GEKKO - 使用自定义目标函数的参数估计 - 错误代码 -13 [英] GEKKO - parameter estimation with custom objective function - error code -13
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问题描述
我使用 Gekko 教程中介绍的相同技术(线性和非线性回归)成功地执行了稳态参数估计.代码如下:
I have been successful in performing a steady-state parameter estimation employing the same techniques presented in the Gekko tutorials (linear and non-linear regression). Below is the code:
# -*- coding: utf-8 -*-
"""
Spyder Editor
This is a temporary script file.
"""
from io import StringIO
from gekko import GEKKO
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
# Tb, Dy, Ho, Y, H, (HA)2
# measurements of the initial and final aqueous concentrations
aqueous_species = ["Tb", "Dy", "Ho", "Y"]
# import dataframe
x_i_a_Lns_v = (
pd.read_csv(
StringIO(
"""Tb,Dy,Ho,Y
0.19538,1.22628,0.2242,3.39823
0.28462,1.83411,0.32435,4.90551
0.34769,2.1979,0.39609,5.89209
0.37692,2.39495,0.43794,6.57722
0.41231,2.62232,0.47382,7.09791
0.43538,2.78906,0.5052,7.45418
0.44462,2.88,0.51866,7.89266
0.46,2.92548,0.52912,7.83785
0.46615,2.94064,0.5351,7.97488
0.45846,2.91032,0.52613,7.94747
0.46,2.86485,0.52613,7.89266
0.46769,2.92548,0.53659,8.00228
0.45692,2.92548,0.5351,8.02969
0.47385,2.98611,0.55005,8.13931
0.47385,2.97095,0.54108,8.1119"""
)
)
/ 1000
)
x_i_a_H_v = pd.read_csv(
StringIO(
"""H
10.01809
7.28346
5.16795
3.62036
2.50218
1.7173
1.17411
0.80491
0.54616
0.37078
0.25406
0.16932
0.11262
0.07574
0.04455"""
)
)
x_i_o_HA2_v = pd.read_csv(
StringIO(
"""(HA)2
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746"""
)
)
x_f_a_Lns_v = (
pd.read_csv(
StringIO(
"""Tb,Dy,Ho,Y
0.19231,1.23234,0.21822,3.34342
0.26923,1.74316,0.31239,4.60405
0.33692,2.13727,0.38862,5.72766
0.37385,2.33432,0.41253,6.05652
0.40923,2.50106,0.43196,5.97431
0.38769,2.1979,0.3363,3.91892
0.33692,1.53095,0.19431,1.91287
0.24,0.82307,0.0852,0.77282
0.12,0.33347,0.03139,0.27679
0.05846,0.14552,0.01495,0.1151
0.02615,0.06669,0.00598,0.05207
0.01231,0.03032,,0.02466
,,,0.00548
0.00615,0.01364,,0.01096
,,,0.00548"""
)
)
/ 1000
)
# Issue with NaN (missing value)
idx = x_f_a_Lns_v[x_f_a_Lns_v["Ho"] >= 0].index
# idx = x_f_a_Lns_v.index
#### Solution
# create model to be solved locally
m = GEKKO(remote=False)
x_i_a_Lns = [m.Param(value=x_i_a_Lns_v.loc[idx, v].values) for v in aqueous_species]
x_i_a_H = m.Param(value=x_i_a_H_v.loc[idx, "H"].values)
x_i_o_HA2 = m.Param(value=x_i_o_HA2_v.loc[idx, "(HA)2"].values)
x_f_a_Lns = [m.CV(value=x_f_a_Lns_v.loc[idx, v].values) for v in aqueous_species]
for i in range(len(aqueous_species)):
x_f_a_Lns[i].FSTATUS = 1
# equilibrium constants
K_eqs = [m.FV(value=1) for i in range(len(aqueous_species))]
for i in range(len(x_f_a_Lns_v.columns)):
K_eqs[i].STATUS = 1
# equilibrium model
x_f_a_H = m.Intermediate(
x_i_a_H
+ 3 * m.sum([(x_i_a_Lns[i] - x_f_a_Lns[i]) for i in range(len(aqueous_species))])
)
x_f_o_HA2 = m.Intermediate(
x_i_o_HA2
- 3 * m.sum([(x_i_a_Lns[i] - x_f_a_Lns[i]) for i in range(len(aqueous_species))])
)
m.Equations(
[
K_eqs[i]
== ((x_i_a_Lns[i] - x_f_a_Lns[i]) * x_f_a_H ** 3)
/ (x_f_a_Lns[i] * x_f_o_HA2 ** 3)
for i in range(len(aqueous_species))
]
)
# regression
m.options.IMODE = 2
m.options.EV_TYPE = 2
m.solve(disp=True)
for i, v in enumerate(aqueous_species):
print(f"K_eq_{v}: {K_eqs[i][0]}")
输出:
K_eq_Tb: 5.7327051238
K_eq_Dy: 15.581534791
K_eq_Ho: 27.414244408
K_eq_Y: 46.925190325
这符合预期.
然而，当我尝试实现一个特定的目标函数时，我遇到了一个错误 (-13).我曾尝试为 Var(s) 设置下限，但随后我收到了不同的错误代码 (-1).我希望有人能指出我在哪里犯了错误.我认为目标函数是这样的，应该找到类似的结果.
However, when I try and implement a specific objective function I encounter an error (-13). I have tried setting a lower bound for the Var(s) however I then receive a different error code of (-1). I hope that that somebody can identify where I have made an error. I am of the view that the objective function is such that a similar result should be found.
# -*- coding: utf-8 -*-
"""
Spyder Editor
This is a temporary script file.
"""
from io import StringIO
from gekko import GEKKO
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
# Tb, Dy, Ho, Y, H, (HA)2
# measurements of the initial and final aqueous concentrations
aqueous_species = ["Tb", "Dy", "Ho", "Y"]
# import dataframes
x_i_a_Lns_v = (
pd.read_csv(
StringIO(
"""Tb,Dy,Ho,Y
0.19538,1.22628,0.2242,3.39823
0.28462,1.83411,0.32435,4.90551
0.34769,2.1979,0.39609,5.89209
0.37692,2.39495,0.43794,6.57722
0.41231,2.62232,0.47382,7.09791
0.43538,2.78906,0.5052,7.45418
0.44462,2.88,0.51866,7.89266
0.46,2.92548,0.52912,7.83785
0.46615,2.94064,0.5351,7.97488
0.45846,2.91032,0.52613,7.94747
0.46,2.86485,0.52613,7.89266
0.46769,2.92548,0.53659,8.00228
0.45692,2.92548,0.5351,8.02969
0.47385,2.98611,0.55005,8.13931
0.47385,2.97095,0.54108,8.1119"""
)
)
/ 1000
)
x_i_a_H_v = pd.read_csv(
StringIO(
"""H
10.01809
7.28346
5.16795
3.62036
2.50218
1.7173
1.17411
0.80491
0.54616
0.37078
0.25406
0.16932
0.11262
0.07574
0.04455"""
)
)
x_i_o_HA2_v = pd.read_csv(
StringIO(
"""(HA)2
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746"""
)
)
x_f_a_Lns_v = (
pd.read_csv(
StringIO(
"""Tb,Dy,Ho,Y
0.19231,1.23234,0.21822,3.34342
0.26923,1.74316,0.31239,4.60405
0.33692,2.13727,0.38862,5.72766
0.37385,2.33432,0.41253,6.05652
0.40923,2.50106,0.43196,5.97431
0.38769,2.1979,0.3363,3.91892
0.33692,1.53095,0.19431,1.91287
0.24,0.82307,0.0852,0.77282
0.12,0.33347,0.03139,0.27679
0.05846,0.14552,0.01495,0.1151
0.02615,0.06669,0.00598,0.05207
0.01231,0.03032,,0.02466
,,,0.00548
0.00615,0.01364,,0.01096
,,,0.00548"""
)
)
/ 1000
)
# Issue with NaN (missing value)
idx = x_f_a_Lns_v[x_f_a_Lns_v["Ho"] >= 0].index
# idx = x_f_a_Lns_v.index
#### Solution
# create model to be solved locally
m = GEKKO(remote=False)
x_i_a_Lns = [m.Param(value=x_i_a_Lns_v.loc[idx, v].values) for v in aqueous_species]
x_i_a_H = m.Param(value=x_i_a_H_v.loc[idx, "H"].values)
x_i_o_HA2 = m.Param(value=x_i_o_HA2_v.loc[idx, "(HA)2"].values)
x_f_a_Lns_m = [m.Param(value=x_f_a_Lns_v.loc[idx, v].values) for v in aqueous_species]
x_f_a_Lns = [m.Var() for v in aqueous_species]
# x_f_a_Lns = [m.Var(lb=1e-12) for v in aqueous_species]
# equilibrium constants
K_eqs = [m.FV(value=1) for i in range(len(aqueous_species))]
for i in range(len(aqueous_species)):
K_eqs[i].STATUS = 1
# equilibrium model
x_f_a_H = m.Intermediate(
x_i_a_H
+ 3 * m.sum([(x_i_a_Lns[i] - x_f_a_Lns[i]) for i in range(len(aqueous_species))])
)
x_f_o_HA2 = m.Intermediate(
x_i_o_HA2
- 3 * m.sum([(x_i_a_Lns[i] - x_f_a_Lns[i]) for i in range(len(aqueous_species))])
)
m.Equations(
[
K_eqs[i]
== ((x_i_a_Lns[i] - x_f_a_Lns[i]) * x_f_a_H ** 3)
/ (x_f_a_Lns[i] * x_f_o_HA2 ** 3)
for i in range(len(aqueous_species))
]
)
m.Obj(
m.sum(
[
((x_f_a_Lns_m[i] - x_f_a_Lns[i]) / x_f_a_Lns_m[i]) ** 2
for i in range(len(aqueous_species))
]
)
)
# regression
m.options.IMODE = 2
# m.options.EV_TYPE = 2
m.solve(disp=True)
for i, v in enumerate(aqueous_species):
print(f"K_eq_{v}: {K_eqs[i][0]}")
推荐答案
输出中报告了求解器 IPOPT 的错误代码 -13 描述:
Error code -13 description for solver IPOPT is reported in the output:
EXIT: Invalid number in NLP function or derivative detected.
An error occured.
The error code is -13
这意味着模型中的某处可能被零除.另一个错误 -1 是当检测到不可行的解决方案时.由于无法满足方程和变量约束，额外的约束会阻止求解器找到解.
This means that there is likely divide by zero somewhere in the model. The other error -1 is when an infeasible solution is detected. The additional constraints prevent the solver from finding a solution because the equations and variable constraints can't be satisfied.
要找到成功的解决方案，需要做两件事:
Two things are needed to find a successful solution:
- 重新排列方程以避免被零除.
m.Equations(
[
K_eqs[i]
== ((x_i_a_Lns[i] - x_f_a_Lns[i]) * x_f_a_H ** 3)
/ (x_f_a_Lns[i] * x_f_o_HA2 ** 3)
for i in range(len(aqueous_species))
]
)
重新排列为:
m.Equations(
[
K_eqs[i] * (x_f_a_Lns[i] * x_f_o_HA2 ** 3)
== ((x_i_a_Lns[i] - x_f_a_Lns[i]) * x_f_a_H ** 3)
for i in range(len(aqueous_species))
]
)
- 切换到 APOT 求解器:
m.options.SOLVER=1
现在有一个成功的解决方案.
There is now a successful solution.
# -*- coding: utf-8 -*-
"""
Spyder Editor
This is a temporary script file.
"""
from io import StringIO
from gekko import GEKKO
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
# Tb, Dy, Ho, Y, H, (HA)2
# measurements of the initial and final aqueous concentrations
aqueous_species = ["Tb", "Dy", "Ho", "Y"]
# import dataframes
x_i_a_Lns_v = (
pd.read_csv(
StringIO(
"""Tb,Dy,Ho,Y
0.19538,1.22628,0.2242,3.39823
0.28462,1.83411,0.32435,4.90551
0.34769,2.1979,0.39609,5.89209
0.37692,2.39495,0.43794,6.57722
0.41231,2.62232,0.47382,7.09791
0.43538,2.78906,0.5052,7.45418
0.44462,2.88,0.51866,7.89266
0.46,2.92548,0.52912,7.83785
0.46615,2.94064,0.5351,7.97488
0.45846,2.91032,0.52613,7.94747
0.46,2.86485,0.52613,7.89266
0.46769,2.92548,0.53659,8.00228
0.45692,2.92548,0.5351,8.02969
0.47385,2.98611,0.55005,8.13931
0.47385,2.97095,0.54108,8.1119"""
)
)
/ 1000
)
x_i_a_H_v = pd.read_csv(
StringIO(
"""H
10.01809
7.28346
5.16795
3.62036
2.50218
1.7173
1.17411
0.80491
0.54616
0.37078
0.25406
0.16932
0.11262
0.07574
0.04455"""
)
)
x_i_o_HA2_v = pd.read_csv(
StringIO(
"""(HA)2
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746
0.4746"""
)
)
x_f_a_Lns_v = (
pd.read_csv(
StringIO(
"""Tb,Dy,Ho,Y
0.19231,1.23234,0.21822,3.34342
0.26923,1.74316,0.31239,4.60405
0.33692,2.13727,0.38862,5.72766
0.37385,2.33432,0.41253,6.05652
0.40923,2.50106,0.43196,5.97431
0.38769,2.1979,0.3363,3.91892
0.33692,1.53095,0.19431,1.91287
0.24,0.82307,0.0852,0.77282
0.12,0.33347,0.03139,0.27679
0.05846,0.14552,0.01495,0.1151
0.02615,0.06669,0.00598,0.05207
0.01231,0.03032,,0.02466
,,,0.00548
0.00615,0.01364,,0.01096
,,,0.00548"""
)
)
/ 1000
)
# Issue with NaN (missing value)
idx = x_f_a_Lns_v[x_f_a_Lns_v["Ho"] >= 0].index
# idx = x_f_a_Lns_v.index
#### Solution
# create model to be solved locally
m = GEKKO(remote=False)
x_i_a_Lns = [m.Param(value=x_i_a_Lns_v.loc[idx, v].values) for v in aqueous_species]
x_i_a_H = m.Param(value=x_i_a_H_v.loc[idx, "H"].values)
x_i_o_HA2 = m.Param(value=x_i_o_HA2_v.loc[idx, "(HA)2"].values)
x_f_a_Lns_m = [m.Param(value=x_f_a_Lns_v.loc[idx, v].values) for v in aqueous_species]
x_f_a_Lns = [m.Var() for v in aqueous_species]
# x_f_a_Lns = [m.Var(lb=1e-12) for v in aqueous_species]
# equilibrium constants
K_eqs = [m.FV(value=1) for i in range(len(aqueous_species))]
for i in range(len(aqueous_species)):
K_eqs[i].STATUS = 1
# equilibrium model
x_f_a_H = m.Intermediate(
x_i_a_H
+ 3 * m.sum([(x_i_a_Lns[i] - x_f_a_Lns[i]) for i in range(len(aqueous_species))])
)
x_f_o_HA2 = m.Intermediate(
x_i_o_HA2
- 3 * m.sum([(x_i_a_Lns[i] - x_f_a_Lns[i]) for i in range(len(aqueous_species))])
)
m.Equations(
[
K_eqs[i] * (x_f_a_Lns[i] * x_f_o_HA2 ** 3)
== ((x_i_a_Lns[i] - x_f_a_Lns[i]) * x_f_a_H ** 3)
for i in range(len(aqueous_species))
]
)
m.Obj(
m.sum(
[
((x_f_a_Lns_m[i] - x_f_a_Lns[i]) / x_f_a_Lns_m[i]) ** 2
for i in range(len(aqueous_species))
]
)
)
# regression
m.options.IMODE = 2
m.options.SOLVER=1
# m.options.EV_TYPE = 2
m.solve(disp=True)
for i, v in enumerate(aqueous_species):
print(f"K_eq_{v}: {K_eqs[i][0]}")
Successful solution
---------------------------------------------------
Solver : APOPT (v1.0)
Solution time : 0.06899999999999999 sec
Objective : 0.32417750846391324
Successful solution
---------------------------------------------------
K_eq_Tb: 5.489166594
K_eq_Dy: 15.245134386
K_eq_Ho: 30.529918885
K_eq_Y: 54.83667882
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