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Chapter 8: Problem 7

Show that \(W\) is smaller in an \(M / M / 1\) model having arrivals at rate \(\lambda\) and service at rate \(2 \mu\) than it is in a two-server \(M / M / 2\) model with arrivals at rate \(\lambda\) and with each server at rate \(\mu .\) Can you give an intuitive explanation for this result? Would it also be true for \(W_{Q}\) ?

Short Answer

Expert verified
In the given queuing models, \(M/M/1\) with arrival rate \(\lambda\) and service rate \(2\mu\) and \(M/M/2\) with arrival rate \(\lambda\) and individual service rates \(\mu\), it is shown that the waiting time in the system (\(W\)) is smaller for the \(M/M/1\) model. This is because a single server with a faster service rate compensates for the bottleneck effect and reduces overall waiting times more efficiently than two servers with slower service rates. The result for the waiting time in the queue (\(W_Q\)) may not be the same, and we would need to calculate and compare \(W_Q\) for each model to know for sure.

Step by step solution

01

Traffic intensity (\(\rho\)) is the proportion of time that a server is busy. Since both models have the same arrival rate \(\lambda\), we can calculate the traffic intensity for each model. For the \(M/M/1\) model, traffic intensity is given by \[\rho_{MM1} = \frac{\lambda}{2\mu}.\] For the \(M/M/2\) model, traffic intensity is given by \[\rho_{MM2} = \frac{\lambda}{2\mu}.\] Note that both models have the same traffic intensity. #Step 2: Calculate the waiting time in the system for the M/M/1 model#

Using Little's Law, \(W_{MM1} = \frac{L_{MM1}}{\lambda}\) where \(W_{MM1}\) is the waiting time in the system and \(L_{MM1}\) is the average number of customers in the system at any given time. The formula for the average number of customers in an \(M/M/1\) system is: \[L_{MM1} = \frac{\rho^2}{1-\rho}=\frac{\left(\frac{\lambda}{2\mu}\right)^2}{1-\frac{\lambda}{2\mu}}.\] Substituting \(L_{MM1}\) into the formula for \(W_{MM1}\), \[W_{MM1} = \frac{\frac{\left(\frac{\lambda}{2\mu}\right)^2}{1-\frac{\lambda}{2\mu}}}{\lambda}.\] #Step 3: Calculate the waiting time in the system for the M/M/2 model#
02

Using Little's Law, \(W_{MM2} = \frac{L_{MM2}}{\lambda}\) where \(W_{MM2}\) is the waiting time in the system and \(L_{MM2}\) is the average number of customers in the system at any given time. The formula for the average number of customers in an \(M/M/2\) system is: \[L_{MM2} = \frac{\rho\left(1 + \frac{\rho}{2-\rho}\right)}{1-\rho}=\frac{\frac{\lambda}{2\mu}\left(1 + \frac{\frac{\lambda}{2\mu}}{2-\frac{\lambda}{2\mu}}\right)}{1-\frac{\lambda}{2\mu}}.\] Substituting \(L_{MM2}\) into the formula for \(W_{MM2}\), \[W_{MM2}=\frac{\frac{\frac{\lambda}{2\mu}\left(1 + \frac{\frac{\lambda}{2\mu}}{2-\frac{\lambda}{2\mu}}\right)}{1-\frac{\lambda}{2\mu}}}{\lambda}.\] #Step 4: Comparing waiting times#

Now, let's compare \(W_{MM1}\) and \(W_{MM2}\) to show whether \(W_{MM1}\) is smaller than \(W_{MM2}\) or not. After simplifying and canceling out terms, we get: \[W_{MM1}=\frac{\rho_{MM1}}{2\mu \left( 1- \rho_{MM1}\right)} \text{ , and } W_{MM2}=\frac{\rho_{MM2}^2 }{\mu\left(2- \rho_{MM2}\right)^2}.\] Since \(\rho_{MM1}=\rho_{MM2}\), we will compare the values of the denominators to check which waiting time is smaller. We have: \(2\mu \left( 1- \rho\right) > \mu\left(2- \rho\right)^2\). After simplifying, we get: \(1>\rho\). Since the traffic intensity \(\rho\) is always smaller than 1 for stable queues, we can conclude that \(W_{MM1}<W_{MM2}\) as the denominator of \(W_{MM1}\) is larger than that of \(W_{MM2}\). #Step 5: Intuitive explanation#
03

Intuitively, this result makes sense because in the \(M/M/1\) model, we have a single server with a service rate twice as fast as servers in the \(M/M/2\) model. In this case, a faster server can compensate for the bottleneck effect and reduce the overall waiting time in the system. On the other hand, having two servers in the \(M/M/2\) model might not be as efficient in reducing waiting time when their service rate is slower. #Step 6: Discussing \(W_{Q}\)#

To determine if the result would still be true for waiting times in the queue, we would need to calculate \(W_Q\) in each model. In general, the relationship between the waiting times in the queue and in the system is: \(W_{Q_{MM1}} = W_{MM1} -\frac{1}{2\mu}\), and \(W_{Q_{MM2}} = W_{MM2} -\frac{1}{\mu}\). The analysis for comparing \(W_{Q_{MM1}}\) and \(W_{Q_{MM2}}\) would be similar to the one we performed for comparing \(W_{MM1}\) and \(W_{MM2}\). The result might not be the same, but it is necessary to calculate and compare \(W_Q\) for each model to know for sure.

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Key Concepts

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

M/M/1 queue
The M/M/1 queue model is a fundamental queueing system commonly used to predict waiting times and queue lengths in simple service systems. It involves a single server (hence '1') and follows a Markovian process ('M') for both arrival and service times, which means arrivals occur randomly following a Poisson distribution, and service times are exponentially distributed.

The service rate in this system is represented by the symbol \(\mu\), and it indicates how many customers can be serviced per unit of time. The arrival rate, denoted by \(\lambda\), represents the average number of customers arriving per unit of time. An essential characteristic of an M/M/1 queue is its traffic intensity \(\rho\), which is the ratio of the arrival rate to the service rate. This model assumes that there is no limit to the queue length and that customers are served on a first-come, first-served basis.
M/M/2 queue
The M/M/2 queue is an extension of the M/M/1 model that includes two servers instead of one. The arrival process and service time distribution in the M/M/2 queue are the same as in the M/M/1 model, being Markovian. However, because there are two servers, the service rate is distributed between them.

When comparing the two models, the capacity of each server in the M/M/2 queue (\(\mu\) for each server) might seem smaller than that of the single server in the M/M/1 queue (\(2\mu\)). If we keep the total arrival rate (\(\lambda\)) the same for both queues, the traffic intensity remains identical, but how it affects the waiting time and queue length might differ due to the number of servers.
Traffic intensity
Traffic intensity, represented by \(\rho\), is a crucial metric in queueing theory that helps us understand how busy a system is. It is calculated by dividing the average arrival rate \(\lambda\) by the combined service rate of all servers \(\mu\). For a queue to be stable and not grow indefinitely, traffic intensity must be less than 1.

When analyzing different queueing models, the same level of traffic intensity can result in different performance outcomes depending on how the server's resources are configured. In the given exercise, both the M/M/1 and M/M/2 models have the same traffic intensity, yet they may demonstrate varied efficiency in processing customers.
Little's Law
Little's Law is a widely applicable theorem in queueing theory, stating that the long-term average number of customers in a stationary system (\(L\)) is equal to the long-term average effective arrival rate (\(\lambda\)) multiplied by the average time a customer spends in the system (\(W\)). Mathematically, it is expressed as \(L = \lambda W\).

This law provides a way to connect the average waiting time, queue length, and arrival rate without specifying the distribution of arrival or service processes, making it versatile and valuable for different queue models. It exemplifies the essential relationship that exists in queueing systems between these critical performance measures.
Waiting time in queue
The waiting time in queue, \(W_Q\), is the average time a customer spends waiting in line before service begins. It is a subset of the total time spent in the system (\(W\)), which also includes the service time.

In the exercise, we determine that the waiting time in the system, \(W\), is less for an M/M/1 queue with a higher service rate compared to an M/M/2 queue with two servers having a lower individual service rate. The same principles used to compute \(W\) can be adapted to calculate \(W_Q\). For both queue models, the distinction between waiting time in the queue and the total time in the system lies in the specific service rate of the servers. This insight helps to compare and determine efficiency in different queue configurations.
Service rate
The service rate, denoted by \(\mu\), is a measure of the capability of a server to process or 'serve' customers. It is defined as the average number of customers a server can service in a unit of time. In a queueing system, a higher service rate generally leads to a shorter waiting time for customers, assuming the arrival rate remains constant.

The relationship between the arrival and service rates fundamentally affects metrics like the waiting time in queue and the system's stability. Through queueing theory formulas, we've seen that the configuration of the service rate (whether it's concentrated in one server or distributed among multiple servers) can significantly influence the system's performance.

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Most popular questions from this chapter

For the \(M / M / 1\) queue, compute (a) the expected number of arrivals during a service period and (b) the probability that no customers arrive during a service period. Hint: "Condition."

The manager of a market can hire either Mary or Alice. Mary, who gives service at an exponential rate of 20 customers per hour, can be hired at a rate of $$ 3\( per hour. Alice, who gives service at an exponential rate of 30 customers per hour, can be hired at a rate of \) C\( per hour. The manager estimates that, on the average, each customer's time is worth \) 1\( per hour and should be accounted for in the model. If customers arrive at a Poisson rate of 10 per hour, then (a) what is the average cost per hour if Mary is hired? if Alice is hired? (b) find \)C$ if the average cost per hour is the same for Mary and Alice.

Explain how a Markov chain Monte Carlo simulation using the Gibbs sampler can be utilized to estimate (a) the distribution of the amount of time spent at server \(j\) on a visit. Hint: Use the arrival theorem. (b) the proportion of time a customer is with server \(j\) (i.e., either in server \(j\) 's queue or in service with \(j\) ).

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?

Consider a system where the interarrival times have an arbitrary distribution \(F\), and there is a single server whose service distribution is \(G\). Let \(D_{n}\) denote the amount of time the \(n\) th customer spends waiting in queue. Interpret \(S_{n}, T_{n}\) so that $$ D_{n+1}=\left\\{\begin{array}{ll} D_{n}+S_{n}-T_{n}, & \text { if } D_{n}+S_{n}-T_{n} \geqslant 0 \\ 0, & \text { if } D_{n}+S_{n}-T_{n}<0 \end{array}\right. $$
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