We remind ourselves about the general problem we want to solve $$ \begin{align*} &\mathrm{min}_{x}\hspace{0.2cm} \frac{1}{2}\boldsymbol{x}^T\boldsymbol{P}\boldsymbol{x}+\boldsymbol{q}^T\boldsymbol{x},\\ \nonumber &\mathrm{subject\hspace{0.1cm} to} \hspace{0.2cm} \boldsymbol{G}\boldsymbol{x} \preceq \boldsymbol{h} \wedge \boldsymbol{A}\boldsymbol{x}=f. \end{align*} $$
Let us show how to perform the optmization using a simple case. Assume we want to optimize the following problem $$ \begin{align*} &\mathrm{min}_{x}\hspace{0.2cm} \frac{1}{2}x^2+5x+3y \\ \nonumber &\mathrm{subject to} \\ \nonumber &x, y \geq 0 \\ \nonumber &x+3y \geq 15 \\ \nonumber &2x+5y \leq 100 \\ \nonumber &3x+4y \leq 80. \\ \nonumber \end{align*} $$ The minimization problem can be rewritten in terms of vectors and matrices as (with \( x \) and \( y \) being the unknowns) $$ \frac{1}{2}\begin{bmatrix} x\\ y \end{bmatrix}^T \begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix}3\\ 4 \end{bmatrix}^T \begin{bmatrix}x \\ y \end{bmatrix}. $$ Similarly, we can now set up the inequalities (we need to change \( \geq \) to \( \leq \) by multiplying with \( -1 \) on bot sides) as the following matrix-vector equation $$ \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ -1 & -3 \\ 2 & 5 \\ 3 & 4\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix} \preceq \begin{bmatrix}0 \\ 0\\ -15 \\ 100 \\ 80\end{bmatrix}. $$ We have collapsed all the inequalities into a single matrix \( \boldsymbol{G} \). We see also that our matrix $$ \boldsymbol{P} =\begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix} $$ is clearly positive semi-definite (all eigenvalues larger or equal zero). Finally, the vector \( \boldsymbol{h} \) is defined as $$ \boldsymbol{h} = \begin{bmatrix}0 \\ 0\\ -15 \\ 100 \\ 80\end{bmatrix}. $$
Since we don't have any equalities the matrix \( \boldsymbol{A} \) is set to zero The following code solves the equations for us
# Import the necessary packages
import numpy
from cvxopt import matrix
from cvxopt import solvers
P = matrix(numpy.diag([1,0]), tc=’d’)
q = matrix(numpy.array([3,4]), tc=’d’)
G = matrix(numpy.array([[-1,0],[0,-1],[-1,-3],[2,5],[3,4]]), tc=’d’)
h = matrix(numpy.array([0,0,-15,100,80]), tc=’d’)
# Construct the QP, invoke solver
sol = solvers.qp(P,q,G,h)
# Extract optimal value and solution
sol[’x’]
sol[’primal objective’]