In a least-squares, or linear regression, problem, we have measurements \(A \in \mathcal{R}^{m \times n}\) and \(b \in \mathcal{R}^m\) and seek a vector \(x \in \mathcal{R}^{n}\) such that \(Ax\) is close to \(b\). Closeness is defined as the sum of the squared differences:

\[\sum_{i=1}^m (a_i^Tx - b_i)^2,\]

also known as the \(\ell_2\)-norm squared, \(\|Ax - b\|_2^2\).

For example, we might have a dataset of \(m\) users, each represented by \(n\) features. Each row \(a_i^T\) of \(A\) is the features for user \(i\), while the corresponding entry \(b_i\) of \(b\) is the measurement we want to predict from \(a_i^T\), such as ad spending. The prediction is given by \(a_i^Tx\).

We find the optimal \(x\) by solving the optimization problem

\[\begin{array}{ll} \mbox{minimize} & \|Ax - b\|_2^2. \end{array}\]

Let \(x^\star\) denote the optimal \(x\). The quantity \(r = Ax^\star - b\) is known as the residual. If \(\|r\|_2 = 0\), we have a perfect fit.


In the following code, we solve a least-squares problem with CVXPY.

# Import packages.
import cvxpy as cp
import numpy as np

# Generate data.
m = 20
n = 15
A = np.random.randn(m, n)
b = np.random.randn(m)

# Define and solve the CVXPY problem.
x = cp.Variable(n)
cost = cp.sum_squares(A @ x - b)
prob = cp.Problem(cp.Minimize(cost))

# Print result.
print("\nThe optimal value is", prob.value)
print("The optimal x is")
print("The norm of the residual is ", cp.norm(A @ x - b, p=2).value)
The optimal value is 7.005909828287484
The optimal x is
[ 0.17492418 -0.38102551  0.34732251  0.0173098  -0.0845784  -0.08134019
  0.293119    0.27019762  0.17493179 -0.23953449  0.64097935 -0.41633637
  0.12799688  0.1063942  -0.32158411]
The norm of the residual is  2.6468679280023557