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Chapter 8: Problem 5

It follows from Exercise 4 that if, in the \(M / M / 1\) model, \(W_{Q}^{*}\) is the amount of time that a customer spends waiting in queue, then $$ W_{Q}^{*}=\left\\{\begin{array}{ll} 0, & \text { with probability } 1-\lambda / \mu \\ \operatorname{Exp}(\mu-\lambda), & \text { with probability } \lambda / \mu \end{array}\right. $$ where \(\operatorname{Exp}(\mu-\lambda)\) is an exponential random variable with rate \(\mu-\lambda\). Using this, find \(\operatorname{Var}\left(W_{Q}^{*}\right)\)

Short Answer

Expert verified
The variance of the waiting time in the queue, \(W_{Q}^{*}\), for the M/M/1 model is \(\operatorname{Var}\left(W_{Q}^{*}\right) = \frac{\lambda}{\mu^2(\mu-\lambda)}\).

Step by step solution

01

Finding the Mean of \(W_{Q}^{*}\)

The mean of \(W_{Q}^{*}\), denoted as \(E(W_{Q}^{*})\), can be found using the following formula considering the given probabilities and values: \[E(W_{Q}^{*}) = P(W_{Q}^{*}=0)\times 0 + P(W_{Q}^{*}=\operatorname{Exp}(\mu-\lambda))\times E(\operatorname{Exp}(\mu-\lambda))\] Substitute the given probability values, and recall that for an exponential random variable with rate \(\mu-\lambda\), the expectation is \(\frac{1}{\mu-\lambda}\): \[E(W_{Q}^{*}) = (1-\frac{\lambda}{\mu})\times 0 + \frac{\lambda}{\mu}\times \frac{1}{\mu-\lambda}\] Now, simplify the expression: \[E(W_{Q}^{*}) = \frac{\lambda}{\mu(\mu-\lambda)}\]
02

Finding the Mean of \(W_{Q}^{*2}\)

To find the mean of the square of \(W_{Q}^{*}\) (denoted as \(E(W_{Q}^{*2})\)), we use a similar approach as in Step 1: \[E(W_{Q}^{*2}) = P(W_{Q}^{*}=0)\times 0^2 + P(W_{Q}^{*}=\operatorname{Exp}(\mu-\lambda))\times E((\operatorname{Exp}(\mu-\lambda))^2)\] Substitute the given probability values, and recall that for an exponential random variable with rate \(\mu-\lambda\), the expectation of the square is \(\frac{2}{(\mu-\lambda)^2}\): \[E(W_{Q}^{*2}) = (1-\frac{\lambda}{\mu})\times 0^2 + \frac{\lambda}{\mu}\times \frac{2}{(\mu-\lambda)^2}\] Now, simplify the expression: \[E(W_{Q}^{*2}) = \frac{2\lambda}{\mu(\mu-\lambda)^2}\]
03

Finding the Variance of \(W_{Q}^{*}\)

Now that we have the values for \(E(W_{Q}^{*})\) and \(E(W_{Q}^{*2})\), we can use the variance formula to find the variance, denoted as \(Var(W_{Q}^{*})\): \[Var(W_{Q}^{*}) = E(W_{Q}^{*2}) - (E(W_{Q}^{*}))^2\] Substitute the values found in Steps 1 and 2: \[Var(W_{Q}^{*}) = \frac{2\lambda}{\mu(\mu-\lambda)^2} - \left(\frac{\lambda}{\mu(\mu-\lambda)}\right)^2\] Now, simplify the expression to get the final result: \[Var(W_{Q}^{*}) = \frac{\lambda}{\mu^2(\mu-\lambda)}\]

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Key Concepts

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

Queueing Theory

Queueing theory is the study of waiting lines, or queues, and is of great importance in fields such as operations research, computer science, and statistics. In this context, it examines the phenomena that occur when 'customers' (who could be anything from people, data packets, cars, etc.) arrive at a 'service station' and wait for service. Different models within queueing theory help to analyze a variety of systems to predict queue lengths, waiting times, and optimize service capacities.

One of the simplest and most studied models in queueing theory is the M/M/1 queue model. In this model, the 'M's refer to 'Markovian', indicating that both the arrival of customers and the service times are memoryless and follow an exponential distribution. The '1' indicates that there is a single service channel.

In the M/M/1 queue model, several performance measures can be analyzed, such as the mean waiting time in the queue, the probability of a customer having to wait, and the variance of waiting time. This model is applicable to many real-world systems with a single server and a Poisson arrival process, such as bank tellers, ticket counters, and certain telecommunications systems.

Exponential Random Variable

An exponential random variable is a continuous random variable that describes the time between events in a Poisson process. It is one of the key stochastic processes used in queueing theory. A variable X is considered exponential with rate \( \lambda \) if the probability of X being less than or equal to a certain value x is given by the cumulative distribution function:

P(X \leq x) = 1 - e^{-\lambda x}, for x \geq 0

This property results in the 'memoryless' characteristic of the exponential distribution, meaning that the probability of an event occurring in the next interval is independent of how much time has already passed. In the context of the M/M/1 queue model, service times are assumed to be exponentially distributed, which simplifies the analysis thanks to the memorylessness property.

The exponential distribution has a mean of \( \frac{1}{\lambda} \) and a variance of \( \frac{1}{\lambda^2} \), providing critical information for analyzing and understanding the behavior of queues.

Variance of Waiting Time

The variance of waiting time in a queue is a measure of the variability or dispersion of waiting times around the mean. High variance indicates that waiting times can fluctuate significantly, while low variance suggests that waiting times are more consistent.

In the M/M/1 queue model, the variance of the waiting time in queue can be calculated using knowledge of both the average waiting time and the average of the squared waiting time. To determine the variance \(\operatorname{Var}(W_Q^*)\), the solution uses the principle that the variance of a random variable is equal to the expected value of the square of the variable minus the square of the expected value of the variable. Mathematically, it's represented as:

\operatorname{Var}(W_Q^*) = E(W_Q^{*2}) - (E(W_Q^*))^2

This measure is crucial for queueing systems as it helps to assess the 'risk' associated with the waiting times. A high variance might imply that some customers wait significantly longer than others, which can be undesirable in many service systems. Reducing the variance can lead to a more predictable and satisfactory customer experience.

Utilizing these fundamental concepts, queueing theory enables organizations and service providers to optimize their queue management, leading to increased efficiency and improved customer satisfaction.

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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 busy-period 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?

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 two 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 entering customer spends 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?

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 momentgenerating 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.

Machines in a factory break down at an exponential rate of six per hour. There is a single repairman who fixes machines at an exponential rate of eight per hour. The cost incurred in lost production when machines are out of service is \(\$ 10\) per hour per machine. What is the average cost rate incurred due to failed machines?
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