京都大学 情報学研究科 数理工学専攻 2023年8月実施 凸最適化

标签：

• Kyoto-University
• Linear-Programming-Duality
• Karush-Kuhn-Tucker-Conditions

Author

Casablanca, 祭音Myyura

Description

大学公表の原題

日本語版

English Version

Let $c=(c_1,c_2, \ldots , c_n)^\top\in\mathbb R^n$ , where the superscript $\top$ denotes transposition. Consider the following linear programming problem P:

$$\begin{aligned} \text{P} : &\text{Minimize} &\boldsymbol{c}^{\top} \boldsymbol{y}\\ &\text{subject to} &\sum_{i=1}^{n} y_{i} \leq 1 \\ &\text{ } &\boldsymbol{y} \geq 0 \end{aligned}$$

where the decision variable of problem P is the vector $\boldsymbol{y} = (y_1, y_2, \ldots y_n)^\top \in \mathbb{R}^n$.

Answer the following questions (i) and (ii)

(i) write out a dual problem of problem P.

(ii) Show that problem P has an optimal solution.

Let $Y$ be the set of optimal solutions of problem P.

Answer the following quetions (iii) and (iv).

(iii) Show that $Y$ is a convex set.

(iv) Suppose that $c_{1} = c_2 = \cdots = c_n$ and $c_1 < 0$. Consider the following optimization problem Q:

$$\begin{aligned} \text{Q} : &\text{Minimize} &\frac{1}{2}\boldsymbol{x}^{\top} \boldsymbol{x} - \boldsymbol{c}^{\top} \boldsymbol{x}\\ &\text{subject to} &\boldsymbol{x} \in Y \end{aligned}$$

where the decision variable of problem Q is the vector $\boldsymbol{x} \in \mathbb{R}^n$. Obtain an optimal solution of problem Q by using Karush-Kuhn-Tucker conditions.

题目描述

设 $\boldsymbol c=(c_1,c_2,\ldots,c_n)^\top$，其中 $\top$ 表示转置。考虑以 $\boldsymbol y=(y_1,\ldots,y_n)^\top$ 为变量的线性规划

$$\begin{aligned} \mathrm P:\quad &\text{最小化}\quad \boldsymbol c^\top\boldsymbol y\\ &\text{满足}\quad\sum_{i=1}^n y_i\le1,\qquad \boldsymbol y\ge0. \end{aligned}$$

回答：

1. 写出 P 的对偶问题。
2. 证明 P 有最优解。

令 $Y$ 为 P 的最优解集合，继续回答：

3. 证明 $Y$ 是凸集。

4. 假设 $c_1=c_2=\cdots=c_n$ 且 $c_1<0$。考虑

   $$\begin{aligned} \mathrm Q:\quad &\text{最小化}\quad \frac12\boldsymbol x^\top\boldsymbol x- \boldsymbol c^\top\boldsymbol x\\ &\text{满足}\quad\boldsymbol x\in Y, \end{aligned}$$

   其中 $\boldsymbol x\in\mathbb R^n$。利用 KKT 条件求 Q 的一个最优解。

Kai

(i)

We have Lagrangian: $L(y,\lambda,\nu) = c^\top y + \lambda (\mathbf{1}^\top y - 1 ) - \nu^\top y$.

The Lagrange dual function is

$$d(\lambda,\nu)= \begin{cases} -\lambda,&c+\lambda\boldsymbol1-\nu=0,\\ -\infty,&\text{otherwise}. \end{cases}$$

Then we write the dual problem D:

$$\begin{aligned} \text{D} : &\text{Maximize} &-\lambda\\ &\text{subject to} &c + \lambda \mathbf{1} - \nu = 0\\ &\text{ } &\lambda\geq 0,\quad \nu \succeq 0 \end{aligned}$$

(ii)

The feasible region

$$\{y\in\mathbb R^n:y\succeq0,\ \boldsymbol1^\top y\leq1\}$$

is nonempty and compact. Since $c^\top y$ is continuous, P has an optimal solution.

(iii)

For any $y_1,y_2 \in Y$ , we know that for any $\widetilde{y}$ which satisfies the constraints of P,

$$c^\top y_1 \leq c^\top \widetilde{y}, \quad c^\top y_2 \leq c^\top \widetilde{y}$$

for $\theta \in [0,1]$,

$$\theta y_1 + (1 - \theta) y_2 \succeq \mathbf{0}, \quad$$

$$\mathbf{1}^\top (\theta y_1 + (1-\theta) y_2) \leq 1$$

$$c^\top (\theta y_1 + (1-\theta)y_2) = \theta c^\top y_1 + (1-\theta)c^\top y_2 \leq c^\top \widetilde{y}$$

then $\theta y_1 + (1-\theta)y_2 \in Y$, $Y$ is a convex set.

(iv)

Since $c_1 = c_2 = \ldots c_n < 0$, $Y = \{ y | y \succeq \mathbf{0}, \mathbf{1}^\top y = 1 \}$, we can rewrite Q as

$$\begin{aligned} Q:&\text{Minimize} &\frac{1}{2}x^\top x-c^\top x \\ &\text{subject to} &x \succeq \mathbf{0}, \mathbf{1}^\top x = 1 \\ \end{aligned}$$

Thus, consistently with the stated objective, the Lagrangian is

$$L(x,\lambda,\mu)=\frac12x^\top x-c^\top x-\lambda^\top x+\mu(1-\boldsymbol1^\top x).$$

Then the KKT-condition:

$$\text{KKT-conditions: } \left\{ \begin{aligned} x - c - \lambda - \mu \mathbf{1} &= 0 \\ \lambda \succeq 0, -\lambda^\top x &=0 \\ x &\succeq 0\\ \mathbf{1}^\top x &= 1 \end{aligned} \right.$$

it is obviously that

$$x^* = \left[\frac{1}{n} , \frac{1}{n}, \ldots , \frac{1}{n}\right]^\top,\qquad \lambda^* = \mathbf{0},\qquad \mu^* = \frac{1}{n}-c_1$$

satisfy the KKT conditions. Since Q is convex, $x^*$ is its optimal solution.