CODEWITHSUNDEEP
pythongekko
Mam
Mam

Reputation: 23

Gekko error: 2 and -13 showing no solution found

I am trying to use gekko for optimization, but I am getting error - 13 (If I use the commented equation (see below) and an error code 2 if I use different version of that equation). Kindly guide how to solve this. I am unable to get a solution. I tried to add a lower bound, but still it is not working. The two equations are same, one is with log y and other one is with y value.

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

# Data points
xm = np.array([0.122, 0.276, 0.303, 0.356, 0.384, 0.444, 0.662 , 0.779, 0.867, 0.964 ,1.09, 1.17, 1.3, 1.33 ,1.41])
ym = np.array([0.00515, 0.00297, 0.00271, 0.00249, 0.00228, 0.00216, 0.0016, 0.00139, 0.0013, 0.00116, 0.00105, 0.00101, 0.000888, 0.000904, 0.000854])

# Define GEKKO model
# Parameters and variable

m = GEKKO(remote=False)
a= m.FV(lb=-4000.0,ub=-4000.0)
b= m.FV(lb=-4000.0,ub=-4000.0)


lower = 1e-4

x =m.Param(value=xm)
y_measure = m.Param(value =ym)
y_predict=m.Var(lb = lower)

# Parameters and variable options
# 1 = avaiable to optimizer to minimize objective
a.status = 1  
b.status = 1


# Equation


m.Equation(m.log(y_predict) == a + b * m.log (x))

#m.Equation(y_predict == m.exp(a + b * m.log (x)))

# Objective
           
m.Minimize(((y_predict-y_measure)/y_measure)**2)
# application options
m.options.IMODE = 2   # regression mode

# m.options.SOLVER=1

# solve
m.solve() # remote=False for local solve

# show final objective
print('Final SSE Objective: ' + str(m.options.objfcnval))

# print solution
print('Solution')
print('a = ' + str(a.value[0]))
print('b = ' + str(b.value[0]))

Upvotes: 2

Views: 94

Answers (1)

LaGrande Gunnell
LaGrande Gunnell

Reputation: 176

it looks like you accidentally made the upper bounds for the a and b coefficients -4000. By changing them to 4000, the solver now works for either prediction function.

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

# Data points
xm = np.array([0.122, 0.276, 0.303, 0.356, 0.384, 0.444, 0.662 , 0.779, 0.867, 0.964 ,1.09, 1.17, 1.3, 1.33 ,1.41])
ym = np.array([0.00515, 0.00297, 0.00271, 0.00249, 0.00228, 0.00216, 0.0016, 0.00139, 0.0013, 0.00116, 0.00105, 0.00101, 0.000888, 0.000904, 0.000854])

# Define GEKKO model
# Parameters and variable

m = GEKKO(remote=False)
a= m.FV(lb=-4000.0,ub=4000.0)
b= m.FV(lb=-4000.0,ub=4000.0)


lower = 1e-4

x =m.Param(value=xm)
y_measure = m.Param(value =ym)
y_predict=m.Var(lb = lower)

# Parameters and variable options
# 1 = avaiable to optimizer to minimize objective
a.status = 1  
b.status = 1


# Equation


m.Equation(m.log(y_predict) == a + b * m.log (x))

#m.Equation(y_predict == m.exp(a + b * m.log (x)))

# Objective
           
m.Minimize(((y_predict-y_measure)/y_measure)**2)
# application options
m.options.IMODE = 2   # regression mode

# m.options.SOLVER=1

# solve
m.solve() # remote=False for local solve

# show final objective
print('Final SSE Objective: ' + str(m.options.objfcnval))

# print solution
print('Solution')
print('a = ' + str(a.value[0]))
print('b = ' + str(b.value[0]))

----------------------------------------------------------------
 APMonitor, Version 1.0.0
 APMonitor Optimization Suite
 ----------------------------------------------------------------
 
 
 --------- APM Model Size ------------
 Each time step contains
   Objects      :  0
   Constants    :  0
   Variables    :  5
   Intermediates:  0
   Connections  :  0
   Equations    :  2
   Residuals    :  2
 
 Number of state variables:    17
 Number of total equations: -  15
 Number of slack variables: -  0
 ---------------------------------------
 Degrees of freedom       :    2
 
 **********************************************
 Model Parameter Estimation with Interior Point Solver
 **********************************************
  
  
 Info: Exact Hessian

******************************************************************************
This program contains Ipopt, a library for large-scale nonlinear optimization.
 Ipopt is released as open source code under the Eclipse Public License (EPL).
         For more information visit http://projects.coin-or.org/Ipopt
******************************************************************************

This is Ipopt version 3.10.2, running with linear solver mumps.

Number of nonzeros in equality constraint Jacobian...:       45
Number of nonzeros in inequality constraint Jacobian.:        0
Number of nonzeros in Lagrangian Hessian.............:       18

Total number of variables............................:       17
                     variables with only lower bounds:       15
                variables with lower and upper bounds:        2
                     variables with only upper bounds:        0
Total number of equality constraints.................:       15
Total number of inequality constraints...............:        0
        inequality constraints with only lower bounds:        0
   inequality constraints with lower and upper bounds:        0
        inequality constraints with only upper bounds:        0

iter    objective    inf_pr   inf_du lg(mu)  ||d||  lg(rg) alpha_du alpha_pr  ls
   0 7.0504855e+002 9.90e-001 8.09e+002   0.0 0.00e+000    -  0.00e+000 0.00e+000   0
   1 3.9209717e+001 4.16e-001 2.77e+002  -3.1 9.98e-001    -  9.90e-001 8.74e-001f  1
   2 3.6268661e+001 3.79e-001 2.52e+002  -5.1 9.95e-001    -  9.90e-001 9.31e-002h  1
   3 1.2792383e+001 1.39e-001 9.28e+001  -2.3 9.95e-001    -  1.00e+000 1.00e+000f  1
   4 1.2870911e+001 5.13e-002 3.41e+001  -4.2 9.86e-001    -  1.00e+000 1.00e+000h  1
   5 1.2255837e+001 1.89e-002 1.25e+001  -5.6 9.63e-001    -  1.00e+000 1.00e+000h  1
   6 1.0837548e+001 6.92e-003 4.58e+000  -6.9 9.08e-001    -  1.00e+000 1.00e+000h  1
   7 8.4929354e+000 2.52e-003 1.63e+000  -7.9 7.96e-001    -  1.00e+000 1.00e+000h  1
   8 6.2935389e+000 9.00e-004 3.10e+001 -11.0 6.27e-001    -  2.60e-001 1.00e+000h  1
   9 2.8727265e+000 1.70e-004 8.26e+001  -5.1 3.67e-001    -  1.32e-003 1.00e+000f  1
iter    objective    inf_pr   inf_du lg(mu)  ||d||  lg(rg) alpha_du alpha_pr  ls
  10 8.3643158e-001 1.24e-003 2.17e-001  -5.8 6.36e-001    -  1.00e+000 1.00e+000f  1
  11 2.2427664e-001 1.82e-003 2.02e-001  -5.8 4.50e-001    -  1.00e+000 1.00e+000f  1
  12 1.8136589e-002 3.88e-004 3.61e-002  -7.0 1.40e-001    -  1.00e+000 1.00e+000h  1
  13 1.0934780e-002 3.04e-005 2.25e-003  -8.8 4.72e-002    -  1.00e+000 1.00e+000h  1
  14 1.1280858e-002 1.81e-007 1.13e-005 -11.0 3.75e-003    -  1.00e+000 1.00e+000h  1
  15 1.1283202e-002 5.56e-012 3.14e-010 -11.0 2.07e-005    -  1.00e+000 1.00e+000h  1

Number of Iterations....: 15

                                   (scaled)                 (unscaled)
Objective...............:  5.4569096085891645e-004   1.1283201917622072e-002
Dual infeasibility......:  3.1361952979191013e-010   6.4846822354946330e-009
Constraint violation....:  5.5613083582706224e-012   5.5613083582706224e-012
Complementarity.........:  1.0000596947430242e-011   2.0678142529061207e-010
Overall NLP error.......:  3.1361952979191013e-010   6.4846822354946330e-009


Number of objective function evaluations             = 16
Number of objective gradient evaluations             = 16
Number of equality constraint evaluations            = 16
Number of inequality constraint evaluations          = 0
Number of equality constraint Jacobian evaluations   = 16
Number of inequality constraint Jacobian evaluations = 0
Number of Lagrangian Hessian evaluations             = 15
Total CPU secs in IPOPT (w/o function evaluations)   =      0.019
Total CPU secs in NLP function evaluations           =      0.003

EXIT: Optimal Solution Found.

 The solution was found.

 The final value of the objective function is  0.011283201917622072
 
 ---------------------------------------------------
 Solver         :  IPOPT (v3.12)
 Solution time  :  0.022600000000000002 sec
 Objective      :  0.011283201917622073
 Successful solution
 ---------------------------------------------------
 

Final SSE Objective: 0.011283201918
Solution
a = -6.7849216408
b = -0.74185760092
​```

Upvotes: 2

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