计算机网络课程作业

Chapter 1.R1,R3,R8,R11,R12,R13,R15,R16,R18,R19,R23,R25,R26(R4,R7,R10,R14,R20,R22,R24)p2,p5,p7,p8,p9,p12,p13,p22,p24,p26(p3,p6,p10,p11,p14,p15,p17,p18,p19,p20,p21 )R1.None.Servers and clients.Yes.R3.A client program is a program running on one end system that requests and receives a service from a server program running on another end system.Yes.R8.Dial-up modems: up to 56kbps; dedicated.HFC: up to 30 Mbps, 2 Mbps upstream; shared.DSL: up to 1 Mbps upstream, up to 8 Mbps downstream; dedicated.R11.More suitable for real-time services; providing a stable bandwidth; no worry about delay and loss; no need of analysis while data transferMore suitable for digital signal transfer.R12.L/R1+L/R2R13.Because packet switching provides on-demand sharing of resources instead of pre-allocated. While TDM decides the bandwidth of each sender-receiver pair in advance.R15.Tier-1 ISPs are each directly connected only to each of the other tier-1 ISPs, while can be each connected to a large number of tier-2 ISPs.R16.Processing delay, transmission delay, propagation delay, and queuing delay. The queuing delay is variable (depends on congestion level of router) and the other three are all constant.R18.Rate: R Propagation delay: D Packet size: LWhen L/R<D, the sender finish transmitting before the first bit of the packet reaches the receiver; when L/R>D, the first bit of the packet reaches the receiver before the sender finishes transmitting.R19.a. 250kbpsb. 2M*8/250k = 64sc. 200kbps; 80sR23.Application layer, transport layer, network layer, link layer, physical layer.Application: covers the reaction between different applicationsTransport: transferring messages between application endpointsNetwork: moving network-layer packets from one host to anotherLink: routing a datagram through a series of routers between the sources and destinationsPhysical: moving entire frames from one network element to an adjacent network elementR25.They are the packet data units of each layer.R26.Worms are malware that can enter a device without any explicit user interaction.Viruses are malware that require some form of user interaction to infect the user’s device.Trojan horse is malware that is a hidden part of some otherwise useful software.(?) P2.a. 6b. 4P5.a. m/sb. L/Rc. m/s +L/Rd. just left host Ae. a distance of s*L/R far from host Af. has reached host Bg. m = s*L/R = 250M*100/28k = 25,000kmP7.a. 10b. 0.1c. 0.1ⁿ*0.9⁽⁴⁰⁻ⁿ⁾*C₄ⁿⁿd. λ=40*0.1=4, P(X≥11)=0.002840P8.a. 1G/100k = 1000b. λ=p*M, P(X≥N+1)P9.a. L/R₀ + d₀/s₀ + L/R₁ + d₁/s₁b. 2*1000*8bits/1Mbps + 2*0.002second + (6000+3000)km/250000(km/s) =0.056secondP12.a. 3.5*1250*8bits/1Mbps = 0.035secondb. (n*L-x)/RP13.(0 + L/R + 2L/R + … + (N-1)L/R)/N = (N-1)L/2RP22.a.( h = 36000km) dprop = 36000km/2*10⁸(km/s) = 0.00018sb. R ∙ dprop = 10Mbps*0.00018s = 1800Mbc. x = R ∙ dprop = 1800MbP24.a. to the first packet switch: 8∙10⁶bits/2Mbps = 4sto destination host: 3*4s = 12sb. 2000bits/2Mbps = 0.001swhen t = 0.001swhen t = 0.002sc. 4000*0.001s + 0.002s = 4.002scomment: Message segmentation save time while packet switchd. If one of the data segments gets lost, mistake will occur in the received file, so the segments have to be added a order mark.P26.F/S * L/R + L/R = FL/SR + L/RChapter 2.R1,R3,R4,R5,R10,R11,R13,R15,R16,R20,R21,R22,R27,R28P1,P4,P5,P7,P8,P16,P24,P25R1.The Web: HTTP; File transfer: FTP; Electronic mail in the Internet: SMTP; Remote terminal access: Telnet; Streaming multimedia: HTTPR3.From the application developer’s perspective, the network architecture is fixed and provides a specific set of services to applications. The application architecture, on the other hand, is designed by the application developer and dictates how the application is structured over the various end systems.R4.The port number.R5.Not exactly. Although the hosts can all perform as either clients and servers, there is still a difference between the sender and receiver of the file during a communication session. The sender could be seen as the server side and the receiver could be seen as the client side.R10.Transport layer.In particular, if an application wants to use the services of SSL, it needs to include SSL code(existing, highly optimized libraries and classes) in both the client and server sides of the application.R11.A handshaking protocol is a protocol which sends a control message to the receiver before starting to transferring the requested data.(like SMTP)R13.R15.a. Alice invokes her user agent for e-mail, provides Bob’s e-mail address, composes a message, and instructs the user agent to send the message.b. Alice’s user aent sends the message to her m ail server, where it is placed in a message queue.c. The client side of SMTP, running on Alice’s mail server, sees the message in the message queue. It opens a TCP connection to an SMTP server, running on Bob’s mail server.d. After some initial SMTP ha ndshaking, the SMTP client sends Alice’s message into the TCP connection.e. At Bob’s mail server, the server side of SMTP receives the message. Bob’s mail server then places the massage in Bob’s mailbox.f. Bob’s user agent opens a TCP connection to his mail server. With the connection established, the user agent sends a username and a password to authenticate the user, retrieves messages, and updates the mails.R16.R20.Every 30 seconds, a peer randomly selects another peer and then starts sending chunks to it. So when Alice gets chosen by some other peer, she will get her first chunk.R21.Not necessary. Bob chooses 4 peers to transfer files to them referring to the highest rate that he receives every 10 seconds. And Alice is transferring to Bob at t his rate doesn’t mean she will surely appear on Bob’s top4 list.R22.In query flooding, the peers form an abstract, logical network called an overlay network.No, it doesn’t include routers.An edge represents a TCP connection between two certain peers. (If peer X maintains a TCP connection with another peer Y, then we say there is an edge between X and Y.)a) Query message sent over existing TCP connectionsb) Peers forward query messagec) QueryHit sent over reverse pathR27.During TCP connections, as soon as the client is executed, it attempts to initiate a TCP connection with the server. If the TCP server is not running, the connection will fail. While during UDP connections, the client doesn’t initiate connections immediately after execution.R28.Because TCP involves one more welcome socket when establishing connections.2nP1.a. falseb. truec. falsed. falseP4.The header line Connection: close in the HTTP request message can tell the server that it doesn’t want to bother with pers istent connection; it wants the server to close the connection after sending the requested object. Both clients and servers can signal the close of a connection.HTTP provides no encryption services.P5.Access control commands:USER PASS ACT CWD CDUP SMNT REIN QUITTransfer parameter commands:PORT PASV TYPE STRU MODEService commands:RETR STOR STOU APPE ALLO REST RNFR RNTO ABOR DELE RMD MRD PWD LIST NLST SITE SYST STAT HELP NOOPP7.2RTTⁿ + RTT₀ + RTT₁+ … + RTTnP8.a. (3+1)*2 = 8RTTb. 2RTT + 2RTT = 4RTTc. 2RTT + 1RTT = 3RTTP16.Dcs = max{NF/us, F/di}Dp2p = max{F/us, F/di, NF/(us + Nui)}图表1: u = 100Kbps图表2: u = 250Kbps图表3: u = 500KbpsP24.Yes, it’s necessary to change UDPServer.java. The port number needs to be updated.5432In the UDPClient.java , it didn’t take the server host name or port number asarguments before this change. In the UDPServer.java , the port number was 9876 before this change.P25.a. The TCPClient tries to connect with the TCPServer, but it fails finding it. So the connection cannot be established.b. Because the UDPClient doesn’t establish the connection as soon as it runs, errors only occurs when you try to send datagram. The datagram cannot find a destination. Only after you run the UDPServer, the transformation will be accessable.c. For the TCPClient, it will fail the connection or establish connection with a wrong process on the server. For the UDPClient, its datagram may be sent to a wrong process or get lost.Chapter 3R3,R5,R6,R7,R8,R14,R15,R17P1,P2,P5,P18,P19,P21,P22,P23,P24,P30,P31,P35,P36R3.a. Finer application-level control over what data is sent, and when.b. No connection establishment.c. No connection state.d. Small packet header overhead.R5.Yes. The application developer can put reliable data transfer into the application layer protocol.R6.Source port number: y; Destination port number: x.R7.Yes. The process at Host C know it by different source IP addresses and port numbers in the headers of the two segments.R8.No, through different sockets. They do have the same port number 80.The header of segments include source IP address and port numbers. When Host C receives the segment, it examines the header to decide which socket to pass to.R14.a. 20 bytesb. ACK 90R15.a. falseb. falsec. trued. falsee. truef. falseg. falseR17.False. It’s set to one half of the CongWin value.P1.A: 0123 B:3456 C:6789Source port number Destination port numbera. 0123 6789b. 3456 6789c. 6789 0123d. 6789 3456e. Yes, it’s possible.f. No, it’s impossible.P2.Source port number: 80Destination port number: 26145Destination IP: CP5.a. sum = 1100 0111sum‘s 1s complement = 0011 1000b. sum = (1)0110 0011sum’s 1s complement = 1001 1100c. eg. The two numbers are changed into 0100 1011 and 0111 1000. And the sum and its 1s comp lement won’t change.P18.a. k-2, k-1, k;k-1, k, k+1;k, k+1, k+2b. ACK 0 ~ ACK kP19.a. trueb. falsec. trued. trueP21.(k+1)/2 (int)P22.a. Because UDP package the data inside a UDP package.b. UDP immediately passes the segment to the network layer. While TCP has a congestion-control mechanism that throttles the transport layer TCP sender when one or more links between the source and destination hosts become excessively congested.P23.a. L = 2³² = 4Gbitsb. 2³²/1460*8 = 367 719.80273972 So the number of segments n = 367 720.Total size for transmission S = n*66bytes + L = 4 319 236 816Time = S/R = 4 319 236 816 /100M = 43.19secP24.a. sequence number: 409source port number: 1028 destination port number: 80b. ACK 409source port number: 80 destination port number: 1028c. ACK 359d.Host A Host BSeq = 359TimeoutP30.If so, maybe just the congestion or a little delay will easily cause retransmission, and retransmission would be executed too often. Unnecessary retransmission waste the resources.P31.Exactly the same.P35.The doubling of the timeout interval does not prevent a TCP sender from sending a large number of first-time-transmitted packets into the network, even when the path is highly congested. Therefore the window-based congestion-control mechanism is needed to stem the flow of ‚data received from the application above‛ when there are signs of network cong estion.P36.a. 7RTTb. 4MSS/RTTChapter 4R1,R2,R3,R7,R10,R11,R16,R17,R19,R22,R24,R26,R30P1,P2,P4,P7,P10,P11,P12,P14,P15,P16,P17,P22,P24,P29,P30R1.Datagram;Link-layer switches base their forwarding decision on the value in the link-layer field. While routers base their forwarding decision on the value in the network-layer field.R2.Forwarding: When a packet arrives at a router’s input link, the router must move the packet to the appropriate output link.Routing: The network layer must determine the route or path taken by packets as they flow from a sender to a receiver.R3.Datagram network: data transfer, maintain forwarding tablesVC network: VC setup, data transfer, VC teardown.R7.Switching via memory: The simplest, earliest routers were often traditional computers, with switching between input and output ports being done underdirect control of the CPU. Input and output ports functioned as traditional I/O devices in a traditional operating system.Switching via buses: In this approach, the input ports transfer a packet directly to the output port over a shared bus, without intervention by the routing processor.Switching via an interconnection network: One way to overcome the bandwidth limitation of a single, shared bus is to use a more sophisticated interconnection network, such as those that have been used in the past to interconnect processors in a multiprocessor computer architecture.R10.Suppose that the switching fabric is at least n times as fast as the line speeds. In the worst case, if in the time it takes to receive a single packet, n packets arrive at the same output port. Eventually, the number of queued packets can grow large enough to exhaust the memory space at the output port, in which case packets are dropped and packet loss occurs.R11.A queued packet in an input queue must wait for transfer through the fabric(even through its output port is free) because it is blocked by another packet at the head of the line.Input port.R16.Through the Protocol field in the datagram format. The protocol number is the glue that binds the transport and application layers together. The value of this field indicates the specific transport-layer protocol to which the data portion of this IP datagram should be passed.R17.Overhead percentage: 40/(40 + 20) = 66.7%Data percentage: 20/(40 + 20) = 33.3%R19.Yes. With tunneling, the IPv6 node on the sending side of the tunnel takes the entire IPv6 datagram and puts it in the data field of an IPv4 datagram, just like encapsulating a network-layer datagram into a link-layer frame.R22.Not necessary. Since one or more gateway routers in an AS will have the added task of being responsible for forwarding packets to destinations outside the AS, so that other routers within the AS need not to know what protocols the routers in another AS use. And the inter-AS routing algorithm will generate the forwarding between gateway routers.R24.No. This table doesn’t provide a better path for D’s transfer, so the original table will not update.R26.AS-pathsR30.Because the AS’s intra-AS routing algorithm has determined the least –cost path from each internal router to the gateway router, each internal router knows how it should forward the packet. Policy considerations within AS are just as important for forwarding as among AS.P1.a.b. VC network. Because VC network is connection-oriented, which permit the routers to know the traffic characters of all sessons.P2.VC: Incoming interface, incoming VC numbers, outgoing interface, outgoing VC numbers;Datagram: Destination address range, link interface.P4.a. No more VC numbers could be assigned to the new VC.b. (2³ / 4)⁴ = 16P7.a. No. At least 2 times the line speed could satisfy the requirement.b. No either. The same to the reason in (a).P10.Destination Address Range Number Of Addresses1000 0000 through 1011 1111 641100 0000 through 1101 1111 321110 0000 through 1111 1111 32Otherwise 128oP11.ISP’s block220.2.240/20 11011100 00000010 11110000 00000000 Subnet 1 220.2.240/21 11011100 00000010 11110000 00000000 Subnet 2 220.2.248/22 11011100 00000010 11111000 00000000 Subnet 3 220.2.252/22 11011100 00000010 11111100 00000000P12.Prefix Match Link Interface200.23.16/21 0200.23.24/24 1200.23.24/21 2Otherwise 3P14.101.101.101.65ISP’s block101.101.128/17 01100101 01100101 10000000 00000000 Subnet 1 101.101.128/19 01100101 01100101 10000000 00000000 Subnet 2 101.101.160/19 01100101 01100101 10100000 00000000 Subnet 3 101.101.192/19 01100101 01100101 11000000 00000000 Subnet 4 101.101.224/19 01100101 01100101 11100000 00000000 P15.a.ISP’s block214.97.254/23Subnet A 214.97.254.0/24Subnet B 214.97.255.0/25 - 214.97.255.0/29Subnet C 214.97.255.128/25Subnet D 214.97.255.0/30Subnet E 214.97.255.2/31Subnet F 214.97.255.4/31P16.4000bytes – 20bytes = 3980bytes.3980/(400 -20) = 11fragments.Fragment Bytes ID Offset Flag1 380 422 0 12 380 422 48 13 380 422 96 14 380 422 144 15 380 422 192 16 380 422 240 17 380 422 288 18 380 422 336 19 380 422 384 110 380 422 432 111 180 422 480 0P17.3M/(1500 - 20 - 20) = 2155 datagramsP22.Step N’D(s),P(s) D(t),P(t) D(u),P(u) D(v),P(v) D(w),P(w) D(y),P(y) D(z),P(z)0 x ∞∞∞8,x 6,x 6,x ∞1 xw ∞∞14,w 8,x 6,x 6,x ∞2 xwy ∞15,y 14,w 7,y 6,x 6,x 18,y3 xwyv ∞11,v 10,v 7,y 6,x 6,x 18,y4 xwyvu 14,u 11,v 10,v 7,y 6,x 6,x 18,y5 xwyvut 12,t 11,v 10,v 7,y 6,x 6,x 16,t6 xwyvuts 12,t 11,v 10,v 7,y 6,x 6,x 16,t7 xwyvutsz 12,t 11,v 10,v 7,y 6,x 6,x 16,tP24.P29.a. eBGPb. iBGPc. eBGPd. iBGPP30.a. L1. Since the least cost path includes 1d -- 1a -- 1c.b. L2. Since the path through AS2 costs less than the other one.c. L1. Since the AS3--AS4 path will cost less.Chapter 5R1,R2,R3,R4,R7,R8,R9,R12P1,P2,P3,P6,P12,P14,P15,P16,P19,P21,P23,P24R1.a. Framing, link access, reliable delivery, flow control, error detection, error correction, half-duplex and full-duplex.b. Error detection.c. Reliable delivery, flow control, error detection.R2.When a link-layer protocol provides reliable delivery service, it guarantees to move each network-layer datagram across the link without error. But wrong routes or other reasons can also cause the loss of datagram. And the TCP is also used to guarantee that the packets arrive at the host in order. So the TCP is still necessary. R3.Slotted ALOHA: 234Token passing: 1234R4.Yes. At t=dprop, the datagram will arrive at its destination host, but this destination host hasn’t finished receiving the broadcast data. So there will be a collision.R7.Yes, C’s adapter will process these frames. But it won’t pass the IP datagram in these frames to the network layer C. With MAC broadcast address, C’s adapter will pass the IP datagram in these frames to the network layer C.R8.16¹²; 2³²; 2¹²⁸R12.1/32;4*512bits/10Mbps = 0.0002048sP1.16 = 4*4 -> 4+4+1 = 9 bitsP2.Add 00000000, 00000001,……, 00001001: 00101101The 1’s implement: 11010010P3.a. sum = 001101111’s implement = 11001000P6.a. 001b. 101c. 110P12.b.c. IP MACSubnet 1: Router 1: 111.111.111.1 E6-E9-00-17-BB-4BA: 111.111.111.2 74-29-9C-E8-FF-55B: 111.111.111.3 CC-49-DE-D0-AB-7D Subnet 2: Router 1(Left): 122.122.122.1 1A-23-F9-CD-06-9BRouter 2(Right): 122.122.122.2 88-B2-2F-54-1A-0FC: 122.122.122.3 5C-66-AB-90-75-B1D: 122.122.122.4 49-BD-D2-C7-56-2A Subnet 3: Router 2: 133.133.133.1 3D-FF-67-1D-12-00-33 E: 133.133.133.2 80-2A-11-1D-FF-FF-00F: 133.133.133.3 10-1A-49-17-BB-4bd. A and F are in different subnets, so A gives its adapter a datagram containing F’s IP(133.133.133.3). A’s adapter construct a link-layer frame with it ,containing the MAC address of router interface 111.111.111.1, and send it to router 1. Then router interface 122.122.122.1’s adapter send it to router 2, containing the MAC address of router interface 122.122.122.2, according to its forwarding table. Next, router interface 133.133.133.1’s adapter encapsulates t he frame with exactly the MAC address of F, and sends it to F.e. First, A finds that the destination F is a host from another subnet, and decides to send the packet to its gateway first. A sends a ARP query with the IP address of the router and a broadcast MAC address(FF-FF-FF-FF-FF-FF). Router 1 receives the query packet ,replies to A and tells A its MAC address. The following steps are the same as in (d).P14.100*512bits/1Mbps = 0.0512s100*512bits/10Mbps = 0.00512sP15.B schedule its retransmission: t = 273 + 1*512 = 785 bit timesA begin retransmission: t = 273 + 96 = 369 bit timesA’s signal reaches B: t = 369 + 225 = 594 bit timesNo, B doesn’t refrain from transmitting at its scheduled time.(Since 785 > 594)P16.Suppose at time t = 512+64-225 = 351 bit times, B begins transmitting its frame. In this case, A finish transmitting at the same time with B’s signal’s arrival. So, if B begins transmitting before this moment(t = 351 bit times), the answer will be NO; if B begins transmitting after this moment, the answer will be YES.P19.a. (1000 + 20*4)bits/10Mbps + 900m/2∙10⁸m/sec = 0.1125msb. 0.1125*2 = 0.225msc. 0.1125msP21.P23.140MbpsSince the topology decides that each host in the diagram is in a single collision domain, they will not affect each other’s transmission.P24.10MbpsSince all the 14 hosts in the diagram are in a same collision domain and they compete for the total bandwidth. They will affect each other’s transmission.。