/*! 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 13 A supermarket has two exponentia... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

Short Answer

Expert verified
In summary, we can set up the equations for \(P_n\) as follows: - For \(n = 0\), \(\lambda P_0 = \mu P_1\). - For \(n = 1\), \(\lambda P_1 = 2\mu P_2\). - For \(n \geq 2\), the recursive equation is \(P_{n+1} = \frac{n\mu - \lambda}{\lambda}P_n - \frac{(n-1)\mu}{\lambda}P_{n-1}\). After solving the balance equations, we can find the rate of transition: - Rate from 0 to 1: \(\lambda P_0\) - Rate from 2 to 1: \(2\mu P_2\) The proportion of time the stock clerk is checking is calculated as follows: \[ P(\text{Stock clerk checking}) = \sum_{n=2}^{\infty} P_n \]

Step by step solution

01

Set up the equations for \(P_n\)

First, let's set up the equations for \(P_n\). The system contains a Poisson arrival process with rate \(\lambda\), and two checkout counters, both operating with exponential service times having rate \(\mu\). The balance equations for \(P_n\) are given by: - For \(n = 0\), the balance equation is \(\lambda P_0 = \mu P_1\). - For \(n = 1\), the balance equation is \(\lambda P_1 = 2\mu P_2\). - For \(n \geq 2\), the balance equation is \((n - 1)\mu P_{n-1} + \lambda P_n = n\mu P_n + \lambda P_{n+1}\).
02

Solve the equations for \(P_n\)

Now, we can solve the balance equations for \(P_n\). The first two equations can be written as: - \(P_1 = \frac{\lambda}{\mu}P_0\) - \(P_2 = \frac{\lambda}{2\mu}P_1\) We also know that the sum of all probabilities is equal to 1: \[ \Pi_0 : \sum_{n=0}^{\infty} P_n = P_0 + P_1 + P_2 + \dots = 1 \] To solve this problem, we will solve the balance equations for \(n \geq 2\) by using the recursive equation. We can write the recursive equation from the balance equation for \(n \geq 2\) as: \[ P_{n+1} = \frac{n\mu - \lambda}{\lambda}P_n - \frac{(n-1)\mu}{\lambda}P_{n-1} \] Now, we can use the recursive equation and the equations for \(P_1\) and \(P_2\) to find the probability distribution \(P_n\).
03

Find the rate of transition

Now we will find the rate at which the number in the system goes from 0 to 1, and from 2 to 1. To do this, we will use the rates of arrival and service: - Rate from 0 to 1: \(\lambda P_0\) - Rate from 2 to 1: \(2\mu P_2\)
04

Calculate the proportion of time the stock clerk is checking

In order to calculate the proportion of time the stock clerk is checking, we need to consider that he only starts checking when there are 2 or more customers in the system. Therefore, the proportion of time the stock clerk is checking can be calculated as follows: \[ P(\text{Stock clerk checking}) = \sum_{n=2}^{\infty} P_n \] Considering the recursive equation and the probabilities \(P_1\) and \(P_2\) found in Step 2, we can find the proportion of time the stock clerk is checking.

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Key Concepts

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

Exponential Service Times
One key concept in queueing theory is exponential service times. Exponential service times refer to the scenario where the time taken to serve a customer follows an exponential distribution. This is quite common in various service industries, such as supermarkets, banks, or customer service centers with unpredictable service durations.
The exponential distribution is characterized by a continuous probability distribution, where the probability of an event occurring within a certain time frame decreases exponentially as time progresses. The parameter that defines this distribution is the rate \( \mu \), which represents the average number of services completed in a unit of time. This is mathematically expressed as:\[P(T > t) = e^{-\mu t}\]Where \( T \) is the service time, and \( t \) is the specific time point.
  • This implies that the longer you wait, the lower the probability that service will take longer.
  • It models real-world service times effectively where certain random patterns occur.
Poisson Arrival Process
The Poisson arrival process is another fundamental component in queueing theory. It is used to describe scenarios in which events, such as customer arrivals at a checkout counter, occur randomly and independently over time. Essentially, it characterizes the number of arrivals in fixed intervals, making it highly relevant for systems facing random demand.
The Poisson process is defined by its arrival rate \( \lambda \), indicating the average number of arrivals per time unit. The formula for the probability of \( n \) arrivals in time \( t \) is given by:\[P(N(t) = n) = \frac{e^{-\lambda t}(\lambda t)^n}{n!}\]
  • \( N(t) \) is a Poisson random variable representing the number of arrivals in time period \( t \).
  • Each arrival occurs regardless of the time elapsed since the last arrival.
This property is incredibly useful for supermarkets where customer arrivals can be erratic, and operators need to optimize staffing and resources accordingly.
Balance Equations
In the context of the supermarket checkout problem, the balance equations are used to determine the steady-state probabilities of customers in the system. This approach relies on the system reaching an equilibrium state where the rate of arrivals equals the rate of service.
For any state \( n \), the balance equation expresses that the rate at which the system enters the state must equal the rate at which it leaves the state. This is crucial for maintaining a stable operation in the long term.
For small values of \( n \), such as \( n = 0 \) or \( n = 1 \), balance equations can look relatively simple:
  • When \( n = 0 \), the equation is \( \lambda P_0 = \mu P_1 \).
  • When \( n = 1 \), the equation is \( \lambda P_1 = 2\mu P_2 \).
For \( n \geq 2 \), the equations account for both arrival and service rates, creating a recursive relationship:\[(n - 1)\mu P_{n-1} + \lambda P_n = n\mu P_n + \lambda P_{n+1}\]Solving these equations helps find the probability of having \( n \) customers in the system, essential for operational decisions.
Probability Distribution
Probability distribution in queueing systems refers to the likelihood of different outcomes, specifically the number of customers being present in the system at any given time. For the supermarket problem, the distribution gives us the values of \( P_n \) (the probability of having \( n \) customers), essential in determining how often the stock clerk helps the cashier.
To find this distribution, we apply recursive methods utilizing balance equations with initial conditions determined for lower \( n \). We begin with simple known values, such as \( P_0 \) decaying to complex forms for higher \( n \).
These recursive relationships ultimately provide the complete probability distribution:
  • \( P_1 = \frac{\lambda}{\mu}P_0 \)
  • \( P_2 = \frac{\lambda}{2\mu}P_1 \)
This distribution helps predict times of peak customer load, when extra staffing is crucial, ensuring efficiency and customer satisfaction.

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Most popular questions from this chapter

Consider a network of three stations. Customers arrive at stations \(1,2,3\) in accordance with Poisson processes having respective rates, \(5,10,15 .\) The service times at the three stations are exponential with respective rates \(10,50,100\). A customer completing service at station 1 is equally likely to (i) go to station 2, (ii) go to station 3, or (iii) leave the system. A customer departing service at station 2 always goes to station \(3 .\) A departure from service at station 3 is equally likely to either go to station 2 or leave the system. (a) What is the average number of customers in the system (consisting of all three stations)? (b) What is the average time a customer spends in the system?

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.

Show that \(W\) is smaller in an \(M / M / 1\) model having arrivals at rate \(\lambda\) and service at rate \(2 \mu\) than it is in a two-server \(M / M / 2\) model with arrivals at rate \(\lambda\) and with each server at rate \(\mu .\) Can you give an intuitive explanation for this result? Would it also be true for \(W_{Q}\) ?

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 see that $$ D=\left\\{\begin{array}{ll} \text { Exponential }(\mu), & \text { with probability } \lambda / \mu \\ \text { Exponential }(\lambda) * \text { Exponential }(\mu), & \text { with probability } 1-\lambda / \mu \end{array}\right. $$ where Exponential \((\lambda) *\) Exponential \((\mu)\) represents the sum of two independent exponential random variables having rates \(\mu\) and \(\lambda\). Now use momentgenerating functions to show that \(D\) has the required distribution. Note that the preceding 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.

Customers arrive at a single-server station in accordance with a Poisson process with rate \(\lambda .\) All arrivals that find the server free immediately enter service. All service times are exponentially distributed with rate \(\mu .\) An arrival that finds the server busy will leave the system and roam around "in orbit" for an exponential time with rate \(\theta\) at which time it will then return. If the server is busy when an orbiting customer returns, then that customer returns to orbit for another exponential time with rate \(\theta\) before returning again. An arrival that finds the server busy and \(N\) other customers in orbit will depart and not return. That is, \(N\) is the maximum number of customers in orbit. (a) Define states. (b) Give the balance equations. In terms of the solution of the balance equations, find (c) the proportion of all customers that are eventually served. (d) the average time that a served customer spends waiting in orbit.

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