/*! 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 35 35\. Customers arrive at a singl... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 35

35\. Customers arrive at a single-server station in accordance with a Poisson process having rate \(\lambda .\) Each customer has a value. The successive values of customers are independent and come from a uniform distribution on \((0,1)\). The service time of a customer having value \(x\) is a random variable with mean \(3+4 x\) and variance 5 . (a) What is the average time a customer spends in the system? (b) What is the average time a customer having value \(x\) spends in the system?

Short Answer

Expert verified
(a) The average time a customer spends in the system is given by \(W = \frac{\frac{位(5) + \frac{1}{2}位^2(5)}{1-位(5)}}{位}\). (b) The average time a customer having value \(x\) spends in the system is \(3 + 4x\).

Step by step solution

01

Compute the average service time

Let's first compute the average service time for a customer with a given value \(x\). The mean service time is given by \(3 + 4x\). Since the value of a customer is uniformly distributed between 0 and 1, we can compute the average service time as follows: \[\mu = \int_0^1 (3 + 4x) dx\] Now, solve the integral: \[\mu = \left[ 3x + 2x^2 \right]_0^1 = 3(1) + 2(1)^2 - (3(0) + 2(0)^2) = 3 + 2 = 5\] Hence, the average service time is 5 units of time.
02

Compute the average number of customers in the system

To find the average number of customers in the system, we'll use the Pollaczek-Khintchine formula, which relates the average number of customers in the system to the variance of service time and mean arrival rate. The formula is: \[L = \frac{位\mu + \frac{1}{2}位^2蟽^2}{1-位\mu}\] We were given that the variance of service time is 5. Thus, multiply the arrival rate by the service time, and plug these values into the formula: \[L = \frac{位(5) + \frac{1}{2}位^2(5)}{1-位(5)}\]
03

Apply Little's Law to find the average time a customer spends in the system

Now that we have both the average number of customers in the system and the arrival rate, we can apply Little's Law to find the average time a customer spends in the system: \[W = \frac{L}{位} = \frac{\frac{位(5) + \frac{1}{2}位^2(5)}{1-位(5)}}{位}\] (a) Answer: This is the average time a customer spends in the system.
04

Find the average time a customer with value \(x\) spends in the system

The average time a customer having value \(x\) spends in the system can be found by just considering the mean service time for a customer with value \(x\), given by \(3 + 4x\). (b) Answer: The average time a customer having value \(x\) spends in the system is \(3 + 4x\).

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.

Poisson Process
In queueing theory, a Poisson process is used to model random events that occur independently over time, such as the arrival of customers at a service station. It is characterized by a rate \( \lambda \), which represents the average number of events (arrivals) occurring per unit time. The process is memoryless, meaning that future arrivals are independent of past events.

Key properties of Poisson process include:
  • The number of arrivals in a fixed period follows a Poisson distribution.
  • The time between consecutive arrivals follows an exponential distribution.
Understanding Poisson processes is crucial for analyzing and predicting behaviors in systems like customer service stations, where arrivals tend to happen randomly.
Service Time Distribution
The service time distribution describes the variability in how long it takes to serve a customer. In this context, the service time of a customer with a specific value \( x \) is modeled as a random variable with a mean of \( 3 + 4x \). This implies that the average service time depends linearly on the customer's value.

Additionally, the variance of this service time is given as \( 5 \). A variance provides insights into the extent of variation or uncertainty around the average service time. Higher variance indicates more variability, while a lower variance suggests more consistent service times. In analyzing queueing systems, understanding both the mean and variance of service times is essential for accurate modeling of system performance.
Expected Value
The expected value of a random variable is a crucial concept in probability and is often thought of as the "average" value. It provides a summary measure of the central tendency of a probability distribution. In our problem, to find the average service time (expected value of the service times), the integral of the function \( 3 + 4x \) over the interval \( 0 \) to \( 1 \) is used:

\[ \mu = \int_0^1 (3 + 4x) \, dx \]

By solving this integral, we find \( \mu = 5 \), meaning the expected or average service time is \( 5 \) units for a uniformly distributed \( x \). Understanding expected value helps in making informed predictions about average behaviors in stochastic systems.
Uniform Distribution
A uniform distribution is a probability distribution where every outcome in a given interval is equally likely. In our example, customer values are independently drawn from a uniform distribution on \( (0,1) \), meaning every value within this range has an equal chance of occurring.

Key features include:
  • The probability density function is constant across the interval.
  • The mean of a uniform distribution over an interval \( (a, b) \) is \( \frac{a+b}{2} \).
  • For \( (0,1) \), the mean is \( \frac{1}{2} \).
Understanding uniform distribution allows for predicting averaged behaviors and is foundational in developing models that assume equal likelihood of events within specified bounds.

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 group of \(m\) customers frequents a single-server station in the following manner. When a customer arrives, he or she either enters service if the server is free or joins the queue otherwise. Upon completing service the customer departs the system, but then returns after an exponential time with rate \(\theta\). All service times are exponentially distributed with rate \(\mu\). (a) Define states and set up the balance equations. In terms of the solution of the balance equations, find (b) the average rate at which customers enter the station. (c) the average time that a customer spends in the station per visit.

Consider a single-server exponential system in which ordinary customers arrive at a rate \(\lambda\) and have service rate \(\mu .\) In addition, there is a special customer who has a service rate \(\mu_{1}\). Whenever this special customer arrives, she goes directly into service (if anyone else is in service, then this person is bumped back into queue). When the special customer is not being serviced, she spends an exponential amount of time (with mean \(1 / \theta\) ) out of the system. (a) What is the average arrival rate of the special customer? (b) Define an appropriate state space and set up balance equations. (c) Find the probability that an ordinary customer is bumped \(n\) times.

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?

For the \(M / G / 1\) queue, let \(X_{n}\) denote the number in the system left behind by the \(n\) th departure. (a) If $$ X_{n+1}=\left\\{\begin{array}{ll} X_{n}-1+Y_{n}, & \text { if } X_{n} \geqslant 1 \\ Y_{n}, & \text { if } X_{n}=0 \end{array}\right. $$ what does \(Y_{n}\) represent? (b) Rewrite the preceding as $$ X_{n+1}=X_{n}-1+Y_{n}+\delta_{n} $$ where $$ \delta_{n}=\left\\{\begin{array}{ll} 1, & \text { if } X_{n}=0 \\ 0, & \text { if } X_{n} \geqslant 1 \end{array}\right. $$ Take expectations and let \(n \rightarrow \infty\) in Equation (8.64) to obtain $$ E\left[\delta_{\infty}\right]=1-\lambda E[S] $$ (c) Square both sides of Equation (8.64), take expectations, and then let \(n \rightarrow\) \(\infty\) to obtain $$ E\left[X_{\infty}\right]=\frac{\lambda^{2} E\left[S^{2}\right]}{2(1-\lambda E[S])}+\lambda E[S] $$ (d) Argue that \(E\left[X_{\infty}\right]\), the average number as seen by a departure, is equal to \(L\).

Consider a network of three stations. Customers arrive at stations \(1,2,3\) in accordance with Poisson processes having respective rates, \(5,10,15 .\) The service times at the three stations are exponential with respective rates \(10,50,100\). A customer completing service at station 1 is equally likely to (i) go to station 2, (ii) go to station 3, or (iii) leave the system. A customer departing service at station 2 always goes to station \(3 .\) A departure from service at station 3 is equally likely to either go to station 2 or leave the system. (a) What is the average number of customers in the system (consisting of all three stations)? (b) What is the average time a customer spends in the system?
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Calculus

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.