北邮《计算机网络》英文选择题课堂小测

课堂小测验及答疑汇总

2020年 5 月 12 日
Q1. Which is the network address after aggregation of following 3 networks: 200.2.50.0/23,
200.2.52.0/22 and 200.2.56.0/22?
A. 200.2.48.0/20
B. 200.2.50.0/20
C. 200.2.48.0/21
D. 200.2.50.0/21
Answer: A

Q2 An IP Datagram with a 20-byte header and total length of 1500 bytes is to be forwarded
into a network with a MTU of 800 bytes, thus fragmentation is necessary. What is the
offset value in the 2nd fragment?
A. 800
B. 780
C. 100
D. 97
Answer: D.

课堂练习
A company (with network 114.23.5.0/24) has 3 departments A, B and C, with 64,
10 and 30 hosts(routers) respectively. How to allocate addresses to them in the
order of A, B and then C? Write out the network address of each department. Is
there a more efficient way for address allocation?

2020年 5 月 7 日
Q1. Network 191.168.0.0 is divided into 8 subnets with same size,which is the correct
subnet mask of each subnet?
A. 255.255.0.0
B. 255.255.224.0
C. 255.255.240.0
D. 255.255.255.0
Answer: B

Q2. For a network with 100 hosts (including routers), which subnet mask shall be adopted as
a best choice? (按照最少够用原则)
A. /23
B. /24
C. /25
D. /26
Answer: C

2020 年 4 月 28 日
Q1. When the load in the network reaches the network capacity, the packet delay _____
and the network throughput reaches ______.
A. increases sharply; its minimum
B. increases sharply; its maximum
C. decreases sharply; its minimum
D. decreases sharply; its maximum
Answer: B

2020 年 4 月 24 日
Q1. Which statement is NOT true for Link State Routing?
A. Compared with Distance Vector Routing, bandwidth consumed is more
due to flooding.
B. It is based on local knowledge since routers only know part of network topology.
C. It makes use of Dijkstra algorithm to build routing table.
D. Compared with Distance Vector Routing, it converges faster and does not
have the problem of count-to-infinity.
Answer: B

Q2. Which congestion control method is reactive and adopts explicit congestion signaling
with the help of destination host?
A. Choke packet
B. Back pressure
C. Traffic aware routing
D. ECN
Answer: D

 2020 年 4 月 21 日
Q1. About virtual circuit and datagram, which statement is NOT true?
A. Both use packet switching technology
B. Virtual circuit networks provide connection-oriented service, while datagram
networks provide connectionless service.
C. In datagram networks, packets from same source to same destination may
traverse over different paths, while in virtual circuit networks, belonging to
same VC, they will flow over the same path.
D. Datagram routers always use full address, while virtual circuit routers only deal
with VCI.
Answer: D

2020 年 4 月 17 日
Q1. About VLAN, which statement is NOT TRUE?
A. VLAN decouples the logical topology from the physical topology.
B. For computers connected to a HUB,some can be VLAN-aware while others
are not VLAN-aware.
C. VLAN-aware bridges can limit the scope of broadcasting frames while normal bridges cannot.
D. Legacy Ethernet frame does not support VLAN.
Answer: B

Q2. 一台有 1 个 WAN 口、4 个 LAN 口的路由器,可以连接几个网络?
A. 1
B. 2
C. 4
D. 5
Answer: B

Q3. About virtual circuit, which statement is NOT TRUE?
A. A physical connection is set up before the source sends out data packets.
B. A route from source to destination is chosen at connection setup, so all data
packets will arrive in same sequence as they are sent out.
C. Over an outgoing port of a router, 2 virtual connections cannot use the
same VCI.
D. As VCI is local-significant, it may change along the path from source to
destination.
Answer: A

2020年 4 月 14 日
Q1. In an IEEE802.11 network, at some instant, 3 stations sense the channel and find it is idle,
then multiple frames are to be transmitted, including a RTS, a re-transmitted data frame
and an ACK, which ONE will probably be sent?
A. RTS
B. re-transmitted data frame
C. ACK
D. None of them
Answer: C

Q2. In which case(s), a LAN switch will flood an incoming frame?
A. the destination address is unknown
B. the destination address is known, the outgoing port is same with incoming port
C. the destination address is known, the outgoing port is different from incoming
port
D. the destination address is 0xFF-FF-FF-FF-FF-FF
Answer: A,D

Q2. About Ethernet, consider Ture(T) or False(F) for the statements below:

  1. In Gigabit Ethernet, if computers connect to a Hub, they work in half-duplex,
    which means that data could be transmitted in only one direction at a time,
    either from computer to hub, or from hub to computer.
  2. CSMA/CD can be used sometimes in a fully switched Ethernet.
  3. Manchester encoding is not used in Ethernet with data rate higher than 10Mbps.
    Answer: TFT

2020 年 4 月 7 日
Q1.For a CSMA/CD network with bus topology, assume the frame time is T0, and the
propagation delay of the cable is τ, which are the vulnerable period and
contention slot time respectively?
A. T0, 2T0
B. T0, 2τ
C. τ, 2T0
D. τ, 2τ
Answer: D

Q2. About IEEE802 Reference Model, determine True(T) or False(F) for the statements
below:

  1. Different LAN may have different LLC sub-layer protocols.
  2. Different LAN may have different MAC sub-layer protocols.
  3. A LAN protocol may include multiple physical layer protocols.
    Answer: FTT

2020 年 4 月 3 日
Q1. What type of service is provided by Ethernet?
A. connection-oriented service
B. connectionless service
C. connectionless service with explicit ACK
D. None of the above
Answer: B

Q2. Which statement is NOT TRUE for stations using pure ALOHA?
A. A station transmits at will without coordination with other stations
B. A station can only find collision when timer expires
C. After timer expires, the station retransmits the frame at once.
D. The vulnerable period is twice the frame time, i.e. 2T0
Answer: C

2020 年 3 月 31 日
Q1. (多选)For protocol 6, in which cases NAK will be sent while frame-expected=6 and no_nak is true?
A. received a damaged frame with seq=6
B. received a frame with seq=4
C. received a duplicated ACK
D. received a frame with seq=7
Answer: A、B、D

Q2. Consider a 50-kbps satellite channel with a 500-msec round-trip propagation delay, 1000-
bit frames are sent with Go-back-N ARQ and short ACK frames as acknowledgement. If
a line utilization of 50% is required, how many bits at least shall be used for sequence
number?
A. 3
B. 4
C. 5
D. 6
Answer: B

. 2020 年 3 月 27 日
Q1. For a Go-back-N ARQ with 3-bit sequence number, if the seq number in
current receiving windows is 0,while the entity receives an un-damaged frame with r.seq=1, what will the entity do for a proper operation?
(Assume the entity always has a packet to send)
A. accept the frame and send a frame with s.ack=1
B. discard the frame
C. discard the frame and send a frame with s.ack=0
D. discard the frame and send a frame with s.ack=7
Answer:D

. 2020 年 3 月 24 日
Q1. Assume 3-bit sequence number is used in Protocol 4. When a node receives a frame with r.seq=7 and r.ack=7, what will s.seq and s.ack be in the frame s it will send as a response?
A. 7 and 7
B. 7 and 0
C. 0 and 7
D. 0 and 0
Answer:C

Q2. Assume 3-bit sequence number is used in a Go-back-N protocol. Now the
sender has sent out 8 frames with sequence number 0,1,… and 7. When
timing out, the sender only received ACK1 and ACK3, how many frames will the sender retransmit?
A. 2
B. 4
C. 6
D. 8
Answer:B

2020 年 3 月 20 日
Q1. A data link layer sending entity has 3 jobs before delivering bit stream to its physical layer,

  1. bit stuffing
  2. adding header flag and trailer flag
  3. adding checksum bits
    What is the correct sequence of the 3 jobs?
    A. 123
    B. 312
    C. 132
    D. 321
    Answer:B

Q2. 计算题
若 G(x)=x5+x2+x+1,接收方收到的数据是:1001 1011 1110 0101 1000,请判断该帧是否传输出错。
Answer:传输出错。

Q3.附加作业题
The following character encoding is used in a data link protocol: A: 01000111; B: 11100011; FLAG: 01111110; ESC: 11100000
Show the bit sequence transmitted (in binary) for the four-character frame: A B ESC FLAG when each of the following framing methods are used:
(a) Character count.
(b) Flag bytes with byte stuffing.
© Starting and ending flag bytes, with bit stuffing.
(d) Calculate the efficiency (as a percentage of real data per bit sent) of your answers to (a), (b)and © .
Answer:
1)字符计数法:
00000101 01000111 11100011 11100000 01111110
增加了一个长度字节(长度值包括其本身),效率为 4/5=80%
2)字节填充法:
01111110 01000111 11100011 11100000 11100000
11100000 0111 1110 011111110
增加了首尾标志和 2 个转义字符,效率为 4/8=50%
3)比特填充法:
01111110 01000111 110100011 111000000 011111010 01111110
增加了首尾标志和 3 位填充,效率为 32/51=62.75%

2020 年 3 月 17 日
Q1. What type of service is provided to network layer by 10BaseT?
A. connectionless service
B. connectionless service with acknowledgement
C. frame-based connection-oriented service
D. byte-based connection-oriented service
Answer:A

Q2. 计算题
从信道上收到比特序列: 1101 0111 1110 0111 1110 1101 1011 1110 0010 1100 0101 1111 0101 1001 1111 1001,该比特序列中包含一个完整的帧,用十六进制写出该帧的内 容(不包含帧的首尾标志)。
Answer:D B E 5 8 B F 6

2020 年 3 月 13 日
Q1.Which is the maximum data rate over a 4-kHz channel whose signal-to-noise ratio is 30 dB, and QAM-16 modulation is used?
A. 8Kbps
B. 32Kbps
C. 40Kbps
D. 64Kbps
Answer:B

Q2. About routing (hunting for a path) in a circuit switching network, which is true?
A.There is no need for routing in circuit switching network.
B. Routing is required only at connection set up phase.
C. Routing is necessary during the whole procedure of communication.
D. Routing is necessary at circuit termination phase.
Answer:B

2020 年 3 月 10 日
Q1. Which multiplexing scheme is adopted in Internet?
A. Code Division Multiple Access
B. Frequency Division Multiplexing
C. Time Division Multiplexing
D. Wavelength Division Multiplexing
Answer:C

2020 年 3 月 6 日
Q1. Which of the following is not defined in network architecture?
A. Function of each layer
B. Protocols in each layer
C. Services provided by each layer
D. Implementation of certain service in each layer
Answer:D

Q2. Talking about bandwidith in Hz, which is true?
A. Twisted pair < Radio and TV < Coax < Fiber
B. Radio and TV < Twisted pair < Coax < Fiber
C. Radio and TV < Coax < Twisted pair < Fiber
D. Coax < Radio and TV < Twisted pair < Fiber
Answer:A

. 2020 年 3 月 3 日
Q1. For the following 2 statements, decide true or false.

  1. Layer N protocol must be used to implement the service provided by layer N.
  2. There is no need to consider the format of PDU when designing service
    primitives.
    A. True, True
    B. True, False
    C. False, True
    D. False, False
    Answer:A

Q2. In communications between 2 adjacent routers, which layer provides the function of data block exchange?
A. Physical layer
B. Data Link layer
C. Network layer
D. Transport layer
Answer:B

. 2020 年 2 月 28 日
Q1.(不定项) Which statement(s) is/are correct for a wifi network?
A. It is a broadcast network
B. It is a point-to-point network
C. It is an access network
D. It is an edge of 4G mobile communication network
Answer:AC

Q2. Which statement is NOT true?
A. A computer network can be organized as a set of layers.
B. A single function can be performed by the cooperation of adjacent 2 layers, thus the implementation is visible to both layers.
C. Each layer offers a certain service(s) to its upper layer.
D. The upper layer access the service provided through the interface.
Answer: B

. 2020 年 2 月 25 日
Q1. Which of the following is/are NOT a type of computer network?
A. Internet
B. a Wifi network in a Cafetiria
C. World Wide Web
D. 10 computers connecting to a router with cable
Answer:C

本文来自互联网用户投稿,该文观点仅代表作者本人,不代表本站立场。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如若转载,请注明出处:http://www.mzph.cn/pingmian/34753.shtml

如若内容造成侵权/违法违规/事实不符,请联系多彩编程网进行投诉反馈email:809451989@qq.com,一经查实,立即删除!

相关文章

MySQL索引优化解决方案--索引测试--explain(2)

索引测试 通过存储过程往数据库中插入300w 条数据分别测试使用索引和没有使用索引的情况下&#xff0c;where 查询的一个效率对比。 -- 创建表 DROP TABLE IF EXISTS person; CREATE TABLE person( PID int(11) auto_increment comment 编号, PNAME varchar(50) comment 姓名…

正则表达式以及文本三剑客grep、sed、awk

正则表达式匹配的是文本内容&#xff0c;文本三剑客都是针对文本内容。 grep&#xff1a;过滤文本内容 sed&#xff1a;针对文本内容进行增删改查 awk&#xff1a;按行取列 一、grep grep的作用使用正则表达式来匹配文本内容 1、grep选项 -m&#xff1a;匹配几次之后停止…

【论文解读】通过多标记预测建立更好更快的大型语言模型

Meta 的这篇多标记预测论文显示,与当前的下一标记预测器相比,多头预测器内存效率高、性能更好、训练速度更快。 https://arxiv.org/pdf/2404.19737 主要收获: 多标记预测是对 LLM 训练的一种简单而强大的修改,可提高样本效率和各种任务的性能。这种方法在大规模应用中尤为…

从 Hadoop 迁移,无需淘汰和替换

我们仍然惊讶于有如此多的客户来找我们&#xff0c;希望从HDFS迁移到现代对象存储&#xff0c;如MinIO。我们现在以为每个人都已经完成了过渡&#xff0c;但每周&#xff0c;我们都会与一个决定进行过渡的主要、高技术性组织交谈。 很多时候&#xff0c;在这些讨论中&#xff…

安卓 jetpack compose

以下是 Jetpack Compose 中常用的一些组件的列表&#xff1a; 组件名称描述Text用于显示文本内容。Button可点击的按钮组件&#xff0c;常用于触发事件。TextField用于输入文本的文本框组件。Image用于展示图片。Column垂直布局容器&#xff0c;可以在其中垂直排列子组件。Row…

Mac环境 aab包转apks,并安装apks

一、下载下载bundletool工具 Releases google/bundletool GitHub 二、将下载bundletool.jar包、aab、keystore文件全部放到同一个目录下 例如我全部放到download目录下 转换命令行&#xff1a; java -jar bundletool-all-1.16.0.jar build-apks --modeuniversal --bundle…

java运维交接项目逆向工程

​ 背景 有承接过Java项目运维的团队估计都处理过的一件事情&#xff0c;就是同步生产代码跟本地代码&#xff0c;条件再差些甚至要直接基于生产部署包逆向本地源码工程。而哪怕是原运维团队交接了源码&#xff0c;往往也会历史久远的原因&#xff0c;给了一份不太可靠的源码…

MySQL连接

MySQL工具包 MySQL实现简单链接 一 引入工具包 JBDCUtils&#xff0c;无需更改&#xff0c;直接使用即可。 import java.io.IOException; import java.io.InputStream; import java.sql.*; import java.util.Properties;public class JDBCUtil {private static String URL;p…

基于PHP+MySQL组合开发的在线客服小程序源码系统 带完整的安装代码包以及搭建教程

系统概述 源码系统是专门为满足企业在线客服需求而设计的&#xff0c;它集成了多种功能&#xff0c;能够帮助企业实现与用户的实时沟通、问题解答、信息反馈等。通过该系统&#xff0c;企业可以更好地了解用户需求&#xff0c;提升用户体验&#xff0c;增强用户对企业的信任感…

Java面试题:解释观察者模式的工作原理,并通过实例展示如何在Java中实现

观察者模式&#xff08;Observer Pattern&#xff09;是一种行为设计模式&#xff0c;它定义了一种一对多的依赖关系&#xff0c;让多个观察者对象同时监听某一个主题对象。当主题对象发生变化时&#xff0c;它的所有依赖者&#xff08;观察者&#xff09;都会收到通知并自动更…

Linux[高级管理]——Squid代理服务器的部署和应用(传统模式详解)

&#x1f3e1;作者主页&#xff1a;点击&#xff01; &#x1f468;‍&#x1f4bb;Linux高级管理专栏&#xff1a;点击&#xff01; ⏰️创作时间&#xff1a;2024年6月24日11点11分 &#x1f004;️文章质量&#xff1a;95分 目录 ————前言———— Squid功能 Squ…

LabelEncoder 类属性类方法及用法

类 LabelEncoder类属性及类方法示例用法用于大型数据集 类 LabelEncoder LabelEncoder 是 scikit-learn 中的一个预处理工具&#xff0c;用于将类别变量&#xff08;例如字符串标签或离散的整数标签&#xff09;转换为整数。 类 LabelEncoder 在包 sklearn.preprocessing.Lab…

【JavaEE精炼宝库】多线程进阶(1)常见锁策略 | CAS | ABA问题

目录 一、常见的锁策略&#xff1a; 1.1 悲观锁 | 乐观锁&#xff1a; 1.2 重量级锁 | 轻量级锁&#xff1a; 1.3 自旋锁 | 挂起等待锁&#xff1a; 1.4 公平锁 | 非公平锁&#xff1a; 1.5 可重入锁 | 不可重入锁&#xff1a; 1.6 互斥锁 | 读写锁&#xff1a; 1.7 面…

Spring Boot 集成 MinIO 实现文件上传

Spring Boot 集成 MinIO 实现文件上传 一、 Minio 服务准备 MinIO的搭建过程参考 Docker 搭建 MinIO 对象存储。 登录MinIO控制台&#xff0c;新建一个 Bucket&#xff0c;修改 Bucket 权限为公开。 二、MinIO 集成 添加 MinIO 依赖 <!-- https://mvnrepository.com/ar…

【工具测评】ONLYOFFICE——你的下一款桌面编辑器

文章目录 前言一、安装1.1 跳转官网下载安装包1.2 安装步骤 二、功能介绍2.1 功能全面的 PDF 编辑器2.2 PDF 表单2.3 文本文档编辑器的更新2.4 电子表格编辑器的更新2.5 演示文稿编辑器有哪些更新2.6 所有编辑器中的改进内容2.7 从右至左显示 & 新的本地化选项2.8 可用性提…

查找一个有向网络的头节点和尾节点 (70%用例)C卷(JavaPythonC++Node.jsC语言)

给定一个有向图,图中可能包含有环,有向边用两个节点表示。第一个整数表示起始节点,第二个整数表示终止节点,如0 1表示存在从0到1的路径。每个节点用正整数表示,求这个数据的头节点与尾节点,题目给的用例会是一个头节点,但可能存在多个尾节点。同时,图中可能含有环,如果…

重生奇迹MU整理装备技巧

除了注意一些基本的事项外&#xff0c;还有一些技巧可以帮助我们更快更好地整理装备&#xff0c;提升自己在游戏中的表现。 1. 理性地管理装备是取得胜利的关键。在整理装备时&#xff0c;根据不同的属性和需求&#xff0c;可以将装备分类&#xff0c;比如攻击型、防御型、辅助…

CRMEB 多门店后台登录入口地址修改(默认admin)

一、>2.4版本 1、修改后端 config/admin.php 配置文件,为自定义的后缀 2、修改 平台后台前端源码中 view/admin/src/settings.js 文件,修改为和上面一样的配置 3、修改后重新打包前端代码,并且覆盖到后端的 public 目录下&#xff1a;打包方法 4、重启swoole 二、<2.4版…

深度神经网络中的“深度”概念解析

引言 深度神经网络&#xff08;Deep Neural Networks, DNNs&#xff09;是机器学习领域的一大突破&#xff0c;它们在图像识别、自然语言处理、游戏等领域取得了革命性的进展。DNNs的核心特性之一就是其“深度”&#xff0c;这通常指的是网络中层的数量。本文将深入探讨深度神…

云函数-uniapp + uniCloud 订阅消息实战教程(二)

一、创建uniCloud项目 在 HBuilderX 新建项目界面&#xff0c;选择 uni-app项目&#xff0c;输入项目名称&#xff0c;选择默认模板即可&#xff0c;根据个人需要选择 vue版本&#xff0c;并勾选 启用uniCloud&#xff0c;在右侧选择服务供应商&#xff08;支付宝云、阿里云、…