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Chapter 11: Problem 3

Series. In a Moscow supermarket customers queue at the cash desk to pay for the goods they want: then they proceed to a second line where they wait for the goods in question. If customers arrive in the shop like a Poisson process with parameter \(\lambda\) and all service times are independent and exponentially distributed, parameter \(\mu_{1}\) at the first desk and \(\mu_{2}\) at the second, find the stationary distributions of queue lengths, when they exist, and show that, at any given time, the two queue lengths are independent in equilibrium.

Short Answer

Expert verified
Queue lengths follow geometric distributions and are independent.

Step by step solution

01

Understanding the Problem

We have a two-stage queueing system in a Moscow supermarket. Customers arrive according to a Poisson process with parameter \( \lambda \), and they go through two service stages with parameters \( \mu_1 \) and \( \mu_2 \) respectively. We need to determine the stationary distributions for the queue lengths at each stage and check if they are independent in equilibrium.
02

Modeling Each Queue

Each service stage can be modeled as an \( M/M/1 \) queue. This means each stage has a single server, exponential inter-arrival (Poisson) and service times, and a FIFO discipline. For Stage 1, the arrival rate is \( \lambda \) and the service rate is \( \mu_1 \). For Stage 2, the effective arrival rate equals the departure rate from Stage 1, and the service rate is \( \mu_2 \).
03

Calculate the Utilization Factors

The utilization factor \( \rho_1 \) for Stage 1 is \( \rho_1 = \frac{\lambda}{\mu_1} \) and for Stage 2 is \( \rho_2 = \frac{\lambda}{\mu_2} \). For stationarity, both \( \rho_1 \) and \( \rho_2 \) must be less than 1. This condition ensures that each system reaches a steady state where queues do not grow indefinitely.
04

Finding the Stationary Distributions

In an \( M/M/1 \) queue, if \( \rho < 1 \), the stationary distribution for the number of customers in the system is a geometric distribution. For Stage 1, the distribution is \( P(N_1 = n_1) = (1-\rho_1)\rho_1^{n_1} \), and for Stage 2, it is \( P(N_2 = n_2) = (1-\rho_2)\rho_2^{n_2} \), where \( n_1 \) and \( n_2 \) are the number of customers in queues 1 and 2 respectively.
05

Independence of Queue Lengths

Since both queues are modeled as separate \( M/M/1 \) queues and their stationary distributions are derived independently, the joint distribution will be the product of their individual stationary distributions: \( P(N_1 = n_1, N_2 = n_2) = P(N_1 = n_1) \times P(N_2 = n_2) \). This indicates that the queue lengths at the two service stages are independent in equilibrium.

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Key Concepts

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

M/M/1 queue
An M/M/1 queue is a simple and widely used model in queueing theory, ideal for understanding basic queue dynamics. In this system:
  • The first 'M' stands for 'memoryless,' which indicates that arrivals occur based on a Poisson process.
  • The second 'M' suggests that service times are memoryless as well, following an exponential distribution.
  • The '1' represents a single server processing the queued items.
Customers arriving in this queue follow a Poisson distribution, meaning they are as likely to arrive at any time within a fixed interval. Each queue operates under a first-come, first-served discipline, much like a typical line in a store. This model helps analysts predict how queues will behave and aids in designing smoother service processes.
Poisson process
The Poisson process is critical in modeling random arrival times in queues. It captures the way events happen independently over continuous time, which suits dynamic environments like supermarkets. This process is defined by:
  • An arrival rate (\( \lambda \)) denoting the average number of customers entering the system per time unit.
  • Independence, meaning the arrival of one customer doesn鈥檛 affect others.
A Poisson process is perfect for predicting probabilities of customer arrivals, ensuring that queues are modeled with a realistic depiction of uncertainty.
Stationary distribution
The stationary distribution describes the long-run behavior of a queue, providing the likelihood that a system is in any given state. Importantly, it's stable when certain conditions are met:
  • The utilization factor (\( \rho \)) must be below one for each queue so that they do not grow indefinitely.
  • State probabilities do not change over time, ensuring predictability.
For an \( M/M/1 \) queue, this distribution is geometric, which means it becomes easier to analyze and manage queue lengths effectively once this stable state is achieved.
Utilization factor
The utilization factor, denoted as \( \rho \), reflects the efficiency of a service station and helps in identifying the load a system can handle. It measures:
  • The ratio of the arrival rate (\( \lambda \)) to the service rate (\( \mu \)).
  • Whether a system can reach a steady state, critical for operational efficiency.
When \( \rho < 1 \), the queue is stable, allowing us to determine that queues are not excessively long, but are manageable, thus delivering a good service experience to customers.

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Most popular questions from this chapter

Finite waiting room. Consider \(\mathrm{M}(\lambda) \mathrm{M}(\mu) / k\) with the constraint that arriving customers who see \(N\) customers in the line ahead of them leave and never retum. Find the stationary distribution of queue length for the cases \(k=1\) and \(k=2\),

Let \(Q(t)\) be the length of an \(\mathrm{M}(\lambda) / \mathrm{M}(\mu) / 1\) queue at time \(t\), and let \(Z=\left\\{Z_{n}\right]\) be the jump chain of \(Q\). Explain how the stationary distribution of \(Q\) may be derived from that of \(Z\), and vice versa.

Consider an extremely idealized model of a telephone exchange having infinitely many channels available. Calls arrive in the manner of a Poisson process with intensity \(\lambda\), and cach requires one channel for a length of time having the exponential distribution with parameter \(\mu\), independently of the arrival process and of the duration of other calts. Let \(Q(t)\) be the number of calls being handled at time \(t\), and suppose that \(Q(0)=I\). Determine the probability generating function of \(Q(t)\), and deduce \(\mathrm{E}(Q(t)), \mathbb{P}(Q(t)=0)\), and the limiting distribution of \(Q(t)\) as \(t \rightarrow \infty\) Assuming the queue is in equilibrium, find the proportion of time that no channels are occupied, and the mean length of an idle period. Deduce that the mean length of a busy period is \(\left(e^{\lambda / \mu}-1\right) / \lambda\).

Difficult customers. Consider an \(\mathrm{M}(\lambda) / \mathrm{M}(\mu) / 1\) queue modified so that on completion of service the customer leaves with probability \(\delta\), or rejoins the queue with probability \(1-\delta\). Find the distribution of the total time a customer spends being served. Hence show that equilibrium is possible if \(\lambda<\delta \mu\), and find the stationary distribution. Show that, in equilibrium, the departure process is Poisson, but if the rejoining customer goes to the end of the queue, the composite arrival process is not Poisson.

Consider \(\operatorname{GM}(\mu) / 1\), and let \(G\) be the distribution function of \(S-X\) where \(S\) and \(X\) are typical (independent) service and interarrival times. Show that the Wiener-Hopf equation $$ F(x)=\int_{-\infty}^{x} F(x-y) d G(y), \quad x \geq 0 $$ for the limiting waiting-time distribution \(F\) is satisfied by \(F(x)=1-\eta e^{-\mu(1-\eta) x}, x \geq 0 .\) Here, \(\eta\) is the smallest positive root of the equation \(x=M_{X}(\mu(x-1))\), where \(M_{X}\) is the moment generating function of \(X\).
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