# 管理科学代写|决策论代写Management Science Models for Decision Making代考|CBAD292

## 管理科学代写|决策论代写Management Science Models for Decision Making代考|Minimizing Positive Linear Combinations

Let $z(x)=w_1\left|c_0^1+c^1 x\right|+\ldots+w_r\left|c_0^r+c^r x\right|$. Consider the problem:
Minimize $z(x)$
subject to $A x \geq b$,
where the weights $w_1, \ldots, w_r$ are all strictly positive. In this problem the objective function to be minimized, $z(x)$, is a PL convex function, hence this problem can be transformed into an LP. This is based on a result that helps to express the absolute value as a linear function of two additional variables, which we will discuss first.
Result 2.4. Consider the affine function $c_0^k+c^k x$ and its value $\bar{\beta}=c_0^k+c^k \bar{x}$ at some point $\bar{x} \in R^n$. Consider the following LP in two variables $u, v$.
$$\begin{array}{ll} \text { Minimize } & u+v \ \text { subject to } & u-v=\beta \ & u, v \geq 0 \end{array}$$
(2.9) has a unique optimum solution $(\bar{u}, \bar{v})$, which satisfies $\bar{u} \bar{v}=0$, and its optimum objective value $\bar{u}+\bar{v}=|\beta|=\left|c_0^k+c^k \bar{x}\right|$.

Proof. If $\beta \geq 0$, the general solution of $(2.9)$ is $(u, v)=(\beta+\alpha, \alpha)$ for some $\alpha \geq 0$, the objective value of this solution, $\beta+2 \alpha$, assumes its minimum value when $\alpha=0$. So in this case $(\bar{u}, \bar{v})=(\beta, 0)$ satisfying $\bar{u} \bar{v}=0$ and having optimum objective value of $\bar{u}+\bar{v}=\beta=|\beta|$.

If $\beta<0$, the general solution of $(2.9)$ is $(u, v)=(\alpha,|\beta|+\alpha)$ for some $\alpha \geq 0$, the objective value of this solution, $|\beta|+2 \alpha$, assumes its minimum value when $\alpha=0$. So in this case $(\bar{u}, \bar{v})=(0,|\beta|)$ satisfying $\bar{u} \bar{v}=0$ and having optimum objective value of $\bar{u}+\bar{v}=|\beta|$.
So, the result holds in all cases.

## 管理科学代写|决策论代写Management Science Models for Decision Making代考|Minimizing the Maximum of the Absolute Values of Several Affine Functions

Let $z(x)=$ Maximum $\left{\left|c_0^1+c^1 x\right|, \ldots,\left|c_0^r+c^r x\right|\right}$. Consider the problem
Minimize $z(x)$
subject to $A x \geq b$.
In this problem the objective function to be minimized, $z(x)$, is the pointwise supremum of several PL convex functions, and hence is a PL convex function, hence this problem can be transformed into an LP. Combining the ideas discussed above, one LP model for this problem is Model 1 given below.

It can be verified that in this model the property $\left(u_t^{+}\right)\left(u_t^{-}\right)=0$ for all $t$ will hold at every optimum solution for it, so this is a valid model for the problem. But it has one disadvantage that it uses the variables $u_t^{+}, u_t^{-}$representing the positive and negative parts of $c_0^t+c^t x$ in additional constraints in the model (those in the first line of constraints), with the result that the pair of column vectors of the variables $u_t^{+}, u_t^{-}$among the constraints no longer form a linearly dependent set, violating Condition 2 expressed in Cautionary Note $2.1$ above.

It is possible to transform (2.11) into an LP model directly without introducing these $u_t^{+}, u_t^{-}$variables at all. This leads to a better and cleaner LP model for this problem, Model 2 , with only one additional variable $z$.

The constraints specify that $z \geq\left|c_0^t+c^t x\right|$ for all $t$; and as $z$ is minimized in Model 2, it guarantees that if $(\hat{z}, \hat{x})$ is an optimum solution of this Model 2, then $\hat{x}$ is an optimum solution also for (2.11), and $\hat{z}$ is the optimum objective value in (2.11).
We will now discuss important applications of these transformations in meeting multiple targets as closely as possible, and in curve fitting, and provide simple numerical examples for each.

Example 2.7. Meeting targets as closely as possible: Consider the fertilizer maker’s product mix problem with decision variables $x_1, x_2$ (hi-ph, lo-ph fertilizers to be made daily in the next period) discussed in Example $3.4 .1$ of Sect. $3.4$ of Murty (2005b) of Chap. 1 and Example $2.5$ above, with net profit coefficients $\left(c_1, c_2\right)=(15,10)$ in $\$ /$ton of hi-ph, lo-ph fertilizers made. In these examples, we considered only maximizing one objective function, the daily net profit$=15 x_1+10 x_2$with the profit vector given. But in real business applications, companies have to pay attention to many other objective functions in order to survive and thrive in the market place. We will consider two others. The second objective function that we will consider is the companies total market share, usually measured by the companies sales volume as a percentage of the sales volume of the whole market. To keep this example simple, we will measure this by the total daily sales revenue of the company. The sale prices of hi-ph, lo-ph fertilizers are$\$222$, $\$ 107 /$ton, respectively, so this objective function is$222 x_1+107 x_2$. The third objective function that we consider is the hi-tech market share, which is the market share of the company among hi-tech products (in this case hi-ph is the hi-tech product). This influences the public’s perception of the company as a market leader. To keep this example simple, we will measure this by the daily sales revenue of the company from hi-ph sales which is$\$222 x_1$.

So, here we have three different objective functions to optimize simultaneously. Problems like this are called multiobjective optimization problems. One commonly used technique to get a good solution in these problems is to set up a target value for each objective function (based on the companies aspirations, considering the tradeoffs between the various objective functions), and to try to find a solution as close to each of the targets as possible. In our example, suppose that the targets selected for daily net profit, market share, and hi-tech market share are $\$ 12,500,200,000$, and 70,000 , respectively. # 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|最小化正线性组合 让$z(x)=w_1\left|c_0^1+c^1 x\right|+\ldots+w_r\left|c_0^r+c^r x\right|$。考虑这个问题: 最小化$z(x)$受制于$A x \geq b$， 其中$w_1, \ldots, w_r$的权重都是严格正的。在该问题中，要最小化的目标函数$z(x)$是一个PL凸函数，因此该问题可以转化为一个LP。这是基于一个结果，该结果有助于将绝对值表示为两个附加变量的线性函数，我们将首先讨论这两个变量。2.4. 结果考虑仿射函数$c_0^k+c^k x$和它在某一点$\bar{x} \in R^n$处的值$\bar{\beta}=c_0^k+c^k \bar{x}$。考虑下面的LP在两个变量$u, v$. $$\begin{array}{ll} \text { Minimize } & u+v \ \text { subject to } & u-v=\beta \ & u, v \geq 0 \end{array}$$ (2.9)有一个唯一的最优解$(\bar{u}, \bar{v})$，它满足$\bar{u} \bar{v}=0$，它的最优目标值$\bar{u}+\bar{v}=|\beta|=\left|c_0^k+c^k \bar{x}\right|$. 证明。如果$\beta \geq 0$，对于某个$\alpha \geq 0$,$(2.9)$的通解是$(u, v)=(\beta+\alpha, \alpha)$，则该解的目标值$\beta+2 \alpha$在$\alpha=0$时取其最小值。因此，在本例中，$(\bar{u}, \bar{v})=(\beta, 0)$满足$\bar{u} \bar{v}=0$，有最佳的目标值$\bar{u}+\bar{v}=\beta=|\beta|$。 如果$\beta<0$,$(2.9)$的通解对于某个$\alpha \geq 0$是$(u, v)=(\alpha,|\beta|+\alpha)$，则该解的客观值$|\beta|+2 \alpha$在$\alpha=0$时取其最小值。因此，在这种情况下，$(\bar{u}, \bar{v})=(0,|\beta|)$满足$\bar{u} \bar{v}=0$，并有最佳的目标值$\bar{u}+\bar{v}=|\beta|$. 所以，结果在所有情况下都成立 ## 管理科学代写|决策论代写管理科学决策模型代考|最小化几个仿射函数绝对值的最大值 让$z(x)=$最大$\left{\left|c_0^1+c^1 x\right|, \ldots,\left|c_0^r+c^r x\right|\right}$。考虑问题 最小化$z(x)$受制于$A x \geq b$. 在这个问题中，要最小化的目标函数$z(x)$是几个PL凸函数的点上最大值，因此是一个PL凸函数，因此这个问题可以转化为一个LP。结合上面讨论的思想，解决这个问题的一个LP模型是下面给出的模型1 可以验证的是，在这个模型中，属性$\left(u_t^{+}\right)\left(u_t^{-}\right)=0$for all$t$将保持它的每个最优解，因此这是问题的一个有效模型。但它有一个缺点，就是在模型的附加约束(第一行约束)中使用了代表$c_0^t+c^t x$正、负部分的变量$u_t^{+}, u_t^{-}$，导致约束中变量$u_t^{+}, u_t^{-}$的列向量对不再形成线性相关集，违反了上面警告注释$2.1$中表达的条件2。 可以直接将(2.11)转换为LP模型，而完全不引入这些$u_t^{+}, u_t^{-}$变量。对于这个问题，这导致了一个更好、更清晰的LP模型，即模型2，只有一个额外的变量$z$. 约束指定$z \geq\left|c_0^t+c^t x\right|$对于所有$t$;由于在模型2中$z$被最小化，保证了如果$(\hat{z}, \hat{x})$是该模型2的最优解，那么$\hat{x}$也是(2.11)的最优解，并且$\hat{z}$是(2.11)的最优目标值。现在，我们将讨论这些转换在尽可能接近地满足多个目标和曲线拟合中的重要应用，并为每一个提供简单的数值例子 尽可能接近实现目标:考虑化肥制造商的产品组合问题，决策变量$x_1, x_2$(下一阶段每天生产高ph值、低ph值化肥)在第一章$3.4$节的示例$3.4 .1$和上面的示例$2.5$中讨论，在生产$\$/$吨高ph值、低ph值化肥时净利润系数$\left(c_1, c_2\right)=(15,10)$。在这些例子中，我们只考虑最大化一个目标函数，即在给定利润向量的情况下，每日净利润$=15 x_1+10 x_2$。但是在实际的商业应用中，为了在市场中生存和发展，公司必须注意许多其他的目标功能。 我们要考虑的第二个目标函数是公司的总市场份额，通常用公司的销售额占整个市场销售额的百分比来衡量。为了使这个例子简单，我们将用公司的每日总销售收入来衡量。高酸碱度、低酸碱度化肥的销售价格是 $\$ 222$，$\$107 /$ 吨，所以这个目标函数是 $222 x_1+107 x_2$我们考虑的第三个目标函数是高科技市场份额，它是公司在高科技产品中的市场份额(在这种情况下，hi-ph是高科技产品)。这影响了公众对该公司作为市场领导者的看法。为了使这个例子简单，我们将用公司从hi-ph销售中获得的每日销售收入来衡量它 $\$ 222 x_1$. 这里我们有三个不同的目标函数要同时优化。这样的问题被称为多目标优化问题。在这些问题中获得一个好的解决方案的常用技术是为每个目标函数设定一个目标值(基于公司的愿望，考虑各种目标函数之间的权衡)，并试图找到尽可能接近每个目标的解决方案。在我们的例子中，假设为每日净利润、市场份额和高科技市场份额选择的目标分别为$\$12,500,200,000$和70000

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