计算机网络英文答案

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Chapter 5 Problems

Problem 1

The rightmost column and bottom row are for parity bits.

1 0 1 0 0

1 0 1 1 1

0 0 0 1 1

Problem 2

Suppose we begin with the initial two-dimensional parity matrix:

0 0 0 0

1 1 1 1

0 1 0 1

1 0 1 0

With a bit error in row 2, column 3, the parity of row 2 and column 3 is now wrong in the

matrix below:

0 0 0 0

1 1 0 1

0 1 0 1

1 0 1 0

Now suppose there is a bit error in row 2, column 2 and column 3. The parity of row 2 is

now correct! The parity of columns 2 and 3 is wrong, but we can't detect in which rows

the error occurred!

0 0 0 0

1 0 0 1

0 1 0 1

1 0 1 0

The above example shows that a double bit error can be detected (if not corrected).

Problem 3

To compute the Internet checksum, we add up the values at 16-bit quantities:

00000000 00000001

00000010 00000011 00000100 00000101

00000110 00000111

00001000 00001001

-------------------------

00010100 00011001

The one's complement of the sum is 11101011 11100110.

Problem 4

a) To compute the Internet checksum, we add up the values at 16-bit quantities:

00000001 00000010

00000011 00000100

00000101 00000110

00000111 00001000

00001001 00001010

-------------------------

00011001 00011110

The one's complement of the sum is 11100110 11100001.

b) To compute the Internet checksum, we add up the values at 16-bit quantities:

01000001 01000010

01000011 01000100

01000101 01000110

01000111 01001000

01001001 01001010

-------------------------

01011000 01011111

The one's complement of the sum is 10100111 10100000

c) To compute the Internet checksum, we add up the values at 16-bit quantities:

01100001 01100010

01100011 01100100

01100101 01100110

01100111 01100111

01101000 01101001

-------------------------

11111001 11111101

The one's complement of the sum is 00000110 00000010.

Problem 5

If we divide 1001 into 10101010000 we get 10111101, with a remainder of R = 101.

Problem 6

a) If we divide 1001 into 10010001000 we get 10000001, with a remainder of R =

001.

b) If we divide 1001 into we get 10100011000 we get 10110101, with a remainder

of R = 101.

c) If we divide 1001 into 010101010000 we get 010111101, with a remainder of R =

101.

Problem 7

1)1()(NpNppE

21)1)(1()1()('NNpNNppNpE

))1()1(()1(2NpppNN

NppE1*0)('

NNNNNNpENNN11)11()11()11(1*)(11

1)11(limNN eNNN1)11(lim

Thus

epEN1*)(lim

Problem 8

)1(2)1()(NpNppE

)3(2)2(2)1)(1(2)1()('NNpNNppNpE

))1(2)1(()1()3(2NpppNN

121*0)('NppE

)1(2)1211(12*)(NNNNpE

eepEN21121*)(lim

Problem 9

a) (1 – p(a))3 p(A)

where,

p(A) = probability that A succeeds in a slot

p(A) = p(A transmits and B does not and C does not)

= p(A transmits) p(B does not transmit) p(C does not transmit)

= p(1 – p) (1 – p) = p(1 – p)2

Hence, p(A succeeds for first time in slot 4) = (1 – p(a))3 p(A) = (1 – p(1 – p)2)3

p(1 – p)2

b) p(A succeeds in slot 2) = p(1-p)2

p(B succeeds in slot 2) = p(1-p)2

p(C succeeds in slot 2) = p(1-p)2

p(either A or B or C succeeds in slot 2) = 3 p(1-p)2

(because these events are mutually exclusive)

c) p(some node succeeds in a slot) = 3 p(1-p)2

p(no node succeeds in a slot) = 1 - 3 p(1-p)2

Hence, p(first success occurs in slot 4) = p(no node succeeds in first 3 slots) p(some

node succeeds in 4th slot) = (1 - 3 p(1-p)2)3 3 p(1-p)2

d) efficiency = p(success in a slot) = 3 p(1-p)2Problem 10

Efficiency of slotted Aloha00.050.10.150.20.250.30.350.40.4500.10.20.30.40.50.60.70.80.91probabilityEfficiencyN=10N=25N=50

Efficiency of ALOHA01234567800.10.20.30.40.50.60.70.80.91probabilityefficiencyN=10N=25N=50Problem 11

The length of a polling round is

)/(polldRQN.

The number of bits transmitted in a polling round is NQ. The maximum throughput

therefore is

QRdRdRQNNQpollpoll1)/(

Problem 12

a), b), c) See figure below.

1. Forwarding table in A determines that the datagram should be routed to interface

111.111.111.002.

2. The adapter in A creates and Ethernet packet with Ethernet destination address

22-22-22-22-22-22.

3. The first router receives the packet and extracts the datagram. The forwarding

table in this router indicates that the datagram is to be routed to 122.222.222.003.

4. The first router then sends the Ethernet packet with the destination address of 55-55-55-55-55-55 and source address of 33-33-33-33-33-33 via its interface with IP

address of 122.222.222.002.

5. The process continues until the packet has reached Host F.

ARP in A must now determine the LAN address of 111.111.111.002. Host A sends out

an ARP query packet within a broadcast Ethernet frame. The first router receives the

计算机网络英文资料

Network

The Internet is a worldwide, publicly accessible series of interconnected

computer networks that transmit data by packet switching using the

standard Internet Protocol (IP). It is a "network of networks" that consists of

millions of smaller domestic, academic, business, and government

networks, which together carry various information and services, such as

electronic mail, online chat, file transfer, and the interlinked web pages and

other resources of the World Wide Web (WWW).

The Internet and the World Wide Web are not one and the same. The

Internet is a collection of interconnected computer networks, linked by

copper wires, fiber-optic cables, wireless connections, etc. In contrast, the

Web is a collection of interconnected documents and other resources,

linked by hyperlinks and URLs. The World Wide Web is one of the services

计算机网络试题库

1 第一章计算机网络概论

一、选择题

1、下面( )不是网络的拓扑结构。 C、

A. 星型 B. 总线型 C. 立方形 D. 环型

2、以下说法哪一个是不正确的( )。D

A. 计算机网络分为资源子网和通讯子网

B. 在局域网中，如每台计算机在逻辑上都是平等的，不存在主从关系，称为对等网络

C. 在非对等网络中，存在着主从关系，即某些计算机扮演主角的服务器，其余计算机则充当配角的客户机

D. 如果服务器的处理能力强而客户机的处理能力弱，就称为胖客户机/瘦服务器

3、组建计算机网络的目的是为了能够相互共享资源，这里的计算机资源主要是指硬件、软件与( )。D

A. 大型机 B. 通信系统 C. 服务器 D. 数据

4、计算机网络的目标是( )。D

A. 提高计算机安全性 B. 分布处理 C. 将多台计算机连接起来 D. 共享软件、硬件和数据资源

5、一座办公楼内各个办公室中的微机进行联网，这个网络属于( )。B

A. WAN B. LAN C. MAN D. GAN

6、下列哪一项是计算机网络技术发展的一个里程碑，它的研究成果对促进网络技术的发展起到重用的作用，并为Internet的形成奠定了基础( )。A

A. ARPANET B. NSFNET C. Internet D. ATM

7、信息高速公路是指( )。

A. Internet B. 国家信息基础设施B C. 智能化高速公路建设 D. 高速公路的信息化建设

8、计算机网络的基本分类方法主要有两种：一种是根据覆盖范围与规模；另一种是根据( )。C

A. 网络协议 B.网络使用的主机类型 C.网络所使用的传输技术 D.网络服务器类型与规模

计算机网络英文题库(附答案)chapter1

*** *** Chapter 1 Computer Networks and the Internet networks use VC. numbers to forward packets toward their

1． The ( ) is a worldwide computer network, that is, a network that destination.

interconnects millions of computing devices throughout the world. D datagram networks use VC. numbers and virtual-circuit networks

ppt3 use destination addresses to forward packets toward their destination.

A public Internet 9．In the following options, which one is not a guided media? ( )

B Intranet A twisted-pair wire

C switch net B fiber optics

D television net C coaxial cable

2．Which kind of media is not a guided media? ( ) D satellite

A twisted-pair copper wire 10．Processing delay does not include the time to ( ).

B a coaxial cable A examine the packet’s header

C fiber optics B wait to transmit the packet onto the link

计算机网络习题(有答案)

一、填空题

1. 计算机网络是现代通信技术和计算机技术密切结合的产物。

2. 计算机网络的定义为一个互连的自主的计算机集合。

3. 计算机网络按规模分为局域网，城域网，广域网。

4. 按带宽分为窄带网和宽带网。

5. 局域网的英文缩写为 LAN ，城域网的英文缩写为 MAN ，广域网的英文缩写为 WAN 。

6. 在某层上进行通信所使用的规则的集合称为该层的协议。网络各层协议按层次顺序排列而成的协议称为网络的协议栈。

7. 对一个网络进行层次结构的划分时应做到：各层功能明确、相互独立；层间接口清晰、穿越接口的信息量尽可能少。

8. 两个最重要的网络参考模型是__OSI/RM______模型和_TCP/IP_______。

9. OSI参考模型分为__7___层，从低到高依次是物理层、数据链路层、网络层、传输层、会话层、表示层和应用层。

10. ISO的全称是___国际标准化组织 ______，在1984年正式颁布了一个称为__开放系统互联参考模型___。该模型包含七个层次。

11. 计算机网络采用___分层结构 _____。每层完成一定的___功能 _____，每层都向它的上层提供一定的____服务 ____，而将如何实现服务的细节对上层屏蔽，即低层协议对高层而言是透明的。相邻两层之间为___层间接口 _____。对等层用户通话时所必须遵守的规则称为___对等层协议_____。

12. 协议由_语义，语法，时序等三部分组成。

13. TCP/IP协议分为___4__层，从低到高依次是___网络接口层_______，___网间网层_____________， __传输层________，___应用层________________。

14. 数据通信的几个主要指标有___信号传输速率

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计算机网络英文答案

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计算机网络英文资料

计算机网络试题库

计算机网络英文题库(附答案)chapter1

计算机网络习题(有答案)

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