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Trajectory optimization with gekko under speed constraints

PSA: I am very new to gekko, thus I might be missing something very obvious here.
I have been trying to find the solution to an optimal control problem, namely trajectory optimization of a regular vehicle, under certain speed constraints at certain distances along their trip. In order to do this, I tried using a pwl function based on the distance and speed constraint data and using v_max as a constraint to v. As a objective function, I use a Vehicle Specific Power (VSP) approximation.
The computation keeps going until the maximum no. of iterations is reached and cancels. Is there maybe a way to discretize the search space of this problem to make it solvable in acceptable time trading off computation time for accuracy?
goal_dist = The distance that needs to be covered
max_accel = maximum possible acceleration of the vehicle
max_decel = maximum possible deceleration of the vehicle
max_velocity = maximum possible velocity of the vehicle
min_velocity = minimum possible velocity of the vehicle
trip_time = No. of discrete data points (1s apart)
distances = array of length trip_time of discrete distance values based on GPS data points of the desired trip
speed_limits = array of length trip_time of discrete speed limits based on GPS data points of the desired trip
slope = array of length trip_time of discrete angle values
def optimal_trip(goal_dist, max_accel, max_decel, max_velocity, min_velocity, trip_time, distances ,speed_limits, slope):
model = GEKKO(remote=True)
model.time = [i for i in range(trip_time)]
x = model.Var(value=0.0)
v = model.Var(value=0.0, lb = min_velocity, ub = max_velocity)
v_max = model.Var()
slope_var = model.Var()
a = model.MV(value=0, lb=max_decel ,ub=max_accel)
a.STATUS = 1
#define vehicle movement
model.Equation(x.dt()==v)
model.Equation(v.dt()==a)
# path constraint
model.Equation(x >= 0)
#aggregated velocity constraint
model.pwl(x, v_max, distances, speed_limits)
model.Equation(v_max>=v)
#slope is modeled as a piecewise linear function
model.pwl(x, slope_var, distances, slope)
#End state constraints
model.fix(x, pos=trip_time-1,val=goal_dist) # vehicle must arrive at destination
model.fix(v, pos=trip_time-1,val=0) # vehicle must be fully stopped
#VSPI Objective function
obj = (v * (1.1 * a + 9.81 * slope_var + 0.132) +0.0003002*pow(v, 3))
model.Obj(obj)
# solve
model.options.IMODE = 6
model.options.REDUCE = 3
model.solve(disp=True)
return x.value, v.value, obj.value
Could someone shed some light onto this?

Here is a version of the model with sample values that solves successfully:
from gekko import GEKKO
import numpy as np
min_velocity = 0
max_velocity = 10
max_decel = -1
max_accel = 1
distances = np.linspace(0,20,21)
goal_dist = 200
trip_time = 100
# set up PWL functions
distances = np.linspace(0,1000,10)
speed_limits = np.ones(10)*5
speed_limits[5:]=7
slope = np.zeros(10)
slope[3:5]=1; slope[7:9]=-1
model = GEKKO(remote=True)
model.time = [i for i in range(trip_time)]
x = model.Var(value=0.0)
v = model.Var(value=0.0, lb = min_velocity, ub = max_velocity)
v_max = model.Var()
slope_var = model.Var()
a = model.MV(value=0, lb=max_decel ,ub=max_accel)
a.STATUS = 1
#define vehicle movement
model.Equation(x.dt()==v)
model.Equation(v.dt()==a)
# path constraint
model.Equation(x >= 0)
#aggregated velocity constraint
model.pwl(x, v_max, distances, speed_limits)
model.Equation(v_max>=v)
#slope is modeled as a piecewise linear function
model.pwl(x, slope_var, distances, slope)
#End state constraints
model.fix(x, pos=trip_time-1,val=goal_dist) # vehicle must arrive at destination
model.fix(v, pos=trip_time-1,val=0) # vehicle must be fully stopped
#VSPI Objective function
obj = (v * (1.1 * a + 9.81 * slope_var + 0.132) +0.0003002*pow(v, 3))
model.Obj(obj)
# solve
model.options.IMODE = 6
model.options.REDUCE = 3
model.solve(disp=True)
It may be that the values you are using cause an infeasible solution. Here are some suggestions to help the model solve more reliably:
Use upper bounds on variables instead of general inequality constraints when possible.
# remove these lines
#model.Equation(x >= 0)
#x = model.Var(value=0.0)
# put lower bound on x
x = model.Var(value=0,lb=0)
Use soft terminal constraints instead of hard terminal constraints.
#End state constraints
# vehicle must arrive at destination
#model.fix(x, pos=trip_time-1,val=goal_dist)
# vehicle must be fully stopped
#model.fix(v, pos=trip_time-1,val=0)
p = np.zeros_like(model.time); p[-1]=1
final = model.Param(p)
model.Minimize(1e4*final*(v**2))
model.Minimize(1e4*final*((x-goal_dist)**2))
Increase maximum iterations. Sometimes the solver needs more iterations to find a solution.
model.options.MAX_ITER=1000
The final version of the model has these changes. I may help converge to a solution and avoid maximum iterations or an infeasible solution.
from gekko import GEKKO
import numpy as np
min_velocity = 0
max_velocity = 10
max_decel = -1
max_accel = 1
distances = np.linspace(0,20,21)
goal_dist = 200
trip_time = 100
# set up PWL functions
distances = np.linspace(0,1000,10)
speed_limits = np.ones(10)*5
speed_limits[5:]=7
slope = np.zeros(10)
slope[3:5]=1; slope[7:9]=-1
model = GEKKO(remote=True)
model.time = [i for i in range(trip_time)]
x = model.Var(value=0.0, lb=0)
v = model.Var(value=0.0, lb = min_velocity, ub = max_velocity)
v_max = model.Var()
slope_var = model.Var()
a = model.MV(value=0, lb=max_decel ,ub=max_accel)
a.STATUS = 1
#define vehicle movement
model.Equation(x.dt()==v)
model.Equation(v.dt()==a)
#aggregated velocity constraint
model.pwl(x, v_max, distances, speed_limits)
model.Equation(v_max>=v)
#slope is modeled as a piecewise linear function
model.pwl(x, slope_var, distances, slope)
#End state constraints
# vehicle must arrive at destination
#model.fix(x, pos=trip_time-1,val=goal_dist)
# vehicle must be fully stopped
#model.fix(v, pos=trip_time-1,val=0)
p = np.zeros_like(model.time); p[-1]=1
final = model.Param(p)
model.Minimize(1e4*final*(v**2))
model.Minimize(1e4*final*((x-goal_dist)**2))
#VSPI Objective function
obj = (v * (1.1 * a + 9.81 * slope_var + 0.132) +0.0003002*pow(v, 3))
model.Minimize(obj)
# solve
model.options.IMODE = 6
model.options.REDUCE = 3
model.options.MAX_ITER=1000
model.solve(disp=True)
If you ask another question on StackOverflow, don't forget to include a minimal and complete working example that replicates the problem.

Solving Receding Horizon Control in GEKKO

I'm trying to implement a receding horizon control (RHC) scheme using GEKKO in Python, and I'd like to check my formulation. The goal is to solve the OCP over some horizon from t=tk to t=tk+H-1, apply the control solution at tk, and discard the remaining values (u_k+1 to u_k+H-1). The following code appears to give the correct solution, but I want to verify I've used the correct functions in GEKKO, namely when "resetting" the states for the next horizon. I had a few issues trying to use the .VALUE function to reset x1 and x2, e.g. TypeError: 'float' object is not subscriptable.
import numpy as np
import matplotlib.pylab as plt
from gekko import GEKKO
if __name__ == '__main__':
# Instantiate GEKKO
m = GEKKO()
# Constants
nRHC = 21
tRHC = 2
m.time = np.linspace(0, tRHC, nRHC)
# Control
u = m.MV(value=0.0,fixed_initial=False)
u.STATUS = 1
u.DCOST = 0
# Vars
t = m.SV(value=0)
x1 = m.SV(value=1)
x2 = m.SV(value=0)
# Equations
m.Equation(t.dt() == 1)
m.Equation(x1.dt() == x2)
m.Equation(x2.dt() == (1 - x2*x2)*x1 - x2 + u)
# Objective Function
m.Minimize(10*x1**2 + 10*x2**2 + u**2)
# Solve RHC
m.options.IMODE = 6
m.options.NODES = 11
m.options.MV_TYPE = 2
m.options.SOLVER = 3
nTotal = 101
tTotal = np.linspace(0, 10, nTotal)
uStore = np.zeros((1,nTotal))
xStore = np.zeros((2,nTotal))
xStore[:,0] = [1, 0]
for i in range(nTotal):
print('Solving Step: ', i+1, ' of ', nTotal-1)
if i == nTotal-1:
break
# Solve MPC over horizon
m.solve(disp=False)
# Update States
t.VALUE = t[1]
x1.MEAS = x1[1]
x2.MEAS = x2[1]
# Store
uStore[:,i] = u.NEWVAL
xStore[:,i+1] = np.array([x1[1], x2[1]])
# Plot States
f1, axs = plt.subplots(2)
axs[0].plot(tTotal, xStore[0,:])
axs[0].set_ylabel('x')
axs[0].grid()
axs[1].plot(tTotal, xStore[1,:])
axs[1].set_ylabel('x_dot')
axs[1].set_xlabel('time')
axs[1].grid()
# Show Plots
plt.show()
Thank you!

There is no need to update the states because Gekko does this automatically.
# Update States
t.VALUE = t[1]
x1.MEAS = x1[1]
x2.MEAS = x2[1]
The state values are stored in run directory files (see m.path or open with m.open_folder()). The file is ctl.t0. At the next command m.solve(), that file is imported and time shifted to make the values at the next time step the initial conditions. The time shift is adjusted with m.options.TIME_SHIFT=1 (1 is the default). If you do want to override the initial condition, use x1.MEAS=x1.value[1] or x1.value=x1.value[1].

How to do space discretization in Gekko?

The goal is to minimize time to complete the lap with Energy constraint this is why my objective is the integral of the speed over distance, but I can’t seem to figure out how to derive and integrate over distance and not time(dt).

If you don't have time in your problem then you can specify m.time as the distance points for integration. However, your differential equations are based on time such as ds/dt = v in 1D. You need to keep time as the variable because that is defined for each of the differentials.
One way to minimize the lap time is to create a new tlap=FV() and then scale all of the differentials by that new adjustable value.
tlap=FV()
m.Equation(s.dt()==v*tlap)
With this tf value, you can minimize final time to reach a final destination.
m.Minimize(tf*final)
This is similar to the rocket launch problem that minimizes final time and control action.
import numpy as np
import matplotlib.pyplot as plt
from gekko import GEKKO
# create GEKKO model
m = GEKKO()
# scale 0-1 time with tf
m.time = np.linspace(0,1,101)
# options
m.options.NODES = 6
m.options.SOLVER = 3
m.options.IMODE = 6
m.options.MAX_ITER = 500
m.options.MV_TYPE = 0
m.options.DIAGLEVEL = 0
# final time
tf = m.FV(value=1.0,lb=0.1,ub=100)
tf.STATUS = 1
# force
u = m.MV(value=0,lb=-1.1,ub=1.1)
u.STATUS = 1
u.DCOST = 1e-5
# variables
s = m.Var(value=0)
v = m.Var(value=0,lb=0,ub=1.7)
mass = m.Var(value=1,lb=0.2)
# differential equations scaled by tf
m.Equation(s.dt()==tf*v)
m.Equation(mass*v.dt()==tf*(u-0.2*v**2))
m.Equation(mass.dt()==tf*(-0.01*u**2))
# specify endpoint conditions
m.fix_final(s, 10.0)
m.fix_final(v, 0.0)
# minimize final time
m.Minimize(tf)
# Optimize launch
m.solve()
print('Optimal Solution (final time): ' + str(tf.value[0]))
# scaled time
ts = m.time * tf.value[0]
# plot results
plt.figure(1)
plt.subplot(4,1,1)
plt.plot(ts,s.value,'r-',linewidth=2)
plt.ylabel('Position')
plt.legend(['s (Position)'])
plt.subplot(4,1,2)
plt.plot(ts,v.value,'b-',linewidth=2)
plt.ylabel('Velocity')
plt.legend(['v (Velocity)'])
plt.subplot(4,1,3)
plt.plot(ts,mass.value,'k-',linewidth=2)
plt.ylabel('Mass')
plt.legend(['m (Mass)'])
plt.subplot(4,1,4)
plt.plot(ts,u.value,'g-',linewidth=2)
plt.ylabel('Force')
plt.legend(['u (Force)'])
plt.xlabel('Time')
plt.show()

There are a few problems that I fixed with your current solution:
Variables w and st are not used
The STATUS for p_s and s_s should be On (1) to be calculated by the solver
The number of time points (50000) is really long and will create a very large problem that will be hard to solve in one solution. You may consider breaking this into successive solutions that advance one cycle (m.options.TIME_SHIFT=1) or multiple (m.options.TIME_SHIFT=10) for each m.solve() command.
There may be references that can help with the problem formulation. It appears that you are taking a more physics-based approach than a data driven approach.
Switched to the APOPT solver for a successful solution.
from gekko import GEKKO
import numpy as np
import matplotlib.pyplot as plt
m = GEKKO(remote=False)
#Constants
mass = m.Const(77) #mass of the rider
g = m.Const(9.81) #gravity
H = m.Const(1.2) #height of the rider
L = m.Const(value=1.4) #lenght of the wheelbase of the bicycle
E_n = m.Const(value=22000) #Energy that can be used
c_rr = m.Const(value=0.0035) #coefficient of drag
s_max = m.Const(value=0.52) #max steer angle
W_m = m.Const(value=1800) #max power that the rider can produce
vWn = m.Const(value=50) #maximal power output variation
vSn = m.Const(value=0.52) #maximal steer output variation
kv = m.Const(value=0.13) #air drag coefficient
ws = m.Const(value=0) #wind speed
Ix = m.Const(value=77) #inertia
W_c = m.Const(value=440) #critical power(watts)
Wj1 = m.Const(value=0.01) ##weighting factors that scale the penalisation
Wj2 = m.Const(value=0.01) #weighting factors that scale the penalisation
dist = 1000 ##distance that that the rider needs to travel
nt = 100 ##calculation at every 10 meters
m.time = np.linspace(0,dist,nt)
p = np.zeros(nt)
p[-1] = 1.0
final = m.Param(value=p)
slope = np.zeros(nt) #SET THE READ CURVATURE AND SLOPE TO 0 for experimentation later we will import it from real road.
curv = np.zeros(nt) #SET THE READ CURVATURE AND SLOPE TO 0 for experimentation later we will import it from real road.
####Import Road Characterisitc####
k = m.Param(value=curv) ##road_curvature
b = m.Param(value=slope) ##slope angle
###Control Variable###
p_s = m.MV(value=1,lb=-1000,ub=1000); p_s.STATUS = 1 ##power
s_s = m.MV(value=0,lb=-100,ub=100); s_s.STATUS = 1 ##steer
###State Variable###
# Not used
#w = m.Param(value=10,lb=-10000,ub=1800) #power done by the rider (positive:pedaling, negative:braking)
#st = m.Param(value=0,lb=-30,ub=30) ##steer angle
s = m.Var(value=1,lb=1e-4,ub=100) #speed along road
v = m.Var(value=1, lb=0, ub=16) #velocity
n = m.Var(value=0,lb=-4, ub=4) ##displacement fron the center of the road upper bound and lower bound are the road width
h = m.Var(value=0,lb=-10,ub=10) #heading of the bicycle
r = m.Var(0,lb=-0.78, ub=0.78) ##roll
r_dot = m.Var(value=0,lb=-100,ub=100) ##roll_rate
W_n = m.Var(value=0.1,lb=-1, ub=1) ##normalised power
s_n = m.Var(value=0,lb=-1, ub=1) #normalised steer angle
e = m.Var(value=22000, lb=0, ub=22000) #energy remaining
####Equations####
#1 dynamics of travelling speed s(s) along the reference line
m.Equation((1-(n-k))*s.dt()==v*m.cos(h))
#2:dynamics of the longitudinal velocity of the bicycle
c1 = m.Intermediate((v*mass)/W_m,'c1')
m.Equation(c1*s*v.dt()==(W_n
-( (v/W_m) * (mass*g* (c_rr* m.cos(b)+m.sin(b))) )
-((v/W_m) * kv*(v-(ws*h))**2)
)
)
#3: dynamic of the lateral displacement
m.Equation(s*n.dt()==m.sin(k))
#4: heading of the bicycle 𝛼(s):
m.Equation((L*s)*h.dt()==(s_n*s_max)-k*(L*s))
#5&6: dynamics of the roll angle 𝜙 (rad) and its rate of change 𝜙dot(s)
m.Equation(s*r.dt()==(r_dot))
m.Equation(((h**2)*mass+Ix)*(g*L*s)*r_dot.dt()==(H*mass*g)*((v**2)*s_max*s_n+L*g*r))
#7: dynamics of the normalised power output Wn
m.Equation(s*W_n.dt()==p_s)
##8: dynamics of the normalised steering angle 𝛿n
m.Equation(s*s_n.dt()==s_s)
#9: dynamic equation describing the evolution of the anaerobic sources
# use lower bound on W_n instead of m.min2(0,W_n)
m.Equation((s*E_n)*e.dt()==(-(W_n*W_m-W_c) ))
####OBJECTIVE####
m.Minimize(m.integral( (1/s) * (1+(Wj1*((p_s/vWn)**2))+(Wj2*((s_s/vSn)**2))) )*final)
m.options.IMODE = 6 # optimal control
m.options.SOLVER = 1 # solver (APOPT)
m.options.DIAGLEVEL=0
#m.open_folder()
m.solve(disp=True, debug=True) # Solve
With this script, I get a successful solution but I haven't investigated the objective function to see if it is giving a reasonable answer.
----------------------------------------------------------------
APMonitor, Version 0.9.2
APMonitor Optimization Suite
----------------------------------------------------------------
--------- APM Model Size ------------
Each time step contains
Objects : 0
Constants : 16
Variables : 15
Intermediates: 1
Connections : 0
Equations : 12
Residuals : 11
Number of state variables: 2970
Number of total equations: - 2772
Number of slack variables: - 0
---------------------------------------
Degrees of freedom : 198
----------------------------------------------
Dynamic Control with APOPT Solver
----------------------------------------------
Iter Objective Convergence
0 2.51001E+03 1.00000E+00
1 4.36075E+04 5.66676E-01
2 3.43092E+03 5.36156E-01
3 7.36773E+03 4.16203E-01
4 2.75250E+03 9.29407E-02
5 4.12278E+03 1.93521E-02
6 5.80466E+05 7.35244E-02
7 4.99119E+04 1.27246E-01
8 2.11556E+03 4.52552E-01
9 6.32932E+03 2.14605E-01
Iter Objective Convergence
10 8.16639E+01 2.76062E-01
11 6.80002E+02 8.83214E-01
12 4.71706E+01 2.87555E-01
13 1.28152E+02 1.36994E-03
14 1.01698E+01 1.08406E+01
15 1.13082E+01 3.00869E+00
16 1.03199E+01 8.67971E+00
17 1.02638E+01 1.28697E-02
18 1.02636E+01 5.64896E-05
19 1.02636E+01 6.72710E-07
Iter Objective Convergence
20 1.02636E+01 6.72710E-07
Successful solution
---------------------------------------------------
Solver : APOPT (v1.0)
Solution time : 3.1271 sec
Objective : 10.263550885927089
Successful solution
---------------------------------------------------
You may want to create plots to make sure that the equations and solver are giving a correct solution. Here is an animation and source code that shows how to set up a model predictive controller with a finite horizon and that advances in time (or space) for each solve command.
The finite horizon approach is used commonly in industrial control to ensure that the optimizer can finish within the required cycle time and balance that with the length of the horizon to "see" future constraints and opportunities for energy or production optimization.

System of FOPDT equations in GEKKO - use more than one input

I want to use two inputs or more to create a more precise estimation of a variable. I already estimated it using only one input and one FOPDT equation, but when I try to add one more input and the respective k, tau and theta, along with another equation, i get "Solution Not Found" error. Can I create a system of equations this way?
More details about the solver output below. Even though I added more variables than equations to use more than one input, this made my problem have negative degrees of freedom.
--------- APM Model Size ------------
Each time step contains
Objects : 2
Constants : 0
Variables : 15
Intermediates: 0
Connections : 4
Equations : 6
Residuals : 6
Number of state variables: 1918
Number of total equations: - 2151
Number of slack variables: - 0
---------------------------------------
Degrees of freedom : -233
* Warning: DOF <= 0
**********************************************
Dynamic 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 3.12.10, running with linear solver ma57.
Number of nonzeros in equality constraint Jacobian...: 6451
Number of nonzeros in inequality constraint Jacobian.: 0
Number of nonzeros in Lagrangian Hessian.............: 1673
Exception of type: TOO_FEW_DOF in file "IpIpoptApplication.cpp" at line 891:
Exception message: status != TOO_FEW_DEGREES_OF_FREEDOM evaluated false: Too few degrees of freedom (rethrown)!
EXIT: Problem has too few degrees of freedom.
An error occured.
The error code is -10
---------------------------------------------------
Solver : IPOPT (v3.12)
Solution time : 2.279999999154825E-002 sec
Objective : 0.000000000000000E+000
Unsuccessful with error code 0
---------------------------------------------------
Creating file: infeasibilities.txt
Use command apm_get(server,app,'infeasibilities.txt') to retrieve file
#error: Solution Not Found
And here's the code
from gekko import GEKKO
import numpy as np
import pandas as pd
import plotly.express as px
d19jc = d19jc.dropna()
d19jcSlice = d19jc.loc['2019-10-22 05:30:00':'2019-10-22 09:30:00'] #jc22
d19jcSlice.index = pd.to_datetime(d19jcSlice.index)
d19jcSliceGroupMin = d19jcSlice.groupby(pd.Grouper(freq='T')).mean()
data = d19jcSliceGroupMin.dropna()
xdf1 = data['Cond_PID_SP']
xdf2 = data['Front_PID_SP']
ydf1 = data['Cond_Center_Top_TC']
xms1 = pd.Series(xdf1)
xm1 = np.array(xms1)
xms2 = pd.Series(xdf2)
xm2 = np.array(xms2)
yms = pd.Series(ydf1)
ym = np.array(yms)
xm_r = len(xm1)
tm = np.linspace(0,xm_r-1,xm_r)
m = GEKKO()
m.options.IMODE=5
m.time = tm; time = m.Var(0); m.Equation(time.dt()==1)
k1 = m.FV(lb=0.1,ub=5); k1.STATUS=1
tau1 = m.FV(lb=1,ub=300); tau1.STATUS=1
theta1 = m.FV(lb=0,ub=30); theta1.STATUS=1
k2 = m.FV(lb=0.1,ub=5); k2.STATUS=1
tau2 = m.FV(lb=1,ub=300); tau2.STATUS=1
theta2 = m.FV(lb=0,ub=30); theta2.STATUS=1
# create cubic spline with t versus u
uc1 = m.Var(xm1); tc1 = m.Var(tm); m.Equation(tc1==time-theta1)
m.cspline(tc1,uc1,tm,xm1,bound_x=False)
# create cubic spline with t versus u
uc2 = m.Var(xm2); tc2 = m.Var(tm); m.Equation(tc2==time-theta2)
m.cspline(tc2,uc2,tm,xm2,bound_x=False)
x1 = m.Param(value=xm1)
x2 = m.Param(value=xm2)
y = m.Var(value=ym)
yObj = m.Param(value=ym)
m.Equation(tau1*y.dt()+(y-ym[0])==k1 * (uc1-xm1[0]))
m.Equation(tau2*y.dt()+(y-ym[0])==k2 * (uc2-xm2[0]))
m.Minimize((y-yObj)**2)
m.options.EV_TYPE=2
print('solve start')
m.solve(disp=True)
print('k1: ', k1.value[0])
print('tau1: ', tau1.value[0])
print('theta1: ', theta1.value[0])
print('k2: ', k2.value[0])
print('tau2: ', tau2.value[0])
print('theta2: ', theta2.value[0])
df_plot = pd.DataFrame({'DateTime' : data.index,
'Cond_Center_Top_TC' : np.array(yObj),
'Fit Cond_Center_Top_TC - Train' : np.array(y),
figGekko = px.line(df_plot,
x='DateTime',
y=['Cond_Center_Top_TC','Fit Cond_Center_Top_TC - Train'],
labels={"value": "Degrees Celsius"},
title = "(Cond_PID_SP & Front_PID_SP) -> Cond_Center_Top_TC JC only - Train")
figGekko.update_layout(legend_title_text='')
figGekko.show()

You currently have only one variable and two equations.
y = m.Var(value=ym)
yObj = m.Param(value=ym)
m.Equation(tau1*y.dt()+(y-ym[0])==k1 * (uc1-xm1[0]))
m.Equation(tau2*y.dt()+(y-ym[0])==k2 * (uc2-xm2[0]))
You need to create two separate variables y1 and y2 for the two equations.
y1 = m.Var(value=ym)
y2 = m.Var(value=ym)
yObj = m.Param(value=ym)
m.Equation(tau1*y1.dt()+(y1-ym[0])==k1 * (uc1-xm1[0]))
m.Equation(tau2*y2.dt()+(y2-ym[0])==k2 * (uc2-xm2[0]))
You may also need to create two separate measurement vectors for ym1 and ym2 if you have different data for each.
Edit: Multiple Input, Single Output (MISO) System
For a MISO system, you need to adjust the equation as shown in the Process Dynamics and Control course.
y = m.Var(value=ym)
yObj = m.Param(value=ym)
m.Equation(tau*y.dt()+(y-ym[0])==k1 * (uc1-xm1[0]) + k2 * (uc2-xm2[0]))
There is only one time constant in this form. If they need different time constants, you can also add the two outputs together with:
y1 = m.Var(value=ym[0])
y2 = m.Var(value=ym[0])
y = m.Var(value=ym)
yObj = m.Param(value=ym)
m.Equation(tau1*y1.dt()+(y1-ym[0])==k1 * (uc1-xm1[0]))
m.Equation(tau2*y2.dt()+(y2-ym[0])==k2 * (uc2-xm2[0]))
m.Equation(y==y1+y2)
In this case, y is the variable that has data and y1 and y2 are unmeasured states.

Please see the below example code for a FOPDT model that has two inputs and one output. Both the transfer function models have different FOPDT parameters, including different deadtimes as well.
It is still in a dynamic simulation mode (IMODE=4), but you can start from this modifying a little bit toward the dynamic estimation mode (IMODE=5) and MPC mode (IMODE=6) later.
from gekko import GEKKO
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
tf = 100
npt = 101
t = np.linspace(0,tf,npt)
u1 = np.zeros(npt)
u2 = np.zeros(npt)
u1[10:] = 5
u2[40:] = -5
m = GEKKO(remote=True)
m.time = t
time = m.Var(0)
m.Equation(time.dt()==1)
K1 = m.FV(1,lb=0,ub=1); K1.STATUS=1
tau1 = m.FV(5, lb=1,ub=300); tau1.STATUS=1
theta1 = m.FV(10, lb=2,ub=30); theta1.STATUS=1
K2 = m.FV(0.5,lb=0,ub=1); K2.STATUS=1
tau2 = m.FV(10, lb=1,ub=300); tau2.STATUS=1
theta2 = m.FV(20, lb=2,ub=30); theta2.STATUS=1
uc1 = m.Var(u1)
uc2 = m.Var(u2)
tc1 = m.Var(t)
tc2 = m.Var(t)
m.Equation(tc1==time-theta1)
m.Equation(tc2==time-theta2)
m.cspline(tc1,uc1,t,u1,bound_x=False)
m.cspline(tc2,uc2,t,u2,bound_x=False)
yp = m.Var()
yp1 = m.Var()
yp2 = m.Var()
m.Equation(yp1.dt() == -yp1/tau1 + K1*uc1/tau1)
m.Equation(yp2.dt() == -yp2/tau2 + K2*uc2/tau2)
m.Equation(yp == yp1 + yp2)
m.options.IMODE=4
m.solve()
print('K1: ', K1.value[0])
print('tau1: ', tau1.value[0])
print('theta1: ', theta1.value[0])
print('')
print('K2: ', K2.value[0])
print('tau2: ', tau2.value[0])
print('theta2: ', theta2.value[0])
plt.figure()
plt.subplot(2,1,1)
plt.plot(t,u1)
plt.plot(t,u2)
plt.legend([r'u1', r'u2'])
plt.ylabel('Inputs 1 & 2')
plt.subplot(2,1,2)
plt.plot(t,yp)
plt.legend([r'y1'])
plt.ylabel('Output')
plt.xlabel('Time')
plt.savefig('sysid.png')
plt.show()
K1: 1.0
tau1: 5.0
theta1: 10.0
K2: 0.5
tau2: 10.0
theta2: 20.0

Solving ODE numerically with Python

I am solving an ODE for an harmonic oscillator numerically with Python. When I add a driving force it makes no difference, so I'm guessing something is wrong with the code. Can anyone see the problem? The (h/m)*f0*np.cos(wd*i) part is the driving force.
import numpy as np
import matplotlib.pyplot as plt
# This code solves the ODE mx'' + bx' + kx = F0*cos(Wd*t)
# m is the mass of the object in kg, b is the damping constant in Ns/m
# k is the spring constant in N/m, F0 is the driving force in N,
# Wd is the frequency of the driving force and x is the position
# Setting up
timeFinal= 16.0 # This is how far the graph will go in seconds
steps = 10000 # Number of steps
dT = timeFinal/steps # Step length
time = np.linspace(0, timeFinal, steps+1)
# Creates an array with steps+1 values from 0 to timeFinal
# Allocating arrays for velocity and position
vel = np.zeros(steps+1)
pos = np.zeros(steps+1)
# Setting constants and initial values for vel. and pos.
k = 0.1
m = 0.01
vel0 = 0.05
pos0 = 0.01
freqNatural = 10.0**0.5
b = 0.0
F0 = 0.01
Wd = 7.0
vel[0] = vel0 #Sets the initial velocity
pos[0] = pos0 #Sets the initial position
# Numerical solution using Euler's
# Splitting the ODE into two first order ones
# v'(t) = -(k/m)*x(t) - (b/m)*v(t) + (F0/m)*cos(Wd*t)
# x'(t) = v(t)
# Using the definition of the derivative we get
# (v(t+dT) - v(t))/dT on the left side of the first equation
# (x(t+dT) - x(t))/dT on the left side of the second
# In the for loop t and dT will be replaced by i and 1
for i in range(0, steps):
vel[i+1] = (-k/m)*dT*pos[i] + vel[i]*(1-dT*b/m) + (dT/m)*F0*np.cos(Wd*i)
pos[i+1] = dT*vel[i] + pos[i]
# Ploting
#----------------
# With no damping
plt.plot(time, pos, 'g-', label='Undampened')
# Damping set to 10% of critical damping
b = (freqNatural/50)*0.1
# Using Euler's again to compute new values for new damping
for i in range(0, steps):
vel[i+1] = (-k/m)*dT*pos[i] + vel[i]*(1-(dT*(b/m))) + (F0*dT/m)*np.cos(Wd*i)
pos[i+1] = dT*vel[i] + pos[i]
plt.plot(time, pos, 'b-', label = '10% of crit. damping')
plt.plot(time, 0*time, 'k-') # This plots the x-axis
plt.legend(loc = 'upper right')
#---------------
plt.show()

The problem here is with the term np.cos(Wd*i). It should be np.cos(Wd*i*dT), that is note that dT has been added into the correct equation, since t = i*dT.
If this correction is made, the simulation looks reasonable. Here's a version with F0=0.001. Note that the driving force is clear in the continued oscillations in the damped condition.
The problem with the original equation is that np.cos(Wd*i) just jumps randomly around the circle, rather than smoothly moving around the circle, causing no net effect in the end. This can be best seen by plotting it directly, but the easiest thing to do is run the original form with F0 very large. Below is F0 = 10 (ie, 10000x the value used in the correct equation), but using the incorrect form of the equation, and it's clear that the driving force here just adds noise as it randomly moves around the circle.

Note that your ODE is well behaved and has an analytical solution. So you could utilize sympy for an alternate approach:
import sympy as sy
sy.init_printing() # Pretty printer for IPython
t,k,m,b,F0,Wd = sy.symbols('t,k,m,b,F0,Wd', real=True) # constants
consts = {k: 0.1, # values
m: 0.01,
b: 0.0,
F0: 0.01,
Wd: 7.0}
x = sy.Function('x')(t) # declare variables
dx = sy.Derivative(x, t)
d2x = sy.Derivative(x, t, 2)
# the ODE:
ode1 = sy.Eq(m*d2x + b*dx + k*x, F0*sy.cos(Wd*t))
sl1 = sy.dsolve(ode1, x) # solve ODE
xs1 = sy.simplify(sl1.subs(consts)).rhs # substitute constants
# Examining the solution, we note C3 and C4 are superfluous
xs2 = xs1.subs({'C3':0, 'C4':0})
dxs2 = xs2.diff(t)
print("Solution x(t) = ")
print(xs2)
print("Solution x'(t) = ")
print(dxs2)
gives
Solution x(t) =
C1*sin(3.16227766016838*t) + C2*cos(3.16227766016838*t) - 0.0256410256410256*cos(7.0*t)
Solution x'(t) =
3.16227766016838*C1*cos(3.16227766016838*t) - 3.16227766016838*C2*sin(3.16227766016838*t) + 0.179487179487179*sin(7.0*t)
The constants C1,C2 can be determined by evaluating x(0),x'(0) for the initial conditions.