Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 𝜆 is greater than s𝜇 with 𝜇 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 𝜆 > s𝜇

In this case, the arrival rate (𝜆) is greater than the maximum possible service rate (s𝜇) 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 𝜇 while keeping 𝜇 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 Vaia!

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 single repairperson looks after both machines 1 and \(2 .\) Each time it is repaired, machine \(i\) stays up for an exponential time with rate \(\lambda_{i}, i=1,2 .\) When machine \(i\) fails, it requires an exponentially distributed amount of work with rate \(\mu_{i}\) to complete its repair. The repairperson will always service machine 1 when it is down. For instance, if machine 1 fails while 2 is being repaired, then the repairperson will immediately stop work on machine 2 and start on \(1 .\) What proportion of time is machine 2 down?

Each time a machine is repaired it remains up for an exponentially distributed time with rate \(\lambda\). It then fails, and its failure is either of two types. If it is a type 1 failure, then the time to repair the machine is exponential with rate \(\mu_{1}\); if it is a type 2 failure, then the repair time is exponential with rate \(\mu_{2} .\) Each failure is, independently of the time it took the machine to fail, a type 1 failure with probability \(p\) and a type 2 failure with probability \(1-p\). What proportion of time is the machine down due to a type 1 failure? What proportion of time is it down due to a type 2 failure? What proportion of time is it up?

In Example \(6.20\), we computed \(m(t)=E[O(t)]\), the expected occupation time in state 0 by time \(t\) for the two-state continuous-time Markov chain starting in state 0\. Another way of obtaining this quantity is by deriving a differential equation for it. (a) Show that $$ m(t+h)=m(t)+P_{00}(t) h+o(h) $$ (b) Show that $$ m^{\prime}(t)=\frac{\mu}{\lambda+\mu}+\frac{\lambda}{\lambda+\mu} e^{-(\lambda+\mu) t} $$ (c) Solve for \(m(t)\).

Potential customers arrive at a single-server station in accordance with a Poisson process with rate \(\lambda .\) However, if the arrival finds \(n\) customers already in the station, then he will enter the system with probability \(\alpha_{n}\). Assuming an exponential service rate \(\mu\), set this up as a birth and death process and determine the birth and death rates.

The following problem arises in molecular biology. The surface of a bacterium consists of several sites at which foreign molecules-some acceptable and some not-become attached. We consider a particular site and assume that molecules arrive at the site according to a Poisson process with parameter \(\lambda\). Among these molecules a proportion \(\alpha\) is acceptable. Unacceptable molecules stay at the site for a length of time that is exponentially distributed with parameter \(\mu_{1}\), whereas an acceptable molecule remains at the site for an exponential time with rate \(\mu_{2}\). An arriving molecule will become attached only if the site is free of other molecules. What percentage of time is the site occupied with an acceptable (unacceptable) molecule?

See all solutions

Recommended explanations on Math Textbooks

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.

Sign-up for free