Can GEKKO solve Minimax Optimal Control problems? - math

Question

I'm aware that GEKKO can solve minimax optimization problems as well as optimal control problems. However, is there a mechanism on GEKKO to solve a minimax/maximin optimal control problem such as:
More specifically,

Answer 1

There is no issue to combine minimax with optimal control. There is also information on minimizing final time with the Jennings example problem as a benchmark optimal control problem. Here is a Minimax example problem from the APMonitor documentation:
from gekko import GEKKO
m = GEKKO(remote=False)
x1,x2,x3,Z = m.Array(m.Var,4)
m.Minimize(Z)
m.Equation(x1+x2+x3==15)
m.Equations([Z>=x1,Z>=x2,Z>=x3])
m.solve()
print('x1: ',x1.value[0])
print('x2: ',x2.value[0])
print('x3: ',x3.value[0])
print('Z: ',Z.value[0])
Here is the Jennings OCP solution where the final time is minimized:
import numpy as np
from gekko import GEKKO
import matplotlib.pyplot as plt
m = GEKKO()
nt = 501; tm = np.linspace(0,1,nt); m.time = tm
# Variables
x1 = m.Var(value=np.pi/2.0)
x2 = m.Var(value=4.0)
x3 = m.Var(value=0.0)
p = np.zeros(nt)
p[-1] = 1.0
final = m.Param(value=p)
# FV
tf = m.FV(value=1.0,lb=0.1,ub=100.0)
tf.STATUS = 1
# MV
u = m.MV(value=0,lb=-2,ub=2)
u.STATUS = 1
m.Equation(x1.dt()==u*tf)
m.Equation(x2.dt()==m.cos(x1)*tf)
m.Equation(x3.dt()==m.sin(x1)*tf)
m.Equation(x2*final<=0)
m.Equation(x3*final<=0)
m.Minimize(tf)
m.options.IMODE = 6
m.solve()
print('Final Time: ' + str(tf.value[0]))
tm = tm * tf.value[0]
plt.figure(1)
plt.plot(tm,x1.value,'k-',lw=2,label=r'$x_1$')
plt.plot(tm,x2.value,'b-',lw=2,label=r'$x_2$')
plt.plot(tm,x3.value,'g--',lw=2,label=r'$x_3$')
plt.plot(tm,u.value,'r--',lw=2,label=r'$u$')
plt.legend(loc='best')
plt.xlabel('Time')
plt.ylabel('Value')
plt.show()
Combining these two types of problems:
import numpy as np
from gekko import GEKKO
import matplotlib.pyplot as plt
m = GEKKO()
nt = 501
tm = np.linspace(0,1,nt)
m.time = tm
x1 = m.Var(value=np.pi/2.0)
x2 = m.Var(value=4.0)
x3 = m.Var(value=0.0)
p = np.zeros(nt)
p[-1] = 1.0
final = m.Param(value=p)
tf = m.FV(value=1.0,lb=0.1,ub=100.0)
tf.STATUS = 1
m.Minimize(tf)
Z = m.Var()
m.Minimize(Z)
m.Equations([Z>=x1,Z>=x2,Z>=x3])
m.Maximize(x1)
m.Maximize(x2)
m.Maximize(x3)
u = m.MV(value=0,lb=-2,ub=2)
u.STATUS = 1
m.Equation(x1.dt()==u*tf)
m.Equation(x2.dt()==m.cos(x1)*tf)
m.Equation(x3.dt()==m.sin(x1)*tf)
m.Equation(x2*final<=0)
m.Equation(x3*final<=0)
m.options.IMODE = 6
m.solve()
print('Final Time: ' + str(tf.value[0]))
tm = tm * tf.value[0]
plt.figure(1)
plt.plot(tm,x1.value,'k-',lw=2,label=r'$x_1$')
plt.plot(tm,x2.value,'b-',lw=2,label=r'$x_2$')
plt.plot(tm,x3.value,'g--',lw=2,label=r'$x_3$')
plt.plot(tm,u.value,'r--',lw=2,label=r'$u$')
plt.legend(loc='best')
plt.xlabel('Time')
plt.ylabel('Value')
plt.show()