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Chapter 5: Problem 6

In Example \(5.3\) if server \(i\) serves at an exponential rate \(\lambda_{i}, i=1,2\), show that $$ P\\{\text { Smith is not last }\\}=\left(\frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}}\right)^{2}+\left(\frac{\lambda_{2}}{\lambda_{1}+\lambda_{2}}\right)^{2} $$

Short Answer

Expert verified
The probability that Smith is not the last customer is \(P\{\text{Smith is not last}\} = \left(\frac{\lambda_1}{\lambda_1+\lambda_2}\right)^2 + \left(\frac{\lambda_2}{\lambda_1+\lambda_2}\right)^2\).

Step by step solution

01

Find the probability Smith is served by server 1

Since Smith can be served by either server 1 or server 2, the probability of Smith being served by server 1 would be given by the ratio of the server's serving rate to the combined serving rates of both servers as follows: \[\frac{\lambda_1}{\lambda_1+\lambda_2}\]
02

Find the probability Smith is served by server 1 and not the last customer

We want to find the probability that Smith is served by server 1 and not the last customer. If Smith is not last, it means server 2 finishes serving its other customer after server 1 finishes serving Smith. This probability can be obtained by multiplying the probability from step 1 by the probability that server 1 finishes first (before server 2). \[\frac{\lambda_1}{\lambda_1+\lambda_2} \times \frac{\lambda_1}{\lambda_1+\lambda_2} = \left(\frac{\lambda_1}{\lambda_1+\lambda_2}\right)^2\]
03

Find the probability Smith is served by server 2

Similarly, we can find the probability that Smith is served by server 2 as the ratio of server 2's serving rate to the combined serving rate of both servers: \[\frac{\lambda_2}{\lambda_1+\lambda_2}\]
04

Find the probability Smith is served by server 2 and not the last customer

To find the probability that Smith is served by server 2 and not the last customer, we do the same as we did in step 2 for server 1 but now for server 2. We multiply the probability that he is served by server 2 by the probability that server 1 finishes after server 2. \[\frac{\lambda_2}{\lambda_1+\lambda_2} \times \frac{\lambda_2}{\lambda_1+\lambda_2} = \left(\frac{\lambda_2}{\lambda_1+\lambda_2}\right)^2\]
05

Find the probability Smith is not the last customer

Finally, we add the probabilities from step 2 and step 4 to find the total probability that Smith is not the last customer. \[\left(\frac{\lambda_1}{\lambda_1+\lambda_2}\right)^2 + \left(\frac{\lambda_2}{\lambda_1+\lambda_2}\right)^2\] Therefore, the probability that Smith is not the last customer is: \[P\{\text{Smith is not last}\} = \left(\frac{\lambda_1}{\lambda_1+\lambda_2}\right)^2 + \left(\frac{\lambda_2}{\lambda_1+\lambda_2}\right)^2\]

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Key Concepts

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

Server Systems
Server systems are a crucial part of many computing environments. They process data requests and deliver them to clients or end users. In this context, servers can be tailored for different tasks like web hosting, data management, or application serving.
This exercise considers servers' efficiency and handling capacities by using exponential distribution, primarily focusing on how quickly a server can complete a task, measured using a rate parameter, \( \lambda \).
This rate defines the average number of tasks a server can complete in a time unit. Different servers can have different rate parameters, making some faster and some slower at serving requests.
  • Server 1 and Server 2 have different serving rates, \( \lambda_1 \) and \( \lambda_2 \) respectively.
  • The exercise explores probabilities related to these rate parameters to determine task-serving outcomes.
Probability Calculation
Probability calculation is a method to measure the likelihood of an event occurring within a defined setting. In this exercise, you're tasked to compute the probability of Smith not being the last customer served, using servers with different serving rates.
The base principle is to weigh different possibilities proportionately based on each server's serving rate capability. When calculating these probabilities, it's important to note:
  • The probability of an event is calculated as the serving rate of the server divided by the total serving rate. This is because this method efficiently captures the likelihood of a server consuming its queue of requests faster due to a higher rate parameter.
  • Calculations involve understanding that if server 1 serves Smith, the probability that Smith is not the last customer involves server 1 finishing before server 2, represented as \( \left(\frac{\lambda_1}{\lambda_1 + \lambda_2}\right)^2 \).
Poisson Processes
Poisson processes are an integral element in probability and statistics, often used to model the occurrence of events over time or space in various fields, including server systems. In this specific example, the focus is on modeling the serving of tasks by servers over time.
The exponential distribution is frequently applied here as it describes the time between events in a Poisson process. In this case, the event is 'being served' by one of the servers.
  • For server systems, Poisson processes help understand and predict the frequency of events, aiding in resource planning and queue management.
  • Each server is associated with an exponential distribution characterized by rate parameter \( \lambda \).
  • This exercise models the servers as two independent Poisson processes, each with its serving rate.
Mathematical Proofs
Mathematical proofs involve a logical process to demonstrate the truth of a given statement or solve a problem. In this example, the proof is constructed around finding the probability that Smith is not the last customer served.
Through a sequence of steps or logical deductions involving the known properties of exponential distributions and probability principles, you can structure a solution that verifies or calculates the desired outcome:
  • Initially, calculate the probability that Smith is served by each server individually.
  • Then, explore the likelihood that being served by that particular server means Smith is not the last in line.
  • Combine these findings to establish the complete probability of the event, `Smith is not last`.
  • Each step uses key elements like rates \( \lambda_1 \) and \( \lambda_2 \) along with known formulae to conclude.
These calculations and proofs are a foundational method to ascertain probability outcomes in systems like server queues.

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Most popular questions from this chapter

If \(X_{i}, i=1,2,3\), are independent exponential random variables with rates \(\lambda_{i}, i=1,2,3\), find (a) \(P\left\\{X_{1}

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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) \(E\left[X^{2} \mid X>1\right]=(1+E[X])^{2}\)

(a) Let \(\\{N(t), t \geqslant 0]\) be a nonhomogeneous Poisson process with mean value function \(m(t) .\) Given \(N(t)=n\), show that the unordered set of arrival times has the same distribution as \(n\) independent and identically distributed random variables having distribution function $$ F(x)=\left\\{\begin{array}{ll} \frac{m(x)}{m(t)}, & x \leqslant t \\ 1, & x \geqslant t \end{array}\right. $$ (b) Suppose that workmen incur accidents in accordance with a nonhomogeneous Poisson process with mean value function \(m(t)\). Suppose further that each injured man is out of work for a random amount of time having distribution \(F\). Let \(X(t)\) be the number of workers who are out of work at time \(t\). By using part (a), find \(E[X(t)]\).

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 \(I\) 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
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