A linear program is an optimization problem with a linear objective and affine inequality constraints. A common standard form is the following: $$ \begin{array}{ll} \mbox{minimize} & c^Tx \\ \mbox{subject to} & Ax \leq b. \end{array} $$ Here $A \in \mathcal{R}^{m \times n}$, $b \in \mathcal{R}^m$, and $c \in \mathcal{R}^n$ are problem data and $x \in \mathcal{R}^{n}$ is the optimization variable. The inequality constraint $Ax \leq b$ is elementwise.

For example, we might have $n$ different products, each constructed out of $m$ components. Each entry $A_{ij}$ is the amount of component $i$ required to build one unit of product $j$. Each entry $b_i$ is the total amount of component $i$ available. We lose $c_j$ for each unit of product $j$ ($c_j < 0$ indicates profit). Our goal then is to choose how many units of each product $j$ to make, $x_j$, in order to minimize loss without exceeding our budget for any component.

In addition to a solution $x^\star$, we obtain a dual solution $\lambda^\star$. A positive entry $\lambda^\star_i$ indicates that the constraint $a_i^Tx \leq b_i$ holds with equality for $x^\star$ and suggests that changing $b_i$ would change the optimal value.

In the following code, we solve a linear program with CVXPY.

In [36]:

```
# Import packages.
import cvxpy as cp
import numpy as np
# Generate a random non-trivial linear program.
m = 15
n = 10
np.random.seed(1)
s0 = np.random.randn(m)
lamb0 = np.maximum(-s0, 0)
s0 = np.maximum(s0, 0)
x0 = np.random.randn(n)
A = np.random.randn(m, n)
b = A@x0 + s0
c = -A.T@lamb0
# Define and solve the CVXPY problem.
x = cp.Variable(n)
prob = cp.Problem(cp.Minimize(c.T@x),
[A@x <= b])
prob.solve()
# Print result.
print("\nThe optimal value is", prob.value)
print("A solution x is")
print(x.value)
print("A dual solution is")
print(prob.constraints[0].dual_value)
```