.. _dpp:

Disciplined Parametrized Programming
------------------------------------
*Note: DPP requires CVXPY version 1.1.0 or greater.*

:py:class:`Parameters <cvxpy.expressions.constants.parameter.Parameter>` are
symbolic representations of constants. Using parameters lets you modify the
values of constants without reconstructing the entire problem. When your
parametrized problem is constructed according to *Disciplined Parametrized
Programming (DPP)*, solving it repeatedly for different values of the
parameters can be much faster than repeatedly solving a new problem.

You should read this tutorial if you intend to solve a :ref:`DCP <dcp>` or
:ref:`DGP <dgp>` problem many times, for different values of the numerical
data, or if you want to differentiate through the solution map of a DCP or DGP
problem.

What is DPP?
^^^^^^^^^^^^
DPP is a ruleset for producing parametrized DCP or DGP compliant problems that
CVXPY can re-canonicalize very quickly. The first time a DPP-compliant problem
is solved, CVXPY compiles it and caches the mapping from parameters to problem
data. As a result, subsequent rewritings of DPP problems can be substantially
faster. CVXPY allows you to solve parametrized problems that are not DPP, but
you won't see a speed-up when doing so.

The DPP ruleset
^^^^^^^^^^^^^^^

DPP places mild restrictions on how parameters can enter expressions in
DCP and DGP problems. First, we describe the DPP ruleset for DCP problems.
Then, we describe the DPP ruleset for DGP problems.

**DCP problems.**
In DPP, an expression is said to be parameter-affine if it does
not involve variables and is affine in its parameters, and it is parameter-free
if it does not have parameters. DPP introduces two restrictions to DCP:

1. Under DPP, all parameters are classified as affine, just like variables.
2. Under DPP, the product of two expressions is affine when
   at least one of the expressions is constant, or when one of the
   expressions is parameter-affine and the other is parameter-free.

An expression is DPP-compliant if it DCP-compliant subject to these two
restrictions. You can check whether an expression or problem is DPP-compliant
by calling the ``is_dcp`` method with the keyword argument ``dpp=True`` (by
default, this keyword argument is ``False``). For example,

.. code:: python3

    import cvxpy as cp


    m, n = 3, 2
    x = cp.Variable((n, 1))
    F = cp.Parameter((m, n))
    G = cp.Parameter((m, n))
    g = cp.Parameter((m, 1))
    gamma = cp.Parameter(nonneg=True)

    objective = cp.norm((F + G) @ x - g) + gamma * cp.norm(x)
    print(objective.is_dcp(dpp=True))

prints ``True``. We can walk through the DPP analysis to understand why
``objective`` is DPP-compliant. The product ``(F + G) @ x`` is affine under DPP,
because ``F + G`` is parameter-affine and ``x`` is parameter-free. The difference
``(F + G) @ x - g`` is affine because the addition atom is affine and both
``(F + G) @ x`` and  ``- g`` are affine. Likewise ``gamma * cp.norm(x)`` is affine
under DPP because ``gamma`` is parameter-affine and ``cp.norm(x)`` is
parameter-free. The final objective is then affine under DPP because addition is
affine.

Some expressions are DCP-compliant but not DPP-compliant. For example,
DPP forbids taking the product of two parametrized expressions:

.. code:: python3

    import cvxpy as cp


    x = cp.Variable()
    gamma = cp.Parameter(nonneg=True)
    problem = cp.Problem(cp.Minimize(gamma * gamma * x), [x >= 1])
    print("Is DPP? ", problem.is_dcp(dpp=True))
    print("Is DCP? ", problem.is_dcp(dpp=False))

This code snippet prints

::

    Is DPP? False
    Is DCP? True

Just as it is possible to rewrite non-DCP problems in DCP-compliant ways, it is
also possible to re-express non-DPP problems in DPP-compliant ways. For
example, the above problem can be equivalently written as

.. code:: python3

    import cvxpy as cp


    x = cp.Variable()
    y = cp.Variable()
    gamma = cp.Parameter(nonneg=True)
    problem = cp.Problem(cp.Minimize(gamma * y), [y == gamma * x])
    print("Is DPP? ", problem.is_dcp(dpp=True))
    print("Is DCP? ", problem.is_dcp(dpp=False))

This snippet prints

::

    Is DPP? True
    Is DCP? True

In other cases, you can represent non-DPP transformations of parameters
by doing them outside of the DSL, e.g., in NumPy. For example,
if ``P`` is a parameter and ``x`` is a variable, ``cp.quad_form(x, P)`` is not
DPP. You can represent a parametric quadratic form like so:

.. code:: python3

  import cvxpy as cp
  import numpy as np
  import scipy.linalg


  n = 4
  L = np.random.randn(n, n)
  P = L.T @ L
  P_sqrt = cp.Parameter((n, n))
  x = cp.Variable((n, 1))
  quad_form = cp.sum_squares(P_sqrt @ x)
  P_sqrt.value = scipy.linalg.sqrtm(P)
  assert quad_form.is_dcp(dpp=True)

As another example, the quotient ``expr / p`` is not DPP-compliant when ``p`` is
a parameter, but this can be rewritten as ``expr * p_tilde``, where ``p_tilde`` is
a parameter that represents ``1/p``.

**DGP problems.**
Just as DGP is the log-log analogue of DCP, DPP for DGP is the log-log analog
of DPP for DCP. DPP introduces two restrictions to DGP:

1. Under DPP, all positive parameters are classified as log-log-affine, just like positive variables.
2. Under DPP, the power atom ``x**p`` (with base ``x`` and exponent ``p``)
   is log-log affine as long as ``x`` and ``p`` are not both parametrized.

Note that for powers, the exponent ``p`` must be either a numerical constant
or a parameter; attempting to construct a power atom in which the exponent
is a compound expression, e.g., ``x**(p + p)``, where ``p`` is a Parameter,
will result in a ``ValueError``.

If a parameter appears in a DGP problem as an exponent, it can have any
sign. If a parameter appears elsewhere in a DGP problem, *it must be
positive*, i.e., it must be constructed with ``cp.Parameter(pos=True)``.

You can check whether an expression or problem is DPP-compliant
by calling the ``is_dgp`` method with the keyword argument ``dpp=True`` (by
default, this keyword argument is ``False``). For example,

.. code:: python3

    import cvxpy as cp


    x = cp.Variable(pos=True)
    y = cp.Variable(pos=True)
    a = cp.Parameter()
    b = cp.Parameter()
    c = cp.Parameter(pos=True)

    monomial = c * x**a * y**b
    print(monomial.is_dgp(dpp=True))

prints ``True``. The expressions ``x**a`` and ``y**b`` are log-log affine, since
``x`` and ``y`` do not contain parameters. The parameter ``c`` is log-log affine
because it is positive, and the monomial expression is log-log affine because
the product of log-log affine expression is also log-log affine.

Some expressions are DGP-compliant but not DPP-compliant. For example,
DPP forbids taking raising a parametrized expression to a power:

.. code:: python3

    import cvxpy as cp


    x = cp.Variable(pos=True)
    a = cp.Parameter()

    monomial = (x**a)**a
    print("Is DPP? ", monomial.is_dgp(dpp=True))
    print("Is DGP? ", monomial.is_dgp(dpp=False))

This code snippet prints

::

    Is DPP? False
    Is DGP? True

You can represent non-DPP transformations of parameters
by doing them outside of CVXPY, e.g., in NumPy. For example,
you could rewrite the above program as the following DPP-complaint program

.. code:: python3

    import cvxpy as cp


    a = 2.0
    x = cp.Variable(pos=True)
    b = cp.Parameter(value=a**2)

    monomial = x**b

Repeatedly solving a DPP problem
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
The following example demonstrates how parameters can speed-up repeated
solves of a DPP-compliant DCP problem.

.. code:: python3

    import cvxpy as cp
    import numpy
    import matplotlib.pyplot as plt
    import time

    n = 15
    m = 10
    numpy.random.seed(1)
    A = numpy.random.randn(n, m)
    b = numpy.random.randn(n)
    # gamma must be nonnegative due to DCP rules.
    gamma = cp.Parameter(nonneg=True)

    x = cp.Variable(m)
    error = cp.sum_squares(A @ x - b)
    obj = cp.Minimize(error + gamma*cp.norm(x, 1))
    problem = cp.Problem(obj)
    assert problem.is_dcp(dpp=True)

    gamma_vals = numpy.logspace(-4, 1)
    times = []
    new_problem_times = []
    for val in gamma_vals:
        gamma.value = val
        start = time.time()
        problem.solve()
        end = time.time()
        times.append(end - start)
        new_problem = cp.Problem(obj)
        start = time.time()
        new_problem.solve()
        end = time.time()
        new_problem_times.append(end - start)

    plt.rc('text', usetex=True)
    plt.rc('font', family='serif')
    plt.figure(figsize=(6, 6))
    plt.plot(gamma_vals, times, label='Re-solving a DPP problem')
    plt.plot(gamma_vals, new_problem_times, label='Solving a new problem')
    plt.xlabel(r'$\gamma$', fontsize=16)
    plt.ylabel(r'time (s)', fontsize=16)
    plt.legend()

.. image:: re_solving_dpp.png

Similar speed-ups can be obtained for DGP problems.

.. _derivatives:


Sensitivity analysis and gradients
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.. versionadded:: 1.1.0 This feature requires CVXPY version 1.1.0 or greater

An optimization problem can be viewed as a function mapping parameters
to solutions. This solution map is sometimes differentiable. CVXPY
has built-in support for computing the derivative of the optimal variable
values of a problem with respect to small perturbations of the parameters
(i.e., the ``Parameter`` instances appearing in a problem).

The problem class exposes two methods related to computing the derivative.
The :py:func:`derivative <cvxpy.problems.problem.Problem.derivative>` evaluates
the derivative given perturbations to the parameters. This
lets you calculate how the solution to a problem would change
given small changes to the parameters, without re-solving the problem.
The :py:func:`backward <cvxpy.problems.problem.Problem.backward>` method
evaluates the adjoint of the derivative, computing the gradient of the solution
with respect to the parameters. This can be useful when combined with
automatic differentiation software.

The derivative and backward methods are only meaningful when the problem
contains parameters. In order for a problem to be differentiable, it must
be :ref:`DPP-compliant <dpp>`. CVXPY can compute the derivative of any
DPP-compliant DCP or DGP problem. At non-differentiable points, CVXPY
computes a heuristic quantity.

**Example.**

As a first example, we solve a trivial problem with an analytical solution,
to illustrate the usage of the ``backward`` and ``derivative``
functions. In the following block of code, we construct a problem with
a scalar variable ``x`` and a scalar parameter ``p``. The problem
is to minimize the quadratic ``(x - 2*p)**2``.

.. code:: python3

    import cvxpy as cp

    x = cp.Variable()
    p = cp.Parameter()
    quadratic = cp.square(x - 2 * p)
    problem = cp.Problem(cp.Minimize(quadratic))

Next, we solve the problem for the particular value of ``p == 3``. Notice that
when solving the problem, we supply the keyword argument ``requires_grad=True``
to the ``solve`` method.

.. code:: python3

    p.value = 3.
    problem.solve(requires_grad=True)

Having solved the problem with ``requires_grad=True``, we can now use the
``backward`` and ``derivative`` to differentiate through the problem.
First, we compute the gradient of the solution with respect to its parameter
by calling the ``backward()`` method. As a side-effect, the ``backward()``
method populates the ``gradient`` attribute on all parameters with the gradient
of the solution with respect to that parameter.

.. code:: python3

    problem.backward()
    print("The gradient is {0:0.1f}.".format(p.gradient))

In this case, the problem has the trivial analytical solution ``2*p``, and
the gradient is therefore just 2. So, as expected, the above code prints

.. code::

    The gradient is 2.0.

Next, we use the ``derivative`` method to see how a small change in ``p``
would affect the solution ``x``. We will perturb ``p`` by ``1e-5``, by
setting ``p.delta = 1e-5``, and calling the ``derivative`` method will populate
the ``delta`` attribute of ``x`` with the change in ``x`` predicted by
a first-order approximation (which is ``dx/dp * p.delta``).

.. code:: python3

    p.delta = 1e-5
    problem.derivative()
    print("x.delta is {0:2.1g}.".format(x.delta))

In this case the solution is trivial and its derivative is just ``2*p``, so we
know that the delta in ``x`` should be ``2e-5``. As expected, the output is

.. code::

    x.delta is 2e-05.

We emphasize that this example is trivial, because it has a trivial analytical
solution, with a trivial derivative. The ``backward()`` and ``forward()``
methods are useful because the vast majority of convex optimization problems
do not have analytical solutions: in these cases, CVXPY can compute solutions
and their derivatives, even though it would be impossible to derive them by
hand.

**Note.** In this simple example, the variable ``x`` was a scalar, so the
``backward`` method computed the gradient of ``x`` with respect to ``p``.
When there is more than one scalar variable, by default, ``backward``
computes the gradient of the *sum* of the optimal variable values with respect
to the parameters.

More generally, the ``backward`` method can be used to compute the gradient of
a scalar-valued function ``f`` of the optimal variables, with
respect to the parameters. If ``x(p)`` denotes the optimal value of
the variable (which might be a vector or a matrix) for a particular value of
the parameter ``p`` and ``f(x(p))`` is a scalar, then ``backward`` can be used
to compute the gradient of ``f`` with respect to ``p``. Let ``x* = x(p)``,
and say the derivative of ``f`` with respect to ``x*`` is ``dx``. To compute
the derivative of ``f`` with respect to ``p``, before calling
``problem.backward()``, just set ``x.gradient = dx``.

The ``backward`` method can be powerful when combined with software for
automatic differentiation. We recommend the software package
`CVXPY Layers <https://www.github.com/cvxgrp/cvxpylayers>`_, which provides
differentiable PyTorch and TensorFlow wrappers for CVXPY problems.

**backward or derivative?** The ``backward`` method should be used when
you need the gradient of (a scalar-valued function) of the solution, with
respect to the parameters. If you only want to do a sensitivity analysis,
that is, if all you're interested in is how the solution would change if
one or more parameters were changed, you should use the ``derivative``
method. When there are multiple variables, it is much more efficient to
compute sensitivities using the derivative method than it would be to compute
the entire Jacobian (which can be done by calling backward multiple times,
once for each standard basis vector).

**Next steps.** See the `introductory notebook <https://www.cvxpy.org/examples/derivatives/fundamentals.html>`_
on derivatives.
