# 数学代写|运筹学作业代写operational research代考|Applications to Queueing Systems

#### Doug I. Jones

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## 数学代写|运筹学作业代写operational research代考|First Entrance Times

For a birth and death process, a natural question is how long it takes, in expectation, to move from state $i$ to state $j>i$. In other words, what is the expected first entrance time of state $j$ if we start in state $i$ ? Let $T_{i, j}$ be the time to move from state $i$ to state $j$, and let us first consider the expected time required to move from state $i$ to $i+1$. First, the remainder of the time spent in state $i$ before jumping to any other state is exponentially distributed with mean $1 / \nu_i$. If we subsequently jump to state $i+1$, we are done. However, if we jump to state $i-1$ instead, we will additionally incur the expected time to move from state $i-1$ to state $i+1$. Thus,
\begin{aligned} \mathbb{E}\left[T_{i, i+1}\right] & =\frac{1}{\nu_i}+p_{i, i+1} \cdot 0+p_{i, i-1} \mathbb{E}\left[T_{i-1, i+1}\right] \ & =\frac{1}{\lambda_i+\mu_i}+\frac{\mu_i}{\lambda_i+\mu_i}\left(\mathbb{E}\left[T_{i-1, i}\right]+\mathbb{E}\left[T_{i, i+1}\right]\right) . \end{aligned}
Taking $\mathbb{E}\left[T_{i, i+1}\right]$ to the left-hand side gives
$$\mathbb{E}\left[T_{i, i+1}\right]=\frac{1}{\lambda_i}+\frac{\mu_i}{\lambda_i} \mathbb{E}\left[T_{i-1, i}\right] .$$
This recursive relation can be solved starting from $\mathbb{E}\left[T_{0,1}\right]=\frac{1}{\lambda_0}$. Finally, if $j>i$ we have $T_{i, j}=T_{i, i+1}+\cdots+T_{j-1, j}$

## 数学代写|运筹学作业代写operational research代考|The Single-Server Queue

Consider a single-server model where customer arrivals follow a Poisson process with arrival rate $\lambda$. The waiting room for the service station has an infinite capacity. An arriving customer who finds the server busy takes place in the waiting room until it is his turn to be served. The server can only help one customer at a time. The customers’ service times are independent of one another and have a common exponential distribution with expected value $1 / \mu$. The service process is independent of the arrival process. Assume $1 / \mu<1 / \lambda$; that is, a customer’s average service time is less than the average interarrival time between two consecutive customers. This model is called the $M / M / 1$ queue; see Chapter 9 . The following are interesting questions for this model: What is the average queue length? What is the average waiting time per customer? What are the probability distributions of the queue length and the customer’s waiting time?
To answer these questions, we let
$$X(t)=\text { number of customers present at time } t .$$

The stochastic process ${X(t), t \geq 0}$ is a continuous-time Markov chain with state space $I={0,1, \ldots}$. The transition rate diagram is given in Figure 8.5 (verify!). The equilibrium probability $\pi_i$ gives the long-run fraction of the time that $i$ customers are present in the system. Calculating the $\pi_i$ follows the “rate out of state $i$ $=$ rate into state $i “$ principle:
$$\lambda \pi_0=\mu \pi_1 \quad \text { and } \quad(\lambda+\mu) \pi_i=\lambda \pi_{i-1}+\mu \pi_{i+1} \quad \text { for } i=1,2, \ldots$$
This system of linear equations can be rewritten as a recursive relation. Substituting the first equality $\lambda \pi_0=\mu \pi_1$ in the second $(\lambda+\mu) \pi_1=\lambda \pi_0+\mu \pi_2$ gives $\lambda \pi_1=\mu \pi_2$. One can use induction to verify that
$$\mu \pi_i=\lambda \pi_{i-1} \quad \text { for } i=1,2, \ldots .$$
This recursive relation can even be solved explicitly. Repeatedly applying $\pi_i=$ $(\lambda / \mu) \pi_{i-1}$ gives $\pi_i=(\lambda / \mu)^i \pi_0$ for $i=1,2, \ldots$. This last equation also holds for $i=$ 0 . Together with $\sum_{i=0}^{\infty} \pi_i=1$, this leads to $1=\pi_0 \sum_{i=0}^{\infty}(\lambda / \mu)^i=\pi_0 /(1-(\lambda / \mu))$, where the assumption $\lambda / \mu<1$ is used. So $\pi_0=1-\lambda / \mu$. This gives
$$\pi_i=\left(1-\frac{\lambda}{\mu}\right)\left(\frac{\lambda}{\mu}\right)^i \quad \text { for } i=0,1, \ldots$$
So the number of customers present has a geometric distribution. The symbol $\rho$ is generally used for the quotient $\lambda / \mu$ :
$$\rho=\frac{\lambda}{\mu}$$

# 运筹学代考

## 数学代写|运筹学作业代写operational research代考|First Entrance Times

$$\rho=\frac{\lambda}{\mu}$$

## 有限元方法代写

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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