【正文】 he most likely cause of this? A. The portfast feature is not enabled on all switch ports. B. The PCs are in two different VLANs. C. Spanning Tree Protocol is not running on the switches. D. PC2 is down and is not able to respond to the request. E. The VTP versions running on the two switches do not match. Answer: C解釋一下：PC1發(fā)出一個(gè)ARP request的數(shù)據(jù)報(bào)，并且是以廣播的形式發(fā)送出去的。當(dāng)ARP報(bào)文傳到switch 2，交換機(jī)對廣播的流量是以泛洪的形式處理的，報(bào)文就從除了連接PC1的接口外的所有接口都發(fā)出去了。Switch1收到廣播后也泛洪，因此一個(gè)廣播環(huán)路就產(chǎn)生了，所以在感覺網(wǎng)絡(luò)性能很差，因?yàn)閺V播的流量占有了很大的帶寬。而我們阻斷二層環(huán)路是通過生成樹來實(shí)現(xiàn)的，在圖中有環(huán)路存在因此就說明沒有運(yùn)行生成樹了。56. An administrator issues the mand ping from the mand line prompt on a PC. If a reply is received, what does this confirm? A. The PC has connectivity with a local host. B. The PC has connectivity with a Layer 3 device. C. The PC has a default gateway correctly configured. D. The PC has connectivity up to Layer 5 of the OSI model. E. The PC has the TCP/IP protocol stack correctly installed. Answer: E解釋一下：，他是一個(gè)回環(huán)的地址，一般用于測試，測試TCP/IP協(xié)議棧是否起來了。在一臺(tái)PC上能ping 。59. Refer to the exhibit. The network administrator requires easy configuration options and minimal routing protocol traffic. What two options provide adequate routing table information for traffic that passes between the two routers and satisfy the requests of the network administrator? (Choose two.) A. a dynamic routing protocol on InternetRouter to advertise all routes to CentralRouter. B. a dynamic routing protocol on InternetRouter to advertise summarized routes to CentralRouter. C. a static route on InternetRouter to direct traffic that is destined for D. a dynamic routing protocol on CentralRouter to advertise all routes to InternetRouter. E. a dynamic routing protocol on CentralRouter to advertise summarized routes to InternetRouter. F. a static, default route on CentralRouter that directs traffic to InternetRouter. Answer: CF解釋一下：因?yàn)樵谶@個(gè)圖中,internetRouter要訪問內(nèi)網(wǎng) 。同樣內(nèi)網(wǎng)要訪問外部，也只能通過路由器InternetRouter才能到達(dá)，所以也可以在CentralRouter上配置一條缺省路由到外部。 are some of the advantages of using a router to segment the network? (Choose two.) A. Filtering can occur based on Layer 3 information. B. Broadcasts are eliminated. C. Routers generally cost less than switches. D. Broadcasts are not forwarded across the router. E. Adding a router to the network decreases latency. Answer: AD解釋一下：這里問的是用路由器來分割一個(gè)網(wǎng)絡(luò)的好處是什么。路由器是工作在三層的設(shè)備，因此我們可以基于三層的信息來實(shí)現(xiàn)過濾；而且大家知道路由器是可以過濾廣播的。這些應(yīng)該就都是他分割一個(gè)網(wǎng)絡(luò)的好處了。要注意路由器只是能阻斷廣播，讓他不能從一個(gè)域中傳播到另一個(gè)域中，他是沒辦法消除廣播的。61. Refer to the exhibit. What is the meaning of the output MTU 1500 bytes? A. The maximum number of bytes that can traverse this interface per second is 1500. B. The minimum segment size that can traverse this interface is 1500 bytes. C. The maximum segment size that can traverse this interface is 1500 bytes. D. The minimum packet size that can traverse this interface is 1500 bytes. E. The maximum packet size that can traverse this interface is 1500 bytes. F. The maximum frame size that can traverse this interface is 1500 bytes.Answer: E解釋一下：MTU是最小傳輸單元的意思，表示在這個(gè)接口上傳輸?shù)淖畲笞止?jié)為1500，如果超過這個(gè)值，包就需要被分片。62. There are no boot system mands in a router configuration in NVRAM. What is the fallback sequence that the router will use to find an IOS during reload? A. TFTP server, Flash, NVRAM B. ROM, NVRAM, TFTP server C. NVRAM, TFTP server, ROM D. Flash, TFTP server, ROM E. Flash, NVRAM, ROM Answer: D解釋一下：這個(gè)問的是路由器尋找IOS的過程。1．路由器在POST后，先查看寄存器的值，這個(gè)值是一組4個(gè)十六進(jìn)制的數(shù)字，而其中的最后的一位影響啟動(dòng)的過程。2．在NVRAM的配置文件中查看boot system命令，這個(gè)命令告訴引導(dǎo)程序在哪里尋找IOS。在這個(gè)題中說沒有boot system的命令保存在NVRAM中。所以這步跳過。3．如果在NVRAM的配置文件中沒有找到boot system命令，引導(dǎo)程序使用flash中所找到的第一個(gè)有效的IOS鏡像。4．如果flash中沒有有效的IOS鏡像，引導(dǎo)程序?qū)⑸梢粋€(gè)TFTP本地廣播以定位TFTP服務(wù)器。5．如果沒有找到TFTP服務(wù)器，引導(dǎo)程序?qū)⒓虞dROM中的迷你IOS（RXBOOT 模式）6．如果ROM中有迷你IOS，那么迷你IOS在隨后加載并且進(jìn)入RXBOOT模式；否則路由器不是重新試圖尋找IOS鏡像，就是加載ROMMON并且進(jìn)入ROM Monitor模式。這樣看，答案就很明顯了。 which circumstance are multiple copies of the same unicast frame likely to be transmitted in a switched LAN? A. during high traffic periods B. after broken links are reestablished C. when upperlayer protocols require high reliability D. in an improperly implemented redundant topology E. when a dual ring topology is in use Answer: D解釋一下：在一個(gè)LAN中有若干的單播幀的拷貝。在一般來說我們的每個(gè)單播幀都是只有一個(gè)目的地，從而從一個(gè)相關(guān)接口發(fā)送出去就可。如果有若干個(gè)單播幀的拷貝就表示我的交換機(jī)上同這個(gè)目的地址綁定的接口有多個(gè)，而這些都應(yīng)該是不必要的。因?yàn)槲业揭粋€(gè)目的地從一條路走就可以了，如果出現(xiàn)了多條路，那么就應(yīng)該是做冗余的，可是不正確的配置可能導(dǎo)致我的LAN中產(chǎn)生環(huán)路，從而形成在LAN中有同一個(gè)幀的多個(gè)拷貝。 of the following describe private IP addresses? (Choose two.) A. addresses chosen by a pany to municate with the Internet B. addresses that cannot be routed through the public Internet C. addresses that can be routed through the public Internet D. a scheme to conserve public addresses E. addresses licensed to enterprises or ISPs by an Internet registry organization Answer: BD解釋一下：私有IP地址是不能在公網(wǎng)上傳遞的。他只能在一個(gè)單獨(dú)的區(qū)域中使用，如果一個(gè)使用私有地址的設(shè)備需要同外網(wǎng)通訊，可以通過NAT將這個(gè)私有地址轉(zhuǎn)換為公有地址，這樣也可以達(dá)到隱藏地址的目的。外網(wǎng)知道的只能是你通過NAT轉(zhuǎn)換后的公有地址，而無法知道你的正在使用的那個(gè)私有地址的。 to the exhibit. A network administrator is adding two new hosts to SwitchA. Which three values could be used for the configuration of these hosts? (Choose three.) A. host 1 IP address: B. host 1 IP address: C. host 1 default gateway: D. host 2 IP address: E. host 2 default gateway: F. host 2 IP address: Answer: ACF解釋一下：我們可以看路由器上的子接口的配置：接口fa0/ 并被劃分到vlan 10中，接口fa0/ 20中了。接下來我們來看交換機(jī)上的接口的vlan分配：和host A相連的接口f0/6劃分到了vlan 10，而和host B相連的接口f0/9劃分到了vlan 20。因?yàn)橹挥邢嗤瑅lan中數(shù)據(jù)才可以通訊，所以我們應(yīng)該將host A的地址和f0/，而將host B的地址和f0/。并且因?yàn)橹鳈C(jī)是沒有路由的功能的，我們需要給他們指定網(wǎng)關(guān)，而他們的網(wǎng)關(guān)地址應(yīng)該是相應(yīng)VLAN中的路由器的子接口的地址。所以，host （）的地址。Host ()的地址，..。 of the following statements are true regarding bridges and switches? (Choose 3.) A. Switches are primarily software based while bridges are hardware based. B. Both bridges and switches forward Layer 2 broadcasts. C. Bridges are frequently faster than switches. D. Switches have a higher number of ports than most bridges. E. Bridges define broadcast domains while switches define collision domains. F. Both bridges and switches make forwarding decisions