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# 数学代写|运筹学代写Operations Research代考|MATH3202 Optimizing Inventory and Queuing Models

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## 数学代写|运筹学代写Operations Research代考|Optimizing Inventory and Queuing Models

Inventory models, to be discussed in Chap. 12, deal with the problem of determining the optimal size $x$ of an order, so as to minimize some total cost function $T C(x)$, which includes the costs of placing an order and receiving the shipment, on the one hand, and the cost of keeping units in stock. Deferring the discussion of details to the chapter on the subject, we attempt to solve the optimization problem

$$\mathrm{P}: \operatorname{Min} T C(x)=\frac{a_1}{x}+b_1 x$$
where $x$ denotes the order quantity, the parameter $a_1$ is proportional to the unit ordering costs, and the parameter $b_1$ is proportional to the unit holding cost. The unconstrained problem can be solved by differential calculus, resulting in $\bar{x}=$ $\sqrt{a_1 / b_1}$. This is the economic order quantity $(E O Q)$, which will be properly introduced in the chapter on inventory management.

One extension to the basic model allows backorders, which leads to the two-variable unconstrained problem
$$P: \operatorname{Min} z=\frac{a_2}{x}+b_2 \frac{(x-y)^2}{x}+c_2 \frac{y^2}{x},$$
with $y$ denoting the maximal allowable shortage and $a_2, b_2$, and $c_2$ being known constants. Partial differentiation (differentiation with respect to one variable at a time) and setting the expression to zero then determines the optimal solution $\bar{x}=$ $\sqrt{\frac{a_2}{c_2} \frac{b_2+c_2}{b_2}}$ and $\bar{y}=\sqrt{\frac{a_2}{c_2} \frac{b_2}{b_2+c_2}}$. For a full discussion, see Chap. 12 of this volume. A final application is found in the area of queuing. A more detailed discussion is found in Chap. 15 of this book. Suppose that on average, $\lambda$ customers arrive in $1 \mathrm{hr}$ at a single service station. Based on training and other factors, the service station is manned by an agent, who can deal with $x$ customers per hour (the service rate). In order to be feasible, $x>\lambda$ is required. Knowing that the expected number of customers in the system is $\frac{\lambda}{x-\lambda}$, a cost of $a x$ for the service station (capturing the fact that a faster server is more expensive) and unit costs $c$ for keeping a customer in the system for an hour, minimizing the total cost of the system results in the problem
$$\mathrm{P}: \operatorname{Min} T C(x)=a x+\frac{c \lambda}{x-\lambda} \text {, s.t. } x>\lambda .$$
Using again differential calculus, we find that the unique optimal solution is given by $\bar{x}=\lambda+\sqrt{\frac{c \lambda}{a}}$, which also satisfies the single constraint $\bar{x}>\lambda$.

## 数学代写|运筹学代写Operations Research代考|Properties of Nonlinear Programming Problems

In dealing with a nonlinear optimization problem, there are two important questions to be answered.

Given a feasible solution, how can we test it for optimality? and

How do we find an optimal solution?
We will relegate a discussion of the second question to the next section and discuss here how to perform an optimality test. First, consider the unconstrained case, i.e., the case, in which a problem has an objective function but no constraints. For simplicity, assume that we are minimizing a differentiable function $f(x)$ of a single variable $x$. A fundamental result from differential calculus is then that if $\bar{x}$ satisfies $f^{\prime}(\bar{x})=0$ and $f^{\prime \prime}(\bar{x})>0$, then the function $f$ has a local minimum (for a definition, see, e.g., Chap. 17) at $\bar{x}$. As an example, consider the function $f(x)=x^3-3 x$, pictured in Fig. 4.5.

This function has a local minimum at $\bar{x}=1$, for which $f(\bar{x})=f(1)=-2$, but it is apparent that there are values of $x$, for which its functional value is less than $-2$, for instance, $x=-3$, for which $f(-3)=-18$. In fact, the function $f(x)$ does not even have a global minimum (a smallest value that the function $f$ can take), since with ever smaller negative values of $x$, there is no lower bound on the value of $f(x)$. However, for a convex function (i.e., a function, with the property that all points on a linear line segment-the chord-between any of its two points is located above the function itself, see, e.g., Fig. 4.6), any local minimum will necessarily be global. In our case, the function is convex in $[0, \infty[$, where $\bar{x}=1$ is the minimal point.

## 数学代写|运筹学代写Operations Research代考|Optimizing Inventory and Queuing Models

$$\mathrm{P}: \operatorname{Min} T C(x)=\frac{a_1}{x}+b_1 x$$

$$P: \operatorname{Min} z=\frac{a_2}{x}+b_2 \frac{(x-y)^2}{x}+c_2 \frac{y^2}{x},$$

$$\mathrm{P}: \operatorname{Min} T C(x)=a x+\frac{c \lambda}{x-\lambda}, \text { s.t. } x>\lambda .$$

## 数学代写|运筹学代写Operations Research代考|Properties of Nonlinear Programming Problems

$f(x)$. 然而，对于一个凸函数 (即，一个函数，其特性是线性线段上的所有点 – 其任何两点之间的弦 – 都位于函数本鳥之上，例 如，参见图 4.6)，任何㕆部嘬小值必然是全局的。在我们的例子中，函数是凸的 $[0, \infty[$ ，在哪里 $\bar{x}=1$ 是最小点。

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