计算机网络英文试题库(附答案)chapter

计算机网络题库c h a p t e r4(总15页)--本页仅作为文档封面，使用时请直接删除即可----内页可以根据需求调整合适字体及大小--Chapter 4 The Network Layer1．What is the name of a network-layer packet（）A.messageB.segmentC.datagramD.frame2．What are the two most important network-layer functions in a datagram network（）A.forwarding & routingB.forwarding & filteringC.routing & detectingD.routing & checking3．When a packet arrives at a router’s input link, the router must move the packet to the appropriate output link, this action is called ____（）A.ForwardingB.routingC.cut throughD.filtering4．The network layer must determine the path taken by packets as they flow from a sender to a receiver. This action is called____ （）A.ForwardingB.routingC.cut throughD.examine5．A router forwards a packet by examining the value of a field in the arriving packet’s header, and then using this value to index into the router’s forwarding table, this value is（）A.destination IP addressB.source IP addressC.destination MAC addressD.source MAC address6．IP belongs to ____ layer （）A.transportworkC.data linkD.physical7．Which layer provides host-to-host services? （）A.transportworkC.data linkD.physical8．Which layer provides a connection service called virtual-circuit networks and a connectionless service called datagram networks? （）A.transportworkC.data linkD.physical9．IPV4 has a ___ bit address （）A.48B.16C.32D.6410．MAC address is ____ bits （）A.48B.16C.32D.6411．The Internet’s network layer has three major component s, the first component is the IP protocol, the second component is the routing protocol, the final componentis____ （）A.forwardingB.address translationC.checkD.ICMP12．How many bits specify the IP protocol version in a datagram? （）A. 1 bitB. 2 bitsC. 4 bitsD.8 bits13．Most IPV4 datagram do not contain option field, how many bytes a header has in a typical IP datagram? （）A. 4 bytesB.8 bytesC.16 bytesD.20 bytes14．How many bits used in IP datagram express total length of the IP datagram? （）A. 4 bitsB.8 bitsC.16 bitsD.20 bits15．Data-grams are rarely larger than ____（）A.1500 bytesB.65535 bytesC.500 bytesD.1024 bytes16．New version of IP----IPV6 does not allow fragment in ____（）A.RouterB.HostC.Both of aboveD.None of above17．How does the TTL changed by one each time the datagram is processed by a router? （）A.decreaseB.increaseC.no changeD.always 018．Which layer protocol is indicated by the value of protocol field in IP datagram? （）A.applicationB.transportworkD.data link19．IP option is dropped in the ____ header. （）A.IPv6B.IPv4C.TCPD.UDP20．The data field in IP datagram can carry many types of data except ____. （）A.TCPB.UDPC.ICMPD.MAC21．ICMP is used for____（）A.Reliable data transferB.Error reportingC.Flow controlD.Congestion control22．Suppose you receive an IP datagram from one link and check your forwarding table to determine the outgoing link, but this outgoing link has an MTU that is smaller than the length of IP datagram. Which technology will be used （）A.resendB.discardC.fragmentD.none23．The designer of IPv4 decided to put the job of datagram reassembly in____.（）A.r outerB.s witchC.h ubD.e nd system24．IP addressing assigns an address to /29, the network address for this networkis____.（）A.B.C.D.25．If all data-grams arriving at the router from WAN have the same destination IP address, then how does the router know the internal host to which it should forward a given datagram? The trick is to use ____ table at router, and include port numbers as well as IP address in the table entries. （）A.routingB.forwardingC.ARPD.NAT translation26．When running a Telnet, FTP, or HTTP session, you may have encountered an error message such as “Destination network un-reachable”, This message had its origin in ____. （）A.IGMPB.EGPC.ICMPD.BGP27．The well-known ping program sends an (a) ____ type 8 code 0 message to the specified host. The destination hosts seeing the echo request, send back a type 0 code0 ICMP echo reply. （）A.ICMPB.IGMPC.TCPD.UDP28．Tracer-out is implemented with ____ messages. （）A.IGMPB.TCPC.UDPD.ICMP29．The standard Tracer-out program actually sends sets of ____ packets with the same TTL. （）A.oneB.twoC.threeD.four30．IPv6 increases the size of the IP address from 32bit to ____ bit. （）A.64B.128C.256D.51231．How many bytes are there in header of IPv6（）A.8B.20C.40D.3232．IPv6 does not allow for fragmentation and reassembly at inter-media ____.（）A.routersB.switchesC.end systemD.hubs33．From sender to receivers, all packets will take the same path, this is we use a____ service. （）A.datagramB.VCC.circuitD.Ethernet34．____ means moving packets from router’s input port to appropriate router output port（）A.forwardingB.filteringC.routingD.switching35．____ means determining route taken by packets from source to destination? （）A.forwardingB.filteringC.routingD.switching36．A router works in ____ layer? （）A.N etworkingB.D ata linkC.A pplicationD.P hysical37．A switch works in ____ layer（）workingB.Data linkC.ApplicationD.Physical38．Networking layer provides services between two ____.（）A.HostsB.ProcessesC.ApplicationsD.Machines39．Link layer provides services between two ____. （）A.HostsB.ProcessesC.ApplicationsD.Machines40．The internet’s network layer provides a single service----that is ____.（）A.Reliable data transferB.Flow controlC.Congestion controlD.Best-effort-service41．Datagram network provides network-layer____.（）A.connectionless serviceB.connection serviceC.both of aboveD.neither of A and B42．VC network provides network-layer ____. （）A.connectionless serviceB.connection serviceC.both of aboveD.neither of A and B43．In VC network, each packet carries ____. （）A.VC identifierB.destination host addressC.IP addressD.Mac address44．A VC consists of three part, this three parts do not include____. （）A.Path from source to destinationB.VC numbers, one number for each link along pathC.Entries in forwarding tables in routers along pathD.Destination address45.In ____ networking, a series of packet may follow different paths and may arrive out of order? （）A.DatagramB.VCC.TCPD.None of above46．The internet is a (an) ____ network（）A.DatagramB.VCC.Both of aboveD.None of above47．The four components of a router do not include____（）A.Input and output portsB.Switch fabricC.Routing processorD.Switching processor48．There are three kinds of switch fabric for a router normally, those three switch fabric do not includes____（）A.Switching via memoryB.Switching via a busC.Switching via an Interconnection-NetworkD.Packet switching49．Which of the following protocol is used for error reporting（）A.ICMPB.TCPC.IPD.UDP50．Which of the following is not a routing protocol（）A.TCPB.RIPC.OSPFD.BGP51．IPV4 has a ___ bits address? （）A.32B.64C.16D.12852．IP address is identifier for ____. （）A.HostB.Router interfaceC.Both of aboveD.None of above53．“/24”, where the /24 notation, so metimes known as a____. （）A.Subnet maskworking addressC.Host addressD.None of above54．Subnets with an 8-, ____-, and 24-bit subnet addresses were know as class A, B, and C networks respectively. （）A.9B.10C.12D.1655．IPV6 has a ___ bits address（）A.32B.64C.16D.12856．____ means that IPv6 nodes also have a complete IPv4 implementation as well （）A.Dual stackB.TunnelingC.Bridge connectionD.Forwarding57．____ means that IPv6 carried as payload in IPv4 data-gram among IPv4 routers （）A.Dual stackB.TunnelingC.Bridge connectionD.Forwarding58．Typically a host is attached directly to one router, the ____ for the host. （）A.Default routerB.Source routerC.Destination routerD.Core router59．The default router of the source host is known as ____.（）A.Default routerB.Source routerC.Destination routerD.Core router60．The default router of the destination host is known as ____. （）A.Default routerB.Source routerC.Destination routerD.Core router61．A path between the source and destination that has beast cost is known as____. （）A.Least cost pathB.Shortest pathrgest cost pathD.None of above62．The path with the smallest number of links between the source and the destination is known as ____. （）A.Least cost pathB.Shortest pathrgest cost pathD.None of above63．Which of following about DV is not correct? （）A.IterativeB.SynchronousC.DistributedD.Self-terminating64．Which of the following is not intra-AS routing protocol? （）A.RIPB.OSPFC.IGRPD.BGP65．RIP is a kind of ____ algorithm. （）A.DVB.LSC.Both of aboveD.Neither of A and B66．OSPF is a kind of ____ algorithm. （）A.DVB.LSC.Both of aboveD.Neither of A and B67．The router in non-backbone areas and perform only intra-AS routing is known as ____. （）A.Internal routersB.Area border routersC.Backbone routersD.Boundary routers68．____“summarize” distances to nets in own area, advertise to other Area Border routers. （）A.Internal routersB.Area border routersC.Backbone routersD.Boundary routers69．____ runs OSPF routing limited to backbone. （）A.Internal routersB.Area border routersC.Backbone routersD.Boundary routers70．____ connect to other AS’s. （）A.Internal routersB.Area border routersC.Backbone routersD.Boundary routers71．Which of the following is an inter-autonomous system routing protocol? （）A.RIPB.OSPFC.IGRPD.BGP72．____means sending a packet to all other nodes in the network. （）A.BroadcastB.MulticastC.Any-castD.Uni-cast73．A ____ packet is delivered to only a subset of network nodes. （）A.BroadcastB.MulticastC.Any-castD.Uni-cast74．Means that the switch must receive the entire packet before it can begin to transmit the first bit of the packet onto the outbound link. （）A.store-and-forward transmissionB.FDMC.End-to-end connectionD.TDM75．Datagram networks and virtual-circuit networks differ in ____.（）A.Datagram networks are circuit-switched networks, and virtual-circuitnetworks are packet-switched networks.B.Datagram networks are packet-switched networks, and virtual-circuitnetworks are circuit-switched networks.C.Datagram networks use destination addresses and virtual-circuit networksuse VC. Numbers to forward packets toward their destination.D.Datagram networks use VC. Numbers and virtual-circuit networks usedestination addresses to forward packets toward their destination.76．The Internet’s network layer is responsibl e for moving network-layer packets known as ____ from one host to another. （）A.frameB.datagramC.segmentD.message77．The protocols of various layers are called ____. （）A.the protocol stackB.TCP/IPC.ISPwork protocol78．The two important network-layer functions are ____. （）A.multiplexing and de-multiplexingB.routing and forwardingC.lookup and forwardingD.routing and IP addressing79．The virtual circuit includes three identifiable phases but ____. （）A.s etupB.d ata transferC.f orwarding VC numbersD.t eardown80．There are some ways to accomplish the switching in the router, which one is not included（）A.switching via memoryB.switching via a busC.switching via hubD.switching via an interconnection81．In the router, the packet queues can form at ____.（）A.institutional cacheB.switch fabricC.input portD.routing processor82．In IPV4 datagram header, which field is to ensure that datagram do not circulate forever in the network? （）A.type of serviceB.time-to-liveC.header checksumD.version83．Suppose one IP datagram of 5,000 bytes (20 bytes of IP header) and it must be forwarded to a link with an MTU of 1,500 bytes, the offset and flag of the second fragment in header are ____. （）A.185, 0B.185, 1C.370, 1D.370,084．Given the IP address /20, its subnet address is ____. （）A.B.C.D.85．If the subnet mask is , then how many bits indicate the host address? （）A.20B.24C.12D.1686．In the following four fields, which is in IPV6 header but not in IPV4（）A.source addressB.destination addressC.versionD.flow label87．One host is attached directly to one router, the router is ____.（）A.NAT routerB.gateway routerC.first-hop routerD.interval router88．The Internet’s network layer includes the following components but ____.（）A.IP protocolB.the Internet Message Control ProtocolC.Routing ProtocolD.Channel Partitioning Protocol89．In the following protocols, which one is inter-AS routing protocol（）A.RIPB.OSPFC.BGPD.ICMP90．In the following four descriptions about autonomous system, which one is not correct（）A.An autonomous system is a collection of routers under the sameadministrative and technical control.B.All routers in an autonomous system run the same routing protocol.C.The routing algorithm running within an autonomous system is called anintra-AS routing protocol.D.The routers that connect AS-s to each other are called default router. 91．In the loop of the Dijkstra’s algorithm, for node x, add y to N’, and update the cost of y’s neighbor v, then D(v) is ____. （）A.c(x, v)B.min{D(v), D(x)+c(x, v)}C.min{D(v), D(y)+c(y, v)}D.c(y, v)92．Consider the data D is 001, if use even parity checking approach, the parity bit is ( ), if use odd parity checking approach, the parity bit is ____. （）A.0, 0B.0, 1C.1, 0D.1, 193．The ability to determine the interfaces to which a frame should be directed, and then directing the frame to those interfaces is ____.（）A.filteringB.forwardingC.self-learningD.optimal routing94．Given that costs are assigned to various edge in the graph abstraction. In the following four descriptions, which one is not correct（）A.The shortest path is the path with smallest number of links between thesource and destination.B.The least cost path is not always the shortest path.C.If all edges have same cost, the least cost path must be the shortest path.D.If all edges have various costs, the least cost path must not be the shortestpath.95．Given the IP address /20, its broadcast address is ____. （）A.B.C.D.96．If the subnet mask is , then how many bits indicate the host address? （）A.20B.12C.24D.1697．In IPV6 datagram header, the field Next-header is equivalent to the field ( ) in IPV4. （）A.upper-layer protocolB.time-to-liveC.header checksumD.version98．For one host, its source router is also called ____. （）A.NAT routerB.gateway routerC.first-hop routerD.interval router99．In the following four descriptions about link-state (LS) algorithm, which one is not correct（）A.The LS algorithm must be aware of the cost of each link in the network.B.The LS algorithm is a decentralized algorithmC.The LS algorithm computes the least cost path between a source anddestination using complete global knowledge about the networkD.The calculation of LS algorithm can be run at one site or multiple sites 100．In the initialization of the Dijkstra’s algorithm, for node x, if y is not a ne ighbor of x, then D(y) is ____. （）A.0B. 1C.∞D.c(x, y)101．In DV algorithm, when one node updates its distance vector, it must ____.（）A.send the updated routing table to its neighborsB.send the updated routing table to all the nodes in the ASC.send the updated routing table at randomD.do nothing102．Now we had known IP address is /26, Please compute :(1). Network address(2). Broadcast address(3). How many hosts are there in the network(4). Fist host IP address(5). Last host IP address103．For the given topology of the network, use Dijstra’s shortest path algorithm to compute the shortest path from A to all network nodes. Give a shortest path tree and node A’s routing table.104．The topology of a network shown as below. Using Link State routing algorithm to calculate the routing table in the node A. The vectors arrive to node C as bellow: B:( 4, 0, 8, 12, 6, 2 ); D:( 13, 10, 6, 0, 9, 10 ); E:( 7, 6, 3, 9, 0, 14 ). The delays measured from C to B, D, E are 4, 2, and 3 separately. Update router C’s route table please.。

Chapter 5 The Link Layer and Local Area Network1．A ( ) protocol is used to move a datagram over an individual link.A application-layerB transport-layerC network-layerD link-layer2．The units of data exchanged by a link-layer protocol are called ( ).A datagramsB framesC segmentsD messages3．Which of the following protocols is not a link-layer protocol? ( )A EthernetB PPPC HDLCD IP4．In the following four descriptions, which one is not correct? ( )A link-layer protocol has the node-to-node job of moving network-layer datagrams over a single link in the path.B The services provided by the link-layer protocols may be different.C A datagram must be handled by the same link-layer protocols on the different links in the path.D The actions taken by a link-layer protocol when sending and receiving frames include error detection, flow control and random access.5．Which of the following services can not offered by a link-layer protocol? ( )A congestion controlB Link AccessC Error controlD Framing6．( ) protocol serves to coordinate the frame transmissions of the many nodes when multiple nodes share a single broadcast link.A ARPB MACC ICMPD DNS7．In the following four descriptions about the adapter, which one is not correct? ( )A The adapter is also called as NIC.B The adapter is a semi-autonomous unit.C The main components of an adapter are bus interface and the link interface.D The adapter can provide all the link-layer services.8．Consider CRC error checking approach, the four bit generator G is 1011, and suppose that the data D is 10101010, then the value of R is( ).A 010B 100C 011D 1109．In the following four descriptions about random access protocol, which one is not correct? ( )A I n slotted ALOHA, nodes can transmit at random time.B I n pure ALOHA, if a frame experiences a collision, the node will immediately retransmit it with probability p.C T he maximum efficiency of a slotted ALOHA is higher than a pure ALOHA.D I n CSMA/CD, one node listens to the channel before transmitting.10．In the following descriptions about MAC address, which one is not correct? ( )A T he MAC address is the address of one node’s adapter.B N o two adapters have the same MAC address.C T he MAC address doesn’t change no matter where the adapter goes.D M AC address has a hierarchical structure.11．The ARP protocol can translate ( ) into ( ). ( )A h ost name, IP addressB h ost name, MAC addressC I P address, MAC addressD broadcast address, IP address12．The value of Preamble field in Ethernet frame structure is ( )A 10101010 10101010……10101010 11111111B 10101011 10101011……10101011 10101011C 10101010 10101010……10101010 10101011D 10101010 10101010……10101010 1010101013．There are four steps in DHCP, the DHCP server can complete ( ).A DHCP server discoveryB DHCP server offersC DHCP requestD DHCP response14．In CSMA/CD, the adapter waits some time and then returns to sensing the channel. In the following four times, which one is impossible? ( )A 0 bit timesB 512 bit timesC 1024 bit timesD 1028 bit times15．The most common Ethernet technologies are 10BaseT and 100BaseT. “10” and “100” indicate( ).A the maximum length between two adaptersB the minimum length between two adaptersC the transmission rate of the channelD the transmission rate of the node16．The principal components of PPP include but not( ).A framingB physical-control protocolC link-layer protocolD network-layer protocol17．In the following four options, which service can not be provided by switch? ( )A filteringB self-learningC forwardingD optimal routing18．In the following four services, which one was be required in PPP? ( )A packet framingB error detectionC error correctionD multiple types of link19．The ability to determine the interfaces to which a frame should be directed, and then directing the frame to those interfaces is( ).A filteringB forwardingC self-learningD optimal routing20．In ( ) transmission(s), the nodes at both ends of a link may transmit packets at the same time.A full-duplexB half-duplexC single-duplexD both full-duplex and half-duplex21．Consider the data D is 01110010001, if use even parity checking approach, the parity bit is( ① ), if use odd parity checking approach, the parity bit is( ② ). ( )A ①0 ②1B ①0 ②0C ①1 ②1D ①1 ②022．In the following four descriptions about parity checks, which one is correct? ( )A Single-bit parity can detect all errors.B Single-bit parity can correct one errors.C Two-dimensional parity not only can detect a single bit error, but also can correct that error.D Two-dimensional parity not only can detect any combination of two errors, but also can correct them.23．MAC address is ( ) bits long.A 32B 48C 128D 6424．Wireless LAN using protocol ( ).A IEEE 802.3B IEEE 802.4C IEEE 802.5D IEEE 802.1125．The following protocols are belonging to multiple access protocols except for ( ).A channel partitioning protocolsB routing protocolsC random access protocolsD taking-turns protocols26．Which of the following is not belonging to channel partitioning protocols? ( )A CSMAB FDMC CDMAD TDM27．In the following four descriptions about CSMA/CD, which one is not correct? ( )A A node listens to the channel before transmitting.B If someone else begins talking at the same time, stop talking.C A transmitting node listens to the channel while it is transmitting.D With CSMA/CD, the collisions can be avoided completely.28．( ) provides a mechanism for nodes to translate IP addresses to link-layer address.A IPB ARPC RARPD DNS29．A MAC address is a ( )address.A physical-layerB application-layerC link-layerD network-layer30．Which of the following is correct? ( )A No two adapters have the same MAC address.B MAC broadcast address is FF-FF-FF-FF-FF-FF.C A portable computer with an Ethernet card always has the same MAC address, no matter where the computer goes.D All of the above31．In the following four descriptions, which one is not correct? ( )A ARP resolves an IP address to a MAC address.B DNS resolves hostnames to IP addresses.C DNS resolves hostnames for hosts anywhere in the Internet.D ARP resolves IP addresses for nodes anywhere in the Internet.32．In the LAN, ( )protocol dynamically assign IP addresses to hosts.A DNSB ARPC DHCPD IP33．DHCP protocol is a four-step process: ①DHCP request. ②DHCP ACK. ③DHCP server discovery. ④DHCP server offer(s). The correct sequence is ( )A ①②③④B ③②①④C ③④①②D ①④③②34．In the Ethernet frame structure, the CRC field is ( )bytes.A 2B 4C 8D 3235．In the Ethernet frame structure, the Data field carries the ( ).A IP datagramB segmentC frameD message36．In the following four descriptions, which one is not correct? ( )A Ethernet uses baseband transmission.B All of the Ethernet technologies provide connection-oriented reliable service to the network layer.C The Ethernet 10Base2 technology uses a thin coaxial cable for the bus.D The Ethernet 10BaseT technology uses a star topology.37．Ethernet’s multiple access protocol is ( ).A CDMAB CSMA/CDC slotted ALOHAD token-passing protocol38．In the following four descriptions about CSMA/CD, which one is not correct? ( )A An adapter may begin to transmit at any time.B An adapter never transmits a frame when it senses that some other adapter is transmitting.C A transmitting adapter aborts its transmission as soon as it detects that another adapter is also transmitting.D An adapter retransmits when it detects a collision.39．Which of the following descriptions about CSMA/CD is correct? ( )A No slots are used.B It uses carrier sensing.C It uses collision detection.D All of the above.40．The Ethernet 10BaseT technology uses( )as its physical media.A fiber opticsB twisted-pair copper wireC coaxial cableD satellite radio channel41．For 10BaseT, the maximum length of the connection between an adapter and the hub is ( )meters.A 100B 200C 500D 1042．A ( )is a physical-layer device that acts on individual bits rather than on frames.A switchB hubC routerD gateway43．A hub is a ( )device that acts on individual bits rather than on frames.A physical-layerB link-layerC network-layerD ransport-layer44．A switch is a( )device that acts on frame.A physical-layerB link-layerC network-layerD transport-layer45．In the following four descriptions, which one is not correct? ( )A Switches can interconnect different LAN technologies.B Hubs can interconnect different LAN technologies.C There is no limit to how large a LAN can be when switches are used to interconnect LAN segments.D There is restriction on the maximum allowable number of nodes in a collision domain when hubs are used to interconnect LAN segments.46．The ability to determine whether a frame should be forwarded to some interface or should just be dropped is ( ).A f ilteringB f orwardingC s elf-learningD o ptimal routing47．Which of the following devices is not a plug and play device? ( )A hubB routerC switchD repeater48．Which of the following devices is not cut-through device? ( )A hubB routerC switchD repeater49．In the following four descriptions, which one is not correct? ( )A Switches do not offer any protection against broadcast storms.B Routers provide firewall protection against layer-2 broadcast storms.C Both switches and routers are plug and play devices.D A router is a layer-3 packet switch, a switch is a layer-2 packet switch. 50．Which device has the same collision domain? ( )A HubB SwitchC RouterD Bridge51．IEEE802.2 protocol belong to ( )layerA networkB MACC LLCD physical52．IEEE802.11 protocol defines ( )rules.A Ethernet BusB wireless WANC wireless LAND Token Bus53．In data link-layer, which protocol is used to share bandwidth? ( )A SMTPB ICMPC ARPD CSMA/CD54．When two or more nodes on the LAN segments transmit at the same time, there will be a collision and all of the transmitting nodes well enter exponential back-off, that is all of the LAN segments belong to the same( ).A collision domainB switchC bridgeD hub55．( )allows different nodes to transmit simultaneously and yet have their respective receivers correctly receive a sender’s encoded data bits.A CDMAB CSMAC CSMA/CDD CSMA/CA56．Because there are both network-layer addresses (for example, Internet IP addresses) and link-layer addresses (that is, LAN addresses), there is a need totranslate between them. For the Internet, this is the job of ( ).A RIPB OSPFC ARPD IP57．PPP defines a special control escape byte, ( ). If the flag sequence, 01111110 appears anywhere in the frame, except in the flag field, PPP precedes that instance of the flag pattern with the control escape byte.A 01111110B 01111101C 10011001D 1011111058．The device ( ) can isolate collision domains for each of the LAN segment.A modemB switchC hubD NIC59．In the following four descriptions about PPP, which one is not correct? ( )A PPP is required to detect and correct errors.B PPP is not required to deliver frames to the link receiver in the same order in which they were sent by the link sender.C PPP need only operate over links that have a single sender and a single receiver.D PPP is not required to provide flow control.60．In the PPP data frame, the( ) field tells the PPP receivers the upper-layer protocol to which the received encapsulated data belongs.A flagB controlC protocolD checksum61．PPP’s link-control protocols (LCP) accomplish ( ).A initializing the PPP linkB maintaining the PPP linkC taking down the PPP linkD all of the above62．The PPP link always begins in the ( ) state and ends in the ( ) state. ( )A open, terminatingB open, deadC dead, deadD dead, terminating63．For( ) links that have a single sender at one end of the link and a single receiver at the other end of the link.A point-to-pointB broadcastC multicastD all of the above64．With ( )transmission, the nodes at both ends of a link may transmit packets at the same time.A half-duplexB full-duplexC simplex(单工)D synchronous65．With ( ) transmission, a node can not both transmit and receive at the same time.A half-duplexB full-duplexC simplex(单工)D synchronous66．Which of the following functions can’t be implemented in the NIC? ( )A encapsulation and decapsulationB error detectionC multiple access protocolD routing67．Which of the following four descriptions is wrong? ( )A The bus interface of an adapter is responsible for communication with the adapter’s parent node.B The link interface of an adapter is responsible for implementing the link-layer protocol.C The bus interface may provide error detection, random access functions.D The main components of an adapter are the bus interface and the link interface. 68．For odd parity schemes, which of the following is correct? ( )A 011010001B 111000110C 110101110D 00011011069．( )divides time into time frames and further divides each time frame into N time slots.A FDMB TMDC CDMAD CSMA70．With CDMA, each node is assigned a different ( )A codeB time slotC frequencyD link71．Which of the following four descriptions about random access protocol is not correct? ( )A A transmission node transmits at the full rate of the channelB When a collision happens, each node involved in the collision retransmits at once.C Both slotted ALOHA and CSMA/CD are random access protocols.D With random access protocol, there may be empty slots.72．PPP defines a special control escape byte 01111101. If the data is b1b201111110b3b4b5, the value is( )after byte stuffing.A b1b20111110101111110b3b4b5B b1b20111111001111101b3b4b5C b5b4b30111111001111101b2b1D b5b4b30111110101111110b2b173．MAC address is in ( ) of the computer.A RAMB NICC hard diskD cache74．Which of the following is wrong? ( )A ARP table is configured by a system administratorB ARP table is built automaticallyC ARP table is dynamicD ARP table maps IP addresses to MAC addresses75．NIC works in ( )layer.A physicalB linkC networkD transport76．In LAN, if UTP is used, the common connector is( ).A AUIB BNCC RJ-45D NNI77．The modem’s function(s) is(are) ( ).A translates digital signal into analog signalB translates analog signal into digital signalC both translates analog signal into digital signal and translates digital signal into analog signalD translates one kind of digital signal into another digital signal78．( )defines Token-Ring protocol.A IEEE 802.3B IEEE 802.4C IEEE 802.5D IEEE 802.279．( )defines Token-Bus protocol.A IEEE 802.3B IEEE 802.4C IEEE 802.5D IEEE 802.280．( ) defines CSMA/CD protocol.A IEEE 802.3B IEEE 802.4C IEEE 802.5D IEEE 802.281．The computer network that concentrated in a geographical area, such as in a building or on a university campus, is ( )A a LANB a MANC a WAND the Internet82．The MAC address is ( ) bits long.A 32B 48C 128D 25683．Which of the following four descriptions about MAC addresses is wrong? ( )A a MAC address is burned into the adapter’s ROMB No two adapters have the same addressC An adapter’s MAC address is dynamicD A MAC address is a link-layer address84．Which of the following four descriptions about DHCP is correct? ( )A DHCP is C/S architectureB DHCP uses TCP as its underlying transport protocolC The IP address offered by a DHCP server is valid foreverD The DHCP server will offer the same IP address to a host when the host requests an IP address85．The ( )field permits Ethernet to multiplex network-layer protocols.A preambleB typeC CRCD destination MAC address86．For 10BaseT, the maximum length of the connection between an adapter and the hub is ( ) meters.A 50B 100C 200D 50087．An entry in the switch table contains the following information excepts for ( )A the MAC address of a nodeB the switch interface that leads towards the nodeC the time at which the entry for the node was placed in the tableD the IP address of a node88．Consider the 4-bit generator , G is 1001, and suppose that D has the value 101110000. What is the value of R?89．Consider the following graph of the network. Suppose Host A will send a datagram to Host B, Host A run OICQ on port 4000, Host B run OICQ on port 8000. All of ARP tables are up to date. Enumerate all the steps when message “Hello” is sent from host A to host B.。

计算机网络双语试题及答案一、选择题（共20题，每题2分，共40分）1. 当使用HTTP协议进行通信时，下列哪个是无连接的协议？A. TCPB. IPC. UDPD. FTP2. 在计算机网络中，IP地址的作用是什么？A. 标识主机在网络中的唯一地址B. 实现数据的可靠传输C. 进行主机之间的通信D. 提供远程访问服务3. 下列哪个不属于网络拓扑结构的类型？A. 总线型拓扑B. 环形拓扑C. 星型拓扑D. 布线型拓扑4. 在TCP/IP协议中，下列哪个协议用于将IP地址转换为物理地址？B. DHCPC. FTPD. ICMP5. 在计算机网络中，HTTP和HTTPS协议之间的区别是？A. HTTP使用明文传输，HTTPS使用加密传输B. HTTP使用UDP传输，HTTPS使用TCP传输C. HTTP使用IP地址，HTTPS使用域名D. HTTP使用GET请求，HTTPS使用POST请求6. 某计算机的IP地址是192.168.0.1，子网掩码是255.255.255.0，那么该计算机所在的网络号是？A. 192.168.0B. 192.168.0.1C. 192.168.0.255D. 192.168.0.07. 在计算机网络中，下列哪个协议用于将域名解析为IP地址？A. DNSB. FTPC. DHCP8. 在TCP/IP协议中，下列哪个协议用于控制数据传输的可靠性？A. ICMPB. UDPC. FTPD. TCP9. 在计算机网络中，什么是反向代理服务器（Reverse Proxy Server）？A. 将外部网络请求转发给内部服务器的服务器B. 将内部网络请求转发给外部服务器的服务器C. 将HTTP请求转发给HTTPS服务器的服务器D. 将HTTPS请求转发给HTTP服务器的服务器10. 在计算机网络中，下列哪个协议用于电子邮件的发送和接收？A. SMTPB. POP3C. HTTPD. FTP11. 在计算机网络中，下列哪个协议用于文件传输？B. SMTPC. UDPD. TCP12. 以下哪个不是IPv6地址的特点？A. 128位长度B. 冒号分隔的十六进制C. 有固定的网络号和主机号D. 全球唯一的地址13. 在计算机网络中，下列哪个技术不属于无线局域网技术？A. Wi-FiB. BluetoothC. NFCD. 4G14. 在网络中，下列哪个设备用于将不同网段的数据转发到目的主机？A. 集线器B. 路由器D. 网关15. 下列哪个网络拓扑结构具有较高的可容错性和可拓展性？A. 星型拓扑B. 总线型拓扑C. 环形拓扑D. 树型拓扑16. 在计算机网络中，下列哪个协议用于互联网上的主机进行网络配置？A. DHCPB. DNSC. HTTPD. FTP17. 在TCP/IP协议中，下列哪个协议用于检测并纠正数据传输中的错误？A. TCPB. ARPC. ICMPD. UDP18. 下列哪个是网络安全的常见攻击方式之一？A. DDoS攻击B. 数据库攻击C. 剪贴板攻击D. 社会工程学攻击19. 下列哪个不是局域网（LAN）的特点？A. 覆盖范围较小B. 速度较快C. 价格较高D. 结构较简单20. 在计算机网络中，下列哪个协议用于向局域网中的所有主机广播消息？A. UDPB. TCPC. DHCPD. ICMP二、问答题（共5题，每题10分，共50分）1. 请简要介绍HTTP协议的工作原理。

计算机⽹络英⽂版——提供给学⽣部分习题答案Solution of Selected Exercises from the End of Chapter ExercisesChapter 1 - Introduction And Overview1.4 To what aspects of networking does data communications refer?Answer:Data communications refers to the study of low-level mechanisms and technologies used to send information acrossa physical communication medium, such as a wire, radio wave, or light beam.1.5 What is packet-switching, and why is packet switching relevant to the Internet?Answer: Packet switching divides data into small blocks, called packets, and includes an identification of the intended recipient in each packet. Packet switching changed networking in a fundamental way, and provided the basis for the modern Internet. Packet switching allows multiple senders to transmit data over a shared network.1.8 What is a communication protocol? Conceptually, what two aspects of communication does a protocol specify? Answer: A communication protocol refer to a specification for network communication.Major aspects of a protocol are syntax (format) and semantics (meaning) of the protocol.1.9 What is a protocol suite, and what is the advantage of a suite?Answer:protocols are designed in complete, cooperative sets called suites or families, instead of creating each protocol in isolation. Each protocol in a suite handles one aspect of communication; together, the protocols in a suite cover all aspects of communication. The entire suite is designed to allow the protocols to work together efficiently. 1.11 List the layers in the TCP/IP model, and give a brief explanation of each.(See Textbook)1.14 Give a brief explain of the layers in the ISO Open System Interconnection model.(See Textbook)Chapter 3 - Internet Applications And Network Programming3.1 What are the two basic communication paradigms used in the Internet?Answer: There are various approaches, but according to textbook, we can specify them as Stream Paradigm and Message Paradigm.3.2 Give six characteristics of Internet stream communication.(See Textbook)3.3 Give six characteristics of Internet message communication.(See Textbook)3.4 If a sender uses the stream paradigm and always sends 1024 bytes at a time, what size blocks can the Internet deliverto a receiver?Answer: stream paradigm does not provide any guarantees for block sizes, so all depends on individual transfer.3.6 What are the three surprising aspects of the Internet’s message delivery semantics?Answer:The Internet’s message delivery has the followi ng undesirable characteristics:* Messages can be lost* Messages can be duplicated* Messages can be delivered out-of-order3.8 When two applications communicate over the Internet, which one is the server?Answer: T he application that waits for some other applications to contact is called server, and the application that contact other one is called client.3.14 What two identifiers are used to specify a particular server?Answer: A particular server is identified by the following identifiers:* An identifier for the computer on which a server runs (IP Address)* An identifier for a particular service on the computer (Port Number)Chapter 4 - Traditional Internet Applications4.1 What details does an application protocol specify?(See Textbook)4.3 What are the two key aspects of application protocols, and what does each include?(See Textbook)4.6 What are the four parts of a URL, and what punctuation is used to separate the parts?Answer: The URL into four components: a protocol, a computer name, a document name, and parameters. The computer name and protocol port are used to form a connection to the server on which the page resides. And the document name and parameters are used to request a specific page.4.7 What are the four HTTP request types, and when is each used?(See Textbook)4.12 When a user requests an FTP directory listing, how many TCP connections are formed? Explain.Answer: FTP uses two types of connections to perform its functionality, namely* A control connection is reserved for commands. Each time the server needs to download or upload a file, the server opens a new connection.* A data connection is used to transfer files.4.16 List the three types of protocols used with email, and describe each.(See Textbook)4.17 What are the characteristics of SMTP?(See Textbook)4.20 What are the two main email access protocols?Answer: Two major email access protocols are:* Post Office Protocol (POP)* Internet Mail Access Protocol (IMAP)Chapter 6- Information Sources and Signals6.4 State and describe the four fundamental characteristics of a sine wave.(See Textbook)6.9 What is the analog bandwidth of a signal?Answer: Analog bandwidth of signal can be defined as to be the difference between the highest and lowest frequencies of the constituent parts (i.e., the highest and lowest frequencies obtained by Fourier analysis)6.11 Suppose an engineer increases the number of possible signal levels from two to four. How many more bits can be sent in the same amount of time? Explain.Answer: The number of levels that can be represented by n bits is given by 2n . So if number of levels changes from 2→4, it means number of bits goes from 1→2612. What is the definition of baud?Answer: Baud is defined as the number of times that a signal can change per second.6.14 What is the bandwidth of a digital signal? Explain.Answer: According to the definition of analog bandwidth, a digital signal has infinite bandwidth because Fourier analysis of a digital signal produces an infinite set of sine waves with frequencies that grow to infinity.6.18 What is the chief advantage of a Differential Manchester Encoding?Answer: The most important property of differential encoding is that the encoding works correctly even if the two wires carrying the signal are accidentally reversed.6.20 If the maximum frequency audible to a human ear is 20,000 Hz, at what rate must the analog signal from a microphone be sampled when converting it to digital?Answer: The sampling rate = 2 × f max, so the signal should be sampled at 2x20,000 = 40,000 HzChapter 7 - Transmission Media7.2 What are the three energy types used when classifying physical media according to energy used?Answer: Three types of energy used when classifying physical media are electrical, electromechanical (radio), and light7.4 What three types of wiring are used to reduce interference form noise?(See Textbook)7.10 List the three forms of optical fiber, and give the general properties of each.(See Textbook)7.21 What is the relationship between bandwidth, signal levels, and data rate?Answer: If a transmission system uses K possible signal levels and has an anal og bandwidth B, Nyquist’s Theorem states that the maximum data rate in bits per second, D, is: D = 2 B log2K7.22 If two signal levels are used, what is the data rate that can be sent over a coaxial cable that has an analog bandwidthof 6.2 MHz?Answer: Using the D= 2 B log2 K relationship, D = 2*6.2*log22 = 2*6.2*1 = 12.4 Mbps7.24 If a system has an input power level of 9000, and an output power level of 3000, what is the difference when expressed in dB?Answer: Decibel is expressed as 10log10(P out/P in) → 10log10(3,000/9,000) = to be determined by reader7.23 If a system has an average power level of 100, an average noise level of 33.33, and a bandwidth of 100 MHz, whatis the effective limit on channel capacity?Answer: Shannon theorem specify the maximum data rate that could be achieved over a transmission system that experiences noise: C = Blog2 (1 + S/N) = 100,000,000 * log2 (1 + 100/33.33) = 100,000,000 * log24 = 200,000,000 = 200 Mbps7.25 If a telephone system can be created with a signal-to-noise ratio of 40 dB and an analog bandwidth of 3000 Hz, how many bits per second could be transmitted?Answer: First we should convert 40 dB to a real number, namely if 40 = 10 log10S/N→S/N = 10,000 , Using the Shannon’s capacity expression C = B log2(1 + S/N) → C = 3,000 log2 (1+ 10,000) = to be determined by readerCh 8 - Reliability And Channel Coding8.1 List and explain the three main sources of transmission errors.(See Textbook)8.3 In a burst error, how is burst length measured?Answer: For a burst error, the burst size, or length, is defined as the number of bits from the start of the corruption to the end of the corruption.8.4 What is a codeword?Answer: We can define the set of all possible messages to be a set of datawords, and define the set of all possible encoded versions to be a set of codewords. So each possible code sequence is considered to be a codeword.8.8 Compute the Hamming distance for the following pairs: (0000, 0001), (0101, 0001), (1111, 1001), and ( 0001, 1110). (See Textbook)8.11 Generate a RAC parity matrix for a (20, 12) coding of the dataword 100011011111.(See Textbook)8.15 Express the two values in the previous exercise as polynomials.Answer:X10+ X7 + X5 + X3 + XX4+ X2+ 1Ch 9 - Transmission Modes9.1 Describe the difference between serial and parallel transmission.Answer: Transmission modes can be divided into two fundamental categories:* Serial: one bit is sent at a time* Parallel: multiple bits are sent at the same time9.2 What are the advantages of parallel transmission? What is the chief disadvantage?Answer: A parallel mode of transmission has two chief advantages:* High speed: Because it can send N bits at the same time, a parallel interface can operate N times faster than an equivalent serial interface.* Match to underlying hardware: Internally, computer and communication hardware uses parallel circuitry.Thus, a parallel interface matches the internal hardware well.The main disadvantage of parallel transmission is number of cables required, for long distance communication, this is an important consideration.9.4 What is the chief characteristic of asynchronous transmission?Answer:Asynchronous transmission can occur at any time, with an arbitrary delay between the transmission of two data items, it allows the physical medium to be idle for an arbitrary time between two transmissions.Chapter 11 - Multiplexing And Demultiplexing11.2 What are the four basic types of multiplexing?(See Textbook)11.4 What is a guard band?Answer: For proper communication without interference, we should choose a set of carrier frequencies with a gap between them known as a guard band. The guard band reduces or eliminates the possible interference between neighboring carrier signals.11.8 Explain how a range of frequencies can be used to increase data rate.Answer:To increase the overall data rate, a sender divides the frequency range of the channel into K carriers, and sends 1 /K of the data over each carrier.11.12 Suppose N users compete using a statistical TDM system, and suppose the underlying physical transport can sendK bits per second. What is the minimum and maximum data rate that an individual user can experience?Answer: If we neglect the overhead generated by statistical TDM, a system will have two possibilities: * Minimum: If all channels have equal data then the rate will be K/N bps* Maximum: If only one channel active and the others are passive, then rate will be K bpsChapter 13 - Local Area Networks: Packets, Frames, And Topologies13.1 What is circuit switching, and what are its chief characteristics?Answer: The term circuit switching refers to a communication mechanism that establishes a path between a sender and receiver with guaranteed isolation from paths used by other pairs of senders and receivers. The circuit switching has the following main characteristics:* Point-to-point communication* Separate steps for circuit creation, use, and termination* Performance equivalent to an isolated physical path13.3 In a packet switching system, how does a sender transfer a large file?Answer: The packet switching system requires a sender to divide each message into blocks of data that are known as packets . The size of a packet varies; each packet switching technology defines a maximum packet size. So, a large file will be divided into smaller pieces and sent.13.5 What are the characteristics of LANs, MANs, and W ANs?Answer: There are lots of details that can be said and discussed for categorization of network types based on geography, few points are highlighted below:* Local Area Network (LAN): Least expensive; spans a single room or a single building* Metropolitan Area Network (MAN) Medium expense; spans a major city or a metroplex* Wide Area Network (WAN) Most expensive; spans sites in multiple cities13.6 Name the two sublayers of Layer 2 protocols defined by IEEE, and give the purpose of each.Answer: The Layer 2 protocols defined by IEEE defines two sub-layers as mentioned below:* Logical Link Control (LLC) Addressing and demultiplexing* Media Access Control (MAC) Access to shared media13.8 What are the four basic LAN topologies?Answer: The four basic LAN topologies are star, ring, mesh and bus.13.10 In a mesh network, how many connections are required among 20 computers?Answer: The expression to calculate number of connections in a mesh network is given by n (n-1)/2. So for 20 computers then number of connections required will be = 20 (20 – 1)/2 =19013.15 Give a definition of the term frame .Answer: In a packet-switched network, each frame corresponds to a packet processed at data link layer.Chapter 14 - The IEEE MAC Sub-Layer14.1 Explain the three basic approaches used to arbitrate access to a shared medium.(See Textbook)14.3 List the three main types of channelization and the characteristics of each.(See Textbook)14.6 What is a token, and how are tokens used to control network access?Answer: A special control message is called a token. In a token passing system, when no station has any packets to send, the token circulates among all stations continuously. When a station captures the token, it sends its data, and when transmission completed, it releases the token.14.8 Expand the acronym CSMA/CD, and explain each part.Answer: The acronym CSMA/CD stands for Carrier Sense Multi-Access with Collision Detection, which means the following: * Carrier Sense: Instead of allowing a station to transmit whenever a packet becomes ready, Ethernet requires each station to monitor the cable to detect whether another transmission is already in progress.* Multiple Access: The system allows multiple users/hosts to make use of a common/shared media* Collision Detection. A collision can occur if two stations wait for a transmission to stop, find the cable idle, and both start transmitting.14.10 Why does CSMA/CD use a random delay? (Hint: think of many identical computers on a network.)Answer: Randomization is used to avoid having multiple stations transmit simultaneously as soon as the cable is idle.That is, the standard specifies a maximum delay, d, and requires each station to choose a random delay less than d after a collision occurs. In most cases, when two stations each choose a random value, the station that chooses the smallest delay willChapter 15 - Wired LAN Technology (Ethernet And 802.3)15.1 How large is the maximum Ethernet frame, including the CRC?Answer: According to Fig. 15.1 a conventional Ethernet frame has the following fields:* Header: 14 bytes (fixed)* Payload: 46-1500 bytes (there is a minimum frame size because of collision detection)* CRC: 4 bytes (fixed)Accordingly an Ethernet frame will be maximum 1518 bytes and minimum 64 bytes15.3 In an 802.3 Ethernet frame, what is the maximum payload size?Answer: The 802.3 Ethernet makes use of 8-bytes of the original/conventional Ethernet for Logical Link Control / Sub-Network Attachment Point (LLC / SNAP) header instead of extending/increasing the header. This is for sake of backward compatibility. So the maximum pay load is reduced from 1500 bytes to 1492 bytes.15.6 How did a computer attach to a Thicknet Ethernet?Answer: Hardware used with Thicknet was divided into two major parts:* Transceiver: A network interface card (NIC) handled the digital aspects of communication, and a separate electronic device called a transceiver connected to the Ethernet cable and handled carrier detection, conversion of bits into appropriate voltages for transmission, and conversion of incoming signals to bits.* AUI: A physical cable known as an Attachment Unit Interface (AUI) connected a transceiver to a NIC in a computer. A transceiver was usually remote from a computer.15.7 How were computers attached to a Thinnet Ethernet?Answer: Thinnet Ethernet (formally named 10Base2) uses a thinner coaxial cable that was more flexible than Thicknet. The wiring scheme differed dramatically from Thicknet. Instead of using AUI connections between a computer and a transceiver, Thinnet integrates a transceiver directly on the NIC, and runs a coaxial cable from one computer to another.15.8 What is an Ethernet hub, and what wiring is used with a hub?Answer: An electronic device that serves as the central interconnection is known as a hub. Hubs were available in a variety of sizes, with the cost proportional to size. The hubs are becoming old-fashioned, and being replaced with switches.15.3 What category of twisted pair wiring is needed for a 10 Mbps network? 100 Mbps? 1000 Mbps?Answer: The three major categories of Ethernet and their wiring is listed below:* 10 Mbps: 10BaseT (Ethernet) Category 5* 100 Mbps: 100BaseT (Ethernet Fast) Category 5E* 1 Gbps: 1000BaseT (Gigabit Ethernet) Category 6Chapter 20 - Internetworking: Concepts, Architecture, and Protocols20.2 Will the Internet be replaced by a single networking technology? Why or why not?Answer: Incompatibilities make it impossible to form a large network merely by interconnecting the wires among networks. The beauty of the Internet is interconnection of wide range of technologies from various manufacturers.Diversity of the products and solutions is a richness instead of limitation as long as they all adopt the same set of protocols.20.3 What are the two reasons an organization does not use a single router to connect all its networks?Answer:An organization seldom uses a single router to connect all of its networks. There are two major reasons: * Because the router must forward each packet, the processor in a given router is insufficient to handle the traffic passing among an arbitrary number of networks.* Redundancy improves internet reliability. To avoid a single point of failure, protocol software continuously monitors internet connections and instructs routers to send traffic along alternative paths when a network or router fails.20.6 In the 5-layer reference model used with the TCP/IP Internet protocols, what is the purpose of each of the five layers?(See 1.11)Chapter 21- IP: Internet Addressing21.3 In the original classful address scheme, was it possible to determine the class of an address from the address itself? Explain.Answer:Yes, since in the classful addressing scheme initial bit(s) gives indication about the class being used.21.7 If an ISP assigned you a /28 address block, how many computers could you assign an address?Answer: When an organization is assigned /28 CIDR address, it means 28 bits out of 32 bits are fixed, so 32-28 = 4 bits available for user space. So number of users 24-2 = 4, since the all 0s and all 1s address are having special use and can’t be assigned to a user.21.8 If an ISP offers a / 17 address block for N dollars per month and a / 16 address block for 1.5 N dollars per month,which has the cheapest cost per computer?Answer: Number of addresses in /17 block 232-17 = 215Price per address: N /215 = N / 215Number of addresses in /16 block 232-16 = 216Price per address: 1.5N /216 = 0.75N/215 So /16 address block will be cheaper in comparison with the price given for /17 block.21.10 Suppose you are an ISP with a / 24 address block. Explain whether you accommodate a request from a customer who needs addresses for 255 computers. (Hint: consider the special addresses.)Answer: For a/24 address block, number of available addresses will be 232-24 = 28 = 256. However, a suffix with all 0s address is reserved for network ID and a suffix with all 1s address is reserved for broadcast address, so number of addresses that can be assigned to computers/hosts will be 256 -2 = 254.21.11 Suppose you are an ISP that owns a / 22 address block. Show the CIDR allocation you would use to allocateaddress blocks to four customers who need addresses for 60 computers each.Answer: The /22 address block can be assigned as follows:ddd.ddd.ddd.00/26ddd.ddd.ddd.01/26ddd.ddd.ddd.10/26ddd.ddd.ddd.11/26Chapter 22- Datagram Forwarding22.1 What are the two basic communication paradigms that designers consider when designing an internet?Answer:* Connection-oriented service * Connectionless service22.2 How does the Internet design accommodate heterogeneous networks that each have their own packet format?Answer: To overcome heterogeneity, the Internet Protocol defines a packet format that is independent of the underlying hardware. The result is a universal, virtual packet that can be transferred across the underlying hardware intact. The Internet packet format is not tied directly to any hardware. The underlying hardware does not understand or recognize an Internet packet.22.5 What is the maximum length of an IP datagram?In the current version of the Internet Protocol (IP version 4), a datagram can contain at most 64 K (65535) octets, including the header.22.7 If a datagram contains one 8-bit data value and no header options, what values will be found in header fields H.LEN and TOTAL LENGTH?Answer: H. LEN indicated header in 32-quantities, since no options, then this value will be 5. The TOTAL LENGTH indicated the number of bytes in a datagram including the header. This means 5x4 bytes + 1 (8-bits) = 21 bytesChapter 23 - Support Protocols And Technologies23.1 When a router uses a forwarding table to look up a next-hop address, the result is an IP address. What must happenbefore the datagram can be sent?Answer: Each router along the path uses the destination IP address in the datagram to select a next-hop address, encapsulates the datagram in a hardware frame, and transmits the frame across one network. A crucial step of the forwarding process requires a translation: forwarding uses IP addresses, and a frame transmitted across a physical network must contain the MAC address of the next hop.23.2 What term is used to describe the mapping between a protocol address and a hardware address?Answer: Translation from a computer’s IP address to an equivalent hardware address is known as address resolution, and an IP address is said to be resolved to the correct MAC address. The TCP/IP protocol being used for this is called Address Resolution Protocol (ARP). Address resolution is local to a network.23.5 How many octets does an ARP message occupy when used with IP and Ethernet addresses?Answer: According to Fig 23.3 an ARP message has 7-lines of each being 32-bit (4 bytes or octets), therefore,number of octets in an ARP can be determined as 7x4 = 28 octets23.10 What types of addresses are used in layers below ARP?Answer:ARP forms a conceptual boundary in the protocol stack; layers above ARP use IP addresses, and layers below ARP use MAC addresses.23.17 What is the chief difference between BOOTP and DHCP?Answer:The main difference is that the BOOTP protocol required manual administration. So before a computer could use BOOTP to obtain an address, a network administrator had to configure a BOOTP server to know the computer’s I P address. Chapter 24 - The Future IP (IPv6)24.3 List the major features of IPv6, and give a short description of each.(See Textbook)24.4 How large is the smallest IPv6 datagram header?Answer: IPv6 datagram header consists of a base header + zero or more extension header. Since, smallest header is being asked, we assume zero extension header and consider IPv6 will have only base header. If we look at IPv6 header format in Fig. 24.3, it shows that 10x4 bytes = 40 bytes.Chapter 26 - TCP: Reliable Transport Service26.2 List the features of TCP.(See Textbook)26.6 When using a sliding window of size N, how many packets can be sent without requiring a single ACK to be received?Answer: If the size of the window is N, then it means a sender can transmit up to N packets without waiting for an ACK, as long as other controls are in place.26.9 What is the chief cause of packet delay and loss in the Internet?Answer: The main cause of packet delay and loss in the Internet is congestion.Chapter 28 - Network Performance (QoS and DiffServ)28.1 List and describe the three primary measures of network performance.(See Textbook)28.2 Give five types of delay along with an explanation of each.(See Textbook)Chapter 30 - Network Security30.1 List the major security problems on the Internet, and give a short description of each.(See Textbook)30.2 Name the technique used in security attacks.(See Textbook)30.8 List and describe the eight basic security techniques.(See Textbook)。

HUST Examination Answer SheetCourse: Computer Networks (closed-book exam) Jan.2014 Department of Electronics and Information EngineeringSolution[Question 1] Network architecture(1) Draw the architecture diagrams of ISO/OSI reference model and TCP/IP network. (2) State the differences between the two architectures.(3) Mark the names of data unit and network equipment in the bottom 4 layers of ISO/OSI diagram.Solution:（1）ISO/OSI reference model TCP/IP modelDifferences:a) TCP/IP model is the actual architecture in the Internet, while OSI model is a theoreticalarchitecture.b) TCP/IP model does not imply strict layering, while OSI model implies strict layering. c) In TCP/IP model, IP serves as the focal point for the architecture.d) TCP/IP model allows for arbitrarily many different network technologies , ranging fromEthernet to wireless to single point-to-point links.e) TCP/IP model emphasize implementations of proposed protocols. (2)Physical : bit, repeater or hubData link : frame, Ethernet switch or bridge Internet : packet, routerTransport : message, gateway[Question 2] Principles of network designSelect ONE of the following principles, tell its main ideas and provide an example. (1) Keep it simple and stupid(2) Complex edge and simple core (3) Smart sender and dumb receiverSolution:(1) KISS: Keep It Simple and StupidIt means that you should make things simple in the designing. One example following it : Ethernet(2) Complex edge and simple coreIt means that the hosts are very complex and have many functions while the nodes are very simple and have few functions.One example following it : The design of router, or the functions of TCP and IP(3) Smart sender and dumb receiverIt means that the function of sender is more complex than that of the receiver, which is help to improve the robustness and performance of communication protocol. One example following it: The flow control of TCP protocol[Question 3] Error detection(1) Tell the main idea of error detection and error correction in communication.(2) Given the CRC polynomial x 4 + x 3 + 1, if the original message is 10110011010, what is the CRC message to send?(3) Suppose the first bit of the message in (2) is inverted due to the noise in transmission. How can the receiver detect it via CRC verification?Solution:(1)1087431()M x x x x x x x =+++++, 43()1C x x x =++. So k=4.a) Multiply M(x) by 2kto get 141211875()T x x x x x x x =+++++,b) Then divide T(x) by C(x) to get the remainder 0000.c) The message that should be transmitted is 101100110100000.(2)The message received is 001100110100000.Divide it by C(x), then the remainder is 1100.So it is not divisible by C(x).So the receiver knows that an error has occurred.(1) What are the essential components to realize reliable transmission?(2) Suppose two computers are communicated via Stop-and-wait protocol. The link bandwidth is 5kbps, and the one-way propagation delay is 20ms. To reach 80% or higher link utilization, what is the minimal frame size for this communication?(3) If it is upgraded to Sliding-Window protocol. To reach the same goal with (2), what is the minimal window size, how many bits are required to describe the frame sequence in window? (Suppose the frame size is 1 or 100Byte)Solution:(1)ACK, and timerActual_throughput = Data / Total_DelayTotal_Delay = RTT + Data / BWLink_Utilization = 100 * Actual_throughput / BWSo: Data/throughput = RTT + Data/BW(BW/throughput - 1) Data = BW*RTTData = BW*RTT/(1/Utilization -1)For the stop-and-wait protocol, for each RTT only one frame is sent,Thus Frame_size = Data = BW*RTT/(1/Utilization - 1) = 5kbps * 40ms / (1/0.8 - 1)= 200 bit / 0.25 = 800 bit = 100 ByteIf the students ignore the data transmission delay, their answer isframe_size = BW*RTT*Utilization = 5000 bit/sec * 40 / 1000 sec * 0.8 = 160 bit = 20 ByteIn this case, at least -2 score.(3)For the sliding-window protocol, in each RTT the data window can be transmitted at most.Thus Window_size = Data = BW*RTT/(1/Utilization - 1) = 5kbps * 40ms / (1/0.8 - 1)= 200 bit / 0.25 = 800 bit = 100 ByteIf the frame size is 100Byte, 1 frames are allowed.To indicate the frames in both sides of sender and receiver, the sequence number should describe 2 frames, thus the [log2(2) ]=1 bitIf the students ignore the data transmission delay, their answer iswindow_size = BW*RTT*Utilization = 5000 bit/sec * 40 / 1000 sec * 0.8 = 160 bit = 20 ByteIf the frame size is 100Byte, 1 frames are allowed.To indicate the frames in both sides of sender and receiver, the sequence number should describe 2 frames, thus the [log2(2) ]=1 bitIn this case, at least -2 score.(1) What is the main idea of CSMA/CD (Carrier Sense Multiple Access/Collision Detection) in traditional Ethernet? Can it be deployed for wireless local network, why?(2) Suppose one traditional Ethernet has 1km cable, the signal propagation speed is 2*105km/s, and the transmission rate is designed to be 100Mbps. What is the minimal frame size to support carrier sensing?Solution:(1) the main idea of CSMA/CD include two parts. One is carrier sensing, which means the node should detect the channel before sending any data. If the node finds the channel is idle, it begins to transmit. Otherwise, the node stop for next round. The second issues is collision detection, which means the node should detect the channel in the duration of its data transmission. If the node finds the channel become busy, or in other word, there is a collision, it should stop transmitting immediately.CSMA/CD cannot be deployed in wireless LAN, because the wireless radio transmitter and receiver can not work in dual mode. The wireless node cannot detect collision when it is transmitting data.(2)the minimal frame should be transmitted throughout the whole traditional Ethernet.Thus t = frame_size / transmit_rate = 2 * cable_length / prop_speed.Frame_size = 100 * 106 bits/sec * 2 * 103 m / (2 * 108)m/sec = 1000 bits = 125 Byte[Question 6] Switched network(1) What are the differences between circuit switching and packet switching?(2) For the following linear topology network, each link has 2ms propagation delay and 4 Mbps bandwidth.A B C DIf we use circuit switching, circuit setup requires a 1KB message to make one round-trip on the path, which incurs a 1ms delay at each switch after the message has been completely received. Then we can send the file as one contiguous bit stream. What is the delay for circuit switching to transmit n-byte from A to D?(3) If we use packet switching in the network of (2), we can break the file into 1KB packets, which has 24byte header and 1000byte payload. The switch takes 1ms process delay after receiving the packet, and then sent it continuously. What is the condition for packet switching to have less delay performance than circuit switching?Solution:(1)(2)T pkt = Packet_Size / Bandwidth = 1 KB / 4Mbps = 1024 *8 / (4 * 106) = 2.048 ms T p = 2 ms, T s = 1 ms,T t = n B / 4MbpsIn circuit switching, Total time duration:D = Singling_Delay + Transmission_Delay = 2 * Packet_Duration + Transmission_Delay = 2 * (T pkt * 2 + T p * 3 + T s * 2 + T pkt ) + (T p * 3 + T t ) = T pkt * 6 + T p * 9 + T s * 4 + T tThus, D = 2.048 * 6 + 2 * 9 + 1 * 4 + n * 8 bits / 4Mbps = 34.288 (ms) + 2n (us)(3)Main ideaThe reliability provided by endhost The reliability provided by thenetwork Information in packet Every packet has its dest-addrin header Every packet has its temp VCIlocally Forwarding action inswitch Every packet was treatedindependentlyThe packets are processed inthe manner of VCPackets received indestinationNot in sequenceIn sequenceT pkt = Packet_Size / Bandwidth = 1 KB / 4Mbps = 1024 *8 / (4 * 106) = 2.048 msT p = 2 ms,T s = 1 ms,T t = 1.024 * n B / 4MbpsIn packet switching, Total time duration:D = Delay_at_Switch * Switch_Num + Delay_at_last_hop= (T p + T pkt +T s)*2 + (T p + T t )= T pkt * 2 + T p * 3 + T s * 2 + T tThus, D = 2.048 * 2 + 2 * 3 + 1 * 2 + 1.024 * n B / 4Mbps= 12.096 (ms) + 2.048 n (us)In order to make34.288 (ms) + 2n (us) > 12.096 (ms) + 2.048 n (us)Thus 22.192(ms) > 0.048 n(us)n < 22.192 * 1000 / 0.048 = 462333 Byte = 451.5 KB[Question 7] Ethernet Switch(1) What are the differences between hub and switch in Ethernet?(2) Suppose one server and nine clients are connected via hub in 10Mbps Ethernet, what is the maximal bandwidth for the client-server connection?(3) If the hub is upgraded to switch, can the client-server connection obtain more bandwidth? if can, how much is it?Solution:(1)hub works in Layer2, it works as the shared media and relays the frames to all the nodes connecting to hub. Switch works in Layer2, it forwards the frames to the specific node according to the destination address embedded in the frame header.(2)the max bandwidth in client-server connection is 10Mbps/(1+9)= 1Mbps(3)the max bandwidth in client-server connection is 10Mbps/9 = 1.1111MbpsAdditional 0.1111 Mbps is obtained for each connection.[Question 8] Router(1) Somebody says that, ``the only difference between switch and router is that they do switch function based on the address in different layers.’’ Is it correct? Why?(2) If we obtain the following information from one router. What kind of routing protocol does itSolution:(1)it is not fully correct. The part talking about the forwarding function is correct, while it is not the only difference. Another but not the last difference is that, router has more functions on control plane, which do routing and find the paths for packets.(2)The routing protocol is RIP.The routing table is:Destination Next hop Interface192.168.1.0/24 * Fa 0/4192.168.2.0/24 * Fa 0/6192.168.10.0/24 192.168.1.1 Fa 0/4192.168.30.0/24 192.168.2.2 Fa 0/6[Question 9] Routing algorithm(1) State the main differences between distance vector routing and link state routing.(2) For the network given in the following figure, provide the steps of forward search in Dijkstra algorithm for node A finding the shortest path to node ESolution:(1) The idea of distance vector routing is to tell the neighbors about the learned topology to all the nodes in the network. The idea of link state routing is to tell all the nodes in network about the neighborhood topology.Step 1 2 3 4 5 6 7 8 9Confirmed (A,0,-) (A,0,-) (A,0,-)(D,2,D)(A,0,-)(D,2,D)(A,0,-)(D,2,D)(B,4,D)(A,0,-)(D,2,D)(B,4,D)(A,0,-)(D,2,D)(B,4,D)(E,6,D)(A,0,-)(D,2,D)(B,4,D)(E,6,D)(A,0,-)(D,2,D)(B,4,D)(E,6,D)(C,7,D)Tentative (B,5,B)(D,2,D)(B,5,B) (B,4,D)(E,7,D)(E,7,D)(E,6,D)(C,8,D)(C,8,D) (C,7,D)A→E A→D→EA→D→B→EA→D→B→E[Question 10] IP address allocationA company has one C class IP address of 200.1.1.*. It has four departments.(1) If each department has less than 25 computers. Provide a kind of IP address allocation. Give the network address, subnet mask, and the available IP address range for each department. (2) If the four departments have 72, 35, 34, 20 computers respectively. Provide the IP address allocation scheme again.Solution:(1)Since each department has less than 25 computers, even considering the additional two more IP address for gateway and broadcast, the 64-computer subnet is enough for them.One IP address allocation scheme is to even divide the 256 IP addresses into 4 subnets, each subnet allows 64 hosts.Another IP address allocation scheme is to even divide the 256 IP addresses into 4 subnets, each subnet allows 32 hosts.(2) Since one of the department has more requirements than 64, then the even distribution scheme[Question 11] TCP protocol(1) Somebody says that, ``because of the reliable transmission service in layer 2, there is no need to provide such service again within TCP protocol in layer 4’’. Is it correct? Why?(2) State the main rules of TCP connection setup according to the following figure. Explain every word and number in the figure.Solution:(1) It is wrong. TCP is based on the un-reliable IP layer, which only provides best effort service. If TCP wants to provide reliable transmission service, it has to realize this by itself.(2)TCP use three hand-shakes to setup the connection.According to the figure, there are two nodes. The sender is with IP address of 192.168.1.163 and the receiver 192.168.1.165 respectively.●At first, the sender send a request ``SYN’’to the receiver to setup the connection. Thismessage is with the sequence number of 424CF1DC;●Secondly, the receiver reply an acknowledgment message ``SYN/ACK’’ to the sender. Thismessage has two sequence numbers. The seq in the ACK is 424CF1DD, which is to confirm the last ``SYN’’ from sender. The seq in the SYN is 30318555, which is a new message from the receiver.●Thirdly, the sender reply an acknowledgment message ``ACK’’ to the receiver. The seq in theACK is 30318556, which is to confirm the last ``SYN’’ from receiver.In the end, both the sender and receiver knows that the other side is ready for this TCP connection.[Question 12] TCP congestion control(1) Both flow control and congestion control in TCP are realized by window based packet control. How can TCP get the window sizes in these two mechanisms?(2) Assume that TCP implements an extension that allows window sizes much larger than 64 KB. Suppose that you are using this extended TCP over a 1Gbps link with a latency of 150ms. TCP packet size is 1KB, and the max receive window is 1 MB. Suppose there is no real congestion and packet loss in transmission. How many RTTs does it take until slow start opens the send window to 1 MB? How long does it take to send the complete file? ( Suppose file size is 10MB )Solution:(1) In flow control, TCP sender knows the window size by the field of advertise-window replied from the receiver. In congestion control, TCP sender learns the window size adaptively by AIMD( Additive Increase and Multiplicative Decrease) mechanism responding to the packet lossevent.(2)When TCP realizes congestion control mechanism, its effective send window size will be min (CongestionWindow, AdvertizedWindow). In original design of TCP header, the field of AdvertizedWindow is 16 bit, which is 216=26*210=64 KB. So the maximum effective window of original TCP sender is 64KB. The assumption of the first sentence in this question relaxes such constraint for TCP.In slow start, the send window starts from w0=1 packet, which is 1 KB. For each RTT after a successful transmission, the window size will be doubled. After i RTT, it will be 2i * w0. Let 2i * 1KB = 1MB, soi = log2(1MB/1KB) = log2(210) = 10.It will take 10 RTTs to reach 1MB send window.Case 1: if the receiver window remains as 1MBIn the first 10 RTTs, total (1 + 2 + 4 + … + 210) * 1KB has been transmitted.Which is (211 - 1) * 1KB = 2 MB - 1 KB, the rest file is 10MB - (2MB - 1KB) = 8MB + 1KBIn the reset transmission, each RTT can only support 1MB transmission.Thus, additional 9 RTTs are required. Total 19 RTT = 19 * 150ms = 2.85 sCase 2: if the receiver window can be changed.Since there is no congestion and loss, the maximum send window will be the bandwidth * delay for this TCP connection.w max= 1G bps * 150ms = 109 * 150 * 10-3= 150 *106 bit = 18.75 * 106 byte = 17.88 MB.So, this 10 MB file can be transferred before reaching w max. In another word, it can be sent in its slow start phase. Assume x RTT is required to send this file, then:(1 + 2 + 4 + … + 2x) * 1KB ≥ 10 MB2 * 2x–1 ≥ 10 * 1024x ≥ log2(10241) – 1 = 12.3Thus x = 13, it will take 13 RTTs to transfer this file. 13RTT = = sTotal delay = 13RTT + Filesize/BW = 13 * 150ms+ 10MB / 1Gbps =1.95 + 0.08 * 1.024^2=2.03 s11。

Chapter 1 Computer Networks and the Internet1．The ( ) is a worldwide computernetwork, that is, a network that interconnectsmillionsofcomputing devices throughout theworld. ppt3 A public InternetB IntranetC switch netD television net2．Which kind of media is not a guidedmedia ( )A twisted-pair copper wireB a coaxial cableC fiber opticsD digital satellite channel3．Which kind of media is a guided media ( )A geostationary satelliteB low-altitude satelliteC fiber opticsD wireless LAN4．The units of data exchanged by a link-layer protocol are called ( ).A FramesB SegmentsC DatagramsD bit streams5．Which of the following optionbelongs to the circuit-switchednetworks ( )A FDMB TDMC VCnetworksD both A and B 6．( )makes sure that neither sideof a connection overwhelms theother side by sending too manypackets too fast.A Reliable data transferB Flow controlC Congestion controlD Handshaking procedure7．( ) means that the switch mustreceive the entire packet before it can begin to transmit the first bit of the packet onto the outbound link.A Store-and-forward transmissionB FDMC End-to-end connectionD TDM8．Datagramnetworksandvirtual-circuit networks differ in that ( ).A datagram networks are circuit-switched networks, and virtual-circuit networks are packet-switched networks.B datagram networks are packet-switched networks, and virtual-circuit networks are circuit-switched networks.C datagram networks use destination addresses and virtual-circuit networks use VC. numbers to forward packets toward their destination.D datagram networks use VC. numbers and virtual-circuit networks use destination addresses to forward packetstoward their destination.9．In the following options, which one is not a guided media ( )A twisted-pairwireB fiber opticsC coaxial cableD satellite10．Processing delay does not include the time to ( ).A examine the packet ’s headerB wait to transmit the packet onto the linkC determine where to direct thepacketD check bit-error in the packet 11．In the following four descriptions, which one is correct ( )A The traffic intensity must begreater than 1.B The fraction of lost packetsincreases as the trafficintensity decreases.C If the traffic intensity isclose to zero, the averagequeuing delay will be close tozero.D If the traffic intensity isclose to one, the averagequeuing delay will be close toone.12．The Internet’s network layer is responsible for movingnetwork-layer packets known as( ) from one host to another.A frameB datagramC segmentD message13．The protocols of various layersare called ( ).A the protocol stackB TCP/IPC ISPD network protocol14．There are two classes ofpacket-switched networks: ( )networks and virtual-circuitnetworks.A datagramB circuit-switchedC televisionD telephone15．Access networks can be looselyclassified into threecategories: residential access,company access and ( ) access.A cabledB wirelessC campusD city areaQuestion 16～17Suppose, a is the average rate at which packets arrive at the queue,R is the transmission rate, and all packets consist of L bits, then thetraffic intensity is ( 16 ), and it should no greater than ( 17 ).16． A LR ／aB La ／RC Ra ／LD LR ／a 17．A 2B 1C 0D -118．In the Internet, the equivalentconcept to end systems is ( ).A hostsB serversC clientsD routers19．In the Internet, end systems are connected together by ( ).A copper wireB coaxial cableC communication linksD fiber optics 20．End systems access to the Internet through its ( ). A modemsB protocolsC ISPD sockets21．End systems, packet switches,and other pieces of the Internet, run ( ) that control thesending and receiving ofinformation within theInternet.A programsB processesC applicationsD protocols22．There are many private networks,such as many corporate andgovernment networks, whosehosts cannot exchange messages with hosts outside of the private network. These private networks are often referred to as ( ).A internetsB LANC intranetsD WAN23．The internet allows ( ) runningon its end systems to exchangedata with each other.A clients applicationsB server applicationsC P2P applicationsD distributed applications24．The Internet provides two services to its distributed applications: a connectionlessunreliable service and ()service.A flowcontrolB connection-oriented reliableC congestion controlD TCP25．It defines the format and theorder of messages exchangedbetween two or morecommunicating entities, as wellas the actions taken on thetransmission and/or receipt ofa message or other event. Thesentence describes ( ).A InternetB protocolC intranetD network26．In the following options, which does not define in protocol ( )A the format of messagesexchanged between two or morecommunicating entitiesB the order of messagesexchanged between two or morecommunicating entitiesC the actions taken on thetransmission of a message orother eventD the transmission signals aredigital signals or analogsignals27．In the following options, which is defined in protocol ( )A the actions taken on thetransmission and/or receiptof a message or other event B the objects exchanged between communicating entities C the content in the exchangedmessagesD the location of the hosts28．In the following options, whichdoes not belong to the network edge( )A end systemsB routersC clientsD servers29．In the following options, whichbelongs to the network core ( )A end systemsB routersC clientsD servers30．In the following options, whichis not the bundled with theInternet’sconnection-oriented service( )A reliable data transferB guarantee of thetransmission timeC flow controlD congestion-control31．An application can rely on theconnection to deliver all of itsdata without error and in theproper order. The sentencedescribes ( ). A flow controlB congestion-controlC reliable datatransferD connection-oriented service 32．It makes sure that neither sideof a connection overwhelms theother side by sending too manypackets too fast. The sentence describes ( ). A flow controlB congestion-controlC connection-oriented serviceD reliable data transfer33．It helps prevent the Internetfrom entering a state of gridlock. When a packet switchbecomes congested, its bufferscan overflow and packet loss canoccur. The sentence describes( ).A flowcontrolB congestion-controlC connection-oriented serviceD reliable data transfer 34．TheInternet ’sconnection-oriented service has aname, it is ( ).A TCPB UDPC TCP/IPD IP35．In the following options, which service does not be provided to an application by TCP( )A reliable transportB flow controlC video conferencingD congestion control36．The Internet ’s connectionless service is called ( ).A TCPB UDPC TCP/IPD IP37．In the following options, which does not use TCP( )A SMTPB internet telephoneC FTPD HTTP38．In the following options, which does not use UDP( )A InternetphoneB video conferencingC streamingmultimediaD telnet39．There are two fundamentalapproaches to building a network core, ( ) and packet switching.A electrical current switchingB circuit switchingC data switchingD message switching40．In ( ) networks, the resourcesneeded along a path to provide for communication between the end system are reserved for theduration of the communicationsession.A packet-switchedB data-switchedC circuit-switchedD message-switched41．In ( ) networks, the resources are not reserved; a session’smessages use the resources ondemand, and as a consequence,may have to wait for access tocommunication link.A packet-switchedB data-switchedC circuit-switchedD message-switched42．In a circuit-switched network, if each link has n circuits, foreach link used by the end-to-endconnection, the connection gets( ) of the link’s bandwidthfor the duration of the connection.A a fraction 1/nB allC 1/2D n times43．For ( ), the transmission rateof a circuit is equal to theframe rate multiplied by thenumber of bits in a slot.A CDMAB packet-switched networkC TDMD FDM44．( ) means that the switch mustreceive the entire packetbefore it can begin to transmitthe first bit of the packet ontothe outbound link.A Queuing delayB Store-and-forwardtransmissionC Packet lossD Propagation45．The network that forwardspackets according to host destination addresses is called( ) network.A circuit-switchedB packet-switchedC virtual-circuitD datagram46．The network that forwardspackets according tovirtual-circuit numbers iscalled ( ) network.A circuit-switchedB packet-switchedC virtual-circuitD datagram47．In the following entries, whichis not a kind of access network( )A residentialaccessB company accessC wirelessaccessD local access48．Suppose there is exactly onepacket switch between a sendinghost and a receiving host. The transmission rates between thesending host and the switch and between the switch and the receiving host are R 1 and R 2,respectively. Assuming that theswitch uses store-and-forwardpacket switching, what is thetotal end-to-end delay to senda packet of length L (Ignorequeuing delay, propagationdelay, and processing delay.) ( ) A L /R 1＋L /R 2B L /R 1C L /R 2D none of the above49．The time required to examine thepacket ’s header and determinewhere to direct the packet ispart of the ( ). A queuing delayB processing delayC propagation delayD transmission delay50．The time required to propagatefrom the beginning of the linkto the next router is ( ). A queuing delayB processing delayC propagation delayD transmission delay51．Consider sending a packet of3000bits over a path of 5 links. Eachlink transmits at 1000bps. Queuingdelays, propagation delay and processing delay are negligible. (6 points)(1).Suppose the network is apacket-switched virtual circuitnetwork. VC setup time is seconds.Suppose the sending layers add atotal of 500 bits of header to eachpacket. How long does it take to sendthe file from source to destination(2).Suppose the network is a packet-switched datagram networkand a connectionless service is used.Now suppose each packet has 200 bitsof header. How long does it take tosend the file(3).Suppose that the network is acircuit-switched network. Furthersuppose that the transmission rate of the circuit between source anddestination is 200bps. Assumingsetup time and 200 bits of header appended to the packet, how longdoes it take to send the packetS olution:(1).t=5*(3000+500)/1000+=( 2).t=5*(3000+200)/1000=16s( 3).t=(3000+200)/200+=。

Chapter 1 Computer Networks and the Internet1．The ( ) is a worldwide computer network, that is, a network that interconnects millions of computing devices throughout the world.ppt3A public InternetB IntranetC switch netD television net2．Which kind of media is not a guided media? ( )A twisted-pair copper wireB a coaxial cableC fiber opticsD digital satellite channel3．Which kind of media is a guided media? ( )A geostationary satelliteB low-altitude satelliteC fiber opticsD wireless LAN4．The units of data exchanged by a link-layer protocol are called ( ).A FramesB SegmentsC DatagramsD bit streams5．Which of the following option belongs to the circuit-switched networks? ( )A FDMB TDMC VC networksD both A and B6．( )makes sure that neither side of a connection overwhelms the other side by sending too many packets too fast.A Reliable data transferB Flow controlC Congestion controlD Handshaking procedure7．( ) means that the switch must receive the entire packet before it can begin to transmit the first bit of the packet onto the outbound link.A Store-and-forward transmissionB FDMC End-to-end connectionD TDM8．Datagram networks and virtual-circuit networks differ in that ( ).A datagram networks are circuit-switched networks, andvirtual-circuit networks are packet-switched networks.B datagram networks are packet-switched networks, andvirtual-circuit networks are circuit-switched networks.C datagram networks use destination addresses and virtual-circuitnetworks use VC. numbers to forward packets toward theirdestination.D datagram networks use VC. numbers and virtual-circuit networksuse destination addresses to forward packets toward their destination. 9．In the following options, which one is not a guided media? ( )A twisted-pair wireB fiber opticsC coaxial cableD satellite10．Processing delay does not include the time to ( ).A examine the packet’s headerB wait to transmit the packet onto the linkC determine where to direct the packetD check bit-error in the packet11．In the following four descriptions, which one is correct? ( )A The traffic intensity must be greater than 1.B The fraction of lost packets increases as the traffic intensitydecreases.C If the traffic intensity is close to zero, the average queuing delaywill be close to zero.D If the traffic intensity is close to one, the average queuing delaywill be close to one.12．The Internet’s network layer is responsible for moving network-layer packets known as ( ) from one host to another.A frameB datagramC segmentD message13．The protocols of various layers are called ( ).A the protocol stackB TCP/IPC ISPD network protocol14．There are two classes of packet-switched networks: ( ) networks and virtual-circuit networks.A datagramB circuit-switchedC televisionD telephone15．Access networks can be loosely classified into three categories: residential access, company access and ( ) access.A cabledB wirelessC campusD city areaQuestion 16～17Suppose, a is the average rate at which packets arrive at the queue, R is the transmission rate, and all packets consist of L bits, then the traffic intensity is ( 16 ), and it should no greater than ( 17 ).16． A LR／aB La／RC Ra／LD LR／a17．A 2B 1C 0D -118．In the Internet, the equivalent concept to end systems is ( ).A hostsB serversC clientsD routers19．In the Internet, end systems are connected together by ( ).A copper wireB coaxial cableC communication linksD fiber optics20．End systems access to the Internet through its ( ).A modemsB protocolsC ISPD sockets21．End systems, packet switches, and other pieces of the Internet, run ( ) that control the sending and receiving of information within theInternet.A programsB processesC applicationsD protocols22．There are many private networks, such as many corporate and government networks, whose hosts cannot exchange messages withhosts outside of the private network. These private networks are often referred to as ( ).A internetsB LANC intranetsD WAN23．The internet allows ( ) running on its end systems to exchange data with each other.A clients applicationsB server applicationsC P2P applicationsD distributed applications24．The Internet provides two services to its distributed applications: a connectionless unreliable service and () service.A flow controlB connection-oriented reliableC congestion controlD TCP25．It defines the format and the order of messages exchanged between two or more communicating entities, as well as the actions taken on thetransmission and/or receipt of a message or other event. The sentence describes ( ).A InternetB protocolC intranetD network26．In the following options, which does not define in protocol? ( )A the format of messages exchanged between two or morecommunicating entitiesB the order of messages exchanged between two or morecommunicating entitiesC the actions taken on the transmission of a message or other eventD the transmission signals are digital signals or analog signals 27．In the following options, which is defined in protocol? ( )A the actions taken on the transmission and/or receipt of a message orother eventB the objects exchanged between communicating entitiesC the content in the exchanged messagesD the location of the hosts28．In the following options, which does not belong to the network edge? ( )A end systemsB routersC clientsD servers29．In the following options, which belongs to the network core? ( )A end systemsB routersC clientsD servers30．In the following options, which is not the bundled with the Internet’s connection-oriented service? ( )A reliable data transferB guarantee of the transmission timeC flow controlD congestion-control31．An application can rely on the connection to deliver all of its data without error and in the proper order. The sentence describes ( ).A flow controlB congestion-controlC reliable data transferD connection-oriented service32．It makes sure that neither side of a connection overwhelms the other side by sending too many packets too fast. The sentence describes ( ).A flow controlB congestion-controlC connection-oriented serviceD reliable data transfer33．It helps prevent the Internet from entering a state of gridlock. When a packet switch becomes congested, its buffers can overflow and packet loss can occur. The sentence describes ( ).A flow controlB congestion-controlC connection-oriented serviceD reliable data transfer34．The Internet’s connection-oriented service has a name, it is ( ).A TCPB UDPC TCP/IPD IP35．In the following options, which service does not be provided to an application by TCP?( )A reliable transportB flow controlC video conferencingD congestion control36．The Internet’s connectionless service is called ( ).A TCPB UDPC TCP/IPD IP37．In the following options, which does not use TCP?( )A SMTPB internet telephoneC FTPD HTTP 38．In the following options, which does not use UDP?( )A Internet phoneB video conferencingC streaming multimediaD telnet39．There are two fundamental approaches to building a network core, ( ) and packet switching.A electrical current switchingB circuit switchingC data switchingD message switching40．In ( ) networks, the resources needed along a path to provide for communication between the end system are reserved for the duration of the communication session.A packet-switchedB data-switchedC circuit-switchedD message-switched41．In ( ) networks, the resources are not reserved; a session’s messages use the resources on demand, and as a consequence, may have to wait for access to communication link.A packet-switchedB data-switchedC circuit-switchedD message-switched42．In a circuit-switched network, if each link has n circuits, for each link used by the end-to-end connection, the connection gets ( ) of thelink’s bandwidth for the duration of the connection.A a fraction 1/nB allC 1/2D n times43．For ( ), the transmission rate of a circuit is equal to the frame rate multiplied by the number of bits in a slot.A CDMAB packet-switched networkC TDMD FDM44．( ) means that the switch must receive the entire packet before it can begin to transmit the first bit of the packet onto the outbound link.A Queuing delayB Store-and-forward transmissionC Packet lossD Propagation45．The network that forwards packets according to host destination addresses is called ( ) network.A circuit-switchedB packet-switchedC virtual-circuitD datagram46．The network that forwards packets according to virtual-circuit numbers is called ( ) network.A circuit-switchedB packet-switchedC virtual-circuitD datagram47．In the following entries, which is not a kind of access network?( )A residential accessB company accessC wireless accessD local access48．Suppose there is exactly one packet switch between a sending host and a receiving host. The transmission rates between the sending host and the switch and between the switch and the receiving host are R1 and R2,respectively. Assuming that the switch uses store-and-forward packetswitching, what is the total end-to-end delay to send a packet of length L? (Ignore queuing delay, propagation delay, and processing delay.)( )A L/R1＋L/R2B L/R1C L/R2D none of the above49．The time required to examine the packet’s header and determine where to direct the packet is part of the ( ).A queuing delayB processing delayC propagation delayD transmission delay50．The time required to propagate from the beginning of the link to the next router is ( ).A queuing delayB processing delayC propagation delayD transmission delay51．Consider sending a packet of 3000bits over a path of 5 links. Each link transmits at 1000bps. Queuing delays, propagation delay and processing delay are negligible. (6 points)(1).Suppose the network is a packet-switched virtual circuit network. VC setup time is 0.1 seconds. Suppose the sending layers add a total of 500 bits of header to each packet. How long does it take to send the file from source to destination?(2).Suppose the network is a packet-switched datagram network and a connectionless service is used. Now suppose each packet has 200 bits of header. How long does it take to send the file?(3).Suppose that the network is a circuit-switched network. Further suppose that the transmission rate of the circuit between source and destination is 200bps. Assuming 0.02s setup time and 200 bits of header appended to the packet, how long does it take to send the packet?Solution: (1). t=5*(3000+500)/1 000+0.1=17.6s(2). t=5*(3000+200)/1 000=16s(3). t=(3000+200)/200+0.02=16.02s 相对压力一般指表压，是测量系统相对于大气压的压力值。

ANDREW S. TANENBAUM 秒，约533 msec.----- COMPUTER NETWORKS FOURTH EDITION PROBLEM SOLUTIONS 8. A collection of five routers is to be conn ected in a poi nt-to-poi nt sub net.Collected and Modified By Yan Zhe nXing, Mail To: Betwee n each pair of routers, the desig ners may put a high-speed line, aClassify: E aEasy, M ^Middle, H Hard , DaDeleteGree n: Importa nt Red: Master Blue: VI Others:Know Grey:—Unnecessary ----------------------------------------------------------------------------------------------ML V Chapter 1 In troductio nProblems2. An alter native to a LAN is simply a big timeshari ng system with termi nals forall users. Give two adva ntages of a clie nt-server system using a LAN.(M)使用局域网模型可以容易地增加节点。

如果局域网只是一条长的电缆，且不会因个别的失效而崩溃(例如采用镜像服务-------------------------------------------器)的情况下，使用局域网模型会更便宜。