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Chapter 8: Problem 10

A group of \(m\) customers frequents a single-server station in the following manner. When a customer arrives, he or she either enters service if the server is free or joins the queue otherwise. Upon completing service the customer departs the system, but then returns after an exponential time with rate \(\theta\). All service times are exponentially distributed with rate \(\mu\). (a) Find the average rate at which customers enter the station. (b) Find the average time that a customer spends in the station per visit.

Short Answer

Expert verified
(a) The average rate at which customers enter the station is \(\theta\). (b) The average time a customer spends in the station per visit is \(\frac{1}{\mu - \theta}\).

Step by step solution

01

(a) Average entrance rate

To find the average rate at which customers enter the station, we will look at the rate at which customers return to the station after completing their service. Since the time until returning is exponentially distributed with rate \(\theta\), we can infer that the average rate at which customers enter the station is equal to the rate of returning customers, so the average entrance rate is given by \(\theta\).
02

(b) Average time spent in the station per visit

To find the average time a customer spends in the system per visit, we can use Little's Law for an M/M/1 queue. Little's Law states that the average number of customers in the system (L) equals the average arrival rate (\(\lambda\)) times the average time a customer spends in the system per visit (W): \[L = \lambda W\] For an M/M/1 queue, the average time a customer spends in the system per visit (W) is given by: \[W = \frac{1}{\mu - \lambda}\] Since we already know that the average rate at which customers enter the station is given by \(\theta\), we can substitute this for \(\lambda\): \[W = \frac{1}{\mu - \theta}\] Hence, the average time a customer spends in the station per visit is \(\frac{1}{\mu - \theta}\).

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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 fundamental concept in queueing theory. It represents a simple system where arrivals and service times are both exponentially distributed, and there is only one server available. This model is widely used because it provides a clear insight into how queues behave in various settings.
Let's break down what "M/M/1" means:
  • The first "M" stands for "Memoryless," indicating that arrivals follow a Poisson process with an exponential inter-arrival time distribution.
  • The second "M" describes the service time, which is also exponentially distributed.
  • The "1" signifies that there's just one server handling the customers.
This model is quite powerful. It allows us to calculate important performance measures like the average waiting time, the average number of customers in the queue, and the utilization of the server. These metrics are crucial for managing and optimizing any service system that experiences variability in demand.
Little's Law
Little's Law is a very neat concept in queueing theory, relating three key metrics of a queueing system. These are the long-term average number of customers in a stationary system (L), the long-term average arrival rate (\(位\)), and the average time a customer spends in the system (W). The law can be expressed as:L = 位W.
This relationship helps us understand how changes in one aspect of the system will affect others.
For example:
  • If the arrival rate (\(位\)) increases and the average number of people in the system (L) remains constant, then the average time spent (W) in the system must decrease.
  • Conversely, if W increases while 位 remains constant, more customers are present in the system on average.
Little's Law is fundamental as it holds under very general conditions. It's widely used in various fields, from computer networks to manufacturing, to estimate crucial performance indices of queueing systems.
Exponential Distribution
The exponential distribution is key to understanding the M/M/1 queue, as it describes both arrival and service times. It's a continuous probability distribution often used to model the time until an event occurs, such as the next customer arrival or the completion of service.
Key properties include:
  • Memoryless: The process has no memory. That means the probability of an event occurring in the future is independent of any previous events.
  • Mean: The mean or expected value of an exponential distribution is \( 1/位 \), where \(位\) is the rate parameter for arrivals or service times.
  • Variance: Similarly, the variance is \( 1/位^2 \).
This distribution is especially handy in modeling real-world processes where events occur continuously and independently over time. Its simplicity allows for easy analytical treatment and prediction of system performance.
Service Rate
The service rate is a critical factor in determining the efficiency of a queueing system. It's the speed at which the server can process customers in an M/M/1 queue.
Here's how service rate plays into the calculations:
  • It is often denoted by \(渭\), representing the average number of customers that can be served per unit of time.
  • A higher service rate means the system can handle more customers quickly, reducing wait times and potentially minimizing congestion.
Essentially, while the arrival rate determines how fast customers come into the system, the service rate defines how quickly they can be processed. Balancing these two rates is vital to maintain an efficient and customer-friendly operation. If the service rate is less than the arrival rate, queues will build up, leading to longer wait times and potential service quality issues.

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Most popular questions from this chapter

For the \(M / G / 1\) queue, let \(X_{n}\) denote the number in the system left behind by the nth departure. (a) If $$ X_{n+1}=\left\\{\begin{array}{ll} X_{n}-1+Y_{n}, & \text { if } X_{n} \geqslant 1 \\ Y_{n}, & \text { if } X_{n}=0 \end{array}\right. $$ what does \(Y_{n}\) represent? (b) Rewrite the preceding as $$ X_{n+1}=X_{n}-1+Y_{n}+\delta_{n} $$ where $$ \delta_{n}=\left\\{\begin{array}{ll} 1, & \text { if } X_{n}=0 \\ 0, & \text { if } X_{n} \geqslant 1 \end{array}\right. $$ Take expectations and let \(n \rightarrow \infty\) in Equation ( \(8.64\) ) to obtain $$ E\left[\delta_{\infty}\right]=1-\lambda E[S] $$ (c) Square both sides of Equation \((8.64)\), take expectations, and then let \(n \rightarrow \infty\) to obtain $$ E\left[X_{\infty}\right]=\frac{\lambda^{2} E\left[S^{2}\right]}{2(1-\lambda E[S])}+\lambda E[S] $$ (d) Argue that \(E\left[X_{\infty}\right]\), the average number as seen by a departure, is equal to \(L\).

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.

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.

Consider a sequential-service system consisting of two servers, \(A\) and \(B\). Arriving customers will enter this system only if server \(A\) is free. If a customer does enter, then he is immediately served by server \(A\). When his service by \(A\) is completed, he then goes to \(B\) if \(B\) is free, or if \(B\) is busy, he leaves the system. Upon completion of service at server \(B\), the customer departs. Assume that the (Poisson) arrival rate is two customers an hour, and that \(A\) and \(B\) serve at respective (exponential) rates of four and two customers an hour. (a) What proportion of customers enter the system? (b) What proportion of entering customers receive service from B? (c) What is the average number of customers in the system? (d) What is the average amount of time that an entering customer spends in the system?

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. Assume customers arrive at a Poisson rate of 10 per hour (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.
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