1、Computer NetworkingFinal ReviewJinghui QinContentComputer Networks and the InternetApplication Layer Transport LayerThe Network LayerThe Link and Local Area NetworksWireless and Mobile NetworksComputer Networks and the InternetSome basic concepts protocol circuit switching packet switching etc.Delay
2、,Loss and Throughput in Packet-Switched Networks processing delay queuing delay transmission delayComputer Networks and the Internet propagation delay end-to-end delay computation throughput:instantaneous,averageProtocol Layers and Their Service Models Seven-layer ISO OSI reference model Five-layer
3、Internet protocol stack The function of each layerComputer Networks and the InternetProblem 1-P5Computer Networks and the InternetProblem 1-P5Computer Networks and the InternetSolution 1-P5Computer Networks and the InternetProblem 2-P17Computer Networks and the InternetSolution 2-P17Throughput=minRs
4、,Rc,R/MApplication Layer Some basic concepts client server socketTransport Services Provided by Internet TCP services:connection-oriented&reliable data transfer UDP services Application-Layer ProtocolApplication Layer HTTP Non-persistent and Persistent HTTP Message Format Request Message Response Me
5、ssage CookiesApplication Layer FTP two parallel connections out-of-band&in-bande-mail SMTP POP3 IMAPApplication Layer DNS Service provided by DNS DNS Working Mechanism DNS Records and MessageApplication Layer Problem-P2Application Layer Problem-P2Application Layer Problem-P2Application Layer Solutio
6、n-P2a)The document request was http:/gaia.cs.umass.edu/cs453/index.html.The Host:field indicates the servers name and/cs453/index.html indicates the file name.b)The browser is running HTTP version 1.1,as indicated just before hte first pair.c)The browser is requesting a persistent connection,as indi
7、cated by the Connection:keep-alive.Application Layer Solution-P2d)This is a trick question.This information is not contained in an HTTP message anywhere.So there is no way to tell this from looking at the exchange of HTTP messages alone.One would need information from the IP datagrams(that carried t
8、he TCP segment that carried the HTTP GET request)to answer this question.Transport LayerUDP UDP Segment Structure UDP Checksum The advantages of UDPPrinciples of Reliable Data Transfer SR GBNTransport LayerTCP TCP Segment Structure TCP flow control TCP congest control AIMD Slow start Reaction To Tim
9、eout EventsTransport LayerProblem 1-P23Consider transferring an enormous file of L bytes from Host A to Host B.Assume an MSS of 1,460 bytes.a.What is the maximum value of L such that TCP sequence numbers are not exhausted?Recall that the TCP sequence number field has 4 bytes.Transport LayerProblem 1
10、-P23b.For the L you obtain in(a),find how long it takes to transmit the file.Assume that a total of 66 bytes of transport,network,and data-link header are added to each segment before the resulting packet is sent out over a 100 Mbps link.Ignore flow control and congestion control so A can pump out t
11、he segments back to back and continuously.Transport LayerSolution 1-P23Transport LayerSolution 1-P23Transport LayerProblem 2-P34Transport LayerProblem 2-P34Transport LayerProblem 2-P34Transport LayerSolution 2-P34a)TCP slowstart is operating in the intervals 1,6 and 23,26b)TCP congestion advoidance
12、is operating in the intervals 6,16 and 17,22c)After the 16th transmission round,packet loss is recognized by a triple duplicate ACK.If there was a timeout,the congestion window size would have dropped to 1.Transport LayerSolution 2-P34d)After the 22nd transmission round,segment loss is detected due
13、to timeout,and hence the congestion window size is set to 1.e)The threshold is initially 32,since it is at this window size that slowtart stops and congestion avoidance begins.f)The threshold is set to half the value of the congestion window when packet loss is detected.When loss is detected during
14、transmission round 16,the congestion windows size is 42.Hence the threshold is 21 during the 18th transmission round.Transport LayerSolution 2-P34g)The threshold is set to half the value of the congestion window when packet loss is detected.When loss is detected during transmission round 22,the cong
15、estion windows size is 26.Hence the threshold is 13 during the 24th transmission round.i)The congestion window and threshold will be set to half the current value of the congestion window(8)when the loss occurred.Thus the new values of the threshold and window will be 4.Transport LayerSolution 2-P34
16、H)During the 1st transmission round,packet 1 is sent;packet 2-3 are sent in the 2nd transmission round;packets 4-7 are sent in the 3rd transmission round;packets 8-15 are sent in the 4th transmission round;packets15-31 are sent in the 5th transmission round;packets 32-63 are sent in the 6th transmis
17、sion round;packets 64 96 are sent in the 7th transmission round.Thus packet 70 is sent in the 7th transmission round.The Network LayerVirtual-Circuit NetworksDatagram Networks longest prefix matchRouter FabricIP MTU、Fragment CIDR DHCP NATThe Network LayerRouting Algorithms Link-State(LS)、Dijkstra Di
18、stance-Vector(DV),Bellman-FordIntra-AS Routing in the Internet RIP OSPFInter-AS Routing BGPBroadcast and Multicast RoutingThe Network LayerProblem 1-P8Consider a datagram network using 32-bit host addresses.Suppose a router have five links,numbered 0 through 4,and packets are to be forwarded to the
19、link interfaces as follows:The Network LayerProblem 1-P8The Network LayerProblem 1-P8a.Provide a forwarding table that has five entries,uses longest prefix matching,and forwards packets to the correct link interfaces.b.Describe how your forwarding table determines the appropriate link interface for
20、datagrams with destination addresses:11001000 10010001 01010001 0101010111100000 10101101 11000011 0011110011100001 10000000 00010001 01110111The Network LayerSolution 1-P8a.Prefix MatchLink Interface 11100000 00000000011100000 00000001 111100000 2111000013otherwise4The Network LayerSolution 1-P8b.P
21、refix match for first address is 5th entry:link interface 4 Prefix match for second address is 3rd entry:link interface 2 Prefix match for third address is 4th entry:link interface 3 The Network LayerProblem 2-P22Consider the following network.With the indicated link costs,use Dijkstras shorest-path
22、 algorithm to compute the shortest path from x to all network nodes.Show how the algorithm works by computing a table similar to Table 4.3The Network LayerProblem 2-P22The Network LayerSolution 2-P22The Link and Local Area NetworksError-Detection and-Correction Parity Checks Checksumming Methods Cyc
23、lic Redundancy Check(CRC)MAC AddressesAddress Resolution Protocol(ARP)CSMA/CDThe Link and Local Area NetworksProblem 1-P5Consider the 4-bit generator,G,shown in Figure 5.8,and suppose that D has the value 11111010.What is the value of R?The Link and Local Area NetworksSolution 1-P5r=3,G=1001,代入公式：代入
24、公式：R=（D*2r）/G 的余数的余数11111010000除以除以1001，商，商11100110，余，余R=110The Link and Local Area NetworksProblem 2-P12Consider three LANs interconnected by two routers,as shown in Figure 5.38.The Link and Local Area NetworksProblem 2-P12a.Redraw the diagram to include adapters.b.Assign IP addresses to all of the
25、 interfaces.For Subnet 1 use addresses of the form 111.111.111.xxx;for Subnet 2 uses addresses of the form 122.122.122.xxx;and for Subnet 3 use addresses of the form 133.133.133.xxx.c.Assign MAC addresses to all of the adaptersThe Link and Local Area NetworksProblem 2-P12d.Consider sending an IP dat
26、agram from Host A to Host F.Suppose all of the ARP tables are up to date.Enumerate all the steps,as done for the single-router example in Section 5.4.2.e.Repeat(d),now assuming that the ARP table in the sending host is empty(and the other tables are up to date).The Link and Local Area NetworksSoluti
27、on 2-P12RouterRouterLANLANLANABCDFE111.111.111.00100-00-00-00-00-00111.111.111.00311-11-11-11-11-11111.111.111.00222-22-22-22-22-22122.222.222.00233-33-33-33-33-33122.222.222.00144-44-44-44-44-44122.222.222.00466-66-66-66-66-66122.222.222.00355-55-55-55-55-55133.333.333.00288-88-88-88-88-88133.333.3
28、33.00177-77-77-77-77-77133.333.333.00399-99-99-99-99-99The Link and Local Area NetworksSolution 2-P12d）1.A的路由表决定数据报将路由到的路由表决定数据报将路由到111.111.111.0022.A的网络适配器使用目的的网络适配器使用目的MAC地址地址22-22-22-22-22-22封装以太网数据报封装以太网数据报3.第一个路由器接收到报文后，根据路由表，把包转发到第一个路由器接收到报文后，根据路由表，把包转发到122.222.222.003，并重新封装报文的链路层协议，并重新封装报文的链路
29、层协议，把目的把目的MAC改为改为55-55-55-55-55-55，源，源MAC改为改为33-33-33-33-33-33.4.这个过程将会持续到它到达这个过程将会持续到它到达FThe Link and Local Area NetworksSolution 2-P12e）A必须知道必须知道111.111.111.002的的MAC地址，地址，因此它以以太网广播帧广播发送因此它以以太网广播帧广播发送ARP请求报文。请求报文。第一个路由器接收到第一个路由器接收到ARP请求后，把请求后，把ARP响应数响应数据包返回给据包返回给A，而这个，而这个ARP响应数据包目的响应数据包目的MAC地址为地址为0
30、0-00-00-00-00-00。The Link and Local Area NetworksProblem 3-P15The Link and Local Area NetworksSolution 3-P15Wireless and Mobile NetworksBase station&infrastructure modeWireless Network Characteristics path loss interference from other sources multipath propagationCDMACSMA/CARTS&CTSMobility in the Sa
31、me IP SubnetWireless and Mobile NetworksProblem-P8Consider the scenario shown in Figure 6.32,in which there are four wireless nodes,A,B,C,and D.The radio coverage of the four nodes is shown via the shaded ovals;all nodes share the same frequency.When A transmits,it can only be heard/received by B;wh
32、en B transmits,both A and C can hear/receive form B;when C transmits,both B and D can hear/receive form C;when C transmits,only C can hear/receive form D;Wireless and Mobile NetworksProblem-P8Wireless and Mobile NetworksProblem-P8Suppose now that each node has an infinite supply of messages that it
33、wants to send to each of the other nodes.If a messages destination is not an immediate neighbor,then the message must be relayed.For example,if A wants to send to D,a message from A must first be sent to B,which then sends message to C,which then sends the message to D.Wireless and Mobile NetworksPr
34、oblem-P8Time is slotted,with a message transmission time taking exactly one time slot,e.g.,as in slotted Aloha.During a slot,a node can do one of the following:(i)send a message(if it has a message to forward towards D);(ii)receive a message(if exactly one message is being sent to it);(iii)remain si
35、lent.Wireless and Mobile NetworksProblem-P8As always,if a node hears two or more simultaneous transmissions,a collision occurs and none of the transmitted messages are received successfully.You can assume here that there are no bit-level errors,and thus if exactly one message is sent,it will be reei
36、ved correctly by those within the transmission radius of the sender.Wireless and Mobile NetworksProblem-P8Wireless and Mobile NetworksSolution-P8a)报文从报文从C传到传到B，再从，再从B传到传到A，需要两个时隙，需要两个时隙，因此最大速率为因此最大速率为1 message/2 slotsb)A向向B发送报文与发送报文与D向向C发送报文互不影响，因此发送报文互不影响，因此一时隙内可同时传送报文，最大速率为：一时隙内可同时传送报文，最大速率为：2 mes
37、sages/slotc)由于由于C向向D传输报文时，传输报文时，B能接收到，能接收到，A如果这时如果这时向向B传输就会发生冲突。因此，至少输入两个时传输就会发生冲突。因此，至少输入两个时隙传输这两个报文，速率为：隙传输这两个报文，速率为：1 messages/slotWireless and Mobile NetworksSolution-P8d)1）1 messages/slot2）2 messages/slot3）2 messages/slote)1)Message：A-B-C，ACK：C-B-A，共需，共需4个时隙，速率：个时隙，速率：1 messages/4slotsWireless
38、and Mobile NetworksSolution-P82）slot 1:Message AB,message DC slot 2:Ack BA slot 3:Ack CD=2 messages/3 slots3)slot 1:Message CD slot 2:Ack DC,message AB slot 3:Ack BA=2 messages/3 slotsRelated ProblemsCH1:P5-P9,P17-P20CH2:R14-R19;P1-P3,P7,P8CH3:R3-R6;P4,P5,P19,P23,P25,P34,P38CH4:R24-R27;P8-P11,P22,P24,P29CH5:R4,R12,P1-P6,P11-P15CH6:R1,R3,R8-R11,R21;P4,P6,P8,P13Thank You!Good Luck!
