计算机网络课后作业以(中英文对照)
计算机网络英文版习题答案
P63 #5 Consider sending a packet of F bits over a path of Q links. Each link transmits at R bps. The network is lightly loaded so that there are no queuing delays. Propagation delay is negligible.a.Suppose the network is a packet-switched virtual-circuit network. Denote the VC setup time by t s seconds. Suppose the sending layers add a total of h bits of header to the packet. How long does it take to send the file from source to destination?t s+[(F+h)/R]Qb.Suppose the network is a packet-switched datagram network and a connectionless service is used. Now suppose each packet has 2h bits of header. How long does it take to send the packet?[(F+2h)/R]Qc.Finally, suppose that the network is a circuit-switched network. Further suppose that the transmission rate of the circuit between source and destination is R bps. Assuming ts setup time and h bits of header appended to the packet, how long does it take to send the packet?t s+(F+h)/RP64 #6 This elementary problem begins to explore propagation delay and transmission delay, two central concepts in data networking. Consider two hosts, A and B, connected by a single link of rate R bps. Suppose that the two hosts are separated by m meters and suppose that the propagation speed along the link is s meters/sec. Host A sends a packet of size L bits to host B.[a] Express the propagation delay, d prop, in terms of m and s.[b] Determine the transmission time of the packet, d trans, in terms of L and R.[c] Ignoring processing and queueing delays, obtain an expression for the end-to-end delay.[d] Suppose Host A begins to transmit the packets at time t=0. At time t=d trans, where is the last bit of the packet?[e] Suppose d prop is greater than d trans. At time t=d trans, where is the first bit of the packet?[f] Suppose d prop is less than d trans. At time t=d trans, where is the first bit of the packet?[g] Suppose s=2.5 x 108, L=100 bits and R=28kbps. Find the distance m so that d prop = d trans.[a] d prop = m/s[b] d trans = L/R[c] end-to-end delay = d prop + d trans=m/s+L/R[d] The beginning position of the link.[e] On the channel between A and B.[f] On the host B.[g] m/s = L/R = > m = sL/R = > m = 892.86 kmP65 #10 Consider the queueing delay in a router buffer. Suppose that all packets are L bits, the transmission rate is R bps, and that N packets simultaneously arrive at the buffer every LN/R seconds. Find the average queueing delay of a packet (in terms of L, R and N). (Hint: The queueing delay for the first packet is zero; for the second packet L/R; for the third packet 2L/R. The Nth packet has already been transmitted when the second batch of packets arrives.)As the Nth packet has already been transmitted when the next batch of packets arrive, we only need to consider the delay for a single batch of packets.Average delay = Total delay / Number of packetsDelay for 1st packet = 0Delay for 2nd packet = L/RDelay for 3rd packet = 2L/R......Delay for Nth packet = (N-1)L/RTotal delay for N packets = (0 + 1 + 2 ... +(N-1) ) * (L/R)Using the formulas for sum of integer series, this can be written as: Total delay for N packets = (N-1) * (N/2) * (L/R)Therefore, average delay for N packets = ((N-1) * L) / 2RP170 #12 What is the difference between persistent HTTP with pipelining and persistent HTTP without pipelinning? Which of the two is used by HTTP/1.1?For the persistent connection without pipelining, the client issues a new request only when the previous has been received. In this case, the client experiences one RTT in order to request and receive each of the referenced objects.For the persistent connection with pipelining, the client issues a request as soon as it encounters a reference. It is possible for only RTT to be expended for all the referenced objects.P170 #14 Telnet into a Web server and send a multiline request message. Include in the request message theIf-modified-since: header line to force a response message with the 304 Not Modified status code.GET/somedir/exp.html HTTP/1.1Host: Connection: closeUser-agent: Mozilla/4.0If-Modified-Since: Thu, 30 May 2007 12:00:00 GMTAccept-language: frP172 #6 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 a DNS look-up is necessary to obtain the IP address. Suppose that n DNS servers are visited before your host receives the IP address from DNS; the successive visits incur an RTT (Round Trip Time) of RTT1, ... RTTn. Further suppose that the web page associated with the link contains exactly one object, consisting of a small amount of HTTP text. Let RTT0 denote the RTT between the local host and the remote 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 object? (Hint: read pages 90 .. 93)Time to visit DNS servers and get IP address = RTT1 + RTT2 + ... + RTTnTime to establish TCP connection (SYN and SYNACK) = RTT0Time to send HTTP request and receive reply = RTT0Total time = 2 * RTT0 + (RTT1 + RTT2 + ... + RTTn)P171 #16 Suppose Alice with a Web-based e-mail account (such as Yahoo! Mail or Hotmail) sends a message to Bob, who accesses his mail from his mail server using POP3. Discuss how the message gets from Alice’s host to Bob’s host. Be sure to list the series of application-layer protocols that are used to move the message between the two hosts.The series of application-layer protocols: HTTP、SMTP、POP3Suppose that you send an e-mail message whose only data is a Microsoft Excel attachment. What might the header lines (including MIME lines) look like?From:***********To:***********Subject: helloMIME-Version: 1.0Content-Transfer-Encoding: base64Content-Type: Application/MS-ExcelP286 #5 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 has sequence number 110.a.How much data is in the first segment?a.20 bytesb.Suppose that the first segment is lost but the second segment arrives at B. In the acknowledgement that host B sends to host A, what will be the acknowledgement number?b.ACK90P291 #27 Consider the following plot of TCP window size as a function of time. (reproduced below for you) 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.a. Identify the intervals of time when TCP slow start is operating.b. Identify the intervals of time when TCP congestion avoidance is operating.c. After the 16th transmission round, is segment loss detected by a tripleduplicate ACK or by a timeout?d. After the 22nd transmission round, is segment loss detected by a triple duplicate ACK or by a timeout?e. What is the initial value of Threshold at the first transmission round?f. What is the value of Threshold at the 18th transmission round?g. What is the value of Threshold at the 24th transmission round?h. During what transmission round is the 70th segment sent?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?Solution:a.1-6, 23-26b.6-16, 17-22c.a triple duplicate ACKd.timeoute.32f.21g.13h.7i.4, 4P293 #34 Consider sending an object of size O = 100 Kbytes from server to client. Let S = 536 bytes and RTT = 100 msec. suppose the transport protocol uses static windows with window size W. (See Section 3.7.2)a.For a transmission rate of 28 kbps, determine the minimum possible latency. Determine the minimum window size that achieves this latency.b.Repeat (a) for 100 kbps.tency=28.8s W=2tency=8.2s W=4P405 #8 Consider a datagram network using 8-bit host addresses. Suppose a router uses longest prefix matching and has t he following forwarding table:-----------------------------------------------------Prefix Match Interface-----------------------------------------------------00 001 110 211 3-----------------------------------------------------For each of the four interfaces, give the associated range of destination host addresses and the number of addresses in the range.6P407 #15 Consider sending a 3000-byte datagram into a link that has a MTU of 500 bytes. Suppose the original datagram is stamped with the identification number 422. How many fragments are generated? What are their characteristics?there are「2980/480」=7 fragments be generatedP408 #22 Consider the network shown in Problem 21 (reproduced below). Using Dijkstra’s algorithm, and showing your work using a table similar to Table 4.3, do the following:a. Compute the shortest path from s to all network nodesSteps D(t),P(t) D(u),P(u)D(v),P(v)D(w),P(w)D(x),P(x)D(y),P(y)D(z),P(z)0 1,s 4,s ∞∞∞∞∞1 3.t 10,t ∞∞5,t 3,t2 4,u 6,u ∞5,t 3,t3 4,u 6,u ∞5,t4 5,v 7,v 5,v5 6,w 5,v6 6,wPlease fill in the following tables using DV algorithm:For the node Z in the graph shown in the 22nd topic (P408), please fill in the following routing table in the router z about the initial distance-vector Destination node Next hop Current shortest distancevalue-DzS —∞T T 2U —∞V —∞W —∞X —∞Y Y 14Z Z 0following rout-ing table in the node z to update this routing tableDestination node Currentdistance-DyDestination node Current distance-DtS 5 S 1 T 4 T 0 U 2 U 2P493 #7 How big is the MAC address space？The IPv4 address space？The IPv6 address space?MAC address: 6 bytes, MAC address space 2^48IPV4 address: 4 bytes, IPV4 address space 2^32IPV6 address: 16 bytes, IPV6 address space 2^128P494 #4 Consider the 4-bit generator, G, shown in Figure 5.8, and suppose the D has the value 10101010. What is the value of R?G=1001, D=10101010, R=101。
(完整版)计算机网络课后作业以及答案(中英文对照)
Chapter11-11.What are two reasons for using layered protocols?(请说出使用分层协议的两个理由)答：通过协议分层可以把设计问题划分成较小的易于处理的片段。
分层意味着某一层的协议的改变不会影响高层或低层的协议。
1-13. What is the principal difference between connectionless communication and connection-oriented communication?(在无连接通信和面向连接的通信两者之间，最主要的区别是什么？)答：主要的区别有两条。
其一：面向连接通信分为三个阶段，第一是建立连接，在此阶段，发出一个建立连接的请求。
只有在连接成功建立之后，才能开始数据传输，这是第二阶段。
接着，当数据传输完毕，必须释放连接。
而无连接通信没有这么多阶段，它直接进行数据传输。
其二：面向连接的通信具有数据的保序性，而无连接的通信不能保证接收数据的顺序与发送数据的顺序一致。
1-20. A system has an n-layer protocol hierarchy. Applications generate messages of length M bytes. At each of the layers, an h-byte header is added. What fraction of the network bandwidth is filled with headers?(一个系统有n层协议的层次结构。
应用程序产生的消息的长度为M字节。
在每一层上需要加上一个h字节的头。
请问，这些头需要占用多少比例的网络带宽)答：hn/(hn+m)*100%1-28. An image is 1024 x 768 pixels with 3 bytes/pixel. Assume the image is uncompressed. How long does it take to transmit it over a 56-kbps modem channel? Over a 1-Mbps cable modem? Over a 10-Mbps Ethernet? Over 100-Mbps Ethernet?(一幅图像的分辨率为1024 x 768像素，每个像素用3字节来表示。
Computer network中英文对照简介
Computer network计算机网络A computer network is a group of more computers connected to each electronically. This means that the computers can "talk" to each other and that every computer in the network can send information to the others. Usually, this means that the speed of the connection is fast - faster than a normal connection to the Internet.计算机网络是一组多台计算机连接到每个电子。
这意味着计算机可以“讲”给对方，并且每一个在网络上的计算机可以将信息发送到其他。
通常情况下，这意味着连接速度快 - 速度比正常的连接到互联网。
Some basic types of computer networks include:▪ A local area network (often called a LAN) connects two or more computers, and may be called a corporate network in an office or business setting.▪An "internetwork", sometimes called a Wide Area Network (because of the wide distance between networks) connects two or more smaller networks together. The largest internetwork is called the Internet.计算机网络的一些基本类型包括：局域网络（通常被称为一个LAN）连接两个或更多的计算机，并可能被称为一个企业网络，在办公室或业务设置。
计网课后习题答案(英文)
Review Questions第一章1.Which characteristics of a multiterminal system make it different from a computer network?A processing power of multiterminal systems remained fully centralized, whilecomputer networks have a distributed processing power.一个多端系统处理能力仍然完全集中，而计算机网络具有分布式处理能力。
2.When were the first important results achieved in the field of joining computers using long-haul links?In the late 60s在60年代后期3.What is ARPANET?A.A network of supercomputers belonging to military organizations and research institutes in the United StatesB.An international scientific research networkC.The technology of creating WANsA is correct4.When did the first network operating systems appear?In the late 60s60年代后期5.In what order did the events listed here take place?A.The invention of WebB.The development of standard LAN technologiesC.The start of voice transmission in digital form through telephone networksThe invention of World Wide Web互联网的诞生6.Which of the events stimulated LAN development?Appearance of large-scale integrated circuits (LSI devices) resulted in invention of microcomputers which in its turn stimulated a research activity in the area of local computer networks (LANs). The adoption of personal computers was a powerful incentive for the development of LANs.大规模集成电路（LSI器件出现）导致这反过来刺激了研究活动在本地计算机网络（LAN）的微型计算机的发明。
计算机网络中英文互译
DWDM (Deｎｓe WDM) 密 集 波 分 复 用
AＲQ (Autｏmaｔic Rｅpeａt reQｕest) 自 动 请 求 重 发
ATM (Ａsynchronous Ｔransｆeｒ Moｄe) 异 步 传 递 方 式
ATU (Aｃceｓs Tｅrminatioｎ Unit) 接 入 端 接 单 元
ATU-C (Accesｓ Ｔermination Unit Ceｎtraｌ Ofｆicｅ ) 端 局 接 入 端 接 单 元
ＣCIＴT (Cｏnsultatiｖe Commｉｔtee, Intｅrnａtional Tｅlｅgraph and Tｅlｅpｈone）国际电报电话咨询委员会
CCＰ 通信控制处理机
CＤＭ (Cｏde Division Mulｔiplexinｇ) 码 分 复 用
CDMA (Ｃoｄe Ｄivisioｎ Ｍuｌtipｌex Aｃｃｅｓs) 码 分 多 址 ﻫCNNＩC (Neｔwｏｒｋ Information Centeｒ of China) 中 国 互 联 网 络 信 息 中 心
CＡP (Cａｒrierｌｅss Amplituｄe Phaｓe) 无 载 波 振 幅 相 位 调 制 ﻫCATV (Comｍunｉty Antenna TV, CAble ＴV) 有 线 电 视
CＢＲ ( Constant Bit Ｒatｅ ) 恒 定 比 特 率 ﻫCCIR （Ｃｏｎsultatiｖe Ｃommiｔｔｅe，Ｉnternational Rａdio） 国 际 无 线 电 咨 询 委 员 会
计算机网络专业词汇中英对照
《计算机网络》中英文对照Chapter 1End system P28 端系统Modem P29 调制解调器（俗称：猫）Base station P29 基站Communication link P30 通信链路Physical media P30 物理介质Coaxial cable P30 同轴电缆Fiber optics P30 光纤Radio spectrum P30 射频频谱Transmission rate P30 传输速率Packets P30 （数据）包，或分组Routers P30 路由器Link-layer switches P30 链路层交换机Path P30 路径ISP (Internet Service Provider) P30 网络服务提供商TCP (Transmission Control Protocol) P31 传输控制协议IP ( Internet Protocol) P31 网际协议Intranets P31 内网API (Application Programming Interface) P32 应用程序编程接口Network edge P35 网络边缘Access Networks P38 接入网Ethernet P42 以太网Network core P48 网络核心Circuit Switching P50 电路转换Packet Switching 分组交换FDM (frequency-division multiplexing) P50 频分多路复用TDM (time-division multiplexing) P50 时分多路复用Statistical Multiplexing 统计复用Store-and-forward 存储转发Queuing delays P53 排队延迟Transmission delay P60 传输延迟，或发送延迟Propagation delay P60 传播延迟Throughput P59 吞吐量Internet backbone P57 骨干网Delay P59 延迟，或时延Loss P59 丢包Packet-Switched Network P59 分组交换网络Nodal processing delay P60 节点处理延迟End-to-end delay P66 端到端延迟Instantaneous throughput P68 瞬时吞吐量Network interface card P74 网络接口卡（即网卡）Message P75 消息，或报文Segment P75 （报文）段Datagram P75 数据报Frames P75 帧Packet sniffer P82 数据包监听器Protocol Stack 协议栈Peer entities 对等实体Chapter 2 应用层Server farm P110 服务器集群Infrastructure P110 基础设施，或基础架构Self-scalability P111 自扩展性Timing P114 实时性Bandwidth-sensitive applications P115带宽敏感应用Connection-oriented service P117 面向连接的服务Directory service P121 目录服务Base HTML 基本HTML文件Stateless protocol P124 无状态协议RTT (round-trip time ) P126 往返时间Web proxy caches P128 网页代理缓存Status line P130 状态行Out-of-band P141 （频）带外（的）In-band P141 带内（的）User agents P144 用户代理Mail servers P144 邮件服务器Pull protocol P148 拉式协议Push protocol p148 推式协议Host aliasing P158 主机别名Canonical hostname P158 规范主机名Mail server aliasing P158 邮件服务器别名Load distribution P158 负载分配Top-level domain (TLD) servers P161 顶级域名服务器Authoritative DNS servers P161 权威域名服务器Iterative queries P168 迭代查询Resource records (RRs) P165 资源记录Overlay network P179 覆盖网Nonpersistent HTTP 非持久HTTP，或非坚持HTTP Persistent HTTP 持久性HTTP，或坚持的HTTP Peer-to-Peer (P2P) Network 对等网络Socket programming 套接字编程Chapter 3 传输层Multiplexing and demultiplexing P226 复用与分用Unidirectional data transfer P241 单向数据传送Finite-state machine (FSM) P242 有限状态机Positive acknowledgments P243 肯定确认Negative acknowledgments P243 否定确认Countdown timer P250 （倒数）计时器Cumulative acknowledgment P258 累积确认Receive buffer P269 接收缓冲区，或接收缓存Resource-management cells 资源管理单元Source (port number) 源端口号Destination (port number) 目的端口号Checksum 校验与Pipelined protocols 流水线（型）协议Go-back-N 回退NSelective Repeat 选择重传Timeout （定时器）超时Fast Retransmit 快速重传Flow Control 流量控制Three way handshake 三次握手sequence number 序列号（简写为seq）acknowledgement number 确认号（简写为ack；注意与大小的ACK不同）Congestion Control 拥塞控制additive increase, multiplicative decrease 加性增乘性减Slow Start 慢启动congestion-avoidance 拥塞避免fast recovery 快速恢复duplicate (ACK) 冗余（ACK）Random Early Detection 随机早期检测Chapter 4 网络层Forwarding table P338 转发表Virtual-circuit networks P343 虚电路网络Datagram networks P343 数据报网络Signaling message P346 信令报文Content Addressable Memory P354 内容可寻址存储器Crossbar switch P356 纵横开关Active queue management 主动队列管理Head-of-the-line (HOL) 队头Classless interdomain routing (CIDR) P371 无类域间路由Plug-and-play P376 即插即用Anycast P386 任播Interior gateway protocols P414 内部网关协议Routing information Protocol P414 路由信息协议（RIP）Open shortest Path First OSPF P414 开放最短路径优先Area border routers P419 区域边界路由器Sequence-number-controlled flooding P430 序列号控制的洪泛，或带序列号的受控洪泛Reverse path forwarding (RPF) P431 逆向路径转发Rendezvous point P433 汇聚点Longest prefix matching 最长前缀匹配Scheduling 调度Fragmentation 分片，或分段Fragment Offset 报文段偏移量Network Address Translation (NAT) 网络地址转换NAT traversal NAT穿越Multicast 组播，或多播Unicast 单播Tunneling 隧道技术Link-State Routing Algorithm 链路状态路由算法Distance Vector Routing Algorithm 距离向量路由算法Count to Infinity Problem 无穷计数问题Hierarchical Routing 分层路由autonomous systems 自治系统BGP (Border Gateway Protocol) 边界网关协议in-network duplication 网内复制broadcast storm 广播风暴spanning tree 生成树redundant packets 冗余数据包Chapter 5 数据链路层，或链路层Broadcast channels P461 广播信道Trailer fields P464 尾部字段Link access P464 链路接入，或链路访问Network interface card P466 网络接口卡（即网卡）Parity checks P469 奇偶校验Forward error correction (FEC) P471 前向纠错Cyclic Redundancy Check 循环冗余校验Polynomial code P472 多项式码（即CRC码）Multiple access P475 多路接入Random access protocols P477 随机接入协议CSMA/CD P484 带冲突检测的载波侦听多路访问CSMA/CA 带冲突避免的载波侦听多路访问Token passing protocol P487 令牌传递协议ARP P491 地址解析协议Preamble P497 前导（字段）Exponential backoff P502 指数回退，或指数退避Repeater P504 中继器Virtual-channel identifier P520 虚拟信道标识Cell-loss priority P520 信元丢失优先权Label-switched router P524 标签交换路由器Framing （封装）成帧error detection 误差检测，或检错Channel Partitioning 信道分割式（MAC协议）Taking turns MAC protocol 轮流式MAC协议Collision 冲突，或碰撞Time Slot 时隙Slotted ALOHA 时隙ALOHAUnslotted ALOHA 无时隙ALOHA Nonpersistent CSMA 非坚持CSMA1-persistent CSMA 1坚持CSMAp-persistent CSMA p坚持CSMAToken Ring 令牌环(Wireless) LAN （无线）局域网Hub 集线器Collision domain 冲突域Bridge 网桥。
计算机网络中英文对照外文翻译文献
中英文资料外文翻译计算机网络计算机网络，通常简单的被称作是一种网络，是一家集电脑和设备为一体的沟通渠道，便于用户之间的沟通交流和资源共享。
网络可以根据其多种特点来分类。
计算机网络允许资源和信息在互联设备中共享。
一．历史早期的计算机网络通信始于20世纪50年代末，包括军事雷达系统、半自动地面防空系统及其相关的商业航空订票系统、半自动商业研究环境。
1957年俄罗斯向太空发射人造卫星。
十八个月后，美国开始设立高级研究计划局(ARPA)并第一次发射人造卫星。
然后用阿帕网上的另外一台计算机分享了这个信息。
这一切的负责者是美国博士莱德里尔克。
阿帕网于来于自印度，1969年印度将其名字改为因特网。
上世纪60年代，高级研究计划局(ARPA)开始为美国国防部资助并设计高级研究计划局网(阿帕网)。
因特网的发展始于1969年，20世纪60年代起开始在此基础上设计开发，由此，阿帕网演变成现代互联网。
二．目的计算机网络可以被用于各种用途:为通信提供便利:使用网络，人们很容易通过电子邮件、即时信息、聊天室、电话、视频电话和视频会议来进行沟通和交流。
共享硬件:在网络环境下，每台计算机可以获取和使用网络硬件资源，例如打印一份文件可以通过网络打印机。
共享文件：数据和信息: 在网络环境中，授权用户可以访问存储在其他计算机上的网络数据和信息。
提供进入数据和信息共享存储设备的能力是许多网络的一个重要特征。
共享软件:用户可以连接到远程计算机的网络应用程序。
信息保存。
安全保证。
三．网络分类下面的列表显示用于网络分类：3．1连接方式计算机网络可以据硬件和软件技术分为用来连接个人设备的网络，如：光纤、局域网、无线局域网、家用网络设备、电缆通讯和G.hn（有线家庭网络标准）等等。
以太网的定义，它是由IEEE 802标准，并利用各种媒介，使设备之间进行通信的网络。
经常部署的设备包括网络集线器、交换机、网桥、路由器。
无线局域网技术是使用无线设备进行连接的。
计算机网络英文版——提供给学生部分习题答案
计算机⽹络英⽂版——提供给学⽣部分习题答案Solution of Selected Exercises from the End of Chapter ExercisesChapter 1 - Introduction And Overview1.4 To what aspects of networking does data communications refer?Answer:Data communications refers to the study of low-level mechanisms and technologies used to send information acrossa physical communication medium, such as a wire, radio wave, or light beam.1.5 What is packet-switching, and why is packet switching relevant to the Internet?Answer: Packet switching divides data into small blocks, called packets, and includes an identification of the intended recipient in each packet. Packet switching changed networking in a fundamental way, and provided the basis for the modern Internet. Packet switching allows multiple senders to transmit data over a shared network.1.8 What is a communication protocol? Conceptually, what two aspects of communication does a protocol specify? Answer: A communication protocol refer to a specification for network communication.Major aspects of a protocol are syntax (format) and semantics (meaning) of the protocol.1.9 What is a protocol suite, and what is the advantage of a suite?Answer:protocols are designed in complete, cooperative sets called suites or families, instead of creating each protocol in isolation. Each protocol in a suite handles one aspect of communication; together, the protocols in a suite cover all aspects of communication. The entire suite is designed to allow the protocols to work together efficiently. 1.11 List the layers in the TCP/IP model, and give a brief explanation of each.(See Textbook)1.14 Give a brief explain of the layers in the ISO Open System Interconnection model.(See Textbook)Chapter 3 - Internet Applications And Network Programming3.1 What are the two basic communication paradigms used in the Internet?Answer: There are various approaches, but according to textbook, we can specify them as Stream Paradigm and Message Paradigm.3.2 Give six characteristics of Internet stream communication.(See Textbook)3.3 Give six characteristics of Internet message communication.(See Textbook)3.4 If a sender uses the stream paradigm and always sends 1024 bytes at a time, what size blocks can the Internet deliverto a receiver?Answer: stream paradigm does not provide any guarantees for block sizes, so all depends on individual transfer.3.6 What are the three surprising aspects of the Internet’s message delivery semantics?Answer:The Internet’s message delivery has the followi ng undesirable characteristics:* Messages can be lost* Messages can be duplicated* Messages can be delivered out-of-order3.8 When two applications communicate over the Internet, which one is the server?Answer: T he application that waits for some other applications to contact is called server, and the application that contact other one is called client.3.14 What two identifiers are used to specify a particular server?Answer: A particular server is identified by the following identifiers:* An identifier for the computer on which a server runs (IP Address)* An identifier for a particular service on the computer (Port Number)Chapter 4 - Traditional Internet Applications4.1 What details does an application protocol specify?(See Textbook)4.3 What are the two key aspects of application protocols, and what does each include?(See Textbook)4.6 What are the four parts of a URL, and what punctuation is used to separate the parts?Answer: The URL into four components: a protocol, a computer name, a document name, and parameters. The computer name and protocol port are used to form a connection to the server on which the page resides. And the document name and parameters are used to request a specific page.4.7 What are the four HTTP request types, and when is each used?(See Textbook)4.12 When a user requests an FTP directory listing, how many TCP connections are formed? Explain.Answer: FTP uses two types of connections to perform its functionality, namely* A control connection is reserved for commands. Each time the server needs to download or upload a file, the server opens a new connection.* A data connection is used to transfer files.4.16 List the three types of protocols used with email, and describe each.(See Textbook)4.17 What are the characteristics of SMTP?(See Textbook)4.20 What are the two main email access protocols?Answer: Two major email access protocols are:* Post Office Protocol (POP)* Internet Mail Access Protocol (IMAP)Chapter 6- Information Sources and Signals6.4 State and describe the four fundamental characteristics of a sine wave.(See Textbook)6.9 What is the analog bandwidth of a signal?Answer: Analog bandwidth of signal can be defined as to be the difference between the highest and lowest frequencies of the constituent parts (i.e., the highest and lowest frequencies obtained by Fourier analysis)6.11 Suppose an engineer increases the number of possible signal levels from two to four. How many more bits can be sent in the same amount of time? Explain.Answer: The number of levels that can be represented by n bits is given by 2n . So if number of levels changes from 2→4, it means number of bits goes from 1→2612. What is the definition of baud?Answer: Baud is defined as the number of times that a signal can change per second.6.14 What is the bandwidth of a digital signal? Explain.Answer: According to the definition of analog bandwidth, a digital signal has infinite bandwidth because Fourier analysis of a digital signal produces an infinite set of sine waves with frequencies that grow to infinity.6.18 What is the chief advantage of a Differential Manchester Encoding?Answer: The most important property of differential encoding is that the encoding works correctly even if the two wires carrying the signal are accidentally reversed.6.20 If the maximum frequency audible to a human ear is 20,000 Hz, at what rate must the analog signal from a microphone be sampled when converting it to digital?Answer: The sampling rate = 2 × f max, so the signal should be sampled at 2x20,000 = 40,000 HzChapter 7 - Transmission Media7.2 What are the three energy types used when classifying physical media according to energy used?Answer: Three types of energy used when classifying physical media are electrical, electromechanical (radio), and light7.4 What three types of wiring are used to reduce interference form noise?(See Textbook)7.10 List the three forms of optical fiber, and give the general properties of each.(See Textbook)7.21 What is the relationship between bandwidth, signal levels, and data rate?Answer: If a transmission system uses K possible signal levels and has an anal og bandwidth B, Nyquist’s Theorem states that the maximum data rate in bits per second, D, is: D = 2 B log2K7.22 If two signal levels are used, what is the data rate that can be sent over a coaxial cable that has an analog bandwidthof 6.2 MHz?Answer: Using the D= 2 B log2 K relationship, D = 2*6.2*log22 = 2*6.2*1 = 12.4 Mbps7.24 If a system has an input power level of 9000, and an output power level of 3000, what is the difference when expressed in dB?Answer: Decibel is expressed as 10log10(P out/P in) → 10log10(3,000/9,000) = to be determined by reader7.23 If a system has an average power level of 100, an average noise level of 33.33, and a bandwidth of 100 MHz, whatis the effective limit on channel capacity?Answer: Shannon theorem specify the maximum data rate that could be achieved over a transmission system that experiences noise: C = Blog2 (1 + S/N) = 100,000,000 * log2 (1 + 100/33.33) = 100,000,000 * log24 = 200,000,000 = 200 Mbps7.25 If a telephone system can be created with a signal-to-noise ratio of 40 dB and an analog bandwidth of 3000 Hz, how many bits per second could be transmitted?Answer: First we should convert 40 dB to a real number, namely if 40 = 10 log10S/N→S/N = 10,000 , Using the Shannon’s capacity expression C = B log2(1 + S/N) → C = 3,000 log2 (1+ 10,000) = to be determined by readerCh 8 - Reliability And Channel Coding8.1 List and explain the three main sources of transmission errors.(See Textbook)8.3 In a burst error, how is burst length measured?Answer: For a burst error, the burst size, or length, is defined as the number of bits from the start of the corruption to the end of the corruption.8.4 What is a codeword?Answer: We can define the set of all possible messages to be a set of datawords, and define the set of all possible encoded versions to be a set of codewords. So each possible code sequence is considered to be a codeword.8.8 Compute the Hamming distance for the following pairs: (0000, 0001), (0101, 0001), (1111, 1001), and ( 0001, 1110). (See Textbook)8.11 Generate a RAC parity matrix for a (20, 12) coding of the dataword 100011011111.(See Textbook)8.15 Express the two values in the previous exercise as polynomials.Answer:X10+ X7 + X5 + X3 + XX4+ X2+ 1Ch 9 - Transmission Modes9.1 Describe the difference between serial and parallel transmission.Answer: Transmission modes can be divided into two fundamental categories:* Serial: one bit is sent at a time* Parallel: multiple bits are sent at the same time9.2 What are the advantages of parallel transmission? What is the chief disadvantage?Answer: A parallel mode of transmission has two chief advantages:* High speed: Because it can send N bits at the same time, a parallel interface can operate N times faster than an equivalent serial interface.* Match to underlying hardware: Internally, computer and communication hardware uses parallel circuitry.Thus, a parallel interface matches the internal hardware well.The main disadvantage of parallel transmission is number of cables required, for long distance communication, this is an important consideration.9.4 What is the chief characteristic of asynchronous transmission?Answer:Asynchronous transmission can occur at any time, with an arbitrary delay between the transmission of two data items, it allows the physical medium to be idle for an arbitrary time between two transmissions.Chapter 11 - Multiplexing And Demultiplexing11.2 What are the four basic types of multiplexing?(See Textbook)11.4 What is a guard band?Answer: For proper communication without interference, we should choose a set of carrier frequencies with a gap between them known as a guard band. The guard band reduces or eliminates the possible interference between neighboring carrier signals.11.8 Explain how a range of frequencies can be used to increase data rate.Answer:To increase the overall data rate, a sender divides the frequency range of the channel into K carriers, and sends 1 /K of the data over each carrier.11.12 Suppose N users compete using a statistical TDM system, and suppose the underlying physical transport can sendK bits per second. What is the minimum and maximum data rate that an individual user can experience?Answer: If we neglect the overhead generated by statistical TDM, a system will have two possibilities: * Minimum: If all channels have equal data then the rate will be K/N bps* Maximum: If only one channel active and the others are passive, then rate will be K bpsChapter 13 - Local Area Networks: Packets, Frames, And Topologies13.1 What is circuit switching, and what are its chief characteristics?Answer: The term circuit switching refers to a communication mechanism that establishes a path between a sender and receiver with guaranteed isolation from paths used by other pairs of senders and receivers. The circuit switching has the following main characteristics:* Point-to-point communication* Separate steps for circuit creation, use, and termination* Performance equivalent to an isolated physical path13.3 In a packet switching system, how does a sender transfer a large file?Answer: The packet switching system requires a sender to divide each message into blocks of data that are known as packets . The size of a packet varies; each packet switching technology defines a maximum packet size. So, a large file will be divided into smaller pieces and sent.13.5 What are the characteristics of LANs, MANs, and W ANs?Answer: There are lots of details that can be said and discussed for categorization of network types based on geography, few points are highlighted below:* Local Area Network (LAN): Least expensive; spans a single room or a single building* Metropolitan Area Network (MAN) Medium expense; spans a major city or a metroplex* Wide Area Network (WAN) Most expensive; spans sites in multiple cities13.6 Name the two sublayers of Layer 2 protocols defined by IEEE, and give the purpose of each.Answer: The Layer 2 protocols defined by IEEE defines two sub-layers as mentioned below:* Logical Link Control (LLC) Addressing and demultiplexing* Media Access Control (MAC) Access to shared media13.8 What are the four basic LAN topologies?Answer: The four basic LAN topologies are star, ring, mesh and bus.13.10 In a mesh network, how many connections are required among 20 computers?Answer: The expression to calculate number of connections in a mesh network is given by n (n-1)/2. So for 20 computers then number of connections required will be = 20 (20 – 1)/2 =19013.15 Give a definition of the term frame .Answer: In a packet-switched network, each frame corresponds to a packet processed at data link layer.Chapter 14 - The IEEE MAC Sub-Layer14.1 Explain the three basic approaches used to arbitrate access to a shared medium.(See Textbook)14.3 List the three main types of channelization and the characteristics of each.(See Textbook)14.6 What is a token, and how are tokens used to control network access?Answer: A special control message is called a token. In a token passing system, when no station has any packets to send, the token circulates among all stations continuously. When a station captures the token, it sends its data, and when transmission completed, it releases the token.14.8 Expand the acronym CSMA/CD, and explain each part.Answer: The acronym CSMA/CD stands for Carrier Sense Multi-Access with Collision Detection, which means the following: * Carrier Sense: Instead of allowing a station to transmit whenever a packet becomes ready, Ethernet requires each station to monitor the cable to detect whether another transmission is already in progress.* Multiple Access: The system allows multiple users/hosts to make use of a common/shared media* Collision Detection. A collision can occur if two stations wait for a transmission to stop, find the cable idle, and both start transmitting.14.10 Why does CSMA/CD use a random delay? (Hint: think of many identical computers on a network.)Answer: Randomization is used to avoid having multiple stations transmit simultaneously as soon as the cable is idle.That is, the standard specifies a maximum delay, d, and requires each station to choose a random delay less than d after a collision occurs. In most cases, when two stations each choose a random value, the station that chooses the smallest delay willChapter 15 - Wired LAN Technology (Ethernet And 802.3)15.1 How large is the maximum Ethernet frame, including the CRC?Answer: According to Fig. 15.1 a conventional Ethernet frame has the following fields:* Header: 14 bytes (fixed)* Payload: 46-1500 bytes (there is a minimum frame size because of collision detection)* CRC: 4 bytes (fixed)Accordingly an Ethernet frame will be maximum 1518 bytes and minimum 64 bytes15.3 In an 802.3 Ethernet frame, what is the maximum payload size?Answer: The 802.3 Ethernet makes use of 8-bytes of the original/conventional Ethernet for Logical Link Control / Sub-Network Attachment Point (LLC / SNAP) header instead of extending/increasing the header. This is for sake of backward compatibility. So the maximum pay load is reduced from 1500 bytes to 1492 bytes.15.6 How did a computer attach to a Thicknet Ethernet?Answer: Hardware used with Thicknet was divided into two major parts:* Transceiver: A network interface card (NIC) handled the digital aspects of communication, and a separate electronic device called a transceiver connected to the Ethernet cable and handled carrier detection, conversion of bits into appropriate voltages for transmission, and conversion of incoming signals to bits.* AUI: A physical cable known as an Attachment Unit Interface (AUI) connected a transceiver to a NIC in a computer. A transceiver was usually remote from a computer.15.7 How were computers attached to a Thinnet Ethernet?Answer: Thinnet Ethernet (formally named 10Base2) uses a thinner coaxial cable that was more flexible than Thicknet. The wiring scheme differed dramatically from Thicknet. Instead of using AUI connections between a computer and a transceiver, Thinnet integrates a transceiver directly on the NIC, and runs a coaxial cable from one computer to another.15.8 What is an Ethernet hub, and what wiring is used with a hub?Answer: An electronic device that serves as the central interconnection is known as a hub. Hubs were available in a variety of sizes, with the cost proportional to size. The hubs are becoming old-fashioned, and being replaced with switches.15.3 What category of twisted pair wiring is needed for a 10 Mbps network? 100 Mbps? 1000 Mbps?Answer: The three major categories of Ethernet and their wiring is listed below:* 10 Mbps: 10BaseT (Ethernet) Category 5* 100 Mbps: 100BaseT (Ethernet Fast) Category 5E* 1 Gbps: 1000BaseT (Gigabit Ethernet) Category 6Chapter 20 - Internetworking: Concepts, Architecture, and Protocols20.2 Will the Internet be replaced by a single networking technology? Why or why not?Answer: Incompatibilities make it impossible to form a large network merely by interconnecting the wires among networks. The beauty of the Internet is interconnection of wide range of technologies from various manufacturers.Diversity of the products and solutions is a richness instead of limitation as long as they all adopt the same set of protocols.20.3 What are the two reasons an organization does not use a single router to connect all its networks?Answer:An organization seldom uses a single router to connect all of its networks. There are two major reasons: * Because the router must forward each packet, the processor in a given router is insufficient to handle the traffic passing among an arbitrary number of networks.* Redundancy improves internet reliability. To avoid a single point of failure, protocol software continuously monitors internet connections and instructs routers to send traffic along alternative paths when a network or router fails.20.6 In the 5-layer reference model used with the TCP/IP Internet protocols, what is the purpose of each of the five layers?(See 1.11)Chapter 21- IP: Internet Addressing21.3 In the original classful address scheme, was it possible to determine the class of an address from the address itself? Explain.Answer:Yes, since in the classful addressing scheme initial bit(s) gives indication about the class being used.21.7 If an ISP assigned you a /28 address block, how many computers could you assign an address?Answer: When an organization is assigned /28 CIDR address, it means 28 bits out of 32 bits are fixed, so 32-28 = 4 bits available for user space. So number of users 24-2 = 4, since the all 0s and all 1s address are having special use and can’t be assigned to a user.21.8 If an ISP offers a / 17 address block for N dollars per month and a / 16 address block for 1.5 N dollars per month,which has the cheapest cost per computer?Answer: Number of addresses in /17 block 232-17 = 215Price per address: N /215 = N / 215Number of addresses in /16 block 232-16 = 216Price per address: 1.5N /216 = 0.75N/215 So /16 address block will be cheaper in comparison with the price given for /17 block.21.10 Suppose you are an ISP with a / 24 address block. Explain whether you accommodate a request from a customer who needs addresses for 255 computers. (Hint: consider the special addresses.)Answer: For a/24 address block, number of available addresses will be 232-24 = 28 = 256. However, a suffix with all 0s address is reserved for network ID and a suffix with all 1s address is reserved for broadcast address, so number of addresses that can be assigned to computers/hosts will be 256 -2 = 254.21.11 Suppose you are an ISP that owns a / 22 address block. Show the CIDR allocation you would use to allocateaddress blocks to four customers who need addresses for 60 computers each.Answer: The /22 address block can be assigned as follows:ddd.ddd.ddd.00/26ddd.ddd.ddd.01/26ddd.ddd.ddd.10/26ddd.ddd.ddd.11/26Chapter 22- Datagram Forwarding22.1 What are the two basic communication paradigms that designers consider when designing an internet?Answer:* Connection-oriented service * Connectionless service22.2 How does the Internet design accommodate heterogeneous networks that each have their own packet format?Answer: To overcome heterogeneity, the Internet Protocol defines a packet format that is independent of the underlying hardware. The result is a universal, virtual packet that can be transferred across the underlying hardware intact. The Internet packet format is not tied directly to any hardware. The underlying hardware does not understand or recognize an Internet packet.22.5 What is the maximum length of an IP datagram?In the current version of the Internet Protocol (IP version 4), a datagram can contain at most 64 K (65535) octets, including the header.22.7 If a datagram contains one 8-bit data value and no header options, what values will be found in header fields H.LEN and TOTAL LENGTH?Answer: H. LEN indicated header in 32-quantities, since no options, then this value will be 5. The TOTAL LENGTH indicated the number of bytes in a datagram including the header. This means 5x4 bytes + 1 (8-bits) = 21 bytesChapter 23 - Support Protocols And Technologies23.1 When a router uses a forwarding table to look up a next-hop address, the result is an IP address. What must happenbefore the datagram can be sent?Answer: Each router along the path uses the destination IP address in the datagram to select a next-hop address, encapsulates the datagram in a hardware frame, and transmits the frame across one network. A crucial step of the forwarding process requires a translation: forwarding uses IP addresses, and a frame transmitted across a physical network must contain the MAC address of the next hop.23.2 What term is used to describe the mapping between a protocol address and a hardware address?Answer: Translation from a computer’s IP address to an equivalent hardware address is known as address resolution, and an IP address is said to be resolved to the correct MAC address. The TCP/IP protocol being used for this is called Address Resolution Protocol (ARP). Address resolution is local to a network.23.5 How many octets does an ARP message occupy when used with IP and Ethernet addresses?Answer: According to Fig 23.3 an ARP message has 7-lines of each being 32-bit (4 bytes or octets), therefore,number of octets in an ARP can be determined as 7x4 = 28 octets23.10 What types of addresses are used in layers below ARP?Answer:ARP forms a conceptual boundary in the protocol stack; layers above ARP use IP addresses, and layers below ARP use MAC addresses.23.17 What is the chief difference between BOOTP and DHCP?Answer:The main difference is that the BOOTP protocol required manual administration. So before a computer could use BOOTP to obtain an address, a network administrator had to configure a BOOTP server to know the computer’s I P address. Chapter 24 - The Future IP (IPv6)24.3 List the major features of IPv6, and give a short description of each.(See Textbook)24.4 How large is the smallest IPv6 datagram header?Answer: IPv6 datagram header consists of a base header + zero or more extension header. Since, smallest header is being asked, we assume zero extension header and consider IPv6 will have only base header. If we look at IPv6 header format in Fig. 24.3, it shows that 10x4 bytes = 40 bytes.Chapter 26 - TCP: Reliable Transport Service26.2 List the features of TCP.(See Textbook)26.6 When using a sliding window of size N, how many packets can be sent without requiring a single ACK to be received?Answer: If the size of the window is N, then it means a sender can transmit up to N packets without waiting for an ACK, as long as other controls are in place.26.9 What is the chief cause of packet delay and loss in the Internet?Answer: The main cause of packet delay and loss in the Internet is congestion.Chapter 28 - Network Performance (QoS and DiffServ)28.1 List and describe the three primary measures of network performance.(See Textbook)28.2 Give five types of delay along with an explanation of each.(See Textbook)Chapter 30 - Network Security30.1 List the major security problems on the Internet, and give a short description of each.(See Textbook)30.2 Name the technique used in security attacks.(See Textbook)30.8 List and describe the eight basic security techniques.(See Textbook)。
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Chapter11-11.What are two reasons for using layered protocols?(请说出使用分层协议的两个理由)答：通过协议分层可以把设计问题划分成较小的易于处理的片段。
分层意味着某一层的协议的改变不会影响高层或低层的协议。
1-13. What is the principal difference between connectionless communication and connection-oriented communication?(在无连接通信和面向连接的通信两者之间，最主要的区别是什么？)答：主要的区别有两条。
其一：面向连接通信分为三个阶段，第一是建立连接，在此阶段，发出一个建立连接的请求。
只有在连接成功建立之后，才能开始数据传输，这是第二阶段。
接着，当数据传输完毕，必须释放连接。
而无连接通信没有这么多阶段，它直接进行数据传输。
其二：面向连接的通信具有数据的保序性，而无连接的通信不能保证接收数据的顺序与发送数据的顺序一致。
1-20. A system has an n-layer protocol hierarchy. Applications generate messages of length M bytes. At each of the layers, an h-byte header is added. What fraction of the network bandwidth is filled with headers?(一个系统有n层协议的层次结构。
应用程序产生的消息的长度为M字节。
在每一层上需要加上一个h字节的头。
请问，这些头需要占用多少比例的网络带宽)答：hn/(hn+m)*100%1-28. An image is 1024 x 768 pixels with 3 bytes/pixel. Assume the image is uncompressed. How long does it take to transmit it over a 56-kbps modem channel? Over a 1-Mbps cable modem? Over a 10-Mbps Ethernet? Over 100-Mbps Ethernet?(一幅图像的分辨率为1024 x 768像素，每个像素用3字节来表示。
假设该图像没有被压缩。
请问，通过56kbps的调制解调器信道来传输这幅图像需要多长时间？通过1Mbps的电缆调制解调器呢？通过10Mbps的以太网呢？通过100Mbps的以太网呢?)答：The image is 1024*768*3 bytes or 2359296 bytes.This is 18874368 bit. At 56,000 bits/sec, it takes about 337.042 sec. At 1,000,000 bits/sec, it takes about 18.874 sec. At 10,000,000 bits/sec, it takes about 1.887 sec. At 100,000,000 bits/sec, it takes about 0.189 sec.Chapter22-2. A noiseless 4-kHz channel is sampled every 1 msec. What is the maximum data rate?(一条无噪声4kHz信道按照每1ms一次进行采样，请问最大数据传输率是多少？)答：无噪声信道最大数据传输率公式：最大数据传输率=2Hlog2V b/s。
因此最大数据传输率决定于每次采样所产生的比特数，如果每次采样产生16bits，那么数据传输率可达128kbps；如果每次采样产生1024bits，那么可达8.2Mbps。
注意这是对无噪声信道而言的，实际信道总是有噪声的，其最大数据传输率由香农定律给出。
2-4. If a binary signal is sent over a 3-kHz channel whose signal-to-noise ratio is 20 dB, what is the maximum achievable data rate?(如果在一条3kHz的信道上发送一个二进制信号，该信道的信噪比为20dB，则最大可达到的数据传输率为多少？)答：信噪比为20 dB 即 S/N =100由于 log101≈6.658，由香农定理，该信道2的信道容量为3log(1+100)=19.98kbps。
2又根据乃奎斯特定理，发送二进制信号的3kHz 信道的最大数据传输速率为2*3*log22=6kbps。
所以可以取得的最大数据传输速率为6kbps。
2-5. What signal-to-noise ratio is needed to put a T1 carrier on a 50-kHz line?(在50kHz的线路上使用1T线路需要多大的信噪比？)答：为发送T1 信号，我们需要所以，在50kHz 线路上使用T1 载波需要93dB 的信噪比。
2-34. A signal is transmitted digitally over a 4-kHz noiseless channel with one sample every 125 µsec. How many bits per second are actually sent for each of these encoding methods?(a) CCITT 2.048 Mbps standard.(b) DPCM with a 4-bit relative signal value.(c) Delta modulation.（一个信号在4kHz的无噪声信道上以数字方式进行传输，没125us采样一次。
请问，按照以下的编码方法，每秒钟实际发送多少位？（a）CCITT 2.048Mbps标准。
（b）有4位相对信号值的DPCM。
（c）增量调制。
）答：a．CCITT 2.048Mbps 标准用32 个8 位数据样本组成一个125的基本帧，30 个信道用于传信息，2 个信道用于传控制信号。
在每一个4kHz 信道上发送的数据率就是8*8000=64kbps。
b．差分脉码调制（DPCM）是一种压缩传输信息量的方法，它发送的不是每一次抽样的二进制编码值，而是两次抽样的差值的二进制编码。
现在相对差值是4 位，所以对应每个4kHz 信道实际发送的比特速率为4*8000=32bps。
c．增量调制的基本思想是：当抽样时间间隔s t 很短时，模拟数据在两次抽样之间的变化很小，可以选择一个合适的量化值作为阶距。
把两次抽样的差别近似为不是增加一个就是减少一个。
这样只需用1bit 二进制信息就可以表示一次抽样结果，而不会引入很大误差。
因此，此时对应每个4kHz 信道实际发送的数据速率为1*8000=8kHz。
2-43. Suppose that x bits of user data are to be transmitted over a k-hop path in a packet-switched network as a series of packets, each containing p data bits and h header bits,with x >> p+h.The bit rate of the lines is b bps and the propagation delay is negligible.What value of p minimizes the total delay?(假定x位用户数据将以一系列分组形式，在一个分组交换网络中沿着一条共有k跳的路径向前传播，每个分组包含p位数据和h位的头，这里x>>p+h。
线路的传输率为b bps，传播延迟忽略不计。
请问，什么样的p值使总延迟最小？) 答：所需要的分组总数是x /p，因此总的数据加上头信息交通量为(p+h)*x/p 位。
源端发送这些位需要时间为(p+h)*x/pb中间的路由器重传最后一个分组所花的总时间为(k-1)*(p+h)/ b因此我们得到的总的延迟为对该函数求p 的导数，得到令得到因为p＞0，所以故时能使总的延迟最小。
2-53. A CDMA receiver gets the following chips: (-1 +1 -3 +1 -1 -3 +1 +1). Assuming the chip sequences defined in , which stations transmitted, and which bits did each one send?(一个CDMA接收器得到了下面的时间片：(-1 +1 -3 +1 -1 -3 +1 +1)。
假设时间片序列如图2.45（b）中所定义，请问哪些移动站传输了数据？每个站发送了什么位?)答：Just compute the four normalized inner products:（1+1 3+1 1 3+1+1）d（1 1 1+1+1 1+1+1）/8=1（1+1 3+1 1 3+1+1）d（1 1+1 1+1+1+1 1）/8=1（1+1 3+1 1 3+1+1）d（1+1 1+1+1+1 1 1）/8=0（1+1 3+1 1 3+1+1）d（1+1 1 1 1 1+1 1）/8=1The result is that A and D sent 1 bits, B sent a 0 bit, and C was silent. Chapter33-3.The following data fragment occurs in the middle of a data stream for which the byte-stuffing algorithm described in the text is used: A B ESC C ESC FLAG FLAG D. What is the output after stuffing?(数据片断（A B ESC C ESC FLAG FLAG D）出现在一个数据流的中间，而成帧方式采用的是本章介绍的字节填充算法，请问经过填充之后的输出时什么？) 答：After stuffing, we get A B ESC ESC C ESC ESC ESC FLAG ESC FLAG D.3-15. A bit stream 10011101 is transmitted using the standard CRC method described in the text. The generator polynomial is x3 + 1. Show the actual bit string transmitted. Suppose the third bit from the left is inverted during transmission. Show that this error is detected at the receiver's end.(利用本章中介绍的标准多项式CRC方法来传输位流1001101。