下载文档原格式
HCIP题库四HCIP-H12-222练习题习题 1IP报⽂中⽤Tos字段进⾏Qos的标记，Tos字段中是使⽤前6bit来标记DSCP的。
A. 正确B. 错误答案: A习题 2传统的丢包策略采⽤尾部丢弃（Tail-Drop）的⽅法，这种丢弃⽅法会导致TCP全局同步现象。
A. 正确B. 错误答案: A习题 3下列选项中，属于Agile Controller的业务编排的亮点的是？A. 基于三层GRE隧道进⾏编排，业务设备的组⽹⽅式、部署位置更加灵活B. 通过拓扑可视化⽅式进⾏业务编排，配置简单，管理便捷C. 业务设备的增删，不改变现⽹转发路由，不改变现⽹物理拓扑D. ⽆需⼈⼯维护，⾃动对业务进⾏分析编排答案: ABC习题 4下列应⽤场景中，不属于Agile Controller的访客接⼊管理的场景的是？A. 客户沟通、交流、参观等接⼊企业⽹络，访问企业公共资源或InternetB. 普通民众通过公共事业单位提供的⽹络访问InternetC. 客户在企业消费，连接企业⽹络访问InternetD. 公司员⼯出差到分公司，连接分公司⽹络访问公司⽹络答案: D习题 5NFV中的VIM管理模块的主要功能包括资源发现、资源分配、资源管理以及？A. 资源调度B. 资源监控C. 资源回收D. 故障处理答案: D习题 6SDN主要技术流派主张SDN采⽤分层的开放架构，那么倡导、定义集中式架构和OpenFlow的是？A. ONFB. IEIFC. ETSID. ITU答案: A习题 7为避免TCP全局同步现象，可使⽤的拥塞避免机制有:A. REDB. WREDC. Tail-DropD. WFQ答案: ABRandom Early Detection；Weighted Random Early Detection习题 8下⾯哪项为配置SDN控制器侦听地址的命令？A. OpenFlow Listening-ip 1.1.1.1B. Sdn Controller souce-address 1.1.1.1C. Controller-ip 1.1.1.1D. Sdn Listening-ip 1.1.1.1答案: A习题 9DHCP服务器分配给客户端的动态IP地址，通常都有⼀定的租借期限，那么关于租借期限的描述，错误的是？A. 租期更新定时器为总租期的50%，当“租期更新定时器”到期时，DHCP客户端必须进⾏IP地址的更新B. 重绑定定时器为总租期的87.5%C. 若“重绑定定时器”到期，但客户端还没有收到服务器的响应，则会⼀直发送DHCP REQUEST报⽂给之前分配过IP地址的DHCP服务器，直到总租期到期D. 在租借期限内，如果客户端收到DHCP NAK报⽂，客户端就会⽴即停⽌使⽤此IP地址，并返回到初始化状态，重新申请新的IP地址答案: C习题 10NFV常常部署在下列哪些应⽤环境中？A. 数据中⼼C. ⽤户接⼊侧D. 客户机/服务器答案: ABC⽹络功能虚拟化（Network Functions Virtualization）；习题 11ASPF（Applicaion Specific Packet Filter）是⼀种基于应⽤层的包过滤，它会检查应⽤层协议信息并且监控连接的应⽤层协议状态，并通过Server-map表实现了特殊的安全机制。
TCP Buffering And Performance Over An ATM NetworkPurdue Technical Report CSD-TR94-026Douglas er and John C.LinDepartment of Computer SciencesPurdue UniversityWest Lafayette,Indiana47907March16,1994AbstractThis paper reports a series of experiments tomeasure TCP performance when transferring datathrough an Asynchronous Transfer Mode(ATM)switch.The results show that TCP buffer sizes andthe ATM interface maximum transmission unithave a dramatic impact on throughput.We ob-serve a throughput anomaly in which an increasein the receiver’s buffer size decreases throughputsubstantially.For example,when using a16Koctet send buffer and ATM Adaptation Layer5ona100megabit per second(Mb/s)ATM path,themean throughput for a bulk transfer drops from15.05Mb/s to0.322Mb/s if the receiver’s buffersize is increased from16K octets to24K octets.This paper analyzes the performance,explains theanomalous behavior,and describes a solution thatprevent the anomaly from occurring.[16]to operate over ATM.A user who connects to an ATM network can run existing applications that use the Transmission Control Protocol(TCP) [17]or the User Datagram Protocol(UDP)[15] without modification.Users who share a conventional network(e.g., a10Mb/s Ethernet)expect dramatic increases in performance from a dedicated ATM connection that operates an order of magnitude faster.How-ever,measurement offile transfers using FTP [19]showed a surprising result:the same ftp program that performs well over a10Mb/sec Ethernet can perform worse over an100Mb/s ATM path.For example,when transferring a4.4 megabyte datafile between two hosts connected to the same Ethernet,ftp reports a mean throughput of1.313Mb/s.However,using the same soft-ware and computers to transfer thefile across a 100Mb/s ATM path produce a mean throughput of only0.366Mb/s.Furthermore,the ATM net-work management software reports no cell lost. The low throughput prompted us to investigate the effects of TCP buffering on its performance. Experiments revealed the sizes of the sender’s and receiver’s buffers have a dramatic effect on per-formance.The remainder of this paper is organized as follows.Section2describes the ATM network configuration used to conduct the experiments, the tool used to measure TCP performance,and the experimental procedures.Section3presents the results of the experiments and identifies a throughput anomaly in which an increases in the receiver’s buffer size decreases TCP throughput significantly.Section4explains the cause of the throughput anomaly and describes a solution that prevents the anomaly from occurring.Section?? summarizes the paper.2Measuring TCP Performance Over ATMFigure1illustrates the network configuration for conducting the experiments.Two multi-homed Sun Microsystems’SPARCstation IPCs,A and B, running SunOS4.1.11are used to measure TCP performance over ATM.Each host has two net-work interfaces:one connects to an Ethernet and the other connects to a Fore Systems’ASX-100 ATM switch via a100Mb/s multi-modefiber link. Each host uses a Fore SBA-200ATM adapter card to interface with the ATM switch.The adapter card embeds a dedicated RISC processor and spe-cial purpose hardware to handle ATM Adaptation Layer5(AAL5)[6,9].The Maximum Transmis-sion Unit(MTU)on the ATM interface and the Ethernet interface is9188octets and1500octets, respectively.2.1Measurement ToolASX-100ATM SwitchHost A Host B SBA-200Adapter 100 Mb/s100 Mb/s SBA-200Adapter 10 Mb/s Ethernet(Sparc IPC)(Sparc IPC)Figure 1:Configuration of an ATM network used to measure TCP throughputHost BHost Attcp(source)(sink)ttcpTCP/IP (AAL5)Figure 2:Using ttcp to measure TCP throughput We used a public domain C program called ttcp 2to measure TCP throughput.A ttcp running on host A uses the BSD socket interface provided by SunOS to communicate with another ttcp on host B .As Figure 2illustrates,we configured one ttcp as a source and the other as a sink.Once a user has specified the amount of data to transmit,the source ttcp continuously transmits the data to the destination ttcp (sink)until all the data are transmitted;the destination ttcp simply discards the data it ers also can specify the sizes of the sending TCP’s send buffer and the receiving TCP’s receive buffer by using an option2: Shaded area indicates abnormal TCP throughput.Table 1: TCP buffer sizes and mean throughputS e n d B u f f e r S i z e (o c t e t )Receive Buffer Size (octet)Note 1: Throughput numbers are in megabits per second (Mb/s).minutes [2].We artificially established an ATM connection before each experiment.Thus,the reported throughput does not include connection setup time.2.2Experimental ProceduresWe conducted 100experiments to investigate the effect of send and receive buffer sizes on TCP throughput when transferring bulk data over a 100Mb/s ATM path.The send and receive buffer sizes range from 16K octets to 51K octets in a 4K increments.The minimum buffer size of 16K was selected because the ATM interface has an MTU of 9188octets and the SunOS kernel is configured to use a default send and receive buffer size of 16K when installing the driver software for the ATM adapter cards.The maximum buffer size of 51K was selected because SunOS 4.1.1TCP restrictsthe buffer size to 52428octets.Each experiment consists of 50independent throughput measurements.In each measurement,the source transmits 32megabytes of data to the sink.There is a delay of 5seconds between mea-surements.All the experiments use AAL5to en-capsulate IP datagrams.3ResultsTable 1shows the mean throughput measured for each experiment;Figures 9to 13in Appendix plot the throughput data of each experiment.As Ta-ble 1shows,in general,TCP throughput increases as the sender’s and receiver’s buffer sizes increase.Some experiments,however,show a decrease in mean throughput when send and/or receive buffer sizes increase.For example,when a sending TCP141516171819202122162024283236404448T h r o u g h p u t (M b /s )Buffer Size (kilooctet)51Figure 3:TCP mean throughput when sender and receiver use the same buffer sizeuses a 16K buffer,the mean throughput decreases about 10%when the receiving TCP increases the receiver buffer size from 16K to 20K.Also,as Figure 3shows,when the sender and receiver use the same buffer size,TCP performs better with a 16K buffer than with a 20K buffer.Surprisingly,as the shaded entries in Table 1shows,certain combinations of unequal send and receiver buffer sizes cause exceptional low throughput.Furthermore,the exceptional low throughput occurs when a receiving TCP in-creases its receive buffer size.For example,when using a 16K send buffer,TCP mean throughput decreases from 15.05Mb/s to 0.322Mb/s if the receive buffer size is increased from 16K to 24K.The next section explains the anomalous behavior and describes a solution to prevent it from occur-ring.4Analysis of the ResultsAlthough network analyzers can be used to cap-ture data on a shared access network (e.g.,an Eth-ernet),the technique does not work well in a point-to-point,non-shared access ATM network.Thus,we use kernel probing [8,14]to study the ob-served TCP throughput anomaly over ATM.The technique uses a data structure in the kernel ad-dress space and inserts probing code at various locations of the TCP source modules to gather relevant data.An application program reads the gathered data from the kernel address space foroff-line analysis.4.1Analysis of the Throughput AnomalyBy analyzing the gathered data,we conclude that the interaction of the following items causes the dramatic throughput decrease:1.The sender’s send buffer size 2.The receiver’s receive buffer size 3.The MTU of the ATM interface 4.The TCP maximum segment size (MSS)5.The way user data is added to the TCP send buffer6.Sender side Silly Window Syndrome avoid-ance (Nagle’s algorithm)7.The receiver side delayed acknowledgmentalgorithmThefirst two items are configurable by applica-tions in BSD derived UNIX by using setsockopt system call.The MTU of Fore Systems’ATM in-terface card is9188octets for IP over ATM when using AAL5.SunOS4.1.1calculates TCP MSS as9148octets(i.e.,9188minus the default TCP and IP header sizes)in our ATM network config-uration.Items5to7are implementation related; we describe how SunOS4.1.1implements them below.4.1.1Adding Data to TCP BufferThe SunOS4.1.1implements TCP buffers as a list of mbuf s[12].Each mbuf can store up to112 octets of data or contain a pointer to a1K octet memory block for storing large messages.During bulk data transfer,if the send buffer is larger than 4K,SunOS adds user data in blocks of4K octets, then invokes a TCP routine to transmit the data; if the available space in the send buffer is smaller than4K,SunOS adds data in multiples of1K octet block.4.1.2Sender Side Silly Window SyndromeAvoidance(Nagle’s Algorithm)Silly Window Syndrome(SWS)[7]is character-ized as a situation in which a steady pattern of small TCP window increments results in small data segments being transmitted.Sending small data segments lowers TCP throughput because TCP and IP headers consume network bandwidth. To avoid SWS,both sender and receiver must im-plement SWS avoidance algorithm[3].On the receiver side,it must avoid advancing the right window edge in small increments when it receives small data segments.On the sender size,it must avoid sending small data segments to the receiver even if the receiver has space available to accept them.For applications that are character-oriented (e.g.,remote login,TELNET[18]),every charac-ter generated by the applications must be pushed explicitly by the application or TCP to avoid dead-locks.If a TCP transmits every pushed data,the result is a stream of one octet data segments.To better utilize network resources,a TCP tries to buffer segments that are small compared to the size of TCP and IP headers.However,to avoid deadlock,TCP must not buffer a data segment that needs immediate delivery.Nagle’s algorithm[13] provides a simple solution to the dilemma:if there is unacknowledged data(i.e.,the connection is not idle),TCP buffers all data(even if the PUSH bit is set)until TCP can send an MSS segment or until all the outstanding data has been acknowledged [3,13].Note that Nagle’s algorithm also pro-vides sender side SWS avoidance.SunOS4.1.1Figure4:Illustration of a sender’s usable windowTCP uses the following two conditions to avoid sending small data segments:S1:If1,then transmita segment with1*MSS octet of data.isthe amount of data to be transmitted,is theusable window[7](i.e.,the available receivebuffer space in the receiver)as illustrated inFigure4.S2:If an ACK from the peer acknowledges all the outstanding data and there are octets ofdata waiting in the send buffer,then transmita segment with octets of data.In condition S1,if a sender has at least1*MSS octets of data to be transmitted and the receiver has space to receive an MSS segment,TCP trans-mits an MSS segment.Condition S2requires TCP to transmit buffered data when the peer acknowl-edges all the outstanding data.When transmitting data,TCP always checks condition S1before con-dition S2.Thus,is always less than1*MSS. When there is unacknowledged data,conditions Figure5:Illustration of how a TCP determines when to send data over an ATM pathS1and S2allow TCP to buffer small data seg-ments until it can send an MSS segment or all the unacknowledged data have been acknowledged. Note that when there is no unacknowledged data, TCP transmits any data queued in the send buffer even the amount of data is less than1*MSS. Figure5illustrates how a sending TCP with 16K send buffer uses conditions S1and S2to de-cide when to transmit data segments over an ATM path.SunOS4.1.1TCP transmits4K octets in thefirst segment because the connection was idle. Because TCP MSS is9148octets,after SunOS finishes adding the fourth4K octet block,TCP transmits a second data segment with9148octet of data leaving3140(12K-9148)octets of data queued in the send buffer(condition S1).When the peer acknowledges all the unacknowledgeddata in the third segment,TCP transmits a3140 octet data segment(condition S2).4.1.3Receiver Side Delayed ACKA receiving TCP can increase TCP throughput, reduce protocol processing at both ends,and gen-erate less traffic by using delayed ACK[7].A receiving TCP implements delayed ACK by gen-erating fewer than one ACK per data segment received.A TCP should implement delayed ACK [3],but should not excessively delay an ACK be-cause TCP uses ACKs to estimate packet round-trip time and determine how much more data to transmit[3,11].RFC-1122recommends that in a stream of full-sized segments there should be an ACK for at least every second segment.SunOS 4.1.1TCP implements delayed ACK and uses it by default.The following two conditions deter-mine when a SunOS4.1.1TCP should transmit an ACK if delayed ACK is used:R1:If the receive buffer is empty and the receive sequence space has advanced at least2*MSSoctets,then transmit an ACK.R2:If the receive sequence space has advanced at least35%of the total receive buffer space,then transmit an ACK.Note that the receive sequence space advances as an application extracts data from the receive buffer,and TCP checks condition R1before condi-tion R2.Condition R1guarantees TCP acknowl-edges the peer after every two MSS segments re-ceived by an application.For a receiving TCP with a small receive buffer,as compared to TCP MSS,condition R2generates ACKs to allow the sending TCP to transmit more data.RFC-1122mandates that a TCP must ACK the peer within500milliseconds(ms)after receiv-ing data.Observe that the above two conditions do not guarantee that a receiving TCP will meet the requirement.For example,if an application extracts data from the receive buffer too slowly, TCP can delay sending an ACK for more than500 ms.Therefore,SunOS4.1.1TCP schedules a de-layed ACK timer event every200ms to check for possible delayed ACKs[12];an ACK is transmit-ted when the timer expires and ACKs have been delayed.4.1.4Anomalous BehaviorTCP experiences low throughput while a sending TCP(A)has a buffer of16K octets and a receiving TCP(B)has a buffer of24K octets.As Figure6 illustrates,after A sends segments1and2to B, it reaches a steady state consists of two data seg-ments from A and an immediate ACK from B. In segment A1,the delayed ACK timer generates an ACK for segment21;the ACK allows SunOS to add4K octets of data to the send buffer.Af-232422432121N o r m a l S t a t eA b n o r m a l S t a t eLabel Seg.....data: 4096 data: 9148data: 3140data: 9148(repeat 8 times)data: 8192ACK 23(200 ms later)ACK 24ACK 21ACK 3, 4ACK 1, 2data: 3140data: 9148ACK 22data: 7236(200 ms later)(repeat till the end of trabsmission)(Sender)TCP A (Receiver)TCP B ACK generated by delayed ACK timerNote: 1. The send and receive buffer sizes are 16K and 24K, respectively.2. S1, S2, R1, and R2 are condition labels.S1 (3140 octets waiting in buffer)R2 (13244 > 0.35 * 24K)S2 (3140 = 12K - 9148)S1 (3140 octets left in buffer)R2 (12288 > 0.35 * 24K)CommentS2 (3140 = 12K - 9148)S1 (3140 octets waiting in buffer)ACK generated by delayed ACK timer R2 (7236 octets waiting in buffer)S2 (7236 = 3140 + 4096)R1, R2 (7236 < 0.35 * 24K)ACK generated by delayed ACK timer R1, R2 (8192 < 0.35 * 24K)A1A2Figure 6:Illustration of the segment exchange between two TCP that leads to a deadlock state ter A receives segment A2that acknowledges all the outstanding data,by condition S2,it immedi-ately sends a data segment with 7236(3140+4096)octets to B .Because 7236is less than 2*MSS and also less than 35%of 24K,B delays acknowledg-ing A and expects A to send more data.In the mean time,A adds only 8K octets of data to the send buffer even although the send buffer has 9148octet space available.Because 8K is less than 1*MSS and does not satisfy condition S2,A waits for an ACK from B before transmitting the 8K data.A circular-wait situation has occurred:A is waiting for an ACK fromB before it sends more data,and B is waiting for more data from A be-fore it sends an ACK.Finally,B ’s delayed ACK timer expires and causes B to send an ACK that breaks the circular-wait.After A responds to the ACK from B by sending an 8K data segment,the circular-wait situation occurs again.Thus,a lockstep interaction in which a circular-wait fol-lowed by an ACK that breaks the circular-wait has established.Because the delayed ACK timer generates one ACK per 200ms,the sender expe-riences unnecessarily long delay before sending additional data.Therefore,two hosts connected via a high-speed network waits to send data while the ATM network remains idle.As a result,TCP throughput decreases dramatically.ACK(200 ms later)data: 3140data: 9148(repeat till the end of transmission)Receive buffer size:40K, 44K, 48K, 51K24K, 28K, 32K, 36K Receive buffer size: 40K, 44K ACKdata: 3140(200 ms later)ACKdata: 9148data: 9148data: 7236Send buffer size: 16KSend buffer size: 20KACKdata: 7236(200 ms later)Receive buffer size: 48K, 51K Send buffer size: 24Kdata: 9148Receive buffer size: 36K, 40K, 44K, 48K, 51K Send buffer size: 16KACKReceive buffer size: 48K, 51K data: 9148data: 7236data: 9148(200 ms later)data: 3140ACK(200 ms later)Send buffer size: 20KLockstep pattern of transmissions:data: 8192(200 ms later)ACK(repeat till the end of transmission)Lockstep pattern of transmissions:(repeat till the end of transmission)Lockstep pattern of transmissions:(repeat till the end of transmission)(repeat till the end of transmission)Lockstep pattern of transmissions:Lockstep pattern of transmissions:Triggered by delayed ACK timer? Yes Triggered by delayed ACK timer? No Triggered by delayed ACK timer? No Triggered by delayed ACK timer? No Triggered by delayed ACK timer? Yes Throughput lower bound: 0.315 Mb/sThroughput lower bound: 0.469 Mb/s.Throughput lower bound: 1.094 Mb/sThroughput lower bound: 0.547 Mb/sThroughput lower bound: 0.625 Mb/sFigure 7:Summary of the lockstep transmission behavior that decreases TCP throughput dramatically4.1.5DiscussionWhen a sender with16K octet buffer communi-cates with a receiver with24K buffer,as Figure6 illustrates,an ACK segment(segment A1)from B that acknowledges a data segment with3140 octets causes A to enter a circular-wait and then a steady state of lockstep transmissions that low-ers TCP throughput significantly.Because the delayed ACK timer generates the ACK,the time at which thefirst circular-wait occurs depends on when the delayed ACK timer will generate such an ACK.The longer the data transfer takes,the more likely a transition to the steady state of lock-step transmissions becomes.Once in the steady state of lockstep transmissions,TCP throughput is approximately8K octets per200ms(or0.3125 Mb/s).Certain combinations of send and receive buffer sizes cause the circular-wait situation to occur shortly after connection establishment.For ex-ample,when a sender with16K buffer commu-nicates with a receiver with a36K buffer,the first circular-wait occurs after the sender transmits the fourth data segment(see Figure6)because 12288(3140+9148)is less than18296(2*MSS) and12903(35%of36K).Then,a steady state of lockstep transmissions that yields a throughput of 12288octets per200ms(or0.46875Mb/s)occurs. However,if the delayed ACK timer generates an ACK for segment3,a different lockstep pattern of transmissions that yields a throughput of8K octets per200ms(or0.3125Mb/s)occurs. Figure7summarizes the behavior of lockstep transmissions that reduces TCP throughput sub-stantially for various combinations of send and receive buffer sizes;it shows the segment ex-change pattern to be repeated till the end of the transmission,the throughput achieved when the steady state of lockstep transmissions occurs.It also indicates whether the delayed timer triggers the transition to the steady state of lockstep trans-missions or not.4.1.6Preventing Anomalous BehaviorIf an implementation of TCP that uses delayed ACKs follows the recommendation of RFC-1122 to generate an ACK each time it receives at least two MSS segments,then a sending TCP can pre-vent the anomalous behavior from occurring,re-gardless of the receiver buffer size,by using a send buffer no smaller than3*MSS octets.When the send buffer contains3*MSS octets,TCP ei-ther allows at least2*MSS octets of data to be outstanding(in case the receive buffer can hold 3*MSS octets or more),or allows at least35%of the receive buffer size to be outstanding(in case the receive buffer is smaller than3*MSS octets). Thus,the receiver will always acknowledge the sender promptly(conditions R1and R2).Send buffer: 16K, receive buffer: 16Kdata: 8192data: 8192(wait for ACK)data: 8192(repeat for many times)State (a)data: 4096n)n+1)ACK n, n+1n)n+1)ACK n-1ACK n (repeat for many times)State (b)Send buffer: 16K, receive buffer: 20K(wait for ACK)(wait for ACK)(wait for ACK)data: 8192data: 8192data: 3140data: 9148(repeat for many times)n)n+1)ACK n, n+1(repeat for many times)State (d)State (c)n)n+1)ACK n ACK n+1(i)(ii)Figure 8:Illustration of main states observed during a data transfer4.2Other ObservationTable 1and Figure 9(a)in Appendix show thata sending TCP with 16K buffer achieves betterperformance if a receiver reduces its buffer sizefrom 20K to 16K.The main reason for the ob-served unintuitive result is because SunOS 4.1.1TCP uses the following condition,in addition toconditions S1and S2,to determine when to senda data segment:S3:If is the largest receive windowthe peer has offered.Because SunOS 4.1.1TCP checks conditions S1and S2before condition S3,is always less then 1*MSS.When a receiver’s buffer space is small (less than 2*MSS),condition S3allows TCP to send data segments that are smaller than 1*MSS.To see how SunOS 4.1.1TCP uses con-dition S3to send data,consider a sending TCP with 16K buffer communicates with a receiving TCP with 16K buffer over an ATM path.Because the receiving TCP uses a 16K buffer,the largestreceive window it offers to the peer is 16K (i.e.,sender’sticularly annoying and surprising to users of ahigh-speed ATM LAN when they discover thatthe same ftp program they use to transferfileson an10Mb/s Ethernet can perform much worseon an ATM connection with100Mb/s interfacehardware.Large MTU size,as compared with the TCPbuffer sizes,and mismatched TCP send and re-ceive buffer sizes are the main cause of the anoma-lous behavior.We conclude that a TCP can pre-vent such a behavior from occurring,regardlessof the receiver buffer size,by using a send buffersize no smaller than3*MSS.The solution is espe-cially effective in a heterogeneous distributed en-vironment because a sending TCP does not havecontrol over the receive buffer size chosen by thepeer.However,a sending TCP knows the TCPMSS and has control over its send buffer size.Finally,it is worth noting that the significantthroughput decrease observed in this paper is notrestricted to ATM.Any environment with anal-ogous combinations of TCP send and receivebuffer sizes can experience poor throughput.Forexample,two hosts attached to the same Ether-net running a standard4BSD-derived TCP(e.g.,SunOS4.1.1TCP)will experience poor perfor-mance when the sending TCP uses a2K octetsend buffer and the receiving TCP uses a6K octetreceive buffer.3TCP will perform poorly when a sending TCP uses1K octetsend buffer and a receiving TCP uses a3K octet receivebuffer.[9]Fore Systems,Incorporated.ForeRunnerSBA-100/-200ATM SBus Adapter User’smanual,1993.[10]The ATM Forum.ATM User-Network In-terface Specification Version3.0.Prentice-Hall,Englewood Cliffs,New Jersey,1993.[11]Van Jacobson.Congestion Avoidance andControl.In Proceedings of ACM SIGCOMM’88,pages314–328,Aug.1988.[12]S.J.Leffler,M.K.McKusick,M.J.Karels,and J.S.Quarterman.The Design and Im-plementation of the4.3BSD UNIX Operat-ing System.Addison-Wesley,Reading,Mas-sachusetts,1990.[13]John Nagle.RFC-896:Congestion Controlin IP/TCP Internetworks.Request For Com-ments,work InformationCenter.[14]C.Papadopoulos and G.M.Parulkar.Ex-perimental Evaluation of SunOS IPCand TCP/IP Protocol Implementation.IEEE/ACM Transactions on Networking,1(2),April1993.[15]J.Postel.RFC-768:User Datagram Pro-tocol.Request For Comments,Aug.1980.Network Information Center.[16]J.Postel.RFC-791:Internet Protocol.Re-quest For Comments,workInformation Center.[17]J.Postel.RFC-793:Transmission ControlProtocol.Request For Comments,Septem-work Information Center. [18]J.Postel and J.Reynolds.RFC-854:TelnetProtocol specification.Request For Com-ments,work InformationCenter.[19]J.Postel and J.Reynolds.RFC-959:FileTransfer Protocol.Request For Comments,work Information Center. [20]Martin De Prycker.Asynchronous TransferMode:Solution for Broadband ISDN(2ndedition).Ellis Horwood,Chichester,Eng-land,1993.02468101214161820225101520253035404550T h r o u g h p u t (M b /s )Experiment Send Buffer: 16K 51K 48K 44K 40K 36K 32K 28K 24K 20K 16K02468101214161820225101520253035404550T h r o u g h p u t (M b /s )Experiment Send Buffer: 20K 51K 48K 44K 40K 36K 32K 28K 24K 20K 16K AppendixReceive Buffer (a)Receive Buffer (b)Figure 9:TCP throughput vs.buffer sizes02468101214161820225101520253035404550T h r o u g h p u t (M b /s )Experiment Send Buffer: 24K 51K 48K 44K 40K 36K 32K 28K 24K 20K 16K 02468101214161820225101520253035404550T h r o u g h p u t (M b /s )Experiment Send Buffer: 28K 51K 48K 44K 40K 36K 32K 28K 24K 20K 16K Receive Buffer Receive Buffer (a)(b)Figure 10:TCP throughput vs.buffer sizes (con’t)02468101214161820225101520253035404550T h r o u g h p u t (M b /s )Experiment Send Buffer: 32K 51K 48K 44K 40K 36K 32K 28K 24K 20K 16K 02468101214161820225101520253035404550T h r o u g h p u t (M b /s )Experiment Send Buffer: 36K 51K 48K 44K 40K 36K 32K 28K 24K 20K 16K Receive Buffer Receive Buffer (a)(b)Figure 11:TCP throughput vs.buffer sizes (con’t)02468101214161820225101520253035404550T h r o u g h p u t (M b /s )Experiment Send Buffer: 40K 51K 48K 44K 40K 36K 32K 28K 24K 20K 16K 02468101214161820225101520253035404550T h r o u g h p u t (M b /s )Experiment Send Buffer: 44K 51K 48K 44K 40K 36K 32K 28K 24K 20K 16K Receive Buffer Receive Buffer(a)(b)Figure 12:TCP throughput vs.buffer sizes (con’t)02468101214161820225101520253035404550T h r o u g h p u t (M b /s )Experiment Send Buffer: 48K 51K 48K 44K 40K 36K 32K 28K 24K 20K 16K 02468101214161820225101520253035404550T h r o u g h p u t (M b /s )Experiment Send Buffer: 51K 51K 48K 44K 40K 36K 32K 28K 24K 20K 16K Receive Buffer Receive Buffer (a)(b)Figure 13:TCP throughput vs.buffer sizes (con’t)。
使用LabVIEW中的TCP/IP和UDP协议前言互联网络协议(IP)，用户数据报协议(UDP)和传输控制协议(TCP)是网络通信的基本的工具。
TCP与IP的名称来自于一组最著名的因特网协议中的两个--传输控制协议和互联网络协议。
你能使用TCP/IP来进行单一网络或者互连网络间的通信。
单独的网络会被大的地理距离分隔。
TCP/IP把数据从一个子网网络或者因特网连接的计算机发送到另一个上。
因为TCP/IP 在大多数计算机上是可用的，它能在多样化的系统中间传送信息。
LabVIEW和TCP/IP你能在所有平台上的LabVIEW中使用TCP/IP。
LabVIEW包含了TCP和UDP程序还有能让你建立客户端或者服务器程序的功能。
IPIP执行低层次的计算机间的数据传送。
在组成部分里的IP数据包称为数据报。
一个数据报包含表明来源和目的地地址的数据和报头字。
IP为通过网络或者因特网把数据发送到指定的目的地的数据报确定正确的路径。
IP协议并不能保证发送。
事实上，如果数据报在传输中被复制，IP可能多次传送一个单独的数据报。
所以，程序很少用IP而是用TCP或者UDP代替。
UDPUDP在计算机进程中提供简单而低层次的通信。
进程通过把数据报发送到一个目的地计算机或者端口进行通信。
一个端口是你发送数据的位置。
IP处理计算机对计算机的发送。
在数据报到达目的地计算机后，UDP把数据报移动到其目的端口。
如果目的端口不是开放的，UDP 将删除数据报。
UDP将发生IP的同样的发送问题。
应用程序的UDP的可靠性不强。
例如，一项应用程序能经常把大量信息的数据传送到目的地而丢失少量的数据是肯定的。
在LabVIEW中使用UDP协议因为UDP不是一个TCP似的一个以连接为基础的协议，在你发送或者收到数据之前，你不需要和目的地建立一种连接。
相反，当你每发送一个数据报时，由你指定数据目的地。
操作系统不会报告传输差错使用UDP打开功能在一个端口上打开一个UDP插口。
计算机络英文复习题-CAL-FENGHAL-(YICAI)-Company One 1.英译汉(10分)TCP(Transmission Control Protocol) IP (Internet Protocol)RFC(Requests for comments)SMTP(Simple Mail Transfer Protocol) Congestion-control Flow controlUDP (User Datagram Protocol)川八数据报协议FTP(File Transfer Protocol) 文件传输协议HTTP( Hyper-Text Transfer Protocol) 超文本传输协议TDM 时分复用 FDM频分复用ISP(Internet Service Provider) DSL(Digital Subscriber Line) DNS(Domain Name System) ARQ(Automatic Repeat Request)ICMP(Internet Control MessageProtocol) 网间控制报文协议AS(Autonomous Systems) RIP(Routing Information Protocol)\ OSPF(Open Shortest Path First) BGP (Border Gateway Protocol) 边界网关协议HFC光纤同轴电缆混合网22. CRC(Cyclic Redundancy Check) 循环冗余检验23. CSMA/CD 带冲突检测的载波侦听多路存取24. ARP 地址解析协议 25. RARP 反向地址解析协议 26.DHCP动态主机配置协议循环时间互联网工程任务组 统一资源定位 应用程序编程接口 多用途互联网邮件扩展32. MTU(P328)最大传输单元二、单项选择题(每小题1分，共30分)1. DSL divides the communication link between the home and the ISP into three nonoverlapping frequency bands, a upstream channel is in _A ______ ・A)50 kHz to 1 MHz band B) 1 MHz to 2MHz band C)4 kHz to 50kHz bandD)Oto 4kHz band2. As a data packet moves from the upper to the lower layers, headers are A .A) Added; B) subtracted; C) rearranged; D) modified1. 2. 3. 4. 5. 6.7. 8. 9. 10. 11. 12.13. 14. 15. 16. 17.18.传输控制协议 互联网协议 请求评议简单邮件传输协议 拥塞控制 流控制互联网服务提供商 数字用户线路 域名系统自动重发请求 自制系统路由信息协议 开放最短路径优先27. RTT 28. IETF(P5) 29. URL(P88) 30. API 31. MIME3.What is the main function of the network layer DA) node-to-node delivery; B) process-to-process message deliveryC) synchronization; D) updating and maintenance of routing tables4.Which of the following is the default mask for the address 168.0.46.201A) 255.0.0.0; B) 255.255.0.0; C) 255.255.255.0; D) 255.255.255.2555. A router reads the __ address on a packet to determine the next hop. AA) IP ; B) MAC; C) source; D)ARP6 . Which device can't isolates the departmental collision domains. AA) Hub; B) switch; C) router; D) A and B7.Input port of a router don't pei-fonn ______ D ______ f imctions.A) the physical layer functions B) the data link layer functionsC) lookup and forwarding function D) network management8.HTTP has a mechanism that allows a cache to verify that its objects are up to date・ The mechanism is DA) persistent connections B) cookies C) Web Caching D) conditional GET9. A protocol layer can be implemented in ______ D ______ ・A) software B) hardware C) a combination of the software and hardware D) All of the above10.A protocol has three important factors, they are_A ___________ ・A) syntax, semantics, order B) syntax, semantics, layerC) syntax, semantics, packet D) syntax , layer, packet11.There are two broad classes of packet-switched networks: datagram networks andvirtual-circuit networks. The virtual-circuit networks forward packets in their switches use D .A) MAC addresses B) IP addressesC) e-mail addresses D) virtual-circuit numbers12.TCP service model doesn't provide _______ D _________ service・A) reliable transport service B) flow control serviceC) congestion-control service D) guarantee a minimum transmission rate service ・ually elastic applications don^t include ________ B ________ ・A) Electronic mail B) Internet telephonyC) file transfer D) Web transfer14.A user who uses a user agent on his local PC receives his mail sited in a mailserver by using _B ______ protocol.A)SMTP B) POP3C)SNMP D) FTP15.Considering sliding-window protocol, if the size of the transmitted window is N and the size of the receiving window is l,the protocol isA) stop-and-wait protocol B) Go-Back-N protocolC) selective Repeat protocol D) alternating-bit protocol16.which IP address is effective ______ B ________ ・A) 202,131,45,61 B) 126.0.0.1C) 192.268.0.2 D) 290.25」35」217.if IP address is 202.130.191.33, subnet mask is 255.255.255.0,then subnet prefixis_D _______A) 202.130.0.0 B) 202.0.0.0C) 202.130.191.33 D)202.130.191.018.The command Ping s implemented with _B ___________ messagesA) DNS B) ICMPC) IGMP D) RIP19.Which layer-function is mostly implemented in an adapter _A __________________A) physical layer and link layer B) network layer and transport layerC)physical layer and network layer D) transport layer and application layer20.If a user brings his computer from Chengdu to Peking, and accesses Internet again.Now, _B __________ of his computer needs to be changed・A) MAC address B) IP addressC) e-mail address D) user address1._____________________________________ ・traceroute is implemented with _B_______________________________________________________ messages.A) DNS B) ICMPC) ARP D) RIP2. A router reads the A address on a packet to determine the next hop・A. IP ;B. MAC;C. source;D.ARP3.There are two broad classes of packet-switched networks: datagram networksand virtual-circuit networks. The virtual-circuit networks forward packets in their switchesuse _____________________ D ____ ・A) MAC addresses B) IP addressesC) e-mail addresses D) virtual-circuit numbers4.About subnet, which underlying description isif t right ____________A)device interfaces with same subnet part of IP addressB)can" physically reach each other without intervening a router.C)all of the devices on a given subnet having the same subnet address・D) A portion of an interface's IP addi'ess must be detennined by the subnet to which it is connected・5.if IP address is 102.100.100.32, subnet mask is 255.255.240.0,then subnetprefix is ___ A ____A) 102」00.96.0 B) 102.100.0.0C) 102.100.48.0 D) 102.100.112.06 If a user brings his computer from chengdu to beijing, and accesses Internetagain・ Now, ____ B_ of his computer needs to be changed・A) MAC address B) IP addressC) e-mail address D) user address7.Input port of a router don^ pei'fonii ________ D ____ functions.A) the physical layer functions B) the data link layer functionsC) lookup and forwarding function D) network management8.switching fabric is at the heart of a router, switching can be accomplished in a number of ways,donit include^ D _A)Switching via memory B)Switching via crossbarC)Switching via a bus D) Switching via buffer9.if a host wants to emit a datagram to all hosts on the same subnet, then the datagram'sdestination IP address is __________________ B_A)255.255.255.0 B) 255.255.255.255C)255.255.255.254 D) 127.0.0.1lO.The advantage of Circuit switching does not include ________________ ・A) small transmission delay B)small Processing costC) high link utilization D)no limited to format of message1 ・a n ARP query sent to _A_A) local network B) all over the Internet.2..packet-switching technologies that use virtual circuits include_B___________ :A) X.25, ATM, IP B) X.25, ATM, frame relay.C) IPX, IP, ATM D) IPX, IP, TCP3・ In Internet, _ D _ protocol is used to report error and provide the information for un-normal cases・A) IP B) TCP C)UDP D) ICMP1 ・A is a Circuit-switched network・A. TDMB. Datagram networkC. InternetD. virtualcircuit network2.The store-and-forward delay is __ D_______A. processing delayB. queuing delay C・ propagation delay D.transmission delay3・ Which is not the function of connection-oriented service DA. flow controlB. congestion control C・ error correction D.reliable data transfer4. The IP protocol lies in CA. application layerB. transport layerC. network layerD. link layer5. Which of the following is the PDU for application layer _B ________A. datagram;B. message; C・ frame; D・ segment6.bandwidth is described in _B_A) Bytes per second B) Bits per secondC) megabits per millisecond D) centimeters7. A user who uses a user agent on his local PC receives his mail sited in a mail server by using _A_ protocol.A)SMTP B) POP3C)SNMP D) FTP8・ As a data packet moves from the lower to the upper layers, headers arc B ,A)Addcd; B. subtracted: C. rearranged; D. modified三. 填空题(每空1分，共22分(注意：所有填空题不能写中文，否则中文答 案对的情况下扣0.5分)1 ・ link-layer address is variously called a LAN address, a MAC address, or a physical address,2 In 什le layered architecture of computer networking, n layer is the user of n-1 layer and the service provider of n+1 layer ・ A) n B) n+3 C) n+1 D) n-1MechanismUseChecksum Used to detect bit errors in a transmitted packet. Sequence number Used for sequential numbering of packets of dataflowing from sender to receiver ・ Acknowledgmen t (或 ACK) Used by the receiver to tell the sender that a packet orset of packets has been received correctly ・ Countdown timer Used to timeout/retransmit a packet, possibly because the packet (or its ACK)was lost ・ Window,pipelinl ing The sender may be restricted to sending only packetswith sequence numbers that fall within a given range ・1. please fill in the types of delay in a router.四. 判断题(每小题1分，共10分)1 ・ J The sernces of TCP's reliable data transfer founded on the seivices of the unreliable datatransfer.2. J Any protocol that performs handshaking between the communication entitiesbefore transferring data is a conncction-oricntcd sen-ice ・3. x HOL blocking occur in output ports of router ・4. J Socket is globally unique ・5. J SMTP require multimedia data to be ASCII encoded before transfer.6. xThe transmission delay is a function of the distance between the two routers ・7. xIP address is associated with the host or router. SO one device only have one IPaddress ・8. J In packet-switched nehvorks, a session's messages use the resources on demand, and Internet makes its best effort to deliver packets in a timely manner.② nodelProcessionPropagation©queueingrouterTransmission9.x UDP is a kind of unreliable transmission layer protocol, so there is not any checksum field in UDP datagram header・10.J Forwarding tabic is configured by both Intra and Inter-AS routing algorithmIP is a kind of reliable transmission protocol. F8.Forwarding table is configured by both Intra and Inter-AS routing algorithm・T9.Distance vector routing protocol use Isa to advertise the network which router knows.F10.RIP and OS PF are Intra-AS routing protocols T11 ・ Packet switching is suitable for real-time services, and offers better sharing of bandwidththan circuit switching F五、计算题(28 poin⑸1 ・Consider the following net\vork. With the indicated link costs, use Dijkstnf s shortest-path algorithm to compute the shortest path from X to all network nodes・2 Given: an organization has been assigned the network number 198」.1.0/24 and it needs todefine six subnets・ The largest subnet is required to support 25 hosts. Please:• Defining the subnet mask; (2 分)27bits or 255.255.255.224•Defining each of the subnet numbers; which are starting from 0# （4 分）198.1.1.0/27198.1.1.32/27 198.1.1.64/27 198.1.1.96/27 198.1.1.128/27 198.1.1.160/27 198.1.1.192/27 19& 1.1.224/27•Defining the subnet 2¥s broadcast address.（2 分）198.1 ・ 1.95/27•Defining host addresses scope for subnet 2#. （2 分）198.1 ・ 1.65/27— 198.1.1.94/273.Consider sending a 3,000-byte datagram into a link that has an MTU of1500bytes.Suppose the original datagram is stamped with the identification number 422 ・Assuming a 20-byte IP header^How many fragments are generated What are their characteristics （10 分）。
XXX年固网产品技术大比武试题第一部分固网基础〔150分〕（一）、可服务性研究部分〔15分〕一、单项选择题〔每题2分，共2小题，合计4分〕1、在产品可爱护特性中，开发远程爱护功能、提供错误的自动复原机制以减少停机时刻、减少现场的软件支持版本数目等特性属于以下哪一个可爱护特性的内容〔〕A、便于软件安装B、便于日常爱护C、故障诊断处理D、减少爱护成本答案：D知识点：可服务性设计提升试题出处：产品可服务性能力提升宣传20050817.ppt难度：中2、安全性指的是在安装和爱护过程中操作人员和设备〔设备包括机器本身和承载内容等方面〕双方的安全，同时产品的安全性也极大的阻碍了安装爱护过程的效率。
以下哪项属于设备安全性设计有缺陷：〔〕A、设备边角有毛刺，搬运和安装过程中容易造成人员受伤。
B、电源触点外露，容易使人触电。
C、设备的电源开关设计在容易触碰的位置，爱护时容易触碰阻碍设备的正常运行。
D、电源指示灯不符合规范。
答案：C知识点：可服务性设计提升试题出处：产品可服务性能力提升宣传20050817.ppt难度：高二、多项选择题〔每题3分，必须全部选正确，共2小题，合计6分〕1、产品可服务性通常指的是：〔〕A、产品的可安装性B、产品的可爱护性C、产品的可靠性D、产品的可支持性答案：A、B、D知识点：可服务性设计提升试题出处：产品可服务性能力提升宣传20050817.ppt难度：中2、全球技术服务部收集可服务性建议的渠道有：〔〕A、SUPPORT网站B、需求承诺电子流C、产品技术&服务产品建议库D、客户需求电子流答案：A、C知识点：可服务性建议收集标准方法试题出处：«关于加强国内、海外固网产品可服务性建议收集的通知»以及相关附件〔公布位置：全球技术服务部文件夹库〕难度：低三、简答题〔每题5分，共1小题，合计5分〕1、请简述产品可服务性设计的目的：答案：〔答对5条以上得总分值，答对1条得1分〕1)满足最终用户对设备安装、爱护的各方面需求2)满足企业内部用户对设备安装、爱护等方面的需求3)提高产品的竞争力4)降低爱护成本5)提高客户中意度知识点：可服务性设计提升试题出处：产品可服务性能力提升宣传20050817.ppt难度：中（二）、交换机基础〔25分〕一、单项选择题〔每题2分，共4小题，合计8分〕1、ISUP信令IAM消息前向呼叫表示语参数字段比特D〔互通表示语〕，当D＝1时，表示：〔〕A、前面通过的路由差不多上NO7信号。
No1(2分)单选题使用超级终端对2609进行配置，应使用COM口的每秒位数（波特率）是When using HyperTerminal to configure the 2609, each second bits (baud rate) of the COM interface should be ( )2400960019200115200No2(2分)单选题IP地址与它的掩码取反相与，所得的非零点分十进制数，是此IP地址的（） The IP address is in inverse phase with its mask and the acquired non-zero decimal numeral is the ( ) of this IP addressA类地址Class A address主机地址Host address网络地址Network address解析地址Resolution addressNo3(2分)单选题在以太网中，工作站在发送数据之前，要检查网络是否空闲，只有在网络不阻塞时，工作站才能发送数据，是采用了（）机制 In the Ethernet, before transmitting data, a workstation needs to check if the networkisavailable. Only when the network is not blocked, the workstation can transmit data,the mechanism adopted here is ( ) .IPTCPICMP载波侦听与冲突检测 CSMA/CD Carrier sense multiple access(CSMA) with collision detectionNo4(2分)单选题给您分配一个B类IP网络172.16.0.0，子网掩码255.255.255.192，则您可以利用的网络数为_______，每个网段最大主机数为______ You are allocated a Class-B IP network 172.16.0.0 with the subnet mask of 255.255.255.192. Then you can make use of ( ) networks and maximum ( ) hosts in each network segment512， 1261022， 621024， 62256， 254192， 254No5(2分)单选题一个包含有等多厂商设备的交换网络，其VLAN中Trunk的标记一般应选For a switching network composed of many vendors’ devices, the Trunk tag in its VLAN usually adopts ( )IEEE 802.1qISLVLT以上都可以All the aboveNo6(2分)单选题一个B类网络,有5位掩码加入缺省掩码用来划分子网,每个子网最多（）台主机 In the Class-B network, a 5Bit mask is added to the default mask for subnet allocation. Then each subnet has maximum ( ) hosts51051210222046No7(2分)单选题能正确描述了数据封装的过程的是 The correct data encapsulation process is ( )数据段->数据包->数据帧->数据流->数据Data segment > datapacket > data frame > data stream > data数据流->数据段->数据包->数据帧->数据Data stream > datasegment > data packet > data frame > data数据->数据包->数据段->数据帧->数据流Data > data packet > datasegment > data frame > data stream数据->数据段->数据包->数据帧->数据流Data > data segment >data packet > data frame > data streamNo8(2分)单选题（）是一个网络层的协议，它提供了错误报告和其它回送给源点的关于 IP 数据包处理情况的消息 ( ) is a network-layer protocol. It provides the error report and other IP data packet processing messages returned to the source point.TCPUDPICMPIGMPNo9(2分)单选题对于这样几个网段：172.128.12.0，172.128.17.0，172.128.18.0，172.128.19.0，最好用下列哪些网段实现路由汇总 For the followingnetwork segments: 172.128.12.0, 172.128.17.0, 172.128.18.0, and 172.128.19.0. Which of the following network segments should be used to implement route aggregation?172.128.0.0/21172.128.17.0/21172.128.12.0/22172.128.20.0/20No10(2分)单选题目前，我国应用最为广泛的LAN标准是基于的以太网标准 At present, the most widely-used LAN standard in China is the Ethernet standard based on ( )IEEE 802.1IEEE 802.2IEEE 802.3IEEE 802.5No11(2分)单选题T64E查看CPU利用率的命令是 The command used by the T64E to view CPU utilization rate is ( )show processorshow upcshow rpushow mpuNo12(2分)单选题术语ARP代表什么？ What does term ARP indicate?Address Resolution PhaseARP Resolution ProtocolAddress Resolution ProtocolAddress Recall ProtocolNo13(2分)单选题选出下列关于802.1Q标签头的错误叙述 Which of the following statements about the 802.1Q label header is wrong?802.1Q标签头长度为4字节The length of the 802.1Q label headeris 4 bytes802.1Q标签头包含了标签协议标识和标签控制信息The 802.1Q labelheader contains the label protocol ID and label control information802.1Q标签头的标签协议标识部分包含了一个固定的值0x8200Thelabel protocol ID part of the 802.1Q label header contains a fixed value: 0x8200802.1Q标签头的标签控制信息部分包含的VLAN Identifier( VLANID )是一个12Bit的域The label control information part of the802.1Q label header contains VLAN Identifier (VLAN ID), which isa 12Bit domainNo14(2分)单选题规划一个C类网,需要将网络分为9个子网,每个子网最多15台主机,下列哪个是合适的子网掩码? For the planning of a Class-C network, it is necessary to divide it into 9 subnets; each of them has maximum 15 hosts.Which of the following subnet masks is suitable?255.255.224.0255.255.255.224255.255.255.240没有合适的子网掩码There is no suitable subnet maskNo15(2分)单选题哪些是会话层的功能（） Which is the function of the session layer?提供加密/解密Provide encryption/decryption提供数据转换及格式Provide data conversion and format在不同主机间建立连接Set up connection among different hosts建立、维护和终止会话Set up, maintain and terminate sessionsNo16(2分)单选题有一台正在运行的2626，它的9－16口没有接网线，但以太口上面的指示灯左面是亮的，为什么？ Ports 9-16 of a running 2626 are not connected to network cables but the left side of the indicator on the Ethernet interface is on. What is the cause?这些端口坏掉了These ports break down是端口设置为强制半双工 The ports are set as forced half duplex是端口设置成强制全双工 The ports are set as forced full duplex是端口设置为自适应端口The ports are set as adaptive portsNo17(2分)单选题PING命令使用ICMP的哪一种code类型： Which code type of ICMP is used by command PING ? ( )RedirectEcho replySource quenchDestination Unreachable在以太网中，是根据（）地址来区分不同的设备的 In the Ethernet, different devices are differentiated according to the ( ) addressIP地址IP addressMAC地址MAC addressIPX地址IPX addressLLC地址LLC addressNo19(2分)多选题对缺省路由描述正确的是( )Which of the following statements correctly describe the default route? ( )缺省路由是一条特殊的静态路由The default route is a specialstatic route缺省用来转发那些路由表中没有明确表示该如何转发的数据包的路由The default route is used to forward the data packets whoseforwarding mode is not specified in the route table缺省路由是一条最佳路由The default route is an optimal route缺省路由最后执行The default route is the last to be implementedNo20(2分)多选题广域网协议HDLC的点到点通讯方式有以下特点( )The point-to-point communication mode of WAN protocol HDLC has the following features ( )HDLC是运行在同步接口上的HDLC runs on the synchronousinterface对于任意的比特流，能够实现透明传输It is able to realize transparent transmission for any bitstream面向比特的协议Bit-oriented protocol可以运行在异步接口上It can run on the asynchronous interface以下关于端口汇聚的说法正确的是（）Which of the following statements about port convergency are correct? ( )能够增加网络带宽Increase network bandwidth降低了网络连接的可靠性Reduce reliability of networkconnection可以支持两个以上设备的应用Support applications of more than 2devices不能工作在半双工下Unable to work in the half-duplex mode捆绑端口的速率可以不同The port binding rate can be differentNo22(2分)多选题2826E支持哪些接口（）Which of the following interfaces are supported by the 2826E? ( )1000Base-X100Base-X1000Base-T10/100 Base-T以太网口No23(2分)多选题与动态路由协议相比，静态路由有哪些优点（）Compared to the dynamic routing protocol, the static route has the following merits ( )带宽占用少Less bandwidth occupation简单Simple路由器能自动发现网络拓扑变化The router can automaticallydiscover network topology changes路由器能自动计算新的路由The router can automatically calculate new routesNo24(2分)多选题以下关于帧中继说法正确的是（）Which of the following statements about the frame relay are correct? ( )帧中继的帧长是确定的The frame length of the frame relay iscertain帧中继是一个面向连接的第二层传输协议The frame relay is a connection-oriented L2 transmission protocol信息传输的效率高Information transmission efficiency is high适合有突发流量的场合It is applicable to burst flowNo25(2分)多选题用VLAN对网络进行分段，有以下哪些好处（）What are the advantages of using VLAN to segment the network? ( )减少了广播的数据量Reduce the broadcast data volume增加了网络带宽Broaden the network bandwidth增加了安全性Enhance security增加网络的可靠性Enhance network reliabilityNo26(2分)多选题下面有关传输层协议的描述，正确的是（）Which of the following describe the transport-layer protocol correctly? ( )传输层位于应用层和网络层之间，为终端主机提供端到端连接Thetransport layer is located between the application layer andnetwork layer, providing end-to-end connection to terminal hosts传输层协议包括TCP协议和UDP协议Transport-layer protocols include TCP and UDPTCP协议提供可靠的面向连接的通信服务，UDP协议提供不可靠的无连接的通信服务The TCP protocol provides reliableconnection-oriented communication service; the UDP protocolprovides unreliable connectionless communication serviceTCP协议提供不可靠的面向连接的通信服务，UDP协议提供可靠的面向连接的通信服务The TCP protocol provides unreliableconnection-oriented communication service while the UDP protocol provides reliable connection-oriented communication serviceNo27(2分)多选题低端以太网交换机可以在一起做集群管理的设备有（）What are the devices that can work with low-end Ethernet switch for clustermanagement? ( )181626182826E2818SNo28(2分)多选题对于这样一个地址,192.168.19.255/20,下列说法正确的是（）For address 192.168.19.255/20, which of the following statements are correct? ( )这是一个广播地址It is broadcast address这是一个私有地址It is a private address这是一个主机地址it is a host address地址在192.168.16.0网段上It is in the network section of192.168.16.0No29(2分)多选题TCP/IP协议栈中网络层的主要功能为（）The main functions in network layer of TCP/IP protocol stack are ( )提供对物理层的控制，完成对数据流检错、纠错、同步等措施Providecontrol over the physical layer and complete data stream errordetection/correction and synchronization etc.检查网络拓扑，确定报文传输的最佳路由，实现数据转发Checknetwork topology, determine the optimal route for messagetransmission, and implement data forwarding为两台主机间的应用程序提供端到端通讯Provide end-to-end communication to the application programs between the two hosts提供分组路由、避免拥塞等措施Provide packet routing and congestion prevention etc.No30(2分)多选题某ZXR10路由器的两个以太网口E0和E1分别连接了两个网段：E0（IP 地址10.1.1.1，子网掩码255.255.255.0）连接的是网络A，E1（IP地址10.1.2.1，子网掩码255.255.255.0）连接的是网络B。
1、快速检测技术可以尽早地检测到与相邻设备间的通信故障,以便系统能够及时采取措施,保证业务不中断（正确）2 复杂流分类是指根据五元组〔源/目的地址、源眉的端口号、协议类型〕等信息对报文进行分类,通常应用在网络的核心位置（错误）3 使用nat技术,只可以对数据报文中的网络层信息(ip地址)进行交换。
（错误）4 BFD（双向转发检测)技术属于快速检测技术,但它较为复杂,需要特殊厂商设备支持（错误）5 镜像端口能够实现将镜像端口上特定业务流的报文,传送到监控设备进行分析和监控的功能。
（正确）6 SDH传送网中的硬件检测机制,可以很快发现故障,且适用于所有介质(错误)7 在不使用BFD检测机制的情况下,通过以太网链路建立邻居关系的OSPF路由器,在链路故障最后最长需要40秒才会中断邻居关系(正确)8 都得是由于属于同一个流的数据包的端到端时延不相等造成的。
(正确)9 BFD单臂回声功能可以用于非直连的2台设备（错误）10 一台VRRP虚拟路由器只能拥有虚拟IP地址（错误）11 传统的丢包策略采用尾部丢弃〔Tail-Drop)的方法,这种丢弃方法会导致TCP全局同步现象（正确）12 SDN控制器可以根据网络状态智能调整流量路径,以达到提升整网香吐的目的(正确)13 BFD机制使用TCP建立连接,其目的端口号为3784。
（错误）14 拥塞管理的中心内容是通过制定调度策略,来决定数据包处理的先后顺序（对）15 在网络中采用Qos,提高某类业务的服务质量的同时,肯定损害其他业务的服务质量（对）16 BFD只能与网络层和数据链路层的协议模块结合使用（错误）17 VXLAN用户可通过VXLAN接口访问Internet (正确)18 Agile Controller作为—个网络资源自动化控制系统,可以提供统一的策略引擎,在整个组织内实施统一的访冋策略,实现基于用户身份,接入时间、接入地址、接入类型、接入方式（5H1W)的认证与授权（对）19 VRRP的接口IP地址和虚拟|P地址可以相同。
