使用 GEKKO（二阶微分方程）的估计参数
[英]Estimation parameters using GEKKO (a second-order differential equation)
问题描述
I am using a second-order parallel model:
我使用的是二阶并行 model：
It has four parameters (initial values Cf0 and Cs0, kf and ks) to be estimated using GEKKO and six experiments.它有四个参数（初始值 Cf0 和 Cs0、kf 和 ks）要使用 GEKKO 和六个实验进行估计。 However I am not achieving a good fit, only experiments 1 and 4 are moving near the experiment data.但是我没有达到很好的拟合效果，只有实验 1 和 4 接近实验数据。 Maybe I have some error in the code.也许我在代码中有一些错误。 Despite this, any guide-lines or link to get more information on controlling boundaries and the estimation method would be very helpful.尽管如此，任何获得有关控制边界和估计方法的更多信息的指南或链接都会非常有帮助。
from gekko import GEKKO
import numpy as np
import matplotlib.pyplot as plt
import math as math
import pandas as pd
tm1 = [0, 0.0667,0.5,1,4]
mca1 = [5.68, 3.48, 3.24, 3.36, 2.96]
tm2 = [0, 0.08333,0.5,1,4.25]
mca2 = [5.68, 4.20, 4.04, 4.00, 3.76]
tm3 = [0,0.08333,0.5,1,4.33]
mca3 = [5.68, 4.64, 4.52, 4.56, 4.24]
tm4 = [0,0.08333,0.5,1,4.0833]
mca4 =[18.90,15.4,14.3,15.10,13.50]
tm5 = [0,0.08333,0.5,1,4.5]
mca5 =[18.90, 15.5, 16.30, 16, 14.70]
tm6 = [0,0.08333,0.5,1,4.6667]
mca6 = [18.90, 15.8, 11.70,15.5,12]
df1=pd.DataFrame({'time':tm1,'x1':mca1})
df2=pd.DataFrame({'time':tm2,'x2':mca2})
df3=pd.DataFrame({'time':tm3,'x3':mca3})
df4=pd.DataFrame({'time':tm4,'x4':mca4})
df5=pd.DataFrame({'time':tm5,'x5':mca5})
df6=pd.DataFrame({'time':tm6,'x6':mca6})
df1.set_index('time',inplace=True)
df2.set_index('time',inplace=True)
df3.set_index('time',inplace=True)
df4.set_index('time',inplace=True)
df5.set_index('time',inplace=True)
df6.set_index('time',inplace=True)
#simulation time points
dfx = pd.DataFrame({'time':np.linspace(0,5,101)})
dfx.set_index('time',inplace=True)
#merge dataframes
dfxx=dfx.join(df1,how='outer')
dfxxx=dfxx.join(df2,how='outer')
dfxxxx=dfxxx.join(df3,how='outer')
dfxxxxx=dfxxxx.join(df4,how='outer')
dfxxxxxx=dfxxxxx.join(df5,how='outer')
df=dfxxxxxx.join(df6,how='outer')
# get True (1) or False (0) for measurement
df['meas1']=(df['x1'].values==df['x1'].values).astype(int)
df['meas2']=(df['x2'].values==df['x2'].values).astype(int)
df['meas3']=(df['x3'].values==df['x3'].values).astype(int)
df['meas4']=(df['x4'].values==df['x4'].values).astype(int)
df['meas5']=(df['x5'].values==df['x5'].values).astype(int)
df['meas6']=(df['x6'].values==df['x6'].values).astype(int)
#replace NaN with zeros
df0=df.fillna(value=0)
m = GEKKO()
m.time = df0.index.values
meas1 = m.Param(df0['meas1'].values)
meas2 = m.Param(df0['meas2'].values)
meas3 = m.Param(df0['meas3'].values)
meas4 = m.Param(df0['meas4'].values)
meas5 = m.Param(df0['meas5'].values)
meas6 = m.Param(df0['meas6'].values)
#adjustable Parameters
cnf01=m.FV(1.3,lb=0.01,ub=10)
cns01=m.FV(1.3,lb=0.01,ub=10)
kf=m.FV(1.3,lb=0.01,ub=10)
ks=m.FV(1.3,lb=0.01,ub=10)
cnf02=m.FV(value=cnf01*0.5,lb=cnf01*0.5, ub=cnf01*0.5)
cns02=m.FV(value=cns01*0.5,lb=cns01*0.5, ub=cns01*0.5)
cnf03=m.FV(value=cnf01*0.25,lb=cnf01*0.25, ub=cnf01*0.25)
cns03=m.FV(value=cns01*0.25,lb=cns01*0.25, ub=cns01*0.25)
cnf04=m.FV(value=cnf01,lb=cnf01, ub=cnf01)
cns04=m.FV(value=cns01,lb=cns01, ub=cns01)
cnf05=m.FV(value=cnf01*0.5,lb=cnf01*0.5, ub=cnf01*0.5)
cns05=m.FV(value=cns01*0.5,lb=cns01*0.5, ub=cns01*0.5)
cnf06=m.FV(value=cnf01*0.25,lb=cnf01*0.25, ub=cnf01*0.25)
cns06=m.FV(value=cns01*0.25,lb=cns01*0.25, ub=cns01*0.25)
#Variables
c1 = m.Var(value=mca1[0])
c2 = m.Var(value=mca2[0])
c3 = m.Var(value=mca3[0])
c4 = m.Var(value=mca4[0])
c5 = m.Var(value=mca5[0])
c6 = m.Var(value=mca6[0])
cm1 = m.Param(df0['x1'].values)
cm2 = m.Param(df0['x2'].values)
cm3 = m.Param(df0['x3'].values)
cm4 = m.Param(df0['x4'].values)
cm5 = m.Param(df0['x5'].values)
cm6 = m.Param(df0['x6'].values)
m.Minimize((meas1*(c1-cm1)**2)+(meas2*(c2-cm2)**2)+(meas3*(c3-cm3)**2)+(meas4*(c4-cm4)**2)+(meas5*(c5-cm5)**2)+(meas6*(c6-cm6)**2))
cnf1=m.Var(value=cnf01)
cns1=m.Var(value=cns01)
cnf2=m.Var(value=cnf02)
cns2=m.Var(value=cns02)
cnf3=m.Var(value=cnf03)
cns3=m.Var(value=cns03)
cnf4=m.Var(value=cnf04)
cns4=m.Var(value=cns04)
cnf5=m.Var(value=cnf05)
cns5=m.Var(value=cns05)
cnf6=m.Var(value=cnf06)
cns6=m.Var(value=cns06)
#Equations
t = m.Param(value=m.time)
m.Equation(cnf1.dt()==-kf*c1*cnf1)
m.Equation(cns1.dt()==-ks*c1*cns1)
m.Equation(c1.dt()==cnf1.dt()+cns1.dt())
m.Equation(cnf2.dt()==-kf*c2*cnf2)
m.Equation(cns2.dt()==-ks*c2*cns2)
m.Equation(c2.dt()==cnf2.dt()+cns2.dt())
m.Equation(cnf3.dt()==-kf*c3*cnf3)
m.Equation(cns3.dt()==-ks*c3*cns3)
m.Equation(c3.dt()==cnf3.dt()+cns3.dt())
m.Equation(cnf4.dt()==-kf*c4*cnf4)
m.Equation(cns4.dt()==-ks*c4*cns4)
m.Equation(c4.dt()==cnf4.dt()+cns4.dt())
m.Equation(cnf5.dt()==-kf*c5*cnf5)
m.Equation(cns5.dt()==-ks*c5*cns5)
m.Equation(c5.dt()==cnf5.dt()+cns5.dt())
m.Equation(cnf6.dt()==-kf*c6*cnf6)
m.Equation(cns6.dt()==-ks*c6*cns6)
m.Equation(c6.dt()==cnf6.dt()+cns6.dt())
m.Equation(ks>0)
m.Equation(kf>0)
m.Equation(cnf01>0)
m.Equation(cns01>0)
#Options
m.options.SOLVER = 3 #IPOPT solver
m.options.IMODE = 5 #Dynamic Simultaneous - estimation = MHE
m.options.EV_TYPE = 2 #absolute error
m.options.NODES = 3 #collocation nodes (2,5)
m.solve(disp=True)
if True:
kf.STATUS=1
ks.STATUS=1
cnf01.STATUS=1
cns01.STATUS=1
cnf02.STATUS=1
cns02.STATUS=1
cnf03.STATUS=1
cns03.STATUS=1
cnf04.STATUS=1
cns04.STATUS=1
cnf05.STATUS=1
cns05.STATUS=1
cnf06.STATUS=1
cns06.STATUS=1
print('Final SSE Objective: ' + str(m.options.objfcnval))
print('Solution')
print('cnf01 = ' + str(cnf01.value[0]))
print('cns01 = ' + str(cns01.value[0]))
print('kf = ' + str(kf.value[0]))
print('ks = ' + str(ks.value[0]))
print('cns02 = '+ str(cns02.value[0]))
print('cnf02 = '+ str(cnf02.value[0]))
print('cns03 = '+ str(cns03.value[0]))
print('cnf03 = '+ str(cnf03.value[0]))
print('cns04 = '+ str(cns04.value[0]))
print('cnf04 = '+ str(cnf04.value[0]))
print('cns05 = '+ str(cns05.value[0]))
print('cnf05 = '+ str(cnf05.value[0]))
print('cns06 = '+ str(cns06.value[0]))
print('cnf06 = '+ str(cnf06.value[0]))
plt.figure(1,figsize=(8,5))
plt.plot(m.time,c1.value,'b',label='Predicted c1')
plt.plot(m.time,c2.value,'r',label='Predicted c2')
plt.plot(m.time,c3.value,'g',label='Predicted c3')
plt.plot(m.time,c4.value,'b',label='Predicted c4')
plt.plot(m.time,c5.value,'r',label='Predicted c5')
plt.plot(m.time,c6.value,'g',label='Predicted c6')
plt.plot(tm1,mca1,'bo',label='Meas c1')
plt.plot(tm2,mca2,'ro',label='Meas c2')
plt.plot(tm3,mca3,'go',label='Meas c3')
plt.plot(tm4,mca4,'bo',label='Meas c4')
plt.plot(tm5,mca5,'ro',label='Meas c5')
plt.plot(tm6,mca6,'go',label='Meas c6')
plt.xlabel('time (h)')
plt.ylabel('Concentration (mg/L)')
plt.legend(loc='best')
1 个解决方案
解决方案1
0 2021-02-09 12:02:39
The STATUS=1 needs to be before the m.solve() to make kf and ks adjustable by the solver. STATUS=1需要在m.solve()之前，以使求解器可以调整kf和ks 。
if True:
kf.STATUS=1
ks.STATUS=1
The additional inequality constraint aren't needed because they already have lower bounds set elsewhere.不需要额外的不等式约束，因为它们已经在别处设置了下限。
m.Equation(ks>0)
m.Equation(kf>0)
m.Equation(cnf01>0)
m.Equation(cns01>0)
The initial conditions are calculated by including fixed_initial=False .初始条件是通过包含fixed_initial=False来计算的。
cnf1=m.Var(value=1.3,lb=0.0,ub=10,fixed_initial=False)
cns1=m.Var(value=1.3,lb=0.0,ub=10,fixed_initial=False)
cnf2=m.Var(value=1.3,lb=0.0,ub=10,fixed_initial=False)
cns2=m.Var(value=1.3,lb=0.0,ub=10,fixed_initial=False)
cnf3=m.Var(value=1.3,lb=0.0,ub=10,fixed_initial=False)
cns3=m.Var(value=1.3,lb=0.0,ub=10,fixed_initial=False)
cnf4=m.Var(value=1.3,lb=0.0,ub=10,fixed_initial=False)
cns4=m.Var(value=1.3,lb=0.0,ub=10,fixed_initial=False)
cnf5=m.Var(value=1.3,lb=0.0,ub=10,fixed_initial=False)
cns5=m.Var(value=1.3,lb=0.0,ub=10,fixed_initial=False)
cnf6=m.Var(value=1.3,lb=0.0,ub=10,fixed_initial=False)
cns6=m.Var(value=1.3,lb=0.0,ub=10,fixed_initial=False)
One way to shorten the model is to use Arrays.缩短 model 的一种方法是使用 Arrays。
cnf1,cnf2,cnf3,cnf4,cnf5,cnf6 = m.Array(m.Var,6,\
value=1.3,lb=0.0,ub=10,fixed_initial=False
cns1,cns2,cns3,cns4,cns5,cns6 = m.Array(m.Var,6,\
value=1.3,lb=0.0,ub=10,fixed_initial=False
The IPOPT solver fails to converge after 250 iterations. IPOPT 求解器在 250 次迭代后无法收敛。 Switching to the APOPT solver gives a successful solution.切换到 APPT 求解器可提供成功的解决方案。
m.options.SOLVER=1
from gekko import GEKKO
import numpy as np
import matplotlib.pyplot as plt
import math as math
import pandas as pd
tm1 = [0, 0.0667,0.5,1,4]
mca1 = [5.68, 3.48, 3.24, 3.36, 2.96]
tm2 = [0, 0.08333,0.5,1,4.25]
mca2 = [5.68, 4.20, 4.04, 4.00, 3.76]
tm3 = [0,0.08333,0.5,1,4.33]
mca3 = [5.68, 4.64, 4.52, 4.56, 4.24]
tm4 = [0,0.08333,0.5,1,4.0833]
mca4 =[18.90,15.4,14.3,15.10,13.50]
tm5 = [0,0.08333,0.5,1,4.5]
mca5 =[18.90, 15.5, 16.30, 16, 14.70]
tm6 = [0,0.08333,0.5,1,4.6667]
mca6 = [18.90, 15.8, 11.70,15.5,12]
df1=pd.DataFrame({'time':tm1,'x1':mca1})
df2=pd.DataFrame({'time':tm2,'x2':mca2})
df3=pd.DataFrame({'time':tm3,'x3':mca3})
df4=pd.DataFrame({'time':tm4,'x4':mca4})
df5=pd.DataFrame({'time':tm5,'x5':mca5})
df6=pd.DataFrame({'time':tm6,'x6':mca6})
df1.set_index('time',inplace=True)
df2.set_index('time',inplace=True)
df3.set_index('time',inplace=True)
df4.set_index('time',inplace=True)
df5.set_index('time',inplace=True)
df6.set_index('time',inplace=True)
#simulation time points
dfx = pd.DataFrame({'time':np.linspace(0,5,101)})
dfx.set_index('time',inplace=True)
#merge dataframes
dfxx=dfx.join(df1,how='outer')
dfxxx=dfxx.join(df2,how='outer')
dfxxxx=dfxxx.join(df3,how='outer')
dfxxxxx=dfxxxx.join(df4,how='outer')
dfxxxxxx=dfxxxxx.join(df5,how='outer')
df=dfxxxxxx.join(df6,how='outer')
# get True (1) or False (0) for measurement
df['meas1']=(df['x1'].values==df['x1'].values).astype(int)
df['meas2']=(df['x2'].values==df['x2'].values).astype(int)
df['meas3']=(df['x3'].values==df['x3'].values).astype(int)
df['meas4']=(df['x4'].values==df['x4'].values).astype(int)
df['meas5']=(df['x5'].values==df['x5'].values).astype(int)
df['meas6']=(df['x6'].values==df['x6'].values).astype(int)
#replace NaN with zeros
df0=df.fillna(value=0)
m = GEKKO()
m.time = df0.index.values
meas1 = m.Param(df0['meas1'].values)
meas2 = m.Param(df0['meas2'].values)
meas3 = m.Param(df0['meas3'].values)
meas4 = m.Param(df0['meas4'].values)
meas5 = m.Param(df0['meas5'].values)
meas6 = m.Param(df0['meas6'].values)
#adjustable Parameters
kf=m.FV(1.3,lb=0.01,ub=10)
ks=m.FV(1.3,lb=0.01,ub=10)
#Variables
c1 = m.Var(value=mca1[0])
c2 = m.Var(value=mca2[0])
c3 = m.Var(value=mca3[0])
c4 = m.Var(value=mca4[0])
c5 = m.Var(value=mca5[0])
c6 = m.Var(value=mca6[0])
cm1 = m.Param(df0['x1'].values)
cm2 = m.Param(df0['x2'].values)
cm3 = m.Param(df0['x3'].values)
cm4 = m.Param(df0['x4'].values)
cm5 = m.Param(df0['x5'].values)
cm6 = m.Param(df0['x6'].values)
m.Minimize((meas1*(c1-cm1)**2)+(meas2*(c2-cm2)**2)\
+(meas3*(c3-cm3)**2)+(meas4*(c4-cm4)**2)\
+(meas5*(c5-cm5)**2)+(meas6*(c6-cm6)**2))
cnf1,cnf2,cnf3,cnf4,cnf5,cnf6 = m.Array(m.Var,6,\
value=1.3,lb=0.0,ub=10,fixed_initial=False)
cns1,cns2,cns3,cns4,cns5,cns6 = m.Array(m.Var,6,\
value=1.3,lb=0.0,ub=10,fixed_initial=False)
#Equations
t = m.Param(value=m.time)
m.Equation(cnf1.dt()==-kf*c1*cnf1)
m.Equation(cns1.dt()==-ks*c1*cns1)
m.Equation(c1.dt()==cnf1.dt()+cns1.dt())
m.Equation(cnf2.dt()==-kf*c2*cnf2)
m.Equation(cns2.dt()==-ks*c2*cns2)
m.Equation(c2.dt()==cnf2.dt()+cns2.dt())
m.Equation(cnf3.dt()==-kf*c3*cnf3)
m.Equation(cns3.dt()==-ks*c3*cns3)
m.Equation(c3.dt()==cnf3.dt()+cns3.dt())
m.Equation(cnf4.dt()==-kf*c4*cnf4)
m.Equation(cns4.dt()==-ks*c4*cns4)
m.Equation(c4.dt()==cnf4.dt()+cns4.dt())
m.Equation(cnf5.dt()==-kf*c5*cnf5)
m.Equation(cns5.dt()==-ks*c5*cns5)
m.Equation(c5.dt()==cnf5.dt()+cns5.dt())
m.Equation(cnf6.dt()==-kf*c6*cnf6)
m.Equation(cns6.dt()==-ks*c6*cns6)
m.Equation(c6.dt()==cnf6.dt()+cns6.dt())
if True:
kf.STATUS=1
ks.STATUS=1
#Options
m.options.SOLVER = 1 #IPOPT solver
m.options.IMODE = 5 #Dynamic Simultaneous - estimation = MHE
m.options.EV_TYPE = 2 #absolute error
m.options.NODES = 3 #collocation nodes (2,5)
m.solve(disp=True)
print('Final SSE Objective: ' + str(m.options.objfcnval))
print('Solution')
print('kf = ' + str(kf.value[0]))
print('ks = ' + str(ks.value[0]))
if True:
print('cnf01 = ' + str(cnf1.value[0]))
print('cns01 = ' + str(cns1.value[0]))
print('cns02 = '+ str(cns2.value[0]))
print('cnf02 = '+ str(cnf2.value[0]))
print('cns03 = '+ str(cns3.value[0]))
print('cnf03 = '+ str(cnf3.value[0]))
print('cns04 = '+ str(cns4.value[0]))
print('cnf04 = '+ str(cnf4.value[0]))
print('cns05 = '+ str(cns5.value[0]))
print('cnf05 = '+ str(cnf5.value[0]))
print('cns06 = '+ str(cns6.value[0]))
print('cnf06 = '+ str(cnf6.value[0]))
plt.figure(1,figsize=(8,5))
plt.plot(m.time,c1.value,'b',label='Predicted c1')
plt.plot(m.time,c2.value,'r',label='Predicted c2')
plt.plot(m.time,c3.value,'g',label='Predicted c3')
plt.plot(m.time,c4.value,'b',label='Predicted c4')
plt.plot(m.time,c5.value,'r',label='Predicted c5')
plt.plot(m.time,c6.value,'g',label='Predicted c6')
plt.plot(tm1,mca1,'bo',label='Meas c1')
plt.plot(tm2,mca2,'ro',label='Meas c2')
plt.plot(tm3,mca3,'go',label='Meas c3')
plt.plot(tm4,mca4,'bo',label='Meas c4')
plt.plot(tm5,mca5,'ro',label='Meas c5')
plt.plot(tm6,mca6,'go',label='Meas c6')
plt.xlabel('time (h)')
plt.ylabel('Concentration (mg/L)')
plt.legend(loc='best')
plt.show()
---------------------------------------------------
Solver : APOPT (v1.0)
Solution time : 20.3781000000017 sec
Objective : 33.5882065490541
Successful solution
---------------------------------------------------
Final SSE Objective: 33.588206549
Solution
kf = 0.01
ks = 1.0397291201
cnf01 = 0.0
cns01 = 2.7570924931
cns02 = 1.9055347775
cnf02 = 0.0
cns03 = 1.2943612345
cnf03 = 0.58537442202
cns04 = 3.9482253816
cnf04 = 3.0425925867
cns05 = 2.8833138483
cnf05 = 2.3309239496
cns06 = 4.6060598019
cnf06 = 4.5472709683
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