[語言類考試復習資料大全]中級網(wǎng)絡工程師上午試題分類模擬24

1、書山有路勤為徑，學海無涯苦作舟。祝愿天下莘莘學子：學業(yè)有成，金榜題名！語言類考試復習資料大全中級網(wǎng)絡工程師上午試題分類模擬24中級網(wǎng)絡工程師上午試題分類模擬24單項選擇題 Traditional network layer packet forwarding relies on the information provided by network layer _ protocols, or static routing, to make an independent forwarding decision at each _ within the network. The forwardin

2、g decision is based solely on the destination _ IP address. All packets for the same destination follow the same path across the network, if no other equal-cost _ exist. Whenever a router has two equal-cost paths toward a destination, the packets toward the destination might take one or both of them

3、, resulting in some degree of load sharing. Enhanced Interior Gateway Routing Protocol (EIGRP) also supports non-equal-cost _ sharing although the default behavior of this protocol is equal-cost. You must configure EIGRP variance for non-equal-cost load balancing. 1.A.switchingB.signalingC.routingD.

4、session答案:C2.A.switchB.hopC.hostD.customer答案:B3.A.connectionB.transmissionC.broadcastD.customer答案:D4.A.pathsB.distanceC.broadcastD.session答案:A5.A.loanB.loadC.contentD.constant答案:B解析 傳統(tǒng)的網(wǎng)絡層數(shù)據(jù)包的轉(zhuǎn)發(fā)依賴于網(wǎng)絡層路由協(xié)議提供的信息，或者靜態(tài)路由，獨立決定網(wǎng)絡內(nèi)每一跳的轉(zhuǎn)發(fā)決策。轉(zhuǎn)發(fā)決策僅僅基于目的客戶機的IP地址。如果不存在等價的路徑，則相同目的地的數(shù)據(jù)包將沿相同的路徑在網(wǎng)絡中轉(zhuǎn)發(fā)。每當路由器有兩條等價的路徑

5、通往目的地，數(shù)據(jù)包可能會選擇其中的一條或者兩條到達目的地，在一定程度上均衡了負載。增強內(nèi)部網(wǎng)關路由協(xié)議(EIGRP)也支持非等價的負載分擔，盡管這個協(xié)議默認是等價的。你必須配置EIGRP非等價負載均衡的方差。 The traditional way of allocating a single channel, such as a telephone think, among multiple competing users is to chop up its _ by using one of the multiplexing schemesr, such as FDM. If there

6、are N users, the bandwidth is divided into N equal-sized portions, with each user being assigned one portion. Since each user has private frequency _, there is now no interference among users. When there is only a small and constant number of users, each of which has a steady stream or heavy load of

7、 _ this division is a simple and efficient allocation mechanism. A wireless example is FM radio stations. Each station gets a portion of the FM band and uses it most of the time to broadcast its signal. However when the number of senders is large and varying or the traffic is _, FDM presents some pr

8、oblems. If the spectrum is cut up into N regions and fewer than N users are currently interested in communicating, a large piece of valuable spectrum will be wasted. And if more than N users want to communicate, some of them will be denied _ for lack of bandwidth, even if some of the users who have

9、been assigned a frequency band ever transmit or receive anything. 6.A.capabilityB.capacityC.abilityD.power答案:B7.A.bandB.rangeC.domainD.assignment答案:B8.A.trafficB.dataC.informationD.communications答案:A9.A.continuousB.steadyC.busyD.flow答案:C10.A.allowanceB.connectionC.percussionD.permission答案:D解析 傳統(tǒng)的分配信

10、號的方法，比如電話中繼，多個競爭用戶使用一種多路復用方案分配信道的容量，如FDM。如果有N個用戶，帶寬將被分成N個同樣大小的部分，每個用戶將分得一部分。既然每個用戶都有自己的頻率范圍，因此用戶之間沒有干擾。當用戶數(shù)量少且不變，每個用戶將保持穩(wěn)定的數(shù)據(jù)流和通信負載，這種劃分是一種簡單而高效的分配機制。FM廣播電臺是一個無線的例子。每個電臺分配有一個FM頻帶，大部分時間用它來廣播信號。但是，當發(fā)送者的數(shù)量很多且是變化的，或者通信很忙，F(xiàn)DM將會有一些問題。如果頻譜被分成N個區(qū)域，而不足N個用戶在通信，大量寶貴的頻譜會被浪費。如果超過N個用戶想要通信，他們中一些人的通信將會因為帶寬的不足而被拒絕，即

11、使一些用戶已經(jīng)分配了某一頻率的帶寬參與過發(fā)送或接收數(shù)據(jù)。 CDMA for cellular systems can be described as follows. As with FDMA, each cell is allocated a frequency _, which is split into two parts: half for reverse (mobile unit to base station) and half for _ (base station to mobile unit). For full-duplex _, a mobile unit uses bo

12、th reverse and forward channels. Transmission is in the form of direct-sequence spread _ which uses a chipping code to increase the data rate of the transmission, resulting in an increased signal bandwidth. Multiple access is provided by assigning _ chipping codes to multiple users, so that the rece

13、iver can recover the transmission of an individual unit from multiple transmissions. 11.A.waveB.signalC.bandwidthD.domain答案:C12.A.forwardB.reverseC.backwardD.ahead答案:A13.A.connectionB.transmissionC.compromiseD.communication答案:B14.A.structureB.spectrumC.streamD.strategy答案:B15.A.concurrentB.orthogonal

14、C.higherD.lower答案:B解析 蜂窩系統(tǒng)中的CDMA可以如下描述：采用FDMA，每個蜂窩分配有一頻率帶寬，該頻率帶寬被分成兩部分，一部分用于反向傳輸(移動單元到基站)，一部分用于正向傳輸(基站到移動單元)。為了實現(xiàn)雙工通信，一個移動單元要使用正向和反向兩個信道。傳輸是以直接序列擴頻的形式進行的，并使用碼片序列來增加數(shù)據(jù)傳輸?shù)乃俾?，進而導致信號帶寬增加。采用多路存取技術將正交的碼片序列分配給多個用戶，因此接收者可以從多路傳輸中恢復自己的傳輸單元。 Traditional IP packet forwarding analyzes the _ IP address contained

15、in the network layer header of each packet as the packet travels from its source to its final destination. A router analyzes the destination IP address independently at each hop in the network. Dynamic _ protocols or static configuration builds the database needed to analyze the destination IP addre

16、ss (the routing table). The process of implementing traditional IP routing also is called hop-by-hop destination-based _ routing. Although successful, and obviously widely deployed, certain restrictions, which have been realized for some time, exist for this method of packet forwarding that diminish

17、 its _. New techniques are therefore required to address and expand the functionality of an IP-based network infrastructure. This first chapter concentrate on identifying these restrictions and presents a new architecture, known as multipleprotocol _ switching, that provides solutions to some of the

18、se restrictions. 16.A.datagramB.destinationC.connectionD.service答案:B17.A.routingB.forwardingC.transmissionD.management答案:A18.A.anycastB.multicastC.broadcastD.unicast答案:D19.A.reliabilityB.flexibilityC.stabilityD.capability答案:B20.A.constB.castC.markD.label答案:D解析 當數(shù)據(jù)包從其來源到其最終目的地址時，傳統(tǒng)的IP報文轉(zhuǎn)發(fā)分析包含在每一個數(shù)據(jù)包的

19、網(wǎng)絡層報頭的目的IP地址。路由器獨立地分析網(wǎng)絡中的每一跳目標IP地址。動態(tài)路由協(xié)議或靜態(tài)配置生成所需的數(shù)據(jù)庫來分析目標IP地址(路由表)。傳統(tǒng)的IP路由實現(xiàn)的過程也被稱為基于逐跳目的地的單播路由。雖然成功地廣泛采用，但是一些已經(jīng)實現(xiàn)一段時間的限制仍存在，此方法存在報文的轉(zhuǎn)發(fā)減少其靈活性。因此，需要新的技術，以處理和擴展一個基于IP的網(wǎng)絡基礎設施的功能。第一章著重識別這些限制，并提出了一種新的架構，被稱為多協(xié)議標簽交換，來提供解決這些限制的方法。 The de facto standard application program interface (API) for TCP/IP applic

20、ations is the sockets interface. Although this API was developed for _ in the early 1980s. it has also been implemented on a wide variety of non-Unix systems. TCP/IP _ written using the sockets API have in the past enjoyed a high degree of portability and we would like the same _ with IPv6 applicati

21、ons. But changes are required to the sockets API to support IPv6 and this memo describes these changes. These include a new socket address structure to carry IPv6 _, new address conversion functions, and some new socket options. These extensions are designed to provide Access to the basic IPv6 featu

22、res required by TCP and UDP applications, including multicasting, while introducing a minimum of change into the system and providing complete _ for existing IPv4 applications. 21.A.WindowsB.LinuxC.UnixD.DoS答案:C22.A.applicationsB.networksC.protocolsD.systems答案:A23.A.portabilityB.availabilityC.capabi

23、lityD.reliability答案:A24.A.connectionsB.protocolsC.networksD.addresses答案:D25.A.availabilityB.compatibilityC.capabilityD.reliability答案:B解析 對于TCP/IP應用，事實上的應用程序接口(API)標準是“套接字”口。雖然這個API是在20世紀80年代早期為Unix開發(fā)的，但是也廣泛地在各種非Unix系統(tǒng)中得到了實現(xiàn)。以前采用套接字API編寫的TCP/IP應用具有高度的兼容性，因而我們也希望對IPv6應用也具有同樣的兼容性。為了支持IPv6，需要對套接字API做出某些

24、改變，這個便箋就是描述這些變化的。這些改變包括一種新的用于支持IPv6地址的套接字地址結構、新的地址轉(zhuǎn)換功能以及新的套接字選項。這些擴展可以滿足TCP和UDPI立用訪問IPv6基本功能(包括組播)時的需要，但是只對系統(tǒng)進行了最小的改變，而且與現(xiàn)有的IPv4應用是完全兼容的。 The TCP protocolis a _ layer protocol. Each connection connects two TCPs that may be just one physical network apart or located on opposite sides of the globe. In

25、 other words, each connection creates a _ with a length that may be totally different from another path created by another connection. This means that TCP cannot use the same retransmission time for all connections. Selecting a fixed retransnussion time for all connections can result in serious cons

26、equences. If the retransmission time does not allow enough time for a _ to reach the destination and an acknowledgment to reach the source, it can result in retransmission of segments that are still on the way. Conversely, if the retransmission time is longer than necessary for a short path, it may

27、result in delay for the application programs. Even for one single connection, the retransmission time should not be fixed. A connection may be able to send segments and receive _ faster during nontraffic period than during congested periods. TCP uses the dynamic retransmission time, a transmission t

28、ime is different for each connection and which may be changed during the same connection. Retransmission time can be made _ by basing it on the round-trip time (RTT). Several formulas are used for this purpose. 26.A.physicalB.networkC.transportD.application答案:C27.A.pathB.windowC.responseD.process答案:

29、A28.A.processB.segmentC.programD.user答案:B29.A.connectionsB.requestsC.acknowledgmentsD.datagrams答案:C30.A.longB.shortC.fixedD.dynamic答案:D解析 TCP是一種傳輸層協(xié)議，每個TCP連接都連接著兩個TCP，這兩個TCP可能是在一個物理網(wǎng)絡里，但是是分開的或者位于全局網(wǎng)絡的對立面。換句話說，每個連接創(chuàng)建一個路徑的長度可能完全不同于由另外一個連接創(chuàng)建的路徑。這意味著TCP不能為所有的連接使用相同的轉(zhuǎn)發(fā)時間。為所有的連接選擇一個固定的轉(zhuǎn)發(fā)時間可能會導致嚴重的后果。如果這個轉(zhuǎn)

30、發(fā)時間并不為一個段提供足夠的時間去到達目的地以確認獲取來源，將導致段轉(zhuǎn)發(fā)的部分仍在路上。相反，如果轉(zhuǎn)發(fā)時間超過了必要的短路徑，它可能會導致延遲的應用程序，甚至一個單獨的連接，轉(zhuǎn)發(fā)時間也應該是不固定的，相比于擁擠的時期，一個連接在不擁擠的時期可能能夠更快地發(fā)送和接收確認。TCP使用動態(tài)轉(zhuǎn)發(fā)時間，每個連接的傳輸時間是不同的，而且可能在相同的連接中變化。轉(zhuǎn)發(fā)時間可能是動態(tài)的基于往返時間，有幾個準則用于這一目的。 Let us now see how randomization is done when a collision occurs. After a _, time is divided in

31、to discrete slots whose length is equal to the worst-case round-trip propagation time on the ether (2t). To accommodate the longest path allowed by Ethernet, the slot tome has been set to 512 bit times, or 51.2sec. After the first collision, each station waits either 0 or 1 _ times before trying aga

32、in. If two stations collide and each one picks the same random number, they will collide again. After the second collision, each one picks either 0, 1, 2, or 3 at random and waits that number of slot times. If a third collision occurs (the probability of this happening is 0.25), then the next time t

33、he number of slots to wait is chosen at _ from the interval 0 to 23-1. In general, after i collisions, a random number between 0 and 2i-1 is chosen, and that number of slots is skipped. However, after ten collisions have been reached, the randomization _ is frozen at a maximum of 1023 slots. After 1

34、6 collisions, the controller throws in the towel and reports failure back to the computer. Further recovery is up to _ layers. 31.A.datagramB.collisionC.connectionD.service答案:B32.A.slotB.switchC.processD.fire答案:A33.A.restB.randomC.onceD.odds答案:B34.A.unicastB.multicastC.broadcastD.interval答案:D35.A.lo

35、calB.nextC.higherD.lower答案:C解析 現(xiàn)在讓我們觀察沖突發(fā)生時如何做隨機處理。沖突發(fā)生后，時間被劃分成離散的長度，等于最壞的往返傳播時間的時槽。為了容納以太網(wǎng)允許的最長路徑，沖突時槽縮小為5.12s。 第一次沖突后，每個站在再次嘗試前需要等待0或1時槽。如果每一站發(fā)生沖突，且每一個挑選相同的隨機數(shù)，它們將再次發(fā)生沖突。第2次沖突之后，每一站隨機選擇0、1、2或3，即等待時槽的個數(shù)。如果第3次沖突發(fā)生(發(fā)生的概率為0.25)，則下次時槽的等待數(shù)量都是在023-1之間隨機挑選的。 總體來說，第i次沖突后，將在021-1之間挑選隨機數(shù)，而且那個時槽數(shù)是被略過的。然而，當沖突次

36、數(shù)達到10次，隨機間隔被鎖定在最高1023時槽。當發(fā)生16次沖突后，控制器丟棄報告而不再返回計算機。進一步恢復已經(jīng)達到更高的層次。 Border Gateway Protocol (BGP) is inter-autonomous system _ protocol. BGP is based on a routing method called path vector routing. Distance vector routing is not a good candidate for inter-autonomous system routing because there are oc

37、casions on which the route with the smallest _ count is not the preferred route. For example, we may not want a packet through an autonomous system that is not secure even though it is shortest route. Also, distance vector routing is unstable due to the fact that the routers announce only the number

38、 of hop counts to the destination without defining the path that leads to that _. A router that receives a distance vector advertisement packet may be fooled if the shortest path is actually calculated through the receiving router itself. Link _ routing is also not a good candidate for inner-autonom

39、ous system routing because an internet is usually too big for this routing method. To use link state routing for the whole internet would require each router to have a huge link state database. It would also take a long time for each router to calculate its routing _ using the Dijkstra algorism. 36.

40、A.routingB.switchingC.transmittingD.receiving答案:A37.A.pathB.hopC.routeD.packet答案:B38.A.connectionB.windowC.sourceD.destination答案:D39.A.statusB.searchC.stateD.research答案:C40.A.tableB.stateC.metricD.cost答案:A解析 邊界網(wǎng)關協(xié)議(BGP)是自治系統(tǒng)間的路由協(xié)議。BGP是基于路由的方法稱為距離矢量路由。距離矢量路由是自己系統(tǒng)路由的很好的候選者，因為跨自治系統(tǒng)路由場合上最小的跳數(shù)的路由不一定是最合適的

41、路由。例如，我們可能不希望通過一個自治系統(tǒng)不安全的數(shù)據(jù)包，即使它是最短的路線。此外，距離矢量路由是不穩(wěn)定的，路由器只宣布到目的地的跳數(shù)，但不指出到達目的地的路徑。實際上，如果是通過接收路由器本身計算的最短路徑，接收距離矢量通告報文的路由器可能被愚弄。鏈路狀態(tài)路由在跨自治系統(tǒng)中也不是一個好的候選者，因為對于這種方法互聯(lián)網(wǎng)是通常過大。要對整個互聯(lián)網(wǎng)使用鏈接狀態(tài)路由，就需要每個路由器有一個巨大的鏈路狀態(tài)數(shù)據(jù)庫。它也需要每個路由器用很長的時間來使用的Dijkstra算法計算自己的路由表。 A transport layer protocol usually has several responsibilties. One is to create a process-to-process communication UDP uses _ numbers to accomplish this. Another responsibili