Exercise 8.9

Answers

(a): Since $u_{0}$ is optimal for (8.10), so we have $c - a^{T} u_{0} \leq 0$ , the maximum value that can be achieved with $α \geq 0$ is thus $\max_{α \geq 0} α (c - a^{T} u_{0}) = 0$ .
(b): We now prove that $u_{1}$ is feasible for (8.10). Assume the contrary that $c - a^{T} u_{1} > 0$ , then $\max_{α \geq 0} α (c - a^{T} u_{0}) = \infty$ and the objective in (8.11) goes to infinite. However, when $u = u_{0}$ , we have a finite objective in (8.11) as $\max_{α \geq 0} α (c - a^{T} u_{0}) = 0$ . This contradicts the assumption that $u_{1}$ is optimal for (8.11). So we have $c - a^{T} u_{1} \leq 0$ and $u_{1}$ is feasible for (8.10).
(c): For the objective in (8.11), we know from problem (a),(b) that for both $u_{0}, u_{1}$ , we have $c - a^{T} u_{0} \leq 0$ and $c - a^{T} u_{1} \leq 0$ . Then we have $\max_{α \geq 0} α (c - a^{T} u_{1}) = \max_{α \geq 0} α (c - a^{T} u_{1}) = 0 .$
Thus we see that $u_{1}$ actually minimizes $\min_{u \in R^{L}} \frac{1}{2} u^{T} Qu + p^{T} u$ , on the other hand, by definition, $u_{0}$ also minimize the objective in (8.10), which is the same objective here. So we must have $\frac{1}{2} u_{0}^{T} Q u_{0} + p^{T} u_{0} = \frac{1}{2} u_{1}^{T} Q u_{1} + p^{T} u_{1}$ .
(d): Let $u^{∗}$ be any optimal solution for (8.11), then by problem (b) and (a), we have $\max_{α \geq 0} α (c - a^{T} u^{∗}) = 0,$
if the maximum is attained at $α^{∗}$ , then $α^{∗} (c - a^{T} u^{∗}) = 0$ , so we either have $c - a^{T} u^{∗} = 0$ or $α^{∗} = 0$ .

niuers

2021-12-08 10:09

Exercise 8.9

Answers

Comments

Add answer