/*! 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 17 Customers arrive at a two-server... [FREE SOLUTION] | 91Ó°ÊÓ

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Customers arrive at a two-server station in accordance with a Poisson process with a rate of two per hour. Arrivals finding server 1 free begin service with that server. Arrivals finding server 1 busy and server 2 free begin service with server 2. Arrivals finding both servers busy are lost. When a customer is served by server 1 , she then either enters service with server 2 if 2 is free or departs the system if 2 is busy. A customer completing service at server 2 departs the system. The service times at server 1 and server 2 are exponential random variables with respective rates of four and six per hour. (a) What fraction of customers do not enter the system? (b) What is the average amount of time that an entering customer spends in the system? (c) What fraction of entering customers receives service from server \(1 ?\)

Short Answer

Expert verified
The short version of the answer for the given question is: (a) The fraction of customers who do not enter the system is approximately 0.4. (b) The average amount of time an entering customer spends in the system is approximately 0.5 hours. (c) The fraction of entering customers that receive service from server 1 is approximately 0.2.

Step by step solution

01

Draw State-Transition Diagram and Identify Variables

First, we draw a state-transition diagram for this problem. In this diagram, the states represent the number of customers in the system where state 0 corresponds to both server 1 and server 2 being idle, state 1 corresponds to server 1 being busy and server 2 being idle, and state 2 corresponds to both server 1 and server 2 being busy. Next, we identify the given variables and assign appropriate letters to them: - Arrival rate (λ) = 2 customers per hour - Service rate at server 1 (μ1) = 4 customers per hour - Service rate at server 2 (μ2) = 6 customers per hour
02

Analyze the Equilibrium Equations and Probabilities

Now we need to analyze the equilibrium equations and find the probabilities. From the state-transition diagram, we can derive the following equations: 1. λ * P0 = μ1 * P1 2. (λ + μ1) * P1 = λ * P0 + μ2 * P2 3. μ1 * P1 = λ * P2 To find P0, P1, and P2, we can solve these equations simultaneously. We know that the sum of all probabilities must equal to 1, i.e., P0 + P1 + P2 = 1.
03

Solve for Probabilities

We can solve the above equations simultaneously to find P0, P1, and P2: 1. P1 = (λ/μ1) * P0 2. P2 = ((λ + μ1) / μ2) * P1 - (λ/μ2) * P0 3. P0 + P1 + P2 = 1 Substituting the values of λ, μ1, and μ2, we get: 1. P1 = (2/4) * P0 = 0.5 * P0 2. P2 = ((2 + 4) / 6) * P1 - (2/6) * P0 = (1/3) * P1 - (1/3) * P0 3. P0 + 0.5 * P0 + ((1/3) * 0.5 * P0 - (1/3) * P0) = 1 Solve for P0: P0 (1 + 0.5 + 0.5/3 - 1/3) = 1 P0 ≈ 0.4 Now, solve for P1 and P2: P1 ≈ 0.5 * P0 ≈ 0.2 P2 ≈ (1/3) * P1 - (1/3) * P0 ≈ 0.4
04

Use Probabilities to Answer Questions

Now we can use the found probabilities to answer the given questions: (a) The fraction of customers who do not enter the system is equal to the probability of both servers being busy (P2). So, we have: Fraction not entering system ≈ P2 ≈ 0.4 (b) The expected time a customer spends in the system is given by the Little's Law: E(T) = E(N) / λ, where E(N) is the expected number of customers in the system. We can find E(N) using the found probabilities: E(N) = 1 * P1 + 2 * P2 ≈ 1 * 0.2 + 2 * 0.4 ≈ 1 Now, use Little's Law: E(T) ≈ E(N) / λ ≈ 1 / 2 ≈ 0.5 hours (c) The fraction of entering customers that receive service from server 1 is equal to the probability of server 1 being busy (P1). So, we have: Fraction receiving service from server 1 ≈ P1 ≈ 0.2

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Key Concepts

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

Exponential random variables
Understanding exponential random variables is pivotal when studying Poisson processes, which are common in describing queuing systems like the one in our exercise. An exponential random variable is used to model the time between events in a Poisson process. This type of variable is characterized by a memoryless property, meaning the probability of an event occurring in the future is independent of any prior events.

For example, in the scenario of customer service at two-server stations, the service times are represented as exponential random variables with rates. These rates, four and six customers per hour for server 1 and server 2 respectively, indicate the average number of customers that can be serviced in a unit of time. Mathematically, if 'X' is an exponential random variable with rate \(\lambda\), the probability that 'X' exceeds a certain time 't' is given by \(P(X > t) = e^{-\lambda t}\), highlighting the continuous and decaying nature of this probability over time.

Grasping the concept of exponential random variables helps students solve part (b) of the exercise, where they need to calculate the average amount of time an entering customer spends in the system—key knowledge for nuanced fields such as operations research and network theory.
State-transition diagram
A state-transition diagram is a visual representation of a system that showcases possible states and the transitions between them. In the context of queuing theory, the states typically represent the number of entities (like people or data packets) within the system at any given time, and transitions represent the arrival or departure of these entities.

For step 1 of our solution, the state-transition diagram is drawn to illustrate two states - where state 0 is both servers being idle, state 1 is server 1 being busy and server 2 idle, and state 2 means both servers are busy. Arrows indicate customer transitions based on availability. Visualizing the problem in this manner aids to digest the system’s dynamics and sets the stage to derive the equilibrium equations necessary to solve the subsequent parts of the exercise. Students are usually encouraged to perfect this skill as it transcends many disciplines from computer science to economics.
Equilibrium equations
Equilibrium equations, sometimes termed 'balance equations', are mathematical representations that describe the long-term behavior of a stochastic or random process. In our exercise, equilibrium equations are used to determine the steady-state probabilities of the system's states.

These equations ensure that, for each state in the system, the rate of entering the state equals the rate of leaving the state, suggesting a balanced system in the long run. Specifically, step 2 shows the equilibrium equations derived from the state-transition diagram. Solving these simultaneously with the normalization condition (the sum of all probabilities equals 1) provides us with the steady-state probabilities (\(P0, P1, P2\)).

By understanding how to formulate and solve equilibrium equations, students can answer questions regarding the fraction of customers not entering the system (part a) and the fraction receiving service from server 1 (part c). These solutions are fundamental to operations management, detailing how often resources are fully utilized—or which scarcities may lead to service denials, a crucial aspect for optimizing service delivery.

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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.

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 \(1, j=0, \ldots, n\).

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 \(\operatorname{Var}\left(W_{Q}^{*}\right)\)

Consider a system where the interarrival times have an arbitrary distribution \(F\), and there is a single server whose service distribution is \(G\). Let \(D_{n}\) denote the amount of time the \(n\) th customer spends waiting in queue. Interpret \(S_{n}, T_{n}\) so that $$ D_{n+1}=\left\\{\begin{array}{ll} D_{n}+S_{n}-T_{n}, & \text { if } D_{n}+S_{n}-T_{n} \geqslant 0 \\ 0, & \text { if } D_{n}+S_{n}-T_{n}<0 \end{array}\right. $$

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.

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