## 管理科学代写|决策论代写Management Science Models for Decision Making代考|Minimizing a Separable PL Convex Function Subject

The negative of a concave function is convex. Maximizing a concave function is the same as minimizing its negative, which is a convex function. Using this, the techniques discussed here can also be used to solve problems in which a separable PL concave function is required to be maximized subject to linear constraints.

A real-valued function $z(x)$ of decision variables $x=\left(x_1, \ldots, x_n\right)^T$ is said to be a separable function if it can be expressed as the sum of $n$ different functions, each one involving only one variable, that is, has the form $z(x)=z_1\left(x_1\right)+z_2\left(x_2\right)+$ $\ldots+z_n\left(x_n\right)$. This separable function is also a PL convex function if $z_j\left(x_j\right)$ is a PL convex function for each $j=1$ to $n$.

Result 2.3. Let $\theta(\lambda)$ be the PL convex function of $\lambda \in R^1$ defined over $\lambda \geq 0$ shown in the following table:

where $\lambda_1<\lambda_2<\ldots<\lambda_{r-1}$ and $c_1<c_2<$ ldots $<c_r$ (conditions for $\theta(\lambda)$ to be convex). Then for any $\bar{\lambda} \geq 0, \theta(\bar{\lambda})$ is the minimum objective value in the following problem.
$$\begin{array}{ll} \text { Minimize } z= & c_1 \mu_1+\ldots+c_r \mu_r \ \text { subject to } & \mu_1+\ldots+\mu_r=\bar{\lambda} \ & 0 \leq \mu_t \leq \lambda_t-\lambda_{t-1} t=1, \ldots, r \end{array}$$
Proof. Problem (2.5) can be interpreted this way: Suppose we want to purchase exactly $\bar{\lambda}$ units of a commodity for which there are $r$ suppliers. For $k=1$ to $r$, $k$ th supplier’s rate is $c_k$ /unit and can supply up to $\lambda_k-\lambda_{k-1}$ units only. $\mu_k$ in the problem represents the amount purchased from the $k$ th supplier, it is $\geq 0$, but is bounded above by the length of the $k$ th interval in which the slope of $\theta(\lambda)$ is $c_k . z$ to be minimized is the total expense to acquire the required $\bar{\lambda}$ of the commodity.
Clearly, to minimize $z$, we should purchase as much as possible from the cheapest supplier, and when he cannot supply any more go to the next cheapest supplier, and continue the same way until the required quantity is acquired. As the cost coefficients satisfy $c_1<c_2<\ldots<c_r$ by the convexity of $\theta(\lambda)$, the cheapest cost coefficient corresponds to the leftmost interval beginning with 0 , the next cheapest corresponds to the next interval just to the right of it, and so on. Because of this, the optimum solution $\bar{\mu}=\left(\bar{\mu}1, \ldots, \overline{\mu_r}\right)$ of $(2.5)$ satisfies the following special property. Special property of optimum solution $\bar{\mu}$ of (2.5) that follows from convexity of $\dot{\theta}(\lambda)$ : If $p$ is such that $\lambda_p \leq \bar{\lambda} \leq \lambda{p+1}$, then $\bar{\mu}t=\lambda_t-\lambda{t-1}$, the upper bound of $\mu_t$ for all $t=1$ to $p, \bar{\mu}_{p \mid 1}=\bar{\lambda}-\lambda_p$, and $\bar{\mu}_t=0$ for all $t \geq p+2$.

This property says that in the optimum solution of (2.5) if any $\mu_k>0$, then the value of $\mu_t$ in it must be equal to the upper bound on this variable for any $t<k$. Because of this, the optimum objective value in (2.5) is $=c_1 \bar{\mu}_1+\ldots+c_r \bar{\mu}_r \theta(\bar{\lambda})$.

## 管理科学代写|决策论代写Management Science Models for Decision Making代考|Min-max, Max-min Problems

As discussed earlier, a PL convex function in variables $x=\left(x_1, \ldots, x_n\right)^T$ can be expressed as the pointwise maximum of a finite set of affine functions. Minimizing a function like that subject to some constraints is appropriately known as a min-max problem.

Similarly, a PL concave function in $x$ can be expressed as the pointwise minimum of a finite set of affine functions. Maximizing a function like that subject to some constraints is appropriately known as a max-min problem. Both min-max and maxmin problems can be expressed as LPs using just one additional variable, if all the constraints are linear constraints.

If the PL convex function $f(x)=\operatorname{maximum}\left{c_0^t+c^t x: t=1, \ldots, r\right}$, then $-f(x)=\operatorname{minimum}\left{-c_0^t-c^t x: t=1, \ldots, r\right}$ is PI. concave and conversely. Using this, any min-max problem can be posed as a max-min problem and vice versa. So, it is sufficient to discuss max-min problems. Consider the max-min problem
Maximize $z(x)=\operatorname{Minimum}\left{c_0^1+c^1 x, \ldots, c_0^r+c^r x\right}$
subject to $A x=b$
$x \geq 0$.
To transform this problem into an LP, introduce the new variable $x_{n+1}$ to denote the value of the objective function $z(x)$ to be maximized. Then the equivalent LP with additional linear constraints is
$$\begin{array}{ll} \operatorname{Maximize} & x_{n+1} \ \text { subject to } & x_{n+1} \leq c_0^1+c^1 x \ & x_{n+1} \leq c_0^2+c^2 x \end{array}$$ \begin{aligned} x_{n+1} & \leq c_0^r+c^r x \ A x &=b \ x & \geq 0 \end{aligned}
The fact that $x_{n+1}$ is being maximized and the additional constraints together imply that if $\left(\bar{x}, \bar{x}{n+1}\right)$ is an optimum solution of this LP model, then $\bar{x}{n+1}=$ $\min \left{c_0^1+c^1 \bar{x}, \ldots, c_0^r+c^r \bar{x}\right}=z(\bar{x})$, and that $\bar{x}_{n+1}$ is the maximum value of $z(x)$ in the original max-min problem.

# 决策论代考

## 管理科学代写|决策论代写管理科学决策模型代考|最小化可分离PL凸函数主题

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: