Find optimal solution x and its derivative with respect to u of min{ $\frac{1}{2} x^T Q x + c^T x$ subject to $a^T x = u$}

Question

consider the convex quadratic program min{$\frac{1}{2} x^T Q x + c^T x$ subject to $a^T x = u$} , Q is n by n, $c \in R^n $ , $a \in R^n $ , $ a \neq 0$, $u \in R $, if Q is positive definite, compute the optimal solution x(u) and its derivative with respect to u

my thinking is to use KKT, $[ Q , a ; a^T, 0] [ x , y]^T = [-c , u]^T$, or say $ Q x + a y = -c$ & $a^T x = u$, but then I am not sure how can I write out x explicitly, thank you for any help.

Answer 1

The Lagrangian is:

$L(x,\lambda) = \frac{1}{2} x^{\top}Qx + c^{\top}x + \lambda(a^{\top}x-u)$

where $\lambda \in \mathbf{R}$. Since $Q$ is PD, it is non-singular. Stationarity condition will give:

$\nabla L(x,\lambda) = Qx^{*} + c + \lambda^{*}a=0 \\ \Rightarrow x^{*} = Q^{-1} (-c - \lambda^{*}a). \qquad - \ (1)$

Primal feasibility condition will give:

$a^{\top}x^{*}-u=0.\qquad - \ (2)$

From this point onwards, it is a matter of combining $(1)$ and $(2)$ to find $x^{*}$ and $\lambda^{*}$.