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Chapter 7: Problem 44

An airport shuttle bus picks up all passengers waiting at a bus stop and drops them off at the airport terminal; it then returns to the stop and repeats the process. The times between returns to the stop are independent random variables with distribution \(F\), mean \(\mu\), and variance \(\sigma^{2} .\) Passengers arrive at the bus stop in accordance with a Poisson process with rate \(\lambda\). Suppose the bus has just left the stop, and let \(X\) denote the number of passengers it picks up when it returns. (a) Find \(E[X]\). (b) Find \(\operatorname{Var}(X)\). (c) At what rate does the shuttle bus arrive at the terminal without any passengers? Suppose that each passenger that has to wait at the bus stop more than \(c\) time units writes an angry letter to the shuttle bus manager. (d) What proportion of passengers write angry letters? (e) How does your answer in part (d) relate to \(F_{e}(x) ?\)

Short Answer

Expert verified
The expected number of passengers picked up by the bus when it returns, denoted by \(E[X]\), is equal to the product of the average rate of passengers arriving at the bus stop and the average time between the bus's returns to the stop, given by \(\lambda \mu\). The variance in the number of passengers, denoted by \(\operatorname{Var}(X)\), can be found using Little's law and is equal to \(\lambda(\sigma^2 + \mu^2)\). The rate at which the shuttle bus arrives at the terminal without passengers can be found using the cumulative distribution function of the exponential distribution, given by \(F(\tau) = 1 - e^{-\lambda \tau}\). The proportion of passengers who write angry letters due to waiting more than a certain time is obtained by integrating the exponential distribution function over the entire time range, given by \(\int_0^{\infty} e^{-\lambda(c-t)} dF(t)\). The derivative of the equilibrium distribution of passenger waiting times \(F_e(x)\) at \(x=c\) gives the relation between the proportion of angry letter-writers and \(F_e(x)\), denoted by \(P(W>c) = \frac{\lambda e^{-\lambda(c)}}{\mu}\).

Step by step solution

01

(a) Find E[X] (Expected number of passengers picked up by the bus)

We need to find the expected number of passengers picked up by the bus when it returns. To do this, we can consider the average rate of passengers arriving at the bus stop (位) and the average time between the bus's returns to the stop (碌). The expected number of passengers picked up by the bus, E[X], can be calculated as: \(E[X] = \lambda \mu\)
02

(b) Find Var(X) (Variance in the number of passengers picked up by the bus)

To find the variance of X, we need to consider the rate at which passengers arrive (位) and the variance in the time between returns to the stop (蟽虏). Use Little's law, a fundamental result in queueing theory: \(\operatorname{Var}(X)=\lambda(\sigma^2 + \mu^2)\)
03

(c) Rate at which the shuttle bus arrives at the terminal without passengers

The shuttle bus arrives at the terminal without any passengers when no one is waiting at the bus stop. In that case, the time between bus returns is less than or equal to the time it takes for one passenger to arrive. Let T be the time between returns to the stop. Then, the rate we need to find is: \(P(T \leq \tau)\), where 蟿 is the time it takes for one passenger to arrive. Since passengers arrive as a Poisson process with rate 位, we have: \(P(T \leq \tau) = F(\tau) = 1 - e^{-\lambda \tau}\), using the cumulative distribution function of the exponential distribution.
04

(d) Proportion of passengers who write angry letters due to waiting more than c time units

Let W be the waiting time of a passenger at the bus stop. We need to find the probability such that W>c (given as waiting time for writing an angry letter) The proportion of passengers who write angry letters is given by: \(P(W>c) = \int_0^{\infty} P(W > c | T=t) dF(t)\) \(= \int_0^{\infty} e^{-\lambda(c-t)} dF(t)\)
05

(e) Relation between the answer in part (d) and Fe(x)

The proportion of passengers who write angry letters (from part d) can be written as: \(P(W>c) = \int_0^{\infty} e^{-\lambda(c-t)} dF(t)\) The term Fe(x) represents the equilibrium distribution of the waiting time for passengers, \(F_e(x) = \frac{1}{\mu} \int_0^{x} e^{-\lambda(u)} du\) Taking the derivative of Fe(x) with respect to x, and evaluating at x=c, we get the relation between the answer in part (d) and Fe(x): \(\frac{dF_e(x)}{dx} = \frac{\lambda e^{-\lambda(x)}}{\mu}\) \(P(W>c) = \frac{\lambda e^{-\lambda(c)}}{\mu}\)

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Key Concepts

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

Poisson Process
When studying bus transportation systems, one key concept is the Poisson process. It's a model used to describe events that occur randomly over a period of time but at a constant average rate. In our scenario with the airport shuttle bus, the arrival of passengers at the bus stop can be modeled as a Poisson process.

Poisson processes are characterized by the fact that the number of events (in this case, passenger arrivals) in non-overlapping intervals is independent. In simpler terms, how many passengers arrive in one period doesn't affect how many will arrive in the next. The average rate of arrival, denoted by \( \lambda \), is crucial as it allows us to predict various aspects of the system, like the expected number of passengers (\(E[X]\)), or the variance in the number of passengers (\(\operatorname{Var}(X)\)). In queuing theory, these metrics help in managing wait times and resource allocation efficiently.
Queueing Theory
Queueing theory dives into the analysis of lines or queues. It pertains to the study of congestion and delay in systems where resources (like buses) must be shared and thus cannot serve everyone at once.

For queueing theory to be applied to our bus system, we consider factors like arrival rate of passengers, service rate of buses, number of buses, and the queue discipline, which is the rule determining which passenger boards next. The theory provides several performance measures, such as waiting time in the queue, probability of passengers having to wait longer than a certain threshold (this could trigger the writing of 'angry letters' mentioned in the exercise), and the overall length of the queue. Little's law, a renowned result in this theory, connects the average number of users in the system to the average waiting time and has applications in calculating the variance of passengers picked up by the bus.
Expected Value
The expected value is a fundamental concept in probability and statistics, representing the average outcome if an experiment (like passengers boarding a bus) is repeated many times. The expected value is referred to by \(E[X]\), which in our case is the expected number of passengers the shuttle will pick up.

Mathematically, \(E[X]\) is calculated based on the average rate of passenger arrival and the average time it takes for the bus to return to the stop. This number gives us an idea of how 'busy' the bus will typically be on its return to the stop. By understanding the expected value, managers can plan resources accordingly, ensuring that the bus size and schedule are adequate to handle the average load.
Variance
While the expected value gives us an average, the variance tells us about the spread of a distribution. In other words, it measures how much the number of passengers picked up by the bus can differ from the expected value.

Variance, represented by \(\operatorname{Var}(X)\), captures the uncertainty and potential variability in passenger arrivals. It takes into account not just the average rate \(\lambda\) but also the variability in the time between the bus鈥檚 returns, \(\sigma^2\). In practical terms, a higher variance means passengers experience a less predictable waiting time, and service may appear less reliable. By reducing the variance, we can improve the consistency of the waits and potentially reduce the number of 'angry letters' written due to long wait times.
Probability Distribution
A probability distribution details how probabilities are assigned to different possible outcomes of a random variable, like the number of passengers waiting at a bus stop or the time until the next bus arrives.

In the context of bus transportation, several distributions are important. The Poisson distribution describes the probability of observing a certain number of arrivals (passengers) within a fixed interval, assuming they follow a Poisson process. Meanwhile, the exponential distribution, often associated with the Poisson process, can model the time between bus arrivals or passenger arrivals. Understanding these distributions helps in answering service-related questions, such as the probability that a bus arrives empty, or the chance that a passenger waits longer than a certain amount of time and decides to write an angry letter.

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Most popular questions from this chapter

Wald's equation can also be proved by using renewal reward processes. Let \(N\) be a stopping time for the sequence of independent and identically distributed random variables \(X_{i}, i \geqslant 1\) (a) Let \(N_{1}=N\). Argue that the sequence of random variables \(X_{N_{1}+1}, X_{N_{1}+2}, \ldots\) is independent of \(X_{1}, \ldots, X_{N}\) and has the same distribution as the original sequence \(X_{i}, i \geqslant 1\) Now treat \(X_{N_{1}+1}, X_{N_{1}+2}, \ldots\) as a new sequence, and define a stopping time \(\mathrm{N}_{2}\) for this sequence that is defined exactly as \(\mathrm{N}_{1}\) is on the original sequence. (For instance, if \(N_{1}=\min \left(n: X_{n}>0\right\\}\), then \(\left.N_{2}=\min \left[n: X_{N_{1}+n}>0\right\\} .\right)\) Similarly, define a stopping time \(N_{3}\) on the sequence \(X_{N_{1}+N_{2}+1}, X_{N_{1}+N_{2}+2}, \ldots\) that is identically defined on this sequence as \(N_{1}\) is on the original sequence, and so on. (b) Is the reward process in which \(X_{i}\) is the reward earned during period \(i\) a renewal Ireward process? If so, what is the length of the successive cycles? (c) Derive an expression for the average reward per unit time. (d) Use the strong law of large numbers to derive a second expression for the average reward per unit time. (e) Conclude Wald's equation.

Satellites are launched according to a Poisson process with rate \(\lambda .\) Each satellite will, independently, orbit the earth for a random time having distribution \(F\). Let \(X(t)\) denote the number of satellites orbiting at time \(t\). (a) Determine \(P\\{X(t)=k\\}\). Hint: Relate this to the \(M / G / \infty\) queue. (b) If at least one satellite is orbiting, then messages can be transmitted and we say that the system is functional. If the first satellite is orbited at time \(t=0\), determine the expected time that the system remains functional. Hint: \(\quad\) Make use of part (a) when \(k=0\).

Consider a single-server bank for which customers arrive in accordance with a Poisson process with rate \(\lambda .\) If a customer will enter the bank only if the server is free when he arrives, and if the service time of a customer has the distribution \(G\), then what proportion of time is the server busy?

A system consists of two independent machines that each function for an exponential time with rate \(\lambda .\) There is a single repairperson. If the repairperson is idle when a machine fails, then repair immediately begins on that machine; if the repairperson is busy when a machine fails, then that machine must wait until the other machine has been repaired. All repair times are independent with distribution function \(G\) and, once repaired, a machine is as good as new. What proportion of time is the repairperson idle?

Consider a single-server queueing system in which customers arrive in accordance with a renewal process. Each customer brings in a random amount of work, chosen independently according to the distribution \(G\). The server serves one customer at a time. However, the server processes work at rate \(i\) per unit time whenever there are \(i\) customers in the system. For instance, if a customer with workload 8 enters service when there are three other customers waiting in line, then if no one else arrives that customer will spend 2 units of time in service. If another customer arrives after 1 unit of time, then our customer will spend a total of \(1.8\) units of time in service provided no one else arrives. Let \(W_{i}\) denote the amount of time customer \(i\) spends in the system. Also, define \(E[W]\) by $$ E[W]=\lim _{n \rightarrow \infty}\left(W_{1}+\cdots+W_{n}\right) / n $$ and so \(E[W]\) is the average amount of time a customer spends in the system. Let \(N\) denote the number of customers that arrive in a busy period. (a) Argue that $$ E[W]=E\left[W_{1}+\cdots+W_{N}\right] / E[N] $$ Let \(L_{i}\) denote the amount of work customer \(i\) brings into the system; and so the \(L_{i}, i \geqslant 1\), are independent random variables having distribution \(G\). (b) Argue that at any time \(t\), the sum of the times spent in the system by all arrivals prior to \(t\) is equal to the total amount of work processed by time \(t .\) Hint: Consider the rate at which the server processes work. (c) Argue that $$ \sum_{i=1}^{N} W_{i}=\sum_{i=1}^{N} L_{i} $$ (d) Use Wald's equation (see Exercise 13\()\) to conclude that $$ E[W]=\mu $$ where \(\mu\) is the mean of the distribution \(G .\) That is, the average time that customers spend in the system is equal to the average work they bring to the system.
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