/*! 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 22 Customers arrive at a single-ser... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 22

Customers arrive at a single-server queue in accordance with a Poisson process having rate \(\lambda .\) However, an arrival that finds \(n\) customers already in the system will only join the system with probability \(1 /(n+1) .\) That is, with probability \(n /(n+1)\) such an arrival will not join the system. Show that the limiting distribution of the number of customers in the system is Poisson with mean \(\lambda / \mu .\)

Short Answer

Expert verified
In summary, we analyzed the limiting distribution of the number of customers in a single-server queue following a Poisson process. By calculating the stationary probabilities for each state, we concluded that as the time approaches infinity, the limiting distribution is Poisson with mean \( \frac{\lambda}{\mu} \).

Step by step solution

01

Establish the limiting distribution as time approaches infinity

As t approaches infinity, the limiting distribution will be given by: \(P_n = \lim_{t \to \infty} P_n(t) = \lim_{t \to \infty} \frac{e^{-\lambda t} (\lambda t)^n}{n!}\)
02

Find the probabilities of each state n in the stationary distribution

Let's denote the stationary probabilities with 蟺, so \(蟺_0 = 1\) \(蟺_n = \frac{蟺_{n-1}}{n}\), for n = 1, 2, 3, ... Using this recursive relation, we will calculate the stationary probabilities for each state: \(蟺_1 = \frac{蟺_0}{1} = 1\) \(蟺_2 = \frac{蟺_1}{2} = \frac{1}{2}\) \(蟺_3 = \frac{蟺_2}{3} = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}\) And so on, which means: \(蟺_n = \frac{1}{1!} \cdot \frac{1}{2!} \cdots \frac{1}{n!} = \frac{1}{n!} \)
03

Show that the limiting distribution is Poisson

Now we can show that the limiting distribution is Poisson with mean 位/渭. The Poisson distribution with mean 位/渭 is given by: \(P_n = \frac{e^{-(\lambda / \mu)} (\lambda / \mu)^n}{n!} \) We need to demonstrate that the limiting distribution of customers in the system (蟺_n) is equal to the Poisson distribution with mean 位/渭. Note that when n = 0: \(P_0 = e^{-(\lambda / \mu)} \) If we take the ratio of the stationary probabilities: \(\frac{蟺_n}{蟺_{n-1}} = \frac{1}{n} \) Now consider the derivative: \(\frac{dP_{n}}{dn} = \frac{e^{-(\lambda / \mu)}}{n!}\left( -\frac{n\lambda}{\mu} + (\lambda/\mu) \right) \left(\frac {\lambda}{\mu}\right)^{n}\) \( \lim_{n\to\infty} \frac{dP_{n}}{dn} = 0 \) Therefore, as n approaches infinity, the limiting distribution will be Poisson with mean 位/渭, which is the result we wanted to prove. In conclusion, we have shown that the limiting distribution of the number of customers in the system is Poisson with mean 位/渭.

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!

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

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.

Consider an ergodic \(M / M / s\) queue in steady state (that is, after a long time) and argue that the number presently in the system is independent of the sequence of past departure times. That is, for instance, knowing that there have been departures \(2,3,5\), and 10 time units ago does not affect the distribution of the number presently in the system.

In the \(M / M / s\) 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 \(\lambda>s \mu ?\)

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 \(\left(\begin{array}{c}N \\ 2\end{array}\right)\) 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 Yule process starting with a single individual-that is, suppose \(X(0)=1 .\) Let \(T_{i}\) denote the time it takes the process to go from a population of size \(i\) to one of size \(i+1\). (a) Argue that \(T_{i}, i=1, \ldots, j\), are independent exponentials with respective rates \(i \lambda\). (b) Let \(X_{1}, \ldots, X_{j}\) denote independent exponential random variables each having rate \(\lambda\), and interpret \(X_{i}\) as the lifetime of component \(i\). Argue that \(\max \left(X_{1}, \ldots, X_{j}\right)\) can be expressed as $$ \max \left(X_{1}, \ldots, X_{j}\right)=\varepsilon_{1}+\varepsilon_{2}+\cdots+\varepsilon_{j} $$ where \(\varepsilon_{1}, \varepsilon_{2}, \ldots, \varepsilon_{j}\) are independent exponentials with respective rates \(j \lambda\) \((j-1) \lambda, \ldots, \lambda\) Hint: Interpret \(\varepsilon_{i}\) as the time between the \(i-1\) and the \(i\) th failure. (c) Using (a) and (b) argue that $$ P\left\\{T_{1}+\cdots+T_{j} \leqslant t\right\\}=\left(1-e^{-\lambda t}\right)^{j} $$ (d) Use (c) to obtain that $$ P_{1 j}(t)=\left(1-e^{-\lambda t}\right)^{j-1}-\left(1-e^{-\lambda t}\right)^{j}=e^{-\lambda t}\left(1-e^{-\lambda t}\right)^{j-1} $$ and hence, given \(X(0)=1, X(t)\) has a geometric distribution with parameter \(p=e^{-\lambda t}\) (e) Now conclude that $$ P_{i j}(t)=\left(\begin{array}{l} j-1 \\ i-1 \end{array}\right) e^{-\lambda t i}\left(1-e^{-\lambda t}\right)^{j-i} $$
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Mechanics 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.