/*! 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 27 In the \(\mathrm{M} / \mathrm{M}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 27

In the \(\mathrm{M} / \mathrm{M} / \mathrm{s}\) queue if you allow the service rate to depend on the number in the system (but in such a way so that it is ergodic), what can you say about the output process? What can you say when the service rate \(\mu\) remains unchanged but \(\lambda>s \mu ?\)

Short Answer

Expert verified
In an M/M/s queue with a service rate that depends on the number of customers in the system in such a way that it remains ergodic, the output process will still have the Markov property and memoryless property. However, if the arrival rate 饾渾 is greater than 蝉饾渿 with 饾渿 remaining unchanged, the queue will become unstable, leading to indefinitely increasing waiting times and an unpredictable output process.

Step by step solution

01

Understand the M/M/s queue model

An M/M/s queue is a queueing model where arrivals follow a Poisson process (Markovian inter-arrival time), service times follow an exponential distribution (Markovian service time), and there are "s" servers in the system.
02

Analyze the service rate depending on the number in the system

Assuming the service rate depends on the number of customers in the system in such a way that the process remains ergodic, the output process will still have the Markov property. Thus, while the service time distribution may vary with the number of customers, the output process will continue to have memoryless property.
03

Analyze the situation when 饾渾 > 蝉饾渿

In this case, the arrival rate (饾渾) is greater than the maximum possible service rate (蝉饾渿) for all customers in the system. Therefore, the queue will become unstable, and the average number of customers in the system and the waiting time in the system will indefinitely increase. In this case, the output process is unpredictable, as customers may not ever leave the system. In conclusion, when the service rate depends on the number of customers in the system, the output process will still follow the Markov property and will be ergodic if the conditions are maintained. However, when the arrival rate is higher than the product of the number of servers s and the service rate 饾渿 while keeping 饾渿 unchanged, the queue will become unstable, and it's impossible to predict the output process.

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.

M/M/s queue
An M/M/s queue is a foundational concept in queueing theory. It's a way to model systems where waiting lines (queues) are involved, such as customer service desks, server requests, or any situation with a sequence of arrivals that needs to be processed. The notation M/M/s indicates three important aspects:

  • M for "Memoryless" arrival process: Arrivals happen according to a Poisson process.
  • M for "Memoryless" service process: Service times are exponentially distributed.
  • s is the number of servers available to serve the customers or tasks.
This model is very flexible and widely used because it assumes that both the time between arrivals and the service time are random and follow specific memoryless distributions, making mathematical analysis convenient. Because of these assumptions, it's also known as an M/M/1 model when there is just one server handling requests.
Poisson process
The Poisson process is used to describe random events occurring over time. In queueing theory, it models how entities arrive at the system, like customers arriving at a shop or data packets hitting a server. A Poisson process has these distinguishing features:

  • Events occur independently of each other.
  • The average number of events in a given time period is known, but the exact number in that period is still random.
  • It represents a "Memoryless" process, meaning the time between consecutive events follows the exponential distribution.
The Poisson process is crucial for modeling arrival rates in an M/M/s queue. It helps predict likelihoods such as how many people might show up at a certain time, which aids in effectively managing resources.
exponential distribution
The exponential distribution is a probability distribution used to model the time between consecutive events in a Poisson process. In queueing systems, it's essential for modeling service times. The key features of an exponential distribution include:

  • Its "memoryless" property, meaning that the probability of an event happening in the future is independent of past events.
  • Characterized by its rate parameter, \( \lambda \,\), which is the reciprocal of the mean.
  • Offers a simple model where the variance equals the mean, simplifying many calculations.
This distribution indicates that service times are random, but they occur around a central average rate. Thus, in an M/M/s queue, as long as service intervals simulate this randomness, the system can remain effective and predictable under normal conditions.
Markov property
The Markov property is essential for understanding how systems are modeled and predicted in queueing and many other applications. It implies that the future state of a process only depends on the present state, not the sequence of events that preceded it. In simpler terms, the process is "memoryless."

In an M/M/s queue:

  • It ensures that knowing the current number of customers or jobs is enough to forecast future dynamics.
  • No need to know how long customers have been waiting or how many have previously been served.
  • Processes with the Markov property are generally easier to model mathematically, leading to simpler calculations and predictions.
Having the Markov property means that despite the complexity and randomness of individual actions within a system, overarching behavior can still be reliably anticipated and managed. This simplicity is why Markovian models, like those in M/M/s queues, are favored in analyzing and designing efficient service systems.

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

A population of organisms consists of both male and female members. In a small colony any particular male is likely to mate with any particular female in any time interval of length \(h\), with probability \(\lambda h+o(h) .\) Each mating immediately produces one offspring, equally likely to be male or female. Let \(N_{1}(t)\) and \(N_{2}(t)\) denote the number of males and females in the population at \(t .\) Derive the parameters of the continuous-time Markov chain \(\left\\{N_{1}(t), N_{2}(t)\right\\}\), i.e., the \(v_{i}, P_{i j}\) of Section \(6.2\).

Let \(Y\) denote an exponential random variable with rate \(\lambda\) that is independent of the continuous-time Markov chain \(\\{X(t)\\}\) and let $$ \bar{P}_{i j}=P[X(Y)=j \mid X(0)=i\\} $$ (a) Show that $$ \bar{P}_{i j}=\frac{1}{v_{i}+\lambda} \sum_{k} q_{i k} \bar{P}_{k j}+\frac{\lambda}{v_{i}+\lambda} \delta_{i j} $$ where \(\delta_{i j}\) is 1 when \(i=j\) and 0 when \(i \neq j\) (b) Show that the solution of the preceding set of equations is given by $$ \overline{\mathbf{P}}=(\mathbf{I}-\mathbf{R} / \lambda)^{-1} $$ where \(\overline{\mathbf{P}}\) is the matrix of elements \(\bar{P}_{i j}, \mathbf{I}\) is the identity matrix, and \(\mathbf{R}\) the matrix specified in Section \(6.8\). (c) Suppose now that \(Y_{1}, \ldots, Y_{n}\) are independent exponentials with rate \(\lambda\) that are independent of \(\\{X(t)\\}\). Show that $$ P\left[X\left(Y_{1}+\cdots+Y_{n}\right)=j \mid X(0)=i\right\\} $$ is equal to the element in row \(i\), column \(j\) of the matrix \(\overline{\mathbf{p}}^{n}\). (d) Explain the relationship of the preceding to Approximation 2 of Section \(6.8\).

There are \(N\) individuals in a population, some of whom have a certain infection that spreads as follows. Contacts between two members of this population occur in accordance with a Poisson process having rate \(\lambda .\) When a contact occurs, it is equally likely to involve any of the \(\left(\begin{array}{c}\mathrm{N} \\ 2\end{array}\right)\) pairs of individuals in the population. If a contact involves an infected and a noninfected individual, then with probability \(p\) the noninfected individual becomes infected. Once infected, an individual remains infected throughout. Let \(X(t)\) denote the number of infected members of the population at time \(t\). (a) Is \(\\{X(t), t \geqslant 0\\}\) a continuous-time Markov chain? (b) Specify its type. (c) Starting with a single infected individual, what is the expected time until all members are infected?

Consider a graph with nodes \(1,2, \ldots, n\) and the \(\left(\begin{array}{l}n \\\ 2\end{array}\right) \operatorname{arcs}(t, j), i \neq j, i, j,=1, \ldots, n\) (See Section 3.6.2 for appropriate definitions.) Suppose that a particle moves along this graph as follows: Events occur along the arcs \((i, j)\) according to independent Poisson processes with rates \(\lambda_{i j} .\) An event along arc \((i, j)\) causes that arc to become excited. If the particle is at node \(i\) at the moment that \((i, j)\) becomes excited, it instantaneously moves to node \(j, i, j=1, \ldots, n .\) Let \(P_{j}\) denote the proportion of time that the particle is at node \(j .\) Show that $$ P_{j}=\frac{1}{n} $$ Hint: Use time reversibility.

Four workers share an office that contains four telephones. At any time, each worker is either "working" or "on the phone." Each "working" period of worker \(i\) lasts for an exponentially distributed time with rate \(\lambda_{i}\), and each "on the phone" period lasts for an exponentially distributed time with rate \(\mu_{i}, i=1,2,3,4\). (a) What proportion of time are all workers "working"? Let \(X_{i}(t)\) equal 1 if worker \(i\) is working at time \(t\), and let it be 0 otherwise. Let \(\mathrm{X}(t)=\left(X_{1}(t), X_{2}(t), X_{3}(t), X_{4}(t)\right)\) (b) Argue that \(\\{\mathrm{X}(t), t \geqslant 0\\}\) is a continuous-time Markov chain and give its infinitesimal rates. (c) Is \(\\{\mathrm{X}(t)\\}\) time reversible? Why or why not? Suppose now that one of the phones has broken down. Suppose that a worker who is about to use a phone but finds them all being used begins a new "working" period. (d) What proportion of time are all workers "working"?
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Statistics

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.