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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 M/M/1 queuing model, \(Var(W_Q^*)\), is given by the formula: \[ \operatorname{Var}\left(W_{Q}^{*}\right) = \frac{\lambda(\mu^2 + \lambda^2 - 2\mu\lambda)}{\mu^2(\mu - \lambda)^2} \]

Step by step solution

01

Find the expectation \(\mathbb{E}(W_Q^{*})\)

To find the expectation of \(W_Q^*\), we will use the definition of expectation for a random variable that takes on different values with different probabilities. In this case, \(W_Q^*\) takes on the value 0 with probability \((1 - \frac{\lambda}{\mu})\) and follows an exponential distribution with rate \((\mu - \lambda)\) with probability \(\frac{\lambda}{\mu}\). Thus, the expectation can be written as: \[ \mathbb{E}(W_Q^{*}) = 0 \times (1 - \frac{\lambda}{\mu}) + \frac{1}{\mu - \lambda} \times \frac{\lambda}{\mu} \]
02

Simplify the expression and calculate \(\mathbb{E}(W_Q^{*})\)

Now, let's simplify the expression for \(\mathbb{E}(W_Q^{*})\): \[ \mathbb{E}(W_Q^{*}) = \frac{\lambda}{\mu(\mu - \lambda)} \]
03

Find the expectation \(\mathbb{E}(W_Q^{*2})\)

To find the expectation of the square of the waiting time, \(\mathbb{E}(W_Q^{*2})\), we can similarly use the definition of expectation. The square of the waiting time is 0 with probability \(1 - \frac{\lambda}{\mu}\) and follows an exponential distribution with rate \((\mu-\lambda)\) squared with probability \(\frac{\lambda}{\mu}\). Thus, the expectation can be written as: \[ \mathbb{E}(W_Q^{*2}) = 0 \times (1 - \frac{\lambda}{\mu}) + \frac{2}{(\mu - \lambda)^2} \times \frac{\lambda}{\mu} \]
04

Simplify the expression and calculate \(\mathbb{E}(W_Q^{*2})\)

Now, let's simplify the expression for \(\mathbb{E}(W_Q^{*2})\): \[ \mathbb{E}(W_Q^{*2}) = \frac{2\lambda}{\mu(\mu - \lambda)^2} \]
05

Compute the variance \(\operatorname{Var}\left(W_{Q}^{*}\right)\)

Now we are ready to compute the variance using the formula \(\operatorname{Var}\left(W_{Q}^{*}\right) = \mathbb{E}(W_Q^{*2}) - (\mathbb{E}(W_Q^{*})^2\). Plugging in the values from previous steps, we get: \[ \operatorname{Var}\left(W_{Q}^{*}\right) = \frac{2\lambda}{\mu(\mu - \lambda)^2} - \left(\frac{\lambda}{\mu(\mu - \lambda)}\right)^2 \]
06

Simplify the expression to obtain the final answer

Now, let's simplify the expression for the variance: \[ \operatorname{Var}\left(W_{Q}^{*}\right) = \frac{\lambda(\mu^2 + \lambda^2 - 2\mu\lambda)}{\mu^2(\mu - \lambda)^2} \] Now we have the answer for the variance of the waiting time in an M/M/1 queuing model: \[ \operatorname{Var}\left(W_{Q}^{*}\right) = \frac{\lambda(\mu^2 + \lambda^2 - 2\mu\lambda)}{\mu^2(\mu - \lambda)^2} \]

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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 mathematical study of waiting lines, or queues. At its core, this discipline analyzes processes through which objects鈥攍ike people, vehicles, or data packets鈥攓ueue up to receive some type of service.

One of the simplest and most commonly used models in queueing theory is the M/M/1 queue model, which represents systems with a single service channel (1), Poisson arrival patterns (M for 'Markovian'), and exponential service times (the second M). This model is an invaluable tool for understanding and optimizing a variety of service and operational systems, from customer service desks to network routers.

Understanding the behavior of queues helps organizations to manage resources efficiently, reduce waiting times, and improve customer satisfaction. Queueing theory provides metrics such as average waiting time, queue length, and the probability of finding the system busy, which are essential in designing and managing service systems.

The calculation of such metrics often involves understanding the nature of the arrivals and service times, which leads us to an integral part of M/M/1 and queueing theory: the exponential distribution.
Exponential Distribution
The exponential distribution is crucial in queueing theory, particularly in the context of the M/M/1 queue model. It characterizes the time between events in a Poisson process, where events occur continuously and independently at a constant average rate.

In the context of queues, these 'events' can be interpreted as customer arrivals or service completions. The exponential distribution is memoryless, which means that the probability of an event occurring in the future is independent of any past events. This property is particularly suited to model random arrival and service patterns typical of many queuing systems, as it simplifies the analysis significantly.

For instance, in our given problem, the service times are explicitly modeled as an exponential random variable with a rate of \(\mu - \lambda\). This rate is the difference between the service rate \(\mu\) and the arrival rate \(\lambda\), indicating the system's capacity to handle the workload. The probability associated with the time a customer spends waiting in the queue, therefore, follows an exponential distribution with the specified rate, giving us insight into how long customers might expect to wait before being served.
Variance of Waiting Time
The variance of waiting time is a statistical measure of how much the length of time that customers wait in queue varies. It's important because it reflects the consistency of the waiting experience. A high variance indicates that waits can fluctuate wildly, while a low variance means that they're relatively stable and predictable.

In the M/M/1 queue model, the variance of the waiting time in the queue \(\operatorname{Var}(W_{Q}^{*})\) provides valuable insight into the variability that customers may experience. Variance is calculated using the expectation of the waiting time and the expectation of the waiting time squared. The step-by-step solution for calculating these expectations and the final variance gives us a clear picture of what customers might endure.

The final expression for variance can be somewhat complex, as seen in the exercise, but it encapsulates important factors like the rates of arrival and service. Understanding this variance helps businesses and service providers to identify strategies for managing queues better鈥攖hrough staffing adjustments, process improvements, or customer communication鈥攊n order to minimize variability in service and enhance the overall customer experience.

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Most popular questions from this chapter

Consider the priority queueing model of Section \(8.6 .2\) but now suppose that if a type 2 customer is being served when a type 1 arrives then the type 2 customer is bumped out of service. This is called the preemptive case. Suppose that when a bumped type 2 customer goes back in service his service begins at the point where it left off when he was bumped. (a) Argue that the work in the system at any time is the same as in the nonpreemptive case. (b) Derive \(W_{Q}^{1}\) Hint: How do type 2 customers affect type 1 s? (c) Why is it not true that $$ V_{Q}^{2}=\lambda_{2} E\left[S_{2}\right] W_{Q}^{2} $$ (d) Argue that the work seen by a type 2 arrival is the same as in the nonpreemptive case, and so \(W_{Q}^{2}=W_{Q}^{2}\) (nonpreemptive) \(+E[\) extra time \(]\) where the extra time is due to the fact that he may be bumped. (e) Let \(N\) denote the number of times a type 2 customer is bumped. Why is $$ E[\text { extra time } \mid \mathrm{N}]=\frac{N E\left[S_{1}\right]}{1-\lambda_{1} E\left[S_{1}\right]} $$ Hint: When a type 2 is bumped, relate the time until he gets back in service to a "busy period." (f) Let \(S_{2}\) denote the service time of a type \(2 .\) What is \(E\left[N \mid S_{2}\right] ?\) (g) Combine the preceding to obtain $$ W_{Q}^{2}=W_{Q}^{2} \text { (nonpreemptive) }+\frac{\lambda_{1} E\left[S_{1}\right] E\left[S_{2}\right]}{1-\lambda_{1} E\left[S_{1}\right]} $$

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 \(S_{1}>Y\)

In a two-class priority queueing model suppose that a cost of \(C_{i}\) per unit time is incurred for each type \(i\) customer that waits in queue, \(i=1,2 .\) Show that type 1 customers should be given priority over type 2 (as opposed to the reverse) if $$ \frac{E\left[S_{1}\right]}{C_{1}}<\frac{E\left[S_{2}\right]}{C_{2}} $$

Potential customers arrive to a single-server hair salon according to a Poisson process with rate \(\lambda .\) A potential customer who finds the server free enters the system; a potential customer who finds the server busy goes away. Each potential customer is type \(i\) with probability \(p_{i}\), where \(p_{1}+p_{2}+p_{3}=1\). Type 1 customers have their hair washed by the server; type 2 customers have their hair cut by the server; and type 3 customers have their hair first washed and then cut by the server. The time that it takes the server to wash hair is exponentially distributed with rate \(\mu_{1}\), and the time that it takes the server to cut hair is exponentially distributed with rate \(\mu_{2}\). (a) Explain how this system can be analyzed with four states. (b) Give the equations whose solution yields the proportion of time the system is in each state. In terms of the solution of the equations of (b), find (c) the proportion of time the server is cutting hair; (d) the average arrival rate of entering customers.

A group of \(n\) customers moves around among two servers. Upon completion of service, the served customer then joins the queue (or enters service if the server is free) at the other server. All service times are exponential with rate \(\mu .\) Find the proportion of time that there are \(j\) customers at server \(1, j=0, \ldots, n\).
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