Created
August 6, 2014 22:45
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Direct single shooting with CasADi
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from casadi import * | |
from numpy import * | |
import matplotlib.pyplot as plt | |
N = 20 # Control discretization | |
T = 10.0 # End time | |
# Declare variables (use scalar graph) | |
u = SX.sym("u") # control | |
x = SX.sym("x",2) # states | |
# System dynamics | |
xdot = vertcat( [(1 - x[1]**2)*x[0] - x[1] + u, x[0]] ) | |
qdot = x[0]**2 + x[1]**2 + u**2 | |
f = SXFunction([x,u],[xdot,qdot]) | |
f.init() | |
# RK4 with M steps | |
U = MX.sym("U") | |
X = MX.sym("X",2) | |
M = 10; DT = T/(N*M) | |
XF = X | |
QF = 0 | |
for j in range(M): | |
[k1, k1_q] = f([XF, U]) | |
[k2, k2_q] = f([XF + DT/2 * k1, U]) | |
[k3, k3_q] = f([XF + DT/2 * k2, U]) | |
[k4, k4_q] = f([XF + DT * k3, U]) | |
XF += DT/6*(k1 + 2*k2 + 2*k3 + k4) | |
QF += DT/6*(k1_q + 2*k2_q + 2*k3_q + k4_q) | |
F = MXFunction([X,U],[XF,QF]) | |
F.init() | |
# Formulate NLP (use matrix graph) | |
nv = N | |
v = MX.sym("v",nv) | |
# Objective function | |
J=0 | |
# Get an expression for the cost and state at end | |
X = MX([0,1]) | |
for k in range(N): | |
[X, qf] = F([X,v[k]]) | |
J += qf | |
# Terminal constraints: x_0(T)=x_1(T)=0 | |
g = X | |
# Allocate an NLP solver | |
nlp = MXFunction(nlpIn(x=v),nlpOut(f=J,g=g)) | |
solver = NlpSolver("ipopt", nlp) | |
solver.init() | |
# Set bounds and initial guess | |
solver.setInput( 0., "x0") | |
solver.setInput(-1., "lbx") | |
solver.setInput( 1., "ubx") | |
solver.setInput( 0., "lbg") | |
solver.setInput( 0., "ubg") | |
# Solve the problem | |
solver.evaluate() | |
# Retrieve the solution | |
u_opt = solver.getOutput("x") | |
# Plot the results | |
plt.figure(1) | |
plt.clf() | |
plt.step(linspace(0,T,N),u_opt,'-.') | |
plt.title("Van der Pol optimization - single shooting") | |
plt.xlabel('time') | |
plt.legend(['u trajectory']) | |
plt.grid() | |
plt.show() | |
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