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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
The short answer based on the given step-by-step solution is: The average number of customers at Server 1: \(L_1 = 4 \times P(\text{state 1}) \) The average number of customers at Server 2: \(L_2 = 3 - L_1\) The service completion rate for Server 1: \(\lambda_1 = L_1 \times \mu_1\) The service completion rate for Server 2: \(\lambda_2 = L_2 \times \mu_2\)

Step by step solution

01

Identify the system states

Our system (queueing network) consists of two servers and two customers. The customers can be in any of the following configurations: 1. Both customers at server 1. 2. One customer at server 1 and one customer at server 2. 3. Both customers at server 2. These three configurations represent the possible states of our system. STEP 2: Create the state transition diagram
02

Create the state transition diagram

Now that we've identified the states, we can create a state transition diagram. This will help us visualize the different possible transitions between states. Each state will be represented by a node, and directed edges will represent the possible transitions accompanied by the probabilities of such transitions. In our case, the transitions among states 1 and 3 will always involve passing through state 2. There are only two transitions that directly involve state 1 and state 3, so we can denote the transitions as shown: 1 --(1/2,μ₁)--> 2 --(1/2,μ₂)--> 3 2 --(1/2,μ₂)--> 1 --(1/2,μ₁)--> 2 Now that we have our state transition diagram, we can proceed to find the average number of customers at each server and the service completion rates. STEP 3: Calculate the average number of customers at each server:
03

Calculate the average number of customers at each server

To calculate the average number of customers at each server, we consider the different states of our system and calculate the proportion of time that each state occurs using the balance equations. Let L₁ and L₂ denote the averages number of customers at Server 1 and Server 2. Then, L₁ = P(state 1) × 2 + P(state 2) × 1 L₂ = P(state 2) × 1 + P(state 3) × 2 Since the system is a closed queueing network with an equal probability of customers going to either server, then P(state 1) = P(state 3). So, we obtain the following balance equations: L₁ = 2 × P(state 1) + P(state 2) L₂ = P(state 2) + 2 × P(state 3) Since the three probabilities P(state 1), P(state 2) and P(state 3) must add up to 1, we can solve the equations above to get: L₁ = 4 × P(state 1) L₂ = 3 - L₁ This will give us the average number of customers at each server. STEP 4: Calculate the service completion rates
04

Calculate the service completion rates

Service completion rate for each server is related to the service rate μ, which is given as μ₁ for server 1 and μ₂ for server 2. Since the system has an exponential service rate and is in steady-state, the service completion rate equals the service arrival rate. Thus, Service completion rate for Server 1: λ₁ = L₁ × μ₁ Service completion rate for Server 2: λ₂ = L₂ × μ₂ This will give us the service completion rates for each server.

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Key Concepts

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

Closed Queueing Network
Imagine a miniature ecosystem where a fixed number of customers visit various service stations, never leaving the system and constantly moving from one service station to another. This is the essence of a closed queueing network. In this closed system, the total number of customers remains constant; whenever a customer leaves a service station, they are reassigned to another, maintaining the network's integrity.

This type of network is particularly useful in scenarios where resources are limited and need to be allocated efficiently, such as in computer networks, manufacturing systems, or even amusement park designs where cars or seats are continuously cycled through different rides or stages of a process.
Exponential Service Rate
The exponential service rate is a critical concept in queueing theory that describes the time between services completed at a station. It follows an exponential probability distribution, which is commonly associated with the memoryless property.

It means that the probability of a service being completed in the next instant is independent of how long the current service has already taken. This simplifies many calculations in queueing theory and the analysis of systems. In the context of our exercise, the service rate at each server is exponential, signified with the symbol \(\mu_i\) for the ith server.
State Transition Diagram
To visualise the labyrinth of potential states and transitions in a queueing system, we use the state transition diagram. This diagram serves as a map, marking each state as a node and the pathways between these states as edges, with each path assigned a probability and rate at which transitions occur.

By mapping out the landscape of the network through this state transition diagram, we gain a visual and mathematical tool to explore and determine how often each state occurs, which is essential for uncovering metrics like the average number of customers at each server or the service completion rates.
Average Number of Customers in a Queue
Students often grapple with the question of how to ascertain the average number of customers in a queue. This number is pivotal, as it reflects the load on the service station and can indicate the level of congestion or efficiency.

To compute this average, we assess the likelihood and duration of the system being in various states. Each state corresponds to a different number of customers at the server, and by finding a weighted sum of these numbers with respect to the probabilities of the states, we arrive at the average number we cross paths with in each queue.
Service Completion Rate
Moving on from the number of customers is the service completion rate. This tells us how rapidly the servers are dealing with their lines, which is pivotal for understanding the system's flow and capacity management.

To calculate this, we look at the service rates and the average number of customers at each server. Since the service times are exponentially distributed, the completion rate is simply the product of the server's service rate and the average number of customers at that server. This can be used to gauge productivity and identify bottlenecks within the network.
Balance Equations
Imagine a balanced scale, representing the harmony that must exist in a queueing system. The balance equations achieve just that for a closed network by ensuring that the flow into each state matches the flow out.

These equations keep track of the probabilities of each state, guaranteeing that the sum of all probabilities equals 1, since the system must be in one of the states at all times. Balancing these equations is akin to solving a puzzle: correctly placing each piece to create a cohesive, functioning whole.
Steady-State Probabilities
The final piece of our theoretical framework is the steady-state probabilities. These probabilities are the bedrock of queueing theory, representing the long-term behavior of the system.

They tell us the likelihood that the system will be found in any given state after it has been operating for a significant amount of time, having reached a state of equilibrium. These probabilities are crucial for calculating many performance measures of the system, such as the average number of customers or service rates, which remain stable over time in a steady-state scenario.

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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) Find the average rate at which customers enter the station. (b) Find the average time that a customer spends in the station per visit.

For the \(M / M / 1\) queue, compute (a) the expected number of arrivals during a service period and (b) the probability that no customers arrive during a service period. Hint: "Condition."

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

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 ayerage 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. (c) In terms of the solution of the balance equations, what is the average number of customers in the system.

Consider the priority queueing 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 \(W_{Q}^{1}\) Hint: How do type 2 customers affect type 1 s? (c) Why is it not true that $$ V_{Q}^{2}=\lambda_{2} E\left[S_{2}\right] W_{Q}^{2} $$ (d) Argue that the work seen by a type 2 arrival is the same as in the nonpreemptive case, and so \(W_{Q}^{2}=W_{Q}^{2}\) (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 $$ E[\text { extra time } \mid \mathrm{N}]=\frac{N E\left[S_{1}\right]}{1-\lambda_{1} E\left[S_{1}\right]} $$ 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 $$ W_{Q}^{2}=W_{Q}^{2} \text { (nonpreemptive) }+\frac{\lambda_{1} E\left[S_{1}\right] E\left[S_{2}\right]}{1-\lambda_{1} E\left[S_{1}\right]} $$

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