具有自由终止时间的最优控制 (gekko)

Question

我正在制作一个数值问题作为示例，并尝试使用 gekko 为以下问题找到最佳控制：

minimize the integral of a*x(t) from 0 to T, where T is the first time x(t) is 0, i.e., it is a random time. The constraints are such that x(t) follows some dynamic f(x(t),u(t)), x(t) >= 0, and u(t) is between 0 and 1.

我按照 GEKKO 网站和 youtube 上的教程进行了固定的最终时间，但我找不到任何关于随机最终时间的信息。以下是我当前的代码，但我如何才能从固定的最终时间移动到随机的最终时间？任何帮助，将不胜感激！谢谢！

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint
from gekko import GEKKO

# Initial conditions
xhh0 = 3; xhi0 = 0; 
xvh0 = 30; xvi0 = 0;
hin0 = 0; vin0 = 0;
tt0 = 0

# Parameters
a1 = 0.1; a2 = 0.1;
b1 = 0.01; b2 = 0.5;
delta1 = 0.1; delta2 = 0.5;
rho1 = 0.3; rho2 = 0.01
mu = 1

# Gekko
m = GEKKO()

# Control variable
u = m.MV(0.5, lb = 0, ub = 1)

# Final time <------------------------ currently a fixed final time
T = 10

# Initialize
xhh, xhi, xvh, xvi, Ah, Av = m.Array(m.Var, 6)
xhh.value = xhh0; xhi.value = xhi0;
xvh.value = xvh0; xvi.value = xvi0;
Ah.value = hin0; Av.value = vin0; 

# System dynamics
m.Equations([xhh.dt() == -a1*xhh - mu*u - b1*xhi*xhh,\
             xhi.dt() == a1*xhh + b1*xhi*xhh - delta1*xhi - rho1*xhi,\
             xvh.dt() == -a2*xvh - mu*(1-u) - b2*xvi*xvh,\
             xvi.dt() == a2*xvh + b2*xvi*xvh - delta2*xvi - rho2*xvi,\
             Ah.dt() == a1*xhh,\
             Av.dt() == a2*xvh])

# Time space
t = np.linspace(0, T, 101) 

m.time = t

# initialize with simulation
m.options.IMODE = 7
m.options.NODES = 3
m.solve(disp = False)

# optimization
m.options.IMODE = 6
xhh.LOWER = 0; xhi.LOWER = 0; xvh.LOWER = 0; xvi.LOWER = 0
u.STATUS = 1
m.options.SOLVER = 3
xhh.value = xhh.value.value
xhi.value = xhi.value.value
xvh.value = xvh.value.value
xvi.value = xvi.value.value
Ah.value = Ah.value.value
Av.value = Av.value.value

# Objective function
m.Minimize(Ah + Av)
m.solve()

Answer 1

当每个微分除以T时，最终时间可以用T = m.FV()和T.STATUS=1调整。这将问题扩展到 t = np.linspace(0,1).

时的任意最终时间

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint
from gekko import GEKKO

# Initial conditions
xhh0 = 3; xhi0 = 0; 
xvh0 = 30; xvi0 = 0;
hin0 = 0; vin0 = 0;
tt0 = 0

# Parameters
a1 = 0.1; a2 = 0.1;
b1 = 0.01; b2 = 0.5;
delta1 = 0.1; delta2 = 0.5;
rho1 = 0.3; rho2 = 0.01; mu = 1

# Gekko
m = GEKKO()

# Control variable
u = m.MV(0.5, lb = 0, ub = 1)

# Final time
T = m.FV(10,lb=1e-2,ub=100); T.STATUS = 1

# Initialize
xhh, xhi, xvh, xvi, Ah, Av = m.Array(m.Var, 6)
xhh.value = xhh0; xhi.value = xhi0;
xvh.value = xvh0; xvi.value = xvi0;
Ah.value = hin0; Av.value = vin0; 
xhh.LOWER = 0; xhi.LOWER = 0; xvh.LOWER = 0; xvi.LOWER = 0
u.STATUS = 1

# System dynamics
m.Equations([xhh.dt()/T == -a1*xhh - mu*u - b1*xhi*xhh,\
             xhi.dt()/T == a1*xhh + b1*xhi*xhh - delta1*xhi - rho1*xhi,\
             xvh.dt()/T == -a2*xvh - mu*(1-u) - b2*xvi*xvh,\
             xvi.dt()/T == a2*xvh + b2*xvi*xvh - delta2*xvi - rho2*xvi,\
             Ah.dt()/T == a1*xhh,\
             Av.dt()/T == a2*xvh])

# Time space
t = np.linspace(0, 1, 101) 
m.time = t

# optimization
m.options.IMODE = 6
m.options.SOLVER = 3

# Objective function
m.Minimize(Ah + Av)
m.solve()

print('Final time: ', T.value[0])

可能缺少约束条件或一些其他信息，因为最佳最终时间总是达到下限。 Jennings problem 是一个具有可变最终时间的相关示例。

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