GEKKO 多相优化：无解，-1 自由度

Question

我现在正在学习 GEKKO，并尝试编写一个简单的 1D 控制示例，其中包含动态块

dx/dt = v
dv/dt = u

与：-5 <= u <= 5

其中 x 是位置，v 是速度，u 是施加的驱动必须在 1 个时间单位内从位置 0 移动到 1，初始速度和最终速度为 0。目标是：

最小绝对值（u）

它与单相一起工作，结果就是这种Bang-Off-Bang 控制。

我现在想尝试将其实现为具有 3 个链接阶段的多阶段轨迹：

1. 你 = 5
2. 你 = 0
3. u = -5

其中每个阶段的时间是控制变量。

问题是求解器找不到解决方案。 自由度为 -1，在理论上（总的最终时间）对于给定的目标应该为 1，或者如果目标被等式约束替换，则为 0（这导致求解器声明更少的自由度，大约 - 100)。

这是我的代码：

import numpy as np
import matplotlib.pyplot as plt
from gekko import GEKKO

m = GEKKO(remote=True)  # Model

n = 3  # Number of phases: Bang-Off-Bang
max_u = 5.0  # Control can either be max_u, 0 or -max_u

# Options
m.options.NODES = 4  # Two intermediate nodes between collocation points
m.options.SOLVER = 1  # APOPT
m.options.IMODE = 6  # MPC Direct Collocation
m.options.MAX_ITER = 500  # To prevent endless loops
m.options.MV_TYPE = 0  # MVs are constant between endpoints
m.options.DIAGLEVEL = 1  # Show some diagnostics

# Time from 0 to 1 in 31 collocation points for every phase each
m.time = np.linspace(0, 1, 31)

tf = [m.FV(value=0.3) for i in range(n)]
for i in range(n):
    tf[i].STATUS = 1  # make final time controllable

# Collocation variables x and v
x = [m.Var(value=0, fixed_initial=False) for i in range(n)]
v = [m.Var(value=0, fixed_initial=False) for i in range(n)]
u = [max_u, 0, -max_u]  # For the three phases

# Fix start- and endpoint (Inbetween points can be fixed with m.fix())
m.fix_initial(x[0], val=0)
m.fix_initial(v[0], val=0)
m.fix_final(x[n-1], val=1)
m.fix_final(v[n-1], val=0)

# Differential equations describing the system scaled by tf
for i in range(n):
    m.Equation(x[i].dt() == v[i]*tf[i])
    m.Equation(v[i].dt() == u[i]*tf[i])

# Connect phases at endpoints
for i in range(n-1):
    m.Connection(x[i+1], x[i], 1, 'end', 1, 'end')
    m.Connection(x[i+1],'calculated', pos1=1, node1=1)
    m.Connection(v[i+1], v[i], 1, 'end', 1, 'end')
    m.Connection(v[i+1],'calculated', pos1=1, node1=1)

# Make final time equal to 1
m.Minimize((m.sum(tf)-1)**2)
#m.Equation(m.sum(tf) == 1)  # Could be used instead of objective

m.open_folder()

# Run optimization with Diagnostics
m.solve(disp=True)

我希望有人可以帮助我，提前谢谢你！

Answer 1

软终端约束通常更适合收敛到解决方案。 这是硬终端约束的替换：

f = np.zeros(31); f[-1]=1; final=m.Param(f)
m.Minimize(final*(x[n-1]-1)**2)
m.Minimize(final*v[n-1]**2)

#m.fix_final(x[n-1], val=1)
#m.fix_final(v[n-1], val=0)

这给出了三个链接阶段的最佳解决方案。

import numpy as np
import matplotlib.pyplot as plt
from gekko import GEKKO

m = GEKKO(remote=True)  # Model

n = 3  # Number of phases: Bang-Off-Bang
max_u = 5.0  # Control can either be max_u, 0 or -max_u

# Options
m.options.NODES = 4  # Two intermediate nodes between collocation points
m.options.SOLVER = 1  # APOPT
m.options.IMODE = 6  # MPC Direct Collocation
m.options.MAX_ITER = 500  # To prevent endless loops
m.options.MV_TYPE = 0  # MVs are constant between endpoints
m.options.DIAGLEVEL = 0  # Show some diagnostics

# Time from 0 to 1 in 31 collocation points for every phase each
m.time = np.linspace(0, 1, 31)

tf = [m.FV(value=0.3) for i in range(n)]
for i in range(n):
    tf[i].STATUS = 1  # make final time controllable

# Collocation variables x and v
x = [m.Var(value=0, fixed_initial=False) for i in range(n)]
v = [m.Var(value=0, fixed_initial=False) for i in range(n)]
u = [max_u, 0, -max_u]  # For the three phases

# Fix start- and endpoint (Inbetween points can be fixed with m.fix())
m.fix_initial(x[0], val=0)
m.fix_initial(v[0], val=0)

f = np.zeros(31); f[-1]=1; final=m.Param(f)
m.Minimize(final*(x[n-1]-1)**2)
m.Minimize(final*v[n-1]**2)

#m.fix_final(x[n-1], val=1)
#m.fix_final(v[n-1], val=0)

# Differential equations describing the system scaled by tf
for i in range(n):
    m.Equation(x[i].dt() == v[i]*tf[i])
    m.Equation(v[i].dt() == u[i]*tf[i])

# Connect phases at endpoints
for i in range(n-1):
    m.Connection(x[i+1], x[i], 1, 'end', 1, 'end')
    m.Connection(x[i+1],'calculated', pos1=1, node1=1)
    m.Connection(v[i+1], v[i], 1, 'end', 1, 'end')
    m.Connection(v[i+1],'calculated', pos1=1, node1=1)

# Make final time equal to 1
m.Minimize((m.sum(tf)-1)**2)
#m.Equation(m.sum(tf) == 1)  # Could be used instead of objective

#m.open_folder()

# Run optimization with Diagnostics
m.solve(disp=True)

# Generate plot
t = [m.time*tf[i].value[0] for i in range(3)]
t[1] += t[0][-1]
t[2] += t[1][-1]
for i in range(3):
    plt.plot(t[i],x[i].value)
    plt.plot(t[i],v[i].value)
plt.xlabel('Time')
plt.show()    

缺少 2 个自由度（-1 DOF）是因为当m.fix_final(x[n-1], val=1)和m.fix_final(v[n-1], val=0)被使用。 这是 Gekko 的一个已知问题以及如何定义变量。 如果您确实需要硬约束，还有另一种方法（创建连接到端点的固定 FV）。