Computer Networking A Top-Down Approach, 5th Edition 第6章
ccf计算机软件能力认证 参考书籍
ccf计算机软件能力认证参考书籍计算机软件能力认证（Computer Software Capability Certification，简称CCF）是一项针对计算机软件人员的职业能力认证。
通过参加CCF考试并取得认证资格，可以证明个人在软件开发、架构设计、系统维护以及相关领域的专业能力。
为了帮助考生充分准备CCF考试，下面将介绍一些参考书籍，供考生参考。
1. 《软件工程导论》（Introduction to Software Engineering）《软件工程导论》是一本系统介绍软件工程概念、原理、方法和实践的经典教材。
这本书从软件工程的基本概念出发，讲解了软件开发的整个过程，包括需求分析、设计、开发、测试、部署和维护等各个环节。
对于CCF认证考试来说，掌握软件工程的基本原理是非常重要的，这本书可以帮助考生建立系统的软件工程知识体系。
2. 《设计模式：可复用面向对象软件的基础》（Design Patterns: Elements of Reusable Object-Oriented Software）《设计模式：可复用面向对象软件的基础》是由四位软件工程师合著的一本著名书籍。
这本书总结了23种常见的设计模式，包括创建型、结构型和行为型模式，对于面向对象软件开发非常有指导意义。
CCF考试中，设计模式也是一个重要的考点，考生需要熟悉各种设计模式的定义、特点和适用场景。
3. 《代码大全》（Code Complete）《代码大全》是一本关于软件构建的经典书籍，着重介绍了如何编写高质量的软件代码。
这本书包含了大量的编程技巧、实用建议和规范，旨在帮助开发人员提高代码质量和开发效率。
CCF认证考试除了要求考生具备良好的软件设计能力，还要求考生具备扎实的编程能力，这本书可以帮助考生提升编码水平。
4. 《操作系统导论》（Operating System Concepts）《操作系统导论》是一本广泛使用的操作系统教材，介绍了操作系统的基本概念、原理和实现技术。
计算机网络自顶向下方法，第7版——习题解答
计算机⽹络⾃顶向下⽅法，第7版——习题解答⽬录前⾔本⽂包含了Computer Networking A Top-Down Approach, 7th Edition中部分回顾性习题的问题与解答，主要参考了英⽂第7版和中⽂第7版的正⽂内容，欢迎各位的交流与指正！CHAPTER 1SECTION 1.1R1. What is the difference between a host and an end system? List several different types of end systems. Is a Web server anend system?在计算机⽹络中，⼆者的含义是相同的。
端系统列举：⼿机、平板电脑、环境传感器等⽹络服务器是端系统R2. The word protocol is often used to describe diplomatic relations. How does Wikipedia describe diplomatic protocol?There are two meanings of the word "protocol". In the legal sense, it is defined as an international agreement that supplements or amends a treaty. In the diplomatic sense, the term refers to the set of rules, procedures, conventions and ceremonies that relate to relations between states. In general, protocol represents the recognized and generally accepted system of international courtesy.“协议”⼀词有两种含义。
计算机网络安全论文参考文献
计算机网络安全论文参考文献以下是一篇关于计算机网络安全的论文参考文献：[1] Stallings, W. (2013). Network security essentials: Applications and standards (Vol. 1). Pearson.这本书探讨了计算机网络安全的基本概念、原理和技术。
它提供了对各种网络安全协议、算法和策略的详细介绍，以及实际应用和标准的示例。
[2] Kurose, J. F., & Ross, K. W. (2010). Computer networking: a top-down approach. Pearson Education.这本书是计算机网络的经典教材，它详细介绍了计算机网络体系结构和协议的工作原理。
它还包括网络安全的基本原理和技术，如加密、身份验证和访问控制。
[3] Anderson, R. (2008). Security engineering: a guide to building dependable distributed systems. John Wiley & Sons.这本书通过分析实际系统的安全问题和解决方案，提供了构建安全分布式系统的指南。
它涵盖了许多关键安全概念和技术，如访问控制、防火墙、入侵检测和身份管理。
[4] Pfleeger, C. P., & Pfleeger, S. L. (2007). Security in computing. Prentice Hall.该书介绍了计算机安全的基本概念、原理和技术，如密码学、访问控制、网络安全和软件安全。
它还提供了关于安全策略和风险管理的实际指南。
[5] Schneier, B. (2015). Applied cryptography: protocols,algorithms, and source code in C. John Wiley & Sons.这本书是密码学的经典教材，它详细介绍了密码学的基本概念、原理和算法。
计算机网络_自顶向下方法_(中文版课件)第八章_网络安全.TopDownV3-8
+ KB
网络安全 18
RSA: 加密，解密
0. 给定(n,e)和(n,d)如上面所计算 1. 为加密比特模式, m, 计算
c = m e mod n (即当 me被n相除时的余数)
(m mod n) d mod n = m edmod n = m
e
ed mod (p-1)(q-1)
1
(using number theory result above)
mod n
= m mod n
(因为我们选择ed(p-1)(q-1) 相除具有余数1 )
= m
网络安全 21
RSA: 另一个重要性质
Alice的 IP 地址
加密的 “I’m Alice” 口令
记录并重放 仍然有效!
Alice的 IP地址
OK
Alice的 IP地址
加密的 “I’m Alice” 口令
网络安全
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鉴别:另一种尝试
目标：避免重放攻击 不重数(Nonce): 数字(R)一生仅用 一次 ap4.0: 为了证实 Alice “活跃”, Bob向Alice发送不重数 。 Alice必须返回 R, 用共享的秘密密钥加密 “I am Alice” R KA-B(R) 实效，缺点?
网络安全 3
什么是网络安全?
机密性: 仅发送方，希望的接收方应当“理解” 报文内容 发送方加密报文 接收方解密报文 鉴别: 发送方、接收方要证实彼此的身份 报文完整性: 发送方、接收方要确保报文(在传输 或以后)不在不知不觉中被篡改 访问和可用性: 服务必须可访问和为用户可用
[计算机网络：自顶向下方方法](中文版课件)第四章
第四章](https://img.taocdn.com/s3/m/b9321b242af90242a895e543.png)
IP 数据报格式
IP协议版本号 首部长度 (字节) 数据的“类型‖ 剩余跳的最大数 (在每台路由器减1) ver 首部 32 bits 服务 类型 总数据报长度 (字节) 长度 段偏移 标志 互联网检查和 对分段/重装
长度
16-bit标识符 高层 寿命
32 bit源IP地址 较高层协议 交付的负载 32 bit目的IP地址 选项 (如果有的话)
网络层
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网络层服务模型:
网络体系结构 服务模型 带宽保证 无丢失保证 排序 定时 拥塞指示
因特网
尽力而为
无
无
任何可能 不维持 的顺序
无
ATM
CBR
保证恒定 是 速率
有序
维持
拥塞不出 现
ATM
ABR
保证 无 最小速率
有序
不维持
提供拥塞 指示
网络层
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第4章 网络层
4. 1 概述 4.2 虚电路和数据报网络 4.3 路由器的构成 4.4 IP: 网际协议 数据报格式 IPv4编址 NAT ICMP IPv6 4.5 选路概念
application transport network data link physical
网络层 数据链路 层 物理层
network data link physical network data link physical
network data link physical
network data link physical
一条VC由下列组成:
1. 2. 3.
从源到目的地的路径 VC号是标识沿路径每条链路的号码 沿路径路由器中转发表中的项
计算机网络自顶向下方法(第四版)ppt第1章
采用分组交换技术
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分组交换(packet switching)技术
发送端将要发送的数据分成若干较小的块，添加首部形 成分组（包packet） ，分别发送到目的端，再组装恢复 原数据。
路径（route或path）：一个分组从发送端系统传输到接收
端系统，所经过的一系列通信链路和分组交换机。
• 端系统之间通信的路径不专用。 • 多个通信端系统同时共享一条路径或一部分。 第一个分组交换网络：
如Telnet（远程注册）、SMTP（电子邮件）、
FTP（文件传输）和HTTP（Web）等。
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2、无连接服务
两个端系统之间交换数据时，不需要“握手过 程”，可直接发送分组，数据传递更快。
特点： 不可靠：源主机不能确定分组是否已经到达目的地。 无流控制或拥塞控制的功能。 用户数据报协议UDP：因特网的无连接服务。
对内容提供者提供接入：如发布信息。 低层次的ISP通过国家、国际的高层ISP互联：
实现世界范围的通信。高层ISP由一些用高速光纤链 路互联的高速路由器组成。
每个ISP独立管理，运行ISP协议：遵从一定的命名和
地址规则。
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4、协议（protocol）
控制网络中信息接收和发送的一组软件。每个端系 统、路由器和其他因特网部件都要运行。
应用程序模式：客户机 /服务器、对等共享、混合等。
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客户机/服务器模式
因特网应用程序广泛采用。如电子邮件、Web服 务等。
是分布式应用程序：客户机程序和服务器程序在端系
统分别上运行。
描述服务和被服务的关系。
客户机程序(client program)：服务请求方。发出请
计算机网络自顶向下 第七版 第七章答案
Computer Networking: A Top-Down Approach,7th Edition计算机网络自顶向下第七版Solutions to Review Questions and ProblemsChapter 7 Review Questions1.In infrastructure mode of operation, each wireless host is connected to the largernetwork via a base station (access point). If not operating in infrastructure mode, a network operates in ad-hoc mode. In ad-hoc mode, wireless hosts have noinfrastructure with which to connect. In the absence of such infrastructure, the hosts themselves must provide for services such as routing, address assignment, DNS-like name translation, and more.2.a) Single hop, infrastructure-basedb) Single hop, infrastructure-lessc) Multi-hop, infrastructure-basedd) Multi-hop, infrastructure-less3.Path loss is due to the attenuation of the electromagnetic signal when it travelsthrough matter. Multipath propagation results in blurring of the received signal at the receiver and occurs when portions of the electromagnetic wave reflect off objects and ground, taking paths of different lengths between a sender and receiver. Interference from other sources occurs when the other source is also transmitting in the samefrequency range as the wireless network.4.a) Increasing the transmission powerb) Reducing the transmission rate5.APs transmit beacon frames. An AP’s beacon frames will be transmitted over one ofthe 11 channels. The beacon frames permit nearby wireless stations to discover and identify the AP.6.False7.APs transmit beacon frames. An AP’s beacon frames will be transmitted over one ofthe 11 channels. The beacon frames permit nearby wireless stations to discover and identify the AP.8.False9.Each wireless station can set an RTS threshold such that the RTS/CTS sequence isused only when the data frame to be transmitted is longer than the threshold. This ensures that RTS/CTS mechanism is used only for large frames.10.No, there wouldn’t be any advantage. Suppose there are two stations that want totransmit at the same time, and they both use RTS/CTS. If the RTS frame is as long asa DATA frames, the channel would be wasted for as long as it would have beenwasted for two colliding DATA frames. Thus, the RTS/CTS exchange is only useful when the RTS/CTS frames are significantly smaller than the DATA frames.11.Initially the switch has an entry in its forwarding table which associates the wirelessstation with the earlier AP. When the wireless station associates with the new AP, the new AP creates a frame with the wireless station’s MAC address and broadcasts the frame. The frame is received by the switch. This forces the switch to update itsforwarding table, so that frames destined to the wireless station are sent via the new AP.12.Any ordinary Bluetooth node can be a master node whereas access points in 802.11networks are special devices (normal wireless devices like laptops cannot be used as access points).13.False14.“Opportunistic Scheduling” refers to matching the physical layer protocol to channelconditions between the sender and the receiver, and choosing the receivers to which packets will be sent based on channel condition. This allows the base station to make best use of the wireless medium.15.UMTS to GSM and CDMA-2000 to IS-95.16.The data plane role of eNodeB is to forward datagram between UE (over the LTEradio access network) and the P-GW. Its control plane role is to handle registration and mobility signaling traffic on behalf of the UE.The mobility management entity (MME) performs connection and mobility management on behalf of the UEs resident in the cell it controls. It receives UE subscription information from the HHS.The Packet Data Network Gateway (P-GW) allocates IP addresses to the UEs and performs QoS enforcement. As a tunnel endpoint it also performs datagram encapsulation/decapsulation when forwarding a datagram to/from a UE.The Serving Gateway (S-GW) is the data-plane mobility anchor point as all UE traffic will pass through the S-GW. The S-GW also performs charging/billing functions and lawful traffic interception.17.In 3G architecture, there are separate network components and paths for voice anddata, i.e., voice goes through public telephone network, whereas data goes through public Internet. 4G architecture is a unified, all-IP network architecture, i.e., both voice and data are carried in IP datagrams to/from the wireless device to severalgateways and then to the rest of the Internet.The 4G network architecture clearly separates data and control plane, which is different from the 3G architecture.The 4G architecture has an enhanced radio access network (E-UTRAN) that is different from 3G’s radio access network UTRAN.18.No. A node can remain connected to the same access point throughout its connectionto the Internet (hence, not be mobile). A mobile node is the one that changes its point of attachment into the network over time. Since the user is always accessing theInternet through the same access point, she is not mobile.19.A permanent address for a mobile node is its IP address when it is at its homenetwork. A care-of-address is the one its gets when it is visiting a foreign network.The COA is assigned by the foreign agent (which can be the edge router in theforeign network or the mobile node itself).20.False21.The home network in GSM maintains a database called the home location register(HLR), which contains the permanent cell phone number and subscriber profileinformation about each of its subscribers. The HLR also contains information about the current locations of these subscribers. The visited network maintains a database known as the visitor location register (VLR) that contains an entry for each mobile user that is currently in the portion of the network served by the VLR. VLR entries thus come and go as mobile users enter and leave the network.The edge router in home network in mobile IP is similar to the HLR in GSM and the edge router in foreign network is similar to the VLR in GSM.22.Anchor MSC is the MSC visited by the mobile when a call first begins; anchor MSCthus remains unchanged during the call. Throughout the call’s duration and regardless of the number of inter-MSC transfers performed by the mobile, the call is routed from the home MSC to the anchor MSC, and then from the anchor MSC to the visitedMSC where the mobile is currently located.23.a) Local recoveryb) TCP sender awareness of wireless linksc) Split-connection approachesChapter 7 ProblemsProblem 1Output corresponding to bit d 1 = [-1,1,-1,1,-1,1,-1,1]Output corresponding to bit d 0 = [1,-1,1,-1,1,-1,1,-1]Problem 2Sender 2 output = [1,-1,1,1,1,-1,1,1]; [ 1,-1,1,1,1,-1,1,1]Problem 3181111)1()1(111111)1()1(1112=⨯+⨯+-⨯-+⨯+⨯+⨯+-⨯-+⨯=d 181111)1()1(111111)1()1(1122=⨯+⨯+-⨯-+⨯+⨯+⨯+-⨯-+⨯=dProblem 4Sender 1: (1, 1, 1, -1, 1, -1, -1, -1)Sender 2: (1, -1, 1, 1, 1, 1, 1, 1)Problem 5a) The two APs will typically have different SSIDs and MAC addresses. A wirelessstation arriving to the café will associate with one of the SSIDs (that is, one of the APs). After association, there is a virtual link between the new station and the AP. Label the APs AP1 and AP2. Suppose the new station associates with AP1. When the new station sends a frame, it will be addressed to AP1. Although AP2 will alsoreceive the frame, it will not process the frame because the frame is not addressed to it. Thus, the two ISPs can work in parallel over the same channel. However, the two ISPs will be sharing the same wireless bandwidth. If wireless stations in different ISPs transmit at the same time, there will be a collision. For 802.11b, the maximum aggregate transmission rate for the two ISPs is 11 Mbps.b) Now if two wireless stations in different ISPs (and hence different channels) transmitat the same time, there will not be a collision. Thus, the maximum aggregatetransmission rate for the two ISPs is 22 Mbps for 802.11b.Problem 6Suppose that wireless station H1 has 1000 long frames to transmit. (H1 may be an AP that is forwarding an MP3 to some other wireless station.) Suppose initially H1 is the onlystation that wants to transmit, but that while half-way through transmitting its first frame, H2 wants to transmit a frame. For simplicity, also suppose every station can hear every other station’s signal (that is, no hidden terminals). Before transmitting, H2 will sense that the channel is busy, and therefore choose a random backoff value.Now suppose that after sending its first frame, H1 returns to step 1; that is, it waits a short period of times (DIFS) and then starts to transmit the second frame. H1’s second frame will then be transmitted while H2 is stuck in backoff, waiting for an idle channel. Thus, H1 should get to transmit all of its 1000 frames before H2 has a chance to access the channel. On the other hand, if H1 goes to step 2 after transmitting a frame, then it too chooses a random backoff value, thereby giving a fair chance to H2. Thus, fairness was the rationale behind this design choice.Problem 7A frame without data is 32 bytes long. Assuming a transmission rate of 11 Mbps, the time to transmit a control frame (such as an RTS frame, a CTS frame, or an ACK frame) is (256 bits)/(11 Mbps) = 23 usec. The time required to transmit the data frame is (8256 bits)/(11 Mbps) = 751DIFS + RTS + SIFS + CTS + SIFS + FRAME + SIFS + ACK= DIFS + 3SIFS + (3*23 + 751) usec = DIFS + 3SIFS + 820 usecProblem 8a) 1 message/ 2 slotsb) 2 messages/slotc) 1 message/slota)i) 1 message/slotii) 2 messages/slotiii) 2 messages/slotb)i) 1 message/4 slotsii) slot 1: Message A→ B, message D→ Cslot 2: Ack B→ Aslot 3: Ack C→ D= 2 messages/ 3 slotsiii)slot 1: Message C→ Dslot 2: Ack D→C, message A→ BRepeatslot 3: Ack B→ A= 2 messages/3 slotsProblem 10a)10 Mbps if it only transmits to node A. This solution is not fair since only A is gettingserved. By “fair” it m eans that each of the four nodes should be allotted equal number of slots.b)For the fairness requirement such that each node receives an equal amount of dataduring each downstream sub-frame, let n1, n2, n3, and n4 respectively represent the number of slots that A, B, C and D get.Now,data transmitted to A in 1 slot = 10t Mbits(assuming the duration of each slot to be t)Hence,Total amount of data transmitted to A (in n1 slots) = 10t n1Similarly total amounts of data transmitted to B, C, and D equal to 5t n2, 2.5t n3, and t n4 respectively.Now, to fulfill the given fairness requirement, we have the following condition:10t n1 = 5t n2 = 2.5t n3 = t n4Hence,n2 = 2 n1n3 = 4 n1n4 = 10 n1Now, the total number of slots is N. Hence,n1+ n2+ n3+ n4 = Ni.e. n1+ 2 n1 + 4 n1 + 10 n1 = Ni.e. n1 = N/17Hence,n2 = 2N/17n3 = 4N/17n4 = 10N/17The average transmission rate is given by:(10t n1+5t n2+ 2.5t n3+t n4)/tN= (10N/17 + 5 * 2N/17 + 2.5 * 4N/17 + 1 * 10N/17)/N= 40/17 = 2.35 Mbpsc)Let node A receives twice as much data as nodes B, C, and D during the sub-frame.Hence,10tn1 = 2 * 5tn2 = 2 * 2.5tn3 = 2 * tn4i.e. n2 = n1n3 = 2n1n4 = 5n1Again,n1 + n2 + n3 + n4 = Ni.e. n 1+ n1 + 2n1 + 5n1 = Ni.e. n1 = N/9Now, average transmission rate is given by:(10t n1+5t n2+ 2.5t n3+t n4)/tN= 25/9 = 2.78 MbpsSimilarly, considering nodes B, C, or D receive twice as much data as any other nodes, different values for the average transmission rate can be calculated.Problem 11a)No. All the routers might not be able to route the datagram immediately. This isbecause the Distance Vector algorithm (as well as the inter-AS routing protocols like BGP) is decentralized and takes some time to terminate. So, during the time when the algorithm is still running as a result of advertisements from the new foreign network, some of the routers may not be able to route datagrams destined to the mobile node.b)Yes. This might happen when one of the nodes has just left a foreign network andjoined a new foreign network. In this situation, the routing entries from the oldforeign network might not have been completely withdrawn when the entries from the new network are being propagated.c)The time it takes for a router to learn a path to the mobile node depends on thenumber of hops between the router and the edge router of the foreign network for the node.Problem 12If the correspondent is mobile, then any datagrams destined to the correspondent would have to pass through the correspondent’s home agent. The foreign agent in the network being visited would also need to be involved, since it is this foreign agent thatnotifies the correspondent’s home agent of the location of the correspondent. Datagrams received by the correspondent’s home agent would need to be encapsulated/tunneled between the correspondent’s home agent and for eign agent, (as in the case of the encapsulated diagram at the top of Figure 6.23.Problem 13Because datagrams must be first forward to the home agent, and from there to the mobile, the delays will generally be longer than via direct routing. Note that it is possible, however, that the direct delay from the correspondent to the mobile (i.e., if the datagram is not routed through the home agent) could actually be smaller than the sum of the delay from thecorrespondent to the home agent and from there to the mobile. It would depend on the delays on these various path segments. Note that indirect routing also adds a home agent processing (e.g., encapsulation) delay.Problem 14First, we note that chaining was discussed at the end of section 6.5. In the case of chaining using indirect routing through a home agent, the following events would happen: •The mobile node arrives at A, A notifies the home agent that the mobile is now visiting A and that datagrams to the mobile should now be forwarded to thespecified care-of-address (COA) in A.•The mobile node moves to B. The foreign agent at B must notify the foreign agent at A that the mobile is no longer resident in A but in fact is resident in Band has the specified COA in B. From then on, the foreign agent in A willforward datagrams it receives that are addressed to the mobile’s COA in A to t he mobile’s COA in B.•The mobile node moves to C. The foreign agent at C must notify the foreign agent at B that the mobile is no longer resident in B but in fact is resident in C and has the specified COA in C. From then on, the foreign agent in B will forwarddatagrams it receives (from the foreign agent in A) that are addressed to themobile’s COA in B to the mobile’s COA in C.Note that when the mobile goes offline (i.e., has no address) or returns to its home network, the datagram-forwarding state maintained by the foreign agents in A, B and C must be removed. This teardown must also be done through signaling messages. Note that the home agent is not aware of the mobile’s mobility beyond A, and that the correspondent is not at all aware of the mobil e’s mobility.In the case that chaining is not used, the following events would happen: •The mobile node arrives at A, A notifies the home agent that the mobile is now visiting A and that datagrams to the mobile should now be forwarded to thespecified care-of-address (COA) in A.•The mobile node moves to B. The foreign agent at B must notify the foreign agent at A and the home agent that the mobile is no longer resident in A but infact is resident in B and has the specified COA in B. The foreign agent in A can remove its state about the mobile, since it is no longer in A. From then on, thehome agent will forward datagrams it receives that are addressed to the mobile’sCOA in B.•The mobile node moves to C. The foreign agent at C must notify the foreign agent at B and the home agent that the mobile is no longer resident in B but in fact is resident in C and has the specified COA in C. The foreign agent in B canremove its state about the mobile, since it is no longer in B. From then on, thehome agent will forward datagrams it receives that are addressed to the mobile’sCOA in C.When the mobile goes offline or returns to its home network, the datagram-forwarding state maintained by the foreign agent in C must be removed. This teardown must also bedone through signaling messages. Note that the home agent is always aware of the mobile’s cu rrent foreign network. However, the correspondent is still blissfully unaware of the mobile’s mobility.Problem 15Two mobiles could certainly have the same care-of-address in the same visited network. Indeed, if the care-of-address is the address of the foreign agent, then this address would be the same. Once the foreign agent decapsulates the tunneled datagram and determines the address of the mobile, then separate addresses would need to be used to send the datagrams separately to their different destinations (mobiles) within the visited network.Problem 16If the MSRN is provided to the HLR, then the value of the MSRN must be updated in the HLR whenever the MSRN changes (e.g., when there is a handoff that requires the MSRN to change). The advantage of having the MSRN in the HLR is that the value can be provided quickly, without querying the VLR. By providing the address of the VLR Rather than the MSRN), there is no need to be refreshing the MSRN in the HLR.。
数据通信的书
数据通信的书数据通信是一个广泛而深入的领域，有很多书籍涵盖了不同方面的知识，从基础概念到高级技术。
以下是一些涵盖数据通信领域不同方面的经典书籍：《数据通信与网络》（Data Communications and Networking） - 作者：Behrouz A. Forouzan这本书全面介绍了数据通信和网络的基本概念，包括协议、网络拓扑、传输媒体、网络层、传输层等。
《计算机网络：自顶向下方法》（Computer Networking: A Top-Down Approach） - 作者：James F. Kurose, Keith W. Ross该书以自顶向下的方法介绍计算机网络，包括了网络协议、传输层、应用层等，适合初学者。
《TCP/IP详解》（TCP/IP Illustrated）- 作者：W. Richard Stevens这是一系列的书籍，深入讲解了 TCP/IP 协议栈。
每一本书都涵盖了某个特定领域，如《TCP/IP协议卷1：协议》、《TCP/IP协议卷2：实现》等。
《高性能TCP/IP网络设计》（High-Performance TCP/IP Networking） - 作者：Mahbub Hassan, Raj Jain该书重点关注高性能网络设计和优化，适合对网络性能有较高要求的专业人员。
《数字通信》（Digital Communications） - 作者：John G. Proakis, Masoud Salehi该书介绍了数字通信的基本原理，包括调制解调、信道编码、调制解调器设计等。
《通信网络基础》（Communication Networks: A First Course） - 作者：Jean Walrand, Pravin Varaiya该书适合初学者，涵盖了通信网络的基础概念和原理。
计算机网络(自顶向下)
在数据传输前，不必建立端到端的连接 只要下一个节点空闲，即可传输
信道的使用率较高
数据的传输采用存储转发，延时不可估计 数据中必须包含地址域 接收到的分组不一定按序，可能还需重组 适合传输文本型数据
计算机网络和因特网概述
7
分组时延的4种来源
1. 节点处理: 检查比特差错 决定输出链路 2. 排队 等待输出链路传输的时间 取决于路由器拥塞的等级
考试时间：2014-04-27（周日）
下午1.30-3.10（100分钟）
1.考试题型
选择题（10*2）
名词性解释（10*2） 简答题（5*8） 综合题（2*10）
第1章 计算机网络和因特网概述
Computer Networks and the Internet
计算机网络：自顶向下方法 (原书第四版) 陈鸣译，机械工业出版社，2010年
邮件 服务器 用户 代理
输出 报文队列 用户邮箱
用户 代理
SMTP
用户代理 亦称为 “邮件阅读器” 写作、编辑、阅读邮件报文 例如Outlook, Apple Mail, Netscape Messenger
存储在服务器上的报文的达到
SMTP
SMTP
邮件 服务器
用户 代理 用户 代理
23
FTP: 文件传输协议
FTP FTP 用户接口 客户机 主机上 的用户 本地文件 系统 文件传输 FTP 服务器 远程文件系统
传输文件到远程主机或从远程主机获取文件
客户机/服务器模型
客户机: 发起传输一侧(到/来自远程之一) 服务器: 远程主机
ftp: RFC 959 ftp 服务器: 端口 21
Computer Networking A Top-Down Approach, 5th Edition 第5章
But maybe errors nonetheless?
Data Link Layer 5-13
Checksumming: Cyclic Redundancy Check
view data bits, D, as a binary number choose r+1 bit pattern (generator), G goal: choose r CRC bits, R, such that
error correction:
half-duplex and full-duplex
receiver identifies and corrects bit error(s) without resorting to retransmission
with half duplex, nodes at both ends of link can transmit, but not at same time
reliable delivery between adjacent nodes
we learned how to do this already (chapter 3)! seldom used on low bit-error link (fiber, some twisted pair) wireless links: high error rates • Q: why both link-level and end-end reliability?
widely used in practice (Ethernet, 802.11 WiFi, ATM)
instantiation and implementation of various link layer technologies
计算机网络自顶向下方法第四版版
? 从 GSM 演化而来 ? 数据在多个信道发送(如果可能的话)
? Enhanced Data rates for Global Evolut(iEoDnGE)
? 也从GSM 演化而来,使用增强调制 ? 数据率高达 384K
? 节点将它们组织成为一 个网络：在它们自己中 选路
无线网和移动网
9
无线链路的特征
不同于有线链路….
? 衰减的信号强度：当无线电信号传播通过物质时， 信号削弱(路径损失)
? 来自其他源的干扰:标准的无线网络频率 (如2.4 GHz) 由其他设备共享 (如电话);设备(发动机) 干扰
? 多径传播:无线电信号反射离开物体，以稍微不同 的时间到达目的地
AP BSS 1基站= 接入点(AP)
? 基本服务集 (BSS) (又称为
“单元”)在基础设施模式中
包含:
hub, switch or 路由器
? 无线主机
? 接入点(AP):基站
? 自组织模式:仅有主机
AP
BSS 2
无线网和移动网 17
802.11:信道,关联
思想: 允许发送方“预约”信道，而不是数据帧的随机
访问：避免长数据帧的碰撞
? 发送方首先使用CSMA 向BS 传输小请求发送(RTS)分组
? RTS 仍可能与其他碰撞(但它们较短) ? BS 为响应RTS 广播允许发送CTS ? RTS 为所有站点听到
? 发送方传输数据帧 ? 其他站点推迟传输
使用小预约分组，避免数据帧完全碰撞!
? 6.3 IEEE 802.11无线局域网 (“wi-fi”) ? 6.4蜂窝因特网接入
计算机网络-1
计算机网络
江西师范大学软件学院
2-24
1.2.1 企业和家庭接入
局域网连接端用户与边缘路由器 常见名词：网线、路由器、交换机 以太网：使用双绞铜线或同轴电缆，将一 些端系统彼此连接起来，并与边缘路由器 连接
计算机网络
江西师范大学软件学院
2-25
1.2.1 无线接入
两类无线接入方式： 1）无线局域网： 无线—基站（几十米），基站为无线接入点， 与有线的因特网相连 2）广域无线接入网：无线—基站(几万米） 无线-基站（数万米），蜂窝网，3G、4G 4G：中国移动推出4G套餐，40元包300兆流 量，按照每秒百兆的速率，这个套餐3秒就用 完了，3秒40元，一个小时就是48000元。如 果晚上忘了关闭4G连接，一觉醒来，你的房 子都快成移动公司的了。（李国庆）
江西师范大学软件学院 2-10
计算机网络
课程情况
教学课时：64学时 实验课时：16学时 考核方式：平时15%，实验15%， 20%分 组工程实践，期末考试50% 工程实践：3人小组，形式：课程、网络小 程序、网络动画
计算机网络
江西师范大学软件学院
2-11
•分组交换 •因特网体系结构
第1章 计算机网络和因特网
网络核心:
1 接入网
边缘路由器：端系统到任何其他远程端系 统的路径上的第一台路由器 接入网：将端系统连接到其边缘路由器的 物理链路 三种类型：家庭接入、企业接入、广域无 线接入
计算机网络
江西师范大学软件学院
2-23
1.2.1 家庭接入
家庭接入：将家庭端系统与边缘路由器相连接。 数字用户线、电缆、光纤 数字用户线接入：普通模拟电话线用拨号调制解 调器与住宅ISP相连，同时承载数据和电话信号， 不冲突。图1-5 调制解调器：将PC输出的数字信号与电话线上的 模拟信号相互转换 电缆因特网接入（宽带网络）：目前主要使用数 字用户线和光纤同轴电缆，通过划分为不同频段 实现电话和上网的并行，使用电缆调制解调器 光纤到户FTTH
ComputerNetworkingATopDownApproach第六版课程设计
Computer Networking: A Top-Down Approach 第六版课程设计课程简介该课程以《计算机网络 - 自顶向下方法第六版》（Computer Networking: A Top-Down Approach, 6th Edition）为主要教材，针对计算机网络的基本概念、原理、协议、技术等方面进行教学。
该课程旨在让学生了解与掌握计算机网络的基础知识和核心技术，从而为学生今后从事计算机网络相关领域的研究与实践打下坚实的基础。
课程目标1.理解计算机网络的基本概念与架构，掌握计算机网络体系结构的发展历程及主要特点。
2.了解计算机网络的传输媒介、传输技术与网络协议，包括物理层、数据链路层、网络层、传输层、应用层等各个方面。
3.掌握TCP/IP协议的基本原理与实现，以及常用的网络应用协议，如HTTP、FTP、DNS等。
4.熟悉计算机网络的性能分析与QoS管理技术，了解网络安全、管理与优化等方面的知识。
主要内容本课程的主要内容按照《计算机网络 - 自顶向下方法第六版》的章节顺序组织，涵盖以下方面：1.网络概述与体系结构2.应用层3.运输层4.网络层5.数据链路层6.其他网络技术与协议7.网络安全、管理、优化与性能分析课程作业1.利用Wireshark等网络抓包工具对HTTP协议进行抓包分析，分析HTTP协议的请求与响应报文格式、状态码、缓存机制等关键信息。
2.设计并实现一个简单的FTP服务器与FTP客户端，要求具备相应的文件传输功能、断点续传功能和权限控制等基本功能。
3.编写Python程序调用socket库模仿ping和traceroute命令，实现对目标主机的网络连通性测试、最短路径跟踪等功能。
实验环境1.操作系统：Windows 10 / Ubuntu 18.04 LTS2.虚拟机：VirtualBox / VMware3.网络模拟器：GNS3 / Packet Tracer4.编程语言：Python / C语言 / Java5.网络抓包工具：Wireshark / tcpdump实验指导书每次实验前均提供详细的实验指导书，包括实验设计、实验步骤、实验注意事项、实验结果分析等内容。
Computer Networking A Top Down Approach (8)
A more sophisticated encryption approach
n substitution ciphers, M1,M2,…,Mn cycling pattern:
e.g., n=4: M1,M3,M4,M3,M2; M1,M3,M4,M3,M2; ..
Network Security 8-3
What is network security?
confidentiality: only sender, intended receiver should
“understand” message contents sender encrypts message receiver decrypts message
key need not be just n-bit pattern
Network Security
Computer Networking: A Top Down Approach
6th edition Jim Kurose, Keith Ross Addison-Wesley March 2012
Chapter 8: Network Security
Chapter goals:
understand principles of network security:
Network Security
8-6
There are bad guys (and girls) out there!
Q: What can a “bad guy” do? A: A lot! See section 1.6 eavesdrop: intercept messages actively insert messages into connection impersonation: can fake (spoof) source address in packet (or any field in packet) hijacking: “take over” ongoing connection by removing sender or receiver, inserting himself in place denial of service: prevent service from being used by others (e.g., by overloading resources)
Computer Networking A Top-Down Approach Featuring the Internet 3rd
McGraw-Hill©The McGraw-Hill Companies, Inc., 2000Figure 1-1Internet todayFigure 2-1OSIModelMcGraw-Hill©The McGraw-Hill Companies, Inc., 2000Figure 2-3An exchange using the OSI modelFigure 2-4Physical LayerMcGraw-Hill ©The McGraw-Hill Companies, Inc., 2000Figure 2-5Data Link Layer Figure 2-6Node-to-node delivery McGraw-Hill ©The McGraw-Hill Companies, Inc., 2000Figure 2-7Network LayerFigure 2-8End-to-end deliveryMcGraw-Hill©The McGraw-Hill Companies, Inc., 2000Figure 2-9Transport LayerFigure 2-10Reliable end-to-end delivery of a messageMcGraw-Hill©The McGraw-Hill Companies, Inc., 2000Figure 2-11Session LayerFigure 2-12Presentation LayerMcGraw-Hill©The McGraw-Hill Companies, Inc., 2000Figure 2-13Application LayerFigure 2-14Summary of layersMcGraw-Hill©The McGraw-Hill Companies, Inc., 2000Figure 2-15TCP/IP and OSI modelFigure 2-16Addresses in TCP/IPMcGraw-Hill©The McGraw-Hill Companies, Inc., 2000Figure 2-17Relation-shipoflayersandaddressesin TCP/IPFigure 3-2CSMA/CDMcGraw-Hill©The McGraw-Hill Companies, Inc., 2000Figure 3-26Connecting devicesFigure 3-27RepeaterMcGraw-Hill©The McGraw-Hill Companies, Inc., 2000Figure 3-28HubsFigure 3-29BridgeMcGraw-Hill©The McGraw-Hill Companies, Inc., 2000Figure 3-30Routing exampleFigure 4-2Occupation of the address spaceMcGraw-Hill©The McGraw-Hill Companies, Inc., 2000Figure 4-3Finding the class in binary notationFigure 4-4Finding the address classMcGraw-Hill©The McGraw-Hill Companies, Inc., 2000Figure 4-5Finding the class in decimal notationFigure 4-6Netid and hostidMcGraw-Hill ©The McGraw-Hill Companies, Inc., 2000Figure 4-7Blocks in class A Figure 4-8Blocks in class B McGraw-Hill ©The McGraw-Hill Companies, Inc., 2000Figure 4-9Blocks in class CFigure 4-10Masking conceptMcGraw-Hill©The McGraw-Hill Companies, Inc., 2000Figure 4-11AND operationFigure 4-14Example of direct broadcast addressMcGraw-Hill©The McGraw-Hill Companies, Inc., 2000Figure 4-18Example of loopback addressFigure 4-19Sample internet。
计算机网络_自顶向下方法_第9章_网络管理.TopDownV3-9
网络管理的架构
定义:
管理实体
管理实体 数据
代理
数据
被管设备 代理 数据
网络管理协议
被管设备包含被管对 象，它们的数据被收 集在管理信息库(MIB)
中
被管设备 代理 代理 数据 被管设备 数据
被管设备
网络管理
5
网络管理标准
OSI CMIP 公共管理信息协议 于二是世纪80年代设计: 统一的网络管理标准 标准化过程极为缓慢 SNMP: 简单网络管理协议 来源于因特网(SGMP) 出发点是简单 迅速地被展开、采用 成长：规模，复杂性 当前：SNMP V3 事实上的网络管理标准
网络管理 13
OSI
对象标识 树
核对 www.alvestrand.no/harald/objectid/top.html
网络管理 14
SNMP协议
传递MIB信息、命令的两种方式:
管理实体 管理实体
请求
响应
代理
数据
Trap报文
代理 数据
被管设备
被管设备
请求/响应模式
ห้องสมุดไป่ตู้
trap模式
网络管理 15
SNMP协议：报文类型
OBJECT-TYPE: ipInDelivers ipInDelivers OBJECT TYPE SYNTAX Counter32 MAX-ACCESS read-only STATUS current DESCRIPTION “The total number of input datagrams successfully delivered to IP userprotocols (including ICMP)” ::= { ip 9} MODULE-IDENTITY: ipMIB
有关信息技术的书籍
有关信息技术的书籍以下是一些关于信息技术的经典书籍：1.《计算机网络：自顶向下方法》（Computer Networking: A Top-Down Approach）- James F. Kurose 和 Keith W. Ross这是一本经典的计算机网络教材，涵盖了计算机网络的各个层次，从应用层到物理层。
2.《计算机组成与设计：硬件/软件接口》（Computer Organization and Design: The Hardware/Software Interface）- David A. Patterson 和 John L. Hennessy这本书探讨了计算机系统的硬件和软件之间的交互关系，介绍了计算机的组织和设计。
3.《算法导论》（Introduction to Algorithms）- Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest 和 Clifford Stein这是一本经典的算法教材，涵盖了算法设计和分析的基本知识，适合深入了解算法和数据结构。
4.《人工智能：现代方法》（Artificial Intelligence: A Modern Approach）- Stuart Russell 和 Peter Norvig这本书会介绍人工智能的各个方面，包括搜索算法、知识表示、机器学习、自然语言处理等内容。
5.《信息系统分析与设计》（Systems Analysis and Design）- Alan Dennis, Barbara Haley Wixom 和 Roberta M. Roth这本书介绍了信息系统分析和设计的方法和技术，包括需求分析、系统设计、数据库设计等内容。
6.《软件工程：现代方法》（Software Engineering: A Practitioner's Approach）- Roger S. Pressman这本书详细介绍了软件工程的原理和实践，包括需求工程、软件设计、软件测试等方面的内容。
《计算机网络A》课程教学大纲
《计算机网络A》课程教学大纲一、课程基本信息1．课程代码：219210012．课程中文名称：计算机网络A课程英文名称：Computer Networks A3．面向对象：软件工程专业本科二年级学生4．开课学院（课部）、系（中心、室）：信息工程学院、软件工程系5．总学时数：56讲课学时数：44 ，实验学时数：126．学分数： 3.57．授课语种：中文，考试语种：中文8．教材：Jim Kurose and Keith Ross,Computer Networking: A Top-Down Approach, 5th Edition 2010.二、课程内容简介本课程以广泛使用的Internet为例，系统介绍网络基本概念和核心内容。
课程使得学生对网络重要的核心内容有个基本了解，并能使用所学知识，解决涉及网络各个层次的问题。
内容主要包括应用层及其程序设计模式客户机服务器模型、P2P模型，传输层及如何保证可靠传输，网络层及分布式寻路算法，链路层及物理层所涉及的相关问题。
三、课程的地位、作用和教学目标课程涵盖面广，课程涉及一些大作业的项目（用PYTHON语言编写）四、与本课程相联系的其他课程面向对象程序设计，概率论五、教学基本要求通过理论讲解从整体上理解计算机网络及互联网作为一个大的软件工程的结构及核心思想和算法。
对各种基本互联网名词及概念有一个基本的了解。
深刻理解INTERNET的主要思想，包括：可靠性及容错：在不可靠的环境下，构建可靠的传输。
分布式算法：在没有全局地图的情形下，多个设备根据算法，协同设计出到达目的地路径。
并通过设计出相关的算法，调试等实践环节，真正理解计算机网络的核心思想。
同时了解计算机网络的最新进展。
六、考核方式与评价结构比例闭卷考试结合平时作业和大项目七、教学参考资料•W. Richard Stevens, Bill Fenner, and Andrew M. Rudoff,UNIX Network Programming, Volume 1: The Sockets Networking API, 3rd Edition Addison-Wesley, 2003.•W. Richard Stevens，TCP/IP Illustrated, Volume 1: The Protocols, 1993•Andrew Tanenbaum, Computer Networks, 4th Edition 2002.•谢希仁,计算机网络, 第5版，电子工业出版社2008.•谢希仁,计算机网络释疑与习题解答, 电子工业出版社，2011.•Luiz Andre Barroso and Urs Holzle, An Introduction to the Design of Warehouse-Scale Machiness, 2009.。
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3G 2G
Outdoor
50-200m
Mid-range outdoor
200m – 4 Km
Long-range outdoor
5Km – 20 Km
Wireless, Mobile Networks 6-7
Elements of a wireless network
infrastructure mode base station connects mobiles into wired network handoff: mobile changes base station providing connection into wired network
architecture standards (e.g., GSM)
Mobility 6.5 Principles: addressing and routing to mobile users 6.6 Mobile IP 6.7 Handling mobility in cellular networks 6.8 Mobility and higherlayer protocols 6.9 Summary
network infrastructure
Wireless, Mobile Networks
6-6
Characteristics of selected wireless link standards
200 Data rate (Mbps) 54 5-11 4 1 .384 .056 Indoor
used in several wireless broadcast channels (cellular, satellite, etc) standards unique “code” assigned to each user; i.e., code set partitioning all users share same frequency, but each user has own “chipping” sequence (i.e., code) to encode data encoded signal = (original data) X (chipping sequence) decoding: inner-product of encoded signal and chipping sequence allows multiple users to “coexist” and transmit simultaneously with minimal interference (if codes are “orthogonal”)
Wireless, Mobile Networks 6-3
Elements of a wireless network
wireless hosts laptop, PDA, IP phone run applications may be stationary (non-mobile) or mobile
• SNR may change with mobility: dynamically adapt physical layer (modulation technique, rate)
BER
10-4 10-5 10-6 10-7
10
20
30
40
SNR(dB) QAM256 (8 Mbps) QAM16 (4 Mbps) BPSK (1 Mbps)
space
Signal attenuation:
B, A hear each other B, C hear each other A, C can not hear each other interfering at B
Wireless, Mobile Networks 6-13
Code Division Multiple Access (CDMA)
Wireless, Mobile Networks
6-9
Wireless network taxonomy
single hop
infrastructure (e.g., APs)
multiple hops
host may have to relay through several wireless nodes to connect to larger Internet: mesh net no base station, no connection to larger Internet. May have to relay to reach other a given wireless node MANET, VANET
10-1 10-2 10-3
SNR versus BER tradeoffs given physical layer:
increase power -> increase SNR->decrease BER given SNR: choose physical layer that meets BER requirement, giving highest thruput
Wireless, Mobile Networks 6-10
Wireless Link Characteristics (1)
Differences from wired link …. decreased signal strength: radio signal attenuates as it propagates through matter (path loss) interference from other sources: standardized wireless network frequencies (e.g., 2.4 GHz) shared by other devices (e.g., phone); devices (motors) interfere as well multipath propagation: radio signal reflects off objects ground, arriving ad destination at slightly different times …. make communication across (even a point to point) wireless link much more “difficult”
10-30m 802.15 802.11n 802.11a,g 802.11b 802.11a,g point-to-point 802.16 (WiMAX) UMTS/WCDMA-HSPDA, CDMA2000-1xEVDO
data 3G cellular enhanced
UMTS/WCDMA, CDMA2000 IS-95, CDMA, GSM
A C B
A’s signal strength C’s signal strength
B
C
A
Hidden terminal problem
B, A hear each other B, C hear each other A, C can not hear each other means A, C unaware of their interference at B
laptops, Internet-enabled phones promise anytime untethered Internet access
two important (but different) challenges
wireless: communication over wireless link mobility: handling the mobile user who changes point
Computer Networks
Dr. Guifeng Zheng (郑贵锋) gfzheng@
Computer Networking: A Top Down Approach
5th edition. Jim Kurose, Keith Ross Addison-Wesley, April 2009.
wireless does not always mean mobility
network infrastructure
Wireless, Mobile Networks
6-4
Elements of a wireless network
base station typically connected to wired network relay - responsible for sending packets between wired network and wireless host(s) in its “area” e.g., cell towers, 802.11 access points
network infrastructure
Wireless, Mobile Networks
6-5
Elements of a wireless network
wireless link typically used to connect mobile(s) to base station also used as backbone link multiple access protocol coordinates link access various data rates, transmission distance
Wireless, Mobile Networks 6-11
Wireless Link Characteristics (2)
SNR: signal-to-noise ratio larger SNR – easier to extract signal from noise (a “good thing”)