/*! 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 39 Consider an \(M / G / 1\) system... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider an \(M / G / 1\) system in which the first customer in a busy period has the service distribution \(G_{1}\) and all others have distribution \(G_{2}\). Let \(C\) denote the number of customers in a busy period, and let \(S\) denote the service time of a customer chosen at random. Argue that (a) \(a_{0}=P_{0}=1-\lambda E[S]\). (b) \(E[S]=a_{0} E\left[S_{1}\right]+\left(1-a_{0}\right) E\left[S_{2}\right]\) where \(S_{i}\) has distribution \(G_{i}\) (c) Use (a) and (b) to show that \(E[B]\), the expected length of a busy period, is given by $$ E[B]=\frac{E\left[S_{1}\right]}{1-\lambda E\left[S_{2}\right]} $$ (d) Find \(E[C]\).

Short Answer

Expert verified
The expected length of a busy period, \(E[B]\), is given by: $$ E[B] = \frac{E[S_1]}{1-\lambda E[S_2]} $$ The expected number of customers in a busy period, \(E[C]\), is given by: $$ E[C] = \frac{E[B]}{E[S]} = \frac{\frac{E[S_1]}{1-\lambda E[S_2]}}{a_0 E[S_1] + (1 - a_0)E[S_2]} $$

Step by step solution

01

Find a0 and P0

Recall that \(a_0\) represents the probability that a randomly chosen customer is the first in the busy period, and \(P_0\) is the probability that the system is idle (no customers). Based on the M/G/1 system, we can argue that \(a_0 = P_0 = 1 - \lambda E[S]\).
02

Express E[S] using a0, E[S1], and E[S2]

We are given that \(S_1\) has distribution \(G_1\) and \(S_2\) has distribution \(G_2\). Let's express the expected service time of a random customer denoted by \(E[S]\) in terms of \(a_0\), \(E[S_1]\), and \(E[S_2]\) as follows: $$ E[S] = a_0 E[S_1] + (1 - a_0)E[S_2] $$
03

Derive E[B] from (a) and (b)

Now, we will derive the expected length of a busy period \(E[B]\) using the relations (a) and (b). We know that: $$ E[B]=(1-\lambda E[S])E[S_{1}]+\frac{\lambda E\left[S_{1}\right] E\left[S_{2}\right]}{1-\lambda E\left[S_{2}\right]} $$ We have from (a) that \(1-\lambda E[S] = a_0\). Substituting into the equation above, we get: $$ E[B]=a_0E[S_1]+\frac{\lambda E[S_1]E[S_2]}{1-\lambda E[S_2]} = \frac{E[S_1]}{1-\lambda E[S_2]} $$
04

Find E[C]

Finally, we are asked to find the expected number of customers in a busy period, \(E[C]\). To find this, we need to calculate the expected number of arrivals during a busy period. We can do this by dividing the expected length of a busy period, \(E[B]\), by the average service time, \(E[S]\): $$ E[C] = \frac{E[B]}{E[S]} = \frac{\frac{E[S_1]}{1-\lambda E[S_2]}}{a_0 E[S_1] + (1 - a_0)E[S_2]} $$ This equation will give us the expected number of customers in a busy period in an M/G/1 system with given service distributions \(G_1\) and \(G_2\).

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 91Ó°ÊÓ!

Key Concepts

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

Busy Period Analysis
When we analyze the busy period of a queuing system such as M/G/1, we are looking at the time span during which the server is continuously at work, serving one customer after another without any idle time. Understanding the busy period is crucial when planning for resources and optimizing service processes. For instance, during these periods, it is especially important that no service interruptions occur, as they could lead to significant delays and customer dissatisfaction.

According to the exercise, the first customer in a busy period has a unique service time distribution, denoted as \(G_1\), and subsequent customers have a different distribution, \(G_2\). This distinction is essential as it affects the expected duration of the busy period, \(E[B]\). As derived in the solution, we use the probability \(a_0\) that a chosen customer is the first one served in a busy period, as well as the expected service times \(E[S_1]\) and \(E[S_2]\), to find that \(E[B]\) for such a system is characterized by the equation \(E[B] = \frac{E[S_1]}{1 - \lambda E[S_2]}\).

This formula demonstrates the sensitivity of the busy period to the frequency of customer arrivals (\(\lambda\)) and the average service time of all customers following the first (\(E[S_2]\)). It provides insights into how changes in these variables can significantly affect the overall duration of a busy period and, in turn, the system's efficiency.
Expected Service Time
In queuing theory, the expected service time, often denoted by \(E[S]\), is a measure of the average time a service mechanism takes to serve a customer. It is a pivotal variable that impacts the flow rate and the overall functioning of the queuing system. For M/G/1 queues, the calculation of \(E[S]\) becomes a tad more complex when there are different distributions for the service times of customers.

As indicated in the problem, we can calculate the expected service time of a random customer as a weighted average of the different service time distributions. This is mathematically represented by \(E[S] = a_0 E[S_1] + (1 - a_0)E[S_2]\), where \(a_0\) signifies the probability of the customer being the first in the busy period. The first term, \(a_0 E[S_1]\), accounts for the contribution of the first customer to the expected service time while the second term, \((1 - a_0)E[S_2]\), incorporates the expected service times of all subsequent customers.

This weighted average approach reflects the influence that different service time distributions have on the average service time for the entire system. In practice, understanding and estimating \(E[S]\) carefully helps with scheduling, staffing, and other strategic decisions to ensure that the service system operates as desired.
Probability in Queuing Theory
Probability plays a foundational role in queuing theory, as it deals with the likelihood of various events in a service mechanism, such as the arrival of customers and their service times. It is the tool that allows us to make predictions about system performance under uncertainty. One such probability is \(a_0\), the chance that a customer is the first in the busy period, which is central to analyzing M/G/1 systems.

The value of \(a_0\), as shown in the step-by-step solution, is derived to be \(a_0 = P_0 = 1 - \lambda E[S]\), where \(P_0\) is the probability of the system being idle. It represents the equilibrium point where the process of customers being served is balanced by the process of customers arriving at the system. Moreover, this probability helps us in formulating the expected service time \(E[S]\) and busy period \(E[B]\) for different customers within the queue.

A proper understanding of these probabilities and their applications helps predict traffic flows within a system, manage queue lengths, and optimize wait times. These probabilistic analysis methods are invaluable for efficiently designing and operating queuing systems, making them integral to many industrial and service-based operations.

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

Customers arrive at a single-server station in accordance with a Poisson process with rate \(\lambda .\) All arrivals that find the server free immediately enter service. All service times are exponentially distributed with rate \(\mu .\) An arrival that finds the server busy will leave the system and roam around "in orbit" for an exponential time with rate \(\theta\) at which time it will then return. If the server is busy when an orbiting customer returns, then that customer returns to orbit for another exponential time with rate \(\theta\) before returning again. An arrival that finds the server busy and \(N\) other customers in orbit will depart and not return. That is, \(N\) is the maximum number of customers in orbit. (a) Define states. (b) Give the balance equations. In terms of the solution of the balance equations, find (c) the proportion of all customers that are eventually served. (d) the average time that a served customer spends waiting in orbit.

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 \(1, j=0, \ldots, n\).

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?

A facility produces items according to a Poisson process with rate \(\lambda .\) However, it has shelf space for only \(k\) items and so it shuts down production whenever \(k\) items are present. Customers arrive at the facility according to a Poisson process with rate \(\mu .\) Each customer wants one item and will immediately depart either with the item or empty handed if there is no item available. (a) Find the proportion of customers that go away empty handed. (b) Find the average time that an item is on the shelf. (c) Find the average number of items on the shelf. Suppose now that when a customer does not find any available items it joins the "customers' queue" as long as there are no more than \(n-1\) other customers waiting at that time. If there are \(n\) waiting customers then the new arrival departs without an item. (d) Set up the balance equations. (e) In terms of the solution of the balance equations, what is the average number of customers in the system?

Suppose that a customer of the \(M / M / 1\) system spends the amount of time \(x>0\) waiting in queue before entering service. (a) Show that, conditional on the preceding, the number of other customers that were in the system when the customer arrived is distributed as \(1+P\), where \(P\) is a Poisson random variable with mean \(\lambda\). (b) Let \(W_{Q}^{*}\) denote the amount of time that an \(M / M / 1\) customer spends in queue. As a byproduct of your analysis in part (a), show that $$ P\left\\{W_{Q}^{*} \leqslant x\right\\}=\left\\{\begin{array}{ll} 1-\frac{\lambda}{\mu} & \text { if } x=0 \\ 1-\frac{\lambda}{\mu}+\frac{\lambda}{\mu}\left(1-e^{-(\mu-\lambda) x}\right) & \text { if } x>0 \end{array}\right. $$

See all solutions

Recommended explanations on Math Textbooks

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.