思科認(rèn)證考試題庫

..CCNA640-802V13題庫試題分析題庫講解：吳老師〔艾迪飛CCIE實(shí)驗(yàn)室首發(fā)網(wǎng)站：1.Whataretworeasonsthatanetworkadministratorwoulduseaccesslists?<Choosetwo.>A.tocontrolvtyaccessintoarouterB.tocontrolbroadcasttrafficthrougharouterC.tofiltertrafficasitpassesthrougharouterD.tofiltertrafficthatoriginatesfromtherouterE.toreplacepasswordsasalineofdefenseagainstsecurityincursionsAnswer:AC解釋一下：在VTY線路下應(yīng)用ACL,可以控制從VTY線路進(jìn)來的telnet的流量。也可以過濾穿越一臺(tái)路由器的流量。2.AdefaultFrameRelayWANisclassifiedaswhattypeofphysicalnetwork?A.point-to-pointB.broadcastmulti-accessC.nonbroadcastmulti-accessD.nonbroadcastmultipointE.broadcastpoint-to-multipointAnswer:C解釋一下：在默認(rèn)的情況下,幀中繼為非廣播多路訪問鏈路。但是也可以通過子接口來修改他的網(wǎng)絡(luò)的類型。Refertotheexhibit.Howmanybroadcastdomainsexistintheexhibitedtopology?A.oneB.twoC.threeD.fourE.fiveF.sixAnswer:C解釋一下：廣播域的問題,在默認(rèn)的情況下,每個(gè)交換機(jī)是不能隔離廣播域的,所以在同一個(gè)區(qū)域的所有交換機(jī)都在同一個(gè)廣播域中,但是為了減少廣播的危害,將廣播限制在一個(gè)更小的范圍,有了VLAN的概念,VLAN表示的是一個(gè)虛擬的局域網(wǎng),而他的作用就是隔離廣播。所以被VLAN隔離了的每個(gè)區(qū)域都表示一個(gè)單獨(dú)的廣播域,這樣一個(gè)VLAN中的廣播的流量是不能傳到其他的區(qū)域的,所以在上題中就有3個(gè)廣播域了。4.Asingle802.11gaccesspointhasbeenconfiguredandinstalledinthecenterofasquareoffice.Afewwirelessusersareexperiencingslowperformanceanddropswhilemostusersareoperatingatpeakefficiency.Whatarethreelikelycausesofthisproblem?<Choosethree.>A.mismatchedTKIPencryptionB.nullSSIDC.cordlessphonesD.mismatchedSSIDE.metalfilecabinetsF.antennatypeordirectionAnswer:CEF6.Thecommandframe-relaymapip102broadcastwasenteredontherouter.Whichofthefollowingstatementsistrueconcerningthiscommand?A.Thiscommandshouldbeexecutedfromtheglobalconfigurationmode.B.TheIPaddressisthelocalrouterportusedtoforwarddata.C.102istheremoteDLCIthatwillreceivetheinformation.D.ThiscommandisrequiredforallFrameRelayconfigurations.E.Thebroadcastoptionallowspackets,suchasRIPupdates,tobeforwardedacrossthePVC.Answer:E解釋一下：關(guān)于命令frame-relaymapip102broadcast,這個(gè)命令用于手工靜態(tài)添加一條映射,到達(dá)的流量封裝一個(gè)DLCI號(hào)為102,而且這條PVC是支持廣播的流量的,比如RIP的更新包。因?yàn)樵谀J(rèn)的情況下,幀中繼的網(wǎng)絡(luò)為非廣播的,而RIP在其上是無法發(fā)包的。8．WhichofthefollowingareassociatedwiththeapplicationlayeroftheOSImodel?<Choosetwo.>A.pingB.TelnetC.FTPD.TCPE.IPAnswer:BC解釋一下：在OSI7層模型中位于應(yīng)用層的應(yīng)用有telnet和ftp這兩種應(yīng)用。9.Forsecurityreasons,thenetworkadministratorneedstopreventpingsintothecorporatenetworksfromhostsoutsidetheinternetwork.Whichprotocolshouldbeblockedwithaccesscontrollists?A.IPB.ICMPC.TCPD.UDPAnswer:B解釋一下：PING命令利用ICMP協(xié)議的echo,和echo-replay兩個(gè)報(bào)文來檢測鏈路是否連通的。所以如果要阻止PING的流量到網(wǎng)絡(luò),就只要過濾掉ICMP的應(yīng)用就可以了。10．Refertotheexhibit.ThenetworkadministratorhascreatedanewVLANonSwitch1andaddedhostCandhostD.TheadministratorhasproperlyconfiguredswitchinterfacesFastEthernet0/13throughFastEthernet0/24tobemembersofthenewVLAN.However,afterthenetworkadministratorcompletedtheconfiguration,hostAcouldcommunicatewithhostB,buthostAcouldnotcommunicatewithhostCorhostD.Whichcommandsarerequiredtoresolvethisproblem?A.Router<config>#interfacefastethernet0/1.3Router<config-if>#encapsulationdot1q3Router<config-if>#ipaddressB.Router<config>#routerripRouter<config-router>#networkRouter<config-router>#networkRouter<config-router>#networkC.Switch1#vlandatabaseSwitch1<vlan>#vtpv2-modeSwitch1<vlan>#vtpdomainciscoSwitch1<vlan>#vtpserverD.Switch1<config>#interfacefastethernet0/1Switch1<config-if>#switchportmodetrunkSwitch1<config-if>#switchporttrunkencapsulationislAnswer:A解釋一下：這是一個(gè)多VLAN間通訊的問題,雖然都同在一臺(tái)交換機(jī)上,但是由于處在不同的VLAN中,而導(dǎo)致了不同VLAN中的主機(jī)是不能通訊的。這時(shí)我們就需要借助與trunk和三層的路由功能了,在交換機(jī)和路由器之間封裝TRUNK,這樣可以允許交換機(jī)間的二層的通訊,但是由于兩個(gè)VLAN是劃分到不同的網(wǎng)段中的,因此需要借助路由器的路由功能來實(shí)現(xiàn)三層的可達(dá),可以將VLAN中的主機(jī)的網(wǎng)關(guān)指定為路由器與該VLAN相連的子接口的地址,這樣VLAN中的數(shù)據(jù)包就都會(huì)發(fā)往網(wǎng)關(guān),而由網(wǎng)關(guān)來進(jìn)行進(jìn)一步的轉(zhuǎn)發(fā)。在這個(gè)題中,題目給出了路由器的的子接口的網(wǎng)段,而又給出了VLAN2與路由器相連的接口的IP地址,所以剩下的一個(gè)網(wǎng)段就是給VLAN3的了,所以要在路由器上將與一個(gè)子接口劃分到VLAN3,并給其分配另一個(gè)網(wǎng)段中的IP地址。這樣就可以了。11．Whataretworecommendedwaysofprotectingnetworkdeviceconfigurationfilesfromoutsidenetworksecuritythreats?<Choosetwo.>A.AllowunrestrictedaccesstotheconsoleorVTYports.B.Useafirewalltorestrictaccessfromtheoutsidetothenetworkdevices.C.AlwaysuseTelnettoaccessthedevicecommandlinebecauseitsdataisautomaticallyencrypted.D.UseSSHoranotherencryptedandauthenticatedtransporttoaccessdeviceconfigurations.E.Preventthelossofpasswordsbydisablingpasswordencryption.Answer:BD解釋一下：要確保外部的安全的站點(diǎn)才可以訪問我的網(wǎng)絡(luò),這就涉及到了安全的問題了,我們可以使用防火墻來限制外網(wǎng)中來的設(shè)備；也可以通過SSH或加密和認(rèn)證來控制。12．Refertotheexhibit.TheaccesslisthasbeenconfiguredontheS0/0interfaceofrouterRTBintheoutbounddirection.Whichtwopackets,ifroutedtotheinterface,willbedenied?<Choosetwo.>access-list101denytcp2.15anyeqtelnetaccess-list101permitipanyanyA.sourceipaddress:;destinationport:21B.sourceipaddress:,7destinationport:21C.sourceipaddress:,1destinationport:21D.sourceipaddress:,6destinationport:23.46;destinationport:23F.sourceipaddress:,9destinationport:23Answer:DE解釋一下：這個(gè)訪問列表定義了兩個(gè)語句：access-list101denytcp2.15anyeqtelnetaccess-list101permitipanyany在訪問列表中匹配的順序是從上到下,如果匹配了某一句,就退出訪問列表,如果沒有就一直往下匹配,在訪問列表中有一句隱含的拒絕所有。所以不管怎么樣都有一句是能被匹配的。在上題中,他定義的第一句是拒絕到從發(fā)出的任何的telnet的流量,然后第二句定義的就是允許所有的IP流量。而且要明確telnet的流量使用的是端口23,所以這個(gè)題的答案就很明確了。Refertotheexhibit.Switch1hasjustbeenrestartedandhaspassedthePOSTroutine.HostAsendsitsinitialframetoHostC.Whatisthefirstthingtheswitchwilldoasregardspopulatingtheswitchingtable?A.Switch1willaddtotheswitchingtable.B.Switch1willadd2totheswitchingtable.C.Switch1willadd000A.8AD.Switch1willadd000B.DB95.2EE9totheswitchingtable.Answer:C解釋一下：交換機(jī)重新啟動(dòng)了,這個(gè)時(shí)候交換機(jī)的MAC地址表是空的,當(dāng)主機(jī)A發(fā)送數(shù)據(jù)給主機(jī)C而經(jīng)過交換機(jī)時(shí),交換機(jī)根據(jù)他的工作的原理他要進(jìn)行原MAC地址學(xué)習(xí),而因?yàn)閷?duì)于這個(gè)目的MAC地址無記錄,而將這個(gè)流量從除收到的這個(gè)接口外的所有接口泛洪出去。所以在最開始的一步中,交換機(jī)是記錄下主機(jī)A的MAC地址到他的MAC地址表中。14.heuserofHost1wantstopingtheDSLmodem/routerat54.BasedontheHost1ARPtablethatisshownintheexhibit,whatwillHost1do?A.sendaunicastARPpackettotheDSLmodem/routerB.sendunicastICMPpacketstotheDSLmodem/routerC.sendLayer3broadcastpacketstowhichtheDSLmodem/routerrespondsD.sendaLayer2broadcastthatisreceivedbyHost2,theswitch,andtheDSLmodem/routerAnswer:B解釋一下：在下面的表中我們可以看到ARP表中有關(guān)于的ARP條目,所以在這主機(jī)都只需要發(fā)送單播的ICMP包到DSLmodem/router即可。15.Refertotheexhibit.WhatisthemostefficientsummarizationthatR1canusetoadvertiseitsnetworkstoR2?DAnswer:C解釋一下：這還是一個(gè)關(guān)于匯總的問題。要求R1將所有的網(wǎng)段用匯總的條目發(fā)送給R2,因?yàn)檫@些條目的網(wǎng)絡(luò)位是相同的都為172.1,所以在這需要匯總的只是第3個(gè)八位,將4,4,5,6,7這些寫成二進(jìn)制的形式,然后找出相同的位數(shù),則有相同位數(shù)的字節(jié)就是他們的掩碼的位數(shù),而最小的有相同位的最小的數(shù)字就是他們的基數(shù)位,所以R1通告出去匯總的條目為。16.Refertotheexhibit.Assumethatallrouterinterfacesareoperationalandcorrectlyconfigured.Inaddition,assumethatOSPFhasbeencorrectlyconfiguredonrouterR2.HowwillthedefaultrouteconfiguredonR1affecttheoperationofR2?A.AnypacketdestinedforanetworkthatisnotdirectlyconnectedtorouterR1willbedropped.B.AnypacketdestinedforanetworkthatisnotdirectlyconnectedtorouterR2willbedroppedimmediately.C.AnypacketdestinedforanetworkthatisnotdirectlyconnectedtorouterR2willbedroppedimmediatelybecauseofthelackofagatewayonR1.D.ThenetworksdirectlyconnectedtorouterR2willnotbeabletocommunicatewiththe,28,and4subnetworks.E.AnypacketdestinedforanetworkthatisnotreferencedintheroutingtableofrouterR2willbedirectedtoR1.R1willthensendthatpacketbacktoR2andaroutingloopwilloccur.Answer:E解釋一下：在R1上產(chǎn)生了一個(gè)OSPF的缺省路由,出接口指定為S0/0,這條缺省路由以5類LSA的形式通告給了R2,于是R2上也有了一條標(biāo)記為O*E2.0/0出接口為Serial0/0的路由。所以R2收到任何路由表中沒有的目的網(wǎng)段時(shí),就將指定給R1,而R1根據(jù)缺省路由的出接口又將數(shù)據(jù)包發(fā)往R2,這樣就形成了一個(gè)路由的環(huán)路。17.Anetworkinterfaceporthascollisiondetectionandcarriersensingenabledonasharedtwistedpairnetwork.Fromthisstatement,whatisknownaboutthenetworkinterfaceport?A.Thisisa10Mb/sswitchport.B.Thisisa100Mb/sswitchport.C.ThisisanEthernetportoperatingathalfduplex.D.ThisisanEthernetportoperatingatfullduplex.E.ThisisaportonanetworkinterfacecardinaPC.Answer:C解釋一下：一個(gè)接口有沖突檢測和載波偵聽,而且是使用雙絞線的網(wǎng)絡(luò),那么對(duì)于這個(gè)接口我們可以推測出他是以太接口,而且是工作在半雙工的模式下。20.Refertothetopologyandrouterconfigurationshowninthegraphic.AhostontheLANisaccessinganFTPserveracrosstheInternet.Whichofthefollowingaddressescouldappearasasourceaddressforthepacketsforwardedbytheroutertothedestinationserver?.1B..2C.3D.7E.7F.8Answer:D解釋一下：這是個(gè)NAT地址轉(zhuǎn)換的題目,在這f0/0接口連接下的為私有的地址,這些地址是不能同外網(wǎng)進(jìn)行通訊的,這時(shí)就借助NAT,將內(nèi)網(wǎng)的私有地址轉(zhuǎn)換為可以在公網(wǎng)上通訊的地址,我們看到NATPOOL中定義的轉(zhuǎn)換后的公有地址為到,則表示這段地址是我轉(zhuǎn)換后的內(nèi)網(wǎng)全局地址,所以HOST想要穿過INTERNET訪問FTP服務(wù)器,則需要轉(zhuǎn)換為公有地址到之內(nèi)的地址,在上面的答案中只有地址滿足條件,所以答案就是D了。21.AcompanyisinstallingIPphones.Thephonesandofficecomputersconnecttothesamedevice.Toensuremaximumthroughputforthephonedata,thecompanyneedstomakesurethatthephonetrafficisonadifferentnetworkfromthatoftheofficecomputerdatatraffic.Whatisthebestnetworkdevicetowhichtodirectlyconnectthephonesandcomputers,andwhattechnologyshouldbeimplementedonthisdevice?<Choosetwo.>A.hubB.routerC.switchD.STPE.subinterfacesF.VLANAnswer:CF解釋一下：公司的語音設(shè)備和辦公的設(shè)備都連在相同的設(shè)備上,還要確保語音的數(shù)據(jù)流在不同與公司的辦公的數(shù)據(jù)流量,最好的網(wǎng)絡(luò)設(shè)備當(dāng)然是交換機(jī)了,然后利用VLAN的技術(shù)就完全可以滿足所有的要求了。22.Refertotheexhibit.WhichstatementdescribesDLCI17?A.DLCI17describestheISDNcircuitbetweenR2andR3.B.DLCI17describesaPVConR2.ItcannotbeusedonR3orR1.C.DLCI17istheLayer2addressusedbyR2todescribeaPVCtoR3.D.DLCI17describesthedial-upcircuitfromR2andR3totheserviceprovider.Answer:C解釋一下：DLCI是在Frame-relay中的描述二層信息的地址,他的地位等同于以太網(wǎng)中的MAC地址。我們以R2上的DLCI17來看,DLCI17描述的是：從這個(gè)接口出去的目的地為R3的接口的這條PVC的二層的地址為17。23.Whichroutingprotocolbydefaultusesbandwidthanddelayasmetrics?A.RIPB.BGPC.OSPFD.EIGRPAnswer:D解釋一下：在我們的路由協(xié)議中使用復(fù)合度量的協(xié)議只有IGP和EIGPR,而他們?cè)谀J(rèn)的情況下是使用帶寬和延時(shí)來計(jì)算度量的。25.IntheimplementationofVLSMtechniquesonanetworkusingasingleClassCIPaddress,whichsubnetmaskisthemostefficientforpoint-to-pointseriallinks?A.B.40C.48D.52E.54Answer:D解釋一下：在點(diǎn)到點(diǎn)的鏈路上因?yàn)橹恍枰峙鋬蓚€(gè)地址給兩端就可以了,所以加上網(wǎng)絡(luò)地址和廣播地址,這個(gè)網(wǎng)段也就只需要有4個(gè)地址了,所以網(wǎng)絡(luò)位需要匹配30位,掩碼就為．26.swillR1forwardtoR2?<Choosetwo.>A.60B.1C.D.55E.F.5Answer:BE解釋一下：這個(gè)題其實(shí)就是考察的匯總的問題,他說的意思是R2發(fā)送了一個(gè)匯總的路由給R1,哪兩個(gè)包文的目的地R1仍將轉(zhuǎn)發(fā)給R2。這還是匯總的問題的一個(gè)反向的考察,根據(jù)21位的掩碼位數(shù)可以推斷在第3個(gè)八位字節(jié)的前5位是相同的,不同的是后面的3位,而將176寫成二進(jìn)制的形式為10110000,所以可以看出來明細(xì)的路由可以是176-183,所以在上面的答案中可以很容易看到答案B和E是我們的明細(xì)路由。27.Refertotheexhibit.Switch-1needstosenddatatoahostwithaMACaddressof00b0.d056.efa4.WhatwillSwitch-1dowiththisdata?A.Switch-1willdropthedatabecauseitdoesnothaveanentryforthatMACaddress.B.Switch-1willfloodthedataoutallofitsportsexcepttheportfromwhichthedataoriginated.C.Switch-1willsendanARPrequestoutallitsportsexcepttheportfromwhichthedataoriginated.D.Switch-1willforwardthedatatoitsdefaultgateway.Answer:B解釋一下：首先Switch1需要發(fā)送一個(gè)數(shù)據(jù)到MAC地址為的主機(jī),了解到目的地后,就查看他的MAC地址表,然后發(fā)現(xiàn)在MAC地址表中沒有這個(gè)MAC地址的條目存在。交換機(jī)在收到未知的單播,組播和廣播時(shí),都采用的是泛洪的方式,往除收到數(shù)據(jù)的這個(gè)接口外的所有接口都發(fā)送。所以在這兒,Switch1也采取的上泛洪的方式。28.woroutersnamedAtlantaandBrevardareconnectedbytheirserialinterfacesasshownintheexhibit,butthereisnodataconnectivitybetweenthem.TheAtlantarouterisknowntohaveacorrectconfiguration.Giventhepartialconfigurationsshownintheexhibit,whatistheproblemontheBrevardrouterthatiscausingthelackofconnectivity?A.Aloopbackisnotset.B.TheIPaddressisincorrect.C.Thesubnetmaskisincorrect.D.Theseriallineencapsulationsareincompatible.E.Themaximumtransmissionunit<MTU>sizeistoolarge.F.Thebandwidthsettingisincompatiblewiththeconnectedinterface.Answer:B解釋一下：很明顯的錯(cuò)誤啊,兩臺(tái)路由器的串行接口的地址配置錯(cuò)誤,不是在相同的網(wǎng)段,從而導(dǎo)致了不能通訊。29.WhichtwovaluesareusedbySpanningTreeProtocoltoelectarootbridge?<Choosetwo.>A.amountofRAMB.bridgepriorityC.IOSversionD.IPaddressE.MACaddressF.speedofthelinksAnswer:BE解釋一下：生成樹的選舉的問題,根橋的選舉是通過比較ＢＩＤ的,而ＢＩＤ由橋優(yōu)先級(jí)和ＭＡＣ地址組成的．所以在選根橋的時(shí)候需要比較的是橋優(yōu)先級(jí)和ＭＡＣaddress。30.Refertotheexhibit.Whichswitchprovidesthespanning-treedesignatedportroleforthenetworksegmentthatservicestheprinters?Switch1B.Switch2C.Switch3D.Switch4Answer:C解釋一下：這是個(gè)關(guān)于生成樹選舉的問題,我們首先需要找到根橋,而根橋的選舉是通過比較橋ID的,而且是越小越優(yōu)先,橋ID的組成為橋優(yōu)先級(jí)和MAC地址。所以我們通過上圖可以找到根橋?yàn)閟witch1。然后在非根橋上選出根端口,通過比較到根橋的花費(fèi)來選舉的,花費(fèi)最小的就是根端口。因?yàn)樯蠄D中沒有表示出鏈路的帶寬,所以無法比較他們的花費(fèi)。下一步我們來選舉指派端口。每條鏈路都需要有一個(gè)DP,先是比較花費(fèi),如果花費(fèi)相同則比較BID〔橋優(yōu)先級(jí),仍是越小越優(yōu)先,根據(jù)上圖的表識(shí),我們可以找到每條鏈路上的DP,而連Printers的鏈路上的DP就為Switch3,因?yàn)樗懈〉腗AC地址。32.Refertotheexhibit.WhywouldthenetworkadministratorconfigureRAinthismanner?A.togivestudentsaccesstotheInternetB.topreventstudentsfromaccessingthecommandpromptofRAC.topreventadministratorsfromaccessingtheconsoleofRAD.togiveadministratorsaccesstotheInternetE.topreventstudentsfromaccessingtheInternetF.topreventstudentsfromaccessingtheAdminnetworkAnswer:B解釋一下：在這兒,將ＡＣＬ應(yīng)用到ＶＴＹ線路下,而且是ＩＮ的方向,表示凡是被我的ＡＣＬ允許的才能telnet到我．在ＲＡ上配置的是permit根據(jù)隱式的denyany允許Ａdmin的網(wǎng)段中的用戶可以telnet到他,所以Ｓtudent的網(wǎng)段中的用戶是被拒絕的．33.InordertoallowtheestablishmentofaTelnetsessionwitharouter,whichsetofcommandsmustbeconfigured?A.router<config>#lineconsole0router<config-line>#enablepasswordciscoB.router<config>#lineconsole0router<config-line>#enablesecretciscorouter<config-line>#loginC.router<config>#lineconsole0router<config-line>#passwordciscorouter<config-line>#loginD.router<config>#linevty0router<config-line>#enablepasswordciscoE.router<config>#linevty0router<config-line>#enablesecretciscorouter<config-line>#loginF.router<config>#linevty0router<config-line>#passwordciscorouter<config-line>#loginAnswer:F解釋一下：telnet是一個(gè)應(yīng)用層的應(yīng)用,他使用的是vty線路,而且在默認(rèn)的情況下,是需要訪問的線路下設(shè)有密碼的。而在VTY線路下設(shè)置密碼的命令為passworkstring,而VTY線路下的另一個(gè)命令login則是默認(rèn)的,可寫也可不寫。如果想Telnet時(shí)在VTY線路下不設(shè)置密碼也可以訪問這個(gè)線路,可以在該VTY線路下輸入命令nologin。34.Refertotheexhibit.ThetwoexhibiteddevicesaretheonlyCiscodevicesonthenetwork.Theserialnetworkbetweenthetwodeviceshasamaskof52.Giventheoutputthatisshown,whatthreestatementsaretrueofthesedevices?<Choosethree.>A.TheManchesterserialaddressis.1.B.TheManchesterserialaddressis.2.C.TheLondonrouterisaCisco2610.D.TheManchesterrouterisaCisco2610.E.TheCDPinformationwasreceivedonportSerial0/0oftheManchesterrouter.F.TheCDPinformationwassentbyportSerial0/0oftheLondonrouter.Answer:ACE解釋一下：ＣＤＰ是ＣＩＳＣＯ私有的一個(gè)二層的協(xié)議,但是他卻可以發(fā)現(xiàn)三層的ＩＰ信息的．通過ＣＤＰ可以發(fā)現(xiàn)的鄰居的信息有：設(shè)備的名稱,ＩＰ地址,端口,能力,平臺(tái),對(duì)端的holddowntime．在上圖的showcdpentry*命令的顯示可以看到的信息有：設(shè)備名稱：Ｌondon；ＩＰ地址：.2；平臺(tái)：cisco2610；能力：Router；端口：s0/１；holdtime：１２５Ｓ．Ｍanchesteter收到這個(gè)ＣＤＰ信息的接口為S0/0．綜合一下,這個(gè)題目的答案就出來了．35.Anetworkadministratorhasconfiguredtwoswitches,namedLondonandMadrid,touseVTP.However,theswitchesarenotsharingVTPmessages.Giventhecommandoutputshowninthegraphic,whyaretheseswitchesnotsharingVTPmessages?A.TheVTPversionisnotcorrectlyconfigured.B.TheVTPoperatingmodeisnotcorrectlyconfigured.C.TheVTPdomainnameisnotcorrectlyconfigured.D.VTPpruningmodeisdisabled.E.VTPV2modeisdisabled.F.VTPtrapsgenerationisdisabled.Answer:C解釋一下：交換機(jī)間不能共享VTP的信息,我們就需要檢查VTP的狀態(tài),首先需要檢查的是VTP的域名,只有同一個(gè)域中的才可能相互學(xué)習(xí),再來檢查VTP的模式,必須有一個(gè)server模式才能有VTP學(xué)習(xí)的過程的,默認(rèn)的情況下VTP的模式為Server的。然后我們檢查圖題目給出的信息,可以看到兩臺(tái)交換機(jī)的VTPdomain是不一致的,所以這個(gè)就是問題的所在了。36.Host1istryingtocommunicatewithHost2.Thee0interfaceonRouterCisdown.Whichofthefollowingaretrue?<Choosetwo.>A.RouterCwilluseICMPtoinformHost1thatHost2cannotbereached.B.RouterCwilluseICMPtoinformRouterBthatHost2cannotbereached.C.RouterCwilluseICMPtoinformHost1,RouterA,andRouterBthatHost2cannotbereached.D.RouterCwillsendaDestinationUnreachablemessagetype.E.RouterCwillsendaRouterSelectionmessagetype.F.RouterCwillsendaSourceQuenchmessagetype.Answer:AD解釋一下：連Host2的接口E0/0down了,那么最直接的反映就發(fā)生在路由器C上,C的路由表中的這個(gè)條目就消失了,因此當(dāng)Host1想要跟Host2建立連接的時(shí)候,RouterC就發(fā)送一個(gè)目的網(wǎng)段不可達(dá)的消息；如果是使用ping命令,那么RouterC就使用ICMP的包文告訴Host1,Host2是不可打的。37.Refertotheexhibit.Assumingthattherouterisconfiguredwiththedefaultsettings,whattypeofrouterinterfaceisthis?A.EthernetB.FastEthernetC.GigabitEthernetD.asynchronousserialE.synchronousserialAnswer:B解釋一下：這個(gè)題是需要根據(jù)圖中提供的信息來判斷接口的類型?？梢钥吹浇涌诘腗AC地址,表示這個(gè)接口肯定不是串行接口,所以可以排除D和E的選項(xiàng)。看帶寬BW100000Kbit,表示的是100M的帶寬,所以這是個(gè)FastEthernet接口。38.Onpoint-to-pointnetworks,OSPFhellopacketsareaddressedtowhichaddress?A.B.C.D.E.F.55Answer:E解釋一下：在OSPF中Hello包發(fā)向的是和這兩個(gè)地址的。大家在做OSPF實(shí)驗(yàn)的時(shí)候,用debug命令是可以看到這兩個(gè)個(gè)地址的。39.Whiletroubleshootingaconnectivityproblem,anetworkadministratornoticesthataportstatusLEDonaCiscoCatalystseriesswitchisalternatinggreenandamber.Whichconditioncouldthisindicate?A.Theportisexperiencingerrors.B.Theportisadministrativelydisabled.C.Theportisblockedbyspanningtree.D.Theporthasanactivelinkwithnormaltrafficactivity.Answer:A解釋一下：CISCO交換機(jī)的端口狀態(tài)指示燈是閃爍的綠色和淺黃色,表示端口有操作的問題——也許是過量的錯(cuò)誤或連接的問題。40.Refertotheexhibit.ThenetworkshownintheexhibitisrunningtheRIPv2routingprotocol.Thenetworkhasconverged,andtheroutersinthisnetworkarefunctioningproperly.TheFastEthernet0/0interfaceonR1goesdown.Inwhichtwowayswilltheroutersinthisnetworkrespondtothischange?<Choosetwo.>A.Allrouterswillreferencetheirtopologydatabasetodetermineifanybackuproutestothenetworkareknown.B.RoutersR2andR3marktherouteasinaccessibleandwillnotacceptanyfurtherroutingupdatesfromR1untiltheirhold-downtimersexpire.C.Becauseofthesplit-horizonrule,routerR2willbepreventedfromsendingerroneousinformationtoR1aboutconnectivitytothenetwork.D.WhenrouterR2learnsfromR1thatthelinktothenetworkhasbeenlost,R2willrespondbysendingaroutebacktoR1withaninfinitemetrictothenetwork.E.R1willsendLSAstoR2andR3informingthemofthischange,andthenallrouterswillsendperiodicupdatesatanincreasedrateuntilthenetworkagainconverges.Answer:CD解釋一下：這涉及到RIP關(guān)于環(huán)路避免的幾種機(jī)制了。在這里R1的直連的鏈路發(fā)生了變化,立即觸發(fā)更新〔觸發(fā)更新,發(fā)送flashupdate出去,將這個(gè)條目置為possibledown,設(shè)置最大跳數(shù)〔路由毒性,R2收到這個(gè)flashupdate后,也回復(fù)一個(gè)flashupdate包〔毒性逆轉(zhuǎn),同時(shí)將這個(gè)條目也置為possibledown,設(shè)置最大跳數(shù)。42.WhichofthefollowingdescribetheprocessidentifierthatisusedtorunOSPFonarouter?<Choosetwo.>A.Itislocallysignificant.B.Itisgloballysignificant.C.ItisneededtoidentifyauniqueinstanceofanOSPFdatabase.D.ItisanoptionalparameterrequiredonlyifmultipleOSPFprocessesarerunningontherouter.E.AllroutersinthesameOSPFareamusthavethesameprocessIDiftheyaretoexchangeroutinginformation.Answer:AC解釋一下：OSPF的進(jìn)程號(hào)只在本地有效。在一臺(tái)路由器上需要為每個(gè)進(jìn)程維護(hù)各自的OSPF數(shù)據(jù)庫。43.Refertotheexhibit.TheFMJmanufacturingcompanyisconcernedaboutunauthorizedaccesstothePayrollServer.TheAccounting1,CEO,Mgr1,andMgr2workstationsshouldbetheonlycomputerswithaccesstothePayrollServer.Whattwotechnologiesshouldbeimplementedtohelppreventunauthorizedaccesstotheserver?<Choosetwo.>A.accesslistsB.encryptedrouterpasswordsC.STPD.VLANsE.VTPF.wirelessLANsAnswer:AD解釋一下：需要控制只允許哪些組可以訪問服務(wù)器,組中的哪些用戶可以訪問,使用的技術(shù)當(dāng)然有ACL和VLAN了。44.Whichtwostatementsaretrueaboutthecommandiproute?<Choosetwo.>A.Itestablishesastaticroutetothenetwork.B.Itestablishesastaticroutetothenetwork.C.Itconfigurestheroutertosendanytrafficforanunknowndestinationtothenetwork.D.Itconfigurestheroutertosendanytrafficforanunknowndestinationouttheinterfacewiththeaddress.E.Itusesthedefaultadministrativedistance.F.Itisaroutethatwouldbeusedlastifotherroutestothesamedestinationexist.Answer:AE解釋一下：命令是靜態(tài)指定一條路由：通過接口可以到達(dá)網(wǎng)段。在這條命令后沒有指定管理距離,就表示使用默認(rèn)的管理距離,為1.45.Thenetworkshowninthediagramisexperiencingconnectivityproblems.Whichofthefollowingwillcorrecttheproblems?<Choosetwo.>A.ConfigurethegatewayonHostAas.1.B.ConfigurethegatewayonHostBas.254.C.ConfiguretheIPaddressofHostAas.2.D.ConfiguretheIPaddressofHostBas.2.E.Configurethemasksonbothhoststobe24.F.Configurethemasksonbothhoststobe40.Answer:BD解釋一下：主機(jī)A到他的指定網(wǎng)關(guān)的這條鏈路是沒有問題的,因?yàn)镠OSTA,接口VLAN1和路由器的f0/0.1網(wǎng)段是相同的,且都是處于VLAN1的。而HOSTB的VLAN2到交換機(jī)是沒有相同的VLAN接口和他通訊的,所以HOSTB發(fā)出的數(shù)據(jù)到交換機(jī)上就被丟棄了。所以需要在交換機(jī)上指定一個(gè)處于VLAN2的接口,并將SVI地址配置為和路由器POP的f0/0.2相同網(wǎng)段的地址。因?yàn)槁酚善鞯慕涌诘牡刂贩峙涞氖蔷W(wǎng)段.0/24,所以我們的HOSTB的地址應(yīng)該也分派一個(gè)的地址,并且網(wǎng)關(guān)也指定為路由器POP的f0/0.2的地址。46.WhichthreestatementsarecorrectaboutRIPversion2?<Choosethree.>A.Ithasthesamemaximumhopcountasversion1.B.Itusesbroadcastsforitsroutingupdates.C.Itisaclasslessroutingprotocol.D.IthasalowerdefaultadministrativedistancethanRIPversion1.E.Itsupportsauthentication.F.Itdoesnotsendthesubnetmaskinupdates.Answer:ACE解釋一下：關(guān)于RIPv2,首先要了解他是一個(gè)無類的路由協(xié)議,在發(fā)送路由更新的時(shí)候是攜帶掩碼的。他的metric的計(jì)算方式和RIPv1的相同,仍然是根據(jù)跳數(shù)的,但是他的跳數(shù)范圍擴(kuò)大了,RIPv1的為16跳,而RIPv2的為255跳。RIPv1是以廣播的形式發(fā)送更新的,在RIPv2中采用的是廣播,地址為。RIPv2是支持認(rèn)證的,而在RIPv1中是沒有這個(gè)功能的。RIPv2是可以關(guān)閉自動(dòng)匯總的,而在RIPv1中是不能關(guān)閉的。49.Refertotheexhibit.Router1wasjustsuccessfullyrebooted.IdentifythecurrentOSPFrouterIDforRouter1.A.0B.62C.94D.0Answer:C解釋一下：這是個(gè)關(guān)于OSPF的RID的選舉的問題。在OSPF中,RID的選舉過程是這樣的：如果通過命令router-id來指定一個(gè)RID,那么就采用手工指定的這個(gè)RID；如果沒有手工指定,則在可以使用的接口中來選舉,他是優(yōu)先采用回環(huán)口的,如果只有一個(gè)回環(huán)口,就采用這個(gè)回環(huán)口的IP作為RID,如果有多個(gè)回環(huán)口,就采用這多個(gè)回環(huán)口中IP地址最大的作為RID；如果沒有回環(huán)口,就采用物理接口中IP地址最大的接口IP作為RID。在上面的圖中可以看到有兩個(gè)回環(huán)口,而Loopback1的IP更大,所以就做為RID了。51.WhatcananetworkadministratorutilizebyusingPPPLayer2encapsulation?<Choosethree.>A.VLANsupportB.compressionC.authenticationD.slidingwindowsE.multilinksupportF.qualityofserviceAnswer:BCE解釋一下：PPP協(xié)議是能支持認(rèn)證的,包括PAP和CHAP；PPP還支持壓縮功能和差錯(cuò)校驗(yàn),還可實(shí)現(xiàn)多鏈路捆綁。而他們的這些功能都是HDLC所沒有的。52.Refertotheexhibit.Whatisthemeaningofthetermdynamicasdisplayedintheoutputoftheshowframe-relaymapcommandshown?A.TheSerial0/0interfaceispassingtraffic.B.TheDLCI100wasdynamicallyallocatedbytherouter.C.TheSerial0/0interfaceacquiredtheIPaddressoffromaDHCPserver.D.TheDLCI100willbedynamicallychangedasrequiredtoadapttochangesintheFrameRelaycloud.E.ThemappingbetweenDLCI100andtheendstationIPaddresswaslearnedthroughInverseARP.Answer:E解釋一下：這是個(gè)關(guān)于MAP的知識(shí)。在圖中可以看到這個(gè)MAP是dynamic的,因此是通過inverseARP學(xué)習(xí)到的。而ipdlci100表示的是DLCI100映射的地址為。就像是以太網(wǎng)中的MAC和IP的映射一樣,通過DLCI100可以找到。53.WhatisthefunctionoftheCiscoIOScommandipnatinsidesourcestatic.5?A.ItcreatesaglobaladdresspoolforalloutsideNATtransactions.B.Itestablishesadynamicaddresspoolforaninsidestaticaddress.C.ItcreatesdynamicsourcetranslationsforallinsidelocalPATtransactions.D.Itcreatesaone-to-onemappingbetweenaninsidelocaladdressandaninsideglobaladdress.E.Itmapsoneinsidesourceaddresstoarangeofoutsideglobaladdresses.Answer:D解釋一下：ipnatinsidesourcestatic.5這條命令是靜態(tài)創(chuàng)建一個(gè)一對(duì)一的地址轉(zhuǎn)換。他把內(nèi)部本地地址轉(zhuǎn)換為全局地址。55.Refertotheexhibit.WhenPC1sendsanARPrequestfortheMACaddressofPC2,networkperformanceslowsdramatically,andtheswitchesdetectanunusuallyhighnumberofbroadcastframes.Whatisthemostlikelycauseofthis?A.Theportfastfeatureisnotenabledonallswitchports.B.ThePCsareintwodifferentVLANs.C.SpanningTreeProtocolisnotrunningontheswitches.D.PC2isdownandisnotabletorespondtotherequest.E.TheVTPversionsrunningonthetwoswitchesdonotmatch.Answer:C解釋一下：PC1發(fā)出一個(gè)ARPrequest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switch2,交換機(jī)對(duì)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.AnadministratorissuesthecommandpingfromthecommandlinepromptonaPC.Ifareplyisreceived,whatdoesthisconfirm?A.ThePChasconnectivitywithalocalhost.B.ThePChasconnectivitywithaLayer3device.C.ThePChasadefaultgatewaycorrectlyconfigured.D.ThePChasconnectivityuptoLayer5oftheOSImodel.E.ThePChastheTCP/IPprotocolstackcorrectlyinstalled.Answer:E解釋一下：地址是一個(gè)私有的保留地址段,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通說明這個(gè)PC的TCP/IP協(xié)議棧是正確安裝的。59.Refertotheexhibit.Thenetworkadministratorrequireseasyconfigurationoptionsandminimalroutingprotocoltraffic.Whattwooptionsprovideadequateroutingtableinformationfortrafficthatpassesbetweenthetworoutersandsatisfytherequestsofthenetworkadministrator?<Choosetwo.>A.adynamicroutingprotocolonInternetRoutertoadvertiseallroutestoCentralRouter.B.adynamicroutingprotocolonInternetRoutertoadvertisesummarizedroutestoCentralRouter.D.adynamicroutingprotocolonCentralRoutertoadvertiseallroutestoInternetRouter.E.adynamicroutingprotocolonCentralRoutertoadvertisesummarizedroutestoInternetRouter.F.astatic,defaultrouteonCentralRouterthatdirectstraffictoInternetRouter.Answer:CF解釋一下：因?yàn)樵谶@個(gè)圖中,internetRouter要訪問內(nèi)網(wǎng)只能通過路由器CentralRouter.所以只需要在InternetRouter上配置一條通過CentralRouter到達(dá)的網(wǎng)段就可以了。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。60.Whataresomeoftheadvantagesofusingaroutertosegmentthenetwork?<Choosetwo.>A.FilteringcanoccurbasedonLayer3information.B.Broadcastsareeliminated.C.Routersgenerallycostlessthanswitches.D.Broadcastsarenotforwardedacrosstherouter.E.Addingaroutertothenetworkdecreaseslatency.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.Refertotheexhibit.WhatisthemeaningoftheoutputMTU1500bytes?A.Themaximumnumberofbytesthatcantraversethisinterfacepersecondis1500.B.Theminimumsegmentsizethatcantraversethisinterfaceis1500bytes.C.Themaximumsegmentsizethatcantraversethisinterfaceis1500bytes.D.Theminimumpacketsizethatcantraversethisinterfaceis1500bytes.E.Themaximumpacketsizethatcantraversethisinterfaceis1500bytes.F.Themaximumframesizethatcantraversethisinterfaceis1500bytes.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.TherearenobootsystemcommandsinarouterconfigurationinNVRAM.WhatisthefallbacksequencethattherouterwillusetofindanIOSduringreload?A.TFTPserver,Flash,NVRAMB.ROM,NVRAM,TFTPserverC.NVRAM,TFTPserver,ROMD.Flash,TFTPserver,ROME.Flash,NVRAM,ROMAnswer: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