当前位置: 首页 > news >正文

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

相关文章:

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&#xff0c; 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 Q…...

kafka 集群安全认证机制的设计实现

kafka 集群安全认证机制的实现 Kafka 提供了多种安全认证机制来保护数据传输的安全性,包括加密、身份认证和授权。这些机制确保 了 Kafka 集群的通信安全和数据访问控制。以下是 Kafka 安全认证机制的实现和配置指南。 1. 安全认证机制概述 Kafka 支持以下几种主要的安全机…...

pandas保存成CSV格式时数据换行:SQL的REPLACE函数过滤掉数据的回车符

在使用Pandas保存数据到CSV文件时&#xff0c;如果数据中包含换行符&#xff08;例如\n&#xff09;&#xff0c;这可能会导致数据在CSV文件中被分割成多行&#xff0c;影响数据的完整性和可读性。为了解决这个问题&#xff0c;你可以在保存CSV之前使用Pandas的replace函数来替…...

紧贴国家大战略需求,聚焦当前行业热点-海云安D10入选第四届香蜜湖金融科技创新奖拟奖项目

近日&#xff0c;第四届香蜜湖金融科技创新奖终审总结会在深圳市福田区湾区国际金融科技城成功举办&#xff0c;活动现场&#xff0c;专家评审委员会最终揭晓25个拟奖项目。海云安“开发者安全助手系统”项目&#xff0c;实力入选第四届香蜜湖金融科技创新奖-优秀项目奖。 1、香…...

构建 Linux 内核指南

该指南列出了用于构建 Linux 内核的各种目标及其说明。 清理目标&#xff1a; clean - 删除大多数生成的文件&#xff0c;但保留配置文件和足以构建外部模块的构建支持。 mrproper - 删除所有生成的文件 配置文件 各个备份文件。 distclean - mrproper 删除编辑器备份和补丁…...

Elasticsearch的快照

ES的快照是什么&#xff1f; snapshot是一个ES集群或者某个指定索引的备份&#xff0c;快照一般用在 不停机的状态下对ES集群进行备份当硬件故障时恢复集群数据用于跨集群的数据迁移对冷数据或冻结数据做快照以降低存储成本&#xff0c;依赖于可搜索的快照。-收费功能 一个快…...

C/C++混合编译说明

C调用C的函数: C的头文件中声明&#xff0c;被C调用的函数&#xff0c;是以C的方式编译 C的头文件&#xff1a; #ifdef __cplusplus extern "C" { #endif void func_c(void); #ifdef __cplusplus } #endif C调用C的函数&#xff1a; C的源文件中声明&#xff…...

三种方式实现人车流统计(yolov5+opencv+deepsort+bytetrack+iou)

一、运行环境 1、项目运行环境如下 2、CPU配置 3、GPU配置 如果没有GPU yolov5目标检测时间会比较久 二、编程语言与使用库版本 项目编程语言使用c++,使用的第三方库,onnxruntime-linux-x64-1.12.1,opencv-4.6.0 opencv 官方地址Releases - OpenCV opencv github地址ht…...

(四十四)Vue Router的命名路由和路由组件传参

文章目录 命名路由组件传参query参数方式参数传递参数接收 params参数方式参数传递参数接收 props配置方式布尔值形式对象模式函数模式 上一篇&#xff1a;&#xff08;四十三&#xff09;Vue Router之嵌套路由 命名路由 命名路由是为路由配置项提供一个名称&#xff0c;以便…...

EXCEL表格怎么批量删除日期后的时间?

竞价师最近有点忙了&#xff0c;因为百度新出来一个“线索有效性诊断”功能 一、下载电话、表单、咨询表格 二、选中整列 三、选中ctrlf 进行替换&#xff0c;日期输入空格&#xff0c;时间输入*&#xff0c;替换为空即可&#xff01; 四、整列单元格格式“日期”拉倒底部&…...

乌班图Ubuntu 24.04 SSH Server 修改默认端口重启无效

试用最新的乌班图版本&#xff0c;常规修改ssh端口&#xff0c;修改完毕后重启sshd提示没有找到service&#xff0c;然后尝试去掉d重启ssh后查看状态&#xff0c;端口仍然是默认的22&#xff0c;各种尝试都试了不行&#xff0c;重启服务器后倒是端口修改成功了&#xff0c;心想…...

QT MQTT (二)编译与集成

一、QT MQTT 提供 MQTT 客户端服务的 Qt 专用库基于标准化发布 / 订阅协议&#xff0c;用于在设备和组件之间可靠地共享数据。MQTT 是为保证状态正确性、满足高安全标准和交换最小数据而设计的协议&#xff0c;因此被广泛应用于各种分布式系统和物联网解决方案中。 Qt开发MQT…...

(上位机APP开发)调用华为云属性修改API接口修改设备属性

一、功能说明 通过调用华为云IOT提供的属性修改API接口,给设备下发属性修改消息。 API接口地址:https://support.huaweicloud.com/api-iothub/iot_06_v5_0034.html 此接口支持在线调试:https://console.huaweicloud.com/apiexplorer/#/openapi/IoTDA/doc?api=UpdatePrope…...

爆火的儿童绘本如何用AI制作?一文解锁从制作到变现的全流程!

大家好我是安琪&#xff01; AI绘图发展势头如此猛烈&#xff0c;无论是Stable Diffusion&#xff0c;Midjourney&#xff0c;还是国内百度的文心一格&#xff0c;字节的豆包等&#xff0c;AI绘图技术越来越成熟&#xff0c;风格也越来越多样化。那么问题来了&#xff0c;对于普…...

Go interface{}类型转换

&#x1f49d;&#x1f49d;&#x1f49d;欢迎莅临我的博客&#xff0c;很高兴能够在这里和您见面&#xff01;希望您在这里可以感受到一份轻松愉快的氛围&#xff0c;不仅可以获得有趣的内容和知识&#xff0c;也可以畅所欲言、分享您的想法和见解。 推荐:「stormsha的主页」…...

利用opencv自带的Haar级联分类器模型

OpenCV自带的Haar级联分类器模型&#xff1a; haarcascade_eye.xml: 这个模型用于检测眼睛。 haarcascade_eye_tree_eyeglasses.xml: 这个模型用于检测眼镜。 haarcascade_frontalcatface.xml: 这个模型用于检测猫脸。 haarcascade_frontalcatface_extended.xml: 这个模型用…...

国产USB音频转换芯片CL7016C 支持国美标线控USB Type-C音频编解码器

CL7016C是一款高保真 USB Type-C 兼容音频编解码芯片。可以录制和回放有 24 比特音乐和声音。内置回放通路信号 动态压缩&#xff0c; 最大42db录音通路增益&#xff0c; PDM 数字麦克风&#xff0c;和立体声无需电容耳机驱动放大器。 5V单电源供电。兼容 USB 2.0 全速工业标…...

【linux网络(六)】IP协议详解

&#x1f493;博主CSDN主页:杭电码农-NEO&#x1f493;   ⏩专栏分类:Linux从入门到精通⏪   &#x1f69a;代码仓库:NEO的学习日记&#x1f69a;   &#x1f339;关注我&#x1faf5;带你学更多操作系统知识   &#x1f51d;&#x1f51d; Linux网络 1. 前言2. IP协议报…...

CesiumJS【Basic】- #012添加点线面(entity方式)

文章目录 添加点线面(entity方式)1 目标2 实现2.1 GeometryManager.ts2.2 main.ts添加点线面(entity方式) 1 目标 使用实体方式添加点线面 2 实现 2.1 GeometryManager.ts // src/GeometryManager.tsimport * as Cesium from cesium;export class GeometryManager {pr…...

【Redis】内存回收和内存淘汰机制

1 概念 Redis 所有的数据都是存储在内存中的, 如果不进行任何的内存回收, 那么很容易出现内存爆满的情况。因此&#xff0c;在某些情况下需要对占用的内存空间进行释放。 Redis 中内存的释放主要分为两类 Redis 中内存的释放主要分为两类: 内存回收: 将过期的 key 清除&#…...

PyTorch实战:借助torchviz可视化计算图与梯度传递

文章目录 Tensor计算的可视化&#xff08;线性回归为例&#xff09; 如何使用可视化库torchviz 安装graphviz软件 安装torchviz库使用 torchviz.make_dot() 在学习Tensor时&#xff0c;将张量y用张量x表示&#xff0c;它们背后会有一个函数表达关系&#xff0c;y的 grad_f…...

【软件测试】软件测试入门

软件测试入门 一、什么是软件测试二、软件测试和软件开发的区别三、软件测试在不同类型公司的定位1. 无组织性2. 专职 OR 兼职3. 项目性VS.职能性4.综合型 四、一个优秀的软件测试人员具备的素质1. 技能相关2. 非技能相关 一、什么是软件测试 最常见的理解是&#xff1a;软件测…...

Windows操作防火墙命令

Windows操作防火墙命令 启用防火墙&#xff1a; netsh advfirewall set allprofiles state on禁用防火墙&#xff1a; netsh advfirewall set allprofiles state off添加新的入站规则允许端口80&#xff08;HTTP&#xff09;&#xff1a; netsh advfirewall firewall add r…...

二维数组的知识

二维数组&#xff1a; 1.同种数组类型的集合 2.连续的内存空间 3.由多个一维数组组成 定义方式&#xff1a;   存储类型 数据类型 数组名[常量表达式&#xff08;行数&#xff09;][常量表达式&#xff08;列数&#xff09;]&#xff1…...

HR3.0时代,人力资本效能如何进化?| 易搭云DHR

宏观经济增速放缓、市场竞争激烈&#xff0c;对各行各业、各种岗位都面临更大挑战&#xff0c;如何降本增效还是每个企业主的关注焦点。 企业的主要支出往往是员工成本&#xff0c;总体上超过企业总开支的75%&#xff0c;轻资产类型的企业甚至可能超80%&#xff0c;但裁员、加班…...

R语言做图

目录 1. 图形参数 2. 低级图形 3. 部分高级图形 参考 1. 图形参数 图形参数用于设置图形中各种属性。 有些参数直接用在绘图函数内&#xff0c;如plot函数可以用 pch&#xff08;点样式&#xff09;、col&#xff08;颜色&#xff09;、cex&#xff08;文字符号大小倍数&…...

跟着我一步两步三步,用开源方式将AI带入企业

“AI有开源派与闭源派&#xff0c;你挺哪一派&#xff1f;”这是红帽公司针对媒体所做的一次小调查。结果显示&#xff0c;坚定的开源派占50%&#xff0c;挺闭源的仅有5.56%。如果是你&#xff0c;又会怎样选择&#xff1f; 如何才能让AI在企业中快速平稳落地&#xff0c;并且开…...

天途重磅推出无人机教管平台3.1版及飞课APP

天途无人机教管平台&#xff0c;是一款为院校和培训机构等企业级客户提供的公开版无人机在线培训系统&#xff0c;包含后台管理的【教管平台】和终端的【掌上天途APP】。 天途历经4年上百次调研和迭代打磨&#xff0c;已为一百多家院校和培训机构等企业级客户解决了无人机教学和…...

虚幻引擎 Gerstner Waves -GPU Gems 从物理模型中实现有效的水体模拟

这篇文章重点在于结合GPU Gems一书中有关Gerstner Waves 的数学公式&#xff0c;在虚幻引擎中复现正确的Gerstner Waves和正确的法线 文中内容整理自书中&#xff0c;并附带我的理解&#xff0c;与在虚幻引擎中的实现&#xff0c;可以参考原文看这篇文章&#xff0c;原文网上很…...

Labview_网络流

网络流的介绍 网络流是一种易于配置、紧密集成的动态通信方法&#xff0c;用于将数据从一个应用程序传输到另一个应用程序&#xff0c;其吞吐量和延迟特性可与 TCP 相媲美。但是&#xff0c;与 TCP 不同的是&#xff0c;网络流直接支持任意数据类型的传输&#xff0c;而无需先…...