python - 参数估计和最小二乘法最小化——Python 和 Gekko

Question

我正在尝试估计 4 个参数(kdoc1、kdoc2、S1、S2)，使用 GEKKO 求解代数常方程组，最小化观察值和预测值之间残差的平方和。我在测量数据和预测数据之间没有得到很好的调整，你对我在代码中做错了什么有什么建议吗？
此外，我也在尝试添加另一组测量数据。我希望模型使用两个测量数据集来估计参数，我需要使用这两个数据集来最小化残差之和。您知道如何在代码中执行此操作吗？

我已经使用建议 python - Infrequent measurements in Gekko with extra simulation points - Stack Overflow 添加了额外的点来帮助求解器收敛

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

#TOC=5.93 mg/L
#M2 C1 5mg/L
#time: [0,0.08333,0.5,1,4,22.61667]; C2 measured: [0,9.33e-5,8.55e-5,8.55e-5,7.77e-5,7.00e-5]
#M4 C1 20mg/l
#time: [0,0.08333,0.5,1,4,22.61667]; C2 measured: [0,2.92e-4,2.72e-4,2.92e-4,2.72e-4,2.14e-4]


#measurements
t_data = [0.08333,0.5,1,4,22.61667]
x_data = [9.33e-5,8.55e-5,8.55e-5,7.77e-5,7.00e-5]
#x_data2 = [2.92e-4,2.72e-4,2.92e-4,2.72e-4,2.14e-4]
df1 = pd.DataFrame({'time':t_data,'x':x_data})
df1.set_index('time', inplace=True)
# simulation time points
df2 = pd.DataFrame({'time':np.linspace(0,25,51)}) #(0,25,51)
df2.set_index('time', inplace=True)
# merge dataframes
df = df2.join(df1,how='outer')
# get True (1) or False (0) for measurement
df['meas'] = (df['x'].values==df['x'].values).astype(int)
# replace NaN with zeros
df0 = df.fillna(value=0)

#Estimator Model
m = GEKKO()
m.time = df.index.values

# adjustable parameters
kdoc1 = m.FV() #m-1h-1
kdoc2 = m.FV() #m-1h-1
S1 = m.FV()
S2 = m.FV()

#State variables
# M (mol/L)
C0 = m.Var(value=8.00e-5)
C1 = m.Var(value=1.36e-3)
C2 = m.Var(value=x_data[0])
C2m = m.Param(df0['x'].values) 
meas = m.Param(df0['meas'].values)
m.Minimize(meas*(C2-C2m)**2)
C3 = m.Var(value=0)
C4 = m.Var(value=3.16e-8)
C5 = m.Var(value=3.83e-7)
C6 = m.Var(value=0)
C7 = m.Var(value=0)
C8 = m.Var(value=0)
C9 = m.Var(value=0)
C10 = m.Var(value=0)
C11 = m.Var(value=0)
C12 = m.Var(value=3.21e-3)
TOC = m.Var(value=5.93)#mg-C/L 
cC1 = m.Var(value=0)
cC2 = m.Var(value=0)

#temperature ºC
T = 20 

#Rate constants
k1 = 2040000000*math.exp(-1887/(273.15+T))*3600
k2 = 138000000*math.exp(-8800/(273.15+T))*3600
k3 = 300000*math.exp(-2010/(273.15+T))*3600
k4 = 0.00000065*3600
k6 = 26700*3600
k7 = 167*3600
k8 = 27700*3600
k9 = 8300*3600
k10 = 0
k5 = 37800000000*math.exp(-2169/(273.15+T))*C4+2.52e25*math.exp(-16860/(273.15+T))*C10+0.87*math.exp(-503/(273.15+T))*C9
 

r1 = k1 * C0 * C1
r2 = k2 * C2
r3 = k3 * C0 * C2
r4 = k4 * C3
r5 = k5 * C2 * C2
r6 = k6 * C3 * C1* C4
r7 = k7 * C3 * C5
r8 = k8 * C6 * C3
r9 = k9 * C6 * C2
r10 = k10 * C2 * C3
r11 = kdoc1*S1*TOC*C2/12000
r12 = kdoc2*S2*TOC*C0/12000

t = m.Param(value=m.time)
m.Equation(C0.dt()== -r1 + r2 - r3 + r4 + r8 - r12)
m.Equation(C1.dt()== -r1 + r2 + r5 - r6 + r11)
m.Equation(C2.dt()== r1 - r2 - r3 + r4 - r5 + r6 - r9 - r10 - r11)
m.Equation(C3.dt()== r3 - r4 + r5 - r6 - r7 - r8 - r10)
m.Equation(C4.dt()== 0)
m.Equation(C5 == 1e-14/C4)
m.Equation(C6.dt()== r7 - r8 - r9)
m.Equation(C7 == (3.16e-8*C0)/C4)
m.Equation(C1 == (5.01e-10*C8)/C4)
m.Equation(C11 == (5.01e-11*C10)/C4)
m.Equation(C9 == (5.01e-7*C9)/C4)
m.Equation(C10 == C12 - 2*C11 - C5 + C4)
m.Equation(C12.dt()== 0)
m.Equation(TOC.dt()== 0)
m.Equation(cC1 == 17000*C1)
m.Equation(cC2 == 51500*C2)

#Application options
m.options.SOLVER = 1 #APOPT solver
m.options.IMODE = 5 #Dynamic Simultaneous - estimation
m.options.EV_TYPE = 2 #absolute error
m.options.NODES = 5 #collocation nodes (2,5)

if True:
    kdoc1.STATUS=1
    kdoc2.STATUS=1
    S1.STATUS=1
    S2.STATUS=1

#Solve
m.solve(disp=False)


#Results
data={'Time (h)':t,'C0 (M)':C0.value,'C1 (M)':C1.value, 'C2 (M)':C2.value,'C3 (M)':C3.value,'C4 (M)':C4.value,'C5 (M)':C5.value,'C6 (M)':C6.value,'C7 (M)':C7.value,'C8 (M)':C8.value,'C9 (M)':C9.value,'C10 (M)':C10.value,'C11 (M)':C11.value,'c12 (M)':C12.value}
dfr = pd.DataFrame(data, columns=['Time (h)','C0 (M)','C1 (M)','C2 (M)','C3 (M)','C4 (M)','C5 (M)','C6 (M)','C7 (M)','C8 (M)','C9 (M)','C10 (M)','C11 (M)','C12 (M)'])
#dfr.to_csv(r'C:\dfestimation.csv', index = False)

print('Solution')
print('kdoc1 = ' + str(kdoc1.value[0]))
print('kdoc2 = ' + str(kdoc2.value[0]))
print('S1 = ' + str(S1.value[0]))
print('S2 = ' + str(S2.value[0]))

plt.figure(1,figsize=(12,8))
plt.subplot(3,1,1)
plt.plot(m.time,C2.value,'bo',label='Predicted')
plt.plot(m.time,df['x'].values,'rs',label='Meas')
plt.plot(m.time,C2.value,label ='C2')
plt.legend(loc='best')
plt.subplot(3,1,2)
plt.plot(m.time,cC2.value,label ='C2')
plt.plot(m.time,cC1.value,label ='C1')
plt.legend(loc='best')
plt.subplot(3,1,3)
plt.plot(m.time,C0.value,label ='C0')
plt.plot(m.time,C1.value,label ='C1')
plt.plot(m.time,C3.value,label ='C3')
plt.plot(m.time,C4.value,label ='C4')
plt.plot(m.time,C5.value,label ='C5')
plt.plot(m.time,C6.value,label ='C6')
plt.plot(m.time,C7.value,label ='C7-')
plt.plot(m.time,C8.value,label ='C8')
plt.plot(m.time,C9.value,label ='C9')
plt.plot(m.time,C10.value,label ='C10')
plt.plot(m.time,C11.value,label ='C11')
plt.plot(m.time,C12.value,label ='C12')

plt.xlabel('time (h)')
plt.ylabel('Concentration (mol/L)')
plt.legend(loc='best')

Answer 1

这是更接近的新结果。

我向参数添加了约束并为它们提供了非零起始值 (1)。

kdoc1 = m.FV(1,lb=0.01,ub=10) #m-1h-1
kdoc2 = m.FV(1,lb=0.01,ub=10) #m-1h-1
S1 = m.FV(1,lb=0.01,ub=10)
S2 = m.FV(1,lb=0.01,ub=10)

计算了 C0 和 C1 的初始条件，因为它们在 C2 中产生了巨大的瞬时跳跃。如果这些初始浓度已知，那么您可能需要切换回固定的初始条件。

m.free_initial(C0)
m.free_initial(C1)

缩放了目标函数，因为 1e-4**2 已经是一个小数并且可能已经满足目标收敛标准。原问题的目标函数值很小。我让每个平方值大约等于 1 作为起点。

m.Minimize(meas*(1e4*(C2-C2m))**2)

这提供了一个合理的解决方案，您可以从中获得所需的回归结果。

kdoc1 = 3.7210285146
kdoc2 = 2.5743961631
S1 = 3.7210285146
S2 = 2.5743961631

参数值未在约束条件下结束，因此这意味着此解决方案是最优的。尽管参数约束在最后没有激活，但约束确实有助于在适当的搜索区域引导求解器。

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

#measurements
t_data = [0.08333,0.5,1,4,22.61667]
x_data = [9.33e-5,8.55e-5,8.55e-5,7.77e-5,7.00e-5]
#x_data2 = [2.92e-4,2.72e-4,2.92e-4,2.72e-4,2.14e-4]
df1 = pd.DataFrame({'time':t_data,'x':x_data})
df1.set_index('time', inplace=True)
# simulation time points
df2 = pd.DataFrame({'time':np.linspace(0,25,51)})
df2.set_index('time', inplace=True)
# merge dataframes
df = df2.join(df1,how='outer')
# get True (1) or False (0) for measurement
df['meas'] = (df['x'].values==df['x'].values).astype(int)
# replace NaN with zeros
df0 = df.fillna(value=0)
#Estimator Model
m = GEKKO()
m.time = df.index.values

# adjustable parameters
kdoc1 = m.FV(1,lb=0.01,ub=10) #m-1h-1
kdoc2 = m.FV(1,lb=0.01,ub=10) #m-1h-1
S1 = m.FV(1,lb=0.01,ub=10)
S2 = m.FV(1,lb=0.01,ub=10)

#State variables
# M (mol/L)
C0 = m.Var(value=8.00e-5,lb=0,ub=1e-3)
m.free_initial(C0)
C1 = m.Var(value=1.36e-3,lb=0,ub=2e-3)
m.free_initial(C1)
C2 = m.Var(value=x_data[0])
C2m = m.Param(df0['x'].values,lb=0)
meas = m.Param(df0['meas'].values)
m.Minimize(meas*(1e4*(C2-C2m))**2)
C3 = m.Var(value=0)
C4 = m.Var(value=3.16e-8)
C5 = m.Var(value=3.83e-7)
C6 = m.Var(value=0)
C7 = m.Var(value=0)
C8 = m.Var(value=0)
C9 = m.Var(value=0)
C10 = m.Var(value=0)
C11 = m.Var(value=0)
C12 = m.Var(value=3.21e-3)
TOC = m.Var(value=5.93)#mg-C/L 
cC1 = m.Var(value=0)
cC2 = m.Var(value=0)

#temperature ºC
T = 20 

#Rate constants
k1 = 2040000000*math.exp(-1887/(273.15+T))*3600
k2 = 138000000*math.exp(-8800/(273.15+T))*3600
k3 = 300000*math.exp(-2010/(273.15+T))*3600
k4 = 0.00000065*3600
k6 = 26700*3600
k7 = 167*3600
k8 = 27700*3600
k9 = 8300*3600
k10 = 0
k5 = 37800000000*math.exp(-2169/(273.15+T))*C4\
     +2.52e25*math.exp(-16860/(273.15+T))*C10\
     +0.87*math.exp(-503/(273.15+T))*C9
 
r1 = k1 * C0 * C1
r2 = k2 * C2
r3 = k3 * C0 * C2
r4 = k4 * C3
r5 = k5 * C2 * C2
r6 = k6 * C3 * C1* C4
r7 = k7 * C3 * C5
r8 = k8 * C6 * C3
r9 = k9 * C6 * C2
r10 = k10 * C2 * C3
r11 = kdoc1*S1*TOC*C2/12000
r12 = kdoc2*S2*TOC*C0/12000

t = m.Param(value=m.time)
m.Equation(C0.dt()== -r1 + r2 - r3 + r4 + r8 - r12)
m.Equation(C1.dt()== -r1 + r2 + r5 - r6 + r11)
m.Equation(C2.dt()== r1 - r2 - r3 + r4 - r5 + r6 - r9 - r10 - r11)
m.Equation(C3.dt()== r3 - r4 + r5 - r6 - r7 - r8 - r10)
m.Equation(C4.dt()== 0)
m.Equation(C5 == 1e-14/C4)
m.Equation(C6.dt()== r7 - r8 - r9)
m.Equation(C7 == (3.16e-8*C0)/C4)
m.Equation(C1 == (5.01e-10*C8)/C4)
m.Equation(C11 == (5.01e-11*C10)/C4)
m.Equation(C9 == (5.01e-7*C9)/C4)
m.Equation(C10 == C12 - 2*C11 - C5 + C4)
m.Equation(C12.dt()== 0)
m.Equation(TOC.dt()== 0)
m.Equation(cC1 == 17000*C1)
m.Equation(cC2 == 51500*C2)

#Application options
m.options.SOLVER = 3 #APOPT solver
m.options.IMODE = 5 #Dynamic Simultaneous - estimation
m.options.EV_TYPE = 2 #absolute error
m.options.NODES = 3 #collocation nodes (2,5)

if True:
    kdoc1.STATUS=1
    kdoc2.STATUS=1
    S1.STATUS=1
    S2.STATUS=1

#Solve
m.solve(disp=True)

print('Solution')
print('kdoc1 = ' + str(kdoc1.value[0]))
print('kdoc2 = ' + str(kdoc2.value[0]))
print('S1 = ' + str(S1.value[0]))
print('S2 = ' + str(S2.value[0]))

plt.figure(1,figsize=(8,5))
plt.subplot(2,1,1)
plt.plot(m.time,C2.value,'bo',label='Predicted')
plt.plot(m.time,df['x'].values,'rs',label='Meas')
plt.plot(m.time,C2.value,label ='C2')
plt.legend(loc='best')
plt.subplot(2,1,2)
plt.plot(m.time,cC2.value,label ='C2')
plt.plot(m.time,cC1.value,label ='C1')
plt.legend(loc='best')
plt.xlabel('time (h)')

plt.figure(2,figsize=(12,8))
plt.subplot(4,3,1)
plt.plot(m.time,C0.value,label ='C0')
plt.legend()

plt.subplot(4,3,2)
plt.plot(m.time,C1.value,label ='C1')
plt.legend()

plt.subplot(4,3,3)
plt.plot(m.time,C3.value,label ='C3')
plt.legend()

plt.subplot(4,3,4)
plt.plot(m.time,C4.value,label ='C4')
plt.legend()

plt.subplot(4,3,5)
plt.plot(m.time,C5.value,label ='C5')
plt.legend()

plt.subplot(4,3,6)
plt.plot(m.time,C6.value,label ='C6')
plt.legend()

plt.subplot(4,3,7)
plt.plot(m.time,C7.value,label ='C7-')
plt.legend()

plt.subplot(4,3,8)
plt.plot(m.time,C8.value,label ='C8')
plt.legend()

plt.subplot(4,3,9)
plt.plot(m.time,C9.value,label ='C9')
plt.legend()
plt.xlabel('time (h)')

plt.subplot(4,3,10)
plt.plot(m.time,C10.value,label ='C10')
plt.legend()
plt.xlabel('time (h)')

plt.subplot(4,3,11)
plt.plot(m.time,C11.value,label ='C11')
plt.legend()
plt.xlabel('time (h)')

plt.subplot(4,3,12)
plt.plot(m.time,C12.value,label ='C12')
plt.legend()
plt.xlabel('time (h)')

plt.show()