Time optimal control example GEKKO
I am trying to implement a time optimal control problem in gekko. In particular, I copied this short code snippet. Also reported here for practicality:
from gekko import GEKKO import matplotlib.pyplot as plt import numpy as np # set up the gekko model m = GEKKO() # set up the time (minimize the time with time scaling) m.time = np.linspace(0, 1, 100) # set up the variables POSITION = m.Var(value=0, ub=330, lb=0) VELOCITY = m.Var(value=0, ub=33, lb=0) m.fix_final(VELOCITY, 0) m.fix_final(POSITION, 300) # set up the value we modify over the horizon tf = m.FV(value=500, lb=0.1) tf.STATUS = 1 # set up the MV u = m.MV(integer=True, lb=-2, ub=1) u.STATUS = 1 # set up the equations m.Equation(POSITION.dt() / tf == VELOCITY) m.Equation(VELOCITY.dt() / tf == u) # set the objective m.Obj(tf) # set up the options m.options.IMODE = 6 # optimal control m.options.SOLVER = 3 # IPOPT # solve m.solve(disp=False) # print the time print("Total time taken: " + str(tf.NEWVAL)) # plot the results plt.figure() plt.subplot(211) plt.plot(np.linspace(0,1,100)*tf.NEWVAL, POSITION, label='Position') plt.plot(np.linspace(0,1,100)*tf.NEWVAL, VELOCITY, label='Velocity') plt.ylabel('Z') plt.legend() plt.subplot(212) plt.plot(np.linspace(0,1,100)*tf.NEWVAL, u, label=r'$u$') plt.ylabel('u') plt.xlabel('Time') plt.legend() plt.show()
As is, it works fine, but when I want to remove the constraint on the final value of velocity.
If I comment out the m.fix_final(velocity, 0)
line, the result does not change. Regardless, it seems to assume that the final velocity should be zero. Also, if I change the final velocity from zero to any other number, I get the error from gekko: exception: @error: solution not found
.
The solution should be easy to find, especially if no constraints are imposed on the final speed, the optimal control would be to keep accelerating() throughout the time.
Any help would be greatly appreciated! :)
Correct answer
Change the final constraints from m.fix_final(velocity, 0)
and m.fix_final(position, 300)
changed to:
p = np.zeros(100); p[-1] = 1 last = m.Param(p) m.Equation(last*(POSITION-300)>=0)
This applies an inequality constraint at the last node so that position>=300
, but it could also be an equality constraint. We also sometimes use soft constraints, such as m.minimize(last*(position-300)**2)
if an infeasible solution prevents the solver from achieving the final condition. Instead, it tries to get the solution as close as possible to the final constraints. When the final value is fixed using m.fix_final()
, the derivative is also fixed to zero because that variable is no longer evaluated. This is a known limitation of gekko, as described here.
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