Improving the quality of non-linear regression in Python gekko

Question

I am trying to use the python GEKKO non-linear regression tools to perform system identification of a second order over-damped system using the step response.

My code is as follows:

m = GEKKO()
m_input = m.Param(value=input)
m_time=m.Param(value=time)
m_T1 = m.FV(value=initT1, lb=T1bounds[0], ub=T1bounds[1])
m_T1.STATUS = 1
m_k = m.FV(value=initk,lb=100)
m_k.STATUS = 1

m_T2 = m.FV(value=initT2, lb=T2bounds[0], ub=T2bounds[1])
m_T2.STATUS = 1


m_output = m.CV(value=output)
m_output.FSTATUS=1

m.Equation(m_output==(m_k/(m_T1+m_T2))*(1+((m_T1/(m_T2-m_T1))*m.exp(-m_time/m_T2))-((m_T2/(m_T2-m_T1))*m.exp(-m_time/m_T1)))*m_input)
m.options.IMODE = 2
m.options.MAX_ITER = 10000
m.options.OTOL = 1e-8
m.options.RTOL = 1e-8
m.solve(disp=True)

The results have not been promising. It seems that the optimizer seems to get stuck in local minimas of the objective function leaving the objective function too high

The output of the solver is:

The final value of the objective function is    160453.282142838     
 
 ---------------------------------------------------
 Solver         :  IPOPT (v3.12)
 Solution time  :    7.60390000000189      sec
 Objective      :    160453.282605857     
 Successful solution
 ---------------------------------------------------

What can I do to improve the quality of the fit? Can I place limits on the objective function value?

Answer 1

Try using the equation for an underdamped 2nd order system instead of an overdamped 2nd order system. There is more information on the explicit solution to 2nd order systems on the Process Dynamics and Control website . An even better approach is to not use the explicit solution where it must be pre-determined if it is underdamped, critically damped, or overdamped. If the original 2nd order differential equation is used then the solver can decide if it is underdamped (overshoot with oscillations) or overdamped. Here is example code for fitting a 2nd order system.

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

# Import CSV data file
# Column 1 = time (t)
# Column 2 = input (u)
# Column 3 = output (y)
url = 'http://apmonitor.com/pdc/uploads/Main/data_sopdt.txt'
data = pd.read_csv(url)
t = data['time'].values - data['time'].values[0]
u = data['u'].values
y = data['y'].values

m = GEKKO(remote=False)
m.time = t; time = m.Var(0); m.Equation(time.dt()==1)

K = m.FV(2,lb=0,ub=10);      K.STATUS=1
tau = m.FV(3,lb=1,ub=200);   tau.STATUS=1
theta_ub = 30 # upper bound to dead-time
theta = m.FV(0,lb=0,ub=theta_ub); theta.STATUS=1
zeta = m.FV(1,lb=0.1,ub=3);  zeta.STATUS=1

# add extrapolation points
td = np.concatenate((np.linspace(-theta_ub,min(t)-1e-5,5),t))
ud = np.concatenate((u[0]*np.ones(5),u))
# create cubic spline with t versus u
uc = m.Var(u); tc = m.Var(t); m.Equation(tc==time-theta)
m.cspline(tc,uc,td,ud,bound_x=False)

ym = m.Param(y); yp = m.Var(y); xp = m.Var(y)
m.Equation(xp==yp.dt())
m.Equation((tau**2)*xp.dt()+2*zeta*tau*yp.dt()+yp==K*(uc-u[0]))

m.Minimize((yp-ym)**2)

m.options.IMODE=5
m.solve(disp=False)

print('Kp: ', K.value[0])
print('taup: ',  tau.value[0])
print('thetap: ', theta.value[0])
print('zetap: ', zeta.value[0])

# plot results
plt.figure()
plt.subplot(2,1,1)
plt.plot(t,y,'k.-',lw=2,label='Process Data')
plt.plot(t,yp.value,'r--',lw=2,label='Optimized SOPDT')
plt.ylabel('Output')
plt.legend()
plt.subplot(2,1,2)
plt.plot(t,u,'b.-',lw=2,label='u')
plt.legend()
plt.ylabel('Input')
plt.show()

Please try this with your data to see if it gives a better solution.