/*! 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 15 A service center consists of two... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 15

A service center consists of two servers, each working at an exponential rate of two services per hour. If customers arrive at a Poisson rate of three per hour, then, assuming a system capacity of at most three customers, (a) what fraction of potential customers enter the system? (b) what would the value of part (a) be if there was only a single server, and his rate was twice as fast (that is, \(\mu=4) ?\)

Short Answer

Expert verified
(a) The fraction of potential customers who enter the two-server system is approximately 0.6433. (b) The fraction of potential customers who enter the single-server system with a service rate of 4 services per hour is approximately 0.52.

Step by step solution

01

Part (a): Fraction of potential customers who enter the system

The first step to calculate the fraction of potential customers who enter the system is calculating the traffic intensity 蟻, and the probability P_0 that the system is empty. In M/M/c/K queue, we will use the following formula for P_0: \[P_0 = \frac{1}{\sum_{k=0}^{c}\frac{(c\rho)^k}{k!} + \frac{(c\rho)^{c+1}}{c!(c(1-\rho))}}\] where c is the number of servers, and 位 and 碌 are the Poisson arrival and service rates, respectively. Then, we can calculate 蟻 and P_0 for the given values. 位 = 3 customers per hour 碌 = 2 services per hour c = 2 servers 蟻 = 位 / (c * 碌) = 3 / (2 * 2) = 0.75 Now, calculate P_0: P_0 = 1 / (危(k=0,2)((2 * 0.75)^k / k!) + (2 * 0.75)^3 / (2 * (1 - 0.75))) P_0 鈮� 0.3158 Now, we can find the fraction of potential customers who enter the system using the following formula for the probability P_K that the system is full: \[P_K = \frac{c\rho}{c!}\frac{P_0}{(1-\rho)},\] where K = c and c = 2 servers in this case. P_2 = (2 * 0.75) / (2 * (1 - 0.75)) * 0.3158 P_2 鈮� 0.3567 Finally, we can determine the fraction of potential customers who enter the system. This fraction is equal to 1 - P_2, representing customers who are not turning away when the system is full. Fraction of potential customers entering = 1 - P_2 = 1 - 0.3567 鈮� 0.6433
02

Part (b): Fraction of potential customers if there was a single server, and his rate was twice as fast

First, set new parameters for a single server with a service rate of 4 services per hour: 位 = 3 customers per hour 碌_1 = 4 services per hour c_1 = 1 server 蟻_1 = 位 / (c_1 * 碌_1) = 3 / (1 * 4) = 0.75 Now, calculate P_0 for the single server system: P_0_1 = 1 / (危(k=0,1)((1 * 0.75)^k / k!) + (1 * 0.75)^2 / (1 * (1 - 0.75))) P_0_1 鈮� 0.64 Calculate P_K for the single server system: P_1 = (1 * 0.75) / (1 * (1 - 0.75)) * 0.64 P_1 鈮� 0.48 Finally, we can determine the fraction of potential customers who enter the system. Fraction of potential customers entering for the single server = 1 - P_1 = 1 - 0.48 鈮� 0.52 To summarize our results: (a) The fraction of potential customers who enter the two-server system is approximately 0.6433. (b) The fraction of potential customers who enter the single-server system with a service rate of 4 services per hour is approximately 0.52.

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Key Concepts

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

Exponential Service Rate
The concept of an *exponential service rate* is a fundamental principle in queueing theory. It describes the rate at which servers process customers or tasks, where the time between services follows an exponential probability distribution. This means services occur randomly but have a consistent average rate over time.
Exponential distributions are characterized by their memoryless property. This signifies that the probability of a service being completed in the next instant is independent of how much time has already passed.
  • Formula: The service rate is often represented as \( \mu \), and it is defined as the number of customers a server can serve per time unit.
  • Impact: A higher \( \mu \) implies faster service, meaning shorter waiting times for customers.
Understanding the exponential service rate helps in analyzing server efficiency and predicting queue behavior, making it an essential tool in planning and managing service-based operations.
Poisson Process
The *Poisson process* is a statistical model used to describe random points in time or space, pertinent to events occurring in a fixed interval. In the context of queueing systems, it is often applied to model the random arrival of customers. This process is characterized by its rate \( \lambda \), representing the average number of arrivals per time unit.
Key features of the Poisson process include:
  • Independence: The number of arrivals in separate time intervals are independent.
  • Stationarity: The average rate \( \lambda \) is constant over time.
These properties make the Poisson process a robust model for various real-world systems where arrivals are sporadic yet predictable in the long run. In queueing theory, it simplifies the analysis of systems by efficiently approximating random arrival patterns.
Traffic Intensity
*Traffic intensity*, denoted as \( \rho \), quantifies the load on a queuing system. It is the ratio of the arrival rate to the service rate for each server, calculated using \( \rho = \frac{\lambda}{c \times \mu} \), where \( \lambda \) is the arrival rate, \( \mu \) is the service rate, and \( c \) is the number of servers.
  • Interpretation: \( \rho \) indicates how busy a system is. A \( \rho \) value less than 1 suggests that the system can cope with incoming traffic without excessive delays.
  • Impact: High traffic intensity often leads to congestion, longer queues, and waiting times.
Calculating traffic intensity is critical for evaluating a system's ability to manage current loads and for assessing potential enhancements to handle future demand.
Queue Capacity
*Queue capacity* refers to the maximum number of customers or tasks that a system can accommodate at any given time. In scenario analysis, like the one given, the queue capacity was capped at three customers.
This metric is vital for understanding and managing the flow within service systems.
  • Finite capacity: Limits on queue capacity often lead to potential customers being turned away once the system is full, affecting service levels.
  • Capacity planning: Knowing the queue capacity helps in designing systems that balance customer satisfaction and operational efficiency.
By evaluating queue capacity, businesses can ensure they provide adequate resources to meet demand while also planning for future scalability.

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Most popular questions from this chapter

Suppose that a one-celled organism can be in one of two states-either \(A\) or \(B .\) An individual in state \(A\) will change to state \(B\) at an exponential rate \(\alpha\); an individual in state \(B\) divides into two new individuals of type \(A\) at an exponential rate \(\beta .\) Define an appropriate continuous- time Markov chain for a population of such organisms and determine the appropriate parameters for this model.

There are two machines, one of which is used as a spare. A working machine will function for an exponential time with rate \(\lambda\) and will then fail. Upon failure, it is immediately replaced by the other machine if that one is in working order, and it goes to the repair facility. The repair facility consists of a single person who takes an exponential time with rate \(\mu\) to repair a failed machine. At the repair facility, the newly failed machine enters service if the repairperson is free. If the repairperson is busy, it waits until the other machine is fixed. At that time, the newly repaired machine is put in service and repair begins on the other one. Starting with both machines in working condition, find (a) the expected value and (b) the variance of the time until both are in the repair facility. (c) In the long run, what proportion of time is there a working machine?

Customers arrive at a two-server station in accordance with a Poisson process having rate \(\lambda\). Upon arriving, they join a single queue. Whenever a server completes a service, the person first in line enters service. The service times of server \(t\) are exponential with rate \(\mu_{i}, t=1,2\), where \(\mu_{1}+\mu_{2}>\lambda .\) An arrival finding both servers free is equally likely to go to cither one. Define an appropriate continuous-time Markov chain for this model, show it is time reversible, and find the limiting probabilities.

Consider a taxi station where taxis and customers arrive in accordance with Poisson processes with respective rates of one and two per minute. A taxi will wait no matter how many other taxis are present. However, if an arriving customer does not find a taxi waiting. he leaves. Find (a) the average number of taxis waiting, and (b) the proportion of arriving customers that get taxis.

Each time a machine is repaired it remains up for an exponentially distributed time with rate \(\lambda\). It then fails, and its failure is either of two types. If it is a type 1 failure, then the time to repair the machine is exponential with rate \(\mu_{1}\) if it is a type 2 failure, then the repair time is exponential with rate \(\mu_{2}\). Each failure is, independently of the time it took the machine to fail, a type 1 failure with probability \(p\) and a type 2 failure with probability \(1-p .\) What proportion of time is the machine down due to a type 1 failure? What proportion of time is it down due to a type 2 failure? What proportion of time is it up?
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