我想使用CVXPY在离散时间内解决最小燃料最优控制问题。使用连续时间线性系统的零阶保持，可以将问题转化为具有凸控制约束的线性程序。我已经使用像Yalmip和CVX这样的Matlab环境完成了这个问题的基本表述，但我无法在CVXPY中使用它。我遇到的问题是，即使问题似乎完全编译和解决，输出值显然不满足边界条件，并且当绘制输出时，结果为空。
我附上了代码。该问题应该是双积分器的最小燃料控制;我想要一个严格正负方向的控制信号，每个控制信号取0到umax之间的值。
这里是定义双积分矩阵的类：
import numpy as np
import control as ctrl
class DoubleIntegrator:
def __init__(self, nu):
self.A = np.matrix([[0,1],[0,0]])
# Note, this matrix doesn't really do anything right now
self.C = np.matrix([[1,0],[0,1]])
try:
if nu == 1:
self.B = np.matrix([[0],[1]])
self.D = np.matrix([[0],[1]])
return
elif nu == 2:
self.B = np.matrix([[0,0],[1,-1]])
self.D = np.matrix([[0,0],[1,-1]])
return
elif nu != 1 or nu != 2:
raise InputError("ERROR: Input 'nu' must be either nu = 1 or nu = 2")
except InputError as me:
print me.message
return
def makeStateSpaceSystem(self):
self.sysc = ctrl.matlab.ss(self.A, self.B, self.C, self.D)
def makeDiscreteSystem(self, dt, method):
self.sysd = ctrl.matlab.c2d(self.sysc, dt, method)
class Error(Exception):
pass
class InputError(Error):
def __init__(self, message):
self.message = message
if __name__ == '__main__':
db = DoubleIntegrator(2)
db.makeStateSpaceSystem()
db.makeDiscreteSystem(0.1, 'zoh')
print db.sysd.A[0,1]
这是主要代码：
from DoubleIntegrator import *
import numpy as np
import cvxpy as cvx
import control as ctrl
import matplotlib.pyplot as plt
db = DoubleIntegrator(2)
db.makeStateSpaceSystem()
db.makeDiscreteSystem(0.1,'zoh')
A = db.sysd.A
B = db.sysd.B
Nsim = 20
x = cvx.Variable(2, Nsim+1)
u = cvx.Variable(2, Nsim)
x0 = [1.0,0.0]
xf = [0.0,0.0]
umax = 1.0
states = []
cost = 0
for t in range(Nsim):
cost = cost + u[0,t] + u[1,t]
constr = [x[:,t+1] == A*x[:,t] + B*u[:,t],
0 <= u[0,t], u[0,t] <= umax,
0 <= u[1,t], u[1,t] <= umax]
states.append( cvx.Problem( cvx.Minimize(cost), constr ) )
prob = sum(states)
prob.constraints += [x[0,Nsim] == xf[0], x[1,Nsim] == xf[1], x[0,0] == x0[0], x[1,0] == x0[1]]
prob.solve(verbose = True)
print u.value
print x.value
f = plt.figure()
plt.subplot(411)
plt.plot(x[0,:].value)
plt.subplot(412)
plt.plot(x[1,:].value)
plt.subplot(413)
plt.plot(u[0,:].value)
plt.subplot(414)
plt.plot(u[1,:].value)
plt.show()
非常感谢您的帮助！
答案 0 :(得分：1)
<强> double_integrator.py
import numpy as np
from cvxpy import Variable, sum_entries, Problem, Minimize, ECOS
import control
from time import time
# model matrices
A = np.matrix([[0, 1], [0, 0]])
B = np.matrix([[0, 0], [1, -1]])
C = np.matrix([[1, 0], [0, 1]])
D = np.matrix([[0, 0], [1, -1]])
Ts = 0.1
N = 20
sys_c = control.matlab.ss(A, B, C, D)
sys_d = control.matlab.c2d(sys_c, Ts, 'zoh')
n = A.shape[0] # number of states
m = B.shape[1] # number of inputs
# optimization variables
x = Variable(n, N+1)
u = Variable(m, N)
states = []
x0 = np.array([1, 0])
xf = np.array([0, 0])
u_max = 1
for t in xrange(N):
cost = sum_entries(u[:, t])
constr = [
x[:, t+1] == sys_d.A * x[:, t] + sys_d.B * u[:, t],
u[:, t] >= 0,
u[:, t] <= u_max
]
states.append(Problem(Minimize(cost), constr))
prob = sum(states)
prob.constraints.append(x[:, 0] == x0)
prob.constraints.append(x[:, N] == xf)
exec_start = time()
prob.solve(solver=ECOS)
exec_end = time()
print "status:", prob.status
print "value:", prob.value
print "solving time:", exec_end - exec_start, "s"
print "u[0] =", u[0, :].value
print "u[1] =", u[1, :].value
print "x[0] =", x[0, :].value
print "x[1] =", x[1, :].value
<强>输出：
root@machine:~# python double_integrator.py
status: optimal
value: 19.9999999025
solving time: 0.067615032196 s
u[0] = [[ -4.75426733e-10 -4.55386327e-10 -4.29550365e-10 -3.95761952e-10
-3.50620396e-10 -2.88248891e-10 -1.97290675e-10 -5.21543663e-11
2.21237875e-10 9.88033157e-10 9.99999957e-01 9.99999995e-01
9.99999999e-01 1.00000000e+00 1.00000000e+00 1.00000000e+00
1.00000000e+00 1.00000000e+00 1.00000000e+00 1.00000000e+00]]
u[1] = [[ 1.00000000e+00 1.00000000e+00 1.00000000e+00 1.00000000e+00
1.00000000e+00 1.00000000e+00 1.00000000e+00 9.99999999e-01
9.99999995e-01 9.99999957e-01 9.86520079e-10 2.20404086e-10
-5.26881945e-11 -1.97648318e-10 -2.88502852e-10 -3.50811744e-10
-3.95915025e-10 -4.29680212e-10 -4.55502467e-10 -4.75535165e-10]]
x[0] = [[ 1.00000000e+00 9.95000000e-01 9.80000000e-01 9.55000000e-01
9.20000000e-01 8.75000000e-01 8.20000000e-01 7.55000000e-01
6.80000000e-01 5.95000000e-01 5.00000000e-01 4.05000000e-01
3.20000000e-01 2.45000000e-01 1.80000000e-01 1.25000000e-01
8.00000001e-02 4.50000000e-02 2.00000000e-02 5.00000001e-03
1.48064807e-12]]
x[1] = [[ -1.47624932e-12 -1.00000000e-01 -2.00000000e-01 -3.00000000e-01
-4.00000000e-01 -5.00000000e-01 -6.00000000e-01 -7.00000000e-01
-8.00000000e-01 -9.00000000e-01 -9.99999995e-01 -9.00000000e-01
-8.00000000e-01 -7.00000000e-01 -6.00000000e-01 -5.00000000e-01
-4.00000000e-01 -3.00000000e-01 -2.00000000e-01 -1.00000000e-01
-1.48504278e-12]]