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Chapter 8: Problem 11

Consider a single-server queue with Poisson arrivals and exponential service times having the following variation: Whenever a service is completed a departure occurs only with probability \(\alpha .\) With probability \(1-\alpha\) the customer, instead of leaving, joins the end of the queue. Note that a customer may be serviced more than once. (a) Set up the balance equations and solve for the steady-state probabilities, stating conditions for it to exist. (b) Find the expected waiting time of a customer from the time he arrives until he enters service for the first time. (c) What is the probability that a customer enters service exactly \(n\) times, \(n=\) \(1,2, \ldots ?\) (d) What is the expected amount of time that a customer spends in service (which does not include the time he spends waiting in line)? Hint: Use part (c). (e) What is the distribution of the total length of time a customer spends being served? Hint: Is it memoryless?

Short Answer

Expert verified
In summary: (a) We set up balance equations and solved for steady-state probabilities, \(P_i\). They exist if \(\sum_{i=0}^{\infty} P_i = 1\). (b) The expected waiting time of a customer until entering service for the first time is \(W = \frac{P_1}{\lambda(1 - P_0)}\). (c) The probability that a customer enters service exactly \(n\) times is \(P(n) = (1 - \alpha)^{n-1}\alpha\). (d) The expected amount of time a customer spends in service is \(T_s = \frac{\alpha}{\mu}\). (e) The distribution of the total length of time a customer spends being served is exponential with parameter \(\frac{\alpha}{\mu}\) and is memoryless.

Step by step solution

01

Notations

Let \(P_i\) denote the steady-state probability that there are \(i\) customers in the system. Let \(\lambda\) be the arrival rate of customers and let \(\mu\) be the service rate.
02

Balance equations

For \(i \ge 0\), we can set up balance equations as follows: \(\lambda P_0 = \mu P_1\) \(\lambda P_1 = \mu P_2\) \(\lambda P_2 = \mu P_3 + (1-\alpha) \mu P_0\) \( \cdots \) \(\lambda P_{i} = \mu P_{i+1} + (1-\alpha) \mu P_{i-1}\) for \(i \ge 2\)
03

Solve steady-state probabilities

Using the balance equations, we can express the probabilities \(P_i\) as follows: \(P_1 = \frac{\lambda}{\mu} P_0\) \(P_2 = (\frac{\lambda}{\mu})^2 P_0 - (1 - \alpha) P_0\) For \(i \ge 2\), \(P_{i+1} = \frac{1}{\mu} (\lambda P_i - (1 - \alpha)\mu P_{i-1})\) To find the steady-state probabilities, we need to find the normalization constant \(P_0\). For the steady-state probabilities to exist, the following condition must hold: \(\sum_{i=0}^{\infty} P_i = 1\) (b) Expected waiting time until entering service
04

Expected waiting time

Let \(W\) be the expected waiting time before entering in service. Then, \(W = \frac{P_1}{\lambda(1 - P_0)}\) (c) Probability of entering service exactly n times
05

Probability of entering service n times

Let \(P(n)\) be the probability of entering service exactly n times, then \(P(n) = (1 - \alpha)^{n-1}\alpha\) (d) Expected amount of time in service
06

Expected time in service

Let \(T_s\) be the expected amount of time in service, then using part (c): \(T_s = \sum_{n=1}^{\infty} n (1 - \alpha)^{n-1}\alpha \cdot \frac{1}{\mu}\) \(T_s = \frac{\alpha}{\mu}\) (e) Distribution of total length of time in service
07

Distribution of total time in service

Since the expected time spent in service by a customer, \(T_s\) is equal to \(\frac{\alpha}{\mu}\) which is constant, the distribution of the total length of time a customer spends being served is exponential. The distribution has the parameter \(\frac{\alpha}{\mu}\) and is memoryless.

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Most popular questions from this chapter

Explain how a Markov chain Monte Carlo simulation using the Gibbs sampler can be utilized to estimate (a) the distribution of the amount of time spent at server \(j\) on a visit. Hint: Use the arrival theorem. (b) the proportion of time a customer is with server \(j\) (i.e., either in server \(j\) 's queue or in service with \(j\) ).

In a two-class priority queueing model suppose that a cost of \(C_{i}\) per unit time is incurred for each type \(i\) customer that waits in queue, \(i=1,2\). Show that type 1 customers should be given priority over type 2 (as opposed to the reverse) if $$ \frac{E\left[S_{1}\right]}{C_{1}}<\frac{E\left[S_{2}\right]}{C_{2}} $$

It follows from Exercise 4 that if, in the \(M / M / 1\) model, \(W_{Q}^{*}\) is the amount of time that a customer spends waiting in queue, then $$ W_{Q}^{*}=\left\\{\begin{array}{ll} 0, & \text { with probability } 1-\lambda / \mu \\ \operatorname{Exp}(\mu-\lambda), & \text { with probability } \lambda / \mu \end{array}\right. $$ where \(\operatorname{Exp}(\mu-\lambda)\) is an exponential random variable with rate \(\mu-\lambda\). Using this, find \(\operatorname{Var}\left(W_{Q}^{*}\right)\)

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.

The economy alternates between good and bad periods. During good times customers arrive at a certain single-server queueing system in accordance with a Poisson process with rate \(\lambda_{1}\), and during bad times they arrive in accordance with a Poisson process with rate \(\lambda_{2}\). A good time period lasts for an exponentially distributed time with rate \(\alpha_{1}\), and a bad time period lasts for an exponential time with rate \(\alpha_{2}\). An arriving customer will only enter the queueing system if the server is free; an arrival finding the server busy goes away. All service times are exponential with rate \(\mu\). (a) Define states so as to be able to analyze this system. (b) Give a set of linear equations whose solution will yield the long run proportion of time the system is in each state. In terms of the solutions of the equations in part (b), (c) what proportion of time is the system empty? (d) what is the average rate at which customers enter the system?
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