Quadratic program with box constraints#

This tutorial shows how we can use some of PyProximal solvers to solve a quadratic function with a box constraint:

\[\mathbf{x} = \argmin_\mathbf{x} \frac{1}{2} \mathbf{x}^T \mathbf{A} \mathbf{x} + \mathbf{b}^T \mathbf{x} \quad s.t. \quad \mathbf{x} \in \mathcal{I}_{\operatorname{Box}}\]

More specifically we will consider both the pyproximal.optimization.primal.ProximalGradient algorithm with and without back-tracking.

In the literature you may find that problem of this kind can be solved by the so-called Projected Gradient Descent (PGD) algorithm: this is a edge case of a Proximal gradient solver when used with a constraint that admits a proximal (instead of a soft regularizer).

import numpy as np
import matplotlib.pyplot as plt
import pylops

import pyproximal


Let’s start defining the terms of the quadratic functional

m = np.array([1, 0])
G = np.array([[10., 9.],
              [9., 10.]])
d = np.dot(G, m)

We can now compute the functional within a grid, which we will show together with the evolution of the solution from the proximal gradient algorithm cost function grid

nm1, nm2 = 201, 201
m_min, m_max = (m[0] - 5, m[1] - 5), (m[0] + 5, m[1] + 5)
m1, m2 = np.linspace(m_min[0], m_max[0], nm1), \
         np.linspace(m_min[1], m_max[1], nm2)
m1, m2 = np.meshgrid(m1, m2, indexing='ij')
mgrid = np.vstack((m1.ravel(), m2.ravel()))
J = 0.5 * np.sum(mgrid * np.dot(G, mgrid), axis=0) - np.dot(d, mgrid)
J = J.reshape(nm1, nm2)

We can now define the upper and lower bounds of the box and again we create a grid to display alongside the solution

lower = 1.5
upper = 3
indic = (mgrid > lower) & (mgrid < upper)
indic = indic[0].reshape(nm1, nm2) & indic[1].reshape(nm1, nm2)

We can now define both the quadratic functional and the box

We are now ready to solve our problem. All we need to do is to choose an initial guess for the proximal gradient algorithm

def callback(x):

m0 = np.array([4, 3])

mhist = [m0,]
minv_slow = pyproximal.optimization.primal.ProximalGradient(l2, ind,
                                                            x0=m0, epsg=1.,
mhist_slow = np.array(mhist)

Provided we can estimate the spectral radius (i.e., max eigenvalue) of our operator G, we can choose an optimal step and improve our convergence speed.

L = np.max(np.linalg.eig(G)[0]) # max eigenvalue of G
tau_opt = 1. / L
mhist = [m0,]
minv_opt = pyproximal.optimization.primal.ProximalGradient(l2, ind,
                                                           tau=tau_opt, x0=m0,
                                                           epsg=1., niter=10,
mhist_opt = np.array(mhist)

Alternatively we can use back-tracking to adaptively find the best step at each iteration

mhist = [m0,]
minv_back = pyproximal.optimization.primal.ProximalGradient(l2, ind, tau=None,
                                                            x0=m0, epsg=1.,
mhist_back = np.array(mhist)

Finally let’s visualize the different trajectories and final solutions

fig, ax = plt.subplots(1, 1, figsize=(16, 9))
cs = ax.contour(m1, m2, J, levels=40, colors='k')
cs = ax.contour(m1, m2, indic, colors='k')
ax.clabel(cs, inline=1, fontsize=10)
ax.plot(m[0], m[1], '.k', ms=20)
ax.plot(m0[0], m0[1], '.r', ms=20)
ax.plot(mhist_slow[:, 0], mhist_slow[:, 1], '.-b', ms=30, lw=2)
ax.plot(mhist_opt[:, 0], mhist_opt[:, 1], '.-m', ms=30, lw=4)
ax.plot(mhist_back[:, 0], mhist_back[:, 1], '.-g', ms=10, lw=2)

Total running time of the script: (0 minutes 0.363 seconds)

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