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Potential customers arrive at a single-server station in accordance with a Poisson process with rate \(\lambda .\) However, if the arrival finds \(n\) customers already in the station, then he will enter the system with probability \(\alpha_{n}\). Assuming an exponential service rate \(\mu\), set this up as a birth and death process and determine the birth and death rates.

Short Answer

Expert verified
The birth and death rates for the single-server station with potential customers arriving according to a Poisson process can be determined as follows: - Birth rates (λ_n) are given by: \(λ_n = λα_n\) - Death rates (μ_n) are given by: \(μ_n = μ\)

Step by step solution

01

Understand the system properties

Begin by understanding the properties of the system: - Arrivals occur according to a Poisson process with a rate of λ. - Customers enter the system with a probability α_n, where n is the number of customers already in the station. - The service rate is μ, and service time follows an exponential distribution. - The system operates as a birth and death process.
02

Set up the birth and death process

Next, we need to setup the birth process (birth rates) and the death process (death rates). - Birth rates (λ_n): these represent the rates at which new customers enter the system. In our case, a customer arrives with a rate of λ, and the probability of them entering the system is α_n. Therefore, the birth rate based on the number of customers already in the station is given by: λ_n = λα_n - Death rates (μ_n): these represent the rates at which customers leave the system. In our case, since it's a single-server station and each customer has an exponential service time with a rate of μ, the death rate is given by: μ_n = μ
03

Determine the birth and death rates

Since we have found the birth rates based on α_n and λ, and the death rates based on μ, we can conclude the exercise with: - Birth rates (λ_n) are given by: λ_n = λα_n - Death rates (μ_n) are given by: μ_n = μ

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Key Concepts

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

Poisson process
The Poisson process is a fundamental concept in stochastic processes, particularly useful in modeling random events that occur independently and with a constant average rate over time, such as customers arriving at a station, phone calls coming into a call center, or even radioactive decay. In the context of the exercise at hand, the Poisson process describes the arrival of potential customers to a service station.

The arrival rate, denoted as \( \lambda \) for the Poisson process, is essential as it allows for the calculation of the probability of a certain number of events (in this case, customer arrivals) occurring in a fixed interval of time. This probability distribution is discrete, which means it predicts the likelihood of a number of events happening in a given timeframe.

In practice, when a Poisson process is in place, two arrivals cannot occur at exactly the same time, and the number of arrivals in non-overlapping intervals are independent of each other. Understanding the Poisson process is crucial when setting up a birth and death process as it lays the foundation for creating reliable probability models for systems like the one being analyzed.
Exponential service rate
An exponential service rate is an important concept, especially in queuing theory and service systems. It's characterized by the memoryless property, which implies that the length of time until the next service completion is independent of the amount of time already waited.

In the exercise scenario, the service rate is denoted by \( \mu \), which indicates the rate at which the single server can service customers. If service times are exponentially distributed, the probability of a service being completed in the next infinitesimal time interval is the same, no matter how long the service has already been in progress. This is because the exponential distribution assumes that events (in this case, service completions) occur continuously and independently.

The service rate \( \mu \) directly affects the death rates in a birth and death process, as it represents the rate at which customers are leaving the system. With a steady, single-channel service rate, every customer is expected to receive service within a predictable time frame, on average, which makes modeling the system's behavior much more manageable.
Probability models
In creating effective probability models, it is essential to understand and apply different mathematical tools and concepts to describe random phenomena accurately. Probability models use mathematical functions and distributions to capture the randomness and uncertainty in systems, helping to predict the likelihood of various outcomes.

The birth and death process mentioned in the exercise is a specific type of Markov chain used extensively in queuing theory to model systems where arrivals (births) and services (deaths) occur randomly over time. To set this model up, we examine how likely it is that customers arrive and enter the system (which depends on the arrival rate \(\lambda\) and the entering probability \(\alpha_n\)) and how likely it is for customers to be serviced and leave the system (determined by the service rate \(\mu\)).

By applying the right probability distributions, such as the Poisson distribution for the arrival of customers and the exponential distribution for service times, a birth and death process can accurately depict the behavior of the queuing system under study. This particular probability model helps in determining the system's performance, like expected wait times or the probability of a customer having to wait before being serviced.

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Most popular questions from this chapter

A small barbershop, operated by a single barber, has room for at most two customers. Potential customers arrive at a Poisson rate of three per hour, and the successive service times are independent exponential random variables with mean \(\frac{1}{4}\) hour. (a) What is the average number of customers in the shop? (b) What is the proportion of potential customers that enter the shop? (c) If the barber could work twice as fast, how much more business would he do?

Customers arrive at a service station, manned by a single server who serves at an exponential rate \(\mu_{1}\), at a Poisson rate \(\lambda .\) After completion of service the customer then joins a second system where the server serves at an exponential rate \(\mu_{2} .\) Such a system is called a tandem or sequential queueing system. Assuming that \(\lambda<\mu_{i}\), \(i=1,2\), determine the limiting probabilities. Hint: Try a solution of the form \(P_{n, m}=C \alpha^{n} \beta^{m}\), and determine \(C, \alpha, \beta\).

(a) Show that Approximation 1 of Section \(6.8\) is equivalent to uniformizing the continuous-time Markov chain with a value \(v\) such that \(v t=n\) and then approximating \(P_{i j}(t)\) by \(P_{i j}^{* n}\). (b) Explain why the preceding should make a good approximation. Hint: What is the standard deviation of a Poisson random variable with mean \(n\) ?

Customers arrive at a two-server station in accordance with a Poisson process having rate \(\lambda\). Upon arriving, they join a single queue. Whenever a server completes a service, the person first in line enters service. The service times of server \(i\) are exponential with rate \(\mu_{i}, i=1,2\), where \(\mu_{1}+\mu_{2}>\lambda .\) An arrival finding both servers free is equally likely to go to either one. Define an appropriate continuoustime Markov chain for this model, show it is time reversible, and find the limiting probabilities.

Consider a time reversible continuous-time Markov chain having infinitesimal transition rates \(q_{i j}\) and limiting probabilities \(\left\\{P_{i}\right\\} .\) Let \(A\) denote a set of states for this chain, and consider a new continuous-time Markov chain with transition rates \(q_{i j}^{*}\) given by $$ q_{i j}^{*}=\left\\{\begin{array}{ll} c q_{i j}, & \text { if } i \in A, j \notin A \\ q_{i j}, & \text { otherwise } \end{array}\right. $$ where \(c\) is an arbitrary positive number. Show that this chain remains time reversible, and find its limiting probabilities.

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