Usage¶
SciPy Compatible Interface¶
For simple cases where you do not need the full power of sparse and structured
Jacobians etc, cyipopt provides the function minimize_ipopt which has
the same behaviour as scipy.optimize.minimize, for example:
>>> from scipy.optimize import rosen, rosen_der
>>> from cyipopt import minimize_ipopt
>>> x0 = [1.3, 0.7, 0.8, 1.9, 1.2]
>>> res = minimize_ipopt(rosen, x0, jac=rosen_der)
>>> print(res)
fun: 2.1256746564022273e-18
info: {'x': array([1., 1., 1., 1., 1.]), 'g': array([], dtype=float64), 'obj_val': 2.1256746564022273e-18, 'mult_g': array([], dtype=float64), 'mult_x_L': array([0., 0., 0., 0., 0.]), 'mult_x_U': array([0., 0., 0., 0., 0.]), 'status': 0, 'status_msg': b'Algorithm terminated successfully at a locally optimal point, satisfying the convergence tolerances (can be specified by options).'}
message: b'Algorithm terminated successfully at a locally optimal point, satisfying the convergence tolerances (can be specified by options).'
nfev: 200
nit: 37
njev: 39
status: 0
success: True
x: array([1., 1., 1., 1., 1.])
In order to demonstrate the usage of sparse jacobians, let’s assume we want to minimize the well-known rosenbrock function
subject to some constraints, i.e. we want to solve the constraint optimization problem
We won’t implement the rosenbrock function and its derivatives here, since
all three can be imported from scipy.optimize. The constraint function
\(c\) and the jacobian \(J_c\) are given by
and we can implement the constraint and the sparse jacobian
by means of an scipy.sparse.coo_array like this:
from scipy.sparse import coo_array
def con(x):
return np.array([ 10 -x[1]**2 - x[2], 100.0 - x[4]**2 ])
def con_jac(x):
# Dense Jacobian:
# J = (0 -2*x[1] -1 0 0 )
# (0 0 0 0 -2*x[4] )
# Sparse Jacobian (COO)
rows = np.array([0, 0, 1])
cols = np.array(([1, 2, 4]))
data = np.array([-2*x[1], -1, -2*x[4]])
return coo_array((data, (rows, cols)))
In addition, we would like to pass the hessian of the objective and the constraints. Note that Ipopt expects the hessian \(\nabla^2_x L\) of the lagrangian function
which is given by
Hence, we need to pass the hessian-vector-product of the constraint hessians \(\nabla^2 c_1(x)\) and \(\nabla^2 c_2(x)\) and the lagrangian multipliers \(\lambda\) (also known as dual variables). In code:
def con_hess(x, _lambda):
H1 = np.array([
[0, 0, 0, 0, 0],
[0, -2, 0, 0, 0 ],
[0, 0, 0, 0, 0 ],
[0, 0, 0, 0, 0 ],
[0, 0, 0, 0, 0 ]
])
H2 = np.array([
[0, 0, 0, 0, 0],
[0, 0, 0, 0, 0],
[0, 0, 0, 0, 0],
[0, 0, 0, 0, 0],
[0, 0, 0, 0, -2]
])
return _lambda[0] * H1 + _lambda[1] * H2
Ipopt only uses the lower triangle of the hessian-vector-product under the hood, due to the symmetry of the hessians. Similar to sparse jacobians, it also supports sparse hessians, but this isn’t supported by the scipy interface yet. However, you can use cyipopt’s problem interface in case you need to pass sparse hessians.
Finally, after defining the constraint and the initial guess, we can solve the problem:
from scipy.optimize import rosen, rosen_der, rosen_hess
constr = {'type': 'ineq', 'fun': con, 'jac': con_jac, 'hess': con_hess}
# initial guess
x0 = np.array([1.1, 1.1, 1.1, 1.1, 1.1])
# solve the problem
res = minimize_ipopt(rosen, jac=rosen_der, hess=rosen_hess, x0=x0, constraints=constr)
Algorithmic Differentation¶
Computing derivatives by hand can be quite error-prone. In case you don’t
provide the (exact) objective gradient or the jacobian of the constraint
function, the scipy interface will approximate the missing derivatives by
finite differences similar to scipy.optimize.minimize. However, finite
differences are prone to truncation errors due to floating point arithmetic and
computationally expensive especially for evaluating jacobians. A more efficient
and accurate way to evaluate derivatives is algorithmic differentation (AD).
In this example we use AD by means of the JAX library to compute derivatives and we use cyipopt’s scipy interface to solve an example problem, namely number 71 from the Hock-Schittkowsky test suite [1],
with the starting point,
and the optimal solution,
We start by importing all required libraries:
from jax.config import config
# Enable 64 bit floating point precision
config.update("jax_enable_x64", True)
# We use the CPU instead of GPU und mute all warnings if no GPU/TPU is found.
config.update('jax_platform_name', 'cpu')
import jax.numpy as np
from jax import jit, grad, jacfwd, jacrev
from cyipopt import minimize_ipopt
Then we define the objective and constraint functions:
def objective(x):
return x[0]*x[3]*np.sum(x[:3]) + x[2]
def eq_constraints(x):
return np.sum(x**2) - 40
def ineq_constrains(x):
return np.prod(x) - 25
Next, we build the derivatives and just-in-time (jit) compile the functions
(more details regarding jit, grad and jacfwd can be found in the
JAX autodiff cookbook):
# jit the functions
obj_jit = jit(objective)
con_eq_jit = jit(eq_constraints)
con_ineq_jit = jit(ineq_constrains)
# build the derivatives and jit them
obj_grad = jit(grad(obj_jit)) # objective gradient
obj_hess = jit(jacrev(jacfwd(obj_jit))) # objective hessian
con_eq_jac = jit(jacfwd(con_eq_jit)) # jacobian
con_ineq_jac = jit(jacfwd(con_ineq_jit)) # jacobian
con_eq_hess = jacrev(jacfwd(con_eq_jit)) # hessian
con_eq_hessvp = jit(lambda x, v: con_eq_hess(x) * v[0]) # hessian vector-product
con_ineq_hess = jacrev(jacfwd(con_ineq_jit)) # hessian
con_ineq_hessvp = jit(lambda x, v: con_ineq_hess(x) * v[0]) # hessian vector-product
Finally, we can call minimize_ipopt similar to scipy.optimize.minimize:
# constraints
cons = [
{'type': 'eq', 'fun': con_eq_jit, 'jac': con_eq_jac, 'hess': con_eq_hessvp},
{'type': 'ineq', 'fun': con_ineq_jit, 'jac': con_ineq_jac, 'hess': con_ineq_hessvp}
]
# starting point
x0 = np.array([1.0, 5.0, 5.0, 1.0])
# variable bounds: 1 <= x[i] <= 5
bnds = [(1, 5) for _ in range(x0.size)]
# executing the solver
res = minimize_ipopt(obj_jit, jac=obj_grad, hess=obj_hess, x0=x0, bounds=bnds,
constraints=cons, options={'disp': 5})
Problem Interface¶
In this example we will use cyipopt problem class interface to solve the aforementioned test problem.
Getting started¶
Before you can use cyipopt, you have to import it:
import cyipopt
This problem will also make use of NumPy:
import numpy as np
Defining the problem¶
The first step is to define a class that computes the objective and its gradient, the constraints and its Jacobian, and the Hessian. The following methods can be defined on the class:
cyipopt.Problem.objective()cyipopt.Problem.gradient()cyipopt.Problem.constraints()cyipopt.Problem.jacobian()cyipopt.Problem.hessian()
The cyipopt.Problem.jacobian() and cyipopt.Problem.hessian()
methods should return the non-zero values of the respective matrices as
flattened arrays. The hessian should return a flattened lower triangular
matrix.
The Jacobian and Hessian can be dense or sparse. If sparse,
cyipopt.Problem.jacobian() and cyipopt.Problem.hessian() methods
should return only the non-zero values of the respective matrices and you must
also define:
cyipopt.Problem.jacobianstructure()cyipopt.Problem.hessianstructure()
which should return a tuple of indices (row indices, column indices) that indicate the location of the non-zero values of the Jacobian and Hessian matrices, respectively. If not defined then these matrices are assumed to be dense.
The cyipopt.Problem.intermediate() method is called every Ipopt iteration
algorithm and can be used to perform any needed computation at each iteration.
Define the problem class:
class HS071():
def objective(self, x):
"""Returns the scalar value of the objective given x."""
return x[0] * x[3] * np.sum(x[0:3]) + x[2]
def gradient(self, x):
"""Returns the gradient of the objective with respect to x."""
return np.array([
x[0]*x[3] + x[3]*np.sum(x[0:3]),
x[0]*x[3],
x[0]*x[3] + 1.0,
x[0]*np.sum(x[0:3])
])
def constraints(self, x):
"""Returns the constraints."""
return np.array((np.prod(x), np.dot(x, x)))
def jacobian(self, x):
"""Returns the Jacobian of the constraints with respect to x."""
return np.concatenate((np.prod(x)/x, 2*x))
def hessianstructure(self):
"""Returns the row and column indices for non-zero vales of the
Hessian."""
# NOTE: The default hessian structure is of a lower triangular matrix,
# therefore this function is redundant. It is included as an example
# for structure callback.
return np.nonzero(np.tril(np.ones((4, 4))))
def hessian(self, x, lagrange, obj_factor):
"""Returns the non-zero values of the Hessian."""
H = obj_factor*np.array((
(2*x[3], 0, 0, 0),
(x[3], 0, 0, 0),
(x[3], 0, 0, 0),
(2*x[0]+x[1]+x[2], x[0], x[0], 0)))
H += lagrange[0]*np.array((
(0, 0, 0, 0),
(x[2]*x[3], 0, 0, 0),
(x[1]*x[3], x[0]*x[3], 0, 0),
(x[1]*x[2], x[0]*x[2], x[0]*x[1], 0)))
H += lagrange[1]*2*np.eye(4)
row, col = self.hessianstructure()
return H[row, col]
def intermediate(self, alg_mod, iter_count, obj_value, inf_pr, inf_du, mu,
d_norm, regularization_size, alpha_du, alpha_pr,
ls_trials):
"""Prints information at every Ipopt iteration."""
msg = "Objective value at iteration #{:d} is - {:g}"
print(msg.format(iter_count, obj_value))
Now define the lower and upper bounds of \(x\) and the constraints:
lb = [1.0, 1.0, 1.0, 1.0]
ub = [5.0, 5.0, 5.0, 5.0]
cl = [25.0, 40.0]
cu = [2.0e19, 40.0]
Define an initial guess:
x0 = [1.0, 5.0, 5.0, 1.0]
Define the full problem using the cyipopt.Problem class:
nlp = cyipopt.Problem(
n=len(x0),
m=len(cl),
problem_obj=HS071(),
lb=lb,
ub=ub,
cl=cl,
cu=cu,
)
The constructor of the cyipopt.Problem class requires:
n: the number of variables in the problem,m: the number of constraints in the problem,lbandub: lower and upper bounds on the variables,clandcu: lower and upper bounds of the constraints.problem_objis an object whose methods implementobjective,gradient,constraints,jacobian, andhessianof the problem.
Setting optimization parameters¶
Setting optimization parameters is done by calling the
cyipopt.Problem.add_option() method, e.g.:
nlp.add_option('mu_strategy', 'adaptive')
nlp.add_option('tol', 1e-7)
The different options and their possible values are described in the ipopt documentation.
Executing the solver¶
The optimization algorithm is run by calling the cyipopt.Problem.solve()
method, which accepts the starting point for the optimization as its only
parameter:
x, info = nlp.solve(x0)
The method returns the optimal solution and an info dictionary that contains the status of the algorithm, the value of the constraints multipliers at the solution, and more.
Accessing iterate and infeasibility vectors in an intermediate callback¶
When debugging an Ipopt solve that converges slowly or not at all, it can be
very useful to track the primal/dual iterate and infeasibility vectors
to get a sense for the variable and constraint coordinates that are causing
a problem. This can be done with Ipopt’s GetCurrentIterate and
GetCurrentViolations functions, which were added to Ipopt’s C interface in
Ipopt version 3.14.0. These functions are accessed in CyIpopt via the
get_current_iterate and get_current_violations methods of
cyipopt.Problem.
These methods should only be called during an intermediate callback.
To access them, we define our problem as a subclass of cyipopt.Problem
and access the get_current_iterate and get_current_violations methods
on self.
In contrast to the previous example, we now define the HS071 problem as a
subclass of cyipopt.Problem:
import cyipopt
import numpy as np
class HS071(cyipopt.Problem):
def objective(self, x):
"""Returns the scalar value of the objective given x."""
return x[0] * x[3] * np.sum(x[0:3]) + x[2]
def gradient(self, x):
"""Returns the gradient of the objective with respect to x."""
return np.array([
x[0]*x[3] + x[3]*np.sum(x[0:3]),
x[0]*x[3],
x[0]*x[3] + 1.0,
x[0]*np.sum(x[0:3])
])
def constraints(self, x):
"""Returns the constraints."""
return np.array((np.prod(x), np.dot(x, x)))
def jacobian(self, x):
"""Returns the Jacobian of the constraints with respect to x."""
return np.concatenate((np.prod(x)/x, 2*x))
def hessianstructure(self):
"""Returns the row and column indices for non-zero vales of the
Hessian."""
# NOTE: The default hessian structure is of a lower triangular matrix,
# therefore this function is redundant. It is included as an example
# for structure callback.
return np.nonzero(np.tril(np.ones((4, 4))))
def hessian(self, x, lagrange, obj_factor):
"""Returns the non-zero values of the Hessian."""
H = obj_factor*np.array((
(2*x[3], 0, 0, 0),
(x[3], 0, 0, 0),
(x[3], 0, 0, 0),
(2*x[0]+x[1]+x[2], x[0], x[0], 0)))
H += lagrange[0]*np.array((
(0, 0, 0, 0),
(x[2]*x[3], 0, 0, 0),
(x[1]*x[3], x[0]*x[3], 0, 0),
(x[1]*x[2], x[0]*x[2], x[0]*x[1], 0)))
H += lagrange[1]*2*np.eye(4)
row, col = self.hessianstructure()
return H[row, col]
def intermediate(self, alg_mod, iter_count, obj_value, inf_pr, inf_du, mu,
d_norm, regularization_size, alpha_du, alpha_pr,
ls_trials):
"""Prints information at every Ipopt iteration."""
iterate = self.get_current_iterate()
infeas = self.get_current_violations()
primal = iterate["x"]
jac = self.jacobian(primal)
print("Iteration:", iter_count)
print("Primal iterate:", primal)
print("Flattened Jacobian:", jac)
print("Dual infeasibility:", infeas["grad_lag_x"])
Now, in the intermediate method of HS071, we call
self.get_current_iterate and self.get_current_violations.
These are implemented on cyipopt.Problem. These methods return dicts
that contain each component of the Ipopt iterate and infeasibility
vectors. The primal iterate and constraint dual iterate can be accessed
with iterate["x"] and iterate["mult_g"], while the primal and
dual infeasibilities can be accessed with infeas["g_violation"]
and infeas["grad_lag_x"]. A full list of keys present in these
dictionaries can be found in the cyipopt.Problem documentation.
We can now set up and solve the optimization problem.
Note that now we instantiate the HS071 class and provide it the arguments
that are required by cyipopt.Problem.
When we solve, we will see the primal iterate and dual infeasibility vectors
printed every iteration:
lb = [1.0, 1.0, 1.0, 1.0]
ub = [5.0, 5.0, 5.0, 5.0]
cl = [25.0, 40.0]
cu = [2.0e19, 40.0]
x0 = [1.0, 5.0, 5.0, 1.0]
nlp = HS071(
n=len(x0),
m=len(cl),
lb=lb,
ub=ub,
cl=cl,
cu=cu,
)
x, info = nlp.solve(x0)
While here we have implemented a very basic callback, much more sophisticated analysis is possible. For example, we could compute the condition number or rank of the constraint Jacobian to identify when constraint qualifications are close to being violated.
Where to go from here¶
Once you feel sufficiently familiar with the basics, feel free to dig into the
reference. For more examples, check the examples/
subdirectory of the distribution.