# Derivatives fundamentals¶

This notebook will introduce you to the fundamentals of computing the derivative of the solution map to optimization problems. The derivative can be used for sensitivity analysis, to see how a solution would change given small changes to the parameters, and to compute gradients of scalar-valued functions of the solution.

In this notebook, we will consider a simple disciplined geometric program. The geometric program under consideration is

$\begin{split}$$\begin{array}{ll} \mbox{minimize} & 1/(xyz) \\ \mbox{subject to} & a(xy + xz + yz) \leq b\\ & x \geq y^c, \end{array}$$\end{split}$

where $$x \in \mathbf{R}_{++}$$, $$y \in \mathbf{R}_{++}$$, and $$z \in \mathbf{R}_{++}$$ are the variables, and $$a \in \mathbf{R}_{++}$$, $$b \in \mathbf{R}_{++}$$ and $$c \in \mathbf{R}$$ are the parameters. The vector

$\begin{split}\alpha = \begin{bmatrix} a \\ b \\ c \end{bmatrix}\end{split}$

is the vector of parameters.

import cvxpy as cp

x = cp.Variable(pos=True)
y = cp.Variable(pos=True)
z = cp.Variable(pos=True)

a = cp.Parameter(pos=True)
b = cp.Parameter(pos=True)
c = cp.Parameter()

objective_fn = 1/(x*y*z)
objective = cp.Minimize(objective_fn)
constraints = [a*(x*y + x*z + y*z) <= b, x >= y**c]
problem = cp.Problem(objective, constraints)

problem.is_dgp(dpp=True)

True


Notice the keyword argument dpp=True. The parameters must enter in the DGP problem acording to special rules, which we refer to as dpp. The DPP rules are described in an online tutorial.

Next, we solve the problem, setting the parameters $$a$$, $$b$$ and $$c$$ to $$2$$, $$1$$, and $$0.5$$.

a.value = 2.0
b.value = 1.0
c.value = 0.5

print(x.value)
print(y.value)
print(z.value)

0.5612147353889386
0.31496200373359456
0.36892055859991446


Notice the keyword argument requires_grad=True; this is necessary to subsequently compute derivatives.

## Solution map¶

The solution map of the above problem is a function

$\mathcal{S} : \mathbf{R}^2_{++} \times \mathbf{R} \to \mathbf{R}^3_{++}$

which maps the parameter vector to the vector of optimal solutions

$\begin{split}\mathcal S(\alpha) = \begin{bmatrix} x(\alpha) \\ y(\alpha) \\ z(\alpha)\end{bmatrix}.\end{split}$

Here, $$x(\alpha)$$, $$y(\alpha)$$, and $$z(\alpha)$$ are the optimal values of the variables corresponding to the parameter vector.

As an example, we just saw that

$\begin{split}\mathcal S((2.0, 1.0, 0.5)) = \begin{bmatrix} 0.5612 \\ 0.3150 \\ 0.3690 \end{bmatrix}.\end{split}$

## Sensitivity analysis¶

When the solution map is differentiable, we can use its derivative

$\mathsf{D}\mathcal{S}(\alpha) \in \mathbf{R}^{3 \times 3}$

to perform a sensitivity analysis, which studies how the solution would change given small changes to the parameters.

Suppose we perturb the parameters by a vector of small magnitude $$\mathsf{d}\alpha \in \mathbf{R}^3$$. We can approximate the change $$\Delta$$ in the solution due to the perturbation using the derivative, as

$\Delta = \mathcal{S}(\alpha + \mathsf{d}\alpha) - \mathcal{S}(\alpha) \approx \mathsf{D}\mathcal{S}(\alpha) \mathsf{d}\alpha.$

We can compute this in CVXPY, as follows.

Partition the perturbation as

$\begin{split}\mathsf{d}\alpha = \begin{bmatrix} \mathsf{d}a \\ \mathsf{d}b \\ \mathsf{d}c\end{bmatrix}.\end{split}$

We set the delta attributes of the parameters to their perturbations, and then call the derivative method.

da, db, dc = 1e-2, 1e-2, 1e-2

a.delta = da
b.delta = db
c.delta = dc
problem.derivative()


The derivative method populates the delta attributes of the variables as a side-effect, with the predicted change in the variable. We can compare the predictions to the actual solution of the perturbed problem.

x_hat = x.value + x.delta
y_hat = y.value + y.delta
z_hat = z.value + z.delta

a.value += da
b.value += db
c.value += dc

problem.solve(gp=True)
print('x: predicted {0:.5f} actual {1:.5f}'.format(x_hat, x.value))
print('y: predicted {0:.5f} actual {1:.5f}'.format(y_hat, y.value))
print('z: predicted {0:.5f} actual {1:.5f}'.format(z_hat, z.value))

a.value -= da
b.value -= db
c.value -= dc

x: predicted 0.55729 actual 0.55734
y: predicted 0.31783 actual 0.31783
z: predicted 0.37179 actual 0.37175


In this case, the predictions and the actual solutions are fairly close.

We can compute gradient of a scalar-valued function of the solution with respect to the parameters. Let $$f : \mathbf{R}^{3} \to \mathbf{R}$$, and suppose we want to compute the gradient of the composition $$f \circ \mathcal S$$. By the chain rule,

$\begin{split}\nabla f(S(\alpha)) = \mathsf{D}^T\mathcal{S}(\alpha) \begin{bmatrix}\mathsf{d}x \\ \mathsf{d}y \\ \mathsf{d}z\end{bmatrix},\end{split}$

where $$\mathsf{D}^T\mathcal{S}$$ is the adjoint (or transpose) of the derivative operator, and $$\mathsf{d}x$$, $$\mathsf{d}y$$, and $$\mathsf{d}z$$ are the partial derivatives of $$f$$ with respect to its arguments.

We can compute the gradient in CVXPY, using the backward method. As an example, suppose

$f(x, y, z) = \frac{1}{2}(x^2 + y^2 + z^2),$

so that $$\mathsf{d}x = x$$, $$\mathsf{d}y = y$$, and $$\mathsf{d}z = z$$. Let $$\mathsf{d}\alpha = \nabla f(S(\alpha))$$, and suppose we subtract $$\eta \mathsf{d}\alpha$$ from the parameter, where $$\eta$$ is a positive constant. Using the following code, we can compare $$f(\mathcal S(\alpha - \eta \mathsf{d}\alpha))$$ with the value predicted by the gradient,

$f(\mathcal S(\alpha - \eta \mathsf{d}\alpha)) \approx f(\mathcal S(\alpha)) - \eta \mathsf{d}\alpha^T\mathsf{d}\alpha.$
problem.solve(gp=True, requires_grad=True)

def f(x, y, z):
return 1/2*(x**2 + y**2 + z**2)

original = f(x, y, z).value

problem.backward()

eta = 0.5

original 0.27513 predicted 0.22709 actual 0.22942