1、网络管理员-计算机专业英语(一)及答案解析(总分：30.00，做题时间：90 分钟)一、单项选择题(总题数：6，分数：30.00)Each machine supporting TCP has a TCP transport entity, either a library procedure, a user process, or part of the kernel. In all case, it manages TCP streams and (1) to the IP layer. A TCP (2) Accepts user date streams from local proc
2、ess, breaks them into pieces not exceeding 64KB, and sends each piece as a separate IP (3) .When datagrams containing TCP date arrive at a machine, they are given to the TCP entity, which reconstructs the original byte streams.The IP layer gives no guarantee that datagrams will be delivered properly
3、, so it is up to TCP to time out and (4) them as need be. Datagrams do arrive may well do so in the wrong order, it is also up to TCP to (5) them into messages in the proper sequence.（分数：5.00）(1).A. calls B. interfaces C. links D. produces（分数：1.00）A.B.C.D.(2).A. connection B. file C. entity D. desti
4、nation（分数：1.00）A.B.C.D.(3).A. dategram B. stream C. connection D. transaction（分数：1.00）A.B.C.D.(4).A. reassemble B. reduce C. re-create D. retransmit（分数：1.00）A.B.C.D.(5).A. reassemble B. reduce C. re-create D. retransmit（分数：1.00）A.B.C.D.A multicast router may receive thousands of multicast (6) every
5、day for different groups. If a router has no knowledge about the membership status of the (7) , it must broadcast all of these packets. This creates a lot of traffic and consumes (8) . A better solution is to keep a list of groups in the network for which there is at least one loyal member. (9) help
6、s the multicast router create and update this list. For each group, there is one router that has the duty of distributing the (10) packets destined for that group. This means that if there are three multicast routers connected to a network, their lists of groupids are mutually exclusive. A host or m
7、ulticast router can have membership in a group.（分数：5.00）(1).A. packets B. errors C. reports D. alarms（分数：1.00）A.B.C.D.(2).A. routers B. network C. packets D. hosts（分数：1.00）A.B.C.D.(3).A. capability B. power C. bandwidth D. address（分数：1.00）A.B.C.D.(4).A. ICMP B. IGMP C. OSPF D. RID（分数：1.00）A.B.C.D.(5
8、).A. anycast B. multicast C. unicast D. broadcast（分数：1.00）A.B.C.D.A transport layer protocol has several responsibilities, One is to create a process-to-process (program-to-program) communication; TCP uses port (11) to accomplish this. Another responsibility of a transport layer protocol is to creat
9、e a (12) and error-control mechanism at the transport levelTCP uses a sliding (13) protocol to achieve flow controlIt uses the acknowledgment packet，time-out，and retransmission to achieve (14) controlThe transport layer is also responsible for providing a connection mechanism for the application pro
10、gramThe application program sends (15) of data to the transport layerIt is the responsibility of the transport layer at the sending station to make a connection with the receiver（分数：5.00）(1).Anumbers Bconnection Cdiagrams Dresources（分数：1.00）A.B.C.D.(2).Aprocedure Bfunction Croute Dflow（分数：1.00）A.B.C
11、.D.(3).Apath Bwindow Cframe Ddiagram（分数：1.00）A.B.C.D.(4).Apacket Btime Cerror Dphase（分数：1.00）A.B.C.D.(5).Aports Bstreams Cpackets Dcells（分数：1.00）A.B.C.D.Both bus and tree topologies are characterized by the use of multipoint (16) For the bus，all stations attach，through appropriate hardware (17) know
12、n as a tap，directly to a linear transmission medium，or busFull-duplex operation between the station and the tap allows data to be transmitted onto the bus and received from the (18) A transmission from any station propagates the length of the medium in both directions and can be received by all othe
13、r (19) At each end of the bus is a (20) ，which absorbs any signal，removing it from the bus（分数：5.00）(1).Amedium Bconnection Ctoken Dresource（分数：1.00）A.B.C.D.(2).Aprocessing Bswitching Crouting Dinterfacing（分数：1.00）A.B.C.D.(3).Atree Bbus Cstar Dring（分数：1.00）A.B.C.D.(4).Arouters Bstations Cservers Dswi
14、tches（分数：1.00）A.B.C.D.(5).Atap Brepeat Cterminator Dconcentrator（分数：1.00）A.B.C.D.For (21) service，we need a virtual-circuit subnetLet US see how that worksThe idea behind virtual circuits is to avoid having to choose a new (22) for every packet sentInstead， when a connection is established，a route f
15、rom the (23) machine to the destination machine is chosen as part of the connection setup and stored in tables inside the (24) That route is used for all traffic flowing over the connection，exactly the same way that the telephone system works When the connection is releasedthe virtual circuit is als
16、o tenninatedWith connection-oriented service，each packet carries an (25) telling which virtual circuit it belongs to（分数：5.00）(1).Aconnectionless Bconnectionoriented Cdatagram Dtelegram（分数：1.00）A.B.C.D.(2).Aprocessor Bdevice Croute Dterminal（分数：1.00）A.B.C.D.(3).Asource Broute Cdestination Dhost（分数：1.
17、00）A.B.C.D.(4).Aconnections Bresources Cbridges Drouters（分数：1.00）A.B.C.D.(5).Aaddress Bidentifier Cinterface Delement（分数：1.00）A.B.C.D.An internet is a combination of networks connected by (26) When a datagram goes from a source to a (27) ，it will probably pass many (28) until it reaches the router a
18、ttached to the destination networkA router receives a (29) from a network and passes it to another networkA router is usually attached to several networksWhen it receives a packet，to which network should it pass the packet? The decision is based on optimization：Which of the available (30) is the opt
19、imum pathway?（分数：5.00）(1).Amodems Brouters Cswitcher Dcomputers（分数：1.00）A.B.C.D.(2).Auser Bhost Ccity Ddestination（分数：1.00）A.B.C.D.(3).Adestinations Brouters Cpackets Dcomputers（分数：1.00）A.B.C.D.(4).Adestination Bresource Cpacket Dsource（分数：1.00）A.B.C.D.(5).Apathways Brouters Cdiagrams Dcalls（分数：1.00
20、）A.B.C.D.网络管理员-计算机专业英语(一)答案解析(总分：30.00，做题时间：90 分钟)一、单项选择题(总题数：6，分数：30.00)Each machine supporting TCP has a TCP transport entity, either a library procedure, a user process, or part of the kernel. In all case, it manages TCP streams and (1) to the IP layer. A TCP (2) Accepts user date streams from lo
21、cal process, breaks them into pieces not exceeding 64KB, and sends each piece as a separate IP (3) .When datagrams containing TCP date arrive at a machine, they are given to the TCP entity, which reconstructs the original byte streams.The IP layer gives no guarantee that datagrams will be delivered
22、properly, so it is up to TCP to time out and (4) them as need be. Datagrams do arrive may well do so in the wrong order, it is also up to TCP to (5) them into messages in the proper sequence.（分数：5.00）(1).A. calls B. interfaces C. links D. produces（分数：1.00）A.B. C.D.解析：(2).A. connection B. file C. ent
23、ity D. destination（分数：1.00）A.B.C. D.解析：(3).A. dategram B. stream C. connection D. transaction（分数：1.00）A. B.C.D.解析：(4).A. reassemble B. reduce C. re-create D. retransmit（分数：1.00）A.B.C.D. 解析：(5).A. reassemble B. reduce C. re-create D. retransmit（分数：1.00）A. B.C.D.解析：解析 每一种支持 TCP 的机器都有一个 TCP 传输实体，或一个库过程
24、，或一个用户进程，或者是一个内核的一部分。在每一种情况下，它都管理着 TCP 流和通向 IP 层的接口。TCP 实体从本地进程接收用户数据流，然后把它们划分成长度不超过 64KB 的段，并把每一段作为一个独立的 IP 数据报进行发送。当包含 TCP 数据的数据包到达机器时，它们会被交给 TCP 实体，重新构造成原来的数据流。IP 层不能保证数据报正确地传输，所以在必要时必须由上层的 TCP 采用超时重传机制。数据报可能会不按顺序到达，也需要由 TCP 把它们按正确的顺序进行重组。A multicast router may receive thousands of multicast (6)
25、every day for different groups. If a router has no knowledge about the membership status of the (7) , it must broadcast all of these packets. This creates a lot of traffic and consumes (8) . A better solution is to keep a list of groups in the network for which there is at least one loyal member. (9
26、) helps the multicast router create and update this list. For each group, there is one router that has the duty of distributing the (10) packets destined for that group. This means that if there are three multicast routers connected to a network, their lists of groupids are mutually exclusive. A hos
27、t or multicast router can have membership in a group.（分数：5.00）(1).A. packets B. errors C. reports D. alarms（分数：1.00）A. B.C.D.解析：(2).A. routers B. network C. packets D. hosts（分数：1.00）A.B.C.D. 解析：(3).A. capability B. power C. bandwidth D. address（分数：1.00）A.B.C. D.解析：(4).A. ICMP B. IGMP C. OSPF D. RID（
28、分数：1.00）A.B. C.D.解析：(5).A. anycast B. multicast C. unicast D. broadcast（分数：1.00）A.B. C.D.解析：解析 组播路由器每天可能接收到成千上万个组播分组。如果路由器不了解组播成员的主机状态，它就必须广播所有的分组。这样就产生了大量的通信，耗费了很多带宽。一个比较好的解决方案是在网络中维护一个组成员表，每个组中至少有一个忠实的成员。IGMP 协议帮助组播路由器生成和更新这个表。对于每一组，要有一个路由器负责发布该组的组播分组。这意味着如果有 3 个组播路由器连接在一个网络中，那么它们的组标识符列表就是相互不包含的。一
29、个主机或组播路由器可以成为多个组的成员。A transport layer protocol has several responsibilities, One is to create a process-to-process (program-to-program) communication; TCP uses port (11) to accomplish this. Another responsibility of a transport layer protocol is to create a (12) and error-control mechanism at the tr
30、ansport levelTCP uses a sliding (13) protocol to achieve flow controlIt uses the acknowledgment packet，time-out，and retransmission to achieve (14) controlThe transport layer is also responsible for providing a connection mechanism for the application programThe application program sends (15) of data
31、 to the transport layerIt is the responsibility of the transport layer at the sending station to make a connection with the receiver（分数：5.00）(1).Anumbers Bconnection Cdiagrams Dresources（分数：1.00）A. B.C.D.解析：(2).Aprocedure Bfunction Croute Dflow（分数：1.00）A.B.C.D. 解析：(3).Apath Bwindow Cframe Ddiagram（分
32、数：1.00）A.B. C.D.解析：(4).Apacket Btime Cerror Dphase（分数：1.00）A.B.C. D.解析：(5).Aports Bstreams Cpackets Dcells（分数：1.00）A.B. C.D.解析：解析 传输层协议有多个职责。一是创建进程到进程(程序到程序)的通信，由 TCP 使用端口号来完成；另一个职责是在传输层创建流和差错控制机制。TCP 利用滑动窗口协议来实现流控制。它使用确认分组、超时、重传机制来完成差错控制。传输层也负责为应用程序提供连接机制，应用程序将数据流发送到传输层，发送端的传输层负责建立于接收端之间的连接。Both bu
33、s and tree topologies are characterized by the use of multipoint (16) For the bus，all stations attach，through appropriate hardware (17) known as a tap，directly to a linear transmission medium，or busFull-duplex operation between the station and the tap allows data to be transmitted onto the bus and r
34、eceived from the (18) A transmission from any station propagates the length of the medium in both directions and can be received by all other (19) At each end of the bus is a (20) ，which absorbs any signal，removing it from the bus（分数：5.00）(1).Amedium Bconnection Ctoken Dresource（分数：1.00）A. B.C.D.解析：
35、(2).Aprocessing Bswitching Crouting Dinterfacing（分数：1.00）A.B.C.D. 解析：(3).Atree Bbus Cstar Dring（分数：1.00）A.B. C.D.解析：(4).Arouters Bstations Cservers Dswitches（分数：1.00）A.B. C.D.解析：(5).Atap Brepeat Cterminator Dconcentrator（分数：1.00）A.B.C. D.解析：解析 总线型和树型拓扑结构都使用多点传输介质。对于总线来说，所有站点都通过被称为接头的硬件直接连接到线性传输介质或总线
36、上。站和接头之间的操作允许数据发送到总线上，也可以从总线上接收数据。任何站点发送的数据都向总线两端传播，并可以被所有其他站点接收到。在总线的两端各有一个终接器，它接收并从总线上移去所有信号。For (21) service，we need a virtual-circuit subnetLet US see how that worksThe idea behind virtual circuits is to avoid having to choose a new (22) for every packet sentInstead， when a connection is establi
37、shed，a route from the (23) machine to the destination machine is chosen as part of the connection setup and stored in tables inside the (24) That route is used for all traffic flowing over the connection，exactly the same way that the telephone system works When the connection is releasedthe virtual
38、circuit is also tenninatedWith connection-oriented service，each packet carries an (25) telling which virtual circuit it belongs to（分数：5.00）(1).Aconnectionless Bconnectionoriented Cdatagram Dtelegram（分数：1.00）A.B. C.D.解析：(2).Aprocessor Bdevice Croute Dterminal（分数：1.00）A.B.C. D.解析：(3).Asource Broute Cd
39、estination Dhost（分数：1.00）A. B.C.D.解析：(4).Aconnections Bresources Cbridges Drouters（分数：1.00）A.B.C.D. 解析：(5).Aaddress Bidentifier Cinterface Delement（分数：1.00）A.B. C.D.解析：解析 对于面向连接的服务，我们需要一个虚拟电路子网。让我们看看它是如何工作的。虚电路的核心思想是避免在发送每个数据时为它们选择一个新的路由。相反，当建立一个连接，将从源计算机向目标计算机选择一条路由作为连接设置的一部分并且存储到内部的路由器表中。这条路由承载着所有
40、连接带来的流量，就跟电话系统的工作原理一样。当连接被释放时，虚电路也终止。在面向连接的服务中，每个数据包携带一个标识符，该标识符告诉它所属的虚电路。An internet is a combination of networks connected by (26) When a datagram goes from a source to a (27) ，it will probably pass many (28) until it reaches the router attached to the destination networkA router receives a (29) f
41、rom a network and passes it to another networkA router is usually attached to several networksWhen it receives a packet，to which network should it pass the packet? The decision is based on optimization：Which of the available (30) is the optimum pathway?（分数：5.00）(1).Amodems Brouters Cswitcher Dcomput
42、ers（分数：1.00）A.B. C.D.解析：(2).Auser Bhost Ccity Ddestination（分数：1.00）A.B.C.D. 解析：(3).Adestinations Brouters Cpackets Dcomputers（分数：1.00）A.B. C.D.解析：(4).Adestination Bresource Cpacket Dsource（分数：1.00）A.B.C. D.解析：(5).Apathways Brouters Cdiagrams Dcalls（分数：1.00）A. B.C.D.解析：解析 互联网是由不同的网络通过路由器互联起来的。当数据包从源地发往目的地时，在到达目的地网络之前它可能经过多个路由器。路由器从一个网络接收到数据包之后就转发到另外一个网络。然而，一个路由器往往与几个网络连接，当路由器接收到一个数据时，它应该将数据包传向哪个网络呢?这将基于路径最优化决策。
