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Linear program

doc/source/examples/basic/linear_program.rst

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Linear program

A linear program is an optimization problem with a linear objective and affine inequality constraints. A common standard form is the following:

.. math::

   \begin{array}{ll}
   \mbox{minimize}   & c^Tx \\
   \mbox{subject to} & Ax \leq b.
   \end{array}

Here :math:A \in \mathcal{R}^{m \times n}, :math:b \in \mathcal{R}^m, and :math:c \in \mathcal{R}^n are problem data and :math:x \in \mathcal{R}^{n} is the optimization variable. The inequality constraint :math:Ax \leq b is elementwise.

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

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

Example

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

.. code:: python

# 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)

.. parsed-literal::

The optimal value is -15.220912604467838
A solution x is
[-1.10131657 -0.16370661 -0.89711643  0.03228613  0.60662428 -1.12655967
  1.12985839  0.88200333  0.49089264  0.89851057]
A dual solution is
[0.         0.61175641 0.52817175 1.07296862 0.         2.3015387
 0.         0.7612069  0.         0.24937038 0.         2.06014071
 0.3224172  0.38405435 0.        ]