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Chapter 8: Problem 1

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."

Short Answer

Expert verified
For the M/M/1 queue system, the expected number of arrivals during a service period is given by \(E[N] = 位 \frac{1}{渭}\). The probability that no customers arrive during a service period is given by \(P_0 = e^{-位 \frac{1}{渭}}\).

Step by step solution

01

Understanding the M/M/1 queue model

The M/M/1 queue system can be represented by the parameters (位, 渭), where 位 represents the average arrival rate of customers (arrivals per unit time), and 渭 is the average service rate (services per unit time). Since the arrivals follow a Poisson distribution, the probability of k arrivals in a unit time is given by \(P_k = \frac{e^{-位} 位^k}{k!}\), whereas service times follow an exponential distribution with parameter 渭. In this exercise, we are interested in computing the expected number of arrivals and the probability of no arrivals during a service period, which is the time taken by the server to serve a customer. To solve this exercise, we need to find the distribution of customers' arrivals during the service period. (a) Expected number of arrivals during a service period
02

Calculate expected service time

The expected service time, denoted as E[T], is the inverse of the service rate (mean service time). Thus: \[E[T] = \frac{1}{渭}\]
03

Calculate the expected number of arrivals

To compute the expected number of arrivals during a service period, we can use the fact that the expected number of arrivals in a given interval is proportional to the length of that interval in a Poisson process. So, given that E[T] is the expected length of a service period: \[E[N] = 位 E[T]\] where E[N] is the expected number of arrivals during a service period. Substituting the expected service time in the formula above: \[E[N] = 位 \frac{1}{渭}\] (b) Probability that no customers arrive during a service period
04

Calculate the probability that no customers arrive

To find the probability that no customers arrive during a service period, we need to apply the Poisson probability formula for k=0 arrivals during an interval of length E[T]: \[P_0 = \frac{e^{-位E[T]} (位E[T])^0}{0!} = e^{-位E[T]}\] Now, substituting the expression for E[T]: \[P_0 = e^{-位 \frac{1}{渭}}\] So, this is the probability that no customers will arrive during a service period. In summary, the expected number of arrivals during a service period is \(E[N] = 位 \frac{1}{渭}\), and the probability that no customers arrive during a service period is \(P_0 = e^{-位 \frac{1}{渭}}\).

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Key Concepts

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

Poisson Distribution
The Poisson distribution is an essential concept in understanding the behavior of random events occurring independently within a fixed interval of time or space. It applies to scenarios where these events happen at a constant average rate and the number of occurrences in a given interval can be counted, but their exact timing is unpredictable.

Within the M/M/1 queue model context, the Poisson distribution describes the probability of a specific number of customers arriving over a defined period. Mathematically, the probability of observing exactly k arrivals is expressed as follows:
\[P_k = \frac{e^{-\lambda} \lambda^k}{k!}\]
where \(e\) is the base of the natural logarithm (approximately equal to 2.71828), \(\lambda\) is the average arrival rate, and k is the number of arrivals. Particularly, when k=0, this formula provides the probability of no arrivals, which is a typical calculation in queue analysis.
Average Arrival Rate
When dissecting the M/M/1 queue model, understanding the concept of average arrival rate, denoted by \(\lambda\), is crucial. This rate measures the number of customers expected to arrive per time unit. In practice, the average arrival rate can be observed by counting arrivals over several periods and calculating the mean. For a service system, a high \(\lambda\) indicates that on average, more customers are arriving to the queue, potentially leading to congestion if the service facility cannot handle them effectively.

In the provided exercise, the average arrival rate helps to determine both the expected number of customers during a service period and their arrival probabilities. This rate plays a significant role in the predictability of the queue鈥檚 length and waiting times, which are vital parameters for designing and managing service systems.
Service Rate
Service rate, denoted as \(\mu\), represents the number of customers that can be served per time unit in the M/M/1 queue model. This is not just any arbitrary number but is a measure of the service facility's efficiency and capability. A higher service rate indicates that the server can handle more customers within the same time period, thus reducing waiting times and potential queue lengths.

Both the average arrival rate and the service rate are fundamental to queue dynamics. If the service rate is higher than the average arrival rate (\(\mu > \lambda\)), the system is generally stable, and customers can be served in a timely manner. Conversely, if \(\mu < \lambda\), it means the service is slower than the rate at which customers are arriving; this leads to longer queues and potential delays. The balance between these two rates is essential for the smooth operation of any service system.

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Most popular questions from this chapter

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 \(E[C]\).

Consider the \(M / M / 1\) system in which customers arrive at rate \(\lambda\) and the server serves at rate \(\mu .\) However, suppose that in any interval of length \(h\) in which the server is busy there is a probability \(\alpha h+o(h)\) that the server will experience a breakdown, which causes the system to shut down. All customers that are in the system depart, and no additional arrivals are allowed to enter until the breakdown is fixed. The time to fix a breakdown is exponentially distributed with rate \(\beta\). (a) Define appropriate states. (b) Give the balance equations. In terms of the long-run probabilities, (c) what is the average amount of time that an entering customer spends in the system? (d) what proportion of entering customers complete their service? (e) what proportion of customers arrive during a breakdown?

Consider a single-server exponential system in which ordinary customers arrive at a rate \(\lambda\) and have service rate \(\mu .\) In addition, there is a special customer who has a service rate \(\mu_{1}\). Whenever this special customer arrives, she goes directly into service (if anyone else is in service, then this person is bumped back into queue). When the special customer is not being serviced, she spends an exponential amount of time (with mean \(1 / \theta\) ) out of the system. (a) What is the average arrival rate of the special customer? (b) Define an appropriate state space and set up balance equations. (c) Find the probability that an ordinary customer is bumped \(n\) times.

There are two types of customers. Type 1 and 2 customers arrive in accordance with independent Poisson processes with respective rate \(\lambda_{1}\) and \(\lambda_{2}\). There are two servers. A type 1 arrival will enter service with server 1 if that server is free; if server 1 is busy and server 2 is free, then the type 1 arrival will enter service with server \(2 .\) If both servers are busy, then the type 1 arrival will go away. A type 2 customer can only be served by server 2 ; if server 2 is free when a type 2 customer arrives, then the customer enters service with that server. If server 2 is busy when a type 2 arrives, then that customer goes away. Once a customer is served by either server, he departs the system. Service times at server \(i\) are exponential with rate \(\mu_{i}, i=1,2\). Suppose we want to find the average number of customers in the system. (a) Define states. (b) Give the balance equations. Do not attempt to solve them. In terms of the long-run probabilities, what is (c) the average number of customers in the system? (d) the average time a customer spends in the system?

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 \(1 ?\)
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