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Chapter 8: Problem 32

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.

Short Answer

Expert verified
(a) The average number of customers at each server is \(\frac{3}{2}\). (b) The service completion rate for Server 1 is \(R_{1} = \frac{3}{2}\times \mu_1\) and for Server 2 is \(R_{2} = \frac{3}{2}\times \mu_2\).

Step by step solution

01

Identify all possible system states

There are two customers who can be at one of the two servers. So, the possible system states are $$ \begin{aligned} &\text {State 1: } (2,0) \quad \text{(both customers at server 1)}\\ &\text {State 2: } (1,1) \quad \text{(one customer at server 1 and one customer at server 2)}\\ &\text {State 3: } (0,2) \quad \text{(both customers at server 2)} \end{aligned} $$
02

Derive the balance equations

The balance equations for a closed queueing network are: \(x_{i,j} = P_{i,i} x_{i-1,j} + P_{j,j} x_{i,j-1}\) i and j denote the number of customers at servers 1 and 2, respectively. Here, since \(P_{1,2}=P_{2,1}=\frac{1}{2}\), the balance equations become: $$ \begin{aligned} &x_{2,0} = \frac{1}{2} x_{1,1}\\ &x_{1,1} = \frac{1}{2} x_{2,0} + \frac{1}{2} x_{0,2}\\ &x_{0,2} = \frac{1}{2} x_{1,1} \end{aligned} $$ Additionally, since the sum of all steady-state probabilities must equal 1, we have: $$ x_{2,0}+x_{1,1}+x_{0,2} = 1 $$
03

Solve the balance equations for steady-state probabilities

Combining the balance equations and the normalization condition: $$ \begin{aligned} &x_{2,0} = \frac{1}{2} x_{1,1}\\ &x_{1,1} = \frac{1}{2} x_{2,0} + \frac{1}{2} x_{0,2}\\ &x_{0,2} = \frac{1}{2} x_{1,1}\\ &x_{2,0}+x_{1,1}+x_{0,2} = 1 \end{aligned} $$ We can solve them to get: $$ \begin{aligned} &x_{2,0} = \frac{1}{4}\\ &x_{1,1} = \frac{1}{2}\\ &x_{0,2} = \frac{1}{4} \end{aligned} $$
04

Compute the average number of customers at each server

The average number of customers at each server can be computed by weighting the number of customers in each state by their steady-state probabilities: $$ \begin{aligned} &\text {Server 1: } N_{1}=\sum_{i, j} i \times x_{i, j}=\, 2 x_{2,0}+\, 1x_{1,1}\, = \, 2\left(\frac{1}{4}\right)+1\left(\frac{1}{2}\right)\, = \frac{3}{2}\\ &\text {Server 2: } N_{2}=\sum_{i, j} j \times x_{i, j}=\, 1 x_{1,1}\, +\, 2 x_{0,2}\, = \, 1\left(\frac{1}{2}\right) + 2\left(\frac{1}{4}\right)\, = \frac{3}{2} \end{aligned} $$ So, the average number of customers at each server is \(\frac{3}{2}\). (a) Answer: The average number of customers at each server is \(\frac{3}{2}\).
05

Compute the service completion rates for each server

The service completion rate for each server can be computed as the product of the average number of customers at the server and the corresponding service rate: $$ \begin{aligned} &\text {Server 1: } R_{1} = N_{1}\times\mu_{1} =\frac{3}{2}\times \mu_1\\ &\text {Server 2: } R_{2} = N_{2}\times\mu_{2} =\frac{3}{2}\times \mu_2 \end{aligned} $$ (b) Answer: The service completion rate for Server 1 is \(R_{1} = \frac{3}{2}\times \mu_1\) and for Server 2 is \(R_{2} = \frac{3}{2}\times \mu_2\).

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Key Concepts

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

Steady-State Probabilities
In a closed queueing network, steady-state probabilities provide a clear picture of the likelihood that the system is in a particular state over time. These probabilities help determine how likely it is for a specific number of customers to be at each server. For example, in the closed queueing network exercise, we have three possible states:
  • Bots customers at server 1: State 1 (2,0)
  • One customer at each server: State 2 (1,1)
  • Both customers at server 2: State 3 (0,2)
For these states, steady-state probabilities
  • The probability of both customers being at server 1 (\(x_{2,0}\)) is \(\frac{1}{4}\).
  • The probability of one customer at each server (\(x_{1,1}\)) is \(\frac{1}{2}\).
  • The probability of both customers being at server 2 (\(x_{0,2}\)) is \(\frac{1}{4}\).
These probabilities tell us the long-term distribution of customers across the servers, helping us understand the system's average performance.
Balance Equations
Balance equations are crucial to solving for steady-state probabilities. They ensure that in each state, the rate of customers entering a state equals the rate of customers leaving that state, maintaining a stable system configuration over time. In our closed queueing network example, the balance equations for the system are structured based on the possibility of customers moving from one server to another. These are:
  • \(x_{2,0} = \frac{1}{2} x_{1,1}\)
  • \(x_{1,1} = \frac{1}{2} x_{2,0} + \frac{1}{2} x_{0,2}\)
  • \(x_{0,2} = \frac{1}{2} x_{1,1}\)
Additionally, all probabilities must sum up to 1, leading to:\(x_{2,0} + x_{1,1} + x_{0,2} = 1\).Solving these equations provides the steady-state probabilities, linking the operational state changes to the overall structure and constraints of the network.
Service Completion Rate
The service completion rate indicates how efficiently each server processes customers. It's the rate at which a server moves customers through its queue, helping identify potential bottlenecks in the network.In this context, the service completion rate for server 1 is given by \(R_{1} = N_{1} \times \mu_{1}\), where \(N_{1}\) is the average number of customers at server 1 and \( \mu_{1}\) is the service rate at server 1. Similarly, for server 2, \(R_{2} = N_{2} \times \mu_{2}\).With an average of \(\frac{3}{2}\) customers at each server and the service rate \( \mu_{i}\), the service completion rate for each server becomes \( \frac{3}{2} \times \mu\). This calculation highlights the direct impact of customer numbers and service rates on overall service efficiency.
Exponential Service Rate
The exponential service rate is a common model used to reflect the service times in queueing networks. This type of service time is memoryless, meaning the probability of a customer being served in the next instant is independent of how long they've already waited. In our exercise, each server has an exponential service rate, denoted by \(\mu_{1}\) for server 1 and \(\mu_{2}\) for server 2. This model helps compute various performance metrics, like average waiting time and queue length, making it easier to predict system behavior over extended periods.Knowing these rates allows for the application of mathematical tools to determine steady-state probabilities and service completion rates, as they are pivotal in calculating how servers perform under normal operational conditions.

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Most popular questions from this chapter

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, she 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, she 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 \(n\) times.

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 facility produces items according to a Poisson process with rate \(\lambda .\) However, it has shelf space for only \(k\) items and so it shuts down production whenever \(k\) items are present. Customers arrive at the facility according to a Poisson process with rate \(\mu .\) Each customer wants one item and will immediately depart either with the item or empty handed if there is no item available. (a) Find the proportion of customers that go away empty handed. (b) Find the average time that an item is on the shelf. (c) Find the average number of items on the shelf. Suppose now that when a customer does not find any available items it joins the "customers' queue" as long as there are no more than \(n-1\) other customers waiting at that time. If there are \(n\) waiting customers then the new arrival departs without an item. (d) Set up the balance equations. (e) In terms of the solution of the balance equations, what is the average number of customers in the system?

The manager of a market can hire either Mary or Alice. Mary, who gives service at an exponential rate of 20 customers per hour, can be hired at a rate of $$ 3\( per hour. Alice, who gives service at an exponential rate of 30 customers per hour, can be hired at a rate of \) C\( per hour. The manager estimates that, on the average, each customer's time is worth \) 1\( per hour and should be accounted for in the model. If customers arrive at a Poisson rate of 10 per hour, then (a) what is the average cost per hour if Mary is hired? if Alice is hired? (b) find \)C$ if the average cost per hour is the same for Mary and Alice.

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?
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