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Chapter 8: Problem 4

For the \(M / M / 1\) queue, show that the probability that a customer spends an amount of time \(x\) or less in quene is given by $$ \begin{aligned} 1-\frac{\lambda}{\mu}, \quad \text { if } x=0 & \\ 1-\frac{\lambda}{\mu}+\frac{\lambda}{\mu}\left(1-e^{-(n-\lambda) r}\right), & \text { if } x>0 \end{aligned} $$

Short Answer

Expert verified
The probability that a customer spends an amount of time \(x\) or less in the M/M/1 queue is given by: - \(P(X \leq 0) = 1 - \frac{\lambda}{\mu}\) for the case when \(x=0\). - \(P(X \leq x) = 1 - \frac{\lambda}{\mu} + \frac{\lambda}{\mu}(1 - e^{-(\mu - \lambda) x})\) for the case when \(x > 0\).

Step by step solution

01

Define the variables

First, we should understand the variables given in the problem. We have a queue described by M/M/1, where: - M: Poisson arrival process - M: Exponential service times - 1: Single server Additionally, we have 位, which represents the arrival rate, and 渭, which represents the service rate.
02

Identify the queue utilization factor

The queue utilization factor, 蟻, represents the proportion of time the server is busy. In the M/M/1 queue, it can be defined as: 蟻 = \(\frac{\lambda}{\mu}\)
03

Determine the probability that the queue is empty

P(0) is the probability that there are no customers in the queue. Assuming steady-state conditions, this is given by: P(0) = \(1 鈭� \frac{\lambda}{\mu}\) This corresponds to the first part of the given formula when x=0.
04

Derive the probability distribution of waiting time in the queue

The distribution of waiting time in the queue can be represented by the function W(x). For the M/M/1 queue, the waiting time follows an exponential distribution with rate (渭 - 位). Therefore, the distribution function of waiting time in the queue can be expressed as: W(x) = \(1 - e^{-(\mu - \lambda) x}\)
05

Show the probability of spending x or less time in the queue

The given function represents the probability of a customer spending x or less time in the queue. When x=0, the probability is equal to P(0), which we have already determined in Step 3. Now, let's determine the probability for the case when x>0. We can use the waiting time distribution (W(x)) and the probability of the queue being empty (P(0)), as the overall probability will be the sum of the probability that the queue is empty and the probability that the waiting time is x or less, multiplied by the queue utilization factor: P(X 鈮� x) = P(0) + 蟻 * W(x) = \(1 - \frac{\lambda}{\mu} + \frac{\lambda}{\mu} (1 - e^{-(\mu - \lambda) x})\) This formula corresponds to the given function when x>0. Thus, we have shown that the given formula represents the probability that a customer spends x or less time in the M/M/1 queue.

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Key Concepts

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

Queueing Theory
Queueing theory is a branch of mathematics that studies the phenomenon of waiting in lines, or queues. It is commonly applied in diverse fields such as telecommunications, traffic engineering, and service facilities like banks, hospitals, and supermarkets.

The primary goal of queueing theory is to derive mathematical models that can predict queue lengths and waiting times. This helps in managing resources efficiently and improving service quality. In our M/M/1 queue example, the theory helps us evaluate the performance of a single-server queue system where customers arrive and depart following specific probability distributions.
Exponential Distribution
The exponential distribution is a continuous probability distribution commonly used to model the time between events in a Poisson process. It is characterized by its memoryless property, which means that the probability of an event occurring in the next moment is the same regardless of how much time has already passed.

In queueing systems, the exponential distribution can be used to describe service times. For our M/M/1 queue, the 'M' relates to this type of distribution, indicating that the time taken to serve a customer is exponentially distributed. The mean service time is determined by the service rate, \(\mu\).
Poisson Arrival Process
The Poisson arrival process is a model used in queueing theory to describe the random nature of arrivals at a facility. It is characterized by the arrivals occurring independently and at a constant average rate. This process is particularly suitable for high-volume and low-density systems.

In our example, the first 'M' in M/M/1 denotes that the customer arrival follows a Poisson distribution. The arrival rate, \(\lambda\), is the average number of customers that arrive per unit of time. Thanks to its mathematical properties, the Poisson distribution simplifies analysis and calculation of probabilities in queueing systems.
Service Rate
The service rate, often denoted as \(\mu\), is a key parameter in any queueing system. It represents the average number of customers that can be served by the system per unit time. A higher service rate usually indicates a more efficient system since the server can handle more customers within the same period, reducing the likelihood of long queues and wait times.

In the M/M/1 context, \(\mu\) is used to calculate the average service time and to determine the queue's efficiency. It is this rate which, when compared with the arrival rate, gives us the queue utilization factor, \(\rho\), crucial for assessing the system's performance.
Arrival Rate
The arrival rate, defined as \(\lambda\), measures the average number of customer arrivals at a queue per time unit. It is critical in determining the intensity of the incoming traffic to the service facility. An increase in the arrival rate generally leads to longer queues and increased waiting times, assuming the service rate remains constant.

Understanding the arrival rate allows us to evaluate the likelihood of the queue being empty and to estimate the waiting times for customers in the M/M/1 queueing system. It's essential for predicting service quality and for planning resource allocations to ensure efficient queue management.

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Most popular questions from this chapter

A supermarket has two exponential checkout counters, each operating at rate \(\mu\). Arrivals are Poisson at rate \(\lambda\). The counters operate in the following way: (i) One queue feeds both counters. (ii) One counter is operated by a permanent checker and the other by a stock clerk who instantaneously begins checking whenever there are two or more customers in the system. The clerk returns to stocking whenever he completes a service, and there are fewer than two customers in the system. (a) Let \(P_{n}=\) proportion of time there are \(n\) in the system. Set up equations for \(P_{n}\) and solve. (b) At what rate does the number in the system go from 0 to \(1 ?\) from 2 to \(1 ?\) (c) What proportion of time is the stock clerk checking? Hint: Be a little careful when there is one in the system.

Consider a system where the interarrival times have an arbitrary distribution \(F\), and there is a single server whose service distribution is \(G\). Let \(D_{n}\) denote the amount of time the \(n\) th customers spends waiting in queue. Interpret \(S_{n}, T_{n}\) so that $$ D_{n+1}=\left\\{\begin{array}{ll} D_{n}+S_{n}-T_{n}, & \text { if } D_{n}+S_{n}-T_{n} \geq 0 \\ 0, & \text { if } D_{n}+S_{n}-T_{n}<0 \end{array}\right. $$

There are two types of customers. Type \(I\) 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?

Consider a closed queueing network consisting of two customers moving among two servers, and suppose that after each service completion the customer is equally likely to go to either server-that is, \(P_{1,2}=P_{2,1}=\frac{1}{2}\). Let \(\mu_{i}\) denote the exponential service rate at server \(i, i=1,2\). (a) Determine the average number of customers at each server. (b) Determine the service completion rate for each server.

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, it 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, the customer 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 \(n\) time.
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