/*! 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 4 Suppose that a customer of the \... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that a customer of the \(M / M / 1\) system spends the amount of time \(x>0\) waiting in queue before entering service. (a) Show that, conditional on the preceding, the number of other customers that were in the system when the customer arrived is distributed as \(1+P\), where \(P\) is a Poisson random variable with mean \(\lambda\). (b) Let \(W_{Q}^{*}\) denote the amount of time that an \(M / M / 1\) customer spends in queue. As a by-product of your analysis in part (a), show that $$ P\left[W_{\mathrm{Q}}^{*} \leqslant x\right]=\left\\{\begin{array}{ll} 1-\frac{\lambda}{\mu} & \text { if } x=0 \\ 1-\frac{\lambda}{\mu}+\frac{\lambda}{\mu}\left(1-e^{-(\mu-\lambda) x}\right) & \text { if } x>0 \end{array}\right. $$

Short Answer

Expert verified
The distribution of the number of customers in the system when a customer arrives is 1+P, where P is a Poisson random variable with mean λx, i.e., Q(n) = \( e^{-\lambda x} \frac{(\lambda x)^{n-1}}{(n-1)!} \). Furthermore, the probability distribution of the waiting time in the queue is: \( P\left[W_{Q}^{*} \leqslant x\right]=\left\\{\begin{array}{ll} 1-\frac{\lambda}{\mu} & \text { if } x=0 \\\ 1-\frac{\lambda}{\mu}+\frac{\lambda}{\mu}\left(1-e^{-(\mu-\lambda) x}\right) & \text { if } x>0 \end{array}\right. \)

Step by step solution

01

Identify Poisson process of arrivals

Since the system operates as an M/M/1 queue, we know that arrivals follow a Poisson process with rate λ.
02

Derive arrival distribution of previous customers

Conditional on the time spent in the queue by a customer, we can calculate the probability of a certain number of arrivals during that time. If there are n arrivals during that time, then the distribution of the number of arrivals (P(n)) will follow a Poisson distribution with mean λx: P(n) = \( e^{-\lambda x} \frac{(\lambda x)^n}{n!} \)
03

Shift the distribution by 1

Since we are given that the customer spends a positive amount of time (x>0) in the queue, there should be at least one customer in the system when this customer arrives. Therefore, we need to shift the distribution by 1 to account for this additional customer. The distribution of the number of customers in the system (Q) is Q(n) = P(n-1) = \( e^{-\lambda x} \frac{(\lambda x)^{n-1}}{(n-1)!} \) #Part b#: Deriving the probability distribution of time spent in queue
04

Obtain complementary cumulative distribution for interarrival time

Recall the system operates using exponential distributions for arrival and service rates with mean interarrival time 1/λ and mean service time 1/μ. The complementary cumulative distribution function (CCDF) of the interarrival time T between customers can be written as \( P(T > x) = e^{-\lambda x} \).
05

Calculate conditional CCDF for time in queue

If the service time S is less than the interarrival time T, the customer will spend no time in queue (x=0). Otherwise, the customer will spend the excess service time in the queue (x = S-T). We can express the conditional CCDF for the time spent in the queue as \( P(W^{*}_Q > x | S > T) = P(T > S - x | T < S) = \frac{ P(T > S-x)^2}{\mu - \lambda} \).
06

Convert the conditional CCDF to CDF for time in queue

The cumulative distribution function (CDF) for the time spent in the queue can be calculated as the complement of the conditional CCDF: \( P(W^{*}_Q \leq x) = 1 - P(W^{*}_Q > x | S > T) \).
07

Obtain the final probability distribution for time in queue

Plugging the values obtained in the previous steps, we find the probability distribution for the time spent in the queue: \( P\left[W_{\mathrm{Q}}^{*} \leqslant x\right]=\left\\{\begin{array}{ll} 1-\frac{\lambda}{\mu} & \text { if } x=0 \\\ 1-\frac{\lambda}{\mu}+\frac{\lambda}{\mu}\left(1-e^{-(\mu-\lambda) x}\right) & \text { if } x>0 \end{array}\right. \)

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Key Concepts

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

Poisson process
The Poisson process is fundamental in queueing theory and is used to model random events happening over time, such as customer arrivals in a queue. This process is characterized by a constant average rate, denoted as \( \lambda \), which represents the expected number of events (arrivals) per unit of time. An important feature of the Poisson process is that the number of events in non-overlapping intervals are independent of each other.
Poisson processes are applicable in various real-world scenarios beyond queueing, such as call arrivals at a call center or cars passing through a toll booth. In the context of an M/M/1 queue, it handles the arrival of customers efficiently since the system assumes a constant arrival rate, making it predictable.
The probability of observing exactly \( n \) events in a time interval \( t \) is given by the Poisson distribution formula:
  • \( P(n; \lambda t) = \frac{e^{-\lambda t}(\lambda t)^n}{n!} \)
This property allows systems like the M/M/1 queue to model arrivals mathematically, estimate waiting times, and manage capacity planning effectively.
Exponential distribution
In the study of the M/M/1 queue, the service times and inter-arrival times are modeled using the exponential distribution. This distribution is essential due to its memoryless property, which means that the probability of an event occurring in the future is independent of how much time has already elapsed.
The exponential distribution is a continuous probability distribution, and it is characterized by its rate parameter \( \lambda \), which is the reciprocal of the mean, \( 1/\lambda \). In the case of service times, the rate parameter \( \mu \) represents the average number of services that can be completed per unit of time.
The probability density function (PDF) of an exponentially distributed random variable \( X \) is given by:
  • \( f(x; \lambda) = \lambda e^{-\lambda x} \) for \( x \geq 0 \)
Furthermore, the cumulative distribution function (CDF), which shows the probability that the variable takes a value less than or equal to \( x \), is:
  • \( F(x; \lambda) = 1 - e^{-\lambda x} \)
These mathematical functions are instrumental in determining system behaviors such as average waiting times and efficiency in queueing models.
M/M/1 queue
The M/M/1 queue is a simple yet foundational model in queueing theory, representing a system with a single server and infinite calling population. The notation M/M/1 indicates a Markovian (memoryless) arrival process (first M), a Markovian service process (second M), and one server (1).
The M/M/1 model assumes that:
  • Arrivals occur according to a Poisson process with rate \( \lambda \).
  • Service times are exponentially distributed with rate \( \mu \).
  • There is a single server handling arrivals.
  • The system has infinite capacity for waiting customers.
A significant aspect of the M/M/1 queue is that it helps compute performance metrics like average number of customers in the system, average time a customer spends in the system, and the probability of having \( n \) customers in the system. Furthermore, because of its simplicity, it serves as a building block for more complex queuing models. In particular, the M/M/1 queue allows analysts to derive waiting time distributions and service efficiency, which are crucial in designing and managing systems effectively.
Probability distribution
Probability distribution is a core concept in defining the likelihood of different outcomes in random experiments. In queueing theory, it determines how random variables, such as the number of customers arriving or the service times, behave. Understanding probability distributions helps model these uncertainties and make informed decisions in system management.
A Poisson distribution is used when modeling the number of events happening in a fixed interval of time, characterized by a single parameter \( \lambda \), the average rate. It is suitable for scenarios where events occur independently, like arrivals in a queue.
An exponential distribution, on the other hand, models the time between events in a Poisson process. It is key for understanding how long a customer might wait or how quickly a service might be completed.
In an M/M/1 queue, these distributions help derive key probabilities, such as:
  • The likelihood that a customer's waiting time in queue will be precisely \( x \).
  • The probability that no other customers are waiting when a new customer arrives.
  • The expected number of customers in the system at any given time.
By applying these distributions, we are able to manage and optimize queueing systems effectively, ensuring that they operate smoothly and efficiently.

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Most popular questions from this chapter

The economy alternates between good and bad periods. During good times customers arrive at a certain single-server queueing system in accordance with a Poisson process with rate \(\lambda_{1}\), and during bad times they arrive in accordance with a Poisson process with rate \(\lambda_{2} .\) A good time period lasts for an exponentially distributed time with rate \(\alpha_{1}\), and a bad time period lasts for an exponential time with rate \(\alpha_{2}\). An arriving customer will only enter the queueing system if the server is free; an arrival finding the server busy goes away. All service times are exponential with rate \(\mu .\) (a) Define states so as to be able to analyze this system. (b) Give a set of linear equations whose solution will yield the long-run proportion of time the system is in each state. In terms of the solutions of the equations in part (b), (c) what proportion of time is the system empty? (d) what is the average rate at which customers enter the system?

In a two-class priority queueing model suppose that a cost of \(C_{i}\) per unit time is incurred for each type \(i\) customer that waits in queue, \(i=1,2 .\) Show that type 1 customers should be given priority over type 2 (as opposed to the reverse) if $$ \frac{E\left[S_{1}\right]}{C_{1}}<\frac{E\left[S_{2}\right]}{C_{2}} $$

Customers arrive at a single-server station in accordance with a Poisson process with rate \(\lambda\). All arrivals that find the server free immediately enter service. All service times are exponentially distributed with rate \(\mu\). An arrival that finds the server busy will leave the system and roam around "in orbit" for an exponential time with rate \(\theta\) at which time it will then return. If the server is busy when an orbiting customer returns, then that customer returns to orbit for another exponential time with rate \(\theta\) before returning again. An arrival that finds the server busy and \(N\) other customers in orbit will depart and not return. That is, \(N\) is the maximum number of customers in orbit. (a) Define states. (b) Give the balance equations. In terms of the solution of the balance equations, find (c) the proportion of all customers that are eventually served; (d) the average time that a served customer spends waiting in orbit.

Consider a single-server queue with Poisson arrivals and exponential service times having the following variation: Whenever a service is completed a departure occurs only with probability \(\alpha .\) With probability \(1-\alpha\) the customer, instead of leaving, joins the end of the queue. Note that a customer may be serviced more than once. (a) Set up the balance equations and solve for the steady-state probabilities, stating conditions for it to exist. (b) Find the expected waiting time of a customer from the time he arrives until he enters service for the first time. (c) What is the probability that a customer enters service exactly \(n\) times, \(n=\) \(1,2, \ldots ?\) (d) What is the expected amount of time that a customer spends in service (which does not include the time he spends waiting in line)? Hint: Use part (c). (e) What is the distribution of the total length of time a customer spends being served? Hint: Is it memoryless?

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 \(1 .\)

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