Scipy.optimize.minimize on trajectory optimization（Cost Functional Minimization）

Question

我在这里发布了一个问题，询问如何解决这个问题：

Trajectory Optimization for "Rocket" using scipy.optimize.minimize

理想情况下，我想最小化最后时间，但我无法让优化器将时间附加到可以正确调整的变量，所以我决定现在尝试最小化u ^ 2。

这是代码：

---

# Code
t_f = 1.0
t = np.linspace(0., t_f, num = 10) # Time array for 1 second into the future with 0.01 increment
u = np.zeros(t.size) + 650
print(u)
g = -650
initial_position = 0
initial_velocity = 0
final_position = 100
final_velocity = 100

def car_dynamics(x):
    # Create time vector
    # t = np.linspace(0., t_f, num = 100) # Time array for 1 second into the future with 0.01 increment


    # Integrate over entire time to find v as a function of t
    a = x + g
    v = int.cumtrapz(a, t, initial = 0) + initial_velocity

    # Integrate v(t) to get s(t)
    s = int.cumtrapz(v, t, initial = 0) + initial_position

    return s, v

def constraint1(x): # Final state constraints (Boundary conditions)
    s, v = car_dynamics(x)
    print('c1', s[0] - initial_position)
    return s[0] - initial_position

def constraint2(x): # Initial state constraints (initial conditions)
    s, v = car_dynamics(x)
    print('c2', v[0] - initial_velocity)
    return v[0] - initial_velocity

def constraint3(x):
    s, v = car_dynamics(x)
    print('c3', s[-1] - final_position)
    return s[-1] - final_position

def constraint4(x):
    s, v = car_dynamics(x)
    print('c4', v[-1] - final_velocity)
    return v[-1] - final_velocity

def constraint5(x):
    return x - 1000

def objective(x):
    u2 = np.square(x)
    return np.sum(u2)

cons = [{'type':'eq', 'fun':constraint1},
                {'type':'eq', 'fun':constraint2},
                {'type':'eq', 'fun':constraint3},
                {'type':'eq', 'fun':constraint4}]
                # {'type':'ineq', 'fun':constraint5}]


result = minimize(objective, u, constraints = cons, method = 'SLSQP', options={'eps':500, 'maxiter':1000, 'ftol':0.001, 'disp':True})
print(result)

代码运行但优化器失败。这是输出中的错误。

        c1 0.0
c2 0.0
c3 -100.0
c4 -100.0
c1 0.0
c2 0.0
c3 -100.0
c4 -100.0
c1 0.0
c1 0.0
c1 0.0
c1 0.0
c1 0.0
c1 0.0
c1 0.0
c1 0.0
c1 0.0
c1 0.0
c1 0.0
c2 0.0
c2 0.0
c2 0.0
c2 0.0
c2 0.0
c2 0.0
c2 0.0
c2 0.0
c2 0.0
c2 0.0
c2 0.0
c3 -100.0
c3 -73.76543209876543
c3 -50.617283950617285
c3 -56.79012345679013
c3 -62.96296296296296
c3 -69.1358024691358
c3 -75.30864197530863
c3 -81.4814814814815
c3 -87.65432098765432
c3 -93.82716049382715
c3 -98.45679012345678
c4 -100.0
c4 -72.22222222222223
c4 -44.44444444444445
c4 -44.44444444444445
c4 -44.44444444444445
c4 -44.444444444444436
c4 -44.44444444444445
c4 -44.44444444444448
c4 -44.44444444444445
c4 -44.44444444444442
c4 -72.22222222222221
Singular matrix C in LSQ subproblem    (Exit mode 6)
            Current function value: 4225000.0
            Iterations: 1
            Function evaluations: 12
            Gradient evaluations: 1
     fun: 4225000.0
     jac: array([1800., 1800., 1800., 1800., 1800., 1800., 1800., 1800., 1800.,
       1800.])
 message: 'Singular matrix C in LSQ subproblem'
    nfev: 12
     nit: 1
    njev: 1
  status: 6
 success: False
       x: array([650., 650., 650., 650., 650., 650., 650., 650., 650., 650.])

似乎在一些迭代中没有满足约束。我应该切换目标函数以包含最终速度和最终位置吗？我尝试过不同的步长，而不是使用相同的退出代码。

是否有更好的方法将此功能用于我想要获得的内容？我试图在从t0到t_f的整个间隔内获得控制向量u（t），这样我就可以将这些命令发送到火箭以实现最佳控制。现在我已经将优化简化为单轴，只是为了学习如何使用该功能。但是你可以看到我没有成功。

类似的例子非常有用，我对其他优化方法持开放态度，只要它们是数字的，并且相对较快，因为我计划最终将其实现为模型预测控制器。

Answer 1

您的模型具有代数和微分方程。您需要一个DAE求解器来解决上面隐式的ODE函数。我知道的一个这样的包是gekko。 （https://github.com/BYU-PRISM/GEKKO）Gekko专注于线性，混合整数和非线性优化问题的动态优化。

下面是一个示例火箭发射问题，可以最大限度地缩短最终时间。可在http://apmonitor.com/wiki/index.php/Apps/RocketLaunch购买

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

# create GEKKO model
m = GEKKO()

# scale 0-1 time with tf
m.time = np.linspace(0,1,101)

# options
m.options.NODES = 6
m.options.SOLVER = 3
m.options.IMODE = 6
m.options.MAX_ITER = 500
m.options.MV_TYPE = 0
m.options.DIAGLEVEL = 0

# final time
tf = m.FV(value=1.0,lb=0.1,ub=100)
tf.STATUS = 1

# force
u = m.MV(value=0,lb=-1.1,ub=1.1)
u.STATUS = 1
u.DCOST = 1e-5

# variables
s = m.Var(value=0)
v = m.Var(value=0,lb=0,ub=1.7)
mass = m.Var(value=1,lb=0.2)

# differential equations scaled by tf
m.Equation(s.dt()==tf*v)
m.Equation(mass*v.dt()==tf*(u-0.2*v**2))
m.Equation(mass.dt()==tf*(-0.01*u**2))

# specify endpoint conditions
m.fix(s, pos=len(m.time)-1,val=10.0)
m.fix(v, pos=len(m.time)-1,val=0.0)

# minimize final time
m.Obj(tf)

# Optimize launch
m.solve()

print('Optimal Solution (final time): ' + str(tf.value[0]))

# scaled time
ts = m.time * tf.value[0]

# plot results
plt.figure(1)
plt.subplot(4,1,1)
plt.plot(ts,s.value,'r-',linewidth=2)
plt.ylabel('Position')
plt.legend(['s (Position)'])

plt.subplot(4,1,2)
plt.plot(ts,v.value,'b-',linewidth=2)
plt.ylabel('Velocity')
plt.legend(['v (Velocity)'])

plt.subplot(4,1,3)
plt.plot(ts,mass.value,'k-',linewidth=2)
plt.ylabel('Mass')
plt.legend(['m (Mass)'])

plt.subplot(4,1,4)
plt.plot(ts,u.value,'g-',linewidth=2)
plt.ylabel('Force')
plt.legend(['u (Force)'])

plt.xlabel('Time')
plt.show()