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

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 \(S_{1}>Y\)

Short Answer

Expert verified
In summary, the covariance between the times spent in the system by the first two customers in an M/M/1 queueing system can be found as follows: 1. Understand the relationship between \(S_1\), \(S_2\), and \(Y\): \(\left(S_1-Y\right)^+ + S_2\) is the total time spent by the second customer in the system. 2. Compute \(E\left[(S-Y)^+\right]\) by conditioning on whether \(S_1>Y\): \(E\left[(S-Y)^+\right] = \int_0^\infty \int_0^{s} (s-y) \lambda e^{-\lambda s} \mu e^{-\mu y} dy ds\). 3. Compute \(E\left[S_1\left(S_1-Y\right)^+\right]\) by conditioning on whether \(S_1>Y\): \(E\left[S_1\left(S_1-Y\right)^+\right] = \int_0^\infty \int_0^{s} s(s-y) \lambda e^{-\lambda s} \mu e^{-\mu y} dy ds\). 4. Calculate the covariance: \(Cov\left(S_1, \left(S_1-Y\right)^+ + S_2\right) = \left( E\left[(S-Y)^+\right] + E\left[S_1\left(S_1-Y\right)^+\right] \right) - \frac{1}{\mu} \left( E\left[(S-Y)^+\right] + \frac{1}{\mu} \right)\).

Step by step solution

01

Understanding the relationship between \(S_1\), \(S_2\), and \(Y\)

The time spent by the second customer in the queue depends on the service time of the first customer (\(S_1\)) and the service time of the second customer. If the arrival time between the two customers (\(Y\)) is less than the service time of the first customer, the second customer will have to wait in the queue. Otherwise, the second customer will not need to wait as the first customer is served. Hence, \(\left(S_1-Y\right)^+\) represents the waiting time of the second customer in the queue. If \(S_1>Y\), the second customer will wait for \(S_1-Y\) time, otherwise, the waiting time is zero. Then, adding the second customer's service time \(S_2\) gives the total time spent by the second customer in the system: \(\left(S_1-Y\right)^+ + S_2\).
02

Calculating the expectation of \(\left(S-Y\right)^+\)

To find the expectation of \(\left(S-Y\right)^+\), we condition on the values of \(S_1\) and \(Y\). The value will be positive if \(S_1>Y\), otherwise zero: \( E\left[(S-Y)^+\right] = E \left[\left(S_1-Y\right) \cdot I\left(S_1>Y\right)\right] \) Where \(I(S_1>Y)\) is an indicator function that is equal to 1 if \(S_1>Y\) and 0 otherwise. Next, we use the joint probability of \(S_1\) and \(Y\), which can be found by multiplying their individual probabilities since they are independent in an M/M/1 queuing system. The expected value can be found as: \( E\left[(S-Y)^+\right] = \int_0^\infty \int_0^{s} (s-y) \lambda e^{-\lambda s} \mu e^{-\mu y} dy ds \)
03

Calculating the expectation of \(S_1\left(S_1-Y\right)^+\)

Using the same approach as in Step 2, we calculate the expectation as follows: \( E\left[S_1\left(S_1-Y\right)^+\right] = E \left[S_1 \cdot (\left(S_1-Y\right)^+ \cdot I\left(S_1>Y\right)) \right] = \int_0^\infty \int_0^{s} s(s-y) \lambda e^{-\lambda s} \mu e^{-\mu y} dy ds \)
04

Calculating the covariance

We now have the values of \(E\left[(S-Y)^+\right]\) and \(E\left[S_1\left(S_1-Y\right)^+\right]\). To find the covariance, we can use the formula: \( Cov(S_1, (S_1-Y)^+ + S_2) = E\left[S_1((S_1-Y)^+ + S_2)\right] - E[S_1] \cdot E[(S_1-Y)^+ + S_2] \) The term \(E\left[S_1((S_1-Y)^+ + S_2)\right]\) is the sum of the expectations calculated in Steps 2 and 3. To find expectations for the service times, we use their individual probability distributions: \( E[S_1] = \frac{1}{\mu} \) \( E[S_2] = \frac{1}{\mu} \) Hence, we can now calculate the covariance: \( Cov\left(S_1, \left(S_1-Y\right)^+ + S_2\right) = \left( E\left[(S-Y)^+\right] + E\left[S_1\left(S_1-Y\right)^+\right] \right) - \frac{1}{\mu} \left( E\left[(S-Y)^+\right] + \frac{1}{\mu} \right) \)

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.

M/M/1 Queue
An M/M/1 queue is a foundational model in queueing theory which is used to represent the queue of items or customers waiting for service. This type of queue has three main characteristics: a single server for processing items, an exponential time distribution for both the arrival of items and their service times, and the rule that the first item in line is the next to be served (First-Come, First-Served or FCFS). It follows the Markovian property, where the next state depends only on the current state, not on the sequence of events that preceded it. Understanding the behavior of M/M/1 queues is important for managing resources efficiently in various service industries like telecommunications, retail, and healthcare.

These models help predict the average waiting time, the average number of items in the queue, and other performance metrics that are vital for optimizing service processes, improving customer satisfaction, and reducing costs associated with over- or under-utilization of resources.
Service Time
Service time in queueing theory refers to the amount of time required to serve a customer or process an item in the system. In an M/M/1 queue, the service time has an exponential probability distribution, which is characterized by a single parameter \( \mu \), known as the service rate. This rate \( \mu \) represents the average number of customers that can be served per unit time.

Understanding and accurately estimating service time is crucial to queue management, as it directly affects how many customers or items can be processed over a period and, consequently, the overall efficiency of the system. Service times, along with arrival rates, are the key inputs for calculating other metrics such as queue length, waiting times, and system utilization.
Covariance Calculation
Covariance is a statistical measure that determines the degree to which two variables change together. In the context of queueing theory, it helps assess the relationship between the service times of different customers. It tells us whether the service times are positively or negatively correlated, or if they are independent of each other. A positive covariance implies that when one service time is longer than average, the other tends to be longer as well, and vice versa. If it's negative, it means they tend to be opposite.

To mathematically compute the covariance between two variables, say \( X \) and \( Y \) in the queue, we use the formula: \[ \text{Cov}(X, Y) = E[XY] - E[X]E[Y] \.\]
To determine covariance as it relates to queues, we consider the service times and the inter-arrival times, and we must often condition our calculations on these variables to reflect the queueing dynamics.
Probability Distribution
Probability distribution in queueing theory describes the likelihood of different outcomes for service times and inter-arrival times. In M/M/1 queues, these times are assumed to follow exponential probability distributions due to the 'memoryless' property. The exponential distribution is defined by its rate parameter, which for arrival times is \( \lambda \) and for service times is \( \mu \).

These distributions help to model and predict complex queue-related processes. The probability functions for exponential distributions allow us to calculate the likelihood of an item's service being completed within a certain timeframe, or the probability of a certain number of arrivals occurring within a given period. It is through these distributions that we calculate various expectations and variances needed to manage and optimize queues effectively.
Expectation
Expectation, commonly known as expected value, is a crucial concept in probability and statistics, including queueing theory. It represents the long-term average outcome of a random variable if an experiment or process were repeated many times. For service times in an M/M/1 queue, the expectation of service time \( S \) given the service rate \( \mu \) is the reciprocal of the rate \( E[S] = 1/\mu \:\).

The expectation allows managers and engineers to predict average wait times and queue lengths over time. This is valuable for capacity planning, staffing requirements, and performance evaluation. Expectation is also used when calculating other metrics, such as covariance and variance, which are integral for understanding variability in queue behavior and for making informed decisions to optimize system performance.

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

A facility produces items according to a Poisson process with rate \(\lambda\). However, it has shelf space for only \(k\) items and so it shuts down production whenever \(k\) items are present. Customers arrive at the facility according to a Poisson process with rate \(\mu\). Each customer wants one item and will immediately depart either with the item or empty handed if there is no item available. (a) Find the proportion of customers that go away empty handed. (b) Find the ayerage time that an item is on the shelf. (c) Find the average number of items on the shelf. Suppose now that when a customer does not find any available items it joins the "customers' queue" as long as there are no more than \(n-1\) other customers waiting at that time. If there are \(n\) waiting customers then the new arrival departs without an item. (d) Set up the balance equations. (c) In terms of the solution of the balance equations, what is the average number of customers in the system.

The economy alternates between good and bad periods. During good times customers arrive at a certain single-server queueing system in accordance with a Poisson process with rate \(\lambda_{1}\), and during bad times they arrive in accordance with a Poisson process with rate \(\lambda_{2} .\) A good time period lasts for an exponentially distributed time with rate \(\alpha_{1}\), and a bad time period lasts for an exponential time with rate \(\alpha_{2}\). An arriving customer will only enter the queueing system if the server is free; an arrival finding the server busy goes away. All service times are exponential with rate \(\mu .\) (a) Define states so as to be able to analyze this system. (b) Give a set of linear equations whose solution will yield the long-run proportion of time the system is in each state. In terms of the solutions of the equations in part (b), (c) what proportion of time is the system empty? (d) what is the average rate at which customers enter the system?

Suppose that a customer of the \(M / M / 1\) system spends the amount of time \(x>0\) waiting in queue before entering service. (a) Show that, conditional on the preceding, the number of other customers that were in the system when the customer arrived is distributed as \(1+P\), where \(P\) is a Poisson random variable with mean \(\lambda\). (b) Let \(W_{Q}^{*}\) denote the amount of time that an \(M / M / 1\) customer spends in queue. As a by-product of your analysis in part (a), show that $$ P\left[W_{\mathrm{Q}}^{*} \leqslant x\right]=\left\\{\begin{array}{ll} 1-\frac{\lambda}{\mu} & \text { if } x=0 \\ 1-\frac{\lambda}{\mu}+\frac{\lambda}{\mu}\left(1-e^{-(\mu-\lambda) x}\right) & \text { if } x>0 \end{array}\right. $$

There are two types of customers. Type 1 and 2 customers arrive in accordance with independent Poisson processes with respective rate \(\lambda_{1}\) and \(\lambda_{2}\). There are two servers. A type 1 arrival will enter service with server 1 if that server is free; if server 1 is busy and server 2 is free, then the type 1 arrival will enter service with server 2\. If both servers are busy, then the type 1 arrival will go away. A type 2 customer can only be served by server \(2 ;\) if server 2 is free when a type 2 customer arrives, then the customer enters service with that server. If server 2 is busy when a type 2 arrives, then that customer goes away. Once a customer is served by either server, he departs the system. Service times at server \(i\) are exponential with rate \(\mu_{i}\), \(i=1,2\) Suppose we want to find the average number of customers in the system. (a) Define states. (b) Give the balance equations. Do not attempt to solve them. In terms of the long-run probabilities, what is (c) the average number of customers in the system? (d) the average time a customer spends in the system?

In the Erlang loss system suppose the Poisson arrival rate is \(\lambda=2\), and suppose there are three servers, each of whom has a service distribution that is uniformly distributed over \((0,2)\). What proportion of potential customers is lost?

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