I am writing the GEKKO equations to determine a vehicle's gear box ratio which depends on the vehicle's previous derivatives. Is there a way to set a variable to the time shifted value of another variable?
Ex:
v=0,[1,2,3,4,5]
shifted_v=[0,1,2,3,4]
where the square bracket is the horizon and v is a state variable defined by equations.
One of the easiest ways to shift data sets is to use the numpy.roll function.
import numpy as np
x = np.linspace(0,5,6)
y = np.roll(x,-1) # shift left
y[-1] = 6
z = np.roll(x,1) # shift right
z[0] = -1
print('x: ' + str(x))
print('y: ' + str(y))
print('z: ' + str(z))
You can apply this strategy using Gekko variables by using the .value property such as:
import numpy as np
from gekko import GEKKO
m = GEKKO()
m.time = np.linspace(0,5,6)
x = m.Param(value=m.time)
y = m.Param()
y.value = np.roll(x.value,-1)
y.value[-1] = 6
z = m.Param()
z.value = np.roll(x.value,1)
z.value[0] = -1
There is also a TIME_SHIFT feature in Gekko that automatically shifts values as if they were advancing in time. The TIME_SHIFT option controls how much the values are shifted with every solve. The time shift happens at the beginning of the solve. Here is a more complete example with a visualization of the result.
import numpy as np
from gekko import GEKKO
import matplotlib.pyplot as plt
m = GEKKO()
m.time = np.linspace(0,5,6)
x = m.Param(value=m.time)
y = m.Param()
y.value = np.roll(x.value,-1)
y.value[-1] = 6
z = m.Param()
z.value = np.roll(x.value,1)
z.value[0] = -1
s = m.Var()
m.Equation(s==x+y-z)
m.options.IMODE=4
m.solve()
plt.subplot(2,1,1)
plt.plot(m.time,x.value,label='x')
plt.plot(m.time,y.value,label='y')
plt.plot(m.time,z.value,label='z')
plt.legend()
# solve a second time
m.options.TIME_SHIFT = 1 # default is 1
m.solve()
plt.subplot(2,1,2)
plt.plot(m.time,x.value,label='x')
plt.plot(m.time,y.value,label='y')
plt.plot(m.time,z.value,label='z')
plt.legend()
plt.show()
From your question, it appears that you need to calculate the previous derivative of a variable. If you need to time shift a value during the calculation, not just in the initialization phase, then I would recommend a discrete state space model with a delay of 1 time step. The link provides an example of how to implement this with 4 steps of delay. You would want to modify the discrete state space matrices to have 1 step of delay between the derivative and gear-box ratio.
Related
When simulating a process using GEKKO (for example, as in Example 15 here), how would I set the initial value of the derivative of a state variable? I am using IMODE=4, but I could also use IMODE=7.
[Edit] I have fitted the parameters of a ODE-model with measured input and output using IMODE=5 and I would like to predict model output beyond measured time points.
Here is a modification of Problem 8 from that same link as a simple example. To initialize the derivative, create a new variable such as dydt and define a new equation that is equal to the derivative.
from gekko import GEKKO
import numpy as np
import matplotlib.pyplot as plt
m = GEKKO()
k = 10
m.time = np.linspace(0,20,100)
y = m.Var(value=5)
dydt = m.Var(value=0)
t = m.Param(value=m.time)
m.Equation(k*dydt==-t*y)
m.Equation(dydt==y.dt())
m.options.IMODE=4
m.solve(disp=False)
plt.plot(m.time,y.value,label='y')
plt.plot(m.time,dydt.value,label='dy/dt')
plt.xlabel('time'); plt.ylabel('y')
plt.legend(); plt.grid(); plt.show()
Unlike other differential algebraic equation (DAE) solvers, Gekko does not require consistent initial conditions for the states and derivatives. Gekko can also solve higher-index DAEs where the index is the number of times that constraints must be differentiated to return to ODE form.
Boundary conditions are
y1(0)=3;
y2(0)=y1(5)
In the interval of [0,5]
Here is one possible implementation in Gekko:
import numpy as np
from gekko import GEKKO
m = GEKKO(remote=False)
m.time = np.linspace(0,5)
x = m.Param(m.time)
y1 = m.Var(3)
y2 = m.Var()
m.Connection(y2,y1,'end',1,'end',1)
m.Equation(y1.dt()+2*y1+y2==m.sin(x))
m.Equation(y2.dt()-4*y1-2*y2==m.cos(x))
m.options.IMODE=6
m.solve()
However, the solver reports that there is no solution. There are some things that are unclear about your problem statement such as whether the derivative is with respect to x or t. If x is calculated, is is a single value or able to vary throughout the time horizon?
Edit in response to comment
You can implement the final conditions y2(5)=y1(5) with m.Connection(y2,y1,'end','end','end','end'). However, there is no feasible solution to this problem unless you create some type of additional degree of freedom (calculated variable). Without the final condition constraint, there is the unique solution as shown in this figure.
When you add the Connection final constraint, the solution to the differential equations cannot change and therefore the solver correctly reports too few degrees of freedom.
import numpy as np
from gekko import GEKKO
m = GEKKO(remote=False)
m.time = np.linspace(0,5)
x = m.Param(m.time)
y1 = m.Var(3)
y2 = m.Var()
m.Equation(y1.dt()+2*y1+y2==m.sin(x))
m.Equation(y2.dt()-4*y1-2*y2==m.cos(x))
m.options.IMODE=6
m.Connection(y2,y1,'end','end','end','end')
m.solve()
import matplotlib.pyplot as plt
plt.plot(x.value,y1.value,'r--',label='y1')
plt.plot(x.value,y2.value,'b-',label='y2')
plt.legend(); plt.xlabel('x'); plt.ylabel('y')
plt.show()
Use Gekko to fit a numerical ODE solution to data.
Hi everyone!
I was wondering, if it is possible to fit coefficients of an ODE using GEKKO.
I unsuccessfully tried to replicate the example given here.
This is what I have come up with (but is flawed – and I should perhaps mention that my math skills are unfortunately rather poor):
import numpy as np
from gekko import GEKKO
tspan = [0, 0.1, 0.2, 0.4, 0.8, 1]
Ca_data = [2.0081, 1.5512, 1.1903, 0.7160, 0.2562, 0.1495]
m = GEKKO(remote=False)
t = m.Param(value=tspan)
m.time = t
Ca_m = m.Param(value=Ca_data)
Ca = m.Var()
k = m.FV(value=1.3)
k.STATUS = 1
m.Equation( Ca.dt() == -k * Ca)
m.Obj( ((Ca-Ca_m)**2)/Ca_m )
m.options.IMODE = 2
m.solve(disp=True)
print(k.value[0]) #2.58893455 is the solution
Can someone help me out here?
Thank you very much,
Martin
(This is my first post here – please be gentle, if I have done something not appropriate.)
Your solution was close but you needed:
More NODES (default=2) to improve the accuracy. Gekko only adds that points that you define. See additional information on collocation.
Define Ca as m.CV() to use built-in error model instead of m.Var() and m.Obj with NODES>=3. Otherwise, the internal nodes of each collocation interval are also matched to the measurements and this gives a slightly wrong answer.
Set EV_TYPE=2 to use a squared error. An absolute value objective EV_TYPE=1 (default) gives a correct but slightly different answer.
import numpy as np
from gekko import GEKKO
m = GEKKO(remote=False)
m.time = [0, 0.1, 0.2, 0.4, 0.8, 1]
Ca_data = [2.0081, 1.5512, 1.1903, 0.7160, 0.2562, 0.1495]
Ca = m.CV(value=Ca_data); Ca.FSTATUS = 1 # fit to measurement
k = m.FV(value=1.3); k.STATUS = 1 # adjustable parameter
m.Equation(Ca.dt()== -k * Ca) # differential equation
m.options.IMODE = 5 # dynamic estimation
m.options.NODES = 5 # collocation nodes
m.options.EV_TYPE = 2 # squared error
m.solve(disp=True) # display solver output
print(k.value[0]) # 2.58893455 is the curve_fit solution
The solution is k=2.5889717102. A plot shows the match to the measured values.
import matplotlib.pyplot as plt # plot solution
plt.plot(m.time,Ca_data,'ro')
plt.plot(m.time,Ca.value,'bx')
plt.show()
There are additional tutorials and course material on parameter estimation with differential and algebraic equation models.
I am using an MPC to run a heater system. Currently I have it take an individual value from my set point array at a given point in time to adjust the process to reach. I would like to be able to give it the current desired value and a couple of points in the future for the set point so that it can better adjust as the set point changes. How can I give gekko a vector in order to have it better adjust to future set points?
this is the part of my code that currently updates my setpoint values.
T1[i] = a.T1
T2[i] = a.T2
TC1.MEAS = T1[i]
TC2.MEAS = T2[i]
DT = .1
TC1.SPHI = sp1[i] + DT #sp1 and sp2 are set point arrays for the two heaters
TC1.SPLO = sp1[i] - DT
TC2.SPHI = sp2[i] + DT
TC2.SPLO = sp2[i] - DT
m.solve(disp=False)
The gekko CV object only uses a scalar value of an array for SP, SPHI, and SPLO so some modification is needed to have the optimizer consider future setpoint changes. A simple MPC application shows how setpoints are used in Gekko.
from gekko import GEKKO
import numpy as np
import matplotlib.pyplot as plt
m = GEKKO()
m.time = np.linspace(0,20,41)
p = m.MV(value=0, lb=0, ub=100) # Declare MV
p.STATUS = 1 # allow optimizer to change
p.DCOST = 0.1 # smooth MV response
p.DMAX = 10.0 # max move each cycle
v = m.CV(value=0) # Declare CV
v.STATUS = 1 # add CV to the objective
m.options.CV_TYPE = 2 # squared error
v.SP = 40 # set point
v.TR_INIT = 1 # set point trajectory
v.TAU = 5 # time constant of trajectory
m.Equation(10*v.dt() == -v + 2*p)
m.options.IMODE = 6 # control
m.solve(disp=False)
# get additional solution information
import json
with open(m.path+'//results.json') as f:
results = json.load(f)
plt.figure()
plt.subplot(2,1,1)
plt.plot(m.time,p.value,'b-',label='MV Optimized')
plt.legend()
plt.ylabel('Input')
plt.subplot(2,1,2)
plt.plot(m.time,results['v1.tr'],'k-',label='Reference Trajectory')
plt.plot(m.time,v.value,'r--',label='CV Response')
plt.ylabel('Output')
plt.xlabel('Time')
plt.legend(loc='best')
plt.show()
There is one setpoint value that is a target of 40 and a reference trajectory is defined with a time constant of 5. The MPC is not able to follow the reference trajectory exactly because of a rate of change constraint with DMAX=10. There are two options if you'd like the optimizer to know about future setpoint changes.
Option 1: Don't use CV, Use Feedforward Parameter
If you don't need the reference trajectory and are okay with a squared error objective then then easiest method to project future setpoint changes is to define your own MPC objective with a feedforward parameter vector of setpoint values. The example problem shows that the optimizer is anticipating the setpoint change and proactively moves before the next setpoint change to minimize the overall sum of squared error. This may be desirable in many circumstances but may be undesirable in manufacturing where there are product grade changes and the end of the production campaign should be in-spec before producing transition material.
from gekko import GEKKO
import numpy as np
import matplotlib.pyplot as plt
m = GEKKO()
m.time = np.linspace(0,20,41)
p = m.MV(value=0, lb=0, ub=100) # Declare MV
p.STATUS = 1 # allow optimizer to change
p.DCOST = 0.1 # smooth MV response
p.DMAX = 10.0 # max move constraint
v = m.Var(value=0)
sp = np.ones(41)*40
sp[20:] = 60
s = m.Param(value=sp)
m.Obj((s-v)**2)
m.Equation(10*v.dt() == -v + 2*p)
m.options.IMODE = 6 # control
m.solve(disp=False)
plt.figure()
plt.subplot(2,1,1)
plt.plot(m.time,p.value,'b-',label='MV Optimized')
plt.legend()
plt.ylabel('Input')
plt.subplot(2,1,2)
plt.plot(m.time,sp,'k-',label='Setpoint')
plt.plot(m.time,v.value,'r--',label='CV Response')
plt.ylabel('Output')
plt.xlabel('Time')
plt.legend(loc='best')
plt.show()
Option 2: Error as CV
If it is desirable to use the reference trajectory and Gekko built-in CV options, then an option is to define a new error variable e and control that instead. The error variable always has a setpoint of zero and the feedforward setpoint is implemented as a feedforward parameter.
from gekko import GEKKO
import numpy as np
import matplotlib.pyplot as plt
m = GEKKO()
m.time = np.linspace(0,20,41)
p = m.MV(value=0, lb=0, ub=100) # Declare MV
p.STATUS = 1 # allow optimizer to change
p.DCOST = 0.1 # smooth MV response
p.DMAX = 10.0 # max move constraint
v = m.Var(value=0)
sp = np.ones(41)*40
sp[20:] = 60
s = m.Param(value=sp)
e = m.CV(value=0) # Declare CV
e.STATUS = 1 # add CV to the objective
m.options.CV_TYPE = 2 # squared error
e.SP = 0 # set point
e.TR_INIT = 1 # error trajectory
e.TAU = 5 # time constant of trajectory
m.Equation(e==s-v)
m.Equation(10*v.dt() == -v + 2*p)
m.options.IMODE = 6 # control
m.solve(disp=False)
plt.figure()
plt.subplot(2,1,1)
plt.plot(m.time,p.value,'b-',label='MV Optimized')
plt.legend()
plt.ylabel('Input')
plt.subplot(2,1,2)
plt.plot(m.time,sp,'k-',label='Setpoint')
plt.plot(m.time,v.value,'r--',label='CV Response')
plt.ylabel('Output')
plt.xlabel('Time')
plt.legend(loc='best')
plt.show()
For every time step, Gekko automatically generates the setpoint in the form of the array for the future control horizon. And, the arrays are usually filled with the single value that you assigned. However, you can give the setpoint as an array as shown below.
sp1 = np.array([[1,2,3,4,5],
[2,3,4,5,6],
[3,4,5,6,7],
[4,5,6,7,8]])
Then, you can assign the every row of your matrix for each time step just as you did in your question.
DT = .1
TC1.SPHI = sp1[i] + DT
Note:
You need to have the same length of setpoint array which means the size of the column in your setpoint matrix with the control horizon (e.g. 'm.time').
You might want to set the setpoint trajectory option '0' if you don't want to filter out your setpoint sequence in your array. (TR_INIT = 0)
I am trying to build a python code for Model Predictive Control using APMonitor. However, I don't want to get the results on an third party online server. Hence, I want to collect the data of the predicted biased and unbiased and plot them on Python myself.
Try this in Python Gekko:
# get additional solution information
import json
with open(m.path+'//results.json') as f:
results = json.load(f)
You can get the unbiased model result by getting the dictionary value of your variable v with v.name. You can get the biased model prediction with v.name+'.bcv'. Here is an example that also shows how to get the raw trajectory information.
This gives you access to the raw data. An example shows how to plot from the JSON data.
from gekko import GEKKO
import numpy as np
import matplotlib.pyplot as plt
m = GEKKO()
m.time = np.linspace(0,20,41)
# Parameters
mass = 500
b = m.Param(value=50)
K = m.Param(value=0.8)
# Manipulated variable
p = m.MV(value=0, lb=0, ub=100)
p.STATUS = 1 # allow optimizer to change
p.DCOST = 0.1 # smooth out gas pedal movement
p.DMAX = 20 # slow down change of gas pedal
# Controlled Variable
v = m.CV(value=0)
v.STATUS = 1 # add the SP to the objective
m.options.CV_TYPE = 2 # squared error
v.SP = 40 # set point
v.TR_INIT = 1 # set point trajectory
v.TAU = 5 # time constant of trajectory
# Process model
m.Equation(mass*v.dt() == -v*b + K*b*p)
m.options.IMODE = 6 # control
m.solve(disp=False,GUI=True)
# get additional solution information
import json
with open(m.path+'//results.json') as f:
results = json.load(f)
plt.figure()
plt.subplot(2,1,1)
plt.plot(m.time,p.value,'b-',label='MV Optimized')
plt.legend()
plt.ylabel('Input')
plt.subplot(2,1,2)
plt.plot(m.time,results['v1.tr'],'k-',label='Reference Trajectory')
plt.plot(m.time,v.value,'r--',label='CV Response')
plt.ylabel('Output')
plt.xlabel('Time')
plt.legend(loc='best')
plt.show()