/*! 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 25 Customers can be served by any o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Customers can be served by any of three servers, where the service times of server \(i\) are exponentially distributed with rate \(\mu_{i}, i=1,2,3 .\) Whenever a server becomes free, the customer who has been waiting the longest begins service with that server. (a) If you arrive to find all three servers busy and no one waiting, find the expected time until you depart the system. (b) If you arrive to find all three servers busy and one person waiting, find the expected time until you depart the system.

Short Answer

Expert verified
(a) The expected time until a customer departs the system when all three servers are busy and no one is waiting: \(E(T_A) = \frac{1}{\mu_1+\mu_2+\mu_3}\) (b) The expected time until a customer departs the system when all three servers are busy and there is one person waiting: \(E(T_B) = \frac{2}{\mu_1+\mu_2+\mu_3}\)

Step by step solution

01

Define the Variables

Let \(X_1, X_2, X_3\) be independent exponential random variables representing the remaining service time for each of the three servers with rates \(\mu_1, \mu_2,\) and \(\mu_3\), respectively. Let \(T_A\) and \(T_B\) represent the expected time until a customer departs the system in case (a) and case (b), respectively.
02

Conditional Expectations for Case (a)

In this case, a customer leaves the system as soon as one of the servers become free. The customer's waiting time \(T_A\) is equal to the minimum of the remaining service times for all three servers: \(T_A = \min(X_1, X_2, X_3)\). We will use the properties of exponential random variables and conditional expectations to find the expected time until departure \(E(T_A)\). Using the conditional expectation formula, we can write: \(E(T_A) = E[T_A | X_i = \min(X_1, X_2, X_3)] P(X_i = \min(X_1, X_2, X_3)) + E[T_A | X_j = \min(X_1, X_2, X_3)] P(X_j = \min(X_1, X_2, X_3)) + E[T_A | X_k = \min(X_1, X_2, X_3)] P(X_k = \min(X_1, X_2, X_3)),\) where \(\{i, j, k\} = \{1, 2, 3\}\).
03

Calculate the Probabilities and Expectations

We will now calculate the probabilities and expectations for case (a): 1. The probability that server \(i\) finishes first: \(P(X_i = \min(X_1, X_2, X_3)) = \frac{\mu_i}{\mu_1 + \mu_2 + \mu_3}\) for \(i = 1, 2, 3\). 2. The expectation of waiting time, given that server \(i\) finishes first: \(E[T_A | X_i = \min(X_1, X_2, X_3)] = \frac{1}{\mu_i}\) for \(i = 1, 2, 3\). Now, substitute these probabilities and expectations into the conditional expectation formula: \(E(T_A) = (\frac{1}{\mu_1}\times\frac{\mu_1}{\mu_1+\mu_2+\mu_3}) + (\frac{1}{\mu_2}\times\frac{\mu_2}{\mu_1+\mu_2+\mu_3}) + (\frac{1}{\mu_3}\times\frac{\mu_3}{\mu_1+\mu_2+\mu_3})\) Simplifying the expression, we get: \(E(T_A) = \frac{1}{\mu_1+\mu_2+\mu_3}\)
04

Calculate Expected Time for Case (b)

For part (b), there is one customer waiting when the new customer joins the line. The new customer has to wait for one other customer to finish before they can leave. We can find the expected time for this case by adding the expected service time of the first customer \(E(T_A)\) (calculated in Step 3) to the expected service time of the new customer joining the line, which is also \(E(T_A)\): \(T_B = T_A + T_A\) Therefore, \(E(T_B) = 2E(T_A) = \frac{2}{\mu_1+\mu_2+\mu_3}\)
05

Final Answer

(a) The expected time until a customer departs the system when all three servers are busy and no one is waiting: \(E(T_A) = \frac{1}{\mu_1+\mu_2+\mu_3}\) (b) The expected time until a customer departs the system when all three servers are busy and there is one person waiting: \(E(T_B) = \frac{2}{\mu_1+\mu_2+\mu_3}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Exponential Distribution
Exponential distribution is pivotal in the realm of probability models, especially in processes that describe the time before an event occurs. It is well-suited for modeling scenarios where events happen continuously and independently over time, such as in this exercise. Here, we have service times for servers that follow an exponential distribution, emphasizing random occurrences.The distinctive property of exponential distribution is its **memoryless** characteristic. This means that the probability of an event occurring in the future is independent of the past, which is mathematically expressed as:\[ P(X > s + t \mid X > t) = P(X > s) \]This property is invaluable in queueing systems, as it allows for a simpler computation of expected service times since the expected remaining time doesn't depend on the elapsed service time. The rate parameter \(\mu\), also known as the rate of occurrence, inversely relates to the mean, giving the formula for the expected value:\[ E(X) = \frac{1}{\mu} \].This shows how frequently events are expected to occur, making it a critical parameter in managing and predicting service times effectively.
Queueing Theory
Queueing theory is the mathematical study of waiting lines or queues. It's foundational when dealing with the logistics of providing services, such as in customer service systems with multiple servers. In this exercise, we delve into a setup where customers are served by any available server, with each server having its own rate. Queueing models typically involve the following elements:
  • **Arrival process:** Determines how customers arrive into the system, often modeled using a Poisson process.
  • **Service mechanism:** Defines the service distribution, which in this case is exponential.
  • **Number of servers:** Indicates how many customers can be served simultaneously.
For example, when all servers are busy, the waiting customer begins service with the first free server. When crafting solutions to such problems, queueing theory provides the rigor to predict metrics like waiting time, queue length, or server utilization effectively. The goal of queueing theory is to understand and analyze systems, thereby facilitating decisions that optimize service efficacy, minimize waiting times, and increase overall satisfaction.
Conditional Expectations
Conditional expectation is a pivotal concept in probability that helps in understanding how probability changes when we know certain information. It involves calculating the expected value of a random variable given a certain condition or event has occurred. In the context of this exercise, we rely on conditional expectations to determine service and departure times.It is used to find the expected waiting times, as shown in the step-by-step solution, where we evaluate the expected time until a server becomes free using the formula:\[ E(X \mid Y = y) \].For the problem at hand, if all three servers are busy, the minimum time needed for any server to finish is based on which finishes first, calculated using:\[ E(T_A) = E[T_A | X_i = \min(X_1, X_2, X_3)] \]These probabilities help in determining which server finishes first and thus how long one might expect to wait, while obtaining insights into the nature of overlapping service times and dependencies.
Exponential Random Variables
Exponential random variables are integral to modeling time-dependent stochastic processes. They feature a continuous probability distribution commonly used to describe the time between events in a Poisson process. In this educational exercise, each server's service time can be in the form of an exponential random variable with a unique rate \(\mu_i\).Some key properties include:
  • **Memoryless Property:** A defining property where the future probability is not influenced by the past, helpful in simplifying queue models.
  • **Rate Parameter (\(\mu\)):** Governs how rapidly occurrences happen, with higher rates indicating more frequent events.
  • **Distribution Function:** The probability density function for an exponential variable is \(f(x;\mu) = \mu e^{-\mu x}\) for \(x \geq 0\).
Understanding these variables forms the backbone of predicting service times in queueing systems as they help in deriving intuitive and mathematically robust models of service dynamics. These models are indispensable for computing waiting times and managing overall system efficiency.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An insurance company pays out claims on its life insurance policies in accordance with a Poisson process having rate \(\lambda=5\) per week. If the amount of money paid on each policy is exponentially distributed with mean \(\$ 2000\), what is the mean and variance of the amount of money paid by the insurance company in a four-week span?

Let \(X, Y_{1}, \ldots, Y_{n}\) be independent exponential random variables; \(X\) having rate \(\lambda\), and \(Y_{i}\) having rate \(\mu\). Let \(A_{j}\) be the event that the \(j\) th smallest of these \(n+1\) random variables is one of the \(Y_{i} .\) Find \(p=P\left[X>\max _{i} Y_{i}\right\\}\), by using the identity $$ p=P\left(A_{1} \cdots A_{n}\right)=P\left(A_{1}\right) P\left(A_{2} \mid A_{1}\right) \cdots P\left(A_{n} \mid A_{1} \ldots A_{n-1}\right) $$ Verify your answer when \(n=2\) by conditioning on \(X\) to obtain \(p\).

A viral linear DNA molecule of length, say, 1 is often known to contain a certain "marked position," with the exact location of this mark being unknown. One approach to locating the marked position is to cut the molecule by agents that break it at points chosen according to a Poisson process with rate \(\lambda .\) It is then possible to determine the fragment that contains the marked position. For instance, letting \(m\) denote the location on the line of the marked position, then if \(L_{1}\) denotes the last Poisson event time before \(m\) (or 0 if there are no Poisson events in \([0, m])\), and \(R_{1}\) denotes the first Poisson event time after \(m\) (or 1 if there are no Poisson events in \([m, 1])\), then it would be learned that the marked position lies between \(L_{1}\) and \(R_{1} .\) Find (a) \(P\left[L_{1}=0\right\\}\), (b) \(P\left(L_{1}x\right\\}, m

Let \(X\) be an exponential random variable. Without any computations, tell which one of the following is correct. Explain your answer. (a) \(E\left[X^{2} \mid X>1\right]=E\left[(X+1)^{2}\right]\) (b) \(E\left[X^{2} \mid X>1\right]=E\left[X^{2}\right]+1\) (c) \(E\left[X^{2} \mid X>1\right]=(1+E[X])^{2}\)

Events occur according to a Poisson process with rate \(\lambda=2\) per hour. (a) What is the probability that no event occurs between \(8 \mathrm{P} . \mathrm{M} .\) and \(9 \mathrm{P.M.?}\) (b) Starting at noon, what is the expected time at which the fourth event occurs? (c) What is the probability that two or more events occur between \(6 \mathrm{P.M}\). and 8 P.M.?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.