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Chapter 8: Problem 10

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.

Short Answer

Expert verified
Define states as the number of customers in the system, denoted as \(n\), where \(n = 0, 1, 2, ..., m\). The balance equations are given by \[\mu P_{n - 1} = (n\theta + \mu)P_n \text{ for } n \ge 1\] and \[m\theta P_m = \mu P_0\]. To find the average rate at which customers enter the station, we compute the arrival rate using \[\lambda = \sum_{n=1}^{m} \theta n P_n\]. To find the average time a customer spends in the station per visit, we first find the average number of customers in the system \(L = \sum_{n=1}^{m} n P_n\), then use Little's Law to get \[W = \frac{L}{\lambda}\].

Step by step solution

01

(a) Define states and set up the balance equations.

We will define the states of the system as the number of customers in the system, denoted as \(n\), where \(n = 0, 1, 2, ..., m\). For the balance equations, we will use the concept of flow-in equals flow-out. Let \(P_n\) be the steady-state probability of having n customers in the system. Then, for \(n \ge 1\), we have the balance equations: \[\mu P_{n - 1} = (n\theta + \mu)P_n\] At \(n = 0\), we have the following equation: \[m\theta P_m = \mu P_0\]
02

(b) Average rate at which customers enter the station.

To find the average rate at which customers enter the station, we can determine the long-term average number of arrivals per unit time. From the balance equations, we can represent \(P_n\) in terms of \(P_0\). Using the balance equations, we get: \[P_n = \frac{\mu}{n\theta + \mu}P_{n - 1}\] If we sum the balance equations over all n, we have: \[\sum_{n=1}^{m} P_n = 1 - P_0\] Now we can compute the rate at which customers enter the system, which is the product of the arrival rate and the probability of the server being busy: \[\lambda = \sum_{n=1}^{m} \theta n P_n\] We need to compute the values of \(P_n\) from the balance equations and plug them in to find the average rate.
03

(c) Average time spent by a customer in the station per visit.

To find the average time a customer spends in the station per visit, we can calculate the average number of customers in the system, and divide it by the average rate at which customers enter the station. We already have the average arrival rate \(\lambda\). The average number of customers in the system can be found using Little's Law: \[L = \lambda W\] Where L is the average number of customers in the system and W is the average time per customer in the system. We can compute L as: \[L = \sum_{n=1}^{m} n P_n\] And from this, we can get the average time spent by a customer in the station per visit: \[W = \frac{L}{\lambda}\] To get the final answer, we need to compute the values of \(\lambda\) and L using the balance equations and plug them in to the above equation for W.

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Key Concepts

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

Single-server Queue
In queueing theory, a single-server queue refers to a system where a group of customers is served by one server. This model is often used to analyze and design systems like ticket counters, customer service desks, or any scenario where service is provided one at a time.
In the single-server queue, when a customer arrives:
  • If the server is free, the customer is immediately served.
  • If the server is busy, the customer waits in a queue until they can be served.
The core idea is to manage how arriving customers get served and how waiting time is minimized. Understanding this helps in optimizing service systems for better efficiency.
Exponential Distribution
The exponential distribution is a probability distribution commonly used to model the time between independent events that happen at a constant average rate. This constant rate is called the arrival rate (\(\theta\) or service rate (\(\mu\) in the context of queueing theory.
Some key properties include:
  • Memoryless: The exponential distribution is memoryless, meaning the probability of an event occurring in the future is independent of the past.

  • Modeling Service Times: Often used to model the time taken to serve a customer in queueing models, the service times are exponentially distributed with rate \(\mu\). This indicates that short service times are more probable, but longer services can occur.
Understanding the exponential distribution helps in predicting how long a customer or event might take in a single-server queue system.
Steady-state Probability
Steady-state probabilities in queueing theory refer to long-term probabilities of the system being in a particular state. For a system in equilibrium, it tells us the probability \(P_n\) of having \(n\) customers in the system for a single-server queue model.
These probabilities emerge when a system has been observed long enough for transient behaviors to stabilize. At this stage, the arrivals and services match over time, so:
  • The rate at which customers arrive and enter service equals the rate at which they complete service and leave.

  • Balance equations: These equations ensure that the flow-in is equal to the flow-out.
By analyzing these probabilities, you can derive metrics like average queue length and average waiting time, which are essential for effective system design and operation.
Little's Law
Little's Law is a fundamental theorem that relates three critical metrics in a queuing system: the average number of customers in the system (\(L\)), the average rate of customer arrival (\(\lambda\)), and the average time a customer spends in the system (\(W\)). The law is mathematically expressed as:
\[L = \lambda W\]
This equation is simple yet powerful because it holds for a wide range of scenarios beyond just queueing models.
A few vital points:
  • It applies to stable systems, where input equals output over time.

  • Helps in evaluating system performance by allowing the calculation of one parameter if two are known.
Using Little's Law alongside balance equations, one can efficiently analyze and infer crucial system performance indicators, which simplify the management and optimization of queueing systems.

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Most popular questions from this chapter

Consider a \(M / G / 1\) system with \(\lambda E[S]<1\). (a) Suppose that service is about to begin at a moment when there are \(n\) customers in the system. (i) Argue that the additional time until there are only \(n-1\) customers in the system has the same distribution as a busy period. (ii) What is the expected additional time until the system is empty? (b) Suppo?e that the work in the system at some moment is \(A\). We are interested in the expected additional time until the system is empty -cail it \(E[T]\). Let \(N\) denote the number of arrivals during the first \(A\) units of time. (i) Compute \(E[T \mid N]\). (ii) Compute \(E[T]\).

Customers arrive at a two-server system at a Poisson rate \(\lambda\). An arrival finding the system empty is equally likely to enter service with either server. An arrival finding one customer in the system will enter service with the idle server. An arrival finding tw? others in the system will wait in line for the first free server. An arrival finding three in the system will not enter. All service times are exponential with rate \(\mu\), and once a customer is served (by either server), he departs the system. (a) Define the states. (b) Find the long-run probabilities. (c) Suppose a customer arrives and finds two others in the system. What is the expected times he spends in the system? (d) What proportion of customers enter the system? (e) What is the average time an cntering customer spends in the system?

Consider a single-server queue with Poisson arrivals and exponential service times having the following variation: Whenever a service is completed a departure occurs only with probability \(\alpha\). With probability \(1-\alpha\) the customer, instead of leaving, joins the end of the queue. Note that a customer may be serviced more than once. (a) Set up the balance equations and solve for the steady-state probabilities, stating conditions for it to exist. (b) Find the expected waiting time of a customer from the time he arrives until he enters service for the first time. (c) What is the probability that a customer enters service exactly \(n\) times, \(n=\) \(1,2, \ldots ?\) (d) What is the expected amount of time that a customer spends in service (which does not include the time he spends waiting in line)? Use part (c). (c) What is the distribution of the total length of time a customer spends being served? 7 Is it memoryless?

Let \(D\) denote the time between successive departures in a stationary \(M / M / 1\) queue with \(\lambda<\mu\). Show, by conditioning on whether or not a departure has left the system empty, that \(D\) is exponential with rate \(\lambda\). Hint: By conditioning on whether or not the departure has left the system empty we see that $$ D=\left\\{\begin{array}{ll} \text { Exponential }(\mu), & \text { with probability } \lambda / \mu \\ \text { Exponential }(\lambda) * \text { Exponential }(\mu), & \text { with probability } 1-\lambda / \mu \end{array}\right. $$ where Exponential \((\lambda) *\) Exponential \((\mu)\) represents the sum of two independent exponential random variables having rates \(\mu\) and \(\lambda\). Now use moment-generating functions to show that \(D\) has the required distribution. Note that the preceding does not prove that the departure process is Poisson. To prove this we need show not only that the interdeparture times are all exponential with rate \(\lambda\), but also that they are independent.

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 quete). 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.
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