Logout Open In App
Open In App
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

Chapter 8: Problem 7

Show that \(W\) is smaller in an \(M / M / 1\) model having arrivals at rate \(\lambda\) and service at rate \(2 \mu\) than it is in a two-server \(M / M / 2\) model with arrivals at rate \(\lambda\) and with each server at rate \(\mu .\) Can you give an intuitive explanation for this result? Would it also be true for \(W_{Q} ?\)

Short Answer

Expert verified
By calculating the average time spent in the system (\(W\)) for both the \(M/M/1\) and \(M/M/2\) queuing models, we find that: \[ W_{M/M/1}=\frac{1}{2\mu-\lambda} < W_{M/M/2}=\frac{1}{\mu}+\frac{\lambda}{2\mu^2-\lambda\mu} \] The intuitive explanation for this result is that, in the \(M/M/1\) model, there is a single, faster server working at rate \(2\mu\), while in the \(M/M/2\) model, two customers can be served simultaneously at a rate of \(\mu\). This leads to more customers spending time in the system in the \(M/M/2\) model, making \(W_{M/M/2}\) larger. However, this conclusion might not hold for the average waiting time in the queue (\(W_Q\)), as the ability to serve two customers simultaneously in the \(M/M/2\) model can reduce the waiting time in the queue compared to the \(M/M/1\) model.

Step by step solution

01

Computing W for the M/M/1 model

For an \( M/M/1 \) queuing model, the formula to calculate \( W \), the average time spent in the system, is given by: \[ W=\frac{1}{\mu-\lambda} \] In the given scenario, the service rate is \( 2\mu \), hence \( W \) would be calculated as: \[ W_{M/M/1}=\frac{1}{2\mu-\lambda} \]
02

Computing W for the M/M/2 model

For an \( M/M/2 \) queuing model, the formula to calculate \( W \) is given by: \[ W=\frac{1}{\mu}+\frac{\lambda}{2\mu(\mu-\lambda)} \] Here, each server has a service rate of \( \mu \), therefore, \( W \) can be calculated as: \[ W_{M/M/2}=\frac{1}{\mu}+\frac{\lambda}{2\mu^2-\lambda\mu} \]
03

Comparing W for both models

To show that \( W \) is smaller in the \( M/M/1 \) model than in the \( M/M/2 \) model given the service rates, compare the two calculated values of \( W \): \[ W_{M/M/1} < W_{M/M/2} \]
04

Providing an intuitive explanation

An intuitive explanation roots in how the queuing works in these systems. In the single server system, even if the server is faster (working at a rate of \( 2\mu \)), there is only one customer being serviced at a time, while others wait. However, in the second system two customers can be serviced simultaneously, resulting in more customers spending more time in the system on an average, thus making \( W_{M/M/2} \) larger.
05

Addressing WQ

For the time spent in the queue only, \( W_{Q} \), the conclusions might not necessarily hold anymore. Since in the \( M/M/2 \) model two customers can be serviced simultaneously, it reduces the waiting time in the queue in comparison to the \( M/M/1 \) model, despite an overall larger total time spent in the system. Thus, it wouldn't necessarily be true that \( W_{Q \; M/M/1} < W_{Q \; M/M/2} \).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

M/M/1 Queue
The M/M/1 queue represents a fundamental model in queueing theory that assumes a single server providing services to arriving customers. Each 'M' refers to a Markovian process, also known as a 'memoryless' process. The first 'M' denotes the arrival times which follow an exponential distribution, and the second 'M' signifies the service times which are also exponentially distributed. The '1' indicates there is just one server.

In this simple yet widely applicable model, the average time a customer spends in the system, often denoted as 'W', is calculated using the formula: \[\begin{equation} W=\frac{1}{\text{service rate} - \text{arrival rate}} \end{equation}\] For instance, if the server is twice as fast as the arrival rate (2\text{service rate}), 'W' in the M/M/1 queue will be shorter compared to a scenario where the service rate is equal to the arrival rate. This demonstrates how service efficiency can dramatically reduce customer wait times.
M/M/2 Queue
Contrary to the M/M/1 queue, the M/M/2 queue model deals with a system having two servers serving the arriving customers. It's a more complex system as it involves additional considerations for how the two servers work together. In this scenario, the formula to compute the average time a customer spends in the system (both waiting and being serviced) is a bit more involved and is given by:\[\begin{equation} W=\frac{1}{\text{service rate}}+\frac{\text{arrival rate}}{2(\text{service rate})^2-\text{arrival rate}\times \text{service rate}} \end{equation}\] Here, the presence of two servers might suggest that the system can handle the customer flow more effectively, but it also introduces the possibility of customers spending more time in the system due to the dynamics of multiple-server coordination and potential idle periods for the servers.
Average Time in the System
The average time a customer spends in the system, denoted as 'W', encompasses both the time spent waiting for the service and the time spent actually receiving the service. This metric is critical for understanding the efficiency and performance of a queuing system. In the context of M/M/1 versus M/M/2 queues, 'W' will vary based on the number of servers and their service rates relative to the arrival rate of customers. Calculating and comparing 'W' for different models can highlight the impact service rates and the number of servers have on overall wait times. Higher service rates and more servers might seem to reduce 'W', but one must consider how those servers interact and their utilization rate to get the complete picture of a system's efficacy.
Service Rate
Service rate, often represented by the symbol 'μ' (mu), is an intrinsic part of the formula used to calculate the average time in the system. It denotes the average number of customers that a server can serve per time unit. The higher the service rate, the faster a customer is likely to be served, which, assuming a constant arrival rate, generally reduces the average time a customer spends in the system. An adept understanding of the service rate is essential when comparing different queuing models or optimizing a system for efficiency. Knowing the service rate allows one to manage and predict the system's capacity to handle varying levels of customer arrivals, ultimately aiming for a balance that minimizes wait times and avoids overburdening the service facility.

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

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 \(1 ?\)

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 $$ W_{Q}^{2}=W_{Q}^{2} \text { (nonpreemptive) }+\frac{\lambda_{1} E\left[S_{1}\right] E\left[S_{2}\right]}{1-\lambda_{1} E\left[S_{1}\right]} $$

Customers arrive at a single-server station in accordance with a Poisson process with rate \(\lambda\). All arrivals that find the server free immediately enter service. All service times are exponentially distributed with rate \(\mu\). An arrival that finds the server busy will leave the system and roam around "in orbit" for an exponential time with rate \(\theta\) at which time it will then return. If the server is busy when an orbiting customer returns, then that customer returns to orbit for another exponential time with rate \(\theta\) before returning again. An arrival that finds the server busy and \(N\) other customers in orbit will depart and not return. That is, \(N\) is the maximum number of customers in orbit. (a) Define states. (b) Give the balance equations. In terms of the solution of the balance equations, find (c) the proportion of all customers that are eventually served; (d) the average time that a served customer spends waiting in orbit.

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 \(S_{1}>Y\)

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?
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

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.

Sign-up for free