CCNA_ENetwork_Chapter_9_答案(中文版)-dengwenhui

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CCNA中英对照题库(0-10.pdf

CCNA中英对照题库(0-10.pdfCCNA （200-120）题库中英对照整理于：2015 年 3 月 31 日版本：1.2 声明：本题库来源互联网（鸿鹄论坛）本题库为英文考试原版左右选择题中文考试为随机抽取英文题的中译版此题库非中文考试题库，而是英文题库手动翻译此题库为手工个人翻译，存在错误和不妥难免，不代表任何官方机构和组织本题库旨在更好的理解英文原本题库，也可以用于中文考试参考用。

如发现任何错误或不当之处，可以自行修改，但请更新版本以免混乱，也可致上传者。

QUESTION 001 Refer to the exhibit. What will Router1 do when it receives the data frame shown? (Choose three.) 参照下图，这个图示显示了Ｒ１接受到如图所示数据帧的时候会怎么做？ A. Router1 will strip off the source MAC address and replace it with the MAC address 0000.0c36.6965. Ｒ１会剥离源ＭＡＣ地址，以 0000：0C36.6965 代替。

B. Router1 will strip off the source IP address and replace it with the IP address 192.168.40.1. Ｒ１会剥离源 IP 地址，1/ 10以 192.168.40.1 代替。

C. Router1 will strip off the destination MAC address and replace it with the MAC address 0000.0c07.4320. Ｒ１会剥离源ＭＡＣ地址，以 0000：0C36.6965 代替。

D. Router1 will strip off the destination IP address and replace it with the IP address of 192.168.40.1. Ｒ１会剥离目的 IP 地址，以 192.168.40.1 代替。

CCNA中英对照题库(选择题)

CCNA（200-120）题库中英对照整理于：2015年3月31日版本：1.2声明：本题库来源互联网（鸿鹄论坛）本题库为英文考试原版左右选择题中文考试为随机抽取英文题的中译版此题库非中文考试题库，而是英文题库手动翻译此题库为手工个人翻译，存在错误和不妥难免，不代表任何官方机构和组织本题库旨在更好的理解英文原本题库，也可以用于中文考试参考用。

如发现任何错误或不当之处，可以自行修改，但请更新版本以免混乱，也可致上传者。

QUESTION001Refer to the exhibit. What will Router1 do when it receives the data frame shown? (Choose three.) 参照下图，这个图示显示了Ｒ１接受到如图所示数据帧的时候会怎么做？Router1 will strip off the source MAC address and replace it with the MACaddress 0000.0c36.6965.Ｒ１会剥离源ＭＡＣ地址，以0000：0C36.6965代替。

Router1 will strip off the source IP address and replace it with the IP address 192.168.40.1. Ｒ１会剥离源IP地址，以192.168.40.1代替。

Router1 will strip off the destination MAC address and replace it with the MAC address 0000.0c07.4320.Ｒ１会剥离源ＭＡＣ地址，以0000：0C36.6965代替。

Router1 will strip off the destination IP address and replace it with the IP address of 192.168.40.1.Ｒ１会剥离目的IP地址，以192.168.40.1代替。

CCNA第一学期各章习题及答案.doc

诶了我累了目录第二章网络通信P2-8第三章应用层功能及协议P9-14 第四章 OSI 传输层 P15-20第五章 OSI 网络层 P21-27第六章网络编址-IPv4 P28-34第七章数据链路层P35-40第八章 OSI 物理层 P41-45第九章以太网P46-51第十章网络规划和布线P52-61 第十一章配置和测试网络P62-70 第一学期末考试P71-91文档末尾返回目录第二章网络通信001 TCP/IP 网络接入层有何作用路径确定和数据包交换数据表示、编码和控制可靠性、流量控制和错误检测详细规定构成物理链路的组件及其接入方法将数据段划分为数据包2下列哪些陈述正确指出了中间设备在网络中的作用（选择三项）确定数据传输路径发起数据通信重新定时和重新传输数据信号发送数据流管理数据流数据流最后的终止点003 下列哪三项陈述是对局域网(LAN) 最准确的描述（选择三项）LAN 通常位于一个地域内。

此类网络由由一个组织管理。

LAN 中的不同网段之间一般通过租用连接的方式连接。

此类网络的安全和访问控制由服务提供商控制。

LAN 为同一个组织内的用户提供网络服务和应用程序访问。

此类网络的每个终端通常都连接到电信服务提供商(TSP)。

004 什么是PDU传输期间的帧损坏在目的设备上重组的数据因通信丢失而重新传输的数据包特定层的封装005 OSI 模型哪两层的功能与TCP/IP 模型的网络接入层相同（选择两项）网络层传输层物理层数据链路层会话层6请参见图示。

所示网络属于哪一类型WANMANLANWLAN7以下哪种特征正确代表了网络中的终端设备管理数据流发送数据流重新定时和重新传输数据信号确定数据传输路径008 第 4 层端口指定的主要作用是什么标识本地介质中的设备标识源设备和目的设备之间的跳数向中间设备标识通过该网络的最佳路径标识正在通信的源终端设备和目的终端设备标识终端设备内正在通信的进程或服务9请参见图示。

CCNA 1 - Chapter 9

CCNA 1 - Chapter 9Posted by beotron at 8:55 PM1. In the graphic, Host A has reached 50% completion in sending a 1 KB Ethernet frame to Host D when Host B wishes to transmit its own frame to Host C. What must Host B do? • Host B can transmit immediately since it is connected on its own cable segment.• Host B must wait to receive a CSMA transmission from the hub, to signal its turn.• Host B must send a request sig nal to Host A by transmitting an interframe gap.• Host B must wait until it is certain that Host A has completed sending its frame.2. Ethernet operates at which layers of the OSI model? (Choose two.)• Network layer• Transport layer• Physical layer• A pplication layer• Session layer• Data-link layer3. Which of the following describe interframe spacing? (Choose two.)• the minimum interval, measured in bit-times, that any station must wait before sending another frame.• the maximum interval, measure d in bit-times, that any station must wait before sending another frame.• the 96-bit payload padding inserted into a frame to achieve a legal frame size• the 96-bit frame padding transmitted between frames to achieve proper synchronization • the time all owed for slow stations to process a frame and prepare for the next frame. • the maximum interval within which a station must send another frame to avoid being considered unreachable4. What three primary functions does data link layer encapsulation provide? (Choose three.) • addressing• error detection• frame delimiting• port identification• path determination• IP address resolution5. When a collision occurs in a network using CSMA/CD, how do hosts with data to transmit respond after the backoff period has expired?• The hosts return to a listen-before-transmit mode. <—• The hosts creating the collision have priority to send data.• The hosts creating the collision retransmit the last 16 frames.• The hosts extend their delay period to allow for rapi d transmission.6. What are three functions of the upper data link sublayer in the OSI model? (Choose three.) • recognizes streams of bits• identifies the network layer protocol.• makes the connection with the upper layers.• identifies the source and de stination applications• insulates network layer protocols from changes in physical equipment.• determines the source of a transmission when multiple devices are transmitting7. What does the IEEE 802.2 standard represent in Ethernet technologies?• MAC s ublayer• Physical layer• Logical Link Control sublayer• Network layer8. Why do hosts on an Ethernet segment that experience a collision use a random delay before attempting to transmit a frame?• A random delay is used to ensure a collision-free link.• A random delay value for each device is assigned by the manufacturer.• A standard delay value could not be agreed upon among networking device vendors.• A random delay helps prevent the stations from experiencing another collision during the transmission.9. Refer to the exhibit. Which option correctly matches the frame field type with the contents that frame field includes?• header field - preamble and stop frame• data field - network layer packet• data field - physical addressing• trailer field - FCS and SoF10. Host A has an IP address of 172.16.225.93 and a mask of 255.255.248.0. Host A needs to communicate with a new host whose IP is 172.16.231.78. Host A performs the ANDing operation on the destination address. What two things will occur? (Choose two.)• Host A will change the destination IP to the IP of the nearest router and forward the packet.• Host A will broadcast an ARP request for the MAC of its default gateway.• A result of 172.16.225.0 will be obtained.• Host A will broadcast an AR P request for the MAC of the destination host.• A result of 172.16.224.0 will be obtained.• A result of 172.16.225.255 will be obtained.11 Which of the following is a drawback of the CSMA/CD access method?• Collisions can decrease network performance.• It is more complex than non-deterministic protocols.• Deterministic media access protocols slow network performance.• CSMA/CD LAN technologies are only available at slower speeds than other LAN technologies.12. Ethernet operates at which layer of the TCP/IP network model?• application• physical• transport• internet• data link• network access13. What is the primary purpose of ARP?• translate URLs to IP addresses• resolve IPv4 addresses to MAC addresses• provide dynamic IP configuration to net work devices• convert internal private addresses to external public addresses14. Refer to the exhibit. The switch and workstation are administratively configured for full-duplex operation. Which statement accurately reflects the operation of this link? • No collisions will occur on this link.• Only one of the devices can transmit at a time.• The switch will have priority for transmitting data.• The devices will default back to half duplex if excessive collisions occur.15. Refer to the exhibit. Host_A is attempting to contact Server_B. Which statements correctly describe the addressing Host_A will generate in the process? (Choose two.) • A packet with the destination IP of Router_B.• A frame with the destination MAC address of Switch_A.• A packet wit h the destination IP of Router_A.• A frame with the destination MAC address of Router_A.• A packet with the destination IP of Server_B.• A frame with the destination MAC address of Server_B.16 Which statements correctly describe MAC addresses? (Choose three.)• dynamically assigned• copied into RAM during system startup• layer 3 address• contains a 3 byte OUI• 6 bytes long• 32 bits long17. Which two features make switches preferable to hubs in Ethernet-based networks? (Choose two.)• reduction in cross-talk• minimizing of collisions• support for UTP cabling• division into broadcast domains• increase in the throughput of communications18. What are the two most commonly used media types in Ethernet networks today? (Choose two.)• coaxial thick net• copper UTP• coaxial thinnet• optical fiber• shielded twisted pair19. Convert the binary number 10111010 into its hexadecimal equivalent. Select the correct answer from the list below.• 85• 90• BA• A1• B3• 1C20. After an Ethernet collision, when the backoff algorithm is invoked, which device has priority to transmit data?• the device involved in the collision with the lowest MAC address• the device involved in the collision with the lowest IP address• any device in the collision domain whose backoff timer expires first• those that began transmitting at the same time。

CCNA认证试题(中文+答案)

CCNA认证试题一（附答案和解析）中文版(一)1、目前，我国应用最为广泛的LAN标准是基于（）的以太网标准.(A) IEEE 802.1(B) IEEE 802.2(C) IEEE 802.3(D) IEEE 802.5答案：C参考知识点：现有标准：IEEE 802.1 局域网协议高层IEEE 802.2 逻辑链路控制IEEE 802.3 以太网IEEE 802.4 令牌总线IEEE 802.5 令牌环IEEE 802.8 FDDIIEEE 802.11 无线局域网记住IEEE802.1-------IEEE802.5的定义以太网是一种计算机局域网组网技术。

IEEE制定的IEEE 802.3标准给出了以太网的技术标准。

它规定了包括物理层的连线、电信号和介质访问层协议的内容。

以太网是当前应用最普遍的局域网技术。

它很大程度上取代了其他局域网标准，如令牌环、FDDI和ARCNET。

以太网的标准拓扑结构为总线型拓扑，但目前的快速以太网（100BASE-T、1000BASE-T标准）为了最大程度的减少冲突，最大程度的提高网络速度和使用效率，使用交换机（Switch）来进行网络连接和组织，这样，以太网的拓扑结构就成了星型，但在逻辑上，以太网仍然使用总线型拓扑的C***A/CD介质访问控制方法。

电气电子工程师协会或IEEE（Institute of Electrical and Electronics Engineers）是一个国际性的电子技术及信息科学工程师的协会。

建会于1963年1月1日。

总部在美国纽约市。

在150多个国家中它拥有300多个地方分会。

目前会员数是36万。

专业上它有35个专业学会和两个联合会。

IEEE发表多种杂志，学报，书籍和每年组织300多次专业会议。

IEEE定义的标准在工业界有极大的影响。

下面列出：IEEE802．3以太网标准802.3--------- 10Base以太网标准802.3u-------- 100Base-T（快速以太网）802.3z-------- 1000Base-X（光纤吉比特以太网）802.3ab-------- 1000Base-T（双绞线吉比特以太网）2、对于这样一个地址,192.168.19.255/20,下列说法正确的是: （）(A) 这是一个广播地址(B) 这是一个网络地址(C) 这是一个私有地址(D) 地址在192.168.19.0网段上(E) 地址在192.168.16.0网段上(F) 这是一个公有地址答案：CE注：IP地址中关键是看她的主机位，将子网掩码划为二进制，1对应上面的地址是网络位，0对应的地址是主机位192.168.19.255/20划为二进制为：11000000.10101000.00010011.1111111111111111.11111111.11110000.00000000主机位变成全0表示这个IP的网络地址主机槐涑扇?表示这个IP的广播地址RFC1918文件规定了保留作为局域网使用的私有地址：10.0.0.0 - 10.255.255.255(10/8 prefix)172.16.0.0 - 172.31.255.255 (1 72.16/12 prefix)192.168.0.0 - 192.168.255.255 (192.168/16 prefix)3、Quidway系列路由器在执行数据包转发时，下列哪些项没有发生变化（假定没有使用地址转换技术）？（）(A) 源端口号(B) 目的端口号(C) 源网络地址(D) 目的网络地址(E) 源MAC地址(F) 目的MAC地址答案：ABCD参考知识点：路由功能就是指选择一条从源网络到目的网络的路径，并进行数据包的转发。

examen9ccna1v5.0

CCNA 1 Chapter 9 v5.0 Exam Answers 20131Refer to the exhibit.How many broadcast domains are there?1234*2How many usable host addresses are there in the subnet 192.168.1.32/27?3230*6416623How many host addresses are available on the network 172.16.128.0 with a subnet mask of 255.255.252.0?5105121022*1024204620484A network administrator is variably subnetting a network. The smallest subnet has a mask of 255.255.255.248. How many host addresses will this subnet provide?46*810125Refer to the exhibit.A company uses the address block of 128.107.0.0/16 for its network. What subnet mask would provide the maximum number of equal size subnets while providing enough host addresses for each subnet in the exhibit?255.255.255.0255.255.255.128*255.255.255.192255.255.255.224255.255.255.2406Refer to the exhibit.The network administrator has assigned the LAN of LBMISS an address range of 192.168.10.0. This address range has been subnetted using a /29 prefix. In order to accommodate a new building, the technician has decided to use the fifth subnet for configuring the new network (subnet zero is the first subnet). By company policies, the router interface is always assigned the first usable host address and the workgroup server is given the last usable host address. Which configuration should be entered into the properties of the workgroup server to allow connectivity to the Internet?IP address: 192.168.10.65 subnet mask: 255.255.255.240, default gateway: 192.168.10.76IP address: 192.168.10.38 subnet mask: 255.255.255.240, default gateway: 192.168.10.33*IP address: 192.168.10.38 subnet mask: 255.255.255.248, default gateway: 192.168.10.33*IP address: 192.168.10.41 subnet mask: 255.255.255.248, default gateway: 192.168.10.46IP address: 192.168.10.254 subnet mask: 255.255.255.0, default gateway: 192.168.10.17How many bits must be borrowed from the host portion of an address to accommodate a router with five connected networks?twothree*fourfive8A company has a network address of 192.168.1.64 with a subnet mask of 255.255.255.192. The company wants to create two subnetworks that would contain 10 hosts and 18 hosts respectively. Which two networks would achieve that? (Choose two.)192.168.1.16/28192.168.1.64/27*192.168.1.128/27192.168.1.96/28*192.168.1.192/289In a network that uses IPv4, what prefix would best fit a subnet containing 100 hosts?/23/24/25*/2610Refer to the exhibit.Given the network address of 192.168.5.0 and a subnet mask of 255.255.255.224, how many addresses are wasted in total by subnetting each network with a subnet mask of255.255.255.224?5660646872*11When developing an IP addressing scheme for an enterprise network, which devices are recommended to be grouped into their own subnet or logical addressing group?end-user clientsworkstation clientsmobile and laptop hostshosts accessible from the Internet*12A network administrator needs to monitor network traffic to and from servers in a data center. Which features of an IP addressing scheme should be applied to these devices?random static addresses to improve securityaddresses from different subnets for redundancypredictable static IP addresses for easier identification*dynamic addresses to reduce the probability of duplicate addresses13Which two reasons generally make DHCP the preferred method of assigning IP addresses to hosts on large networks? (Choose two.)It eliminates most address configuration errors.*It ensures that addresses are only applied to devices that require a permanent address.It guarantees that every device that needs an address will get one.It provides an address only to devices that are authorized to be connected to the network.It reduces the burden on network support staff.*14Refer to the exhibit.A computer that is configured with the IPv6 address as shown in the exhibit is unable to access the internet. What is the problem?The DNS address is wrong.There should not be an alternative DNS address.The gateway address is in the wrong subnet.*The settings were not validated.15When subnetting a /64 IPv6 network prefix, which is the preferred new prefix?/66/70/72*/7416What is the subnet address for the address 2001:DB8:BC15:A:12AB::1/64?2001:DB8:BC15::02001:DB8:BC15:A::0*2001:DB8:BC15:A:1::12001:DB8:BC15:A:12::017Which two notations are useable nibble boundaries when subnetting in IPv6? (Choose two.)/62/64*/66/68*/7018Fill in the blank.In dotted decimal notation, the IP address ”172.25.0.126” is the last host address for the network 172.25.0.64/26.The binary representation of the networkaddress 172.25.0.64 is 10101100.00011001.00000000.01000000,where the last six zeros represent the host part of the address.The last address on that subnet would have the host part equal to 111111,and the last host address would end in 111110. This results in a binaryrepresentation of the last host of the IP address as 10101100.00011001.00000000.01111110,which translates in decimal to 172.25.0.126.19Fill in the blank.In dotted decimal notation, the subnet mask ”255.255.254.0” will accommodate 500 hosts per subnet.If the network has to accommodate 500 hosts per subnet, then we need 9 host bits(2^9 – 2 = 510 hosts). The Class B subnet mask has 16 bits available and ifwe use 9 bits for hosts, we will have 7 network bits remaining.The subnet mask with 9 host bits is 11111111.11111111.11111110.00000000, which corresponds to 255.255.254.0.20Consider the following range of addresses:2001:0DB8:BC15:00A0:0000::2001:0DB8:BC15:00A1:0000::2001:0DB8:BC15:00A2:0000::…2001:0DB8:BC15:00AF:0000::The prefix for the range of addresses is “60” .All the addresses have the part 2001:0DB8:BC15:00Ain common. Each number or letter in the addressrepresents 4 bits, so the prefix is /60.21Fill in the blank.A nibble consists of ”4” bits.A nibble is half of a byte or 4 bits. This is significant becausesubnetting in IPv6 is usually done on a nibble boundary.22Question as presented:Place the options in the following order:- not scored -192.168.1.64/27- not scored -192.168.1.32/27192.168.1.96/2723Question as presented:Place the options in the following order: - not scored -Network C- not scored -Network ANetwork DNetwork B24Open the PT Activity.Perform the tasks in the activity instructions and then answer the question. What issue is causing Host A to be unable to communicate with Host B? The subnet mask of host A is incorrect.Host A has an incorrect default gateway.Host A and host B are on overlapping subnets.*The IP address of host B is not in the same subnet as the default gateway is on.。

CCNA中文版本题库部分

A.主机B将不能访问该服务器在VLAN9，直到电缆连接B.对于不到一分钟，主机B将无法访问服务器VLAN9.然后正常的网络功能将恢复。

C.VLAN和其他VLAN之间的通信，应禁用D.文件从主机B在VLAN9转印到服务器将是显著慢。

注：图中交换机存在环路，有一个端口会处于堵塞状态。

当F0/9 down之后，STP重新计算，在汇聚之前VLAN3不能访问VLAN9，端口汇聚之后恢复正常D说hostB访问VLAN 9中的server会变慢。

不同vlan之间的通信需要通过网关转发，并不是switch1直接转发给switch3，所以这里访问速度不会变。

172、主机A无法ping通HostB的。

假设路由配置是否正确，这可能是这个问题的原因是什么？A.主机A不是在同一子网作为其默认网关B. SwitchA的地址是一个子网地址C.该Fa0 / 0接口在RouterA是不能使用的子网D.路由器的串行接口不在同一子网E.该Fa0 / 0接口在RouterB上使用的是广播地址173、哪个端口状态由快速PVST出台？A. learning（学习）B. listening（侦听）C. discarding（丢弃---Rpvst新增了一种端口角色----discarding）D. forwarding（转发）174、交换机上的两个连接的端口无法运转橙色或绿色亮起。

什么是最有效的步骤来解决这个物理层问题？（选择三项。

）A：确认以太网封装是否匹配，错误。

封装错误不会在端口LED显示B：确认线缆A/B是否为直通线，正确。

异构设备使用直通线C：确认线缆A是否插到trunk口，错误。

D：确认是否通电，正确。

E：重启设备，错误。

F：重置所有电缆，正确。

注：考查交换机端口颜色的定义。

绿色正常，橘黄色有故障。

175、网络管理员试图从主机1 ping通主机2和接收显示的结果。

什么是可能出现的问题？A.主机1和交换机1之间的链路中断。

B. TCP/ IP不正常的主机1。

CCNA 第二学期第九章答案

1 哪两项功能描述了访问控制列表的用途？（请选择两项。

）正确响应您的响应ACL 可以根据路由器上的始发 MAC 地址来允许或拒绝流量。

ACL 可以控制主机能够访问网络中的哪些区域。

ACL 可以帮助路由器确定到目的地的最佳路径。

标准 ACL 可限制对特定应用程序和端口的访问。

ACL 提供基本的网络安全性。

2 访问控制列表在企业网络中的可能用途是下列哪两项？（选择两项。

）正确响应您的响应控制路由器接口的物理状态控制调试输出允许通过路由器过滤第 2 层流量降低路由器上的处理负载控制虚拟终端对路由器的访问3 哪两个特征是标准 ACL 和扩展 ACL 共有的特征？（请选择两项。

）正确您的响应响应两者都可以通过使用描述性名称或号码创建。

两类ACL 都可以根据协议类型进行过滤。

两者都包含隐式 deny 作为最终 ACE。

两者都可以通过端口号允许或拒绝特定服务。

两者都可为特定目的主机 IP 地址过滤数据包。

4下列哪项描述了标准 IPv4 ACL 的特点？正确响应您的响应创建它们时可以使用数字，但不可以使用名称。

它们是在接口配置模式下配置的。

可以根据源 IP 地址和源端口对过滤流量进行配置。

它们仅根据源 IP 地址过滤流量。

5网络管理员需要配置标准 ACL，使得只有 IP 地址为 192.168.15.23 的管理员工作站能够访问主要路由器的虚拟终端。

哪两个配置命令可以完成该任务？（请选择两项。

）正确响应您的响应Router1(config)# access-list 10 permit 192.168.15.23 255.255.255.255Router1(config)# access-list 10 permit 192.168.15.23 0.0.0.0Router1(config)# access-list 10 permit 192.168.15.23 255.255.255.0Router1(config)# access-list 10 permit host 192.168.15.23Router1(config)# access-list 10 permit 192.168.15.23 0.0.0.2556哪一个 IPv4 地址范围包含由 172.16.2.0 指定的、与 ACL 过滤器匹配的、通配符掩码为 0.0.1.255 的所有 IP 地址？正确响应您的响应172.16.2.1 到 172.16.3.254172.16.2.1 到172.16.255.255172.16.2.0 到 172.16.3.255172.16.2.0 到 172.16.2.2557如果路由器有两个接口，并且路由 IPv4 和 IPv6 的流量，那么可以在其中创建并应用多少个 ACL？正确响应您的响应12616488通常认为下列哪三项是安置 ACL 的最佳做法？（请选择三项。

计算机网络课后题答案第九章(最新整理)

第九章无线网络9-01．无线局域网都由哪几部分组成？无线局域网中的固定基础设施对网络的性能有何影响？接入点AP 是否就是无线局域网中的固定基础设施？答：无线局域网由无线网卡、无线接入点(AP)、计算机和有关设备组成，采用单元结构，将整个系统分成许多单元，每个单元称为一个基本服务组。

所谓“固定基础设施”是指预先建立起来的、能够覆盖一定地理范围的一批固定基站。

直接影响无线局域网的性能。

接入点AP 是星形拓扑的中心点，它不是固定基础设施。

9-02．Wi-Fi 与无线局域网WLAN 是否为同义词？请简单说明一下。

答：Wi-Fi 在许多文献中与无线局域网WLAN 是同义词。

802.11 是个相当复杂的标准。

但简单的来说，802.11 是无线以太网的标准，它是使用星形拓扑，其中心叫做接入点AP(Access Point)，在MAC 层使用CSMA/CA 协议。

凡使用802.11系列协议的局域网又称为Wi-Fi(Wireless-Fidelity,意思是“无线保真度”)。

因此，在许多文献中，Wi-Fi 几乎成为了无线局域网WLAN 的同义词。

9-03 服务集标示符SSID 与基本服务集标示符BSSID 有什么区别？答：SSID（Service Set Identifier）AP 唯一的ID 码，用来区分不同的网络，最多可以有32 个字符，无线终端和AP 的SSID 必须相同方可通信。

无线网卡设置了不同的SSID 就可以进入不同网络，SSID 通常由AP 广播出来，通过XP 自带的扫描功能可以相看当前区域内的SSID。

出于安全考虑可以不广播SSID，此时用户就要手工设置SSID 才能进入相应的网络。

简单说，SSID 就是一个局域网的名称，只有设置为名称相同SSID 的值的电脑才能互相通信。

BSS 是一种特殊的Ad-hoc LAN 的应用，一个无线网络至少由一个连接到有线网络的AP 和若干无线工作站组成，这种配置称为一个基本服务装置BSS (Basic Service Set)。

ccna第二学期练习答案中文

ERouting Chapter 1 - CCNA Exploration: 路由协议和概念(版本4.0)1口令可用于限制对Cisco IOS 所有或部分内容的访问。

请选择可以用口令保护的模式和接口。

（选择三项。

）VTY 接口控制台接口特权执行模式2路由器从相连的以太网接口收到消息后，会更改哪项报头地址，再将消息从另一个接口发送出去？第 2 层源地址和目的地址3请参见图示。

网络管理员已经为路由器连接到直连网络的接口配置了如图所示的IP 地址。

从路由器ping 相连网络上的主机口之间相互ping 都会遭到失败。

此问题最可能的原因是什么？必须使用no shutdown命令启用接口。

4请参见图示。

主机 A ping 主机B。

当R4 收到对以太网接口的ping 时，哪两块报头信息包括在内？（选择两项。

）目的IP 地址：192.168.10.134目的MAC 地址：9999.DADC.12345网络管理员刚把新配置输入Router1。

要将配置更改保存到NVRAM，应该执行哪一条命令？Router1# copy running-config startup-config6您需要配置图中所示的串行连接，必须在Sydney 路由器上发出以下哪条配置命令才能与Melbourne 站点建立连接？（选Sydney(config-if)# ip address 201.100.53.2 255.255.255.0Sydney(config-if)# no shutdownSydney(config-if)# clock rate 560007请参见图示。

从路由器的运行配置输出可得出什么结论？显示的命令决定了路由器的当前运行情况。

8请参见图示。

在主机 2 连接到LAN 上的交换机后，主机2 无法与主机 1 通信。

导致此问题的原因是什么？主机 1 和主机2 位于不同的网络中。

9输入以下命令的作用是什么？R1(config)# line vty 0 4R1(config-line)# password check123R1(config-line)# login设置通过Telnet 连接该路由器时使用的口令10以下哪一项正确描述了路由器启动时的顺序？加载bootstrap、加载IOS、应用配置11加载配置文件时的默认顺序是怎样的？NVRAM、TFTP、CONSOLE12请参见图示。

CCNA ENetwork Chapter 9

ENetwork Chapter 9 - CCNA Exploration: Network(2009-05-10 12:32:43)转载分类：CCNA.CISCO.ANSWERS标签：itPosted by Space on Sunday, May 10, 20091In the graphic, Host A has reached 50% completion in sending a 1 KB Ethernet frame to Host D when Host B wishes to transmit its own frame to Host C. What must Host B do?Host B can transmit immediately since it is connected on its own cable segment.Host B must wait to receive a CSMA transmission from the hub, to signal its turn.Host B must send a request signal to Host A by transmitting an interframe gap.**Host B must wait until it is certain that Host A has completed sending its frame.2 Ethernet operates at which layers of the OSI model? (Choose two.)Network layerTransport layer**Physical layerApplication layerSession layer**Data-link layer3 Which of the following describe interframe spacing? (Choose two.)**the minimum interval, measured in bit-times, that any station must wait before sending another framethe maximum interval, measured in bit-times, that any station must wait before sending another framethe 96-bit payload padding inserted into a frame to achieve a legal frame sizethe 96-bit frame padding transmitted between frames to achieve proper synchronization**the time allowed for slow stations to process a frame and prepare for the next framethe maximum interval within which a station must send another frame to avoid being considered unreachable4 What three primary functions does data link layer encapsulation provide? (Choose three.)**addressing**error detection**frame delimitingport identificationpath determinationIP address resolution5 When a collision occurs in a network using CSMA/CD, how do hosts with data to transmit respond after the backoff period has expired?**The hosts return to a listen-before-transmit mode. <—The hosts creating the collision have priority to send data.The hosts creating the collision retransmit the last 16 frames.The hosts extend their delay period to allow for rapid transmission.6 What are three functions of the upper data link sublayer in the OSI model? (Choose three.)recognizes streams of bits**identifies the network layer protocol**makes the connection with the upper layersidentifies the source and destination applications**insulates network layer protocols from changes in physical equipmentdetermines the source of a transmission when multiple devices are transmitting7 What does the IEEE 802.2 standard represent in Ethernet technologies?MAC sublayerPhysical layer**Logical Link Control sublayerNetwork layer8 Why do hosts on an Ethernet segment that experience a collision use a random delay before attempting to transmit a frame?A random delay is used to ensure a collision-free link.A random delay value for each device is assigned by the manufacturer.A standard delay value could not be agreed upon among networking device vendors.**A random delay helps prevent the stations from experiencing another collision during the transmission.98Refer to the exhibit. Which option correctly identifies content that the frame data field may contain?preamble and stop framenetwork layer packetphysical addressingFCS and SoF10Host A has an IP address of 172.16.225.93, a mask of 255.255.248.0, and a default gateway of 172.16.224.1. Host A needs to to a new host whose IP is 172.16.231.78. Host A performs the ANDing operation on its address and subnet mask. What two thin (Choose two.)Host A will get a result of 172.16.224.0 from the AND process.Host A will send on to the media a broadcast frame that contains the packet.Host A will broadcast an ARP request for the MAC of the host 172.16.231.78.Host A will change the destination IP of the packet to 172.16.224.1 and forward the packet.Host A will encapsulate the packet in a frame with a destination MAC that is the MAC address associated with 172.16.224.11 Which of the following is a drawback of the CSMA/CD access method?**Collisions can decrease network performance.It is more complex than non-deterministic protocols.Deterministic media access protocols slow network performance.CSMA/CD LAN technologies are only available at slower speeds than other LAN technologies.12 Ethernet operates at which layer of the TCP/IP network model?applicationphysicaltransportinternetdata link**network access10Ethernet operates at which layers of the OSI model? (Choose two.)Network layerTransport layerPhysical layerApplication layerSession layerData-link layer13 What is the primary purpose of ARP?translate URLs to IP addresses**resolve IPv4 addresses to MAC addressesprovide dynamic IP configuration to network devicesconvert internal private addresses to external public addresses14Refer to the exhibit. The switch and workstation are administratively configured for full-duplex operation. Which statement accurately reflects the operation of this link?**No collisions will occur on this link.Only one of the devices can transmit at a time.The switch will have priority for transmitting data.The devices will default back to half duplex if excessive collisions occur.15Refer to the exhibit. Host_A is attempting to contact Server_B. Which statements correctly describe the addressing Host_A will generate in the process? (Choose two.)A packet with the destination IP of Router_B.A frame with the destination MAC address of Switch_A.A packet with the destination IP of Router_A.**A frame with the destination MAC address of Router_A.**A packet with the destination IP of Server_B.A frame with the destination MAC address of Server_B.16 Which statements correctly describe MAC addresses? (Choose three.)dynamically assigned**copied into RAM during system startuplayer 3 address**contains a 3 byte OUI**6 bytes long32 bits long17 Which two features make switches preferable to hubs in Ethernet-based networks? (Choose two.)reduction in cross-talk**minimizing of collisionssupport for UTP cabling**division into broadcast domainsincrease in the throughput of communications18 What are the two most commonly used media types in Ethernet networks today? (Choose two.)**coaxial thicknetcopper UTPcoaxial thinnetoptical fiber**shielded twisted pair19 Convert the binary number 10111010 into its hexadecimal equivalent. Select the correct answer from the list below.8590**BAA1B31C20 After an Ethernet collision, when the backoff algorithm is invoked, which device has priority to transmit data?the device involved in the collision with the lowest MAC addressthe device involved in the collision with the lowest IP address**any device in the collision domain whose backoff timer expires first those that began transmitting at the same time。

ccna2答案第九章

ERouting Chapter 9 - CCNA Exploration: 路由协议和概念(版本4.0)1 窗体顶端在如图所示的网络中，主机192.168.1.66 无法ping 通主机192.168.1.130。

必须如何配置EIGRP 才能让这通？（选择两项。

）R1(config-router)# network 192.168.1.128R1(config-router)# auto-summaryR1(config-router)# no auto-summaryR2(config-router)# no auto-summaryR2(config-router)# auto-summaryR2(config-router)# network 192.168.1.642 窗体顶端EIGRP DUAL 算法在下列哪些表中存储到目的地的主路由？（选择两项。

）路由拓扑邻居路径最短路径哪两项陈述描述了EIGRP 的特性？（选择两项。

）EIGRP 是一种距离矢量路由协议。

EIGRP 支持无类路由和VLSM。

EIGRP 属于链路状态路由协议。

EIGRP 使用TCP 进行可靠的EIGRP 更新数据包传输。

EIGRP 使用抑制计时器来消除路径环路。

EIGRP 每30 分钟发送一次定期更新。

4 窗体顶端请参见图示。

根据show ip eigrp neighbors 的输出，Router1 和Router2 之间的邻接关系可能存在哪两个问两项。

）路由器上配置的EIGRP 进程ID 不同。

禁用了自动总结。

更改了R1 的hello 计时器。

两台路由器的串行接口位于不同的网络中。

未找到可行后继路由。

窗体底端请参见图示。

这是 2 个直连EIGRP 路由器的调试输出。

这两台路由器并未建立邻接关系。

原因是什么？其中一台路由器是非cisco 路由器它们的自治系统编号不同它们使用不同的序列号它们发送的hello 类型不正确窗体底端6 窗体顶端在运行EIGRP 的路由器上，什么数据库将维护可行后继路由的列表？路由表邻居表拓扑表邻接表窗体底端7 窗体顶端请参见图示。

CCNA考试题库中英文翻译版及答案

CCNA考试题库中英⽂翻译版及答案CCNA考试题库中英⽂翻译版及答案1[1]1、 What are two reasons that a network administrator would use access lists? (Choose two、)1、出于哪两种理由，⽹络管理员会使⽤访问列表?A、 to control vty access into a routerA、控制通过VTY访问路由器B、 to control broadcast traffic through a routerB、控制⼴播流量穿越路由器2、⼀个默认得帧中继WAN被分类为哪种物理⽹络类型?A、 point-to-pointA、点到点B、 broadcast multi-accessB、⼴播多路访问C、 nonbroadcast multi-accessC、⾮⼴播多路访问D、 nonbroadcast multipointD、⾮⼴播多点E、 broadcast point-to-multipointE、⼴播点到多点Answer: C3、 A single 802、11g access point has been configured and installed in the center of a squarA few wireless users are experiencing slow performance and drops while most users are oat peak efficiency、 What are three likely causes of this problem? (Choose three、)3、⼀个802、11接⼊点被部署在⼀个⽅形办公室得中央，当⼤多数⽤户在⼤流量传输数⼀些⽆线⽤户发现⽆线⽹络变得缓慢与出现丢包A、 mismatched TKIP encryptionB、 null SSIDC、 cordless phonesD、 mismatched SSIDE、 metalF、 antenna type or directionAnswer: CEF4、 Refer to the exhibit、 How many broadcast domains exist in the exhibited topology?根据下图，图中得拓扑中存在多少个⼴播域?A、 one A、1B、 two B、2C、 three C、3D、 four D、4E、 five E、5F、 six F、6Answer: C5、 Refer to the exhibit、 What two facts can be determined from the WLAN diagram? (Choose two、) 5、根据下图，WLAN diagram决定了哪两个事实A、 The area of overlap of the two cells represents a basic service set (BSS)、A、两个 cells得overlap得区域描述了⼀个basic service setB、 The network diagram represents an extended service set (ESS)、B、⽹络描述了⼀个extended service setC、 Access points in each cell must be configured to use channel 1、C、再每个CELL中得AP必须被配置成使⽤channel 1D、 The area of overlap must be less than 10% of the area to ensure connectivity、D、为了确保连通性,重叠区域必须⼩于10%E、 The two APs should be configured to operate on different channels、E、两个访问点应该被配置成⼯作在不同得频道Answer: BE6、 The command frame-relay map ip 10、121、16、8 102 broadcast was entered on the router、Which of the following statements is true concerning this command?6、路由器上输⼊命令frame-relay map ip 10、121、16、8 102 broadcast，以下选项正确得就是?A、 This command should be executed from the global configuration mode、A、该命令应该在全局配置模式下被执⾏B、 The IP address 10、121、16、8 is the local router port used to forward data、B、IP地址10、121、16、8就是本地路由器⽤来转发数据得接⼝C、 102 is the remote DLCI that will receive the information、C、102就是远端得DLCI它将接受信息。

CCNA_ENetwork_Chapter_10_答案(中文版)-dengwenhui

1以下哪种设备用于创建或划分广播域？集线器交换机网桥路由器中继器2网络管理员必须在网络中使用电缆长度无需中继器即可达100 米的介质。

所选介质必须价格适中并易于安装，而且需要在布制的现成建筑物内实施安装。

哪种类型的介质最适合这些要求？STPUTP同轴单模光纤多模光纤3请参见图示。

PC1 会将哪个目的IP 地址放入发往PC2 的数据包报头中？192.168.1.1192.168.2.1192.168.2.2192.168.3.1192.168.3.544一家公司计划将其网络分为多个子网，每个子网最多支持27 台主机。

哪一子网掩码可支持所需的主机数，同时让每个子网址最少？255.255.255.0255.255.255.192255.255.255.224255.255.255.240255.255.255.2485“衰减”一词在数据通信中是什么意思？信号强度随着距离的增加而减弱信号到达其目的地的时间信号从一个线对泄露到另一个线对网络设备对信号的强化6请参见图示。

网络管理员决定使用数据包捕获软件评估从学生子网到Internet 之间的所有流量。

要确保捕获所有数据包，应络设备将监控站点连接到R1 和R2 之间的网络？集线器路由器防火墙设备无线接入点7下列哪些因素使光缆比铜缆更适用于连接不同建筑物？（选择三项）每根电缆覆盖的距离更远安装成本更低EMI/RFI 的影响有限耐用的连接潜在带宽更高易于端接8将UTP 以太网端口设置为MDI 或MDIX 运行的两种常用方法是什么？（选择两项）与电缆的颜色代码关联选择和配置电缆使用电缆测试仪确定线序自动检测和协商端口的MDI/MDIX 运行通过电子方式调换发射线对和接收线对9请参见图示。

下列关于图示拓扑的陈述，哪三项是正确的？（选择三项）主机 B 和 C 位于同一个子网中。

存在五个广播域。

主机 B 使用交叉电缆连接到路由器。

存在四个广播域。

显示了五个网络。

主机 B 使用全反电缆连接到路由器。

CCNA网络学院第一学期-第九章以太网-试题及答案

第九章以太网001请参见图示。

Host_A 尝试连接Server_B。

下列哪些陈述正确描述了Host_A 在该过程中生成的地址？（选择两项）A 以Router_B为目的IP地址的数据包B 以Switch_A为目的MAC地址的帧C 以Router_A为目的IP地址的数据包D 以Router_A 为目的MAC 地址的帧E 以Server_B为目的IP地址的数据包F 以Server_B为目的MAC地址的帧002 OSI 模型中的数据链路上子层有哪三项功能？（选择三项）A 识别比特流B 确定网络层协议C 建立与上层的连接D 标识源应用程序和目的应用程序E 使网络层协议不受物理设备变化的影响F 多台设备同时传输时确定传输来源003在图示中，当主机B 希望向主机C 传输帧时，主机A 向主机D 发送1 KB 以太网帧的过程已完成了50%。

主机 B 必须执行什么操作？A主机B可以立即传输，因为它连接到自己的电缆部分中B主机B必须等待接收从集线器传输的CSMA，这表示轮到它传输C主机B 必须通过传输帧间隙向主机A 发送请求信号D主机B 必须等待，直至确定主机A 已完成帧发送004采用CSMA/CD 的网络中发生冲突时，需要传输数据的主机在回退时间到期后做何反应？A 主机恢复传输前侦听模式B 造成冲突的主机优先发送数据C 造成冲突的主机重新传输最后16个帧D 主机延长其延迟时间以便快速传输005发生以太网冲突之后调用回退算法时，哪种设备优先传输数据？A 涉入冲突的设备中MAC 地址最小的设备B 涉入冲突的设备中IP 地址最小的设备C 冲突域中回退计时器首先过期的任何设备D 同时开始传输的设备006以下哪些陈述正确描述了MAC 地址？（选择三项）A 动态分配B 系统启动期间复制到RAM 中C 第3 层地址D 包含3 字节OUIE 长6 个字节F 长32 位007以太网网段中的主机发生冲突时，为何要在尝试传输帧之前随机延迟？A 随机延迟旨在确保无冲突链路B 每台设备的随机延迟值由制造商指定C 网络设备厂商之间无法就标准的延迟值达成一致D 随机延迟有助于防止站点在传输期间再次发生冲突008请参见图示。

CCNA 2 Chapter 9 V4.0 Answers

1. What two actions will the EIGRP DUAL FSM take if a link to a network goes down? (Choose two.)put the route into passive modequery neighbors for a new routesearch routing table for a feasible successorrun the SPF algorithm to find a new successorsearch topology table for a feasible successor2.Host 192.168.1.66 in the network illustrated is unable to ping host 192.168.1.130. How must EIGRP be configured to enable connectivity between the two hosts? (Choose two.)R1(config-router)# network 192.168.1.128R1(config-router)# auto-summaryR1(config-router)# no auto-summaryR2(config-router)# no auto-summaryR2(config-router)# auto-summaryR2(config-router)# network 192.168.1.643.Refer to the exhibit. The company is using EIGRP with an autonomous system number of 10. Pings between hosts on networks that are connected to router A and those that are connected to router B are successful. However, users on the 192.168.3.0 network are unable to reach users on the 192.168.1.32 network. What is the most likely cause of this problem? IP classless is enabled and is causing the packet to drop.The command network 192.168.1.32 was not issued on router C.The routers are not configured in the same EIGRP routing domain. Automatic summarization of the networks is causing the subnetted routes to be dropped.4. What information is maintained in the EIGRP topology database for a destination route? (Choose three.)the routing protocolthe feasible distance of the routethe highest cost of the routethe SRTT value for the routethe route cost as advertised by the neighboring routerthe physical address of the gateway interface5. On a router running EIGRP, what database would maintain a list of feasible successors?routing tableneighbor tabletopology tableadjacency table6.Refer to the exhibit. What is indicated by the P at the beginning of the topology entry?the route is in a stable statethe route is a preferred routeDUAL is searching for a better route to this destinationthe exit interface is in passive mode and EIGRP advertisements are blocked7. In the command router eigrp 20, what is the purpose of the number 20? specifies the administrative distance for all EIGRP routesidentifies the autonomous system number this EIGRP process will advertise determines what metric is added to all advertised routesindicates the number of addresses in the EIGRP routing domain8.Refer to the exhibit. In the topology table, what do the numbers 3011840 and 3128695 represent?the route metric that is applied to those EIGRP routes for this router the trustworthiness of the routing information sourcethe composite of the hop count and bandwidth to that destination network the total metric for that network as advertised by the EIGRP neighbor9. Which two statements describe characteristics of EIGRP? (Choose two.) EIGRP is a distance vector routing protocol.EIGRP supports classless routing and VLSM.EIGRP is classified as a link-state routing protocol.EIGRP uses TCP for reliable delivery of EIGRP update packets.With EIGRP, loop-free paths are achieved through the use of hold-down timers.EIGRP sends a periodic update every 30 minutes.10.Refer to the exhibit. Network 192.168.0.0/28 goes down. What type of packet does Router2 immediately send to Router1 and Router3?a query for network 192.168.0.0/28an acknowledgment packet to 224.0.0.9an update packet that is sent to 255.255.255.255a packet that contains the new routing table for R2unicast update packets to 192.168.1.1 and 192.168.2.111.Refer to the exhibit. Which command will advertise the 192.168.1.64/30 network but not the 192.168.1.32 network on router A?network 192.168.1.0network 192.168.1.0 255.255.255.0network 192.168.1.64 0.0.0.3network 192.168.1.64 0.0.0.7network 192.168.1.64 0.0.0.25512. What administrative distance would a router assign to a default route in EIGRP that is learned from a source external to the autonomous system? 15709017019013. In which of the following tables does the EIGRP DUAL algorithm store the primary route to a destination? (Choose two.)routingtopologyneighborpathshortest path14. Which of the following types of routes will be denoted by EX in EIGRP routing table entries? (Choose two.)routes learned from other routing protocolsroutes learned from any non-adjacent EIGRP routersany route with a hop count metric higher than 224EIGRP routes that originate in different autonomous systemsall passive routes in the routing table15. Which term defines a collection of networks under the administrative control of a single entity that presents a common routing policy to the Internet?autonomous systemcontiguous networksprocess IDBGP16.Refer to the exhibit. EIGRP is the only routing protocol enabled on this network. No static routes are configured on this router. What can be concluded about network 198.18.1.0/24 from the exhibited output?A route to network 198.18.1.0/24 is not listed in the routing table. Packets that are destined for 198.18.1.0/24 will be forwarded to 198.18.10.6.EIGRP will perform equal cost load balancing across two paths when forwarding packets to 198.18.1.0/24.The router with interface 172.16.3.2 is a successor for network198.18.1.0/24.17.Refer to the exhibit. All interfaces have been configured with the bandwidths that are shown in the exhibit. Assuming that all routers are using a default configuration of EIGRP as their routing protocol, what path will packets take from the 172.16.1.0/16 network to the192.168.200.0/24 network?A,B,EA,C,EA,D,EPackets will load balance across the A,B,E and A,C,E paths.Packets will load balance across the A,B,E and A,D,E paths.Packets will load balance across the A,C,E and A,D,E paths.18.By default, which two metrics are used by EIGRP to determine the best path between networks?MTUloaddelaybandwidthreliability19. Which of the following statements describes the bounded updates used by EIGRP?Bounded updates are sent to all routers within an autonomous system. Partial updates are sent only to routers that need the information. The updates are sent to all routers in the routing table.Updates are bounded by the routers in the topology table.20. The show ip eigrp topology command output on a router displays a successor route and a feasible successor route to network 192.168.1.0/24. In order to reduce processor utilization, what does EIGRP do when the primary route to this network fails?The router sends query packets to all EIGRP neighbors for a better route to network 192.168.1.0/24.The DUAL FSM immediately recomputes the algorithm to calculate the next backup route.Packets that are destined for network 192.168.1.0/24 are sent out the default gateway instead.The backup route to network 192.168.1.0/24 is installed in the routing table.21.Refer to the exhibit. Based on the output of show ip eigrp neighbors, what are two possible problems with adjacencies between Router1 and Router2? (Choose two.)The routers are configured with different EIGRP process IDs.Automatic summarization was disabled.The hello timer for R1 was altered.The serial interfaces for both routers are in different networks.No feasible successors were found.22.Refer to the exhibit. This is the debug output from 2 directly connected EIGRP routers. They are not forming an adjacency. What is the cause? one router is a non-cisco routerthey have different autonomous-system numbersthey are using difference sequence numbersthey are sending incorrect hello types。

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