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In an \(M / G / 1\) queue, (a) what proportion of departures leave behind 0 work? (b) what is the average work in the system as seen by a departure?

Short Answer

Expert verified
(a) The proportion of departures that leave behind 0 work is equal to the probability of the system being idle, which can be calculated as: \[P_0 = 1 - \rho\] (b) The average work in the system as seen by a departure can be found by dividing the average number of customers in the system (L) by the arrival rate (\(\lambda\)): \[W = \frac{L}{\lambda}\]

Step by step solution

01

Computing the Utilization Factor

First, calculate the utilization factor, denoted as \(\rho\). The utilization factor represents the proportion of time the server is busy. It can be calculated as: \[\rho = \frac{\lambda}{\mu}\]
02

Applying the Pollaczek-Khintchine Formula

Using the Pollaczek-Khintchine formula, we can find the average number of customers in the system by substituting the values of \(\lambda\), \(\mu\), and \(\sigma^2\) as well as the utilization factor \(\rho\) into the formula: \[L = \rho + \frac{\lambda^2 \sigma^2 + \lambda^2 \mu^2}{2(1-\rho)}\]
03

Answering Question (a) - Proportion of Departures

To find the proportion of departures that leave behind 0 work, we must first understand the concept of an idle period. An idle period occurs when there are no more customers in the system, and the server is idle. The proportion of departures that leave behind 0 work is equal to the probability of the system being idle, which can be calculated as: \[P_0 = 1 - \rho\]
04

Answering Question (b) - Average Work in System

To find the average work in the system as seen by a departure, we must find the average number of customers in the system (L) that we calculated in Step 2. Then, we need to divide the average number of customers (L) by the arrival rate (\(\lambda\)) to get the average work in the system: \[W = \frac{L}{\lambda}\] Now, we have broken down the exercise and explained how to compute the proportion of departures that leave behind 0 work and the average work in the system as seen by a departure for an M/G/1 queueing system.

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Key Concepts

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

M/G/1 Queue
In queueing theory, a common model is the M/G/1 queue. This model represents a single-server queueing system.
Here, 'M' stands for Markovian (or memoryless) arrival process, 'G' denotes a general service time distribution, and '1' indicates a single server. In the M/G/1 model, customers arrive at the queue according to a Poisson process.
This means the time between arrivals is exponentially distributed. However, the service times can take on any distribution.
The flexibility of the service time distribution allows for a more realistic description of many real-world systems.
  • Important: Arrival process is Poisson.
  • Service time can be any distribution.
  • Single-server handles one customer at a time.
Understanding this model helps in analyzing and predicting system performance, such as waiting time and queue length.
Utilization Factor
The utilization factor, denoted by \(\rho\), is a crucial metric in the study of queueing systems. It represents the proportion of time the server is actively working.
Mathematically, it is defined as the ratio of the arrival rate \(\lambda\) to the service rate \(\mu\): \(\rho = \frac{\lambda}{\mu}\).
  • \(\lambda\): Arrival rate (average number of arrivals per unit time).
  • \(\mu\): Service rate (average number of customers served per unit time).
  • \(\rho < 1\): The system is stable.
This factor indicates how busy the server is. If \(\rho\) equals 1 or more, the system may become overloaded, leading to infinite queues.
Therefore, maintaining a utilization factor less than 1 is essential for a stable system.
Pollaczek-Khintchine Formula
The Pollaczek-Khintchine (P-K) formula is a fundamental tool in queueing theory.
It provides a way to calculate the average number of customers in the system, denoted by \(L\), in an M/G/1 queue.
The formula is given by:\[ L = \rho + \frac{\lambda^2 \sigma^2 + \lambda^2 \mu^2}{2(1-\rho)} \]
  • \(\rho\): Utilization factor.
  • \(\lambda\): Arrival rate.
  • \(\sigma^2\): Variance of the service time.
  • \(\mu\): Service rate.
This formula shows that the average number of customers in the system depends on both the variability of the arrival and service processes.
It highlights the importance of service time variability (\(\sigma^2\)) in influencing queue length.
A higher variance indicates more fluctuations, leading to longer queues and wait times.
Idle Period
An idle period in a queueing system is the time when the server is not busy and there are no customers in the queue.
This concept helps in understanding how often a server is completely free from work. In the M/G/1 queue, the probability that the system is idle is:\[ P_0 = 1 - \rho \]
  • This means idle period probability is inversely related to \(\rho\).
  • More idle periods occur when \(\rho\) is small.
  • During idle periods, the server waits for new arrivals.
This probability, \(P_0\), helps to determine the proportion of departures that leave behind zero work.
A high portion of idle time means that many departures occur without leaving customers behind.
Average Work in System
The concept of average work in a system is essential for understanding performance metrics in queueing theory.
In an M/G/1 queue, it refers to the average amount of work or time a customer spends in the system.The average work can be calculated by dividing the average number of customers in the system (\(L\)) by the arrival rate (\(\lambda\)):\[ W = \frac{L}{\lambda} \]
  • \(W\): Average time a customer spends in the system.
  • \(L\): Average number of customers in the system (from P-K formula).
  • \(\lambda\): Arrival rate.
This measure provides insight into system efficiency.
A lower average work time signifies a more efficient system with quicker service and shorter queues.

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Most popular questions from this chapter

Carloads of customers arrive at a single-server station in accordance with a Poisson process with rate 4 per hour. The service times are exponentially distributed with rate 20 per hour. If cach carload contains either 1,2 , or 3 customers with respective probabilities \(\frac{1}{2}, \frac{1}{2}, \frac{1}{4}\) compute the average customer delay in queue.

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 $$ E[\text { extra time } \mid N]=\frac{N E\left[S_{1}\right]}{1-\lambda_{1} E\left[S_{1}\right]} $$ 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 $$ W_{Q}^{2}=W_{Q}^{2} \text { (nonpreemptive) }+\frac{\lambda_{1} E\left[S_{1}\right] E\left[S_{2}\right]}{1-\lambda_{1} E\left[S_{1}\right]} $$

Customers arrive at a two-server station in accordance with a Poisson process with a rate of two per hour. Arrivals finding server 1 free begin service with that server. Arrivals finding server 1 busy and server 2 free begin service with server 2. Arrivals finding both servers busy are lost. When a customer is served by server 1 , she then either enters service with server 2 if 2 is free or departs the system if 2 is busy. A customer completing service at server 2 departs the system. The service times at server 1 and server 2 are exponential random variables with respective rates of four and six per hour. (a) What fraction of customers do not enter the system? (b) What is the average amount of time that an entering customer spends in the system? (c) What fraction of entering customers receives service from server \(1 ?\)

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 according to a Poisson process having rate \(\lambda=5\). An arrival finding server 1 free will begin service with that server. An arrival finding server 1 busy and server 2 free will enter service with server 2. An arrival finding both servers busy goes away. Once a customer is served by either server, he departs the system. The service times at server \(i\) are exponential with rates \(\mu_{i}\), where \(\mu_{1}=4, \mu_{2}=2\) (a) What is the average time an entering customer spends in the system? (b) What proportion of time is server 2 busy?

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