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In the Erlang loss system suppose the Poisson arrival rate is \(\lambda=2\), and suppose there are three servers, each of whom has a service distribution that is uniformly distributed over \((0,2)\). What proportion of potential customers is lost?

Short Answer

Expert verified
The proportion of potential customers who are lost in this Erlang loss system is given by the Erlang B formula. With the given Poisson arrival rate 位 = 2 and uniformly distributed service rate 渭 = 1 over 3 servers, the traffic intensity 蟻 = \(\frac{2}{3}\). The Erlang B formula yields a probability of all servers being busy (P_loss) of \(\frac{1}{2}\), or 50%.

Step by step solution

01

Identify relevant parameters

In this problem, we are given the Poisson arrival rate 位 = 2, the number of servers n = 3, and the service distribution as uniform over the interval (0, 2).
02

Compute the average service rate

Since the service distribution is uniformly distributed over the interval (0, 2), we can obtain the average service rate (碌) as the midpoint, which is \(\frac{0+2}{2}\), giving us 碌 = 1.
03

Compute the traffic intensity (蟻)

Traffic intensity (蟻) is the ratio of arrival rate to the average service rate times the number of servers: \(蟻 = \frac{位}{n \cdot 碌}\). We plug in the given values and compute the traffic intensity: 蟻 = \(\frac{2}{3 \cdot 1}\) = \(\frac{2}{3}\)
04

Calculate the probability of all servers being busy (P_loss)

Using the Erlang B formula for loss systems, we can find the probability of all servers being busy (P_loss) when a customer arrives. The Erlang B formula is given as: \[P_loss(E, n) = \frac{\frac{(E^n)}{n!}}{\sum_{k=0}^n \frac{(E^k)}{k!}}\] where \(E = 蟻 \cdot n\) is the offered load and n is the number of servers. Plug in the values: E = \(\frac{2}{3} \cdot 3\) = 2 Let's compute the numerator and denominator separately: Numerator = \(\frac{2^3}{3!} = \frac{8}{6}\) Denominator = \(\frac{1}{0!} + \frac{2^1}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!} = 1 + 2 + 2 + \frac{8}{6}\) Now, compute P_loss: \[P_{loss} = \frac{\frac{8}{6}}{1 + 2 + 2 + \frac{8}{6}} = \frac{\frac{8}{6}}{5 + \frac{8}{6}} = \frac{8}{16}\] The proportion of potential customers who are lost is P_loss = \(\frac{1}{2}\), or 50%.

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Key Concepts

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

Poisson Arrival Rate
In queuing theory and specifically in the Erlang loss system, the arrival of events (such as customers or calls) over a period of time is typically modeled using a Poisson process. The Poisson arrival rate, denoted by \( \lambda \), is a key parameter in this model. It represents the average number of arrivals per time unit. For example, if \( \lambda = 2 \), it implies an average of two arrivals per time unit.

This parameter is crucial because it helps determine the load on the system. Understanding the Poisson arrival rate is essential as it influences the traffic intensity and subsequently, the blocking probability. It's also important to note that the Poisson process assumes a random, unpredictable arrival pattern, which is often realistic for many real-world systems.
Service Distribution
The service distribution in a queueing system refers to the statistical pattern with which services are provided to arriving customers. In our problem, each server operates under a uniform distribution over the interval \((0, 2)\). Essentially, this means each customer's service time is randomly selected with equal probability from this interval.

Uniform service distribution is straightforward compared to other distributions like exponential or normal because it suggests that any service duration within the given range is just as likely as any other. This is less common in reality but useful in academic settings to simplify the computations and focus on the core concept of variability in service times.

To compute the average service rate \( \mu \) from this distribution, we take the midpoint of the interval, resulting in \( \mu = 1 \). This average rate is a key component when calculating traffic intensity.
Traffic Intensity
Traffic intensity, denoted by \( \rho \), is a measure of the average load on a system compared to its capacity. It is computed using the formula \( \rho = \frac{\lambda}{n \cdot \mu} \), where \( \lambda \) is the Poisson arrival rate, \( n \) is the number of servers, and \( \mu \) is the average service rate.

In our scenario, with \( \lambda = 2 \), \( n = 3 \) servers, and \( \mu = 1 \), we find \( \rho = \frac{2}{3 \times 1} = \frac{2}{3} \). This value indicates the proportion of time the system is busy. A \( \rho \) value less than one typically means that the system has the capacity to handle the incoming traffic without excessive delay.

Traffic intensity is pivotal in queuing theory as it influences the likelihood of customer loss, server idle times, and overall system efficiency.
Erlang B Formula
The Erlang B formula is a mathematical expression used to calculate the probability of blocking in a loss system where arriving customers may be refused service if all servers are busy. This formula is essential in telecommunications and service systems modeling.

The formula calculates the probability \( P_{loss} \) that a new arrival finds all \( n \) servers occupied as: \[ P_{loss}(E, n) = \frac{\frac{E^n}{n!}}{\sum_{k=0}^n \frac{E^k}{k!}} \]where \( E = \rho \times n \) is the offered load.

In our case, with traffic intensity \( \rho = \frac{2}{3} \) and \( n = 3 \), we have \( E = 2 \). Calculating further, we find that the loss ratio, \( P_{loss} \), equals \( \frac{1}{2} \) or 50%. This result implies that half of all potential customers will be unable to receive service when they arrive.

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Most popular questions from this chapter

In an \(M / G / 1\) queue, (a) what proportion of departures leave behind 0 work? (b) what is the average work in the system as seen by a departure?

Consider a single-server exponential system in which ordinary customers arrive at a rate \(\lambda\) and have service rate \(\mu\). In addition, there is a special customer who has a service rate \(\mu_{1}\). Whenever this special customer arrives, she goes directly into service (if anyone else is in service, then this person is bumped back into quete). When the special customer is not being serviced, she spends an exponential amount of time (with mean \(1 / \theta\) ) out of the system. (a) What is the average arrival rate of the special customer? (b) Define an appropriate state space and set up balance equations. (c) Find the probability that an ordinary customer is bumped \(n\) times.

Consider the priority queueing model of Section \(8.6 .2\) but now suppose that if a type 2 customer is being served when a type 1 arrives then the type 2 customer is bumped out of service. This is called the preemptive case. Suppose that when a bumped type 2 customer goes back in service his service begins at the point where it left off when he was bumped. (a) Argue that the work in the system at any time is the same as in the nonpreemptive case. (b) Derive \(W_{Q}^{1}\). Hint: How do type 2 customers affect type 1 s? (c) Why is it not true that $$ V_{Q}^{2}=\lambda_{2} E\left[S_{2}\right] W_{Q}^{2} $$ (d) Argue that the work seen by a type 2 arrival is the same as in the nonpreemptive case, and so \(W_{Q}^{2}=W_{Q}^{2}\) (nonpreemptive) \(+E\) [extra time] where the extra time is due to the fact that he may be bumped. (e) Let \(N\) denote the number of times a type 2 customer is bumped. Why is $$ E[\text { extra time } \mid N]=\frac{N E\left[S_{1}\right]}{1-\lambda_{1} E\left[S_{1}\right]} $$ Hint: When a type 2 is bumped, relate the time until he gets back in service to a "busy period." (f) Let \(S_{2}\) denote the service time of a type 2 . What is \(E\left[N \mid S_{2}\right] ?\) (g) Combine the preceding to obtain $$ W_{Q}^{2}=W_{Q}^{2} \text { (nonpreemptive) }+\frac{\lambda_{1} E\left[S_{1}\right] E\left[S_{2}\right]}{1-\lambda_{1} E\left[S_{1}\right]} $$

Explain how a Markov chain Monte Carlo simulation using the Gibbs sampler can be utilized to estimate (a) the distribution of the amount of time spent at server \(j\) on a visit. Hint: Use the arrival theorem. (b) the proportion of time a customer is with server \(j\) (i.e., either in server \(j\) 's queue or in service with \(j\) ).

Consider a sequential-service system consisting of two servers, \(A\) and \(B\). Arriving customers will eater this system only if server \(A\) is free. If a customer does enter, then he is immediately served by server \(A\). When his service by \(A\) is completed, he then goes to \(B\) if \(B\) is free, or if \(B\) is busy, he leaves the system. Upon completion of service at server \(B\), the customer departs. Assuming that the (Poisson) arrival rate is two customers an hour, and that \(A\) and \(B\) serve at respective (exponential) rates of four and two customers an hour, (a) what proportion of customers enter the system? (b) what proportion of entering customers receive service from \(\mathrm{B}\) ? (c) what is the average number of customers in the system? (d) what is the average amount of time that an entering customer spends in the system?

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