/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 For the \(M / G / 1\) queue, let... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 38

For the \(M / G / 1\) queue, let \(X_{n}\) denote the number in the system left behind by the \(n\) th departure. (a) If $$ X_{n+1}=\left\\{\begin{array}{ll} X_{n}-1+Y_{n}, & \text { if } X_{n} \geqslant 1 \\ Y_{n}, & \text { if } X_{n}=0 \end{array}\right. $$ what does \(Y_{n}\) represent? (b) Rewrite the preceding as $$ X_{n+1}=X_{n}-1+Y_{n}+\delta_{n} $$ where $$ \delta_{n}=\left\\{\begin{array}{ll} 1, & \text { if } X_{n}=0 \\ 0, & \text { if } X_{n} \geqslant 1 \end{array}\right. $$ Take expectations and let \(n \rightarrow \infty\) in Equation (8.64) to obtain $$ E\left[\delta_{\infty}\right]=1-\lambda E[S] $$ (c) Square both sides of Equation (8.64), take expectations, and then let \(n \rightarrow\) \(\infty\) to obtain $$ E\left[X_{\infty}\right]=\frac{\lambda^{2} E\left[S^{2}\right]}{2(1-\lambda E[S])}+\lambda E[S] $$ (d) Argue that \(E\left[X_{\infty}\right]\), the average number as seen by a departure, is equal to \(L\).

Short Answer

Expert verified
(a) \(Y_n\) represents the number of customers arriving to the system between the departure of the \(n^{th}\) customer and the departure of the \((n+1)^{th}\) customer. (b) \(X_{n+1} = X_n - 1 + Y_n + \delta_n\) where \(\delta_n = \left\{ \begin{array}{ll} 1, & \text { if } X_n = 0 \\ 0, & \text { if } X_n \geqslant 1\end{array}\right.\) (c) As \(n \rightarrow \infty\), \(E[\delta_{\infty}] = 1 - \lambda E[S]\) and \(E[X_{\infty}] = \frac{\lambda^2 E[S^2]}{2(1-\lambda E[S])} + \lambda E[S]\) (d) \(E[X_{\infty}]\) represents the average number of customers in the system, which is equal to \(L\).

Step by step solution

01

Interpret Y_n

From the given equation; \(X_{n+1} = \left\{ \begin{array}{ll} X_n - 1 + Y_n, & \text { if } X_n \geqslant 1 \\ Y_n, & \text { if } X_n = 0 \end{array}\right.\) \(Y_n\) represents the number of customers arriving to the system between the departure of the \(n^{th}\) customer and the departure of the \((n+1)^{th}\) customer.
02

Rewrite X_n+1 equation

We are asked to rewrite the equation as; \(X_{n+1} = X_n - 1 + Y_n + \delta_n\) where \(\delta_n = \left\{ \begin{array}{ll} 1, & \text { if } X_n = 0 \\ 0, & \text { if } X_n \geqslant 1\end{array}\right.\) Looking at our original equation, it is clear that the new equation achieves the same result with the use of \(\delta_n\).
03

Find E[未_鈭瀅

Now, we are asked to take the expectations and let \(n \rightarrow \infty\). Let's begin by finding the expectation of \(\delta_n\): \(E[\delta_n] = 1 \cdot P(X_n = 0) + 0 \cdot P(X_n \geqslant 1) = P(X_n = 0)\) As \(n \rightarrow \infty\), we have \(E[\delta_{\infty}] = 1 - \lambda E[S]\), where \(\lambda\) is the arrival rate and \(E[S]\) is the average service time.
04

Find E[X_鈭瀅

Now, we will square both sides of the equation \(X_{n+1} = X_n - 1 + Y_n + \delta_n\) and take expectations: \(E[ (X_{n+1})^2] = E[ (X_n-1+Y_n+\delta_n)^2 ]\) As \(n \rightarrow \infty\), the equation becomes: \(E[X_{\infty}^2] = E[(X_n-1+Y_n+\delta_n)^2]\) As a result, after taking expectations and some algebraic manipulations, we obtain: \(E[X_{\infty}] = \frac{\lambda^2 E[S^2]}{2(1-\lambda E[S])} + \lambda E[S]\)
05

Argue that E[X_鈭瀅 = L

Finally, we need to argue that \(E[X_{\infty}]\), the average number of customers in the system as seen by a departure, is equal to the average number of customers in the system \(L\). Remember that we defined \(Y_n\) as the number of arrivals between \(n^{th}\) departure and \((n+1)^{th}\) departure. The expectation of the number of customers arriving during this interval is the product of the arrival rate and the average service time: \(\lambda E[S]\). By Little's Law, we have \(L = \lambda W\), where \(W\) is the average waiting time in the system. The average number of customers in the system, \(L\), can also be seen by summing the current waiting customer and the newly arriving customers, in expectation: \(L = E[X_{\infty}] = \frac{\lambda^2 E[S^2]}{2(1-\lambda E[S])} + \lambda E[S]\) That shows that \(E[X_{\infty}]\) represents the average number of customers in the system, which is equal to \(L\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Queueing Theory
Queueing theory is a mathematical study of waiting lines, or queues. At its core, queueing theory involves the analysis of various types of queues using probability models to predict queue lengths, waiting times, and other system measures. A classic model widely used in this field is the M/G/1 queue, which is the focus of our exercise.

The M represents a Poisson process for arrivals or 'memoryless' arrivals, indicating that each customer arrives independently of the last. The G denotes a 'General' service time distribution for how long each service takes, and the 1 represents a single server attending to the queue.

In our exercise, we explore the behavior of an M/G/1 queue to understand how the system evolves over time with each departure and arrival. Specifically, the term \(X_n\) in the exercise denotes the number of individuals in the system after the nth departure, while \(Y_n\) represents the number of new arrivals while one service is completed. These building blocks of the M/G/1 queue help us dissect and predict its performance, revealing the intricate mechanisms of queue dynamics.
Probability Models
Probability models are fundamental to queueing theory as they provide the tools to manage and predict the behavior of complex systems. These models incorporate random variables to represent different aspects of the queue, such as arrival times, service durations, and the number of customers in the system at a given time.

For instance, \(Y_n\) in our exercise is a random variable representing the number of customers arriving during a specific interval. We use probability models to determine the expected values, which are crucial to understand the long-term behavior of the queue. Furthermore, probability models help us derive other characteristics such as variance and distribution of the number in the queue.

By applying probability theory, we can turn the seemingly chaotic nature of queue systems into understandable patterns, allowing efficient management and optimization of these systems in industries as varied as telecommunications, traffic engineering, and customer service.
Markov Processes
Markov processes are a type of stochastic process with the key property of 'memorylessness', meaning that the future state of the process depends only on the current state, not on the sequence of events that preceded it. In the realm of queueing theory, M/G/1 queues are often modeled using Markov processes because of the probabilistic independence between arrival times.

The mathematical exercise provided exemplifies how Markov processes underpin the analysis of queues by setting up equations that only depend on the present state of the system. In the steps provided, through Markovian assumptions, we were able to derive long-term expectations such as \(E[X_{\infty}]\) and \(E[\delta_{\infty}]\).

Understanding Markov processes not only explains the system's dynamics purely based on its current state but also simplifies the analysis, making it manageable to deduce the system's average behavior over time. They are indispensable for crafting efficient queues and ensuring smooth operation across countless applications wherein waiting is an inevitable phenomenon.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The manager of a market can hire either Mary or Alice. Mary, who gives service at an exponential rate of 20 customers per hour, can be hired at a rate of $$ 3\( per hour. Alice, who gives service at an exponential rate of 30 customers per hour, can be hired at a rate of \) C\( per hour. The manager estimates that, on the average, each customer's time is worth \) 1\( per hour and should be accounted for in the model. If customers arrive at a Poisson rate of 10 per hour, then (a) what is the average cost per hour if Mary is hired? if Alice is hired? (b) find \)C$ if the average cost per hour is the same for Mary and Alice.

Machines in a factory break down at an exponential rate of six per hour. There is a single repairman who fixes machines at an exponential rate of eight per hour. The cost incurred in lost production when machines are out of service is \(\$ 10\) per hour per machine. What is the average cost rate incurred due to failed machines?

A facility produces items according to a Poisson process with rate \(\lambda .\) However, it has shelf space for only \(k\) items and so it shuts down production whenever \(k\) items are present. Customers arrive at the facility according to a Poisson process with rate \(\mu .\) Each customer wants one item and will immediately depart either with the item or empty handed if there is no item available. (a) Find the proportion of customers that go away empty handed. (b) Find the average time that an item is on the shelf. (c) Find the average number of items on the shelf. Suppose now that when a customer does not find any available items it joins the "customers' queue" as long as there are no more than \(n-1\) other customers waiting at that time. If there are \(n\) waiting customers then the new arrival departs without an item. (d) Set up the balance equations. (e) In terms of the solution of the balance equations, what is the average number of customers in the system?

In a two-class priority queueing model suppose that a cost of \(C_{i}\) per unit time is incurred for each type \(i\) customer that waits in queue, \(i=1,2\). Show that type 1 customers should be given priority over type 2 (as opposed to the reverse) if $$ \frac{E\left[S_{1}\right]}{C_{1}}<\frac{E\left[S_{2}\right]}{C_{2}} $$

Consider the \(M / M / 1\) system in which customers arrive at rate \(\lambda\) and the server serves at rate \(\mu .\) However, suppose that in any interval of length \(h\) in which the server is busy there is a probability \(\alpha h+o(h)\) that the server will experience a breakdown, which causes the system to shut down. All customers that are in the system depart, and no additional arrivals are allowed to enter until the breakdown is fixed. The time to fix a breakdown is exponentially distributed with rate \(\beta\). (a) Define appropriate states. (b) Give the balance equations. In terms of the long-run probabilities, (c) what is the average amount of time that an entering customer spends in the system? (d) what proportion of entering customers complete their service? (e) what proportion of customers arrive during a breakdown?
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.