def test_optimal_logging(capsys):
"""Test logging functions (mainly for code coverage)"""
sys = ct.ss2io(ct.ss(np.eye(2), np.eye(2), np.eye(2), 0, 1))
# Set up the optimal control problem
cost = opt.quadratic_cost(sys, 1, 1)
state_constraint = opt.state_range_constraint(sys, [-np.inf, 1], [10, 1])
input_constraint = opt.input_range_constraint(sys, [-100, -100],
[100, 100])
time = np.arange(0, 3, 1)
x0 = [-1, 1]
# Solve it, with logging turned on (with warning due to mixed constraints)
with pytest.warns(sp.optimize.optimize.OptimizeWarning,
match="Equality and inequality .* same element"):
res = opt.solve_ocp(sys,
time,
x0,
cost,
input_constraint,
terminal_cost=cost,
terminal_constraints=state_constraint,
log=True)
# Make sure the output has info available only with logging turned on
captured = capsys.readouterr()
assert captured.out.find("process time") != -1
def time_steering_bezier_basis(nbasis, ntimes):
# Set up costs and constriants
Q = np.diag([.1, 10, .1]) # keep lateral error low
R = np.diag([1, 1]) # minimize applied inputs
cost = opt.quadratic_cost(vehicle, Q, R, x0=xf, u0=uf)
constraints = [opt.input_range_constraint(vehicle, [0, -0.1], [20, 0.1])]
terminal = [opt.state_range_constraint(vehicle, xf, xf)]
# Set up horizon
horizon = np.linspace(0, Tf, ntimes, endpoint=True)
# Set up the optimal control problem
res = opt.solve_ocp(
vehicle,
horizon,
x0,
cost,
constraints,
terminal_constraints=terminal,
initial_guess=bend_left,
basis=flat.BezierFamily(nbasis, T=Tf),
# solve_ivp_kwargs={'atol': 1e-4, 'rtol': 1e-2},
minimize_method='trust-constr',
minimize_options={'finite_diff_rel_step': 0.01},
# minimize_method='SLSQP', minimize_options={'eps': 0.01},
return_states=True,
print_summary=False)
t, u, x = res.time, res.inputs, res.states
# Make sure we found a valid solution
assert res.success
def time_steering_terminal_constraint(integrator_name, minimizer_name):
# Get the integrator and minimizer parameters to use
integrator = integrator_table[integrator_name]
minimizer = minimizer_table[minimizer_name]
# Input cost and terminal constraints
R = np.diag([1, 1]) # minimize applied inputs
cost = opt.quadratic_cost(vehicle, np.zeros((3, 3)), R, u0=uf)
constraints = [opt.input_range_constraint(vehicle, [8, -0.1], [12, 0.1])]
terminal = [opt.state_range_constraint(vehicle, xf, xf)]
res = opt.solve_ocp(
vehicle,
horizon,
x0,
cost,
constraints,
terminal_constraints=terminal,
initial_guess=bend_left,
log=False,
solve_ivp_method=integrator[0],
solve_ivp_kwargs=integrator[1],
minimize_method=minimizer[0],
minimize_options=minimizer[1],
)
# Only count this as a benchmark if we converged
assert res.success
def test_equality_constraints():
"""Test out the ability to handle equality constraints"""
# Create the system (double integrator, continuous time)
sys = ct.ss2io(ct.ss(np.zeros((2, 2)), np.eye(2), np.eye(2), 0))
# Shortest path to a point is a line
Q = np.zeros((2, 2))
R = np.eye(2)
cost = opt.quadratic_cost(sys, Q, R)
# Set up the terminal constraint to be the origin
final_point = [opt.state_range_constraint(sys, [0, 0], [0, 0])]
# Create the optimal control problem
time = np.arange(0, 3, 1)
optctrl = opt.OptimalControlProblem(sys,
time,
cost,
terminal_constraints=final_point)
# Find a path to the origin
x0 = np.array([4, 3])
res = optctrl.compute_trajectory(x0, squeeze=True, return_x=True)
t, u1, x1 = res.time, res.inputs, res.states
# Bug prior to SciPy 1.6 will result in incorrect results
if NumpyVersion(sp.__version__) < '1.6.0':
pytest.xfail("SciPy 1.6 or higher required")
np.testing.assert_almost_equal(x1[:, -1], 0, decimal=4)
# Set up terminal constraints as a nonlinear constraint
def final_point_eval(x, u):
return x
final_point = [(sp.optimize.NonlinearConstraint, final_point_eval, [0, 0],
[0, 0])]
optctrl = opt.OptimalControlProblem(sys,
time,
cost,
terminal_constraints=final_point)
# Find a path to the origin
x0 = np.array([4, 3])
res = optctrl.compute_trajectory(x0, squeeze=True, return_x=True)
t, u2, x2 = res.time, res.inputs, res.states
np.testing.assert_almost_equal(x2[:, -1], 0, decimal=4)
np.testing.assert_almost_equal(u1, u2)
np.testing.assert_almost_equal(x1, x2)
# Try passing and unknown constraint type
final_point = [(None, final_point_eval, [0, 0], [0, 0])]
with pytest.raises(TypeError, match="unknown constraint type"):
optctrl = opt.OptimalControlProblem(sys,
time,
cost,
terminal_constraints=final_point)
res = optctrl.compute_trajectory(x0, squeeze=True, return_x=True)
def test_flat_cost_constr(self):
# Double integrator system
sys = ct.ss([[0, 1], [0, 0]], [[0], [1]], [[1, 0]], 0)
flat_sys = fs.LinearFlatSystem(sys)
# Define the endpoints of the trajectory
x0 = [1, 0]
u0 = [0]
xf = [0, 0]
uf = [0]
Tf = 10
T = np.linspace(0, Tf, 500)
# Find trajectory between initial and final conditions
traj = fs.point_to_point(flat_sys,
Tf,
x0,
u0,
xf,
uf,
basis=fs.PolyFamily(8))
x, u = traj.eval(T)
np.testing.assert_array_almost_equal(x0, x[:, 0])
np.testing.assert_array_almost_equal(u0, u[:, 0])
np.testing.assert_array_almost_equal(xf, x[:, -1])
np.testing.assert_array_almost_equal(uf, u[:, -1])
# Solve with a cost function
timepts = np.linspace(0, Tf, 10)
cost_fcn = opt.quadratic_cost(flat_sys,
np.diag([0, 0]),
1,
x0=xf,
u0=uf)
traj_cost = fs.point_to_point(
flat_sys,
timepts,
x0,
u0,
xf,
uf,
cost=cost_fcn,
basis=fs.PolyFamily(8),
# initial_guess='lstsq',
# minimize_kwargs={'method': 'trust-constr'}
)
# Verify that the trajectory computation is correct
x_cost, u_cost = traj_cost.eval(T)
np.testing.assert_array_almost_equal(x0, x_cost[:, 0])
np.testing.assert_array_almost_equal(u0, u_cost[:, 0])
np.testing.assert_array_almost_equal(xf, x_cost[:, -1])
np.testing.assert_array_almost_equal(uf, u_cost[:, -1])
# Make sure that we got a different answer than before
assert np.any(np.abs(x - x_cost) > 0.1)
# Re-solve with constraint on the y deviation
lb, ub = [-2, -0.1], [2, 0]
lb, ub = [-2, np.min(x_cost[1]) * 0.95], [2, 1]
constraints = [opt.state_range_constraint(flat_sys, lb, ub)]
# Make sure that the previous solution violated at least one constraint
assert np.any(x_cost[0, :] < lb[0]) or np.any(x_cost[0, :] > ub[0]) \
or np.any(x_cost[1, :] < lb[1]) or np.any(x_cost[1, :] > ub[1])
traj_const = fs.point_to_point(
flat_sys,
timepts,
x0,
u0,
xf,
uf,
cost=cost_fcn,
constraints=constraints,
basis=fs.PolyFamily(8),
)
# Verify that the trajectory computation is correct
x_const, u_const = traj_const.eval(T)
np.testing.assert_array_almost_equal(x0, x_const[:, 0])
np.testing.assert_array_almost_equal(u0, u_const[:, 0])
np.testing.assert_array_almost_equal(xf, x_const[:, -1])
np.testing.assert_array_almost_equal(uf, u_const[:, -1])
# Make sure that the solution respects the bounds (with some slop)
for i in range(x_const.shape[0]):
assert np.all(x_const[i] >= lb[i] * 1.02)
assert np.all(x_const[i] <= ub[i] * 1.02)
# Solve the same problem with a nonlinear constraint type
nl_constraints = [(sp.optimize.NonlinearConstraint, lambda x, u: x, lb,
ub)]
traj_nlconst = fs.point_to_point(
flat_sys,
timepts,
x0,
u0,
xf,
uf,
cost=cost_fcn,
constraints=nl_constraints,
basis=fs.PolyFamily(8),
)
x_nlconst, u_nlconst = traj_nlconst.eval(T)
np.testing.assert_almost_equal(x_const, x_nlconst)
np.testing.assert_almost_equal(u_const, u_nlconst)
def test_terminal_constraints(sys_args):
"""Test out the ability to handle terminal constraints"""
# Create the system
sys = ct.ss2io(ct.ss(*sys_args))
# Shortest path to a point is a line
Q = np.zeros((2, 2))
R = np.eye(2)
cost = opt.quadratic_cost(sys, Q, R)
# Set up the terminal constraint to be the origin
final_point = [opt.state_range_constraint(sys, [0, 0], [0, 0])]
# Create the optimal control problem
time = np.arange(0, 3, 1)
optctrl = opt.OptimalControlProblem(
sys, time, cost, terminal_constraints=final_point)
# Find a path to the origin
x0 = np.array([4, 3])
res = optctrl.compute_trajectory(x0, squeeze=True, return_x=True)
t, u1, x1 = res.time, res.inputs, res.states
# Bug prior to SciPy 1.6 will result in incorrect results
if NumpyVersion(sp.__version__) < '1.6.0':
pytest.xfail("SciPy 1.6 or higher required")
np.testing.assert_almost_equal(x1[:,-1], 0, decimal=4)
# Make sure it is a straight line
Tf = time[-1]
if ct.isctime(sys):
# Continuous time is not that accurate on the input, so just skip test
pass
else:
# Final point doesn't affect cost => don't need to test
np.testing.assert_almost_equal(
u1[:, 0:-1],
np.kron((-x0/Tf).reshape((2, 1)), np.ones(time.shape))[:, 0:-1])
np.testing.assert_allclose(
x1, np.kron(x0.reshape((2, 1)), time[::-1]/Tf), atol=0.1, rtol=0.01)
# Re-run using initial guess = optional and make sure nothing changes
res = optctrl.compute_trajectory(x0, initial_guess=u1)
np.testing.assert_almost_equal(res.inputs, u1)
# Re-run using a basis function and see if we get the same answer
res = opt.solve_ocp(sys, time, x0, cost, terminal_constraints=final_point,
basis=flat.BezierFamily(4, Tf))
np.testing.assert_almost_equal(res.inputs, u1, decimal=2)
# Impose some cost on the state, which should change the path
Q = np.eye(2)
R = np.eye(2) * 0.1
cost = opt.quadratic_cost(sys, Q, R)
optctrl = opt.OptimalControlProblem(
sys, time, cost, terminal_constraints=final_point)
# Turn off warning messages, since we sometimes don't get convergence
with warnings.catch_warnings():
warnings.filterwarnings(
"ignore", message="unable to solve", category=UserWarning)
# Find a path to the origin
res = optctrl.compute_trajectory(
x0, squeeze=True, return_x=True, initial_guess=u1)
t, u2, x2 = res.time, res.inputs, res.states
# Not all configurations are able to converge (?)
if res.success:
np.testing.assert_almost_equal(x2[:,-1], 0)
# Make sure that it is *not* a straight line path
assert np.any(np.abs(x2 - x1) > 0.1)
assert np.any(np.abs(u2) > 1) # Make sure next test is useful
# Add some bounds on the inputs
constraints = [opt.input_range_constraint(sys, [-1, -1], [1, 1])]
optctrl = opt.OptimalControlProblem(
sys, time, cost, constraints, terminal_constraints=final_point)
res = optctrl.compute_trajectory(x0, squeeze=True, return_x=True)
t, u3, x3 = res.time, res.inputs, res.states
# Check the answers only if we converged
if res.success:
np.testing.assert_almost_equal(x3[:,-1], 0, decimal=4)
# Make sure we got a new path and didn't violate the constraints
assert np.any(np.abs(x3 - x1) > 0.1)
np.testing.assert_array_less(np.abs(u3), 1 + 1e-6)
# Make sure that infeasible problems are handled sensibly
x0 = np.array([10, 3])
with pytest.warns(UserWarning, match="unable to solve"):
res = optctrl.compute_trajectory(x0, squeeze=True, return_x=True)
assert not res.success
plot_results(t2, y2, u2, figure=2, yf=xf[0:2])
#
# Approach 3: terminal constraints
#
# We can also remove the cost function on the state and replace it
# with a terminal *constraint* on the state. If a solution is found,
# it guarantees we get to exactly the final state.
#
print("Approach 3: terminal constraints")
# Input cost and terminal constraints
R = np.diag([1, 1]) # minimize applied inputs
cost3 = opt.quadratic_cost(vehicle, np.zeros((3, 3)), R, u0=uf)
constraints = [opt.input_range_constraint(vehicle, [8, -0.1], [12, 0.1])]
terminal = [opt.state_range_constraint(vehicle, xf, xf)]
# Reset logging to its default values
logging.basicConfig(level=logging.DEBUG,
filename="./steering-terminal_constraint.log",
filemode='w',
force=True)
# Compute the optimal control
start_time = time.process_time()
result3 = opt.solve_ocp(
vehicle,
horizon,
x0,
cost3,
constraints,