/*! 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 35 Satellites are launched accordin... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 35

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 \(k=0\).

Short Answer

Expert verified
To determine the probability that there are k satellites orbiting at time t, use the probability distribution of an \(M / G / \infty\) queue: \( P\{X(t)=k\} = e^{-\Lambda(t)} \frac{[\Lambda(t)]^k}{k!} \), where \(\Lambda(t) = \lambda \int_0^t (1 - F(u)) du\) is the mean number of satellites orbiting at time t with the orbit time distribution F. The expected time the system remains functional, E[T], can be found by integrating over all time: \(E[T] = 鈭玙0^鈭� (1-P\{X(t)=0\}) dt = 鈭玙0^鈭� (1 - e^{-\Lambda(t)}) dt\). The value of the integral depends on the orbit time distribution F and the arrival rate 位.

Step by step solution

01

Understand the Poisson process

A Poisson process is a stochastic process representing the number of events that occur within a fixed interval of time. In this problem, the events are the launches of satellites, occurring at a rate 位.
02

Relate the problem to the \(M/G/\infty\) queue

The \(M / G / \infty\) queue represents a queuing system with a Poisson arrival process (rate 位), general service time distribution F, and an infinite number of servers. Since each satellite's orbit time is independent and has the same distribution, we can relate the satellite problem to the \(M / G / \infty\) queue. The number of satellites orbiting at time t, X(t), is equivalent to the number of customers being served in the queuing system at time t.
03

Find the probability \(P\{X(t)=k\}\)

To determine the probability that there are k satellites orbiting at time t, we need to use the probability distribution of an \(M / G / \infty\) queue. \( P\{X(t)=k\} = e^{-\Lambda(t)} \frac{[\Lambda(t)]^k}{k!} \), where \(\Lambda(t) = \lambda \int_0^t (1 - F(u)) du\) is the mean number of satellites orbiting at time t with the orbit time distribution F.
04

Determine the expected time the system remains functional

The system is said to be functional as long as there is at least one satellite orbiting the Earth. We can use the result from part (a) for k=0 to find the probability P{X(t)=0}, which is the probability that there are no satellites orbiting at time t. To find the expected time the system remains functional, we need to calculate the expected time until the first moment when there are no satellites orbiting, or equivalently, when X(t)=0. Let T be the time when X(t)=0 for the first time. We need to find E[T]. E[T] = 鈭玙0^鈭� t dP(T鈮)= 鈭玙0^鈭� P(T>t)dt = 鈭玙0^鈭� P\{X(t)>0\} dt Using the result from part (a) for P\{X(t)=0\}, E[T] = 鈭玙0^鈭� (1-P\{X(t)=0\}) dt = 鈭玙0^鈭� (1 - e^{-\Lambda(t)}) dt. We need to evaluate the integral to find the expected time the system remains functional. Note that the value of the integral depends on the orbit time distribution F and the arrival rate 位.

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Key Concepts

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

Satellite Launches
In the study of satellite launches, we often examine the frequency and timing of these events using mathematical models. A common way to represent the occurrence of satellite launches is through a Poisson process.
This process is ideal for modeling the random nature of satellite launches as it describes the probability of a given number of events occurring in a fixed interval of time, given a known average rate of occurrence, \( \lambda \).
For satellites, \( \lambda \) represents the average number of launches per unit time. Each launch can be seen as an independent event.
This randomness is quite applicable to real-world satellite launches, considering the many variables such as weather and technical readiness. Utilizing Poisson processes helps in not only predicting the number of satellites that will be launched but also in planning resources for subsequent activities.
Probability Distribution
A probability distribution is a fundamental concept used to determine the likelihood of possible outcomes for a random variable.
In the context of satellite launches and their orbits, each satellite's time in orbit can be regarded as a random variable with its own probability distribution, generally denoted by \( F \).
This distribution informs us about the satellite's expected duration in service before deorbiting or becoming non-functional.
For instance, the probability that a satellite is still operational after a certain amount of time can be determined using this distribution.
Knowing \( F \), we can integrate it into our equations to compute the expected number of satellites in orbit, \( X(t) \).
For students aiming to understand how systems behave over time, this probability distribution is a key tool, adapting general statistical concepts into practical applications such as satellite management.
Queuing Theory
Queuing theory plays a vital role in managing systems where events or items line up for processing, much like in a queue at a bank.
In our scenario, this theory is utilized to understand how satellites, once launched, can be seen as customers in a service line.
The satellite launches and their orbiting times reflect in the framework of an \( M / G / \infty \) queue, where each satellite independently and randomly orbits Earth.
Using this theory, each satellite's orbit time relates to service time, and the infinite number of servers suggests there's no waiting time for orbit.
This concept allows us to identify the number of satellites simultaneously in orbit and calculate probabilities associated with their functioning over time.
In addition to satellite management, queuing theory is applicable in many fields, helping optimize resource allocation and system functionality.
Service Time Distribution
The service time distribution is the core concept used in queue-related problems to describe how long each customer, or in this case, satellite, will need to be served.
In satellite orbiting terms, this 'service time' refers to how long a satellite remains operationally functional around the Earth.
The distribution, akin to the probability distribution, represents the variability or randomness in this time for each satellite and is a crucial part of determining satellite system functionality.
It dictates our ability to predict how long the system will be functional, i.e., having at least one satellite orbiting and capable of communication.
Understanding the service time distribution helps in evaluating strategies for future satellite launches and ensuring continuous functionality of the satellite system for communication purposes.

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Most popular questions from this chapter

Consider a miner trapped in a room that contains three doors. Door 1 leads him to freedom. after two days of travel; door 2 returns him to his room after a four-day journiey; and door 3 returns him to his room after a six-day journey. Suppose at all times he is equally likely to choose any of the three doors, and let \(T\) denote the time it takes the miner to become free. (a) Define a sequence of independent and identically distributed random variables \(X_{1}, X_{2} \ldots\) and a stopping time \(N\) such that $$ T=\sum_{i=1}^{N} X_{i} $$ Note: You may have to imagine that the miner continues to randomly choose doors even after he reaches safety. (b) Use Wald's equation to find \(E[T]\). (c) Compute \(E\left[\sum_{i=1}^{N} X_{i} \mid N=n\right]\) and note th芒t it is not equal to \(E\left[\sum_{i=1}^{n} X_{i}\right]\). (d) Use part (c) for a second derivation of \(E[T]\).

Considera 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 seryer 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 $$ E[W]=\lim _{n \rightarrow \infty}\left(W_{1}+\cdots+W_{u}\right) / n $$ 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.

Each of \(n\) skiers continually, and independently, climbs up and then skis down a particular slope. The time it takes skier \(i\) to climb up has distribution \(F_{i}\), and it is independent of her time to ski down, which has distribution \(H_{l}\), \(i=1, \ldots, n\). Let \(N(t)\) denote the total number of times members of this group have skied down the slope by time \(t\). Also, let \(U(t)\) denote the number of skiers climbing up the hill at time \(t\). (a) What is \(\lim _{t \rightarrow \infty} N(t) / t ?\) (b) Find \(\lim _{t \rightarrow \infty} E[U(t)]\). (c) If all \(F_{i}\) are exponential with rate \(\lambda\) and all \(G_{i}\) are exponential with rate \(\mu\), what is \(P\\{U(t)=k\\}\) ?

Consider a renewal process \(\\{N(t), t \geqslant 0]\) having a gamma \((r, \lambda)\) interarrival distribution. That is, the interarrival density is $$ f(x)=\frac{\lambda e^{-\lambda x}(\lambda x)^{r-1}}{(r-1) !}, \quad x>0 $$ (a) Show that $$ P[N(t) \geqslant n\\}=\sum_{i=n r}^{\infty} \frac{e^{-\lambda t}(\lambda t)^{i}}{i !} $$ (b) Show that $$ m(t)=\sum_{i=r}^{\infty}\left[\frac{i}{r}\right] \frac{e^{-\lambda t}(\lambda t)^{i}}{i !} $$ where \([i / r]\) is the largest integer less than or equal to \(i / r\). Hint: Use the relationship between the gamma \((r, \lambda)\) distribution and the sum of \(r\) independent exponentials with rate \(\lambda\), to define \(N(t)\) in terms of a Poisson process with rate \(\lambda\).

If \(A(t)\) and \(Y(t)\) are, respectively, the age and the excess at time \(t\) of a renewal process having an interarrival distribution \(F\), calculate $$ P\\{Y(t)>x \mid A(t)=s\\} $$
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