/*! 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 1 For the \(M / M / 1\) queue, com... [FREE SOLUTION] | 91影视

91影视

For the \(M / M / 1\) queue, compute (a) the expected number of arrivals during a service period and (b) the probability that no customers arrive during a service period. Hint: "Condition."

Short Answer

Expert verified
The expected number of arrivals during a service period in an M/M/1 queue system is given by: \(蟻 = \frac{位}{渭}\) The probability that no customers arrive during a service period is calculated as: \(P_0 ( \frac{1}{渭} ) = e^{ -蟻}\)

Step by step solution

01

Understand the M/M/1 queue model

The M/M/1 queue model is a single-server queueing model, where both the inter-arrival times and service times are exponentially distributed. Here, 位 represents the arrival rate, and 渭 represents the service rate.
02

Define the given parameters

In the problem, we have given the following parameters: - Arrival rate (位): It represents the rate at which customers arrive in the system. - Service rate (渭): This represents the rate at which the server works on the customers in the system.
03

Calculate the expected number of arrivals during a service period (Part a)

For calculating the expected number of arrivals during a service period, we can use the concept of compound Poisson distribution. We must find the parameter of this distribution, 蟻, which is found by dividing the arrival rate, 位, by the service rate, 渭. 1. Calculate 蟻: \( 蟻 = \frac{位}{渭} \) 2. Expected number of arrivals during a service period: Since the arrival process is a Poisson process, the expected number of arrivals during a service period is equal to 蟻.
04

Find the probability that no customers arrive during a service period (Part b)

To find this probability, we can use the property of the Poisson process that states the probability of no arrivals during an interval of time of length t is given by the exponential distribution: \(P_0 (t) = e^{ -位t} \) In this case, the length of the interval is the service time 't,' which is exponentially distributed with parameter 渭. Therefore, the service time's mean duration is given by \( \frac{1}{渭} \). 1. Calculate the probability that no customers arrive during a service period: \(P_0 ( \frac{1}{渭} ) = e^{ -位( \frac{1}{渭} )} = e^{ -蟻} \)

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 paramount to understanding complex systems where resources are shared among various users or tasks, such as customers waiting in line for service. This theory examines the behavior, dynamics, and flow of these queues to make predictions or improve efficiency. In our case, the M/M/1 queue model is a fundamental queueing system with a single-server serving customers one at a time, and both the arrival of customers and the service process follow a random pattern. By analyzing this model, we can draw conclusions about the system's performance, such as how long customers have to wait or the expected number of customers in the system at any given time.

Queues are described by their arrival process (input), service mechanism (how customers are served), the number of servers, and the capacity of the system (if there is a limit to the queue length). The M/M/1 queue is a basic model but serves as a stepping stone to comprehend more complex queueing systems.
Poisson Process
The Poisson process is a stochastic process that represents events occurring randomly over time. These events are independent of each other, and their occurrence rate is constant鈥攖wo fundamental characteristics of the process. In the context of queueing, the Poisson process is often used to model random arrivals of customers to a queue, where the term '位' denotes the average number of arrivals per time unit. This rate is crucial for calculating various metrics of the queue's performance, such as the average wait time or the probability of the queue being empty.

Furthermore, the Poisson process has a key property that the number of events occurring in a fixed interval of time follows a Poisson distribution, allowing us to predict probabilities of certain numbers of arrivals during specified intervals.
Exponential Distribution
The exponential distribution is inherently linked to the Poisson process and is characterized by its 'memoryless' property, meaning future probabilities are not influenced by past events. It is frequently used to model the time between successive events in a Poisson process, such as the time between customer arrivals or service times in a queue. In the context of the M/M/1 queue model, both inter-arrival times (time between arrivals) and service times are exponentially distributed, which simplifies the analysis as the time until the next event is always the same on average, regardless of what happened before.
Expected Value
The expected value is a fundamental concept in probability and statistics, often interpreted as the long-term average or 'mean' value of a random variable after repeating an experiment many times. It is crucial for any probabilistic analysis since it provides the central tendency of a random variable's possible outcomes weighted by their probabilities. In our M/M/1 queue analysis鈥攑art (a) of the exercise鈥攚e calculated the expected number of arrivals during a service period, which gives insight into how busy the system will be on average and helps in assessing the system's capacity requirements.
Probability
Probability is the measure of how likely an event is to occur and lies at the heart of queueing theory and related models such as the M/M/1 queue. In related terms, probability allows us to quantify the uncertainty of various outcomes, such as the likelihood of a customer arriving during a specific period, or in our case鈥攑art (b) of the exercise鈥攖he chance that no customers arrive during a service period. Understanding these probabilities is essential for managers and system designers to make informed decisions concerning resource allocation and service levels.

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

Consider the priority queueing model of Section \(8.6 .2\) but now suppose that if a type 2 customer is being served when a type 1 arrives then the type 2 customer is bumped out of service. This is called the preemptive case. Suppose that when a bumped type 2 customer goes back in service his service begins at the point where it left off when he was bumped. (a) Argue that the work in the system at any time is the same as in the nonpreemptive case. (b) Derive \(W_{Q}^{1}\) Hint: How do type 2 customers affect type 1 s? (c) Why is it not true that $$ V_{Q}^{2}=\lambda_{2} E\left[S_{2}\right] W_{Q}^{2} $$ (d) Argue that the work seen by a type 2 arrival is the same as in the nonpreemptive case, and so \(W_{Q}^{2}=W_{Q}^{2}\) (nonpreemptive) \(+E[\) extra time \(]\) where the extra time is due to the fact that he may be bumped. (e) Let \(N\) denote the number of times a type 2 customer is bumped. Why is $$ E[\text { extra time } \mid \mathrm{N}]=\frac{N E\left[S_{1}\right]}{1-\lambda_{1} E\left[S_{1}\right]} $$ Hint: When a type 2 is bumped, relate the time until he gets back in service to a "busy period." (f) Let \(S_{2}\) denote the service time of a type \(2 .\) What is \(E\left[N \mid S_{2}\right] ?\) (g) Combine the preceding to obtain $$ W_{Q}^{2}=W_{Q}^{2} \text { (nonpreemptive) }+\frac{\lambda_{1} E\left[S_{1}\right] E\left[S_{2}\right]}{1-\lambda_{1} E\left[S_{1}\right]} $$

Consider a \(M / G / 1\) system with \(\lambda E[S]<1\). (a) Suppose that service is about to begin at a moment when there are \(n\) customers in the system. (i) Argue that the additional time until there are only \(n-1\) customers in the system has the same distribution as a busy period. (ii) What is the expected additional time until the system is empty? (b) Suppose that the work in the system at some moment is \(A\). We are interested in the expected additional time until the system is empty - call it \(E[T]\). Let \(N\) denote the number of arrivals during the first \(A\) units of time. (i) Compute \(E[T \mid N]\). (ii) Compute \(E[T]\).

Suppose we want to find the covariance between the times spent in the system by the first two customers in an \(M / M / 1\) queueing system. To obtain this covariance, let \(S_{i}\) be the service time of customer \(i, i=1,2\), and let \(Y\) be the time between the two arrivals. (a) Argue that \(\left(S_{1}-Y\right)^{+}+S_{2}\) is the amount of time that customer 2 spends in the system, where \(x^{+}=\max (x, 0)\) (b) Find \(\operatorname{Cov}\left(S_{1},\left(S_{1}-Y\right)^{+}+S_{2}\right)\). Hint: Compute both \(E\left[(S-Y)^{+}\right]\) and \(E\left[S_{1}\left(S_{1}-Y\right)^{+}\right]\) by conditioning on whether \(S_{1}>Y\)

Customers arrive at a two-server station in accordance with a Poisson process with a rate of two per hour. Arrivals finding server 1 free begin service with that server. Arrivals finding server 1 busy and server 2 free begin service with server 2. Arrivals finding both servers busy are lost. When a customer is served by server 1 , she then either enters service with server 2 if 2 is free or departs the system if 2 is busy. \(\mathrm{A}\) customer completing service at server 2 departs the system. The service times at server 1 and server 2 are exponential random variables with respective rates of four and six per hour. (a) What fraction of customers do not enter the system? (b) What is the average amount of time that an entering customer spends in the system? (c) What fraction of entering customers receives service from server \(1 ?\)

Customers arrive at a two-server system according to a Poisson process having rate \(\lambda=5\). An arrival finding server 1 free will begin service with that server. An arrival finding server 1 busy and server 2 free will enter service with server \(2 .\) An arrival finding both servers busy goes away. Once a customer is served by either server, he departs the system. The service times at server \(i\) are exponential with rates \(\mu_{i}\), where \(\mu_{1}=4, \mu_{2}=2\) (a) What is the average time an entering customer spends in the system? (b) What proportion of time is server 2 busy?

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.