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Chapter 8: Problem 35

Customers arrive at a single-server station in accordance with a Poisson process having rate \(\lambda .\) Each customer has a value. The successive values of customers are independent and come from a uniform distribution on \((0,1)\). The service time of a customer having value \(x\) is a random variable with mean \(3+4 x\) and variance \(5 .\) (a) What is the average time a customer spends in the system? (b) What is the average time a customer having value \(x\) spends in the system?

Short Answer

Expert verified
(a) The average time a customer spends in the system, denoted by \(W\), is given by Little's formula: \[W = \frac{1}{\mu_s - \lambda}\] where \(\mu_s = \frac{1}{5}\) is the service rate and \(\lambda\) is the Poisson arrival rate. (b) The average time a customer with value \(x\) spends in the system, denoted by \(W_x\), is given by: \[W_x = W \cdot \frac{5}{3 + 4x}\] where \(W\) is the average time a customer spends in the system, and \(x\) is the customer's value.

Step by step solution

01

Understanding the Mean Service Time and Service Rate

First, we need to determine the mean service time for the customers. As stated in the problem, the service time of a customer with value \(x\) has a mean of \(3 + 4x\). Since the values of the customers are uniformly distributed on \((0,1)\), we need to compute the mean of the mean service time: \[\mu = E[3 + 4x] = 3 + 4E[x]\] Since \(x\) is uniformly distributed on \((0,1)\), the mean of \(x\) is: \[E[x] = \frac{1}{2}\] Now, we can compute the mean service time: \[\mu = 3 + 4 \left(\frac{1}{2}\right) = 5\] The service rate, denoted by \(\mu_s\), is the inverse of the mean service time: \[\mu_s = \frac{1}{\mu} = \frac{1}{5}\] Now that we have the service rate, we can proceed to find the average time a customer spends in the system.
02

Determine the Traffic Intensity

In order to find the average time a customer spends in the system, we need to calculate the traffic intensity, denoted by \(\rho\). Traffic intensity is the ratio of the arrival rate to the service rate: \[\rho = \frac{\lambda}{\mu_s}\]
03

Find the Average Time a Customer Spends in the System (Part a)

We can use Little's formula to determine the average time a customer spends in the system, denoted by \(W\): \[W = \frac{1}{\mu_s - \lambda}\] By plugging in the values of \(\mu_s\) and \(\lambda\), we can find the average time a customer spends in the system.
04

Find the Average Time a Customer with Value \(x\) Spends in the System (Part b)

To determine the average time a customer with value \(x\) spends in the system, we need to find how the time a customer spends in the system is related to its value. Since the mean service time for a customer with value \(x\) is \(3 + 4x\), we can use the following ratio: \[\frac{5}{3 + 4x}\] Now, we can multiply this ratio by the average time a customer spends in the system, \(W\), to find the average time a customer with value \(x\) spends in the system: \[W_x = W \cdot \frac{5}{3 + 4x}\] By plugging in the values of \(W\), we can find the average time a customer with value \(x\) spends in the system.

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Key Concepts

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

Uniform Distribution
In statistics, a uniform distribution is a type of probability distribution where every outcome in a range is equally likely to occur. For our exercise, the values of the customers are uniformly distributed on the interval \(0,1\). This means that any value a customer might have between 0 and 1 is equally probable.
The mean of a uniformly distributed variable is calculated as the midpoint of the interval. Since our interval is \(0,1\), the mean \(E[x]\) can be computed as:
  • \(E[x] = \frac{1}{2}\)
This concept is important because it helps us calculate the mean service time, which directly influences other aspects of the problem, like service rate and customer wait time.
Mean Service Time
Mean service time refers to the average time taken to serve a customer. In the given exercise, service time depends on a customer's value \(x\). The mean service time for a customer with some value \(x\) is given by the expression:
  • \(3 + 4x\)
To find the average or mean service time for all customers, we take the expected value, considering \(x\) is uniformly distributed from 0 to 1. Hence, the mean becomes:
  • \(E[3 + 4x] = 3 + 4E[x] = 5\)
This computation gets us a clear picture of how long it typically takes to serve a customer, helping us later determine important metrics like the service rate.
Traffic Intensity
Traffic intensity, denoted by \(\rho\), describes how busy a service system is. It is calculated by taking the ratio of the arrival rate \(\lambda\) to the service rate, which is the reciprocal of the mean service time \(\mu_s\):
  • \(\rho = \frac{\lambda}{\mu_s}\)
Traffic intensity is crucial in queueing theory because it impacts the performance of the system. If \(\rho\) approaches 1, the system becomes more congested, leading to longer waiting times. Maintaining a balance by optimizing traffic intensity ensures a smoother operation, reducing delay and increasing efficiency in serving customers.
Little's Formula
Little's Formula is a fundamental principle in queueing theory that relates several system performance metrics to understand better how a queue operates. It states that the average number of customers in a queue \(L\) is equal to the product of the arrival rate \(\lambda\) and the average time a customer spends in the system \(W\):
  • \(L = \lambda W\)
In our exercise, we use Little's Formula to calculate the average time a customer spends in the system. It is rearranged to solve for \(W\):
  • \(W = \frac{1}{\mu_s - \lambda}\)
This relationship helps us understand and predict system behavior, indicating how modifications to arrival or service rates affect customer wait times and service level.

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Most popular questions from this chapter

Customers arrive at a two-server system according to a Poisson process having rate \(\lambda=5\). An arrival finding server 1 free will begin service with that server. An arrival finding server 1 busy and server 2 free will enter service with server \(2 .\) An arrival finding both servers busy goes away. Once a customer is served by either server, he departs the system. The service times at server \(i\) are exponential with rates \(\mu_{i}\), where \(\mu_{1}=4, \mu_{2}=2\) (a) What is the average time an entering customer spends in the system? (b) What proportion of time is server 2 busy?

Consider a queueing system having two servers and no queue. There are two types of customers. Type 1 customers arrive according to a Poisson process having rate \(\lambda_{1}\), and will enter the system if either server is free. The service time of a type 1 customer is exponential with rate \(\mu_{1}\). Type 2 customers arrive according to a Poisson process having rate \(\lambda_{2}\). A type 2 customer requires the simultaneous use of both servers; hence, a type 2 arrival will only enter the system if both servers are free. The time that it takes (the two servers) to serve a type 2 customer is exponential with rate \(\mu_{2}\). Once a service is completed on a customer, that customer departs the system. (a) Define states to analyze the preceding model. (b) Give the balance equations. In terms of the solution of the balance equations, find (c) the average amount of time an entering customer spends in the system; (d) the fraction of served customers that are type \(1 .\)

Consider a closed queueing network consisting of two customers moving among two servers, and suppose that after each service completion the customer is equally likely to go to either server-that is, \(P_{1,2}=P_{2,1}=\frac{1}{2}\). Let \(\mu_{i}\) denote the exponential service rate at server \(i, i=1,2\) (a) Determine the average number of customers at each server. (b) Determine the service completion rate for each server.

Consider a sequential-service system consisting of two servers, \(A\) and \(B\). Arriving customers will enter 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. Assume 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 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?

A group of \(m\) customers frequents a single-server station in the following manner. When a customer arrives, he or she either enters service if the server is free or joins the queue otherwise. Upon completing service the customer departs the system, but then returns after an exponential time with rate \(\theta\). All service times are exponentially distributed with rate \(\mu\). (a) Find the average rate at which customers enter the station. (b) Find the average time that a customer spends in the station per visit.
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