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一维控制示例，其中有一个具有动态特性的块。

dx/dt =v

dv/dt =u

电话：-5 <= u <= 5

其中x是位置，v是速度，u是施加的驱动，在初始和最终速度为0的情况下，从位置0移动到1倍的时间单位。目标是：

分钟abs(u)

它工作在一个单一的阶段，结果是这 Bang Bang控件.

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现在，我想尝试将其作为一个多阶段的轨道来实现，其中包括三个相互关联的阶段：

1. U=5
2. U=0
3. U= -5

每个阶段的时间是控制变量。

问题是求解者找不到解决办法。自由度为-1，理论上为-1(总最后一次)与给定目标对应；如果目标被平等约束所取代，则为0(结果导致求解者表示的自由度更少，约为-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)

我希望有人能帮我，谢谢你提前！

Answer 1

软终端约束通常用于收敛到解决方案。下面是对硬终端约束的替换：

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)

这给出了三个连接相的最优解。

​

​

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自由度(-1自由度)是因为当使用m.fix_final(x[n-1], val=1)和m.fix_final(v[n-1], val=0)时，端点处的导数也是固定的。这是关于Gekko和如何定义变量的一个已知问题。如果您确实需要一个硬约束，还有另一种方法(创建一个连接到端点的固定FV )。