CISCO考試題庫試題翻譯1-86題

1、第一部分計劃和設計design a simple lan using cisco technology使用思科技術設計一個簡單的局域網1、 which of the following devices can an administrator use to segment their lan? (choose all that apply) 下列哪個設備能使管理員分段他們的局域網？a. hubs b. repeaters c. switches d. bridges e. routers f. media converters g. all of the above a 集線器 b 中繼器 c

2、 交換機d 網橋 e 路由器 f 媒體轉換g 所有答案 ：cde 解析 ：routers, switches, and bridges dont transmit broadcasts. they segment a large cumber some network, into multiple efficient networks. 路由器，交換機，和網橋不能傳輸廣播，它們分一個大的網絡，到多個網絡錯誤答案 ：a. hubs is incorrect because a hub doesnt segment a network, it only allows more hosts on o

3、ne. hubs operate at layer one, and is used primarily to physically add more stations to the lan. b. this also incorrect because the job of a repeater is to repeat a signal so it can exceed distance limitations. it also operates at layer one and provides no means for logical lan segmentation. f. this

4、 is incorrect because media converters work by converting data from a different media type to work with the media of a lan. it also operates at layer one and provides no means for logical lan segmentation. a 集線器是不正確的因為不能分段，它只允許許多主機在上面，集線器工作在第一層，它用于首先到自身增加許多位置到局域網b 不正確因為中繼器的工作是反復一個信號，因此它能超越遠距離限制，它也是工

5、作在一層并且提供無意義的局域網邏輯劃分f 它不正確是因為媒體轉換工作通過從一個不同的媒體類型轉換數據到局域網的其他媒體，它也是工作在1 層并提供無意義的局域網劃分。2、routers perform which of the following functions? (select three) 下列哪些是路由器提供的功能？（選3 項）a. packet switching b. collision prevention on a lan segment. c. packet filtering d. broadcast domain enlargement e. broadcast forw

6、arding f. internetwork communication a 包交換 b 防止局域網分段中的沖突c 包過濾d 擴展廣播域e 推進廣播f 網絡通訊答案：acf 解析：a. routers work in layer 3 of the osi model. a major function of the router is to route packets between networks. c. through the use of access lists, routers can permit and deny traffic using layer 3 and layer

7、4 packet information. f. the primary purpose of a router is to route traffic between different networks, allowing for internetworking. 路由器工作在 osi 的 3 層，主要的功能是在網絡中路由數據包通過使用訪問控制列表， 通過使用 3 層和 4 層的包信息路由器能允許和拒絕傳輸數據f 路由器的主要目的是在不同網絡間的路由傳送數據，允許網絡錯誤回答 ：b. while routers can be used to segment lans, which will

8、 reduce the amount of collisions; it can not prevent all collisions from occurring. as long as there are 2 or more devices on a lan segment, the possibility of a collision exists, whether a router is used or not. d. the broadcast domain of a lan is often segmented through the use of a router. this r

9、esults in reducing the size of the broadcast domain. e. routers do not forward broadcast traffic. b 路由能夠用于局域網的分段，它能減少沖突數量，它不能防止所有的沖突事件， 只要在分段中有 2 個或更多的分段，沖突存在的可能性是是否使用路由d 一個 lan 的廣播域是通過使用路由器分段， 這個結論是減少了廣播域的大小e 路由器不是廣播傳輸數據3、which of the following statements most accurately describes the characteristi

10、cs of the above networks broadcast and collision domains? (select the two best answer choices)如圖所示，銷售部門和生產部門通過一個路由器分開，下列哪個陳述正確的描述了關于網絡廣博域和沖突域的特征？a. there are two broadcast domains in the network. b. there are four broadcast domains in the network. c. there are six broadcast domains in the network. d

11、. there are four collision domains in the network. e. there are five collision domains in the network. f. there are seven collision domains in the network. a 網絡中有 2 個廣播域、網絡中有個廣播域c、網絡中有 6 個廣播域d、網絡中有 4 個沖突域e、網絡中有 5 個沖突域f、網絡中有 7 個沖突域答案：af 解析：in this network we have a hub being used in the sales departm

12、ent, and a switch being used in the production department. based on this, we have two broadcast domains: one for each network being separated by a router. for the collision domains, we have 5 computers and one port for e1 so we have 6 collision domains total because we use a switch in the production

13、 department so 5 are created there, plus one collision domain for the entire sales department because a hub is being used. 在這個網絡中我們有一個集線器用于銷售部門，和一個交換機用于生產部門。 基于這些，我們有兩個廣播域：一個通過一個路由器分開。對于這沖突域，我們有5 臺計算機都在 e1中因此一共有 6 個沖突域，因為我們使用了一個交換機在生產部門這樣看來有5 個在一起，在加上整個銷售部門的一個沖突域，因為是用一個集線器連接的4、the testking corporate

14、 lan consists of one large flat network. you decide to segment this lan into two separate networks with a router. what will be the affect of this change? 局域網是由一個大的網絡組成。你打算將這個網絡通過一個路由器分成兩部分，這樣改變的影響是什么？a. the number of broadcast domains will be decreased. b. it will make the broadcasting of traffic be

15、tween domains more efficient between segments. c. it will increase the number of collisions. d. it will prevent segment 1s broadcasts from getting to segment 2.e. it will connect segment 1s broadcasts to segment 2.a、廣播域的數量減少b、使在域中的廣播段的傳輸更有效c、增加沖突域的數目 d、防止段 1 的廣播到段 2 中e、連接段 1 和段 2 的廣播答案：d 解析：a router

16、 does not forward broadcast traffic. it therefore breaks up a broadcast domain, reducing unnecessary network traffic. broadcasts from one segment will not be seen on the other segment. 一個路由器不能傳輸廣播， 它分解廣播域， 減少不必要的網絡傳輸。 一個段中的廣播不能被其他段的看見。錯誤答案 ：a. this will actually increase the number of broadcast doma

17、ins from one to two. b. all link level traffic from segment one to segment two will now need to be routed between the two interfaces of the router. although this will reduce the traffic on the lan links, it does also provide a less efficient transport between the segments. c. since the network size

18、is effectively cut into half, the number of collisions should decrease dramatically. e. broadcasts from one segment will be completely hidden from the other segment. 、實際上這能增加廣播域的數量從個到個、從段到段的所有的標準連接將通過路由器在兩個接口間被路由，在段中它也提供一些有效的小的傳輸c、因為網絡的大小被有效的打斷了一半，沖突域的數目將減少e、來自一個段的廣播將完全隱藏到其他段中5、which of the followin

19、g are benefits of segmenting a network with a router? (select all that apply) 下列哪個是用路由器分段的好處？（選擇所有可能的選項）a. broadcasts are not forwarded across the router. b. all broadcasts are completely eliminated. c. adding a router to the network decreases latency. d. filtering can occur based on layer 3 informa

20、tion. e. routers are more efficient than switches and will process the data more quickly. f. none of the above.a、廣播不能通過路由器運送b、所有廣播將完全除去c、增加路由器將減少延時d、能過濾基于3 層的信息e、路由器比交換機更快更有效的處理數據f、都沒有答案：解析： routers do not forward broadcast messages and therefore breaks up a broadcast domain. in addition, routers ca

21、n be used to filter network information with the use of access lists. 路由器不能傳送廣播信息并因此分割廣播域，另外，通過訪問控制列表路由器能濾波網絡信息錯誤答案 ：b. broadcasts will still be present on the lan segments. they will be reduced, because routers will block broadcasts from one network to the other. c. adding routers, or hops, to any

22、network will actually increase the latency. e. the switching process is faster than the routing process. since routers must do a layer 3 destination based lookup in order to reach destinations, they will process data more slowly than switches. 、在局域網段中廣播仍然存在，它將減少，因為路由器將妨礙從一個網絡到其他網絡的廣播、在任何網絡中增加路由器或跳數實

23、際上都增大延時、交換的過程比路由過程快， 為了到達目的地， 路由器需要通過層目的查找，它將比交換機處理數據更慢6、如圖所示：of the following choices, which ip address should be assigned to the pc host?在下列選項中，哪個ip 地址能夠分配到 pc 主機上？a. 192.168.5.5 b. 192.168.5.32 c. 192.168.5.40 d. 192.168.5.6 e. 192.168.5.75 答案： c 解析：the subnet mask used on this ethernet segment i

24、s /27, which translates to 255.255.255.224. valid hosts on the 192.168.5.33/27 subnet are 192.168.5.33- 192.168.5.62, with 192.168.5.32 used as the network ip address and 192.168.5.63 used as the broadcast ip address. therefore, only choice c falls within the usable ip range. 這 個 以 太 網 段 的 子 網 掩 碼 是

25、 27， 轉 換 為 二 進 制 為 255.255.255.224。 在192.168.5.33/27中有效的子網是 192.168.5.33-192.168.5.62 ， 習慣使用 192.168.5.33作為網絡 ip 地址， 192.168.5.62作為廣播 ip 地址。因此，在可用的ip 范圍中選擇 c 7、如圖所示：based on the diagram above, how many collision domains are present in the testk network?基于這個圖表，在這里有多少個沖突域？a 、1 個 b、2 個 c 、3 個 d、4 個 e 、

26、5 個 f 、6 個 g 、7 個答案： b 解析：since hubs are being used for both ethernet segments, there are a total of two collision domains. routers do not forward broadcast and are used to segment lans, so testkinga consists of one collision domain while testkingb consists of the second collision domain. 因為集線器經常用于以

27、太網段中，有這兩個沖突域的總數。 路由器不能傳送廣播和分割局域網，因此testkinga 是由一個沖突域組成而testkingb 是第二個沖突域8、下列的網絡拓撲圖，注意testking1 交換機和 testking2 集線器which of the devices shown can transmit simultaneously without causing collisions?哪些設備能同時傳輸數據并且不會造成沖突？a. all hosts b. only hosts attached to the switch c. all hosts attached to the hub an

28、d one host attached to the switch d. all hosts attached to the switch and one host attached to the huba、所有主機 b 、只有和交換機連接的主機 c 、所有和集線器相連的主機和一個同交換機相連的主機 d 、所有和交換機相連的主機與一個和集線器相連的主機答案： b 解析：as we know switch is the device which avoids collisions. when two computers communicate through a switch they make

29、 their own collision domain. so, there is no chance of collisions. whenever a hub is included, it supports on half duplex communication and works on the phenomena of csma/cd so, there is always a chance of collision. 正如我們所知道的交換機是避免沖突的設備，當兩個計算機通過一個路由器通訊時它們將增大沖突域。因此，沒有沖突的機會。無論何時包含一個集線器，它將支持一部分雙方的連接并通過

30、csma/cd 工作，因此也總會有沖突的機會9、study the network topology exhibit carefully, in particular the two switches testking1, testking2, and the router testking3. which statements are true in this scenario? select two.仔細研究這個網絡拓撲圖，特別是這兩個交換機testking1 ，testking2 和路由器 testking3 ，下列哪個是正確的？選擇2 個a. all the devices in bo

31、th networks will receive a broadcast to 255.255.255.255 sent by host testkinga. b. only the devices in network 192.168.1.0 will a broadcast to 255.255.255.255 sent by host testkinga. c. all the devices on both networks are members of the same collision domain. d. the hosts on the 192.168.1.0 network

32、 form one collision domain, and the hosts on the 192.168.2.0 network form a second collision domain. e. each host is in a separate collision domain.a、在網絡中的所有的設備都能收到一個255.255.255.255的廣播，這廣播是testkinga 發(fā)送的 b 、 只有在 192.168.1.0 網絡中的設備能夠接收到testkinga發(fā)送的255.255.255.255廣播 c 、在所有網絡中的所有設備在同一個沖突域d、在 192.168.0.1

33、網絡中的所有主機是一個沖突域，并且在192.168.2.0網絡的主機是另外的一個沖突域 e 、所有的主機是單獨的沖突域答案： b e 解析：the switch forms the collision domains. the router divides the broadcast domains and collision domains. the router doesnt forward the broadcasts. so, hosts in networks 192.168.1.0 and 192.168.2.0 are in two different broadcast dom

34、ains. each host is in its own collision domain. b是正確的，而 d不是。如果圖表中是使用集線器而不是用的交換機那就是正確的，這里只有 2 個沖突域， 但是這圖中有交換機。 這作者可能是有意陳述那些正確的廣播域 e是正確的，因為整個網絡是由交換機組成的10、which address represents a unicast address?哪個地址象征一個單播地址？a. 224.1.5.2 b. ffff. ffff. ffff. c. 192.168.24.59/30d. 255.255.255.255 e. 172.31.128.255/18

35、答案： e 解析：172.31.128.255 is the only unicast address. it seems to be a broadcast address, because of 255 in the last octett, the broadcast address for this network is 172.31.131.255. not a: 224.1.5.2 is a multicast address. 172.31.128.255/18是唯一一個單播地址， 它像一個廣播地址， 因為 255 是最后的一位，這個網絡的廣播地址是172.31.128.255

36、a 224.1.5.2 是多播地址11、wit regard to bridges and switches, which of the following statements are true? (choose three.) 關于網橋和交換機，下列哪個陳述是正確的？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

37、 than switches.d. switches typically have a higher number of ports than bridges.e. bridges define broadcast domain while switches define collision domains.f. both bridges and switches make forwarding decisions based on layer 2 addresses. a 交換機是基于軟件的而網橋是基于硬件的b、網橋和交換機都是在2 層傳送廣播的c、網橋比交換機快d、交換機的端口號比網橋高e

38、、網橋定義的是沖突域而交換機是廣播域f、網橋和交換機所做的決定都是基于2 層地址的答案： bdf 12、which layer 1 devices can be used to enlarge the area covered by a single lan segment? select two 在1層設備中，哪個是通過一個單一的局域網段來擴大網絡范圍的？a. switch b. router c. nic d. hub e. repeater f. rj-45 transceiver a、交換機 b、路由器c、網卡 d、集線器e、中繼器f、rj-45收發(fā)器答案： de 解析：both hu

39、b, repeater, router and switch repeat the packet. but only hub and repeater do not segment the network. 集線器，中繼器和交換機都能夠反復數據包，但是只有集線器和中繼器不能分割網絡13、what information is supplied by cdp? select three. cdp所提供的信息是什么？a. device identifiers b. capabilities list c. platform d. route identifier e. neighbour traf

40、fic data a、設備標識符b、性能列表c、平臺 d、路由標識符e、相鄰間傳數據答案： abc 解析cdp 是一個思科私有的協(xié)議， 支持在接口間傳送 cdp 信息，這些接口必須支持 snap首部。所有的局域網接口， hdlc ，幀中繼和 atm 都支持 cdp 。路由器和交換機能夠發(fā)現(xiàn)相鄰路由的詳細的 3層地址甚至沒有配置 3層協(xié)議因為 cdp 不能支持依靠所有的 3層協(xié)議cdp 發(fā)現(xiàn)的相鄰設備的詳細信息：設備標識符代表主機名地址列表網絡地址和數據鏈路地址端口標識符標識端口的文本，一個端口的其他性能列表關于設備作什么的信息例如端口，一個路由器或者交換機平臺在設備中運行的模型和操作系統(tǒng)水平1

41、4、if a host on a network has the address 172.16.45.14/30, what is the address of the subnetwork to which this host belongs? 如果一個網絡中的主機地址是172.16.45.14/30，這個主機的子網是哪個地址？a. 172.16.45.0 b. 172.16.45.4 c. 172.16.45.8 d. 172.16.45.12 e. 172.16.45.18 答案： d 解析：the last octet in binary form is 00001110. only

42、6 bits of this octet belong to the subnet mask. hence the subnetwork is 172.16.45.12. 這最后八位字節(jié)轉換為二進制為00001110，在這八位中只有六個字節(jié)屬于子網號，因此子網是 172.16.45.12 15、how many broadcast domains are shown in the graphic assuming only the default vlan is configured on the switches? 在下列圖里交換機配置的默認vlamn 中顯示有多少廣播域？a、一個 b、兩個

43、c、六個d、十二個答案： a 解析：there is only one broadcast domain because switches and hubs do not switch the broadcast domains. only layer 3 devices can segment the broadcast domains. 這里只有一個廣播域是因為交換機和集線器都不能交換廣播域。只有 3 層設備能分段廣播域design an ip addressing scheme to meet design requirements (55 questions) 設計一個 ip 地址方案

44、用于符合設計要求1、 you have the binary number 10011101. convert it to its decimal and hexadecimal equivalents. (select two answer choices) 你有一個二進制號 10011101，把它改變成同等大小的十進制數和十六進制數a. 158 b. 0 x9d c. 156 d. 157 e. 0 x19 f. 0 x9f 答案： bd 解析：10011101 = 128+0+0+16+8+4+0+1 = 157 for hexadecimal, we break up the bina

45、ry number 10011101 into the 2 parts: 1001 = 9 and 1101 = 13, this is d in hexadecimal, so the number is 0 x9d. we can further verify by taking the hex number 9d and converting it to decimal by taking 16 times 9, and then adding 13 for d (0 x9d = (16x9)+13 = 157). 十進制的轉換： 10011101 = 128+0+0+16+8+4+0+

46、1 = 157對于十六進制，我們把二進制數10011101分成兩部分： 1001=9 1101=13，轉為十六進制是 d ，所以十六進制是 ox9d 。我們可以驗證，通過把十六進制數9d轉為十進制： (0 x9d = (16x9)+13 = 157) 2、the subnet mask on the serial interface of a router is expressed in binary as 11111000 for the last octet. how do you express the binary number 11111000 in decimal? 在一個路由器串

47、口的子網掩碼是用二進制11111000表示的作為最后八位，把它轉換為十進制是？a. 210 b. 224 c. 240 d. 248 e. 252 答案： d 解析：128 + 64+32+16+8 = 248. since this is the last octet of the interface, the subnet mask would be expressed as a /29. 這最后八位組，它的子網可以表示為/29 錯誤答案： a. the number 210 would be 11010010 in binary. b. the number 224 would be 1

48、1100000 in binary. c. the number 240 would be 11110000 in binary e. the number 252 would be 11111100 in binary. this is known as a /30 and is used often in point-point links, since there are only 2 available addresses for use in this subnet. a、210的二進制數為 11010010 b、224的二進制數為 11100000 c、240的二進制數為 1111

49、0000 e、252的二進制數為 11111100，它可以看作是 /30并用于點對點的連接，因為在這個子網中只有兩個可以使用的地址3、 which one of the binary number ranges shown below corresponds to the value of the first octet in class b address range? 下列二進制數的范圍中哪個顯示的是b類前八位地址的范圍？a. 10000000-11101111 b. 11000000-11101111 c. 10000000-10111111 d. 10000000-11111111 e

50、. 11000000-10111111 答案： c 解析： class b addresses are in the range 128.0.0.0 through 191.255.255.255. in binary, the first octet (128 through 191) equates to 10000000-10111111 b類地址的范圍是 128.0.0.0到191.255.255.255 ，轉換為二進制的前八位是10000000-10111111 錯誤答案：a. binary 10000000 does equate to 128 but binary 1110111

51、1 equates to 239 b. binary 11000000 equates to 192 and binary 11101111 equates to 239 d. binary 10000000 does equate to 128 but binary 11011111 equates to 223 e. binary 11000000 equates to 192 but binary 10111111 does equate to 191 a、二機制數 10000000的等值數是 128但是二進制 11101111的等值數是 239 b、二機制數 11000000的等值數是

52、 192并且二進制 11101111的等值數是 239 c、二機制數 10000000的等值數是 128但是二進制 11011111的等值數是 223 e、二機制數 11000000的等值數是 192但是二進制 10111111的等值數是 191 4、how would the number 231 be expressed as a binary number? 231表示成二進制是？a. 11011011 b. 11110011 c. 11100111 d. 11111001 e. 11010011 答案： c 解析：decimal number 231 equates to 111001

53、11 in binary (128+64+32+0+0+4+2+1) 十進制數 231等同的二進制數是 11100111 (128+64+32+0+0+4+2+1) 錯誤答案：a: binary 11011011 equates to 219 (128+64+0+16+8+0+2+1) b: binary 11110011 equates to 243 (128+64+32+16+0+0+2+1) d: binary 11101011 equates to 249 (128+64+32+16+8+0+0+1) e: binary 11010011 equates to 211 (128+64+

54、0+16+0+0+2+1) a、11011011的十進制數是 219 b、11110011的十進制數是 243 、11101011的十進制數是 249 、11010011的十進制數是 211 、how would the number 172 be expressed in binary form? 172如何用二進制表示？a. 10010010b. 10011001c. 10101100d. 10101110 答案：解析： 10101100= 128 + 0 + 32 + 0 + 8 + 4 + 0 + 0 = 172 錯誤答案：a. binary 10010010 = 128+0+0+16

55、+0+0+2+0=146 b. binary 10011001 = 128+0+0+16+8+0+0+1=153 d. binary 10101110 = 128+0+32+0+8+4+2+0= 174 、the mac address for your pc nic is: c9-3f-32-b4-dc-19. what is the address of the oui portion of this nic card, expressed as a binary number? 你的機網卡地址是 c9-3f-32-b4-dc-19。這網卡地址的 oui部分用二進制表示是？a. 11001

56、100-00111111-00011000 b. 11000110-11000000-00011111 c. 11001110-00011111-01100000 d. 11001001-00111111-00110010 e. 11001100-01111000-00011000 f. 11111000-01100111-00011001 答案：解析： the first half of the address identifies the manufacturer of the card. this code, which isassigned to each manufacturer b

57、y the ieee, is called the organizationally uniqueidentifier (oui). in this example, the oui is c9-3f-32. if we take this number andconvert it to decimal form we have: c9 = (12x16) + 9 = 2013f = (3x16) + 15 = 6332 = (3x16) + 2 = 50 so, in decimal we have 201.63.50. if we then convert this to binary,

58、we have: 201 = 1100100163 = 0011111150 = 00110010 so the correct answer is d: 11001001-00111111-00110010 前半部分的地址標識了廠商號，這個代碼都是ieee分配的，也叫做組織的唯一標識（），例如，它的 oui是c93f32。如果我們把它轉換為十進制數是：c9 = (12x16) + 9 = 2013f = (3x16) + 15 = 6332 = (3x16) + 2 = 50 因此我們所得到的十進制數是210.63.50如果把它轉換為二進制是：201 = 1100100163 = 00111

59、11150 = 00110010 所以正確的答案是 d：11001001-00111111-00110010 、how do you express the binary number 10110011 in decimal form? 你怎樣用十進制數表示二進制數10110011？a. 91b. 155c. 179d. 180e. 201f. 227 答案： c 解析：if you arrange the binary number 10110011, against the place value and multiply thevalues, and add them up, you g

60、et the correct answer. 1 0 1 1 0 0 1 1 128 64 32 16 8 4 2 1 128 + 0 + 32 +16 + 0 + 0 +2 +1 = 179如果你排列二進制，依靠位值并增加值，合計，你就可以得到正確的答案、convert the hex and decimal numbers on the left into binary, and match them to their corresponding slot on the right. (not all of the hexadecimal and decimal numbers will