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Chapter 5: Problem 43

Customers arrive at a two-server service station according to a Poisson process with rate \(\lambda .\) Whenever a new customer arrives, any customer that is in the system immediately departs. A new arrival enters service first with server 1 and then with server \(2 .\) If the service times at the servers are independent exponentials with respective rates \(\mu_{1}\) and \(\mu_{2}\), what proportion of entering customers completes their service with server \(2 ?\)

Short Answer

Expert verified
The proportion of entering customers who complete their service with server 2 can be found by the formula: \(P(\text{Completing service with server 2}) = e^{-\frac{\lambda}{\mu_1}} * e^{-\frac{\lambda}{\mu_2}}\) where 位 is the arrival rate, 渭1 and 渭2 are the service rates at server 1 and server 2, respectively.

Step by step solution

01

Calculate the interruption rate

First, we need to calculate the rate at which interruptions occur. An interruption occurs when a new customer arrives while a customer is still in the system. Since the arrival process is a Poisson process with rate 位, we have the interruption rate as 位.
02

Calculate the probability of no interruption during server 1 service

Now we have to calculate the probability of no interruption while a customer is in server 1. Since the service time at server 1 follows an exponential distribution with rate 渭1, we can find the probability of no interruption during the service time using the failure rate formula: \(P(\text{No interruption during server 1 service}) = e^{-\lambda T_1}\) where \(T_1\) is the service time at server 1.
03

Calculate the probability of no interruption during server 2 service

Next, we have to calculate the probability of no interruption while a customer is in server 2. Since the service time at server 2 follows an exponential distribution with rate 渭2, we can find the probability of no interruption during the service time using the failure rate formula: \(P(\text{No interruption during server 2 service}) = e^{-\lambda T_2}\) where \(T_2\) is the service time at server 2.
04

Calculate the proportion of customers completing service with server 2

Finally, we want to find the proportion of entering customers who complete their service with server 2. This is equal to the probability of no interruption during server 1 service multiplied by the probability of no interruption during server 2 service. Thus, the proportion of customers completing their service with server two is: \(P(\text{Completing service with server 2}) = P(\text{No interruption during server 1 service}) * P(\text{No interruption during server 2 service})\) \(P(\text{Completing service with server 2}) = e^{-\lambda T_1} * e^{-\lambda T_2}\) Since \(T_1\) and \(T_2\) follow exponential distributions with mean \(1/\mu_1\) and \(1/\mu_2\), we have \(E[T_1] = \frac{1}{\mu_1}\) and \(E[T_2] = \frac{1}{\mu_2}\) \(P(\text{Completing service with server 2}) = e^{-\lambda E[T_1]} * e^{-\lambda E[T_2]}\) \(P(\text{Completing service with server 2}) = e^{-\frac{\lambda}{\mu_1}} * e^{-\frac{\lambda}{\mu_2}}\) Now you can find the proportion of entering customers who complete their service with server 2 by substituting the given values for 位, 渭1, and 渭2 in the final formula.

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Key Concepts

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

Exponential Distribution
Exponential distribution is a critical concept when dealing with the times between events in a Poisson process. In the context of server systems, it describes how long a server might take to process a customer. The key feature of an exponential distribution is that it is memoryless, meaning the probability of the process taking a certain amount of time only depends on the rate of occurrence and not the past history.

In this exercise, each server's service time follows an independent exponential distribution with respective rates of \( \mu_1 \) and \( \mu_2 \). This means the expected time a customer spends in service with the server is the inverse of the rate: \( \frac{1}{\mu_1} \) for server 1 and \( \frac{1}{\mu_2} \) for server 2.

Understanding this helps in calculating the likelihood of no interruptions during service. This is due to the exponential service time helping determine if a new customer interrupts the service.
Service Rate
The service rate, denoted as \( \mu \), is a crucial parameter that describes how fast a server can process jobs or tasks. In queuing theory, which studies waiting lines or queues, the service rate is used to understand how long a customer might expect to wait before their service begins and completes.

In our exercise, each server has its unique service rate: \( \mu_1 \) for server 1 and \( \mu_2 \) for server 2. These rates inform us about the efficiency and speed of each server. A higher service rate implies quicker service times, while a lower rate indicates slower processing.

By understanding these rates, one can determine the average service time and thus how quickly the entire system can function. It鈥檚 crucial for assessing the performance and reliability of systems that rely on queuing and service structures, helping predict and alleviate potential delays.
Probability of Interruption
Interruption probability is vital in assessing whether a task or service in progress will be disturbed by another event. In the context of Poisson processes as in our exercise, this involves calculating the likelihood that another customer arrives before the current one is served.

Given a Poisson arrival process with rate \( \lambda \), the interruption rate is also \( \lambda \). The exponential service times mean that for no interruptions, the service of a customer must complete before the next arrival. The probability of an uninterrupted service can be computed as \( e^{-\lambda T} \), where \( T \) is the service time following an exponential distribution.

Thus, our goal here is to find the probability that no new customer arrives while another is under service of server 1 and server 2. A low interruption probability across both servers indicates a greater chance that customers complete their entire service without being bumped.
Two-Server Queue Model
The two-server queue model is an interesting system that helps businesses and operations manage workflows efficiently. In such a system, an arriving customer can be processed by two servers consecutively. Understanding the operations of both servers helps in optimizing performance across service channels.

In this exercise, each customer initially arrives and engages with server 1 before continuing to server 2, assuming no interruptions occur. The entire process is dictated by Poisson arrivals and exponential service times at both servers.

The probability of successfully moving from server 1 to server 2 without an interruption is crucial for determining what proportion of customers fully complete their service. This system highlights the importance of computing probabilities with the given rates to improve efficiency.

By using the formula \( e^{-\lambda/\mu_1} \cdot e^{-\lambda/\mu_2} \), you can calculate the expected proportion of customers completing service with server 2, reflecting the model's efficacy in real-world applications.

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Most popular questions from this chapter

I et \(T_{1}, T_{2}\), denote the interarrival times of events of a nonhomngeneous Poisson process having intensity function \(\lambda(t)\). (a) Are the \(T_{i}\) independent? (b) Are the \(T_{i}\) identically distributed? (c) Find the distribution of \(T_{1}\).

Shocks occur according to a Poisson process with rate \(\lambda\), and each shock independently causes a certain system to fail with probability \(p .\) Let \(T\) denote the time at which the system fails and let \(N\) denote the number of shocks that it takes. (a) Find the conditional distribution of \(T\) given that \(N=n\). (b) Calculate the conditional distribution of \(N\), given that \(T=t\), and notice that it is distributed as 1 plus a Poisson random variable with mean \(\lambda(1-p) t\). (c) Explain how the result in part (b) could have been obtained without any calculations.

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 $$ p=P\left(A_{1} \cdots A_{n}\right)=P\left(A_{1}\right) P\left(A_{2} \mid A_{1}\right) \cdots P\left(A_{n} \mid A_{1} \cdots A_{n-1}\right) $$ Verify your answer when \(n=2\) by conditioning on \(X\) to obtain \(p\).

If an individual has never had a previous automobile accident, then the probability he or she has an accident in the next \(h\) time units is \(\beta h+o(h) ;\) on the other hand, if he or she has ever had a previous accident, then the probability is \(\alpha h+o(h) .\) Find the expected number of accidents an individual has by time \(t\).

A doctor has scheduled two appointments, one at 1 P.M. and the other at 1:30 P.M. The amounts of time that appointments last are independent exponential random variables with mean 30 minutes. Assuming that both patients are on time, find the expected amount of time that the \(1: 30\) appointment spends at the doctor's office.
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