计算机网络课后习邈指导 陕西师范大学 名m除成 P22.Consider Figure 1.19(b).Suppose that each link between the server and the client has a packet loss probability p.and the packet loss probabilities for these links are independent.What is the er)is successfully received by the receiver?If a packet is ost inthe pathfrom the server the server ill the average how many times will the server re-transmit the packet in order for the client to successfully receive the packet? 答：成功接收到一个分组的概率为：卫=(1一p)N: 传输在一直进行中，直到客户机成功接收到分组为止，传输的数量是几何随机变量，概率为 ,因此，平均需要传输的数量为：1/B:然后，平均需要重传的数量为1/R-1。 P23.Consider Figure 1.19(a).Assume that we know the bottleneck link along the path from the server to the client is the first link with rate Rs bits/sec.Suppose we send a pair of packets back to back from the server to the client,and there is no other traffic on this path.Assume each packet of size L bits,and both links have the me propagatio n delay d a.What is the packet interarrival time at the destination?That is,how much time elapses from when the last bit of the first packet arrives until the last bit of the second packet arrives b.Now assume that the second link is the bottleneck link (ie..RR).Is it possible that the second packet queues at the input queue of the second link?Explain.Now suppose that the server sends the son packet after sending the first packet.How large must be toensuren queuing befo link?Explain 答：我们把第 个分组称为A,第二个为B .瓶颈链路是第一条链路，所以分组B在第一条链路上等待分组A的传输，分组在目的地 的到达间隔为：L/Rs。 .如果第二条链路是瓶颈链路，两个分组是连续传输的，必须保证：在第一个分组在路由 器传输进入第二条链路之前，第二个分组到达第二条链路的的输入队列，需要满足以下条 R+元+4rom<元+dron+R怎 其中左边表示第二个分组到达传第二条链路的传输队列时间，右边表示第一个分组传输到 第一条路的时间 如果第二个分组在第一个分组T时间间隔之后传输，我们需要确保第二个分组在第二条链 路没有排队时延，需要保证： ++4+T≥克+4om+ 因此，T的最小值为： P24.Suppose you would like to urgently deliver 40 terabytes data from Boston to Los Angeles.You
计算机网络课后习题指导 陕西师范大学 13 { , , ⋯ , } P22. Consider Figure 1.19(b). Suppose that each link between the server and the client has a packet loss probability p, and the packet loss probabilities for these links are independent. What is the probability that a packet (sent by the server) is successfully received by the receiver? If a packet is lost in the path from the server to the client, then the server will re-transmit the packet. On average, how many times will the server re-transmit the packet in order for the client to successfully receive the packet? 答：成功接收到一个分组的概率为： = (1 − ) ； 传输在一直进行中，直到客户机成功接收到分组为止，传输的数量是几何随机变量，概率为 ，因此，平均需要传输的数量为：1⁄；然后，平均需要重传的数量为1⁄ − 1。 P23. Consider Figure 1.19(a). Assume that we know the bottleneck link along the path from the server to the client is the first link with rate Rs bits/sec. Suppose we send a pair of packets back to back from the server to the client, and there is no other traffic on this path. Assume each packet of size L bits, and both links have the same propagation delay dprop. a. What is the packet inter-arrival time at the destination? That is, how much time elapses from when the last bit of the first packet arrives until the last bit of the second packet arrives? b. Now assume that the second link is the bottleneck link (i.e., Rc< Rs). Is it possible that the second packet queues at the input queue of the second link? Explain. Now suppose that the server sends the second packet T seconds after sending the first packet. How large must T be to ensure no queuing before the second link? Explain. 答：我们把第一个分组称为 A，第二个为 B。 a. 瓶颈链路是第一条链路，所以分组 B 在第一条链路上等待分组 A 的传输，分组在目的地 的到达间隔为： ⁄ 。 b. 如果第二条链路是瓶颈链路，两个分组是连续传输的，必须保证：在第一个分组在路由 器传输进入第二条链路之前，第二个分组到达第二条链路的的输入队列，需要满足以下条 件： L + L + < L + + L 其中左边表示第二个分组到达传第二条链路的传输队列时间，右边表示第一个分组传输到 第二条链路的时间。 如果第二个分组在第一个分组 T 时间间隔之后传输，我们需要确保第二个分组在第二条链 路没有排队时延，需要保证： L + L + + T ≥ L + + L 因此，T 的最小值为： − P24. Suppose you would like to urgently deliver 40 terabytes data from Boston to Los Angeles. You
计算机网络课后习题指导 陕西师范大学 have available a 100 Mbps dedicated link for data transfer.Would you prefer to transmit the data 40rB-40×102×8bi=3.2×105s=37days 100Mbps=100×106bits/5 如果使用联邦快递(FedEx overnight delivery),只需要一天，并且少花费100美元，综上可知， 景急传输大容量数据快递公司较好， P25.Suppose two hosts.Aand B.are separated by 20.00kilometers and are conected by a direct link of R=2 Mbps.Suppose the propagation speed over the link is 2.5-108 meters/sec. a Calculate the bandwidth-delay product,R b.Consider sending a file of 800000 bits from Host a to Host b.Suppose the file is sent continuously asone large message.What is the maximum number of bits that will be in the link a any given time c.Provide an interpretation of the bandwidth-delay product. d.What is the width (in meters)of a bit in the link?Is it longer than a foothall field? e.Derive a general expression for the width of a bit in terms of the propagation speed s.the ransmission the length of the linkm. 答：a由题意，传播时延为：dprop=0ooa心=0.08s,则宽带时延积bandwidth-dcla product)为： R·drop=0.08s×2×105bps=160000bits b.在链路上的最大比特数量为160000，小于要传输的800000bits. ©.宽带时延积指的是在任意时刻，链路上能具有的最大比特数量。 d 个比特的宽度即传播距离m与宽带时延积的商，为： 20000×103m 160000bit =125m/bit e.比特宽度为： m R4rnm,题元 P26.Referring to problem P25,suppose we can modify R.For what value of R is the width of a bit as long as the length of the link? 答：由题，比特宽度大小等于链路传插速率大小，即s/R=20000km,则： P27.Consider problem P25 but now with a link of R=I Gbps. a Calculate the bandwidth-delay product.Rdocp b.Consider sending afile of bits from Host Host B.Suppose the file is continuously as one big message.What is the maximum number of bits that will be in the link at any giventime? c.What is the width (in meters)of a bit in the link? 答：1由25题可知，传播时延为0.08秒，则宽带时延积为：
计算机网络课后习题指导 陕西师范大学 14 have available a 100 Mbps dedicated link for data transfer. Would you prefer to transmit the data via this link or instead use FedEx overnight delivery? Explain. 答：如果使用 100 Mbps 专用链路(dedicated link)传输，需要花费的时间为： 40 100 = 40 × 10 × 8 100 × 10/ = 3.2 × 10 = 37days 如果使用联邦快递(FedEx overnight delivery)，只需要一天，并且少花费 100 美元，综上可知， 紧急传输大容量数据快递公司较好。 P25. Suppose two hosts, A and B, are separated by 20,000 kilometers and are connected by a direct link of R = 2 Mbps. Suppose the propagation speed over the link is 2.5 ·108 meters/sec. a. Calculate the bandwidth-delay product, R ·dprop. b. Consider sending a file of 800,000 bits from Host A to Host B. Suppose the file is sent continuously as one large message. What is the maximum number of bits that will be in the link at any given time? c. Provide an interpretation of the bandwidth-delay product. d. What is the width (in meters) of a bit in the link? Is it longer than a football field? e. Derive a general expression for the width of a bit in terms of the propagation speed s, the transmission rate R, and the length of the link m. 答：a. 由题意，传播时延为：d = × ×/ = 0.08，则宽带时延积(bandwidth-delay product)为： R ∙ = 0.08 × 2 × 10 = 160000 b. 在链路上的最大比特数量为 160000，小于要传输的 800000bits。 c. 宽带时延积指的是在任意时刻，链路上能具有的最大比特数量。 d. 一个比特的宽度即传播距离 m 与宽带时延积的商，为： 20000 × 10m 160000 = 125/ e. 比特宽度为： m ∙ = ∙ = P26. Referring to problem P25, suppose we can modify R. For what value of R is the width of a bit as long as the length of the link? 答：由题，比特宽度大小等于链路传播速率大小，即 s/R=20000km，则： R = 20000 = 2.5 × 10 2 × 10 = 12.5 P27. Consider problem P25 but now with a link of R = 1 Gbps. a. Calculate the bandwidth-delay product, R·dprop. b. Consider sending a file of 800,000 bits from Host A to Host B. Suppose the file is sent continuously as one big message. What is the maximum number of bits that will be in the link at any given time? c. What is the width (in meters) of a bit in the link? 答：a. 由 25 题可知，传播时延为 0.08 秒，则宽带时延积为：
计算机网络课后习邈指导 陕西师范大学 R.dm=0.08s×1×109bns=8x107bits b.由于宽带时延积远远大于发送文件的大小，所以在任意给定的时间，链路能具有最大比 持粉为80000hitg .比特宽度为传播距离与宽带时延积的商，即 20000km 8×10bi=0.25m/b P28.Refer again to problem P25 a How long does it take to send the file,assuming it is sent continuously? b.Suppose now the file is broken up into 20 packets with each packet containing 40.000 bits Suppose that each packet is acknowledged by the receiver and the transmission time of an acknowledgment packet is negligible.Finally,assume that the sender cannot send a packet until How long does it take to send the file? c.Compare the ults from (a)and (b). 答：a.由题意总时延为： 800000bits20000×103m dtrans+dproe=x 10bps2.5x10m/s =0.4+0.08=0.48s b.由题意，原来的800000bis大文件分成了20个小分组来传输，并且一个分组在发送方收 到确认之前不会发送下一个分组 且忽 r接收 时间和确认分组的传输 可，每 分组包括三部分，在sender的传输时间，链路的传播时间，确认ACK的传播时间， 一共 要传输20次，时间为： 2n6+2x4】=20×(60p+2×200） 40000b 25×109m3=3.6s P29.Suppose there is a 10 Mbps microwave link between a geostationary satellite and its base station on Earth.Every minute the satellite takes a digital photo and sends it to the base station.Assumea What is the propagation delay of the link b.What is the bandwidth-delay product,Rd? c.Letx denote the size of the photo.What is the minimum value of for the microwave link to be continuously transmitting? 答：a地球同步卫星与地球的距离为36000m,传播时延为： dp=36000×10 =0.15s=150m 2.4×10m/ b.宽带时延积为： R·dprop=(10×10bps)×0.15s=1.5×105bits c,由趣意，每分钟传输的最小的数据大小为： 60R」 =60s×10×105bps=6×10bit P30.Consider the airline travel analogy in our discussion of layeringin Section 15,and the addition of headers to protocol data units as they flow down the protocol stack.Is there an equivalent notion
计算机网络课后习题指导 陕西师范大学 15 R ∙ = 0.08 × 1 × 10 = 8 × 10 b. 由于宽带时延积远远大于发送文件的大小，所以在任意给定的时间，链路能具有最大比 特数为 80000bits。 c. 比特宽度为传播距离与宽带时延积的商，即： 20000km 8 × 10 = 0.25/ P28. Refer again to problem P25. a. How long does it take to send the file, assuming it is sent continuously? b. Suppose now the file is broken up into 20 packets with each packet containing 40,000 bits. Suppose that each packet is acknowledged by the receiver and the transmission time of an acknowledgment packet is negligible. Finally, assume that the sender cannot send a packet until the preceding one is acknowledged. How long does it take to send the file? c. Compare the results from (a) and (b). 答：a. 由题意总时延为： + d = 800 000 2 × 10 + 20000 × 10 2.5 × 10/ = 0.4+ 0.08 = 0.48 b. 由题意，原来的 800000bits 大文件分成了 20 个小分组来传输，并且一个分组在发送方收 到确认之前不会发送下一个分组，且忽略 receiver 接收时间和确认分组的传输时间，每一个 分组包括三部分，在 sender 的传输时间，链路的传播时间，确认 ACK 的传播时间，一共需 要传输 20 次，时间为： 20d + 2 × = 20 × 40000 2 × 10 + 2 × 2 × 10 × 10 2.5 × 10/ = 3.6 c. 将一个文件分解成多个小的分组发送将花费更多时间，因为每一个分组都要传播时间以 及确认 ACK 的传播时间，多出很多时间开销。 P29. Suppose there is a 10 Mbps microwave link between a geostationary satellite and its base station on Earth. Every minute the satellite takes a digital photo and sends it to the base station. Assume a propagation speed of 2.4 · 108 meters/sec. a. What is the propagation delay of the link? b. What is the bandwidth-delay product, R · dprop? c. Let x denote the size of the photo. What is the minimum value of x for the microwave link to be continuously transmitting? 答：a. 地球同步卫星与地球的距离为 36000km，传播时延为： d = 36000 × 10 2.4 × 10/ = 0.15 = 150 b. 宽带时延积为： R ∙ = (10 × 10) × 0.15s = 1.5 × 10 c. 由题意，每分钟传输的最小的数据大小为： 60R = 60s × 10 × 10 = 6 × 10 P30. Consider the airline travel analogy in our discussion of layering in Section 1.5, and the addition of headers to protocol data units as they flow down the protocol stack. Is there an equivalent notion
计算机网络课后习题指导 陕西师范大学 of header information that is added to passengers and baggage as they move down the airline 答：有与首信思等价的概念比如旅客和行李对应到达协议栈顶部的数据单元，当旅 票的时候，他的行李也被检查了，行李和机票被加上标记，当旅客稍后通过安检，通常会另 外添加一个标记，指明该旅客已经通过了安检，之后的行程中，每到一个节点时，票上或行 李上都会做相应的标记以保证旅客和行李的正确运输。正如协议的首部一样，协议的数据单 元向协议的底层流动时，都要增加相应的首部信息使其能够保证数据单元下一步的正确进行 P31.In moder packet-switched networks.including the Intemet,the source host segments long. application-layer messages (for example,an image or a music file)into smaller packets and sends the packets into the network.The receiver then reassembles the packets back into the original message.We refer to this process as message segmentation.figure 1 27 illustrates the end-to-end transpor of a me sagewith and without m ation. Consider a message that is810 bits long that is to be sent from source to destination in Figure 1.27.Suppose each link in the figure is 2 Mbps.Ignore propagation.queuing,and processing delays. a.Source Packet switch Packet switch Destinatio b.Source Packet switch Packet switch Figure 1.27 Endtoend mess segmentation;(b)with messoge segmentation a.Consider sending the message from soure to destination message seg nentation.How long does it take to move the message from the source host to the first packet switch?Keeping in mind that each switch uses store-and-forward packet switching.what is the total time to move themessage from source host to destination host? b.Now suppose that the message is segmented into 800 packets,with each packet being 10.000 bits long How long does it take to move the first packet from source host to the first switch? witch,the second packet is being sent from the source host to the first switch.At what time will the second packet be full received at the first switch? c.How long does it take to move the file from source host to destination host when message segmentation is used?Compare this result with your answer in part (a)and comment d.toreducing delay,what messae? e.Discuss the of me 答：a由题意，忽略传播时延、排队时延和处理时延，在没有分组时，从源主机到第一个 分组交换机的时间即传输时间为：
计算机网络课后习题指导 陕西师范大学 16 of header information that is added to passengers and baggage as they move down the airline protocol stack? 答：有与首部信息等价的概念。比如旅客和行李对应到达协议栈顶部的数据单元，当旅客检 票的时候，他的行李也被检查了，行李和机票被加上标记，当旅客稍后通过安检，通常会另 外添加一个标记，指明该旅客已经通过了安检，之后的行程中，每到一个节点时，票上或行 李上都会做相应的标记以保证旅客和行李的正确运输。正如协议的首部一样，协议的数据单 元向协议的底层流动时，都要增加相应的首部信息使其能够保证数据单元下一步的正确进行。 P31. In modern packet-switched networks, including the Internet, the source host segments long, application-layer messages (for example, an image or a music file) into smaller packets and sends the packets into the network. The receiver then reassembles the packets back into the original message. We refer to this process as message segmentation. Figure 1.27 illustrates the end-to-end transport of a message with and without message segmentation. Consider a message that is 8 · 106 bits long that is to be sent from source to destination in Figure 1.27. Suppose each link in the figure is 2 Mbps. Ignore propagation, queuing, and processing delays. a. Consider sending the message from source to destination without message segmentation. How long does it take to move the message from the source host to the first packet switch? Keeping in mind that each switch uses store-and-forward packet switching, what is the total time to move themessage from source host to destination host? b. Now suppose that the message is segmented into 800 packets, with each packet being 10,000 bits long. How long does it take to move the first packet from source host to the first switch? When the first packet is being sent from the first switch to the second switch, the second packet is being sent from the source host to the first switch. At what time will the second packet be fully received at the first switch? c. How long does it take to move the file from source host to destination host when message segmentation is used? Compare this result with your answer in part (a) and comment. d. In addition to reducing delay, what are reasons to use message segmentation? e. Discuss the drawbacks of message segmentation. 答：a. 由题意，忽略传播时延、排队时延和处理时延，在没有分组时，从源主机到第一个 分组交换机的时间即传输时间为：
计算机网络课后习邈指导 陕西师范大学 8×105b 2x 105bps=4s 一共要传输3次到链路，从源主机到目的主机总时延为 4×3=12 b.报文段一共分成800个组，每个组大小为10000bis,从源主机到第一个分组交换机，第 一个分组的时延为： 10000h 2×10bms=0.005s=5ms 第一个分组发送到第二台交换机，时延为5×2=10ms:第二个分组从源主机发送到第一台交 换机时延为5ms:第二个分组被第一个交换机收到时间为5×2=10ms。 c.由题意，分组没有等待前一个分组ACK就连续发送，时间间隔为5ms,当第一个分组到 达目的地后，经过5s第二个分组就到达，以此类推得到分组发送总时延为： 5×3+799×5=4010ms=4.01s d.如果没有报文分段，则不允许比特错误，如果有 一个比特发生错误要重传整个报文段 如果没有报文分段，巨大的分组（如HD videos)被发送到网络中，路由器必须容纳这些巨大 的分组，小的分组必须排队在这些巨大的分组后面，忍受不公平的时延。 ©.分组在目的地必须按序排列：报文分组在许多小的分组中起作用，对于所有的分组而言， 不管其大小，头部长度一样，报文分组以后在首部会有很大的开销。 P32.Experiment with the Message Segmentation applet at the book's Web site.Do the delays in the applet correspond to the delays in the previous problem?How do link propagation delays affect the overall end-to-end delay for packet switching (with message segmentation)and for message switching? 答是的，在 plt实验中的时延和3引题的时延一致，传播时延影响整个端到端的时延， 对交换机的分组交换和报文段分组影响一样。 P33.Consider sending a large file of F bits from Host A to Host B.There are three links (and two switches)between A and B.and the links are uncongested (that is,no queuing delays).Host A segments the file into segments of Sbits each and adds 80 bits of header to each segment,forming packets ofL+bits.Each link has a tra smission rate ofRbps.Find the alue。 minimizes the delay of moving the filefrom Host Ato Host B.Disregard pro agat 答：由题意，忽略排队时延，处理时延和传搔时延，只需要考虑传输时延。分组个数为F/S, 每个分组报文段大小为(S+80)bits,分组传输进入一条链路时间为(S+80)/R,第一个分组 报文段传输到主机B时延为3×(S+80)/R,以后每经过(S+80)/R时延一个分组报文段到 大目的主机B,总时延为： da加w=(6+80)x3+6+80)×(传-1)=(6+80)×(5+2） 要找到合适的S使时延最小，对上式进行求导得： 解得S=V40F
计算机网络课后习题指导 陕西师范大学 17 8 × 10 2 × 10 = 4 一共要传输 3 次到链路，从源主机到目的主机总时延为： 4 × 3 = 12s b. 报文段一共分成 800 个组，每个组大小为 10000bits，从源主机到第一个分组交换机，第 一个分组的时延为： 10000 2 × 10 = 0.005 = 5 第一个分组发送到第二台交换机，时延为 5×2=10ms；第二个分组从源主机发送到第一台交 换机时延为 5ms；第二个分组被第一个交换机收到时间为 5×2=10ms。 c. 由题意，分组没有等待前一个分组 ACK 就连续发送，时间间隔为 5ms，当第一个分组到 达目的地后，经过 5ms 第二个分组就到达，以此类推得到分组发送总时延为： 5 × 3 + 799 × 5 = 4010ms = 4.01s d. 如果没有报文分段，则不允许比特错误，如果有一个比特发生错误要重传整个报文段； 如果没有报文分段，巨大的分组(如 HD videos)被发送到网络中，路由器必须容纳这些巨大 的分组，小的分组必须排队在这些巨大的分组后面，忍受不公平的时延。 e. 分组在目的地必须按序排列；报文分组在许多小的分组中起作用，对于所有的分组而言， 不管其大小，头部长度一样，报文分组以后在首部会有很大的开销。 P32. Experiment with the Message Segmentation applet at the book’s Web site. Do the delays in the applet correspond to the delays in the previous problem? How do link propagation delays affect the overall end-to-end delay for packet switching (with message segmentation) and for message switching? 答：是的，在 applet 实验中的时延和 31 题的时延一致，传播时延影响整个端到端的时延， 对交换机的分组交换和报文段分组影响一样。 P33. Consider sending a large file of F bits from Host A to Host B. There are three links (and two switches) between A and B, and the links are uncongested (that is, no queuing delays). Host A segments the file into segments of S bits each and adds 80 bits of header to each segment, forming packets of L = 80 + S bits. Each link has a transmission rate of R bps. Find the value of S that minimizes the delay of moving the file from Host A to Host B. Disregard propagation delay. 答：由题意，忽略排队时延，处理时延和传播时延，只需要考虑传输时延。分组个数为F/S， 每个分组报文段大小为(S + 80)bits，分组传输进入一条链路时间为(S + 80)/，第一个分组 报文段传输到主机 B 时延为3 × (S + 80)/，以后每经过(S + 80)/时延一个分组报文段到 达目的主机 B，则总时延为： = (S + 80) × 3 + ( + 80) × − 1 = ( + 80) × + 2 要找到合适的 S 使时延最小，对上式进行求导得： d = 0 解得S = √40