def test_ocp_argument_errors():
sys = ct.ss2io(ct.ss([[1, 1], [0, 1]], [[1], [0.5]], np.eye(2), 0, 1))
# State and input constraints
constraints = [
(sp.optimize.LinearConstraint, np.eye(3), [-5, -5, -1], [5, 5, 1]),
]
# Quadratic state and input penalty
Q = [[1, 0], [0, 1]]
R = [[1]]
cost = opt.quadratic_cost(sys, Q, R)
# Set up the optimal control problem
time = np.arange(0, 5, 1)
x0 = [4, 0]
# Trajectory constraints not in the right form
with pytest.raises(TypeError, match="constraints must be a list"):
res = opt.solve_ocp(sys, time, x0, cost, np.eye(2))
# Terminal constraints not in the right form
with pytest.raises(TypeError, match="constraints must be a list"):
res = opt.solve_ocp(
sys, time, x0, cost, constraints, terminal_constraints=np.eye(2))
# Initial guess in the wrong shape
with pytest.raises(ValueError, match="initial guess is the wrong shape"):
res = opt.solve_ocp(
sys, time, x0, cost, constraints, initial_guess=np.zeros((4,1,1)))
def test_optimal_basis_simple():
sys = ct.ss2io(ct.ss([[1, 1], [0, 1]], [[1], [0.5]], np.eye(2), 0, 1))
# State and input constraints
constraints = [
(sp.optimize.LinearConstraint, np.eye(3), [-5, -5, -1], [5, 5, 1]),
]
# Quadratic state and input penalty
Q = [[1, 0], [0, 1]]
R = [[1]]
cost = opt.quadratic_cost(sys, Q, R)
# Set up the optimal control problem
Tf = 5
time = np.arange(0, Tf, 1)
x0 = [4, 0]
# Basic optimal control problem
res1 = opt.solve_ocp(sys,
time,
x0,
cost,
constraints,
basis=flat.BezierFamily(4, Tf),
return_x=True)
assert res1.success
# Make sure the constraints were satisfied
np.testing.assert_array_less(np.abs(res1.states[0]), 5 + 1e-6)
np.testing.assert_array_less(np.abs(res1.states[1]), 5 + 1e-6)
np.testing.assert_array_less(np.abs(res1.inputs[0]), 1 + 1e-6)
# Pass an initial guess and rerun
res2 = opt.solve_ocp(sys,
time,
x0,
cost,
constraints,
initial_guess=0.99 * res1.inputs,
basis=flat.BezierFamily(4, Tf),
return_x=True)
assert res2.success
np.testing.assert_allclose(res2.inputs, res1.inputs, atol=0.01, rtol=0.01)
# Run with logging turned on for code coverage
res3 = opt.solve_ocp(sys,
time,
x0,
cost,
constraints,
basis=flat.BezierFamily(4, Tf),
return_x=True,
log=True)
assert res3.success
np.testing.assert_almost_equal(res3.inputs, res1.inputs, decimal=3)
def test_finite_horizon_simple():
# Define a linear system with constraints
# Source: https://www.mpt3.org/UI/RegulationProblem
# LTI prediction model
sys = ct.ss2io(ct.ss([[1, 1], [0, 1]], [[1], [0.5]], np.eye(2), 0, 1))
# State and input constraints
constraints = [
(sp.optimize.LinearConstraint, np.eye(3), [-5, -5, -1], [5, 5, 1]),
]
# Quadratic state and input penalty
Q = [[1, 0], [0, 1]]
R = [[1]]
cost = opt.quadratic_cost(sys, Q, R)
# Set up the optimal control problem
time = np.arange(0, 5, 1)
x0 = [4, 0]
# Retrieve the full open-loop predictions
res = opt.solve_ocp(
sys, time, x0, cost, constraints, squeeze=True)
t, u_openloop = res.time, res.inputs
np.testing.assert_almost_equal(
u_openloop, [-1, -1, 0.1393, 0.3361, -5.204e-16], decimal=4)
def time_steering_terminal_cost():
# Define cost and constraints
traj_cost = opt.quadratic_cost(vehicle, None, np.diag([0.1, 1]), u0=uf)
term_cost = opt.quadratic_cost(vehicle, np.diag([1, 10, 10]), None, x0=xf)
constraints = [opt.input_range_constraint(vehicle, [8, -0.1], [12, 0.1])]
res = opt.solve_ocp(
vehicle,
horizon,
x0,
traj_cost,
constraints,
terminal_cost=term_cost,
initial_guess=bend_left,
print_summary=False,
solve_ivp_kwargs={
'atol': 1e-4,
'rtol': 1e-2
},
# minimize_method='SLSQP', minimize_options={'eps': 0.01}
minimize_method='trust-constr',
minimize_options={'finite_diff_rel_step': 0.01},
)
# Only count this as a benchmark if we converged
assert res.success
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_discrete_lqr():
# oscillator model defined in 2D
# Source: https://www.mpt3.org/UI/RegulationProblem
A = [[0.5403, -0.8415], [0.8415, 0.5403]]
B = [[-0.4597], [0.8415]]
C = [[1, 0]]
D = [[0]]
# Linear discrete-time model with sample time 1
sys = ct.ss2io(ct.ss(A, B, C, D, 1))
# Include weights on states/inputs
Q = np.eye(2)
R = 1
K, S, E = ct.dlqr(A, B, Q, R)
# Compute the integral and terminal cost
integral_cost = opt.quadratic_cost(sys, Q, R)
terminal_cost = opt.quadratic_cost(sys, S, None)
# Solve the LQR problem
lqr_sys = ct.ss2io(ct.ss(A - B @ K, B, C, D, 1))
# Generate a simulation of the LQR controller
time = np.arange(0, 5, 1)
x0 = np.array([1, 1])
_, _, lqr_x = ct.input_output_response(lqr_sys, time, 0, x0, return_x=True)
# Use LQR input as initial guess to avoid convergence/precision issues
lqr_u = np.array(-K @ lqr_x[0:time.size]) # convert from matrix
# Formulate the optimal control problem and compute optimal trajectory
optctrl = opt.OptimalControlProblem(sys,
time,
integral_cost,
terminal_cost=terminal_cost,
initial_guess=lqr_u)
res1 = optctrl.compute_trajectory(x0, return_states=True)
# Compare to make sure results are the same
np.testing.assert_almost_equal(res1.inputs, lqr_u[0])
np.testing.assert_almost_equal(res1.states, lqr_x)
# Add state and input constraints
trajectory_constraints = [
(sp.optimize.LinearConstraint, np.eye(3), [-5, -5, -.5], [5, 5, 0.5]),
]
# Re-solve
res2 = opt.solve_ocp(sys,
time,
x0,
integral_cost,
trajectory_constraints,
terminal_cost=terminal_cost,
initial_guess=lqr_u)
# Make sure we got a different solution
assert np.any(np.abs(res1.inputs - res2.inputs) > 0.1)
def test_discrete_lqr():
# oscillator model defined in 2D
# Source: https://www.mpt3.org/UI/RegulationProblem
A = [[0.5403, -0.8415], [0.8415, 0.5403]]
B = [[-0.4597], [0.8415]]
C = [[1, 0]]
D = [[0]]
# Linear discrete-time model with sample time 1
sys = ct.ss2io(ct.ss(A, B, C, D, 1))
# Include weights on states/inputs
Q = np.eye(2)
R = 1
K, S, E = ct.lqr(A, B, Q, R) # note: *continuous* time LQR
# Compute the integral and terminal cost
integral_cost = opt.quadratic_cost(sys, Q, R)
terminal_cost = opt.quadratic_cost(sys, S, None)
# Formulate finite horizon MPC problem
time = np.arange(0, 5, 1)
x0 = np.array([1, 1])
optctrl = opt.OptimalControlProblem(sys,
time,
integral_cost,
terminal_cost=terminal_cost)
res1 = optctrl.compute_trajectory(x0, return_states=True)
with pytest.xfail("discrete LQR not implemented"):
# Result should match LQR
K, S, E = ct.dlqr(A, B, Q, R)
lqr_sys = ct.ss2io(ct.ss(A - B @ K, B, C, D, 1))
_, _, lqr_x = ct.input_output_response(lqr_sys,
time,
0,
x0,
return_x=True)
np.testing.assert_almost_equal(res1.states, lqr_x)
# Add state and input constraints
trajectory_constraints = [
(sp.optimize.LinearConstraint, np.eye(3), [-10, -10, -1], [10, 10, 1]),
]
# Re-solve
res2 = opt.solve_ocp(sys,
time,
x0,
integral_cost,
constraints,
terminal_cost=terminal_cost)
# Make sure we got a different solution
assert np.any(np.abs(res1.inputs - res2.inputs) > 0.1)
def time_steering_integrated_cost():
# Set up the cost functions
Q = np.diag([.1, 10, .1]) # keep lateral error low
R = np.diag([.1, 1]) # minimize applied inputs
quad_cost = opt.quadratic_cost(vehicle, Q, R, x0=xf, u0=uf)
res = opt.solve_ocp(
vehicle,
horizon,
x0,
quad_cost,
initial_guess=bend_left,
print_summary=False,
# solve_ivp_kwargs={'atol': 1e-2, 'rtol': 1e-2},
minimize_method='trust-constr',
minimize_options={'finite_diff_rel_step': 0.01},
)
# Only count this as a benchmark if we converged
assert res.success
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
def test_optimal_doc():
"""Test optimal control problem from documentation"""
def vehicle_update(t, x, u, params):
# Get the parameters for the model
l = params.get('wheelbase', 3.) # vehicle wheelbase
phimax = params.get('maxsteer', 0.5) # max steering angle (rad)
# Saturate the steering input
phi = np.clip(u[1], -phimax, phimax)
# Return the derivative of the state
return np.array([
np.cos(x[2]) * u[0], # xdot = cos(theta) v
np.sin(x[2]) * u[0], # ydot = sin(theta) v
(u[0] / l) * np.tan(phi) # thdot = v/l tan(phi)
])
def vehicle_output(t, x, u, params):
return x # return x, y, theta (full state)
# Define the vehicle steering dynamics as an input/output system
vehicle = ct.NonlinearIOSystem(vehicle_update,
vehicle_output,
states=3,
name='vehicle',
inputs=('v', 'phi'),
outputs=('x', 'y', 'theta'))
# Define the initial and final points and time interval
x0 = [0., -2., 0.]
u0 = [10., 0.]
xf = [100., 2., 0.]
uf = [10., 0.]
Tf = 10
# Define the cost functions
Q = np.diag([0, 0, 0.1]) # don't turn too sharply
R = np.diag([1, 1]) # keep inputs small
P = np.diag([1000, 1000, 1000]) # get close to final point
traj_cost = opt.quadratic_cost(vehicle, Q, R, x0=xf, u0=uf)
term_cost = opt.quadratic_cost(vehicle, P, 0, x0=xf)
# Define the constraints
constraints = [opt.input_range_constraint(vehicle, [8, -0.1], [12, 0.1])]
# Solve the optimal control problem
horizon = np.linspace(0, Tf, 3, endpoint=True)
result = opt.solve_ocp(vehicle,
horizon,
x0,
traj_cost,
constraints,
terminal_cost=term_cost,
initial_guess=u0)
# Make sure the resulting trajectory generate a good solution
resp = ct.input_output_response(vehicle,
horizon,
result.inputs,
x0,
t_eval=np.linspace(0, Tf, 10))
t, y = resp
assert (y[0, -1] - xf[0]) / xf[0] < 0.01
assert (y[1, -1] - xf[1]) / xf[1] < 0.01
assert y[2, -1] < 0.1
# Provide an intial guess (will be extended to entire horizon)
bend_left = [10, 0.01] # slight left veer
# Turn on debug level logging so that we can see what the optimizer is doing
logging.basicConfig(level=logging.DEBUG,
filename="steering-integral_cost.log",
filemode='w',
force=True)
# Compute the optimal control, setting step size for gradient calculation (eps)
start_time = time.process_time()
result1 = opt.solve_ocp(
vehicle,
horizon,
x0,
quad_cost,
initial_guess=bend_left,
log=True,
minimize_method='trust-constr',
minimize_options={'finite_diff_rel_step': 0.01},
)
print("* Total time = %5g seconds\n" % (time.process_time() - start_time))
# If we are running CI tests, make sure we succeeded
if 'PYCONTROL_TEST_EXAMPLES' in os.environ:
assert result1.success
# Extract and plot the results (+ state trajectory)
t1, u1 = result1.time, result1.inputs
t1, y1 = ct.input_output_response(vehicle, horizon, u1, x0)
plot_results(t1, y1, u1, figure=1, yf=xf[0:2])
