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python - How to apply m.connection from Gekko using arrays?

I am trying to solve a mechanistic reaction model using arrays, "for cycle" and following the inputs from How to solve warning message in Gekko due to m.connection?

I am not sure if the initialization of the variables can be done like this. Moreover, I am in doubt if I am using the constraints correctly. Thank you in advance.

Here is the code I am working on.

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

####measured lab data - nh2cl decay
#data set 1 
t_data1 = [0,0.08333,0.5,1,4,22.6167] #nh2cl
x_data1 = [0,4.91e-5,4.57e-5,4.74e-5,4.17e-5,2.76e-5] #8.01e-5
x_data1mgl = [0, 3.48,3.24,3.36,2.96,1.96]#5.68

#data set 2
t_data2 = [0,0.08333,0.5,1,4,22.8167]
x_data2 = [0,5.92e-5,5.7e-5,5.64e-5,5.30e-5,4.6e-5] #8.01e-5
x_data2mgl = [0,4.2,4.04,4,3.76,2.88]#5.68


####Preparing lab data and time dataframe
#combine data and time in the same dataframe
data1 = pd.DataFrame({'time':t_data1,'x1':x_data1,'x1mgl': x_data1mgl})
data2 = pd.DataFrame({'time':t_data2,'x2':x_data2,'x2mgl': x_data2mgl})

data1.set_index('time', inplace=True)
data2.set_index('time', inplace=True)

# merge dataframes
data = data1.join(data2, how='outer')

# simulation time points
dftp = pd.DataFrame({'time':np.linspace(0,5,100)})
dftp.set_index('time', inplace=True)

# merge dataframes
dff = data.join(dftp,how='outer')

z1 = (dff['x1']==dff['x1']).astype(int)
z2 = (dff['x2']==dff['x2']).astype(int)

# replace NaN with zeros
df0 = dff.fillna(value=0)


####Estimator Model
m = GEKKO(remote=False)
#m = GEKKO() 

m.time = df0.index.values

# measurements - Gekko arrays to store measured data
xm = m.Array(m.Param,2)
xm[0].value = df0['x1'].values
xm[1].value = df0['x2'].values

####inital measured values
hocl_init_val=m.Array(m.Var,2)
hocl_init_val[0].value= 8.01e-5
hocl_init_val[1].value= 8.01e-5 

nh3_init_val=m.Array(m.Var,2)
nh3_init_val[0].value= 1.37e-3 
nh3_init_val[1].value= 6.82e-4 

h_init_val=m.Array(m.Var,2)
h_init_val[0].value= 6.31e-8 
h_init_val[1].value= 2e-8

h2co3_init_val=m.Array(m.Var,2)
h2co3_init_val[0].value= 5.39e-4
h2co3_init_val[1].value= 1.88e-4

hco3_init_val=m.Array(m.Var,2)
hco3_init_val[0].value= 3.46e-3
hco3_init_val[1].value= 3.80e-3

co32_init_val=m.Array(m.Var,2)
co32_init_val[0].value= 2.16e-6
co32_init_val[1].value= 7.5e-6

alk_init_val=m.Array(m.Var,2)
alk_init_val[0].value= 3.46e-3
alk_init_val[1].value= 3.82-3

hocl=m.Array(m.Var,2)
nh3=m.Array(m.Var,2)
nh2cl=m.Array(m.Var,2,lb=1e-10)
nh2cl_meas=m.Array(m.Param,2)
nhcl2=m.Array(m.Var,2,lb=1e-10)
h=m.Array(m.Var,2)
oh=m.Array(m.Var,2)
I=m.Array(m.Var,2,lb=1e-10)
ocl=m.Array(m.Var,2)
nh4=m.Array(m.Var,2)
h2co3=m.Array(m.Var,2)
hco3=m.Array(m.Var,2)
co32=m.Array(m.Var,2)
alk=m.Array(m.Var,2)
#DOC1=m.Array(m.Var,2)
#DOC2=m.Array(m.Var,2)
cnh3=m.Array(m.Var,2)
cnh2cl=m.Array(m.Var,2)

r1=m.Array(m.Var,2)
r2=m.Array(m.Var,2)
r3=m.Array(m.Var,2)
r4=m.Array(m.Var,2)
r5=m.Array(m.Var,2)
r6=m.Array(m.Var,2)
r7=m.Array(m.Var,2)
r8=m.Array(m.Var,2)
r9=m.Array(m.Var,2)
r10=m.Array(m.Var,2)

r11=m.Array(m.Var,2)
r12=m.Array(m.Var,2)
r13=m.Array(m.Var,2)
r14=m.Array(m.Var,2)

r15=m.Array(m.Var,2)
r16=m.Array(m.Var,2)



#Define GEKKO variables that determine if time point contains data to be used in regression
#index for objective (0=not measured, 1=measured)
zm = m.Array(m.Param,2)
zm[0].value=z1
zm[1].value=z2

# fit to measurement
x=m.Array(m.Var,2)                         
x[0].value=x_data1[0]
x[1].value=x_data2[0]

meas=m.Array(m.Param,2) 

#### adjustable parameters 

kdoc1 = m.FV(1.2334e6,lb=1,ub=100000000000) #m-1h-1
kdoc2 = m.FV(3.8809e9,lb=1,ub=10000000000) #m-1h-1

x10=3.1684e-5
x20=3.1308e-5
#DOC10 = m.FV(x10,lb=1e-6,ub=1)
#DOC20 = m.FV(x20,lb=1e-6,ub=1)

#constrains
DOC10=m.Array(m.FV,2)
DOC10[0].value=x10
DOC10[1].value=x10*0.5

DOC20=m.Array(m.FV,2)
DOC20[0].value=x20
DOC20[1].value=x20*0.5

#variables connected to the parameters DOC20 and DOC10: DOC10 is DOC1 when t=0 and DOC20 is DOC2 when t=0 
DOC1=m.Array(m.SV,2,fixed_initial=False)
#DOC1[0].value=x10
#DOC1[1].value=x10*0.5

DOC2=m.Array(m.SV,2,fixed_initial=False)
#DOC2[0].value=x20
#DOC2[1].value=x20*0.5

#t = m.Param(value=m.time)

####Rate constants
k1 = m.FV(1.5e10)
k2 = m.FV(7.6e-2)
k3 = m.FV(1e6)
k4 = m.FV(2.3e-3)
k6 = m.FV(2.2e8)
k7 = m.FV(4e5)
k8 = m.FV(1e8)
k9 = m.FV(3e7)
k10 = m.FV(55)
k11 = m.FV(3.16e-8*1e10)
k12 = m.FV(1e10)
k13 = m.FV(5.01e-10*1e10)
k14 = m.FV(1e10)

k5=m.Array(m.Var,2)

for i in range(2):
    m.Connection(DOC2[i],DOC20[i],pos1=1,pos2=1,node1=1,node2=1)
    m.Connection(DOC1[i],DOC10[i],pos1=1,pos2=1,node1=1,node2=1)
    
    #variables - initial values
    hocl[i] = m.Var(value=hocl_init_val[i])
    nh3[i] = m.Var(value=nh3_init_val[i])
    nh2cl[i] = m.Var(value=x[i])
    nh2cl_meas[i] = m.Param(xm[i])
    meas[i] = m.Param(zm[i])
    nhcl2[i] = m.Var(value=0)
    h[i] = m.Var(value=h_init_val[i])
    oh[i] = m.Var(value=1e-14/h_init_val[i])
    I[i] = m.Var(value=0)
    ocl[i]= m.Var(value=0)
    nh4[i] = m.Var(value=0)
    h2co3[i] = m.Var(value=h2co3_init_val[i])
    hco3[i] = m.Var(value=hco3_init_val[i])
    co32[i] = m.Var(value=co32_init_val[i])
    alk[i] = m.Var(value=alk_init_val[i])
    DOC1[i] = m.SV(value=DOC10[i])
    DOC2[i] = m.SV(value=DOC20[i])
    cnh3[i] = m.Var(value=0)
    cnh2cl[i] = m.Var(value=0)
    

###Equations 1   
    m.Equation(k5[i] == 2.5e7*h[i]+4e4*h2co3[i]+800*hco3[i])

###Equations 2 
    m.Equation(r1[i] == k1 * hocl[i] * nh3[i])
    m.Equation(r2[i] == k2 * nh2cl[i])
    m.Equation(r3[i] == k3 * hocl[i] * nh2cl[i])
    m.Equation(r4[i] == k4 * nhcl2[i])
    m.Equation(r5[i] == k5[i] * nh2cl[i] * nh2cl[i])
    m.Equation(r6[i] == k6 * nhcl2[i] * nh3[i]* h[i])
    m.Equation(r7[i] == k7 * nhcl2[i] * oh[i])
    m.Equation(r8[i] == k8 * I[i] * nhcl2[i])
    m.Equation(r9[i] == k9 * I[i] * nh2cl[i])
    m.Equation(r10[i] == k10 * nh2cl[i] * nhcl2[i])
    
    m.Equation(r11[i] == k11*hocl[i])
    m.Equation(r12[i] == k12*h[i]*ocl[i])
    m.Equation(r13[i] == k13*nh4[i])
    m.Equation(r14[i] == k14*h[i]*nh3[i])
    
    m.Equation(r15[i] == kdoc1*DOC1[i]*nh2cl[i])
    m.Equation(r16[i] == kdoc2*DOC2[i]*hocl[i])
    
####Equations 3

    m.Equation(hocl[i].dt() == -r1[i] + r2[i] - r3[i] + r4[i] + r8[i] - r16[i] - r11[i] + r12[i])
      
    m.Equation(nh3[i].dt() == -r1[i] + r2[i] + r5[i] - r6[i] + r13[i] - r14[i])
    m.Equation(nh2cl[i].dt() == r1[i] - r2[i] - r3[i] + r4[i] - r5[i] + r6[i] - r9[i] - r10[i] - r15[i])
    m.Equation(nhcl2[i].dt() == r3[i] - r4[i] + r5[i] - r6[i] - r7[i] - r8[i] - r10[i])
    m.Equation(h[i].dt() == 0)
    
    m.Equation(oh[i] == 1e-14/h[i])
    
    m.Equation(I[i].dt() == r7[i] - r8[i] - r9[i])  
    m.Equation(ocl[i].dt() == r11[i] - r12[i])
    m.Equation(nh4[i].dt() == -r13[i]  + r14[i])
    
    m.Equation(h2co3[i] == (hco3[i]*h[i])/5.01e-7)
    
    m.Equation(hco3[i] == alk[i] - 2*co32[i] - oh[i] + h[i])
    
    m.Equation(co32[i] == (5.01e-11*hco3[i])/h[i])
    
    m.Equation(alk[i].dt() == 0)
       
    m.Equation(DOC1[i].dt() == -r15[i])
    m.Equation(DOC2[i].dt() == -r16[i])
    
    m.Equation(cnh3[i] == 17000*nh3[i])
    m.Equation(cnh2cl[i] == 70900*nh2cl[i])
    
###obj function
    
    m.Minimize(meas[i]*((nh2cl[i]-nh2cl_meas[i]))**2)
        
#Application options
m.options.SOLVER = 1 #IPOPT solver=3 #APOPT solver=1
m.options.IMODE = 8 #Dynamic Simultaneous - estimation = 5 # Dynamic sequential - estimation = 8
m.options.RTOL = 1E-3
m.options.EV_TYPE = 2 #absolute error
m.options.NODES = 2 #collocation nodes (2 up to 6)
#m.options.MAX_ITER = 500
m.open_folder()

if True:
    kdoc1.STATUS=1
    kdoc2.STATUS=1
    DOC10[i].STATUS=1
    DOC20[i].STATUS=1

m.options.TIME_SHIFT = 0
try:
    m.solve(disp=True)
except:
    print("don't stop when not finding Doc10")


print('Final SSE Objective: ' + str(m.options.objfcnval))

print('Solution')
print('kdoc1 = ' + str(kdoc1.value[0]))
print('kdoc2 = ' + str(kdoc2.value[0]))
print('DOC10_1 = ' + str(DOC10[0].value[0]))
print('DOC20_1 = ' + str(DOC20[0].value[0]))
print('DOC10_2 = ' + str(DOC10[1].value[0]))
print('DOC20_2 = ' + str(DOC20[1].value[0]))

####Graphics

plt.figure(1,figsize=(8,5))
plt.subplot(2,1,1)
plt.plot(m.time,nh2cl[0],'m',label='Predicted 1')
plt.plot(m.time,nh2cl[1],'c',label='Predicted 2')

plt.plot(t_data1,x_data1,'mo',label='Meas 1')
plt.plot(t_data2,x_data2,'co',label='Meas 2')

plt.legend(loc='best')
plt.ylabel('mol/L')
plt.legend(loc='center right', bbox_to_anchor=(1.45, 0.5), ncol=2) #shadow=True,

plt.subplot(2,1,2)
plt.plot(m.time,cnh2cl[0].value,'m',label ='C2_M1')
plt.plot(m.time,cnh2cl[1].value,'c',label ='C2_M2')

plt.plot(t_data1,x_data1mgl,'mo',label='Meas 1')
plt.plot(t_data2,x_data2mgl,'co',label='Meas 2')

plt.legend(loc='best')
plt.ylabel('mg Cl2/L')
plt.xlabel('time (h)')
plt.legend(loc='right', bbox_to_anchor=(1.4, 0.5),  ncol=2) #shadow=True,

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

plt.subplot(4,3,2)
plt.plot(m.time,nh3[i].value,label ='nh3')
plt.legend()

plt.subplot(4,3,3)
plt.plot(m.time,nhcl2[i].value,label ='nhcl2')
plt.legend()

plt.subplot(4,3,4)
plt.plot(m.time,h[i].value,label ='h')
plt.legend()

plt.subplot(4,3,5)
plt.plot(m.time,oh[i].value,label ='oh')
plt.legend()

plt.subplot(4,3,6)
plt.plot(m.time,I[i].value,label ='I')
plt.legend()

plt.subplot(4,3,7)
plt.plot(m.time,ocl[i].value,label ='ocl')
plt.legend()

plt.subplot(4,3,8)
plt.plot(m.time,nh4[i].value,label ='nh4')
plt.legend()

plt.subplot(4,3,9)
plt.plot(m.time,h2co3[i].value,label ='h2co3')
plt.legend()
plt.xlabel('time (h)')

plt.subplot(4,3,10)
plt.plot(m.time,hco3[i].value,label ='hco3')
plt.legend()
plt.xlabel('time (h)')

plt.subplot(4,3,11)
plt.plot(m.time,co32[i].value,label ='co32')
plt.legend()
plt.xlabel('time (h)')

plt.subplot(4,3,12)
plt.plot(m.time,alk[i].value,label ='alk')
plt.legend()
plt.xlabel('time (h)')

plt.show()
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1 Answer

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The problem dimension doesn't look correct with 18,546 degrees of freedom. It appears that there should be just the rate constants as the degrees of freedom. You may want to check that you have an equation for every variable and that it solves initially with zero degrees of freedom (number of equations equals number of variables) with all of the rate constants fixed.

 ----------------------------------------------------------------
 APMonitor, Version 1.0.0
 APMonitor Optimization Suite
 ----------------------------------------------------------------
 
 
 --------- APM Model Size ------------
 Each time step contains
   Objects      :  0
   Constants    :  0
   Variables    :  141
   Intermediates:  0
   Connections  :  8
   Equations    :  36
   Residuals    :  36
 
 Variable time shift OFF
 Number of state variables:    30082
 Number of total equations: -  11536
 Number of slack variables: -  0
 ---------------------------------------
 Degrees of freedom       :    18546
 
 ----------------------------------------------
 Dynamic Estimation with APOPT Solver
 ----------------------------------------------
 
 Iter    Objective  Convergence
    0  1.37152E+01  4.08093E-01
    1  2.83795E+01  9.57058E-05
    2  1.36177E-01  1.62209E-05
  247  3.06822E+28  5.05213E-04
  248  3.01561E+28  5.05213E-04
  249  3.04167E+28  5.05213E-04
 
 Iter    Objective  Convergence
  250  3.06822E+28  5.05213E-04
 Maximum iterations
 
 ---------------------------------------------------
 Solver         :  APOPT (v1.0)
 Solution time  :  264.2566 sec
 Objective      :  4.856644896504421E-8
 Unsuccessful with error code  0
 ---------------------------------------------------
 
 Creating file: infeasibilities.txt
 Use command apm_get(server,app,'infeasibilities.txt') to retrieve file
 @error: Solution Not Found

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