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填空1层次型域名系统。
2域3第一，主机名字的唯一性；第二，要便于管理；第三，要便于映射。
4商业组织,教育机构，军事部门5一系列的域名服务器6存储一转发7POP3,IMAP4,SMTP,MIME8用户名，at,邮箱服务器的名字。
9浏览Internet上WWW页面的软件。
IO0从服务端下载文件11客户程序，服务程序12用户就可以像远程主机本地终端一样地进行工作，并可使用远程主机对外开放的全部资源。
13客户机/服务器，TeInet服务器、Te1r1et客户机和通信协议组成14DNS15一是电信的铜缆接入技术；二是基于有线电视CATV网传输设施的电缆调制解调器接入技术；三是基于光缆的宽带光纤接入技术；四是基于无线电传输手段的无线接入技术。
16NA是网络地址转换简称。
该技术可以将内部网络地址转换成指定的IP地址，一个局域网只需使用少量IP地址（甚至是1个）即可实现内部地址网络内与外部网络的通信需求。
判断：1对2错3错4错5对6错7对8错9对简答：IDNS的分层管理机制是采用一个规则的树型结构的名字空间，每个节点都有一个独立的节点名字2一个节点的域由该节点以及该节点以下的名字空间组成。
一个域是树状域名空间的一棵子树，每个域都有一个域名，域名定义了该域在分布式主机数据库中的位置。
3DNS系统采用的是一个联机分布式数据库系统，采用客户机/服务器模式。
4域名解析是由一系列的域名服务器来完成的。
域名服务器是运行在指定主机上的软件，能够完成从域名到IP地址的映射。
5Internet上的电子邮件客户机是电子邮件使用者用来收发、浏览存放在邮件服务器上的电子邮件的工具6当用户编辑好需要发送的邮件后，通过用户接口交给邮件传送程序。
发送信件时，邮件传送程序作为远程目的计算机邮件服务器的客户，与目的主机建立连接，并将邮件传送到目的主机。
接收方计算机的邮件传输程序在收到邮件后，将邮件存放在接收方的邮箱中，等待用户来读取。
由于用户接口的屏蔽作用，用户在发送和接收邮件时看不见邮件传输程序的工作情况。
1对网络电缆采用的连接器不正确可能产生什么后果？会将数据转发到错误的节点。
通过该电缆传输的数据可能发生信号丢失。
将对该电缆中传输的数据采用不正确的信号方法。
该电缆中发送的数据所采用的编码方法将更改，用于补偿不当连接。
2大多数企业LAN 中的双绞线网络电缆使用哪种连接器？BNCRJ-11RJ-45F 型3在网络中传输数据时，物理层的主要作用是什么？创建信号以表示介质上每个帧中的比特为设备提供物理编址确定数据包的网络通路控制数据对介质的访问4数据传输的三种量度标准是什么？（选择三项）实际吞吐量频率幅度吞吐量串扰带宽5UTP 电缆的哪个特征有助于减少干扰的影响？屏蔽层中的金属编织包裹核心的反射涂层电缆中的线对绕绞外皮中的绝缘材料6XYZ 公司正在其数据网络中安装新的电缆。
下列哪两种类型是新架设时最常用的电缆？（选择两项）同轴4 类UTP5 类UTP6 类UTPSTP7使用什么设备可以检查安装的光纤是否存在错误并检查其完整性和介质性能？光注入器OTDRTDR万用表8以下哪项是单模光缆的特征？一般使用LED 作为光源因为有多条光通路，核心相对较粗价格比多模低一般使用激光作为光源9请参见图示。
哪种 5 类电缆用于在主机 A 和主机 B 之间建立以太网连接？同轴电缆全反电缆交叉电缆直通电缆10以下哪项被视为选择无线介质的优点？主机移动更方便安全风险更低减少干扰的影响环境对有效覆盖面积的影响更小11如果在网络中使用非屏蔽双绞线铜缆，导致线对内串扰的原因是什么？相邻线对周围的磁场使用编织线屏蔽相邻线对从电缆远端反射回来的电波因两个节点尝试同时使用介质而导致的冲突12哪种光纤连接器支持全双工以太网？13哪种电缆通常与光缆相关联？主干电缆水平电缆跳线电缆工作区域电缆14在可能存在电气危险或电磁干扰的LAN 安装中，主干布线建议使用哪种类型的介质？同轴光纤5e 类UTP6 类UTP15网络中什么时候使用直通电缆？通过控制台端口连接路由器时连接两台交换机时连接主机与交换机时连接两台路由器时16以下哪些特征描述的是光缆？（选择两项）不受 EMI 或RF I 影响。
C C N A第一学期第五章答案集团文件版本号：（M928-T898-M248-WU2669-I2896-DQ586-M1988）最新CCNA第一学期第五章答案一个 MAC 地址包含 6 个字节。
前 3 个字节用于供应商标识，最后 3 个字节必须在同一 OUI 中分配唯一的值。
MAC 地址在硬件中实施。
网卡需要使用 MAC 地址来通过 LAN 通信。
IEEE 规范了 MAC 地址。
争用方法具有不确定性，没有受控访问方法中遇到的开销。
因为设备不需要轮流访问介质，所以无需跟踪次序。
争用方法在介质使用率高的情况下无法很好地扩展。
逻辑链路控制在软件中实施，能够让数据链路层与协议簇的上层协议通信。
网卡驱动程序软件直接与网卡上的硬件交互，可在 MAC 子层和物理介质之间传输数据。
逻辑链路控制在 IEEE 802.2 标准中指定。
IEEE 802.3 是定义不同以太网类型的一组标准。
MAC（介质访问控制）子层负责定位和检索介质上和介质外的帧。
组播 MAC 地址是一个特殊的十六进制数值，以 01-00-5E 开头。
然后将IP 组播组地址的低 23 位换算成以太网地址中剩余的 6 个十六进制字符，作为组播 MAC 地址的结尾。
MAC 地址的剩余位始终为“0”。
在此例中：224（十进制）=> 01-00-5E34（十进制）= 00100010（二进制）= 22（十六进制）56（十进制）= 00111000（二进制）= 38（十六进制）综合结果：01:00:5E:0B:22:38如果整个路径都基于以太网，则目的和源 MAC 地址将随着每个路由器跳数发生变化。
除非网络地址转换处于活动状态，否则目的 IP 地址在整个路径中保持不变。
请参见图示。
由于 PC1 需要向 PC2 发送数据包，因此 PC1 发出一个ARP 请求。
在这种情况下，接下来将发生什么？当一台网络设备要与相同网络中的另一台设备通信时，它会发送广播ARP 请求。
计算机网络第五章作业参考答案Chapter53.Consider the network of Fig. 5-7, but ignore the weights on thelines. Suppose that it uses flooding as the routing algorithm. If a packet sent by A to D has a maximum hop count of 3, list all the routes it will take. Also tell how many hops worth of bandwidth it consumes. Solution:It uses flooding as the routing algorithm, which means to transmit packets to all ports except the coming one. It will use:A,B,C,D,A,B,C,F,A,B,E,F,A,B,E,G,A,G,E,B,A,G,E,F,A,G,H,D,A,G,H,F.So,24 hops bandwidth are used.5.Consider the network of Fig. 5-12(a). Distance vector routing is used, and the following vectors have just come in to router C: from B: (5, 0, 8, 12, 6, 2); from D: (16, 12, 6, 0, 9, 10); and from E: (7, 6, 3, 9, 0, 4). The cost of the links from C to B, D and E, are 6, 3, and 5, respectively. What is C’s new routing table? Give both the outgoingline to use and the cost.Solution:If C transmit packet via B,A,B,C,D,E,F=(5,0,8,12,6,2)+(6,6,6,6,6,6)=(11,6,14,18,12,8)If C transmit packet g viaD,A,B,C,D,E,F=(16,12,6,0,9,10)+(3,3,3,3,3,3)=(19,15,9, 3, 12,13) If C transmit packet via E,A,B,C,D,E,F= (7,6,3,9,0,4) + (5,5,5,5,5,5) =(12,11, 8,14,5, 9), (11,6,0,3,5,8)So, C’s new routing table(from up to down is from C to A,B,C,D,E) Destination Cost outgoing line(next hop)A 11 BB 6 BC 0 -D 3 DE 5 EF 8 B9.Looking at the network of Fig. 5-6, how many packets are generatedby a broadcast from B,using(a) reverse path forwarding?(b) the sink tree?Solution:(a)Reverse path forwarding: when a broadcast packet arrives at a router, the router checks to see ifthe packet arrived on the link that is normally used for sending packets toward the source of the broadcast. This being the case, the router forward copies of it onto all links except the one it arrived on.DFhop1:2packetshop 2:4 packetshop 3:10 packetshop 4:8packetshop5:4packetstotal broadcast packets=2+4+10+8+4=28packets (b) the sink tree:Level1:2packetsLevel2:4packetsLevel3:5packetsLevel4:3packetsTotal broadcast packets=2+4+5+3=14packets.10.Consider the network of Fig. 5-15(a). Imagine that one new line is added, between F and G,but the sink tree of Fig. 5-15(b) remains unchanged. What changes occur to Fig. 5-15(c)?Solution: (figure come from 谭棋)The new tree built by reverse path forwarding should be:/F15.Describe two major differences between the ECN method and the RED method of congestion avoidance.Solution:(1) ECN(Explicit Congestion Notification)A router can tag any packet it forwards to signal that it is experiencing congestion. When the network delivers the packet, the destination can note that there is congestion and inform the sender when it sends a reply packet. The sender can then throttle its transmissions.(2) RED(Random Early Detection)This method prefer to dealing with congestion when it first starts rather than let it gum up the works and then try to deal with it. The motivation for this idea is that most Internet hosts do not yet get congestion signals from routers in the form of ECN. Instead, the only reliable indication of congestion that hosts get from the network is packet loss. Having routers drop packets early, before the situation has become hopeless can help reduce congestion. The affected sender will notice the loss when there is no acknowledgement, and then the transport protocol will slow down. (3) Their differences:a. Application environment: RED is used when hosts cannot receive explicit signal.b. Congestion notified method: ECN delivers a congestion signal explicitly as a chokepacket while RED notices congestion through packet loss.16.Imagine a flow specification that has a maximum packet size of 1000 bytes, a token bucket rate of 10 million bytes/sec, a token bucket size of 1 million bytes, and a maximumtransmission rate of 50 million bytes/sec. How long can a burst at maximum speed last? Solution:The burst length S sec, the maximum output rate M bytes/sec, thetoken bucket capacity B bytes, the token arrival rate R bytes/sec.B+RS=MS,S=B/(M-R)From this problem, B=1MB, M=50MB/s, R=10MB/sS=B/(M-R)=25ms20.Suppose that host A is connected to a router R1, R1 is connected to another router, R2, and R2 is connected to host B. Suppose that a TCP message that contains 900 bytes of data and 20bytes of TCP header is passed to the IP code at host A for delivery to B. Show the Total length, Identification, DF, MF, and Fragment offset fields of the IP header in each packet transmitted over the three links. Assume that link A-R1 can support a maximum frame size of 1024 bytes including a 14-byte frameheader, link R1-R2 can support a maximum frame size of 512 bytes, including an 8-byte frame header, and link R2-B can support a maximum frame size of 512 bytes including a 12-byte frame header.Solution:We have an IP payload of 920 bytes to send. Assume a 20 byte IPv4 header.The first link can carry IP packets up to 1010 bytes, so there will be no fragmentation. The second link can carry IP packets up to 504 bytes, so there will be fragmentation. There may be up to 484 bytes of data, but fragments must carry a multiple of 8 bytes of data (except the last fragment). So the first fragment will carry 480 bytes of data, and the second fragment will carry 440 bytes.The third link can carry IP packets up to 500 bytes, so both fragments will fit and no other fragmentation will occur. The value of the fields is as following table:512B(8)512B(12)1024B(14)BR1R2ATotal length DF MF Fragment offset Identification(random)A,R1 940 0000 0000 0001 0101 0 0 0R1,R2 500 0000 0000 0001 0101 0 1 0460 0000 0000 0001 0101 0 0 60 (8*60=480)R1,R2 500 0000 0000 0001 0101 0 1 0460 0000 0000 0001 0101 0 0 60 (8*60=480)23.Suppose that instead of using 16 bits for the network part of a class B address originally, 20bits had been used. How many class B networks would there have been? Solution:20 bits had been used for the network part of a class B address，but there are 2 bits isfixed，so there18are only 2 class B networks18the number of networks would have been 2 or 262,144. However, all 0s and all 1s are special, so only 262,142 are available.24.Convert the IP address whose hexadecimal representation is C22F1582 to dotted decimalnotation.Solution:According to: C 2 2 F 1 5 8 2,1100 0010 0010 1111 0001 0101 1000 0010,194 47 21 130,194.47.21.13025.A network on the Internet has a subnet mask of 255.255.240.0. Whatis the maximum number of hosts it can handle?Solution:The third 8-bit group in mask:240=11110000,So, there are network bits: 8+8+4=20bitsHosts bits: 32-20=12bitsThe maximum number of IP addresses it can handle:=4096Because the network address and the broadcast network address, hosts =4094.26.While IP addresses are tried to specific networks, Ethernet addresses are not. Can you think of a good reason why they are not?Answer:IP addresses are hierarchical, unlike Ethernet addresses. Each 32-bit address is comprised of a variable-length network partition in the top bits and a host partition in the bottom bits. Using a hierarchy lets Internet routing scale. However, the IP address of a host depends on where it is located in the network. An Ethernet address can be used anywhere in the world, but every IP address belongs to a specific network, and routers will only be able to deliver packets destined to that address to the network.Or: (from 袁子超)Each Ethernet adapter sold in stores comes hardwired with anEthernet (MAC) address in it. When burning the address into the card, the manufacturer has no idea where in the world the card will be used, making the address useless for routing in whole internet. In contrast, IP addresses are either assigned either statically or dynamically by anISP or company, which knows exactly how to get to the host getting the IP address.27.A large number of consecutive IP addresses are available starting at 198.16.0.0. Suppose that four organizations, A, B, C, and D, request 4000, 2000, 4000, and 8000 addresses, respectively, and in that order. For each of these, give the first IP address assigned, the last IP address assigned, and the mask in the w.x.y.z/s notation.Solution:Organization A: allocate hosts=4096=,host partition=12bits,network partition=20bitsOrganization B: allocate hosts=2048=,host partition=11bits,network partition=21bitsOrganization C: allocate hosts=4096=,host partition=12bits,network partition=20bitsOrganization D: allocate hosts=8192=,host partition=13bits,network partition=19bitsOrganization the first address the last address amount prefixA 198.16.0.0 198.16.15.255 4096 198.16.0.0/20B 198.16.16.0 198.16.23.255 2048 198.16.16.0/21C 198.16.32.0 198.16.47.255 4096 198.16.32.0/20D 198.16.64.0 198.16.95.255 8192 192.16.64.0/1928.A router has just received the following new IP addresses:57.6.96.0/21, 57.6.104.0/21, 57.6.112.0/21, and 57.6.120.0/21. If all of them use the same outgoing line, can they be aggregated? If so, to what? If not, why not?Solution:They can be aggregated.57.6.96.0 , 0110 000057.6.104.0,0110 100057.6.112.0,0111 000057.6.120.0,0111 1000,They can be aggregated by 57.6.96.0/19.29.The set of IP addresses from 29.18.0.0 to 29.18.128.255 has been aggregated to 29.18.0.0/17. However , there is a gap of 1024 unassigned addresses from 29.18.60.0 to 29.18.63.255 that are now suddenly assigned to a host using a different outgoing line. Is it now necessary to split up the aggregate address into its constituent blocks, add the new block to the table, and then see if any re-aggregation is possible? If not, what can be done instead? Solution:No, it is not necessary to split up the aggregate address. In CIDR, prefixes are allowed to overlap. The rule is that packets are sent in the direction of the most specific route, or the longest matching prefix that has the fewest IP addresses. So in this problem, we just need to add the block 29.18.60.0/22 to the table. If there is a math for both29.18.0.0/17 and 29.18.60.0/22, 29.18.60.0/22 will be used to lookup the outgoing line for the packet.30.A router has the following (CIDR) entries in its routing table:Address/mask Next hop135.46.56.0/22 Interface 0135.46.60.0/22 Interface 1192.53.40.0/23 Router 1default Router 2For each of the following IP addresses, what does the router do if a packet with that address arrives?(a) 135.46.63.10(b) 135.46.57.14(c) 135.46.52.2(d) 192.53.40.7(e) 192.53.56.7Solution:(a) 135.46.63.10,135.46.0011 1111.10Prefix : /22 ,135.46.0011 1100.10,135.46.60.0,Interface 1(b) 135.46.57.14,135.46.0011 1001.14Prefix : /22 ,135.46.0011 1000.14,135.46.56.0,Interface 0(c) 135.46.52.2,135.46.0011 0100.2Prefix : /22 ,135.46. 0011 0100.2,135.46.52.0,Router 2(d) 192.53.40.7 ,192.53.0010 1000.7Prefix : /22 ,192.53.0010 1000.7,192.53.40.0,Router 1(e) 192.53. 56.7,192.53.0011 1000.7Prefix : /22 ,192.53. 0011 1000.7,192.53.56.0,Router 231.Many companies have a policy of having two (or more) routers connecting the company to the Internet to provide some redundancy in case one of them goes down. Is this policy still possible with NAT? Explain your answer.Answer: (modified)Yes，it’s still possible. Whenever an outgoing packet enters the NAT box, the 10x.y.z source address is replaced by a global IP address. And the same time, the segment’s source port field isreplaced by an index into the NAT box’s 65536-entry translation table. This table entry contains the original IP address and original source port. When a return packet arrives at the NAT box from the outside, the Source Port in the segment header is extracted and used as an index in to the NATbox’s mapping table. From the entry located, the internal IP address and original Source are extracted and inserted into the packet.So it’s c rucial for all the packets pertaining to a single connection pass in and out of the company via the same router, sincethat is where the mapping is kept.IPv4 ARP 协议报文MAC帧头ARP头协议数据0X080632.You have just explained the ARP protocol to a friend. When you areall done, he says: “I’ve got it. ARP provides a service to the network layer, so it is part of the data link layer.” What do you say to him?Answer: (modified)ARP solves the problem of finding out which physical address corresponds to a given IP address. ARP is part of the network layer because it is related to IP address which is the representation of network layer.However, when data is passed down through OSI model, header will be added from each layer and they must be independent. Add a link header to a packet isn’t the function of network layer. ARPworks without using IP header. From this perspective, ARP belongs to the data link layer. (Pay attention: ARP belongs to 2 or 3 layer is not wrong. In fact, ARP is replaced by ND which is part of network layer completely in IPv6)34.Most IP datagram reassembly algorithms have a timer to avoid having a lost fragment tie up reassembly buffers forever. Suppose that a datagram is fragmented into four fragments. The first three fragments arrive, but the last one is delayed. Eventually, the timer goes off and thethree fragments in the receiver’s memory are discarded. A little later, the last fragment stumbles in. What should be done with it?Answer:To buffer all the pieces until the last fragment arrives and we can know the size. Then build a buffer with the right size, and put the fragments into the buffer, maintaining a bit map with 1 bit per 8 bytes to keep track of which bytes are present in the buffer. When all the bits in the bit map are 1, the datagram is complete.As far as the receiver is concerned, this is a part of new datagram, since no other parts of it are known. It will therefore be queued until the rest show up. If they do not, this one will time out too.37.IPv6 uses 16-byte addresses. If a block of 1 million addresses is allocated every picosecond, how long will the addresses last?My answer:16-byte=128bits128 Total address amount=26 1 million address=1012861213 Time lasts: 2/(10*10)/(365*24*60*60)=1.08×10years(Pay at tention: It’s so big a space that we don’t worry about run-out again.)38.The Protocol field used in the IPv4 header is not present in the fixed IPv6 header. Why not? Answer:The protocol field in IPv4 packet is replaced by next header witch tells which of the (currently) six extension headers, if any, followthis one. But if there is no extension header or next header field is in the last extension header, the meaning of next header field is same as protocol field’s.So, we can say that it’s not delet ed protocol field simply, but extension.43.Use the traceroute (UNIX) or tracert (Windows) programs to trace the route from your computer to various universities on other continents.Make a list of transoceanic links you have discovered. Some sites to try are (California) (Massachusetts)www.vu.nl (Amsterdam) (London).au (Sydney)www.u-tokyo.ac.jp (Tokyo)www.uct.ac.za (Cape Town)For problem 43, please add two sites: , , and copy the result interface (image) of tracert (all).Solution:In every test, I will used a red rectangle to show the beginning of transoceanic links. All the routers after it may form the list of transoceanic links. (California): (Massachusetts)www.vu.nl (Amsterdam) (London)As follows, I used tracert in windows to trace the route(use WiFi:CMCC-HEMC).au (Sydney)www.u-tokyo.ac.jp (Tokyo)www.uct.ac.za (Cape Town)。
计算机网络第5、6章课后测验参考答案与评分标准一、判断题（共20题）参考答案与评分标准：（每小题1分，共20分）1V 2 V 3 V 4X 5X 6X 7 V 8X 9 V 10X二、简答题（共20分）1,计算项目计算结果该IP数据报的总长度？2480字节（1分）可分为几个分片？4个（1分）每个分片数据报的总长度分别是多少？820字节/820字节/820字节/80字节（3分）每个分片数据报片偏移值分别是多少？0/100/200/300 （3分）2、根据：(121) io= (79)护(01111001) 2(248) io= (11111000) 2011110001)主机号为：12)子网地址：192. 168. 5. 1203)子网号的编码范围是：0~31 (5位)(1)将Hl、H2、H3、H4的IP地址分别与它们的子网掩码进行与操作，可以得到4个子网的网络地址，分别是：NET1 202.99.98.16NET2 202.99.98.32NET3 202.99.98.48NET4 202.99.98.64(2)R2的路由表(每项路由1.5分，共6分)目的网络子网掩码下一跳N1202.99.98.16255.255.255.240202.99.98.34N2202.99.98.32255.255.255.240直接N3202.99.98.48255.255.255.240直接N4202.99.98.64255.255.255.24202.99.98.50(3)主机H2向H3发送一个IP数据报的过程如下：A.主机H2首先构造IP数据报，并将其传递给数据链路层。
其中：源IP 地址为202.99.98.35,目的地址为202.99.98.51；B.主机H2再通过ARP协议获得R2路由器0佛口(202.99.98.33)所对应的MAC 地址，并将其作为目的MAC地址填入封装有IP数据报的帧中，然后将该帧通过物理层发送到N2子网上；C.路由器R2收到该帧后，去除帧头与帧尾，得到IP数据报，然后根据IP数据报中的目的IP地址(202.99.98.35)去查找路由表，得到下一跳地址为直接交付；D.路由器R2通过ARP协议得到主机H2的MAC地址，并将其作为目的MAC地址填入封装有IP数据报的帧，然后将该帧发送到N3子网上；E.主机H3将收到该帧，去除帧头与帧尾，并最终得到从主机H2发来的IP数据报。
第五章传输层5—01 试说明运输层在协议栈中的地位和作用，运输层的通信和网络层的通信有什么重要区别为什么运输层是必不可少的答：运输层处于面向通信部分的最高层，同时也是用户功能中的最低层，向它上面的应用层提供服务运输层为应用进程之间提供端到端的逻辑通信，但网络层是为主机之间提供逻辑通信（面向主机，承担路由功能，即主机寻址及有效的分组交换）。
各种应用进程之间通信需要“可靠或尽力而为”的两类服务质量，必须由运输层以复用和分用的形式加载到网络层。
5—02 网络层提供数据报或虚电路服务对上面的运输层有何影响答：网络层提供数据报或虚电路服务不影响上面的运输层的运行机制。
但提供不同的服务质量。
5—03 当应用程序使用面向连接的TCP和无连接的IP时，这种传输是面向连接的还是面向无连接的答：都是。
这要在不同层次来看，在运输层是面向连接的，在网络层则是无连接的。
<5—05 试举例说明有些应用程序愿意采用不可靠的UDP，而不用采用可靠的TCP。
答：VOIP：由于语音信息具有一定的冗余度，人耳对VOIP数据报损失由一定的承受度，但对传输时延的变化较敏感。
有差错的UDP数据报在接收端被直接抛弃，TCP数据报出错则会引起重传，可能带来较大的时延扰动。
因此VOIP宁可采用不可靠的UDP，而不愿意采用可靠的TCP。
5—06 接收方收到有差错的UDP用户数据报时应如何处理答：丢弃5—07 如果应用程序愿意使用UDP来完成可靠的传输，这可能吗请说明理由答：可能，但应用程序中必须额外提供与TCP相同的功能。
5—08 为什么说UDP是面向报文的，而TCP是面向字节流的答：发送方UDP 对应用程序交下来的报文，在添加首部后就向下交付IP 层。
UDP 对应用层交下来的报文，既不合并，也不拆分，而是保留这些报文的边界。
接收方UDP 对IP 层交上来的UDP 用户数据报，在去除首部后就原封不动地交付上层的应用进程，一次交付一个完整的报文。
第五章 RIPv1 CCNA2 Exploration: 路由协议和概念 (2010-12-15 18:07:13) 转载▼标签： 杂谈 分类： 2010｜CCNA2官方试题--路由 1窗体顶端ERouting Chapter 7 - CCNA Exploration: 路由协议和概念 (版本 4.0)以下行显示在 show ip route 命令的输出中。
R 192.168.3.0/24 [120/3] via 192.168.2.2, 00:00:30, Serial0/0该路由度量的值是什么？3122030120 窗体底端2窗体顶端下列关于 RIPv1 特性的陈述中，哪两项是正确的？（选择两项。
）它是一种距离矢量路由协议。
它会在路由更新中通告路由的地址和子网掩码。
RIP 消息的数据部分封装在 TCP 数据段中。
RIP 消息的数据部分封装在 UDP 数据段中。
它每 15 秒广播一次更新。
它最多允许路由域中存在 15 台路由器。
窗体底端3窗体顶端请参见图示。
图中所示的网络运行RIPv1。
不久前其中添加了192.168.10.0/24 网络，该网络将只包含最终要防止向新网络中的最终用户设备发送RIPv1 更新，同时允许将此新网络通告给其它路由器，应该在Rou 什么命令？Router1(config-router)# no router ripRouter1(config-router)# network 192.168.10.0Router1(config-router)# no network 192.168.10.0Router1(config-router)# passive-interface fastethernet 0/0Router1(config-router)# passive-interface serial 0/0/0窗体底端4 窗体顶端请参见图示。
该网络包含多台路由器。
5—01 试说明运输层在协议栈中的地位和作用，运输层的通信和网络层的通信有什么重要区别？为什么运输层是必不可少的？答：运输层处于面向通信部分的最高层，同时也是用户功能中的最低层，向它上面的应用层提供服务运输层为应用进程之间提供端到端的逻辑通信，但网络层是为主机之间提供逻辑通信（面向主机，承担路由功能，即主机寻址及有效的分组交换）。
各种应用进程之间通信需要“可靠或尽力而为”的两类服务质量，必须由运输层以复用和分用的形式加载到网络层。
5—02 网络层提供数据报或虚电路服务对上面的运输层有何影响？答：网络层提供数据报或虚电路服务不影响上面的运输层的运行机制。
但提供不同的服务质量。
5—03 当应用程序使用面向连接的TCP和无连接的IP时，这种传输是面向连接的还是面向无连接的？答：都是。
这要在不同层次来看，在运输层是面向连接的，在网络层则是无连接的。
5—04 试用画图解释运输层的复用。
画图说明许多个运输用户复用到一条运输连接上，而这条运输连接有复用到IP数据报上。
5—05 试举例说明有些应用程序愿意采用不可靠的UDP，而不用采用可靠的TCP。
答：VOIP：由于语音信息具有一定的冗余度，人耳对VOIP数据报损失由一定的承受度，但对传输时延的变化较敏感。
有差错的UDP数据报在接收端被直接抛弃，TCP数据报出错则会引起重传，可能带来较大的时延扰动。
因此VOIP宁可采用不可靠的UDP，而不愿意采用可靠的TCP。
5—06 接收方收到有差错的UDP用户数据报时应如何处理？答：丢弃5—07 如果应用程序愿意使用UDP来完成可靠的传输，这可能吗？请说明理由答：可能，但应用程序中必须额外提供与TCP相同的功能。
5—08 为什么说UDP是面向报文的，而TCP是面向字节流的？答：发送方UDP 对应用程序交下来的报文，在添加首部后就向下交付IP 层。
UDP 对应用层交下来的报文，既不合并，也不拆分，而是保留这些报文的边界。
接收方UDP 对IP 层交上来的UDP 用户数据报，在去除首部后就原封不动地交付上层的应用进程，一次交付一个完整的报文。
第1章检查你的理解1．B、C、E以太网交换机和转发器通常用于LAN 中。
路由器可视为LAN 设备和W AN 设备，用于在公司网络内部路由分组、将分组路由到ISP 以及在自主系统之间路由分组。
接入服务器集中拨入和拨出的用户通信，它可能有模拟和数字接口，可支持数百名用户同时连接到提供商的WAN。
2．D3．电路交换：D分组交换：C面向连接的分组交换：B无连接的分组交换：A4．城域以太网：AX.25：DATM：B帧中继：C5．C DTE 通过W AN 传输来自客户网络或主机计算机的数据。
DTE 通过DCE 连接到本地环路。
路由器是一种DTE 设备，它使用CSU/DSU（DCE 设备）连接到W AN。
6．E租用线是一种专用的点到点链路，通常是向运营商租用的。
7．DLCI 唯一地标识了帧中继虚电路，这确保DTE 设备之间能够进行双向通信。
8．异步传输模式（ATM）技术可通过私有和公共网络传输语音、视频和数据，它是建立在基于信元的架构而不是基于帧的架构的基础之上的。
9．C10．B11．C12．E13．C14．C15．CCisco 企业分支机构架构让能够将园区网中的应用程序和服务延伸到各种远程位置和用户，接入层用于将用户连接到网络，这层通常使用第 2 层和第3 层交换机。
在北美，ISDN PRI 包含23 个64kbit/s 的B 信道和 1 个64kbit/s 的D 信道，总比特率高VPN 使用隧道通过公共网络（如Internet）在两个私有网络之间安全地传输数据。
分布层将工作组放在不同的网段中，从而隔离子网或VLAN 中的网络问题。
VLAN 之间本地环路将用户驻地的CPE 连接到服务提供商的CO，有时也被称为“最后一公里”。
企业远程工作人员架构让雇员能够远程（通常是从家里）连接到网络。
而不管其规模和位置如何。
核心层也称为网络主干，设计用于尽可能快地交换分组。
达1.544Mbit/s。
的路由选择通常是在分布层进行的。
第五章 RIPv1 CCNA2 Exploration: 路由协议和概念(2019-12-15 18:07:13)转载▼标签：杂谈分类： 2010｜CCNA2官方试题--路由1 窗体顶端ERouting Chapter 7 - CCNA Exploration: 路由协议和概念(版本 4.0)以下行显示在 show ip route 命令的输出中。
R 192.168.3.0/24 [120/3] via 192.168.2.2, 00:00:30, Serial0/0该路由度量的值是什么？3122030120 窗体底端2 窗体顶端下列关于 RIPv1 特性的陈述中，哪两项是正确的？（选择两项。
）它是一种距离矢量路由协议。
它会在路由更新中通告路由的地址和子网掩码。
RIP 消息的数据部分封装在 TCP 数据段中。
RIP 消息的数据部分封装在 UDP 数据段中。
它每 15 秒广播一次更新。
它最多允许路由域中存在 15 台路由器。
窗体底端3 窗体顶端请参见图示。
图中所示的网络运行 RIPv1。
不久前其中添加了192.168.10.0/24 网络，该网络将只包含最终用户。
如果要防止向新网络中的最终用户设备发送 RIPv1 更新，同时允许将此新网络通告给其它路由器，应该在 Router1 上输入什么命令？Router1(config-router)# no router ripRouter1(config-router)# network 192.168.10.0Router1(config-router)# no network 192.168.10.0Router1(config-router)# passive-interface fastethernet 0/0Router1(config-router)# passive-interface serial 0/0/0窗体底端4 窗体顶端请参见图示。
