计算机网络英文习题（中文及答案解析）

这篇文章是在面向刚接触网络的新手的一些习题，主要难度是英文题阅读上有些障碍。文章已经经过一些简单的处理，中文为机翻，答案在习题后面，部分有些绕弯的习题有解析。
教材为谢希仁自顶向下。

Chapter 1 Test in 2019(CN & Internet)

1.Network core does not include网络核心不包括
smart phones智能手机

2.End system does not include端系统不包括
А. router路由器

3.The original reason for the Internet to use packet switching is因特网使用分组交换的最初原因是

A.Statistical multiplexing enables efficient sharing of expensive long-haul links.统计多路复用能够有效地共享昂贵的长途链路。

4.Which of the following is not a performance measure in computer networks下列哪项不是计算机网络中的性能度量
packet size数据包大小

5.By which of the following access technology, bandwidth is not shared among the users?下列哪一种访问技术不允许用户共享带宽?
ADSL非对称数字用户线路

6.The full text in English of FTTH is Fiber To The Home… The full text in English of ISP is ().The full text in English of TDM is ().FTTH的英文全文是Fiber To The Home…ISP的英文全文是()。TDM的英文全文是()。
Internet service provider Time-Division Multiplexing

7.There are two fundamental approaches to move data through a network of links and switches: circuit switching and (). And in a circuit Sswitched network, a circuit in a link is implemented with either () or TDM.有两种基本方法可以通过链路和交换机网络移动数据:电路交换和()。在电路交换网络中，链路中的电路用()或时分复用(TDM)实现。
packet switching, FDM

8.Describe the layers in the Internet protocol stack from top to down: application layer, () layer, () layer, () layer and physical layer.从上到下描述Internet协议栈中的各个层:应用层、()层、()层、()层和物理层。
. transport, Network, link

9.OSl reference model have () layers.。OSl参考模型有()层。
7

10.Packet switching allows more users to use network than circuit switching.分组交换允许比电路交换更多的用户使用网络。
Ture

11… There is no difference between a host and an end system.
主机和终端系统之间没有区别。
Ture

12.A host doesn’t process network-layer in the Internet protocol stack主机不处理Internet协议栈中的网络层
False

13.The full text in English of RFC is request for comments.RFC的英文全文正在征求意见。
Ture

14.All communication activity in Internet governed by protocols.Internet上的所有通信活动都由协议控制。
Ture

15… Packet switching is commonly used in traditional telephone networks分组交换是传统电话网络中常用的一种交换方式
False

16.Suppose users share a 2 Mbps link. Also suppose each user transmits continuously at 1 Mbps when transmitting, but each user transmits only 20 percent of the time.假设用户共享一个2mbps链接。假设每个用户在传输时以每秒
1 Mbps的速度连续传输，但是每个用户只传输20%的时间。

a.When circuit switching is used, how many users can be supported? () usersa.当使用电路切换时，可以支持多少用户?()的用户
2 //2 mbps / 1 mbps = 2

b.For the remainder of this problem, suppose packet switching is used. There be a queuing delay if three users transmit at the same time对于这个问题的其余部分，假设使用了分组交换。如果三个用户同时发送，就会出现排队延迟
Ture

find the probability that a given user is transmitting. ()求给定用户发送的概率。()
0.2

c.Suppose now there are three users. Find the probability that at any giventime, all three users are transmitting simultaneously (at the same time）.d.假设现在有三个用户。求任意给定时间，所有三个用户同时(同时)传输的概率。
0.008 //C(3,3)(0.20.20.2)=0.20.20.2 = 0.008

d.Suppose now there are four users. Find the probability that at any given time, three users are transmitting simultaneously(at the same time).．e.假设现在有四个用户。求任意给定时刻三个用户同时(同时)发送的概率。
0.0256 //C(4,3)(0.20.20.2)(1-0.2) = 40.2*0.20.20.8 = 0.0256

17.Consider a packet of length L which begins at end system A and travels over three links to a destination end system.These three links are connected by two packet switches. Assuming no queuing delays, and suppose now the packet is 1,500 bytes, the propagation speed on all three links is 2.5 *10^8 m/s, the transmission rates of all three links are 2 Mbps, the packet switch processing delay is 3 msec, the length of the first link is 5,000 km, the length of the second link is 4,000 km, and the length of the last link is 1,000 km.考虑一个长度为L的包，它从端系统a开始，经过三个链路到达目标端系统。这三个链路由两个分组交换机连接。假设没有排队延误,现在假设包是1500个字节,这三个环节的传播速度是2.5 * 10 ^ 8 m / s的传输速率三个链路的传输速率均为2mbps，分组交换处理延迟为3msec,第一个链接的长度是5000公里,第二个链接是4000公里的长度,和最后一个连杆的长度是1000公里。

For these values, what is the propagation delay? ( )ms对于这些值，传播延迟是多少?()
40 //(5000+4000+1000)1000/2.510^8=40

For these values, what is the transmission delay? ()ms对于这些值，传输延迟是多少?()
18 //1500B/2Mbps * 3=18

For these values, what is the processing delay? ()ms对于这些值，处理延迟是多少?()
6 //3*2

For these values, what is the end-to-end delay? ( )ms对于这些值，端到端延迟是多少?
40+18+6=64

18.Suppose two hosts, A and B, are separated by 10,000 kilometers and are connected by a direct link of R = 2 Mbps. Suppose the propagation speed over the link is 2.510^8 meters/sec.假设两个主机A和B相距1万公里，它们之间的直接连接是R = 2mbps。假设传播速度是2.510^8米/秒。

a. The value of the bandwidth-delay product is () bit. (Notice: the bandwidth-delay product = Rd(prop), and d(prop) is the propagation rate.)a.带宽延迟积的值为()位。(注意:带宽时延积= Rd(prop)， d(prop)为传播速率。)
80000 //R* s/v=210^6 * 10^7 / 2.510^8

b. Consider sending a file of 800,000 bits from Host A to Host B. Suppose the file is sent continuously as one large message. What is the maximumnumber of bits that will be in the link at any given time? ()考虑从主机a向主机b发送一个800,000位的文件。在任何给定的时间，链接中的最大比特数是多少?()
80000 //带宽时延积

c. Consider sending a file of 8000 bits from Host A to Host B. Suppose the file is sent continuously as one large message. What is the maximumnumber of bits that will be in the link at any given time? ()c.考虑从主机a向主机b发送一个8000位的文件。在任何给定的时间，链接中的最大比特数是多少?()
8000 //小于带宽时延积

d.What is the width (in meters) of a bit in the link? ()d.链接中一点的宽度(以米为单位)是多少?()
125 //长度除以带宽时延积

e. Derive a general expression for the width of a bit in terms of the propagation speed s, the transmission rate R, and the lenath of the link
e.根据传播速度s、传输速率R和链路长度推导比特宽度的一般表达式
s/R //d/(R* d/s)

2019Chapter 2 test

1.Which of the following is not Internet application layer Protocol?下列哪个不是Internet应用层协议?
TCP

2.DNS runs on top of () and uses the port number 53.DNS在()之上运行，并使用端口号53。
UDP

3.FTP runs over ().FTP运行在？
TCP

4.which service of the following can be provided by TCP?以下哪些服务可以由TCP提供?
Reliable data transfer可靠的数据传输

5.Which of the following is the default port number of TCP control connection in FTP?以下哪个选项是FTP中TCP控制连接的默认端口号?
21

6.Which tool can capture network packets and display that packet data?哪个工具可以捕获网络数据包并显示数据包数据?
Wireshark

7.Which of the following port number is the default port number of HTTP? 下列哪个端口号是HTTP的默认端口号?
80

8.The application of () is loss-tolerant.什么的应用是容错的。
stored audio/video存储音频/视频

9.The service of TCP is unreliable data transmission.TCP的服务是不可靠的数据传输。
False

10.Status code 404 denotes that requested document does not exist on the server according the HTTP protocol.状态代码404表示根据HTTP协议，服务器上不存在被请求的文档。
Ture

11.Local DNS name servers cache resource records, but discard them after a period of time.本地DNS名称服务器缓存资源记录，但在一段时间后丢弃它们。
Ture

12.For the Web application, the user agent is the user’ s keyboard.对于Web应用程序，用户代理是用户的键盘。
False

13.The Date: header in the HTTP response message indicates when the object in the response was last modified.HTTP响应消息中的Date: header指示响应中的对象最后一次修改的时间。
False

14.Suppose that host A wants to send data over TCP to host B. and host B wants to send data to host A over TCP. Two separate TCP conneco ch direction - are needed.假设主机A希望通过TCP向主机B发送数据，而主机B希望通过TCP向主机A发送数据。需要两个独立的TCP连接端口。
False

15.For a P2P file-sharing application, There is no notion of client and server sides of a communication session.对于P2P文件共享应用程序，没有通信会话的客户端和服务器端概念。
False

16.A user requests a Web page that consists of some text and three images. For this page, the client will send one request message and receive用户请求由一些文本和三个图像组成的Web页面。对于此页，客户端将发送一个请求消息并接收
False

17.Socket programming with UDP, the server process must run before the client process.套接字编程使用UDP，服务器进程必须运行之前的客户端进程。
False

18.The full text in English of HTTP is Hypertext Transfer Protocol.The full text in English of FTP is ().The full text in English of SMTP is ().The full text in English of IMAP is ().The full text in English of POP3 is ().The full text in English of DNS is ().HTTP的英文全文是超文本传输协议。FTP的英文全文是()。SMTP的英文全文是()。IMAP的英文全文是()。POP3的英文全文是()。DNS的英文全文是()。
File Transfer Protocol Simple Mail Transfer Protocol Internet Mail Access Protocol Post Office Protocol Version 3 Domain Name System

19.Consider the following string of ASCIl characters that were captured by Wireshark when the browser sent an HTTP request message (ie., this the actual content of an HTTP request message). The characters are carriage return and line-feed characters. Answer the following questions, indicating where in the HTTP request message below you find the answer.
GET /HTML/intro/ HTTP/1.1 Host: www.nsu.edu.cn User-Agent: Mozilla/5.0 (Windows;U: Windows NT 5.1; en-US; rv:1.7.2) Accept:ext/xml, application/xml, application/xhtml+xml, text/html;q-0.9, text/plain; q=0.8, image/png,/t: q-0.5 Accept-Language: en-us,en;q=0.5scr> Accept-Encoding: zip.deflate Accept-Charset: ISO-8859-1,utf-8q-0.7,*q-0.7 Keep-Alive: 300 Connection:close
考虑一下Wireshark在浏览器发送HTTP请求消息时捕获的下列字符串。，这是HTTP请求消息的实际内容)。字符是回车符和换行符。回答以下问题，指出在下面的HTTP请求消息中您在何处找到答案。

1)What is the method of the HTTP request message? ()HTTP请求消息的方法是什么?()
GET

2)What is the whole URL of the document requested by the browser? ().浏览器请求的文档的整个URL是什么?()。
www.nsu.edu.cn/HTML/intro/

3)What version of HTTP is the browser running?浏览器运行的是什么版本的HTTP ?
1.1

4)The browser request a persistent connection.浏览器请求一个持久连接。
False //Connection:close

5)you can know the IP address of the host on which the browser is running.您可以知道正在运行浏览器的主机的IP地址。
False

20.Suppose within your Web browser you click on a link to obtain a Web page. The IP address for the associated URL is not cached in your local host, so aDNS lookup is necessary to obtain the IP address. Suppose that n DNS servers are visited before your host receives the IP address from DNS thesuccessive visits incur an RTT of RTT1, …, RTTn.假设您在Web浏览器中单击一个链接以获得一个Web页面。关联URL的IP地址没有缓存在本地主机中，因此必须使用aDNS查找来获取IP地址。假设在您的主机接收到来自DNS的IP地址之前访问了n个DNS服务器，则隐性访问的RTT为RTT1，……,RTTn。

1)Further suppose that the Web page associated with the link contains exactly one object, consisting of a small amount of HTML text. Let RTTO denote the RTT between the local host and the server containing the object. Assuming zero transmission time of the object, how much time elapses from when the client clicks on the link until the client receives the obiect?进一步假设与链接关联的Web页面只包含一个对象，由少量HTML文本组成。让RTTO表示本地主机和包含对象的服务器之间的RTT。假设对象的传输时间为零，那么从客户端单击链接到客户端收到obiect需要多少时间?
B //2

2)suppose the HTML file references 4 very small objects on the same server. Neglecting transmission times, how much time elapses witha. Non-persistent HTTP with no parallel TCP connections?假设HTML文件引用同一服务器上的4个非常小的对象。忽略传输时间，a消耗了多少时间?没有并行TCP连接的非持久HTTP ?
D //2+4*2=10

3)Non-persistent HTTP with the browser configured for 2 parallel connections?非持久性HTTP，浏览器配置为2个并行连接?
C //2+2*(4/2)=6 假设是共6个包呢？

4)Persistent HTTP without pipelining?没有管道的持久HTTP ?
C //2+4=6

5)Persistent HTTP with pipelining?d.使用管道的持久HTTP ?
C //2+1=3

21.Consider distributing a file of F=5 Gbits to N peers. The server has an upload rate of us =20Mbps, and each peer has a download rate of di=1Mbps and an upload rate of u=(from the graph above, d1=d2 .=di, and u1=u2- …Eu).Notice, please use 1G = 1000 M = 1000000 K here, and directly remove the decimal part for the final result(for example, 100/3 = 33 and 200/3 = 66).考虑将一个F=5 Gbits的文件分配给N个节点。服务器的上传速率为us =20Mbps，每个端下载速率为di=1Mbps，上传速率为u=(从上图可以看出，d1=d2 .=di, u1=u2- …Eu)。请注意，这里使用1G = 1000m = 1000000 K，并直接去掉最后结果的小数部分(例如，100/3 = 33和200/3 = 66)。
For N-10, u=100Kbps, the minimum distribution time for client-server distribution is ()s.For N=100, u=100Kbps, the minimum distribution time for client-server distribution is ()s.For N=1000, u=100Kbps, the minimum distribution time for client-server distribution is ()s.For N-1000, u-250Kbps, the minimum distribution time for client-server distribution is ()s For N=1000, u=500Kbps, the minimum distribution time for client-server distribution is ()s对于N-10, u=100Kbps，
客户机-服务器分发版的最小分发时间是()s。对于N=100, u=100Kbps，客户机-服务器分发版的最小分发时间是()s。对于N=1000, u=100Kbps，客户机-服务器分发版的最小分发时间是()s。对于N-1000, u- 250kbps，客户端-服务器分布的最小分布时间为()s，对于N=1000, u=500Kbps，客户端-服务器分布的最小分布时间为()s
5000、25000、250000、250000、250000
//NF/us和F/di求max，服务器以us速度上传F*N个包；客户机以di下载F，两者的最大值就是最小时间
For N=10, u=100Kbps, the minimum distribution time for P2P distribution is ()s.For N=100, u=100Kbps, the minimum distribution time for P2P distribution is ()s.For N-1000, u=100Kbps, the minimum distribution time for P2P distribution is ()sFor N-1000, u=250Kbps, the minimum distribution time for P2P distribution is ()sFor N=1000, u=500Kbps, the minimum distribution time for P2P distribution is ()s
对于N=10, u=100Kbps, P2P分发的最小分发时间为()s。对于N=100, u=100Kbps, P2P分发的最小分发时间为()s。对于N-1000, u=100Kbps, P2P分发的最小分发时间为()sn -1000, u=250Kbps, P2P分发的最小分发时间为()sn =1000, u=500Kbps, P2P分发的最小分发时间为()s

550000、100000、55000、9000、7000
//F/us和F/di、NF/(us+Nu)求max，服务器以us速度加上客户机Nu上传F*N个包；客户机以di下载F，服务器以us速度上传F（因为是P2P），三者的最大值就是最小时间

2019Chapter 3 test

1.The application of ( ) is loss-tolerant.应用()是容错的。
stored audio/video存储音频/视频

2.The UDP segment includes ()UDP段包括()
checksum校验和

3.How does an operating system identify the application to which data coming in from the network should be delivered?操作系统如何识别来自网络的数据应该交付给哪个应用程序?
By the port number carried by the transport protocol由传输协议携带的端口号决定

4.The service of TCP is ( )TCP的服务是()
reliable data transmission可靠的数据传输

5.Consider a TCP connection between Host A and Host B. Suppose that the TCP segments traveling from Host A to Host B have source port number xand destination port number y. So the source port numer is () and destination port number is () for the segments traveling from Host B to Host A.考虑之间的TCP连接主机和主机B .假设TCP段旅行从主机到主机B源端口号xand目的端口号y。所以的源端口号码是()和目的端口号是段从主机B()举办。
y、x

6.In TCP, the acknowledgement number that a host puts in a segment is the sequence number of the next byte the host is expecting from the sender.在TCP中，主机放入段中的确认号是主机期望从发送方得到的下一个字节的序列号。
Ture

7.Suppose that host A wants to send data over TCP to host B, and host B wants to send data to host A over TCP. Two separate TCP connections.one for each direction - are needed.假设主机A希望通过TCP向主机B发送数据，而主机B希望通过TCP向主机A发送数据。两个独立的TCP连接。每个方向都需要一个。
False

8.The TCP segment has a field in its header for rwnd.TCP段的头文件中有一个rwnd字段。
Ture

9.Suppose Host A is sending Host B a large file over a TCP connection. The number of unacknowledged bytes that A sends cannot exceed the size of the receive buffer.假设主机A通过TCP连接向主机B发送一个大文件。发送的未确认字节数不能超过接收缓冲区的大小。
Ture

10.Suppose Host A is sending a large file to Host B over a TCP connection. If the sequence number for a segment of this connection is m, then the sequence number for the subsequent segment will necessarily be m + 1假设主机A通过TCP连接向主机B发送一个大文件。如果此连接的一个段的序列号是m，则后续段的序列号必然是m + 1
False

11.Suppose Host A sends one segment with sequence number 38 and 4 bytes of data over a TCP connection to Host B. In this same segment the acknowledgment number is necessarily 42假设主机A通过TCP连接向主机b发送一个序列号为38和4字节的数据段在同一段中，确认号码必然是42
False //没有说明是确认号的方向，另外还有丢包的情况

12.The size of the TCP rwnd never changes throughout the duration of the connection.TCP rwnd的大小在整个连接期间都不会改变。
False

13.Consider congestion control in TCP. When the timer expires at the sender, the value of threshold is set to one half of its previous value.考虑TCP中的拥塞控制。当发送方的计时器过期时，threshold的值被设置为其先前值的一半。
false

14.Suppose you have the following three 8-bit bytes: 01010011, 01100110, 01110100. What is the 1s complement of the sum of these 8-bit•bytes? (Note that although UDP and TCP use 16-bit words in computing the checksum, for this problem you are being asked to consider 8-bit sums.)假设您有以下三个8位字节:01010011、01100110、01110100。这8位•字节之和的1s补数是多少?(注意，虽然UDP和TCP在计算校验和时使用16位的字，但是对于这个问题，您需要考虑8位的和。)
11010001 //求和求补码（+1取反）
Is it possible that a 1-bit error will go undetected?一个1位的错误有可能不被检测到吗?
False
Is it possible that a 2-bit error will go undetected?一个2位的错误有可能不被检测到吗?
Ture

15.Suppose Host A sends two TCP segments back to back to Host B over a TCP connection. The first segment has sequence number 90; the second hassequence number 110假设主机A通过TCP连接将两个TCP段发送回主机B。第一段序号为90;第二个是序列号110

a.How much data is in the first segment?Anser: ()Bytes.a.第一部分有多少数据?()个字节。
20 //110-90

b.Suppose that the first segment is lost but the second segment arrives at B. In the acknowledgment that Host B sends to Host A, what will be theacknowledgment number? ( )假设第一个段丢失了，而第二个段到达了b。在主机b发送给主机A的应答中，应答号是多少?( )
90 //因为第一个段丢失

16.Consider Fiqure below. Assuming TCP Reno is the protocol experiencing the behavior shown above, answer the following questions. In all cases, you should provide a short discussion justifying your answer.下面考虑Fiqure。假设TCP Reno是上述行为的协议，请回答以下问题。在任何情况下，你都应该提供一个简短的讨论来证明你的答案是正确的。

a.Identify the intervals of time when TCP slow start is operating.a.确定TCP慢启动运行的时间间隔。
[1,6]and [23,26]

b.Identify the intervals of time when TCP congestion avoidance is operating.b.识别TCP拥塞避免操作的时间间隔。
[6,16]and[17,22]

c.After the 16th transmission round, is segment loss detected by a triple duplicate ACK or by a timeout?C.在第16轮传输后，段丢失是由三重重复ACK检测到的还是由超时检测到的?
A triple duplicate ACK
在第16轮传输之后，包丢失被一个三重重复ACK识别。如果有超时，拥塞窗口大小将下降到1。

d. After the 22nd transmission round, is segment loss detected by a triple duplicate ACK or by a timeout?d.在第22轮传输后，段丢失是被三重重复ACK检测到还是被超时检测到?
A timeout

e. What is the initial value of threshold at the first transmission round?e. threshold在第一轮传输时的初始值是多少?
32
阈值最初是32，因为正是在这个窗口大小，slowtart停止，开始避免拥塞。

f. What is the value of threshold at the 18th transmission round? ()f.第18轮的阈值值多少?()
22 // 42/2

g. What is the value of threshold at the 24th transmission round? g.第24轮传播的阈值是多少?
14.5 //29/2

h. During what transmission round is the 70th segment sent?h.第70段发送的是哪一轮?
7 //1+2+4+8+16+32=63，所以第70段发送的是第7轮

i. Assuming a packet loss is detected after the 26th round by the receipt of a triple duplicate ACK, what will be the values of the congestion window size and of threshold?Answer: cwnd value is ( ) and the threshold value is ( )。i.假设在第26轮后收到一个三次重复的ACK检测到数据包丢失，那么拥塞窗口大小和阈值是多少?答:cwnd值是()，阈值是()。
8、4 //如图，cwnd为8，阈值8/2=4

2019Chapter 4 test

1.In Internet, the two key network-layer functions are(）在Internet中，两个关键的网络层函数是()
Routing and forwarding路由和转发

2.Two packets A and B are sent in the order A-B and will traverse a connectionless communication network. They will arrive at the destination（）两个包A和B按A-B的顺序发送，将穿越无连接的通信网络。他们将到达目的地()
in any order, after possibly having traversed different routes在可能经过不同的路线之后，以任何顺序

3.Which layer in the Internet protocol stack doesn’t a router process?在互联网协议栈的哪一层没有路由器处理?
the transport layer传输层

4.How many IP addresses does a router have?一个路由器有多少个IP地址?
one address for each interface每个接口一个地址

5.What is the 32-bit binary equivalent of the IP address 223.1.3.27? ()与IP地址223.1.3.27等价的32位二进制是什么?()
11011111.00000001.00000011.00011011

6.In CIDR notation, which of the following IP and host 192.168.14.2 belong to the same subnet? ()在CIDR符号中，下列哪个IP和主机192.168.14.2属于同一子网?()
192.168.11.0/21 14=0000 1110 11=0000 1011

7.Consider sending a 999 byte datagram into a link that has an MTU of 500 bytes,()考虑将一个999字节的数据报发送到一个MTU为500字节的链接中，()
3 fragments are created with offset field value 0, 60,120, respectively分别用偏移字段值0、60,120创建3个片段数据长度=偏移量* 8字节（偏移量=8*分段标志）

8.What is the name of a network-layer packet? ()网络层数据包的名称是什么?()
Datagram

9.In CIDR notation, the host 192.168.11.0/21 and the host 192.168.14.2/21 belong to the same subnet.在CIDR符号中，主机192.168.11.0/21和主机192.168.14.2/21属于同一子网。
Ture

10.IP (Internet Protocol) can guaranteed delivery.IP (Internet Protocol)可以保证交付。
False

11.The routers in both datagram networks and virtual-circuit networks use forwarding tables.数据报网络和虚拟电路网络中的路由器都使用转发表。
Ture

12.Consider a datagram network using 8-bit host addresses. Suppose a router uses longest prefix matching and has the following forwarding table:考虑使用8位主机地址的数据报网络。假设一个路由器使用最长的前缀匹配，有如下的转发表:

A、For the interface 0, the associated range of destination host addresses is () and the number of addresses in the range is ().对于接口0，目标主机地址的关联范围是()，该范围内的地址数量是()。

1100 0000~1101 1111,32

B、For the interface 3(the last one), the number of addresses in the associated range of destination host addresses is ().对于接口3(最后一个)，目标主机地址的关联范围中的地址数是()。

0 0000000—0 1111111, 128 = 2^7

13.Consider the following routing table of a router R.

14.Destination Network ---- Outgoing Link Interface
15.196.80.0.0/12 -------------- 1
16.196.96.0.0/12 -------------- 2
17.64.0.0.0/2 --------------- 3
18.Others --------------- 4
19.考虑以下路由器R的路由表。
目标网络——传出链路接口
196.80.0.0/12 - - - - - - - - - - - - - - - - 1
196.96.0.0/12 - - - - - - - - - - - - - - - - 2
64.0.0.0/2 - - - - - - - - - - - - - - - - - - 3
其他- - - - - - - - - - - - - - - - - - 4

A、Suppase a packet with the destination IP addresses 96.94.19.135 arrives at the router R. Find which outgoing link interface for forwarding he packet.提供一个目的IP地址为96.94.19.135的数据包到达路由器r，查找哪个发送链路接口转发数据包。
3
1100 0100 0101xxxx
1100 0100 0110xxxx
01xxxx
0110 0000 xxxx

B、Suppose a packet with the destination IP addresses 196.100.100.100 arrives at the router R. Find which outgoing link interface for forwarding the packet.假设一个目的IP地址为196.100.100.100的数据包到达路由器r，找出转发该数据包的出站链路接口。
2 //1100 0100 0110 0100 xxxx

C、Suppose a packet with the destination IP addresses 196.94.32.128 arrives at the router R. Find which outgoing link interface for forwarding the packet.假设一个目的IP地址为196.94.32.128的数据包到达路由器r，找出转发该数据包的出站链路接口。
1

Suppose a packet with the destination IP addresses 197.95.32.128 arrives at the router R. Find which outgoing link interface for forwarding the packet.假设一个目的IP地址为197.95.32.128的数据包到达路由器r，找出转发该数据包的传出链路接口。
4

20.Consider sending a 4000-byte datagram（including 20-byte IP header） into a link that has an MTU of 500 bytes. Suppose the original datagram is stamped with the identification number 422.考虑将一个4000字节的数据报(包括20字节的IP报头)发送到一个MTU为500字节的链接中。假设原始数据报盖有标识号422。

A、How many fragments are generated?生成了多少个片段?
9 //4000-20/500-20≈9

B、For the first fragment, please write the value ( ), ( ), ( ) and ( ) respectively for the field of the size of data, ID, offset, MF flag. Note that the size of data is not including the IP header part.对于第一个片段，请分别为数据大小、ID、偏移量、MF标记的字段编写值()、()、()和()。注意，数据的大小不包括IP报头部分。
480、422、0、1 //500-20、标识号422、第一个偏移0、flag=1

C、For the last fragment, please write the value ( ), ( ), ( ) and ( ) respectively for the field of the size of data, ID, offset, MF flag. Note that the size of data is not including the IP header part.对于最后一个片段，请分别为数据大小、ID、偏移量、MF标记的字段写值()、()、()和()。注意，数据的大小不包括IP报头部分。
140、422、480、0 //4000%480-20=140

21.Suppose an ISP owns the block of addresses of the form 128.129.80.128/26. 假设ISP拥有128.129.80.128/26格式的地址块。

A、Suppose the block of address is a subnet. which of the following is its subnet mask?假设地址块是子网。下列哪个是它的子网掩码?
255.255.255.192 // /26

B、Suppose it wants to create four subnets from this block, with each block having the same number of IP addresses. What are the prefixes (of form a.b.c.d/x, such as 128.129.80.128/26) for the four subnets?假设它想从这个块创建四个子网，每个子网具有相同数量的IP地址。a.b.c.形式的前缀是什么?d/x，例如128.129.80.128/26
128.129.80.128/28、128.129.80.144/28、128.129.80.160/28、128.129.80.176/28
//4=2^2 子网掩码为/28
即128.129.80. 10 00/01/10/11 0000四个子网

22.There is a block of addresses 218.94.21.0/24. 有一个地址块218.94.21.0/24。

A、Please denote the block of addresses in binary system.请用二进制表示地址块。
11011010 01011110 00010101 00000000/24

B、Please create four subnets (of form a.b.c.d/x) from the addresses at the same size. the four subnet are ( )、（）、（）and ( ). please use the form of a.b.c.d/x, such as 218.94.21.0/24 for each blank.请从相同大小的地址创建四个子网(表格a.b.c.d/x)。这四个子网是()、()、()和()。请使用a.b.c表格。d/x，如218.94.21 /24为每个空白。
218.94.21.0/26、218.94.21.64/26、218.94.21.128/26、218.94.21.192/26
//4=2^2 子网掩码为/26
即218.94.21.00/01/10/11 00 0000 四个子网

C、What is the range of host id for the sub-network 218.94.21.64/26?子网络218.94.21.64/26的主机id范围是什么?
218.94.21.65~218.94.21.126

2019Chapter 5 test

1.Network core does not include ( ) .网络核心不包括在内
Smart phones

2.Which tool can capture network packets and display that packet data? ( )哪个工具可以捕获网络数据包并显示数据包数据?( )
Wireshark

3.Which of the following is Internet link-layer protocol?下列哪个是Internet链路层协议?
PPP

4.which of the following is not a class of multiple access protocol?下列哪一项不是多址协议的类别?
Flow control //流量控制

5.The function of CRC is ( ).CRC的函数是()。
error detection //循环冗余校验码

6.What is the name of a link-layer packet? （）链路层数据包的名称是什么?()
Frame

7.The mask of a subnet is 255.255.255.248. So the subnet at most can connect ( ) hosts.子网的掩码是255.255.255.248。所以子网最多只能连接()主机。
6 //8-2=6

8.which of the following is not a multiple access protocol?下列哪个不是多址协议?
NAT //网络地址转换

9.Which of the following is a correct subnet mask?下列哪个是正确的子网掩码?
255.255.224.0 //0-255

10.Which of the following is used by the switch to forward frames? ( )下面哪个选项是switch到转发帧使用的?( )
MAC address //mac地址

11.Which of the following is the protocol which is the most widely used for wired LAN at present?以下哪个协议是目前有线局域网使用最广泛的协议?
802.3

12.Which of the following MAC addresses is correct?（）下列哪个MAC地址是正确的?( )
00-06-5B-4F-45-BA //MAC： 48bits，6个字节、16进制表示（16进制: 0~9 ABCDEF）

13.Which size of frames are treated as invalid frames? ( )哪些大小的帧被视为无效帧?( )
frames smaller than 64 //小于64帧

14.Which of the following is the decimal for the IP address 11000000 00110100 00000000 00001111? ( )以下哪个是IP地址11000000 00110100 00000000 00001111的小数?( )
192.52.0.15

15.MAC broadcast address is ( ).MAC广播地址为()。
FF-FF-FF-FF-FF-FF

16.What is stored in the ARP table of the Internet host? ( )因特网主机的ARP表中存储了什么?( )
the mapping of IP addresses to MAC addressses //IP地址到MAC地址的映射

17.Suppose the information content of a packet is the bit pattern 0111000110101011 and an even parity scheme is being used. What would the value of the parity bit? ( )
假设信息包的信息内容是位模式0111000110101011，并且使用了偶校验方案。奇偶校验位的值是多少?( )
1 //9+1,偶校验

18.The two most important network-layer functions in a datagram network is _____________________ and ___________________.在数据报网络中最重要的两个网络层功能是
Forwarding、routing

The full text in English of FTTH is Fiber To The Home.
The full text in English of CDMA is ( ).英文书写

code division multiple access|||Code Division Multiple Access
The full text in English of FTTH is Fiber To The Home.

The full text in English of TDMA is ( ).英文书写
time division multiple access|||Time Division Multiple Access

The full text in English of FTTH is Fiber To The Home.
The full text in English of CSMA is ( ).

Carrier Sense Multiple Access|||carrier sense multiple access
The full text in English of FTTH is Fiber To The Home.
The full text in English of ARP is ( ).英文书写

Address Resolution Protocol|||address resolution protocol
The full text in English of FTTH is Fiber To The Home.

The full text in English of VLAN is ( ).
Virtual Local Area Networks|||virtual local area networks|||Virtual Local Area Network|||virtual local area network

19.Two kinds of communication devices commonly used in computer networks are __________________ and __________________.两种通讯设备中常用的计算机网络是
switches||| switch 、 router|||routers

20.In a TCP / IP network, the physical address is related to the ___________ layer. The logical IP address is related to the _______________ layer, and the port address is related to the _______________ layer.在TCP / IP网络中，物理地址与层相关。逻辑IP地址与层相关，端口地址与层相关。
link|||data link|||Link|||Data Link 、 Network|||network、 transport|||Transport

21.How many bits for a MAC address? ( )
How big is the MAC address space? ( )
How big is the IPv4 address space? ( )
How big is the IPv6 address space? ( ) (幂格式说明：2的18次方，请记为2^18)
一个MAC地址需要多少位?( )
MAC地址空间有多大?( )
IPv4地址空间有多大?( )
IPv6地址空间有多大?( )
48、2^48、232、2^128

22.If all the links in the Internet were to provide reliable delivery service, the TCP reliable delivery service would be redundant.如果Internet中的所有链接都要提供可靠的传递服务，那么TCP可靠的传递服务就是多余的。
False //TCP还会确保数据正确

23.The frame structures for 10BASE-T, 100BASE-T, and Gigabit Ethernet are all same.10BASE-T、100BASE-T和千兆以太网的帧结构都是一样的。
Ture //三种以太网技术具有相同的帧结构。

24.When a conflict occurs in the LAN, the hosts that encounter the collision wait for the same time to resend.当局域网中发生冲突时，遇到冲突的主机等待同一时间重新发送。
False

25.The MAC(Media Access Control) address is the hardware address, expressed as a 48-bit binary number.MAC(媒体访问控制)地址是硬件地址，表示为48位二进制数字。
Ture

A router is an interconnection device that belongs to the data link layer.路由器是属于数据链路层的互连设备。
False //网络层

26.Consider the 4-bit generator, G=1001, and suppose that D has the value 10101010. 考虑4位生成器G=1001，并假设D的值为10101010。

a、The full text in English of CRC is ( ).CRC的英文全文是()。
Cyclic Redundancy Check

b、What is the value of R while using CRC to detect errors?使用CRC检测错误时R的值是多少?
101 //
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