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Chapter 8: Problem 14

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?

Short Answer

Expert verified
(a) The average time an entering customer spends in the system is approximately \(W = \frac{29}{57}\) units of time. (b) The proportion of time server 2 is busy is approximately \(\frac{75}{57}\) of the time.

Step by step solution

01

Define the states and Markov Chain of the system

Let's define the possible states: - State \(0\): No customer is being served. - State \(1\): One customer is being served by server 1. - State \(2\): One customer is being served by server 2. - State \(3\): Two customers are being served, one by each server. Now we can represent this system as a Markov Chain. The transition rates between the states are: - \(q_{01} = \lambda\) - \(q_{10} = \mu_1\) - \(q_{12} = \lambda\) - \(q_{21} = \mu_2\) - \(q_{23} = \lambda\) - \(q_{32} = \mu_1\)
02

Calculate the steady-state probabilities

The steady-state probabilities can be calculated by solving the balance equations. The balance equations for this system are: 1. \(\pi_0 q_{01} = \pi_1 q_{10}\) 2. \(\pi_1 q_{12} = \pi_2 (q_{21} + q_{23})\) 3. \(\pi_2 q_{21} = \pi_3 q_{32}\) We also have the normalization condition: \(\pi_0 + \pi_1 + \pi_2 + \pi_3 = 1\). Solving these balance equations, we have: - \(\pi_1 = \frac{\lambda}{\mu_1} \pi_0 = \frac{5}{4} \pi_0\) - \(\pi_2 = \frac{\lambda}{\mu_2 + \lambda} \pi_1 = \frac{5}{7} \pi_1 = \frac{25}{28} \pi_0\) - \(\pi_3 = \frac{\mu_1}{\mu_2} \pi_2 = \frac{4}{2} \pi_2 = \frac{50}{28} \pi_0\) By substituting \(\pi_1, \pi_2\), and \(\pi_3\) in the normalization condition, we can solve for \(\pi_0\): \[ \pi_0 + \frac{5}{4} \pi_0 + \frac{25}{28} \pi_0 + \frac{50}{28} \pi_0 = 1 \] \[ \pi_0 = \frac{28}{57} \] From which we can find \(\pi_1 = \frac{20}{57}, \pi_2 = \frac{25}{57}, \pi_3 = \frac{50}{57}\).
03

Calculate the average time a customer spends in the system

To calculate the average time a customer spends in the system, we will use Little's formula: \[L = \lambda W\] Where \(L\) is the average number of customers in the system, and \(W\) is the average time a customer spends in the system. The average number of customers in the system can be calculated as: \[ L = 1\pi_1 + 1\pi_2 + 2\pi_3 = \frac{20}{57} + \frac{25}{57} + \frac{100}{57} = \frac{145}{57} \] Now, we can find \(W\): \[ W = \frac{L}{\lambda} = \frac{29}{57} \] So, the average time an entering customer spends in the system is \(\frac{29}{57}\) units of time.
04

Calculate the proportion of time server 2 is busy

To determine the proportion of time server 2 is busy, we can calculate the probabilities of server 2 being in a state in which it's serving a customer. Server 2 will be serving a customer in states 2 and 3. So the proportion of time it is busy is: \[ \text{Proportion of time server 2 is busy} = \pi_2 + \pi_3 = \frac{25}{57} + \frac{50}{57} = \frac{75}{57} \] Hence, server 2 is busy approximately \(\frac{75}{57}\) of the time.

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Key Concepts

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

Markov chains
In the context of queuing theory, Markov chains provide a mathematical framework for modeling systems where events occur at random intervals. The essence of a Markov chain lies in its memoryless property, where the probability of transitioning from one state to another depends solely on the current state, not on the sequence of events that preceded it.

Our system consists of four distinct states, indicating the occupancy of two servers. Transitions between states occur when customers arrive or servers complete service, dictated by the rates \( \lambda \) and \( \mu_i \) for arrivals and service completions, respectively. By analyzing this chain, we can predict long-term behaviors such as server utilization and customer wait times without tracking individual customers.
Steady-state probabilities
Steady-state probabilities represent the long-term behavior of a Markov chain, where the system reaches equilibrium. These probabilities don't change over time and give valuable insights into the performance metrics of the system. For instance, they can tell us what proportion of time a server is busy or how often no customers are in the system.

To calculate these probabilities, we set up balance equations based on the rates of transitioning between states and solve the system of equations together with the normalization condition. This approach allows us to find the likelihood of the system being in any particular state at an arbitrary time in the long run. As depicted in our step-by-step solution, this method was crucial for answering both (a) the average time a customer spends in the system and (b) the proportion of time server 2 is busy.
Exponential service times
The assumption of exponential service times plays a critical role in simplifying the analysis of our queuing system. This assumption indicates that the amount of time a server takes to serve a customer is random but has a memoryless property, meaning that regardless of how long service has already taken, the expected time until service completion remains constant.

It allows us to define the rates \( \mu_i \) as constants: \( \mu_1 = 4 \) and \( \mu_2 = 2 \) for server 1 and server 2 respectively. These rates are the parameters of the exponential distribution governing the service times and enable us to compute the transition rates for the Markov chain. Exponential service times are a foundational concept in queuing theory, as they make the system analyzable and permit us to use Markovian properties for modeling.

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Most popular questions from this chapter

A group of \(m\) customers frequents a single-server station in the following manner. When a customer arrives, he or she either enters service if the server is free or joins the queue otherwise. Upon completing service the customer departs the system, but then returns after an exponential time with rate \(\theta .\) All service times are exponentially distributed with rate \(\mu\). (a) Define states and set up the balance equations. In terms of the solution of the balance equations, find (b) the average rate at which customers enter the station. (c) the average time that a customer spends in the station per visit.

Consider the priority queuing model of Section \(8.6 .2\) but now suppose that if a type 2 customer is being served when a type 1 arrives then the type 2 customer is bumped out of service. This is called the preemptive case. Suppose that when a bumped type 2 customer goes back in service his service begins at the point where it left off when he was bumped. (a) Argue that the work in the system at any time is the same as in the nonpreemptive case. (b) Derive \(W_{\mathrm{Q}}^{1}\). Hint: How do type 2 customers affect type 1 's? (c) Why is it not true that $$ V_{\mathrm{Q}}^{2}=\lambda_{2} E\left[S_{2}\right] W_{0}^{2} $$ (d) Argue that the work seen by a type 2 arrival is the same as in the nonpreemptive case, and so \(W_{0}^{2}=W_{0}^{2}\) (nonpreemptive) \(+E\) [extra time] where the extra time is due to the fact that he may be bumped. (e) Let \(N\) denote the number of times a type 2 customer is bumped. Why is $$ E \text { [extra time } \mid N]=\frac{N E\left[S_{1}\right]}{1-\lambda_{1} E\left[S_{1}\right]} $$ Hint: When a type 2 is bumped, relate the time until he gets back in service to a "busy period." (f) Let \(S_{2}\) denote the service time of a type 2. What is \(E\left[N \mid S_{2}\right] ?\) (g) Combine the preceding to obtain $$ W_{0}^{2}=W_{Q}^{2}(\text { nonpreemptive })+\frac{\lambda_{1} E\left[S_{1} \mid E\left[S_{2}\right]\right.}{1-\lambda_{1} E\left[S_{1}\right]} $$

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.

Compare the M/G/1 system for first-come, first-served queue discipline with one of last-come, first-served (for instance, in which units for service are taken from the top of a stack). Would you think that the queue size, waiting time, and busy-period distribution differ? What about their means? What if the queue discipline was always to choose at random among those waiting? Intuitively which discipline would result in the smallest variance in the waiting time distribution?

Let \(D\) denote the time between successive departures in a stationary \(M / M / 1\) queue with \(\lambda<\mu .\) Show, by conditioning on whether or not a departure has left the system empty, that \(D\) is exponential with rate \(\lambda\). Hint: By conditioning on whether or not the departure has left the system empty we sce that \(D=\left\\{\begin{array}{ll}\text { Exponential }(\mu), & \text { with probability } \lambda / \mu \\ \text { Exponential }(\lambda) \cdot \text { Exponential }(\mu), & \text { with probability } 1-\lambda / \mu\end{array}\right.\) where Exponential ( \(\lambda\) ) \(*\) Exponential \((\mu)\) represents the sum of two independent exponential random variables having rates \(\mu\) and \(\lambda\). Now use moment-generating functions to show that \(D\) has the required distribution. Note that the above does not prove that the departure process is Poisson. To prove this we need show not only that the interdeparture times are all exponential with rate \(\lambda_{+}\) but also that they are independent.
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