/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 A group of \(n\) customers moves... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 7

A group of \(n\) customers moves around among two servers. Upon completion of service, the served customer then joins the queue (or enters service if the server is free) at the other server. All service times are exponential with rate \(\mu .\) Find the proportion of time that there are \(j\) customers at server \(1, j=0, \ldots, n .\)

Short Answer

Expert verified
The proportion of time that there are \(j=0,1,\ldots,n\) customers at server 1 is \(\frac{1}{(n+1)}\).

Step by step solution

01

1. Determine state space

State space for this Markov chain is given by \(M(N, C_1, C_2) \in S = \{ M(N, j, k) : j + k = N , j, k \in \mathbb{N}\}\), where \(j = 0,1, \ldots, n\) and \(k=n-j\) is the number of customers at server 2.
02

2. Write the balance equations

We will use the balance equations for a steady-state Markov chain, which state that the total rate of leaving the state is the same as the total rate of entering the state: $$\mu p_{j-1} = \mu p_{j}$$ where \(p_{j}\) is the steady-state probability of having \(j\) customers in server 1.
03

3. Simplify and obtain equations for steady-state probabilities

We simplify the balance equation to obtain an expression for the steady-state probability \(p_{j}\): $$p_{j} = p_{j-1}$$ This shows that the probabilities are the same for each state, given that a customer moves from one server to the other.
04

4. Calculate normalizing constant

Since the sum of all probabilities must be equal to 1, we can calculate the normalizing constant to obtain: $$\sum_{j=0}^{n} p_{j} = 1$$ Substituting \(p_j = p_{j-1}\) into the sum, we obtain: $$\sum_{j=0}^{n} p_{0} = 1$$ Since the sum represents \((n+1)\) terms equal to \(p_{0}\), we have: $$p_{0}(n+1) = 1 \Rightarrow p_{0} = \frac{1}{(n+1)}$$
05

5. Calculate probability for each state

With the normalizing constant \(p_{0}\) found, we can calculate the probability for each state \(p_{j}\). Since \(p_j = p_{j-1}\), we find that \(p_{j} = p_{0}\) for all \(j = 0, 1, \ldots, n\). Therefore, the probability of having \(j\) customers at server 1 is: $$p_{j} = \frac{1}{(n+1)}$$ for \(j=0,1,\ldots, n\). In conclusion, the proportion of time that there are \(j=0,1,\ldots,n\) customers at server 1 is \(\frac{1}{(n+1)}\).

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Key Concepts

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

Markov Chains
Markov chains are mathematical systems that undergo transitions from one state to another within a finite or countable number of possible states. They are characterized by the Markov property, which states that the probability of transitioning to any future state depends solely on the present state and not on the sequence of events that preceded it.

In the context of the given problem, we are examining a discrete-time Markov chain where the states represent the number of customers at server 1. The changes in state occur when a customer either joins the queue or enters service after being served. Since each customer鈥檚 move does not depend on the previous sequence of moves, the process fits the definition of a Markov chain, allowing us to calculate the probabilities associated with each state.
Exponential Distribution
The exponential distribution is widely used to model the amount of time until a certain event occurs, such as the time between customer arrivals in a queue or service times in a server system. It is characterized by a constant mean arrival rate or service rate, represented by \( \mu \). One key property of the exponential distribution is that it is memoryless, meaning the probability of an event occurring in the next instant is independent of how much time has already elapsed.

In our problem, service times are exponentially distributed with rate \( \mu \), which simplifies the analysis of the transitions between states for our Markov chain. The exponential distribution鈥檚 memoryless property also justifies the use of balance equations to solve for steady-state probabilities.
Balance Equations
Balance equations are fundamental to understanding steady-state conditions in Markov chains. They describe a state of equilibrium where the rate at which probability mass enters a state equals the rate at which it leaves. Simply put, for every state in a continuous-time Markov chain, the probability of leaving the state multiplied by the steady-state probability of being in that state must equal the sum of probabilities of entering that state from all other states.

In the solution of the exercise, balance equations are utilized to express that for each state \( j \), the steady-state probability \( p_j \) of being in state \( j \) must be equal to the steady-state probability \( p_{j-1} \) of being in the previous state \( j-1 \), assuming a constant service rate \( \mu \). This assumption forms the basis for the calculation of steady-state probabilities.
Normalizing Constant
In probability theory, the normalizing constant ensures that the sum of probabilities for all possible outcomes is equal to one. This condition is essential because in a probability distribution, the total probability must account for every possible event and, therefore, should sum up to 100%.

The use of a normalizing constant is illustrated in the solution of the textbook problem where after determining that \( p_j = p_{j-1} \) for all \( j \), it is necessary to find out what \( p_0 \) is such that the sum of the probabilities across all possible states equals one. The computation of the normalizing constant \( p_{0} \) allows us to find a value that, when used to calculate the steady-state probabilities for each state, adheres to the total probability law.

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Most popular questions from this chapter

Poisson \((\lambda)\) arrivals join a queue in front of two parallel servers \(A\) and \(B\), having exponential service rates \(\mu_{A}\) and \(\mu_{B}\). When the system is cmpty, arrivals go into server \(A\) with probability \(\alpha\) and into \(B\) with probability \(1-\alpha\). Otherwise, the head of the queue takes the first free server. (a) Define states and set up the balance equations. Do not solve. (b) In terms of the probabilities in part (a), what is the average number in the system? Average number of servers idle? (c) In terms of the probabilities in part (a), what is the probability that an arbitrary arrival will get serviced in \(A\) ?

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.

Customers arrive at a two-server system according to a Poisson process having rate \(\lambda=5\). An arrival finding server 1 free will begin service with that server. An arrival finding server 1 busy and server 2 free will enter service with server 2. An arrival finding both servers busy goes away. Once a customer is served by either server, he departs the system. The service times at server \(i\) are exponential with rates \(\mu_{i}\), where \(\mu_{1}=4, \mu_{2}=2\). (a) What is the average time an entering customer spends in the system? (b) What proportion of time is server 2 busy?

Carloads of customers arrive at a single-server station in accordance to a Poisson process with rate 4 per hour. The service times are exponentially distributed with rate 20 per hour. If each carload contains cither 1,2 , or 3 customers with respective probabilities \(\frac{1}{8}, \frac{1}{2}, \frac{1}{4}\), compute the average customer delay in queue.

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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