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_constraint_specification(constraint_list):
sys = ct.ss2io(ct.ss([[1, 1], [0, 1]], [[1], [0.5]], np.eye(2), 0, 1))
"""Test out different forms of constraints on a simple problem"""
# Parse out the constraint
constraints = []
for constraint_setup in constraint_list:
if constraint_setup[0] in \
(sp.optimize.LinearConstraint, sp.optimize.NonlinearConstraint):
# No processing required
constraints.append(constraint_setup)
else:
# Call the function in the first argument to set up the constraint
constraints.append(constraint_setup[0](sys, *constraint_setup[1:]))
# Quadratic state and input penalty
Q = [[1, 0], [0, 1]]
R = [[1]]
cost = opt.quadratic_cost(sys, Q, R)
# Create a model predictive controller system
time = np.arange(0, 5, 1)
optctrl = opt.OptimalControlProblem(sys, time, cost, constraints)
# Compute optimal control and compare against MPT3 solution
x0 = [4, 0]
res = optctrl.compute_trajectory(x0, 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=3)
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 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