【计算机类职业资格】网络工程师-计算机专业英语(一)及答案解析.doc

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1、网络工程师-计算机专业英语(一)及答案解析(总分：40.00，做题时间：90 分钟)In low-speed network, it is usually adequate to wait for congestion to occur and then react to it by telling the source of packets to slow down. In high-speed networks, this approach often works poorly, because in the (61) between sending the notification an

2、d notification arriving at the source, thousands of additional (62) may arrive. In ATM network, a major tool for preventing (63) is (64) control. When a host wants a new virtual (65) , it must describe the traffic to be offered and the service expected.（分数：5.00）(1).A. interval B. time C. slot D. del

3、ay（分数：1.00）A.B.C.D.(2).A. packets B. cells C. message D. files（分数：1.00）A.B.C.D.(3).A. collision B. congestion C. drop D. delay（分数：1.00）A.B.C.D.(4).A. flow B. admission C. traffic D. time（分数：1.00）A.B.C.D.(5).A. path B. rout C. circuit D. way（分数：1.00）A.B.C.D.For each blank, choose the best answer from

4、 the four choices and write down on the answer sheet.(11) is a protocol that a host uses to inform a router when it joins Or leaves an Internet multicast group.(12) is an error detection code that most data communication networks use.(13) is an interior gateway protocol that uses a distance vector a

5、lgorithm to propagate routing information.(14) is a transfer mode in which all types of information are organized into fixed form cells on an asynchronous or non-periodic basis over a range of media.(15) is an identifier of a web page.（分数：5.00）(1).A. ICMP B. SMTP C. IGMP D. ARP（分数：1.00）A.B.C.D.(2).A

6、. 4B/SB B. CRC C. Manchester Code D. Huffman Code（分数：1.00）A.B.C.D.(3).A. OSPF B. RIP C. RARP D. BGP（分数：1.00）A.B.C.D.(4).A. ISDN B. X.25 C. Frame Relay D. ATM（分数：1.00）A.B.C.D.(5).A. HTYP B. URL C. HTML D. TAG（分数：1.00）_The purpose of the requirements definition phase is to produce a clear, complete, c

7、onsistent, and testable (6) of the technical requirements for the software product. During the requirements definition phase, the requirements definition team uses an iterative process to expand a broad statement of the system requirements into a complete and detailed specification of each function

8、that the software must perform and each (7) that it must meet. The starting point is usually a set of high level requirements from the (8) that describe the project or problem.In either case, the requirements definition team formulates an overall concept for the system and then defines (9) showing h

9、ow the system will be operated publishes the system and operations concept document and conducts a system concept review(SCR).Following the SCR, the team derives (10) require merits for the system from the high level requirements and the system and operations conceptusing structured or object-orient

10、ed analysis, the team specifies the software functions and algorithms needed to satisfy each detailed requirement.（分数：5.00）(1).A. function B. definition C. specification D. statement（分数：1.00）A.B.C.D.(2).A. criterion B. standard C. model D. system（分数：1.00）A.B.C.D.(3).A. producer B. customer C. progra

11、mmer D. analyser（分数：1.00）A.B.C.D.(4).A. rules B. principles C. scenarios D. scenes（分数：1.00）A.B.C.D.(5).A. cotailed B. outlined C. total D. complete（分数：1.00）A.B.C.D.In the following essay, each blank has four choices. Choose the best answer and write down on the answer sheet.Spread spectrum simply me

12、ans that data is sent in small pieces over a number of the (16) frequencies available for use at any time in the specified range. Devices using (17) spread spectrum(DSSS) communicate by (18) each byte of data into several parts and sending them concurrently on different (19) . DSSS uses a lot of the

13、 available (20) , about 22 megahertz(MHz).（分数：5.00）(1).A. continuous B. high C. low D. discrete（分数：1.00）A.B.C.D.(2).A. direct-sequence B. discrete-sequenceC. duplicate-sequence D. dedicate-sequence（分数：1.00）A.B.C.D.(3).A. splitting B. combining C. packing D. compacting（分数：1.00）A.B.C.D.(4).A. bits B.

14、frequencies C. packets D. messages（分数：1.00）A.B.C.D.(5).A. rate B. velocity C. bandwidth D. period（分数：1.00）A.B.C.D.Networks can be interconnected by different devices in the physical layer networks can be connected by (1) or hubs. Which just move the bits from one network to an identical network. One

15、 layer up we find bridges and switches which operate at data link layer. They can accept (2) examine the MAC address and forward the frames to a different network while doing minor protocol translation in the process in me network layer, we have routers that can connect two networks, If two networks

16、 have (3) network layer, the router may be able to translate between the packer formats. In the transport layer we find transport gateway, which can interface between two transport connections Finally, in the application layer, application gateways translate message (4) . As an example, gateways bet

17、ween Internet e-mail and X.400 e-mail must (5) the e-mail message and change various header fields.（分数：5.00）(1).A. reapers B. relays C. packages D. modems（分数：1.00）A.B.C.D.(2).A. frimes B. packets C. packages D. cells（分数：1.00）A.B.C.D.(3).A. special B. dependent C. similar D. dissimilar（分数：1.00）A.B.C.

18、D.(4).A. syntax B. semantics C. language D. format（分数：1.00）A.B.C.D.(5).A. analyze B. parse C. delete D. create（分数：1.00）A.B.C.D.In the following essay, each blank has four choices. Choose the best answer and write down on the answer sheet.Microwave communication uses high-frequency (26) waves that tr

19、avel in straightlines through the air. Because the waves cannot (27) with the curvature of the earth, they can be (28) only over short distance. Thus, microwave is a good (29) for sendingdata between buildings in a city or on a large college campus. For longer distances, the waves must be relayed by

20、 means of “dishes“ or (30) . These can be installed on towers, high buildings, and mountain tops.（分数：5.00）(1).A. optical B. radio C. electrical D. magnetic（分数：1.00）A.B.C.D.(2).A. reflex B. distort C. bend D. absorb（分数：1.00）A.B.C.D.(3).A. transmitted B. transformed C. converted D. delivered（分数：1.00）A

21、.B.C.D.(4).A. material B. equipment C. medium D. channel（分数：1.00）A.B.C.D.(5).A. repeaters B. radars C. telephones D. antennas（分数：1.00）A.B.C.D.Packet-switching wireless networks are preferable (41) when transmissions are (42) because of the way charges are (43) per packet. Circuit-switched networks a

22、re preferable for transferring large files or for other lengthy transmissions because customers are (44) for the (45) of time they use the net-work.（分数：5.00）(1).A. to B. for C. than D. only（分数：1.00）A.B.C.D.(2).A. long B. short C. large D. small（分数：1.00）A.B.C.D.(3).A. computing B. incurious C. incurv

23、ed D. incurred（分数：1.00）A.B.C.D.(4).A. charged B. fined C. free D. controlled（分数：1.00）A.B.C.D.(5).A. point B. start C. length D. end（分数：1.00）A.B.C.D.In the following essay, each blank has four choices. Choose the most suitable one from the four choices and write down in the answer sheet.Open Shortest

24、 Path First (OSPF) is a (36) routing algorithm that (37) work clone on the OSI IS-IS intradomain routing protocol. This routing, as compared to distance-vector routing, requires (38) processing power. The (39) algorithm is used to calculate routes. OSPF routing table updates only take place when nec

25、essary, (40) at regular intervals.（分数：5.00）(1).A. distance-vector B. link-state C. flow-based D. selective flooding（分数：1.00）A.B.C.D.(2).A. derived from B. deviated from C. was derived from D. was deviated from（分数：1.00）A.B.C.D.(3).A. more B. less C. same D. most（分数：1.00）A.B.C.D.(4).A. Bellman-Ford B.

26、 Ford-Fulkerson C. Dijkstra D. RIP（分数：1.00）A.B.C.D.(5).A. but rather B. rather too C. rather than D. rather that（分数：1.00）A.B.C.D.网络工程师-计算机专业英语(一)答案解析(总分：40.00，做题时间：90 分钟)In low-speed network, it is usually adequate to wait for congestion to occur and then react to it by telling the source of packets

27、 to slow down. In high-speed networks, this approach often works poorly, because in the (61) between sending the notification and notification arriving at the source, thousands of additional (62) may arrive. In ATM network, a major tool for preventing (63) is (64) control. When a host wants a new vi

28、rtual (65) , it must describe the traffic to be offered and the service expected.（分数：5.00）(1).A. interval B. time C. slot D. delay（分数：1.00）A. B.C.D.解析：(2).A. packets B. cells C. message D. files（分数：1.00）A. B.C.D.解析：(3).A. collision B. congestion C. drop D. delay（分数：1.00）A.B. C.D.解析：(4).A. flow B. ad

29、mission C. traffic D. time（分数：1.00）A.B. C.D.解析：(5).A. path B. rout C. circuit D. way（分数：1.00）A.B.C. D.解析：解析 在低速网络中，网络拥塞经常发生，通过通知包的发送者放慢速度来对此做出反应。在高树甲络中，这种方法通常是不起作用的，因为在发送通知与同志到达这段(61)内，成千上万的额外的(62)可能会来到。在 ATM 网络中，改善(63)的主要方法是(64)控制。当一个主机想要一个新的虚拟(65)时，它必须描述线路情况和服务的需要。For each blank, choose the best a

30、nswer from the four choices and write down on the answer sheet.(11) is a protocol that a host uses to inform a router when it joins Or leaves an Internet multicast group.(12) is an error detection code that most data communication networks use.(13) is an interior gateway protocol that uses a distanc

31、e vector algorithm to propagate routing information.(14) is a transfer mode in which all types of information are organized into fixed form cells on an asynchronous or non-periodic basis over a range of media.(15) is an identifier of a web page.（分数：5.00）(1).A. ICMP B. SMTP C. IGMP D. ARP（分数：1.00）A.B

32、.C. D.解析：(2).A. 4B/SB B. CRC C. Manchester Code D. Huffman Code（分数：1.00）A.B. C.D.解析：(3).A. OSPF B. RIP C. RARP D. BGP（分数：1.00）A.B. C.D.解析：(4).A. ISDN B. X.25 C. Frame Relay D. ATM（分数：1.00）A.B.C.D. 解析：(5).A. HTYP B. URL C. HTML D. TAG（分数：1.00）_解析：The purpose of the requirements definition phase is to

33、 produce a clear, complete, consistent, and testable (6) of the technical requirements for the software product. During the requirements definition phase, the requirements definition team uses an iterative process to expand a broad statement of the system requirements into a complete and detailed sp

34、ecification of each function that the software must perform and each (7) that it must meet. The starting point is usually a set of high level requirements from the (8) that describe the project or problem.In either case, the requirements definition team formulates an overall concept for the system a

35、nd then defines (9) showing how the system will be operated publishes the system and operations concept document and conducts a system concept review(SCR).Following the SCR, the team derives (10) require merits for the system from the high level requirements and the system and operations conceptusin

36、g structured or object-oriented analysis, the team specifies the software functions and algorithms needed to satisfy each detailed requirement.（分数：5.00）(1).A. function B. definition C. specification D. statement（分数：1.00）A.B.C. D.解析：(2).A. criterion B. standard C. model D. system（分数：1.00）A. B.C.D.解析：

37、(3).A. producer B. customer C. programmer D. analyser（分数：1.00）A.B. C.D.解析：(4).A. rules B. principles C. scenarios D. scenes（分数：1.00）A. B.C.D.解析：(5).A. cotailed B. outlined C. total D. complete（分数：1.00）A.B. C.D.解析：解析 需求定义阶段的目的就是，需求定义团队使用迭代过程来将系统需求陈述扩展成完整的详细的对软件必须实现的每个功能的规格说明和必须遵守的标准准则。开始点就是一系列高级的从用户对

38、问题和工程的描述中获得的需求说明。在另外一种情况下，需求定义团队必须为系统定义完整的概念，并且定义一种能够展示系统如何工作的情景，发布系统和操作概念文档，并进行系统概念的评审。紧随着 SCR，团队从高级文档以及系统和操作概念中得出详细需求说明。使用结构化或者面向对象的分析，对软件功能和算法进行细化，这些功能和算法需要满足每一个细节要求。In the following essay, each blank has four choices. Choose the best answer and write down on the answer sheet.Spread spectrum simp

39、ly means that data is sent in small pieces over a number of the (16) frequencies available for use at any time in the specified range. Devices using (17) spread spectrum(DSSS) communicate by (18) each byte of data into several parts and sending them concurrently on different (19) . DSSS uses a lot o

40、f the available (20) , about 22 megahertz(MHz).（分数：5.00）(1).A. continuous B. high C. low D. discrete（分数：1.00）A. B.C.D.解析：(2).A. direct-sequence B. discrete-sequenceC. duplicate-sequence D. dedicate-sequence（分数：1.00）A. B.C.D.解析：(3).A. splitting B. combining C. packing D. compacting（分数：1.00）A. B.C.D.解

41、析：(4).A. bits B. frequencies C. packets D. messages（分数：1.00）A.B. C.D.解析：(5).A. rate B. velocity C. bandwidth D. period（分数：1.00）A.B.C. D.解析：解析 扩频，简单的说，就是通过一定范围内的任意时刻的连续的可用频率将数据用小片传送。使用直接序列扩频技术的设备通过将数据分解成几部分并且以不同的频率发送它们，来进行通信。DSSS使用许多可用带宽，大约是 22Mhz。Networks can be interconnected by different devices i

42、n the physical layer networks can be connected by (1) or hubs. Which just move the bits from one network to an identical network. One layer up we find bridges and switches which operate at data link layer. They can accept (2) examine the MAC address and forward the frames to a different network whil

43、e doing minor protocol translation in the process in me network layer, we have routers that can connect two networks, If two networks have (3) network layer, the router may be able to translate between the packer formats. In the transport layer we find transport gateway, which can interface between

44、two transport connections Finally, in the application layer, application gateways translate message (4) . As an example, gateways between Internet e-mail and X.400 e-mail must (5) the e-mail message and change various header fields.（分数：5.00）(1).A. reapers B. relays C. packages D. modems（分数：1.00）A. B

45、.C.D.解析：(2).A. frimes B. packets C. packages D. cells（分数：1.00）A. B.C.D.解析：(3).A. special B. dependent C. similar D. dissimilar（分数：1.00）A.B.C.D. 解析：(4).A. syntax B. semantics C. language D. format（分数：1.00）A. B.C.D.解析：(5).A. analyze B. parse C. delete D. create（分数：1.00）A.B. C.D.解析：解析 可以通过不同的设备实现网络的互连。

46、在物理层，网络可以通过中继器或者 HUB 相连，这些设备仅仅是将比特源目的网络之间传输；在数据链路层，通过交换机和网桥实现网络的互连；它们可以接收帧，检测 MAC 地址，并且能够将帧传向另一个网络，同时，还可以进行少量的协议转换工作。在网络层，使用路由器来连接两个网络。如果两个网络拥有不同的网络层，路由器能够在不同帧格式之间相互转换。在传输层有传输网关，能够在两个传输层之间建立端口连接。最后，应用层有应用网关实现消息语法之间的转化。作为例子，在 Internet e-mail 和 X.400 e-mail 之间的应用网关可以对 e-mail 的消息进行语法分析，并且能够变换消息头。In the

47、 following essay, each blank has four choices. Choose the best answer and write down on the answer sheet.Microwave communication uses high-frequency (26) waves that travel in straightlines through the air. Because the waves cannot (27) with the curvature of the earth, they can be (28) only over shor

48、t distance. Thus, microwave is a good (29) for sendingdata between buildings in a city or on a large college campus. For longer distances, the waves must be relayed by means of “dishes“ or (30) . These can be installed on towers, high buildings, and mountain tops.（分数：5.00）(1).A. optical B. radio C.

49、electrical D. magnetic（分数：1.00）A.B. C.D.解析：(2).A. reflex B. distort C. bend D. absorb（分数：1.00）A.B.C. D.解析：(3).A. transmitted B. transformed C. converted D. delivered（分数：1.00）A. B.C.D.解析：(4).A. material B. equipment C. medium D. channel（分数：1.00）A.B.C. D.解析：(5).A. repeaters B. radars C. telephones D. antennas（分数：1.00）A.B.C.D. 解析：解析 微波通信使用在空气中直线传播的高频电磁波，由于波不能随着地球的曲率弯曲，只能进行短距离的传播。因此，微