
    Jf.STATUS = 1
    m.Connection(Jf,J,pos2='end')
    m.Equation(J.dt() == m.log((A+E*(1-l))*h*Pa0-C*u+E*Pe0+shift)*d)
    
    # maximize U
    m.Maximize(Jf)
    
    # options
    m.options.IMODE = 6  # optimal control
    m.options.NODES = 3  # collocation nodes
    m.options.SOLVER = 3 # solver (IPOPT)
    
    # solve optimization problem
    m.solve()
    
    # print profit
    print('Optimal Profit: ' + str(Jf.value[0]))

案例 3. 将目标函数从 m.log 更改为简单的线性求和。效果很好。

from gekko import GEKKO
import pandas as pd
import numpy as np

dat = {'A0': [23221, 2198, 4296, 104906, 691], 'h': [0.04, 0.25, 0.07, 0, 12.58],'emax': [23221, 2198, 4296, 104906, 691] }
dftemp = pd.DataFrame(data=dat)
na=len(dftemp)

# time points
n=51
year=50

# constants
Pa0 = 3.061 
Pe0 = 10.603 
C0 = 100 
r = 0.05 
k=50 
shift=10000000 # to make positive inside log function
ll=0.15
    
for i in range(0,na):
    # create GEKKO model
    m = GEKKO(remote=False)

    m.time = np.linspace(0,year,n)
    t=m.time
    
    A0=dftemp.loc[i][0]
    h=dftemp.loc[i][1]
    emax=dftemp.loc[i][2]    
    
    A = m.SV(value=A0, lb=0, ub=A0)
    E = m.SV(value=0, lb=0, ub=A0)

    u = m.MV(value=0, lb=-emax, ub=emax) 
    u.STATUS = 1

    t = m.Param(value=m.time)
    C = m.Var(value=C0)
    d = m.Var(value=1)
    l = m.Param(value=ll)
    
    # Equation
    m.Equation(A.dt()==-u)
    m.Equation(E.dt()==u)
    m.Equation(C.dt()==-C/k)
    m.Equation(d==m.exp(-t*r))
    # Objective (Utility)
    J = m.Var(value=0)
    
    # Final objective
    Jf = m.FV()
    Jf.STATUS = 1
    m.Connection(Jf,J,pos2='end')
    m.Equation(J.dt() == ((A+E*(1-l))*h*Pa0-C*u+E*Pe0+shift)*d)
    
    # maximize U
    m.Maximize(Jf)
    
    # options
    m.options.IMODE = 6  # optimal control
    m.options.NODES = 3  # collocation nodes
    m.options.SOLVER = 3 # solver (IPOPT)
    
    # solve optimization problem
    m.solve()
    
    # print profit
    print('Optimal Profit: ' + str(Jf.value[0]))

案例 4. 将目标函数从 m.log 更改为简单求和，并从目标函数中删除“变量”d。发生错误

from gekko import GEKKO
import pandas as pd
import numpy as np

dat = {'A0': [23221, 2198, 4296, 104906, 691], 'h': [0.04, 0.25, 0.07, 0, 12.58],'emax': [23221, 2198, 4296, 104906, 691] }
dftemp = pd.DataFrame(data=dat)
na=len(dftemp)

# time points
n=51
year=50

# constants
Pa0 = 3.061 
Pe0 = 10.603 
C0 = 100 
r = 0.05 
k=50 
shift=10000000 # to make positive inside log function
ll=0.15

    
for i in range(0,na):
    # create GEKKO model
    m = GEKKO(remote=False)

    m.time = np.linspace(0,year,n)
    t=m.time
    
    A0=dftemp.loc[i][0]
    h=dftemp.loc[i][1]
    emax=dftemp.loc[i][2]    
    
    A = m.SV(value=A0, lb=0, ub=A0)
    E = m.SV(value=0, lb=0, ub=A0)

    u = m.MV(value=0, lb=-emax, ub=emax) 
    u.STATUS = 1

    t = m.Param(value=m.time)
    C = m.Var(value=C0)
    d = m.Var(value=1)
    l = m.Param(value=ll)
    
    # Equation
    m.Equation(A.dt()==-u)
    m.Equation(E.dt()==u)
    m.Equation(C.dt()==-C/k)
    m.Equation(d==m.exp(-t*r))
    # Objective (Utility)
    J = m.Var(value=0)
    
    # Final objective
    Jf = m.FV()
    Jf.STATUS = 1
    m.Connection(Jf,J,pos2='end')
    m.Equation(J.dt() == ((A+E*(1-l))*h*Pa0-C*u+E*Pe0+shift))
    
    # maximize U
    m.Maximize(Jf)
    
    # options
    m.options.IMODE = 6  # optimal control
    m.options.NODES = 3  # collocation nodes
    m.options.SOLVER = 3 # solver (IPOPT)
    
    # solve optimization problem
    m.solve()
    
    # print profit
    print('Optimal Profit: ' + str(Jf.value[0]))

案例5.将目标函数从m.log改为简单线性求和，并将shift改为0，出现错误

from gekko import GEKKO
import pandas as pd
import numpy as np

dat = {'A0': [23221, 2198, 4296, 104906, 691], 'h': [0.04, 0.25, 0.07, 0, 12.58],'emax': [23221, 2198, 4296, 104906, 691] }
dftemp = pd.DataFrame(data=dat)
na=len(dftemp)

# time points
n=51
year=50

# constants
Pa0 = 3.061 
Pe0 = 10.603 
C0 = 100 
r = 0.05 
k=50 
shift=0 
ll=0.15
    
for i in range(0,na):
    # create GEKKO model
    m = GEKKO(remote=False)

    m.time = np.linspace(0,year,n)
    t=m.time
    
    A0=dftemp.loc[i][0]
    h=dftemp.loc[i][1]
    emax=dftemp.loc[i][2]    
    
    A = m.SV(value=A0, lb=0, ub=A0)
    E = m.SV(value=0, lb=0, ub=A0)

    u = m.MV(value=0, lb=-emax, ub=emax) 
    u.STATUS = 1

    t = m.Param(value=m.time)
    C = m.Var(value=C0)
    d = m.Var(value=1)
    l = m.Param(value=ll)
    
    # Equation
    m.Equation(A.dt()==-u)
    m.Equation(E.dt()==u)
    m.Equation(C.dt()==-C/k)
    m.Equation(d==m.exp(-t*r))
    # Objective (Utility)
    J = m.Var(value=0)
    
    # Final objective
    Jf = m.FV()
    Jf.STATUS = 1
    m.Connection(Jf,J,pos2='end')
    m.Equation(J.dt() == ((A+E*(1-l))*h*Pa0-C*u+E*Pe0+shift)*d)
    
    # maximize U
    m.Maximize(Jf)
    
    # options
    m.options.IMODE = 6  # optimal control
    m.options.NODES = 3  # collocation nodes
    m.options.SOLVER = 3 # solver (IPOPT)
    
    # solve optimization problem
    m.solve()
    
    # print profit
    print('Optimal Profit: ' + str(Jf.value[0]))

最佳答案

对于其中的许多情况，进行一些更改可能会有所帮助，例如:

  • 重新制定以避免 m.log() 出现负数。中间变量也有帮助。
# Objective (Utility)
J = m.Var(value=0)
rhs = m.Intermediate((A+E*(1-l))*h*Pa0-C*u+E*Pe0+shift)
m.Equation(m.exp(J.dt()/d)==rhs)
  • 优化前初始化。 APOPT 求解器在初始化方面做得更好，而 IPOPT 使用 APOPT 的初始化解决方案做得更好。 article 中讨论了初始化策略:Safdarnejad, S.M.、Hedengren, J.D.、Lewis, N.R.、Haseltine, E.，动态系统、计算机和化学工程优化的初始化策略，2015 年，卷。 78，第 39-50 页，DOI:10.1016/j.compchemeng.2015.04.016。
# solve optimization problem
m.options.NODES = 3  # collocation nodes
m.options.SOLVER = 1 # solver (APOPT)
m.options.IMODE=4
m.solve()

# solve optimization problem
m.options.TIME_SHIFT=0
m.options.SOLVER=3 # solver (IPOPT)
m.options.IMODE=6
m.solve()

案例 2:n=501 的成功解决方案

from gekko import GEKKO
import pandas as pd
import numpy as np

dat = {'A0': [23221, 2198, 4296, 104906, 691], \
       'h': [0.04, 0.25, 0.07, 0, 12.58],\
       'emax': [23221, 2198, 4296, 104906, 691] }
dftemp = pd.DataFrame(data=dat)
na=len(dftemp)

# time points
n=501 
year=50

# constants
Pa0 = 3.061 
Pe0 = 10.603 
C0 = 100 
r = 0.05 
k=50 
shift=10000000 # to make positive inside log function
ll=0.15

for i in range(0,na):
    # create GEKKO model
    m = GEKKO(remote=True)

    m.time = np.linspace(0,year,n)
    t=m.time
    
    A0=dftemp.loc[i][0]
    h=dftemp.loc[i][1]
    emax=dftemp.loc[i][2]    
    
    A = m.SV(value=A0, lb=0, ub=A0)
    E = m.SV(value=0, lb=0, ub=A0)

    u = m.MV(value=0, lb=-emax, ub=emax) 
    u.STATUS = 1

    t = m.Param(value=m.time)
    C = m.Var(value=C0)
    d = m.Var(value=1)
    l = m.Param(value=ll)
    
    # Equation
    m.Equation(A.dt()==-u)
    m.Equation(E.dt()==u)
    m.Equation(C.dt()==-C/k)
    m.Equation(d==m.exp(-t*r))
    # Objective (Utility)
    J = m.Var(value=0)
    rhs = m.Intermediate((A+E*(1-l))*h*Pa0-C*u+E*Pe0+shift)
    m.Equation(m.exp(J.dt()/d)==rhs)
        
    # Final objective
    final = np.zeros_like(m.time)
    final[-1] = 1
    final = m.Param(final)
    
    # maximize U
    m.Maximize(final*J)
        
    # solve optimization problem
    m.options.NODES = 3  # collocation nodes
    m.options.SOLVER = 1 # solver (APOPT)
    m.options.IMODE=4
    print('\n\nInitializing with APOPT')
    m.solve(disp=False)

    # solve optimization problem
    m.options.TIME_SHIFT=0
    m.options.SOLVER=3 # solver (IPOPT)
    m.options.IMODE=6
    m.solve()
        
    # print profit
    print('Optimal Profit: ' + str(J.value[-1]))

请在其他情况下也尝试使用此方法。通常使用 m.sum() 比 sum 更好，因为 Gekko 将求和构造为内置对象而不是长字符串。

编辑:更新的解决方案 这是一个具有正确初始条件的更新解决方案。 IMODE=4 的初始条件不会转移到 IMODE=6。另一种方法是将 IMODE=6 与等效的 COLDSTART=1 一起使用。

Results 

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

dat = {'A0': [23221, 2198, 4296, 104906, 691], \
       'h': [0.04, 0.25, 0.07, 0, 12.58],\
       'emax': [23221, 2198, 4296, 104906, 691] }
dftemp = pd.DataFrame(data=dat)
na=len(dftemp)

# time points
n=101 
year=50

# constants
Pa0 = 3.061 
Pe0 = 10.603 
C0 = 100 
r = 0.05 
k=50 
shift=10000000 # to make positive inside log function
ll=0.15

plt.figure(figsize=(10,5))
for i in range(0,na):
    # create GEKKO model
    m = GEKKO(remote=True)

    m.time = np.linspace(0,year,n)
    t=m.time
    
    A0=dftemp.loc[i][0]
    h=dftemp.loc[i][1]
    emax=dftemp.loc[i][2]    

    print('A (initial): ' + str(A0))
    A = m.Var(value=A0, lb=0, ub=A0)
    E = m.Var(value=0, lb=0, ub=A0)

    u = m.MV(value=0, lb=-emax, ub=emax) 
    u.STATUS = 1

    t = m.Param(value=m.time)
    C = m.Var(value=C0)
    d = m.Var(value=1)
    l = m.Param(value=ll)
    
    # Equation
    m.Equation(A.dt()==-u)
    m.Equation(E.dt()==u)
    m.Equation(C.dt()==-C/k)
    m.Equation(d==m.exp(-t*r))
    # Objective (Utility)
    J = m.Var(value=0)
    rhs = m.Intermediate((A+E*(1-l))*h*Pa0-C*u+E*Pe0+shift)
    m.Equation(m.exp(J.dt()/d)==rhs)
        
    # Final objective
    final = np.zeros_like(m.time)
    final[-1] = 1
    final = m.Param(final)
    
    # maximize U
    m.Maximize(final*J)
        
    # solve optimization problem
    m.options.NODES = 3  # collocation nodes
    m.options.SOLVER = 1 # solver (APOPT)
    m.options.IMODE=6
    m.options.COLDSTART=1
    print('\n\nInitializing with APOPT')
    m.solve(disp=False)

    # solve optimization problem
    m.options.COLDSTART=0
    m.options.TIME_SHIFT=0
    m.options.SOLVER=3 # solver (IPOPT)
    m.options.IMODE=6
    m.solve()
        
    # print profit
    print('Optimal Profit: ' + str(J.value[-1]))

    plt.plot(m.time,A.value,label='A case '+str(i))
    plt.plot(m.time,E.value,label='E case '+str(i))
    plt.plot(m.time,u.value,label='u case '+str(i))
    

plt.legend()
plt.show()

关于python - 更改参数、目标函数和初始条件时使用 Gekko 优化套件出现 "Solution Not Found"错误，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/66977417/

上一篇：typescript - 类型 'never[]' 不可分配给类型 '...'

下一篇：javascript - 性能 - foreach 内部条件 vs 具有相同条件的过滤器和 javascript 中过滤数组的 foreach

相关文章：

python - TypeError : must be real number, 不是 GK_Operators

python-3.x - 为什么延迟参数会影响预测曲线？

python-3.x - Gekko 中间变量，错误 : equation without equality or inequality

python - “self”未定义 - 使用计时器进行线程处理

c++ - VS2008 中的简单 swig python 示例 - 导入错误 - 内部 (py)readline 错误

python - 在使用 dict.values() 或 dict.itervalues() 迭代时修改字典值

python - 在 python 中复制缓冲区时优化内存使用

python - Cython:内联函数不是纯 C

python - 使用 pandas df 分组计算平均值

node.js - Nodejs 代理请求合并

©2024 IT工具网  联系我们