【计算机类职业资格】网络规划设计师-应用数学与财务管理、专业英语及答案解析.doc

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1、网络规划设计师-应用数学与财务管理、专业英语及答案解析(总分：53.00，做题时间：90 分钟)一、单项选择题(总题数：23，分数：53.00)1.局域网争用信道方案的一个缺点是，由于多个站点试图同时访问信道而造成的信道带宽浪费。假设将访问时间分割成离散的时隙，每个时隙中有 n 个站点以概率 p 试图发送数据帧。由于多个站点试图同时发送数据而被浪费的时隙比例是_。A1-np(1-p) n B1-(np+1)(1-p) nC1-p n(1-p)n-n(1-p)n-1 D1-np(1-p) n-1（分数：1.00）A.B.C.D.2.在大多数网络中，数据链路层主要是通过请求重发已损坏的数据帧的办法

2、来解决发送出错问题。如果一个帧被损坏的概率是 P，而且确认信息不会丢失，则发送一帧的平均发送次数是_。A1-p B1/(1-P)C1+p D1/(1+P)（分数：1.00）A.B.C.D.3.某个高层报文被分为 20 帧进行传送，每帧无损坏到达目的地的可能性为 90%，若数据链路层协议不进行差错控制，则该报文平均需要_次才能完整的到达接收方。(注：0.45 103.410 -4，0.9 100.3487，0.45 201.1610 -7，0.9 200.1216)A1.138 B1.535 C2.868 D8.224（分数：1.00）A.B.C.D.4.某个高层报文被分为 5 帧进行传送，帧出

3、错的概率为 0.1，纠错重发以报文为单位，则整个报文的平均发送次数约为_次。A1.243 B1.694 C2.868 D3.221（分数：1.00）A.B.C.D.5.假设数据帧出错的概率为 p，应答帧不会出现错误，允许重传的次数不受限制；成功发送两个数据帧之间的最小间隔时间为 tT。此时链路的最大吞吐量 max=_。Ap(1-p)/2t T B(1-p)/(2pt T)C2p/(1-p)t T) D(1-p)/t T（分数：1.00）A.B.C.D.6.若某个信道误码率为 10-5，数据帧帧长为 10103比特，差错为单个错，则帧出错的概率为_。A0.01 B0.1C(1-10 -5)101

4、03 D1-(1-10 -5)10103（分数：1.00）A.B.C.D.某省 6 个城市(AF)之间的网络通信线路(每条通信线路旁标注了其长度千米数)如图 28-1 所示。如果要将部分千兆通信线路改造成万兆通信线路，以提升各个城市网络之间的通信容量，则至少要改造总计_千米的通信线路，这种总千米数最少的改造方案共有_个。（分数：2.00）(1).A1000 B1 300 C1600 D2000（分数：1.00）A.B.C.D.(2).A1 B2 C3 D4（分数：1.00）A.B.C.D.7.11 个乡镇之间的光缆铺设网络结构及每条光缆长度如图 28-5 所示。从乡镇 s 到乡镇 t 的最短光

5、缆铺设距离为_千米。（分数：1.00）A.B.C.D.8.假设某项网络工程项目由设备 x1、x 2、x 3、x 4、x 5和设备 y1、y 2串联组成。其中，设备 x1必须在左端或右端，且设备 y1、y 2必须相邻，则共有_种排法。A120 B192 C360 D480（分数：1.00）A.B.C.D.9.某省电子政务外网由 r 台骨干路由器通过 2.5Gbps RPR/POS 环网相连组成省级横向网的核心层。现从门台有编号的路由器中取 r(0rn)台，则该 RPR/POS 环共有_种不同的排法。（分数：1.00）A.B.C.D.10.利用 M/M/1 排队论模型分析图 28-6 所示的是一个

6、简单的工作流模型(其中单位时间为 1 小时)。它表示这样一个执行过程：每小时将会有 20 个任务达到 c1，这 20 个任务首先经过处理 task1，再经过处理task2，最终将结果传递到 c3。处理 task1 和处理 task2 相互独立，且每个任务到达间隔服从负指数分布。处理 task1、task2 的平均等待时间分别是_。（分数：1.00）A.B.C.D.11.某信息系统的 LAN、Web 服务器和数据库服务器按照图 28-7 所示串联连接。根据测得的处理时间，LAN、Web 服务器和数据库服务器对单个客户请求的平均处理时间为 30ms、40ms 和 10ms。根据以往的经验可知，每秒

7、平均有 20 个事务，且事务到达间隔服从负指数分布。现利用 M/M/1 排队论模型估算并发连接的客户数量增加时的响应时间。每个队列的平均系统时间(平均等待时间与平均处理时间之和，单位：秒)可用以下公式计算。该信息系统的平均响应时间 R 为_ms。（分数：1.00）A.B.C.D.12.某省广域骨干网的部分网络拓扑结构如图 28-8 所示，图中标识出了各节点之间网络流量的传输能力(单位：Mbps)。从节点到节点的最大网络流量可以达到_Mbps。（分数：1.00）A.B.C.D.13.某网络集成公司项目 A 的利润分析如表 284 所示。设贴现率为 10%，第 3 年的利润净现值是_元。表 28-

8、4 某软件公司项目 A 利润分析表年份 利润值(元) 年份 利润值(元)第 0 年 第 2 年 ￥203800第 1 年 ￥118800 第 3 年 ￥235200A89256.20 B153117.96C176709.24 D246598.00（分数：1.00）A.B.C.D.14.某公司的销售收入状态如表 28-5 所示，就销售收入而言该公司的盈亏平衡点是_百万元人民币。表 28-5 某公司的销售收入状态表 单位：百万元人民币项目 金额 项目 金额销售收入 800 毛利 270材料成本 300 固定销售成本 150分包费用 100 利润 120固定生产成本 130A560 B608C615

9、 D680（分数：1.00）A.B.C.D.某软件企业 2011 年初计划投资 1000 万人民币开发一套统一安全网关(UTM)产品，预计从 2012 年开始，年实现产品销售收入 1500 万元，年市场销售成本 1000 万元。该产品的分析员张工根据财务总监提供的贴现率，制作了如表 28-6 所示的产品销售现金流量表。根据表中的数据，该产品的动态投资回收期是_年，投资收益率是_。表 28-6 某企业 UTM 产品销售现金流量表年度 2011 2012 2013 2014 2015投 资 1000 成 本 1000 1000 1000 1000收 入 1500 1500 1500 1500净现金

10、流量 -1000 500 500 500 500净现值 -925.93428.67396.92367.51340.29（分数：2.00）(1).A1 B2 C2.27 D2.73（分数：1.00）A.B.C.D.(2).A42% B44% C50% D100%（分数：1.00）A.B.C.D.To compete in todays fast-paced competitive environment, organizations are increasingly allowing contractors, partners, visitors and guests to access the

11、ir internal enterprise networks. These users may connect to the network through wired ports in conference rooms or offices, or via wireless access points. In allowing this open access for third parties, LANs become _. Third parties can introduce risk in a variety of ways from connecting with an infe

12、cted laptop to unauthorized access of network resources to _ activity. For many organizations, however, the operational complexity and costs to ensure safe third party network access have been prohibitive. Fifty-two percent of surveyed CISOs state that they currently use a moat and castles security

13、approach, and admit that defenses inside the perimeter are weak. Threats from internal users are also increasingly a cause for security concerns. Employees with malicious intent can launch _ of service attacks or steal _ information by snooping the network. As they access the corporate network, mobi

14、le and remote users inadvertently can infect the network with _ and worms acquired from unprotected public networks. Hackers masquerading as internal users can take advantage of weak internal security to gain access to confidential information.（分数：5.00）(1).A. damageable B. susceptible C. changeable

15、D. vulnerable（分数：1.00）A.B.C.D.(2).A. malicious B. venomous C. felonious D. villainous（分数：1.00）A.B.C.D.(3).Arenounce BvirtUOUS Cdenlal Dtraverse（分数：1.00）A.B.C.D.(4).Areserved Bconfidential Ccomplete Dmysterious（分数：1.00）A.B.C.D.(5).Asickness Bdisease Cgerms Dviruses（分数：1.00）A.B.C.D.WLANs are increasin

16、gly popular because they enable cost-effective connections amongpeople, applications and data that were not possible, or not cost-effective, in the past.For example, WLAN-based applications can enable fine-grained management of supply and distribution _ to improve their efficiency and reduce _. WLAN

17、s can also enable entirely new business processes. To cite but one example, hospitals are using WLAN-enabled point-of-care applications to reduce errors and improve overall _ care.WLAN management solutions provide a variety of other benefits that can be substantial but difficult to measure. For exam

18、ple, they can protect corporate data by preventing _ through rogue access points. They help control salary costs, by allowing IT staffs to manage larger networks without adding staff. And they can improve overall network management by integrating with customers existing systems, such as OpenView. Fo

19、rtunately, it isnt necessary to measure these benefits to justify investing in WLAN management solutions, which can quickly pay for themselves simply by minimizing time-consuming _ and administrative chores.（分数：5.00）(1).A. chores B. chains C. changes D. links（分数：1.00）A.B.C.D.(2).A. personnel B. expe

20、nses C. overhead D. hardware（分数：1.00）A.B.C.D.(3).A. finance B. patient C. affair D. doctor（分数：1.00）A.B.C.D.(4).A. intrusion B. aggression C. inbreak D. infall（分数：1.00）A.B.C.D.(5).A. exploitation B. connection C. department D. deployment（分数：1.00）A.B.C.D.Pharming is a scamming practice in which malici

21、ous code is installed on a personal computer or server, misdirecting users to _ Web sites without their knowledge or consent. Pharming has been called “phishing without a lure“.In phishing, the perpetrator sends out legitimate-_ e-mails, appearing to come from some of the Webs most popular sites, in

22、 an effort to obtain personal and financial information from individual recipients. But in pharming, larger numbers of computer users can be _ because it is not necessary to target individuals one by one and no conscious action is required on the part of the victim.In one form of pharming attack, co

23、de sent in an e-mail modifies local host files on a personal computer. The host files convert URLs into the number strings that the computer uses to access Web sites. A computer with a compromised host file will go to the fake Web site even if a user types in the correct Internet address or clicks o

24、n an affected _ entry. Some spyware removal programs can correct the corruption, but it frequently recurs unless the user changes browsing _（分数：5.00）(1).A. few B. fraudulent C. normal D. structured（分数：1.00）A.B.C.D.(2).A. connecting B. contenttagging C. looking D. binding（分数：1.00）A.B.C.D.(3).A. victi

25、mized B. personate C. identity D. latency（分数：1.00）A.B.C.D.(4).A. hypertext B. computation C. expectation D. bookmark（分数：1.00）A.B.C.D.(5).A. habits signature B. manageabilityC. efficiency D. address（分数：1.00）A.B.C.D.When the system upon which a transport entity is running fails and subsequently reatar

26、ts, the _ information of all active connections is lost. The affected connections become half-open, as the side that did not fail does not yet realize the promble.The still active side of a half-open connections using a _ timer. This timermeasures the time transport machine will continue to await an

27、 _ of a transmitted segment after the segment has been retransmitted the maximum number of times. When the timer _, the transport entity assumes that either the other transport entity or the intervening network has failed. As a result, the timer closes the connection, and signals an abnormal close t

28、o the TS user.In the event that a transport entity fails and quickly restart, half-open connections can be teminated more quickly by the use of the RST segment. The failed side returns an RSTi to every segment i that it receives. When the RSTi reaches the other side, it must be checked for validity

29、dased on the _ number i, as the RST could be in response to an old segment .If the reset is valid, the transport entity performs an abnormal termination.（分数：5.00）(1).A. data B. control C. signal D. state（分数：1.00）A.B.C.D.(2).A. abandon B. give-up C. quit D. connection（分数：1.00）A.B.C.D.(3).A. reset B.

30、acknowledgment C. synchroizer D. sequence（分数：1.00）A.B.C.D.(4).A. expires B. restarts C. stops D. abandons（分数：1.00）A.B.C.D.(5).A. connection B. acknowledgment C. sequence D. message（分数：1.00）A.B.C.D._ is used to ensure the confidentiality, integrity and authenticity of the two end points in the privat

31、e network. _, an application-layer protocol, authenticates each peer in an IPSec transaction. IKE negotiates security policy, determining which algorithm may be used to set up the tunnel. It also handles the exchange of session keys used for that one transaction.Networks that use _ to secure data tr

32、affic can automatically authenticate devices by using by using _, which verify the identities of the two users who are sending information back and forth. IPSec can be ideal way to secure data in large networks that require secure connections among many devices.Users deploying IPSec can _ their netw

33、ork infrastructure without affecting the applications on individual computer. The protocol suite is available as a software-only upgrade to the network infrastructure. This alows security to be implemented without costly changes to each computer. Most important, IPSec allows interoperability among d

34、ifferent network devices, PCs and other computing systems.（分数：5.00）(1).A. Certificate B. EncryptionC. Tunnel D. Presentation（分数：1.00）A.B.C.D.(2).A. IPSec B. SSLC. L2TP D. The Internet Key Exchange（分数：1.00）A.B.C.D.(3).A. authenticity B. IPSec C. confidentiality D. integrity（分数：1.00）A.B.C.D.(4).A. cha

35、racteristics B. associated linksC. digital certificates D. attributes（分数：1.00）A.B.C.D.(5).A. secure B. relation C. script D. strict（分数：1.00）A.B.C.D.The metric assigned to each network depends on the type of protocol. Some simple protocol, like RIP, treats each network as equals. The _ of passing thr

36、ough each network is the same; it is one _ count. So if a packet passes through 10 network to reach the destination, the total cost is 10 hop counts. Other protocols, such as OSPF, allow the administrator to assign a cost for passing through a network based on the type of service required.A _ throug

37、h a network can have different costs (metrics). For example, if maximum _ is the desired type of service, a satellite link has a lower metric than a fiber-optic line. On the other hand, if minimum _ is the desired type of service, a fiber-optic line has a lower metric than a satellite line. OSPF all

38、ow each router to have several routing table based on the required type of service.（分数：5.00）(1).A. cost B. link-state C. connection D. diagram（分数：1.00）A.B.C.D.(2).Aflow Bhop Cprocess Droute（分数：1.00）A.B.C.D.(3).Aadaptability Bscalablity Croute Dsecurity（分数：1.00）A.B.C.D.(4).A1atency Bthroughput Cutili

39、zation Djitter（分数：1.00）A.B.C.D.(5).Abandwidth Befficiency Cavailability Ddelay（分数：1.00）A.B.C.D.The Border Gateway Protocol (BGP) is an interautonomous system. _ protocol. The primary function of a BGP speaking system is to exchange network _ information with other BGP system. This network reachabili

40、ty information includes information on the list of Autonomous System (ASs) that teachability information traverses. BGP-4 provides a new set of mechanisms for supporting _ interdomain routing. These mechanisms include support for advertising an IP _ and eliminate the concept of network class within

41、BGP. BGP-4 also introduces mechanisms that allow aggregation of routes, including _ of AS paths. These changes provide support for the proposed supernettting scheme.（分数：5.00）(1).A. resolving B. connecting C. supernettting D. routing（分数：1.00）A.B.C.D.(2).A. teachability B. scalablity C. availability D

42、. reliability（分数：1.00）A.B.C.D.(3).A. confirmless B. connectionless C. answerless D. classless（分数：1.00）A.B.C.D.(4).A. suffix B. prefix C. reflex D. infix（分数：1.00）A.B.C.D.(5).A. reservation B. utilization C. aggregation D. connection（分数：1.00）A.B.C.D.网络规划设计师-应用数学与财务管理、专业英语答案解析(总分：53.00，做题时间：90 分钟)一、单项选

43、择题(总题数：23，分数：53.00)1.局域网争用信道方案的一个缺点是，由于多个站点试图同时访问信道而造成的信道带宽浪费。假设将访问时间分割成离散的时隙，每个时隙中有 n 个站点以概率 p 试图发送数据帧。由于多个站点试图同时发送数据而被浪费的时隙比例是_。A1-np(1-p) n B1-(np+1)(1-p) nC1-p n(1-p)n-n(1-p)n-1 D1-np(1-p) n-1（分数：1.00）A.B. C.D.解析：解析 由于访问信道的时间被分割成离散的时隙，在一个时隙中所有可能发生的事件共有 3 种：n 个站点中只有 1 个站点传输帧而成功获得信道的访问权，其概率 P1 为 n

44、p(1-p)n；没有站点试图发送帧，其概率 P2为(1-p) n；n 个站点中有 1 个以上的站点试图同时发送数据而造成信道访问的冲突，其概率为 P3。由于Pl+P2+P3=1，因此 P3=1-P1-P2=1-np(1-p)n-(1-p)n=1-(np+1)(1-p)n。2.在大多数网络中，数据链路层主要是通过请求重发已损坏的数据帧的办法来解决发送出错问题。如果一个帧被损坏的概率是 P，而且确认信息不会丢失，则发送一帧的平均发送次数是_。A1-p B1/(1-P)C1+p D1/(1+P)（分数：1.00）A.B. C.D.解析：解析 假设一个帧需要传送 k 次才能传送成功的概率为 Pk，表示

45、的含义是：前 k-1 次均发送失败，其概率为 Pk-1；第 k 次传送成功，其概率为(1-P)，则 pk=Pk-1(1-P)。因此，一个帧需要的平均传输次数是：3.某个高层报文被分为 20 帧进行传送，每帧无损坏到达目的地的可能性为 90%，若数据链路层协议不进行差错控制，则该报文平均需要_次才能完整的到达接收方。(注：0.45 103.410 -4，0.9 100.3487，0.45 201.1610 -7，0.9 200.1216)A1.138 B1.535 C2.868 D8.224（分数：1.00）A.B.C.D. 解析：解析 由于每个帧正确到达接收方的概率为 0.9，整个报文信息正确

46、到达接收方的概率P=0.9200.1216。为了使信息完整的到达接收方，发送一次成功的概率为 P，发送两次成功的概率为P(1-P)，发送 i 次成功的概率为 P(1-P)i-1，因此所需的平均次数 E 为：E=P+2P(1-p)+i(1-P)i-1p+=4.某个高层报文被分为 5 帧进行传送，帧出错的概率为 0.1，纠错重发以报文为单位，则整个报文的平均发送次数约为_次。A1.243 B1.694 C2.868 D3.221（分数：1.00）A.B. C.D.解析：解析 由于帧出错的概率为 0.1，即每个帧正确到达接收方的概率为 0.9，整个报文信息正确到达接收方的概率 P=0.95=0.59

47、049。为了使信息完整的到达接收方，发送一次成功的概率为 P，发送两次成功的概率为 P(1-P)，发送 i 次成功的概率为 P(1-P)i-1，因此所需的平均次数 E 为：5.假设数据帧出错的概率为 p，应答帧不会出现错误，允许重传的次数不受限制；成功发送两个数据帧之间的最小间隔时间为 tT。此时链路的最大吞吐量 max=_。Ap(1-p)/2t T B(1-p)/(2pt T)C2p/(1-p)t T) D(1-p)/t T（分数：1.00）A.B.C.D. 解析：解析 在题意所描述的情况下，正确传送一帧所需的平均时间 tav=tT(1+一个帧的平均重传次数)。由以上新题 4 和新题 5 考

48、点分析可知，每个帧正确到达接收方的概率 P 与出错的概率 p 是对立事件，即P+p=1，则一个帧的平均重传次数 E=1/P=1/(1-p)。进而可得，t av=tT/(1-p)。当传输差错率 P 增大时，平均时间 tav也增大；当无差错(p=0)时，t av=tT。每秒成功发送的最大数据帧就是链路的最大吞吐量 max=1/tav=(1-p)/tT。6.若某个信道误码率为 10-5，数据帧帧长为 10103比特，差错为单个错，则帧出错的概率为_。A0.01 B0.1C(1-10 -5)10103 D1-(1-10 -5)10103（分数：1.00）A.B.C.D. 解析：解析 若设帧出错的概率为 P，P b为误码率，L 为帧长，则 P=1-(1-Pb)L。依题意，P b=10-5，L=1010 3，则 P=1-(1-Pb)L=1-(1-10-5)10103。某省 6 个城市(AF)之间的网络通信线路(每条通信线路旁标