# 经济代写|博弈论代写Game Theory代考|ECON90022

Given $m$ functions $a_1, \ldots, a_m: \mathbb{R}{+}^n \rightarrow \mathbb{R}$ and $m$ scalars $b_1, \ldots$, $b_m \in \mathbb{R}$, the optimization problem $$\max {x \in \mathbb{R}{+}^n} f(x) \quad \text { s.t. } \quad a_1(x) \leq b_1, \ldots, a_m(x) \leq b_m$$ is of type (22) with the $m$ restriction functions $g_i(x)=b_i-a_i(x)$ and has the LAGRANGE function \begin{aligned} L(x, y) &=f(x)+\sum{i=1}^m y_i\left(b_i-a_i(x)\right) \ &=f(x)-\sum_{i=1}^m y_i a_i(x)+\sum_{i=1}^m y_i b_i . \end{aligned}
For an intuitive interpretation of the problem (26), think of the data vector
$$x=\left(x_1, \ldots, x_n\right)$$
as a plan for $n$ products to be manufactured in quantities $x_j$ and of $f(x)$ as the market value of $x$.

Assume that $x$ requires the use of $m$ materials in quantities $a_1(x), \ldots, a_m(x)$ and that the parameters $b_1, \ldots, b_m$ describe the quantities of the materials already in the possession of the manufacturer.

If the numbers $y_1, \ldots, y_m$ represent the market prices (per unit) of the $m$ materials, we find that $L(x, y)$ is the total value of the manufacturer’s assets:
\begin{aligned} L(x, y)=& \text { market value of the production } x \ &+\text { value of the materials left in stock. } \end{aligned}
The manufacturer would like to have that value as high as possible by deciding on an appropriate production plan $x$.

## 经济代写|博弈论代写Game Theory代考|Equilibria of convex LAGRANGE games

Remarkably, the KKT-conditions turn out to be not only necessary but also sufficient for the characterization of equilibria in convex LAGRANGE games with differentiable objective functions. This gives a way to compute such equilibria and hence to solve optimization problems of type $(22)$ in practice : $^9$ :

• Find a solution $\left(x^, y^\right) \in X \times \mathbb{R}{+}^m$ for the $K K T$ inequalities. $\left(x^, y^\right)$ will yield an equilibrium in $\Lambda=$ $\left(X, \mathbb{R}{+}^m, L\right)$ and $x^$ will be an optimal solution for (22). Indeed, one finds: THEOREM 3.3. A pair $\left(x^, y^\right) \in X \times \mathbb{R}{+}^m$ is an equilibrium of the convex Lagrange game $\Lambda=\left(X, \mathbb{R}{+}^m, L\right)$ if and only if $\left(x^, y^\right)$ satisfies the $K K T$-conditions. Proof. From Lemma 3.2, we know that the KKT-conditions are necessary. To show sufficiency, assume that $\left(x^, y^\right) \in X \times \mathbb{R}_{+}^m$ satisfies the KKT-conditions. We must demonstrate that $\left(x^, y^*\right)$ is
• an equilibrium of the convex Lagrange game $\Lambda=\left(X, \mathbb{R}{+}^m, L\right)$, i.e., satisfies $$\max {x \in X} L\left(x, y^\right)=L\left(x^, y^\right)=\min {y \geq 0} L\left(x^, y\right) •$$
• for the function $L(x, y)=f(x)+y^T g(x)$.
• Since $x \mapsto L\left(x, y^\right)$ is a concave function, we have for every $x \in X$, $$L\left(x, y^\right)-L\left(x^, y^\right) \leq \nabla_x L\left(x^, y^\right)\left(x-x^\right) .$$ Because $\left(\mathrm{K}_2\right)$ guarantees $\nabla_x L\left(x^, y^\right)\left(x-x^\right) \leq 0$, we conclude
• $$• L\left(x, y^\right) \leq L\left(x^, y^\right),$$ which implies the first equality in (27). For the second equality, $\left(\mathrm{K}_0\right)$ and $\left(\mathrm{K}_1\right)$ yield $g\left(x^\right) \geq 0$ and $\left(y^\right)^T g\left(x^\right)=0$ and therefore:
• • \begin{aligned} • \min {y \geq 0} L\left(x^, y\right) &=f\left(x^\right)+\min _{y \geq 0} y^T g\left(x^\right)=f\left(x^\right)+0 \ • &=f\left(x^\right)+\left(y^\right)^T g\left(x^\right)=L\left(x^, y^*\right) . • \end{aligned} •

# 博弈论代考

$$\max x \in \mathbb{R}+{ }^n f(x) \quad \text { s.t. } \quad a_1(x) \leq b_1, \ldots, a_m(x) \leq b_m$$

$$L(x, y)=f(x)+\sum i=1^m y_i\left(b_i-a_i(x)\right) \quad=f(x)-\sum_{i=1}^m y_i a_i(x)+\sum_{i=1}^m y_i b_i .$$

$$x=\left(x_1, \ldots, x_n\right)$$

$L(x, y)=$ market value of the production $x \quad$ + value of the materials left in stock.

## 经济代写|博弈论代写Game Theory代考|Equilibria of convex LAGRANGE games

Yeft( $x^{\wedge}, y^{\wedge} \backslash$ right) $\backslash$ in $X \backslash t$ imes $\backslash$ mathbb{R ${+}^{\wedge} m$ 是凸拉格朗日博亦的均衡 $\Lambda=\left(X, \mathbb{R}+{ }^m, L\right)$ 当且仅当 左 $\left(x^{\wedge}, y^{\wedge} \backslash\right.$ 右 $)$ 满足 $K K T$-条件。证明。从引理 3.2，我们知道 KKT 条件是必要的。为了显示充分性， 假设 $\backslash$ left $\left(x^{\wedge}, y^{\wedge} \backslash\right.$ right $\backslash$ in $X \backslash t$ imes $\backslash m a t h b b{R}_{-}{+}^{\wedge} m$ 满足 $K K T$ 条件。我们必须证明 $\left(x^{\prime} y^*\right)$ 是 $y^{\wedge} \backslash$ right $)=\backslash \min {y \backslash$ geq 0$}$ L\eft $\left(x^{\wedge}, y \backslash\right.$ 右 $)$

• $\$ \$$• 对于功能 L(x, y)=f(x)+y^T g(x). • 自从 x \backslash \mapsto L \backslash l left \left(x, y^{\wedge} \backslash\right. right ) 是凹函数，我们对每个 x \in X, 因为 \left(\mathrm{K}2\right) 保证 \backslash nabla x \backslash \backslash left \left(x^{\wedge}, y^{\wedge} \backslash\right. right ) \backslash left \left(x x^{\wedge} \backslash\right. right \backslash \backslash leq 0 , 我们得出结论 • \ \$$ 和 $\left(K_1\right)$ 屈沚 $g_{\bigvee l e f t}\left(x^{\wedge} \backslash\right.$ right $\backslash$ geq 0 和 $\backslash$ left( $\left(y^{\wedge} \backslash \text { right }\right)^{\wedge} T g \backslash l$ eft $\left(x^{\wedge} \backslash r i g h t\right)=0$ 因此:
• $\$ \$$• \backslash begin{对齐 } \left(x^{\wedge} \backslash\right. 右 )+0 \backslash • \& =f \backslash left( x^{\wedge} \backslash right )+\backslash left( \left(y^{\wedge} \backslash \text { right }\right)^{\wedge} T g \backslash left \left(x^{\wedge} \backslash\right. Yright )=\left\lfloor\backslash l\right. left \left(x^{\wedge}, y^{\wedge} \backslash\right. right) 。 • \结束{对齐} • \ \$$

myassignments-help数学代考价格说明

1、客户需提供物理代考的网址，相关账户，以及课程名称，Textbook等相关资料~客服会根据作业数量和持续时间给您定价~使收费透明，让您清楚的知道您的钱花在什么地方。

2、数学代写一般每篇报价约为600—1000rmb，费用根据持续时间、周作业量、成绩要求有所浮动(持续时间越长约便宜、周作业量越多约贵、成绩要求越高越贵)，报价后价格觉得合适，可以先付一周的款，我们帮你试做，满意后再继续，遇到Fail全额退款。

3、myassignments-help公司所有MATH作业代写服务支持付半款，全款，周付款，周付款一方面方便大家查阅自己的分数，一方面也方便大家资金周转，注意:每周固定周一时先预付下周的定金，不付定金不予继续做。物理代写一次性付清打9.5折。

Math作业代写、数学代写常见问题

myassignments-help擅长领域包含但不是全部: