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

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.

Short Answer

Expert verified
We define the state of the system as a tuple (i, j), where 'i' represents the server status and 'j' represents the number of customers in orbit. Using steady-state probabilities, we can write the balance equations for a system with given parameters. To find the proportion of served customers, S, and average time a served customer spends waiting in orbit, W, we can use the following formulas: \(S = \frac{\mu}{\lambda} \sum_{j=0}^{N} P(1,j) \) \(W = \frac{ \sum_{j=1}^{N} jP(1,j)}{S\theta} \)

Step by step solution

01

Define states

We can define the state of the system as a tuple (i, j), where 'i' is the status of the server (0 if the server is free and 1 if the server is busy) and 'j' is the number of customers in orbit (from 0 to N).
02

Balance equations

Let \(P(i, j)\) represent the steady-state probability of being in state (i, j). Using the steady-state probabilities, we can write the balance equations for a system with 1 server, N customers in orbit, and λ, μ, and θ as given parameters. For the case when the server is free (i=0) and there are no customers in orbit (j=0): \( \lambda P(0,0) = \mu P(1,0) \) For the case when the server is busy (i=1) and there are no customers in orbit (j=0): \( (\lambda + \mu)P(1,0) = \lambda P(0,0) + \theta P(1,1) \) For the case when the server is busy (i=1) and there are j customers in orbit (1 ≤ j ≤ N-1): \( (\lambda + \theta + \mu)P(1,j) = \theta P(1,j+1) + \lambda P(0,j) + \theta P(1,j-1) \) For the last customer in orbit (j=N): \( (\lambda + \theta)P(1,N) = \lambda P(0,N) + \theta P(1,N-1) \)
03

Proportion of all customers that are eventually served

To find the proportion of all customers that are eventually served, we denote by S the proportion served. This can be found using the balance equations, where the arrival rate multiplied by served proportion equals the service rate multiplied by the server busy state. \(S = \frac{\mu}{\lambda} \sum_{j=0}^{N} P(1,j) \)
04

Average time a served customer spends waiting in orbit

Let W be the average time a served customer spends waiting in orbit. Using Little's Law, which relates the mean time spent waiting in orbit, the mean number of customers in orbit, and the arrival rate in the orbit, we can find the average time a served customer spends waiting in orbit: \(W = \frac{ \sum_{j=1}^{N} jP(1,j)}{S\theta} \)

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!

Key Concepts

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

Stochastic models
Understanding the complexity of real-world processes often requires sophisticated mathematical approaches like stochastic models. These models are equipped to handle uncertainty and randomness as core aspects of the phenomena they describe. They employ the use of probabilities to forecast the outcomes of a system that is under the influence of random factors. In our exercise, the customers arriving at a service station and the uncertain time they spend in orbit before potentially returning to the server are perfect examples of a scenario that stochastic modeling can clarify. This provides insight into the functioning of the system over time, allowing for predictions and informed decision-making.
Balance equations
The heart of understanding stochastic models often lies in what are known as balance equations. These equations represent equilibrium conditions where the amount of something entering a state is equal to the amount leaving it. In the context of our exercise, balance equations are formulated by considering the rates of customers entering and leaving each state of the system. The steady-state probabilities, which describe the long-term behavior of the system, can be found by solving these balance equations. The significance of achieving balance implies that the system has reached a point where the distribution of probabilities over its states remains constant over time.
Exponential distribution
An integral part of many stochastic models is the exponential distribution, which is widely used in scenarios with a memoryless property – meaning the future is independent of the past. In our textbook example, service times and the time customers spend in orbit are assumed to follow exponential distributions characterized by the rates \( \mu \) and \( \theta \) respectively. The key feature of the exponential distribution is its constant hazard rate, which is particularly useful when modeling the time until an event, such as the time until the next customer arrives or the time until a customer in orbit returns.
Steady-state probability
The notion of steady-state probability is central to understanding complex systems that evolve over time. In a steady-state, the system's properties are unchanging because the processes of entering and leaving states are balanced. In the context of the queuing system, steady-state probabilities tell us the likelihood of finding the system in a particular state (e.g., the server being busy and a certain number of customers in orbit) after a long period. These probabilities are essential for deriving other performance measures of the system, such as the proportion of customers who will eventually be served or the average time customers spend in specific parts of the system.
Little's Law
When analyzing queueing systems, an indispensable tool is Little's Law, a formula that provides a relationship between the average number of items in a queuing system, the average time an item spends in the system, and the average rate at which items arrive. In our exercise, Little's Law is used to calculate the average amount of time a served customer spends waiting in orbit, which involves understanding the flow of customers through the system and how long they tend to stay. The application of Little’s Law offers a simple yet powerful way to quantify the experience of customers within a queueing network, encapsulating complex dynamic interactions in a neat, empirically verifiable equation.

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

Compare the \(M / G / 1\) system for first-come, first-served queue discipline with one of last-come, first-served (for instance, in which units for service are taken from the top of a stack). Would you think that the queue size, waiting time, and busyperiod distribution differ? What about their means? What if the queue discipline was always to choose at random among those waiting? Intuitively, which discipline would result in the smallest variance in the waiting time distribution?

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 \(W_{Q} ?\)

Explain how a Markov chain Monte Carlo simulation using the Gibbs sampler can be utilized to estimate (a) the distribution of the amount of time spent at server \(j\) on a visit. Hint: Use the arrival theorem. (b) the proportion of time a customer is with server \(j\) (i.e., either in server \(j\) 's queue or in service with \(j\) ).

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 \(1 ?\)

A supermarket has two exponential checkout counters, each operating at rate \(\mu .\) Arrivals are Poisson at rate \(\lambda\). The counters operate in the following way: (i) One queue feeds both counters. (ii) One counter is operated by a permanent checker and the other by a stock clerk who instantaneously begins checking whenever there are two or more customers in the system. The clerk returns to stocking whenever he completes a service, and there are fewer than two customers in the system. (a) Find \(P_{n}\), proportion of time there are \(n\) in the system. (b) At what rate does the number in the system go from 0 to \(1 ?\) From 2 to \(1 ?\) (c) What proportion of time is the stock clerk checking? Hint: Be a little careful when there is one in the system.

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