qp_ex1

qp_ex1()

qp_ex1() - Example of quadratic program (QP) optimization.

Example of solving the following QP problem, first using opt_model and opt_model.solve(), then directly using qps_master().

(18)$$\underset{\mathit{x}}{min}\frac{1}{2}{\mathit{x}}^{\mathsf{T}}\underset{―}{\mathit{Q}}\mathit{x}$$

subject to

(19)$$\underset{―}{\mathit{l}}\le \underset{―}{\mathit{A}}\mathit{x}\le \underset{―}{\mathit{u}}$$

(20)$${\underset{―}{\mathit{x}}}_{\min }\le \mathit{x}\le {\underset{―}{\mathit{x}}}_{\max }$$

where

(21)$$\begin{array}{r}\underset{―}{\mathit{Q}}=\left[\begin{array}{cccc}8& 1& -3& -4\\ 1& 4& -2& -1\\ -3& -2& 5& 4\\ -4& -1& 4& 12\end{array}\right]\end{array}$$

(22)$$\begin{array}{r}\underset{―}{\mathit{l}}=\left[\begin{array}{c}4\\ -\mathrm{\infty }\end{array}\right],\underset{―}{\mathit{A}}=\left[\begin{array}{cccc}6& 1& 5& -4\\ 4& 9& 0& 0\end{array}\right],\underset{―}{\mathit{u}}=\left[\begin{array}{c}4\\ 2\end{array}\right]\end{array}$$

(23)$$\begin{array}{r}{\underset{―}{\mathit{x}}}_{\min }=\left[\begin{array}{c}0\\ 0\\ -\mathrm{\infty }\\ -\mathrm{\infty }\end{array}\right],{\underset{―}{\mathit{x}}}_{\max }=\left[\begin{array}{c}\mathrm{\infty }\\ \mathrm{\infty }\\ 0\\ 2\end{array}\right]\end{array}$$