/*! 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 26 Consider an ergodic \(M / M / s\... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

Short Answer

Expert verified
In an ergodic M/M/s queue system in steady state, the number of customers currently in the system is independent of the sequence of past departure times. This is due to the memoryless property of the Poisson process for customer arrivals and exponentially distributed service times for departures. The ergodicity of the system ensures that after a sufficiently long time, the distribution of the number of customers in the system reaches equilibrium regardless of previous events, and therefore is independent of past departure times.

Step by step solution

01

Understand the M/M/s queue system

The M/M/s queue model is a queuing system where arrivals follow a Poisson process with rate λ, service times are exponentially distributed with rate μ, and there are s service channels. Ergodicity refers to the property that the long-run average number of customers in the system equals the time average number of customers in the system.
02

Explain Poisson Process

The Poisson process is a model for events happening at a constant average rate, λ. In the M/M/s queue system, the arrival of customers follows a Poisson process. This process has the property of being memoryless, meaning that the time until the next customer arrives is independent of the time when the previous customer arrived.
03

Establish Ergodicity

An ergodic system implies that after a sufficiently long time, the distribution of the number n of customers in the system in steady state reaches its equilibrium. That is, the probability distribution of the number of customers in the system, P(n), stabilizes and becomes stationary. This occurs regardless of the initial conditions or previous events in the system.
04

Argue independence using Poisson process properties

As we mentioned earlier, arrivals follow a Poisson process and are memoryless. This means that the time until the next arrival is independent of the time at which the previous arrival happened. Similarly, the service times are exponentially distributed with rate μ, which are also memoryless. This means that the time until the next departure is independent of the time when the previous departure happened.
05

Show independence from past departure times

To prove the independence of the number presently in the system from the sequence of past departure times, let's consider the information given about past departures. Knowing that there have been departures 2, 3, 5, and 10 time units ago does not impact the arrival and service rate because as stated earlier, both arrivals and services are memoryless. Due to the Poisson process memoryless property, future customer arrivals and services are independent of past arrival and departure times. Since the distribution of the number of customers currently in the system has reached its steady-state equilibrium, these past departure times do not have any effect on the distribution of the number presently in the system. In conclusion, using the ergodicity of the M/M/s queue system and the memoryless property of the Poisson process for arrivals and services, the number of customers presently in the system is independent of the sequence of past departure times.

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 91Ó°ÊÓ!

Key Concepts

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

M/M/s Queue
An M/M/s queue represents a classic model in queuing theory used to understand systems with multiple servers. "M" stands for memoryless, indicating exponential service time and inter-arrival time distributions. The number "s" denotes the number of parallel service channels or servers.
  • Arrival Process: This system assumes that arrivals occur according to a Poisson process, characterized by a rate \( \lambda \) (lambda), which specifies the average number of arrivals per time unit.
  • Service Process: The service times follow an exponential distribution with a rate \( \mu \) (mu), describing how quickly customers can be served.
  • Multiple Servers: The "s" in M/M/s indicates there are "s" servers available, processing requests, and dispatching customers.
In this framework, customers may wait in line when all servers are busy. The M/M/s queue is particularly useful for scenarios like call centers and computing systems, where resources are allocated based on customer demand.
Poisson Process
A Poisson process is fundamental in modeling random events occurring over time. In queuing theory, it describes how arrivals enter a system. One of its key characteristics is the memoryless property, profoundly impacting system dynamics.
  • Constant Rate: Events occur at an average rate \( \lambda \), unaffected by when the last event occurred. This describes a constant, independent occurrence pattern.
  • Memoryless Nature: The time until the next event is not dependent on any past events. Meaning, even if we know when the last customer arrived, it doesn't give us more information about when the next will.
  • Exponential Inter-arrival Times: The time between each event follows an exponential distribution, providing a constant hazard rate.
These properties make the Poisson process ideal for modeling arrivals to an M/M/s queue, supporting a stable, predictable environment.
Ergodicity
Ergodicity in queuing systems, particularly in M/M/s models, ensures stability over time. A system described as ergodic reaches a point where its time average and long-run average outcomes converge.
  • Equilibrium State: Ergodic systems eventually stabilize, meaning after sufficient time, the probability distribution of the number of items in the queue remains constant.
  • Independence from Initial Conditions: Regardless of how the system began—how many were in the queue initially—over time, it's predicted to mirror the steady-state distribution.
  • Average Calculations: In practice, this allows us to compute reliable average metrics like wait times or queue lengths that align with real observations.
This concept is essential because it justifies the assumption that past events don't affect the system's future state, further validating the M/M/s queue's memoryless nature.
Steady-State Distribution
The steady-state distribution of a queuing system refers to the probability distribution of the number of customers in the system when it has reached stability after a long time.
  • Achieving Equilibrium: Once a system reaches steady-state, the number of customers in the queue and being served follows a consistent distribution pattern. It no longer fluctuates unpredictably.
  • Independence from History: This state implies that the system is unaffected by past arrival or departure times, as both are memoryless due to Poisson and exponential characteristics.
  • Importance in Analysis: Steady-state allows analysts to make accurate, long-term predictions about performance metrics like wait times and utilization without concern for initial disorders or past fluctuations.
In conclusion, a system’s steady-state is crucial for determining its reliability and consistency, ensuring predictions and plans are grounded in stable reality.

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 two machines that are maintained by a single repairman. Machine \(i\) functions for an exponential time with rate \(\mu_{1}\) before breaking down, \(i=1,2\). The repair times (for either machine) are exponential with rate \(\mu\). Can we analyze this as a birth and death process? If so, what are the parameters? If not, how can we analyze it?

Consider a set of \(n\) machines and a single repair facility to service these machines. Suppose that when machine \(i, i=1, \ldots, n\), fails it requires an exponentially distributed amount of work with tate \(\mu_{1}\) to repair it. The repair facility divides its efforts equally among all failed machines in the sense that whenever there are \(k\) failed machines cach one receives work at a rate of \(1 / k\) per unit time. If there are a total of \(r\) working machines, including machine \(i\), then \(i\) fails at an instantaneous rate \(\lambda_{i} / r\). (a) Define an appropriate state space so as to be able to analyze the above system as a continuous-time Markov chain. (b) Give the instantaneous transition rates (that is, give the \(q_{j}\) ). (c) Write the time reversibility equations. (d) Find the limiting probabilities and show that the process is time reversible.

Four workers share an office that contains four telephones. At any time, cach worker is either "working" or "on the phone." Each "working" period of worker \(i\) lasts for an exponentially distributed time with rate \(\lambda_{i}\). and each "on the phone" period lasts for an exponentially distributed time with rate \(\mu_{i}, i=1,2,3,4\), (a) What proportion of time are all workers "working"? Let \(X_{i}(t)\) equal 1 if worker \(i\) is working at time \(t\), and let it be 0 otherwise. Let \(\mathbf{X}(t)=\left(X_{1}(t), X_{2}(t), X_{3}(t), X_{4}(t)\right)\). (b) Argue that \(\\{\mathbf{X}(t), t \geq 0]\) is a continuous-time Markov chain and give its infinitesimal rates. (c) Is \(\\{\mathbf{X}(t) \mid\) time reversible? Why or why not? Suppose now that one of the phones has broken down. Suppose that a worker who is about to use a phone but finds thern all being used begins a new "working'" period. (d) What proportion of time are all workers "working'r?

Consider a graph with nodes \(1,2, \ldots, n\) and the \(\left(\begin{array}{l}n \\\ 2\end{array}\right)\) arcs \((i, j)\). \(l \neq J, i, j,=1, \ldots, n .\) (See Section 3.6.2 for appropriate definitions.) Suppose that a particle moves along this graph as follows: Events occur along the arcs \((i, J)\) according to independent Poisson processes with rates \(\lambda_{U}\). An event along arc \((i, j)\) causes that arc to become excited. If the particle is at node \(i\) at the moment that \((i, j)\) becomes excited, it instantaneously moves to node \(j ; i, j=1, \ldots, n\). Let \(P_{j}\) denote the proportion of time that the particle is at node \(j\). Show that $$ P_{j}=\frac{1}{n} $$

Fach individual in a biological population is assumed to give birth at an exponential rate \(\lambda\), and to die at an exponential rate \(\mu\). In addition, there is an exponential rate of increase \(\theta\) due to immigration. However, immigration is not allowed when the population size is \(N\) or larger. (a) Set this up as a birth and death model. (b) If \(N=3,1=\theta=\lambda, \mu=2\), determine the proportion of time that immigration is restricted.

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.