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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

Potential customers arrive to a single-server hair salon according to a Poisson process with rate \(\lambda .\) A potential customer who finds the server free enters the system; a potential customer who finds the server busy goes away. Each potential customer is type \(i\) with probability \(p_{i}\), where \(p_{1}+p_{2}+p_{3}=1\). Type 1 customers have their hair washed by the server; type 2 customers have their hair cut by the server; and type 3 customers have their hair first washed and then cut by the server. The time that it takes the server to wash hair is exponentially distributed with rate \(\mu_{1}\), and the time that it takes the server to cut hair is exponentially distributed with rate \(\mu_{2}\). (a) Explain how this system can be analyzed with four states. (b) Give the equations whose solution yields the proportion of time the system is in each state. In terms of the solution of the equations of (b), find (c) the proportion of time the server is cutting hair; (d) the average arrival rate of entering customers.

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?

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 $$ E[\text { extra time } \mid \mathrm{N}]=\frac{N E\left[S_{1}\right]}{1-\lambda_{1} E\left[S_{1}\right]} $$ 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 $$ W_{Q}^{2}=W_{Q}^{2} \text { (nonpreemptive) }+\frac{\lambda_{1} E\left[S_{1}\right] E\left[S_{2}\right]}{1-\lambda_{1} E\left[S_{1}\right]} $$

In a queue with unlimited waiting space, arrivals are Poisson (parameter \(\lambda\) ) and service times are exponentially distributed (parameter \(\mu\) ). However, the server waits until \(K\) people are present before beginning service on the first customer; thereafter, he services one at a time until all \(K\) units, and all subsequent arrivals, are serviced. The server is then "idle" until \(K\) new arrivals have occurred. (a) Define an appropriate state space, draw the transition diagram, and set up the balance equations. (b) In terms of the limiting probabilities, what is the average time a customer spends in queue? (c) What conditions on \(\lambda\) and \(\mu\) are necessary?

For the \(M / M / 1\) queue, compute (a) the expected number of arrivals during a service period and (b) the probability that no customers arrive during a service period. Hint: "Condition."
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