計算機網(wǎng)絡(luò)作業(yè)

1、Chapter 111. What are two reasons for using layered protocols?(M)通過協(xié)議分層可以把設(shè)計問題劃分成較小的易于處理的片段分層意味著某一層的協(xié)議的改變不會影響高層或低層的協(xié)議13. What is the principal difference between connectionless communication andconnection-oriented communication?(E) 主要的區(qū)別有兩條。其一：面向連接通信分為三個階段，第一是建立連接，在此階段，發(fā)出一個建立連接的請求。第二階段，只有在連接成功建立之后，保

2、持連接狀態(tài)，才能開始數(shù)據(jù)傳輸。第三階段，當數(shù)據(jù)傳輸完畢，必須釋放連接。而無連接通信沒有這么多階段，它直接進行數(shù)據(jù)傳輸。其二：面向連接的通信具有數(shù)據(jù)的保序性， 而無連接的通信不能保證接收數(shù)據(jù)的順序與發(fā)送數(shù)據(jù)的順序一致。18.Which of the OSI layers handles each of the following? (a) Dividing the transmitted bit stream into frames. (b) Determining which route through the subnet to use. Answer: (a) Data link laye

3、r. (b) Network layer.22. What is the main difference between TCP and UDP?（E）TCP 是面向連接的，而UDP 是一種數(shù)據(jù)報服務(wù)。25. When a file is transferred between two computers, two acknowledgementstrategies are possible. In the first one, the file is chopped up into packets, which are individually acknowledged by the rec

4、eiver, but the file transfer as a whole is not acknowledged. In the second one, the packets are not acknowledged individually,but the entire file is acknowledged when it arrives. Discuss these two approaches.（E）答：如果網(wǎng)絡(luò)容易丟失分組，那么對每一個分組逐一進行確認較好，此時僅重傳丟失的分組。如果網(wǎng)絡(luò)高度可靠，那么在不發(fā)差錯的情況下，僅在整個文件傳送的結(jié)尾發(fā)送一次確認，從而減少了確認的次

5、數(shù)，節(jié)省了帶寬；不過，即使有單個分組丟失，也需要重傳整個文件。（課堂練習）若待發(fā)送數(shù)據(jù)為：1010001101，現(xiàn)要計算CRC校驗和。如果我們選G=110101，1）請給出對應(yīng)的生成多項式G(X)。2）請給出實際發(fā)送出去的數(shù)據(jù)（比特流）。補充題1：試在下列條件下比較電路交換和分組交換。要傳送的報文共 x (bit)，從源站到目的站共經(jīng)過 k 段鏈路，每段鏈路的傳播時延為d 秒，數(shù)據(jù)率為 b (bit/s)。在電路交換時電路的建立時間為 s 秒。在分組交換時分組長度為 p (bit)，且各結(jié)點的排隊等待時間可忽略不計。問在怎樣的條件下，分組交換的時延比電路交換要??？電路交換時延：s+x/b+kd

6、分組交換時延：x/b+kd+(k-1)p/b x/b+kd+(k-1)p/b (k-1)p/b *但前提是：xp, 或分組數(shù)大于鏈路數(shù). 補充題2：在上題的分組交換網(wǎng)中，設(shè)報文和分組長度分別為 x 和 (p+h)(bit)，其中 p 為分組的數(shù)據(jù)部分的長度，而 h 為每個分組所帶的控制信息固定長度，與 p 的大小無關(guān)。通信的兩端共經(jīng)過 k 段鏈路。鏈路的數(shù)據(jù)率為b(bit/s)，但傳播時延和結(jié)點的排隊時延均可忽略不計。若打算使總的時延為最小，問分組的數(shù)據(jù)部分長度 p 應(yīng)該取多大？（2010考研）在下圖所示的采用“存儲-轉(zhuǎn)發(fā)”方式的分組交換網(wǎng)絡(luò)中，所有鏈路的數(shù)據(jù)傳輸速率為100Mbps，分組大小

7、為1000 B，其中分組頭大小為20 B。若主機H1向主機H2發(fā)送一個大小為980 000 B的文件，則在不考慮分組拆裝時間和傳播延遲的情況下，從H1發(fā)送開始到H2接收完為止，至少需要多少時間？b=100Mbps; x=980 000 Bp=(1000-20) B; h=20B; k=3D=80.16 msec分組長度L=1000B=980B+20B分組數(shù)N=980000/980=1000發(fā)送1個分組的時間Ttran=(1000x8)/(100x106) =8x10-5 secTtotal=N x Ttran+2 x Ttran= 80.16 msec某局域網(wǎng)采用CSMA/CD協(xié)議實現(xiàn)介質(zhì)訪問

8、控制，數(shù)據(jù)傳輸速率為10Mbps，主機甲和主機乙之間的距離為2km，信號傳播速度是200 000km/s。請回答下列問題，要求說明理由或?qū)懗鲇嬎氵^程。（1） 若主機甲和主機乙發(fā)送數(shù)據(jù)時發(fā)生沖突，則從開始發(fā)送數(shù)據(jù)時刻起，到兩臺主機均檢測到?jīng)_突時刻止，最短需經(jīng)過多長時間？最長需以過多長時間？（假設(shè)主機甲和主機乙發(fā)送數(shù)據(jù)過程中，其他主機不發(fā)送數(shù)據(jù)）（2） 若網(wǎng)絡(luò)不存在任何沖突與差錯主同甲總是以標準的最長以太網(wǎng)數(shù)據(jù)幀（1518字節(jié)）向主同乙發(fā)數(shù)據(jù)主機乙成功收到一個數(shù)據(jù)幀后立即發(fā)送下一個數(shù)據(jù)幀。此時主機甲的有效數(shù)據(jù)傳輸速率是多少？（不考慮以太網(wǎng)幀的前導(dǎo)碼）（1）當甲乙同時向?qū)Ψ桨l(fā)送數(shù)據(jù)時，兩臺主機均檢測

9、到?jīng)_突所需時間最短：1km/200000km/s2=110(-5)s當一方發(fā)送的數(shù)據(jù)馬上要到達另一方時，另一方開始發(fā)送數(shù)據(jù)，兩臺主機均檢測到?jīng)_突所需時間最長：2km/200000km/s2=210(-5)s（2）發(fā)送一幀所需時間：1518B/10Mbps=1.2144ms數(shù)據(jù)傳播時間：2km/200 000km/s=110(-5)s=0.01ms有效的數(shù)據(jù)傳輸速率=10Mbps1.2144ms/1.2244ms=9.92MbpsChapter 3layer.What is the string actually transmitted after bit stuffing?（E）輸出：6. W

10、hen bit stuffing is used, is it possible for the loss, insertion, or modification ofa single bit to cause an error not detected by the checksum? If not, why not? If so,how? Does the checksum length play a role here?（M）可能。假定原來的正文包含位序列01111110 作為數(shù)據(jù)。位填充之后，這個序列將變成011111010。如果由于傳輸錯誤第二個0 丟失了，收到的位串又變成01111

11、110，被接收方看成是幀尾。然后接收方在該串的前面尋找檢驗和，并對它進行驗證。如果檢驗和是16 位，那么被錯誤的看成是檢驗和的16 位的內(nèi)容碰巧經(jīng)驗證后仍然正確的概率是1/216。如果這種概率的條件成立了，就會導(dǎo)致不正確的幀被接收。顯然，檢驗和段越長，傳輸錯誤不被發(fā)現(xiàn)的概率會越低，但該概率永遠不等于零。15. A bit stream 10011101 is transmitted using the standard CRC method described in the text. The generator polynomial is x3 + 1. Show the actual bi

12、t string transmitted. Suppose the third bit from the left is inverted during transmission. Show that this error is detected at the receivers end.A: The frame isbitbitstreamwith an error in the thirdbitwhich is different from zero. Thus, the receiver detects the error and can ask for a retransmission

13、.16. Data link protocols almost always put the CRC in a trailer rather than in a header. Why?（E）CRC是在發(fā)送期間進行計算的。一旦把最后一位數(shù)據(jù)送上外出線路，就立即把CRC編碼附加在輸出流的后面發(fā)出。如果把CRC 放在幀的頭部，那么就要在發(fā)送之前把整個幀先檢查一遍來計算CRC。這樣每個字節(jié)都要處理兩遍，第一遍是為了計算檢驗碼，第二遍是為了發(fā)送。把CRC放在尾部就可以把處理時間減半。17. A channel has a bit rate of 4 kbps and a propagation del

14、ay of 20 msec. For what range of frame sizes does stop-and-wait give an efficiency of at least 50 percent?（E）當發(fā)送一幀的時間等于信道的傳播延遲的2倍時，信道的利用率為50%?；蛘哒f，當發(fā)送一幀的時間等于來回路程的傳播延遲時，效率將是50%。而在幀長滿足發(fā)送時間大于延遲的兩倍時，效率將會高于50%?，F(xiàn)在發(fā)送速率為4Mb/s，發(fā)送一位需要0.25微秒。只有在幀長不小于160kb 時，停等協(xié)議的效率才會至少達到50%。18. A 3000-km-long T1 trunk is used t

15、o transmit 64-byte frames using protocol 5.If the propagation speed is 6 sec/km, how many bits should the sequence numbers be?（M）為了有效運行，序列空間（實際上就是發(fā)送窗口大?。┍仨氉銐虻拇?，以允許發(fā)送方在收到第一個確認應(yīng)答之前可以不斷發(fā)送。信號在線路上的傳播時間為31. Consider an error-free 64-kbps satellite channel used to send 512-byte dataframes in one direction,

16、 with very short acknowledgements coming back the otherway. What is the maximum throughput for window sizes of 1, 7, 15, and 127? Theearth-satellite propagation time is 270 msec.（M）使用衛(wèi)星信道端到端的傳輸延遲為270ms，以64kb/s 發(fā)送，周期等于604ms。發(fā)送一幀的時間為64ms，我們需要604/64=9 個幀才能保持通道不空。對于窗口值1，每604ms 發(fā)送4096 位，吞吐率為4096/0.604=6.

17、8kb/s。對于窗口值7，每604ms 發(fā)送4096*7 位，吞吐率為4096*7/0.604=47.5kb/s。對于窗口值超過9（包括15、127），吞吐率達到最大值，即64kb/s。32. A 100-km-long cable runs at the T1 data rate. The propagation speed in thecable is 2/3 the speed of light in vacuum. How many bits fit in the cable?（E）在該電纜中的傳播速度是每秒鐘200 000km，即每毫秒200km，因此100km 的電纜將會在0.5m

18、s 內(nèi)填滿。T1 速率125 微秒傳送一個193 位的幀，0.5ms 可以傳送4 個T1 幀，即193*4=772bit。Chapter 43. Consider the delay of pure ALOHA versus slotted ALOHA at low load. Whichone is less? Explain your answer.（E）對于純的ALOHA，發(fā)送可以立即開始。對于分隙的ALOHA，它必須等待下一個時隙。這樣，平均會引入半個時隙的延遲。因此，純ALOHA 的延遲比較小。16. What is the baud rate of the standard 10-

19、Mbps Ethernet?（E）以太網(wǎng)使用曼徹斯特編碼，這就意味著發(fā)送的每一位都有兩個信號周期。標準以太網(wǎng)的數(shù)據(jù)率為10Mb/s，因此波特率是數(shù)據(jù)率的兩倍，即20MBaud。19. A 1-km-long, 10-Mbps CSMA/CD LAN (not 802.3) has a propagation speed of200 m/sec. Repeaters are not allowed in this system. Data frames are 256 bits long,including 32 bits of header, checksum, and other overh

20、ead. The first bit slot after asuccessful transmission is reserved for the receiver to capture the channel in orderto send a 32-bit acknowledgement frame. What is the effective data rate, excludingoverhead, assuming that there are no collisions?(M)依題知一公里的在銅纜中單程傳播時間為1/200000 =510-6 s=5 usec，往返傳播時間為2t

21、 =10 usec，一次完整的傳輸分為6步：發(fā)送者偵聽銅纜時間為10usec，若線路可用發(fā)送數(shù)據(jù)幀傳輸時間為256 bits / 10Mbps = 25.6 usec數(shù)據(jù)幀最后一位到達時的傳播延遲時間為5.0usec接收者偵聽銅纜時間為10 usec，若線路可用接收者發(fā)送確認幀用時3.2 usec確認幀最后一位到達時的傳播延遲時間為5.0 usec總共58.8sec，在這期間發(fā)送了224 bits 的數(shù)據(jù)，所以數(shù)據(jù)傳輸率為3.8 Mbps.21. Consider building a CSMA/CD network running at 1 Gbps over a 1-km cable wi

22、th no repeaters. The signal speed in the cable is 200,000 km/sec. What is the minimum frame size?(E)對于1km 電纜，單程傳播時間為1/200000 =510-6 s=5微秒 ，往返傳播時間為2t=10 微秒。為了能夠按照CSMA/CD 工作，最小幀的發(fā)射時間不能小于10 微秒。以1Gb/s速率工作，10 可以發(fā)送的比特數(shù)等于：（10*10-6）/（1*10-9）=10000因此，最小幀是10000 bit = 1250字節(jié)長。22. An IP packet to be transmitted

23、 by Ethernet is 60 bytes long, including all its headers. If LLC is not in use, is padding needed in the Ethernet frame, and if so, how many bytes?(E)最小的以太網(wǎng)幀是64 bytes，包含了以太網(wǎng)地址幀頭，類型/長度域，以及校驗和。由于幀頭域占用18 bytes，并且分組是60 bytes，總幀長是78 bytes，這已經(jīng)超過了64-byte 的最小限制。因此，不必再填充 z 了。23. Ethernet frames must be at le

24、ast 64 bytes long to ensure that the transmitter isstill going in the event of a collision at the far end of the cable. Fast Ethernet has thesame 64-byte minimum frame size but can get the bits out ten times faster. How is itpossible to maintain the same minimum frame size?(E)將快速以太網(wǎng)的電纜長度至為以太網(wǎng)的1/10即可

25、。24. Some books quote the maximum size of an Ethernet frame as 1518 bytes instead of 1500 bytes. Are they wrong? Explain your answer.（E）以太網(wǎng)一幀中數(shù)據(jù)占用是1500 bytes，但是把目的地地址，源地址，類型/長度域以及校驗和域也算上，幀總長就為1518 bytes26. How many frames per second can gigabit Ethernet handle? Think carefully and take into account

26、all the relevant cases. Hint: the fact that it is gigabit Ethernet matters.(E)最小的以太網(wǎng)幀是64bytes = 512 bits，所以依題1 Gbps 的帶寬可得1,953,125 =2*106 frames/sec，然而，這只是在充滿最小的幀時是這樣，如果沒有充滿幀，填充短幀至4096 bits，這時每秒處理的幀的最大數(shù)量為244,140 bytes，對于最大的幀長12,144 bits，每秒處理的幀的最大數(shù)量為82,345 frames/sec.28. In Fig. 4-27, four stations,

27、A, B, C, and D, are shown. Which of the last two stations do you think is closest to A and why?(E)站C最接近 A。因為C最先聽到A發(fā)出的RTS并且通過插入一個NAV信號作為回應(yīng)。D 對其沒有回應(yīng)，說明它不在A的頻率范圍內(nèi)。29. Suppose that an 11-Mbps 802.11b LAN is transmitting 64-byte framesback-to-back over a radio channel with a bit error rate of 10-7. How m

28、any framesper second will be damaged on average?（E）一幀是64bytes=512 bits，位出錯率為p=10-7 ，所有512 位正確到達的概率為(1- p)512 = 0.9999488，所以幀被破壞的概率約為5*10-5，每秒鐘發(fā)送的幀數(shù)為11* 106/512 = 21,484frames/sec ，將上兩個數(shù)乘一下，大約每秒鐘有一幀被破壞。30. An 802.16 network has a channel width of 20 MHz. How many bits/sec can be sent to a subscriber

29、station?（E）這取決于離子站有多遠。如果子站就在附近，那么使用QAM-64 可得帶寬120Mbps；中等距離時，使用QAM-16 可得帶寬80 Mbps；遠程距離，QPSK 可得帶寬40 Mbps.（原題給出的是20mhz 的帶寬，要求的是數(shù)據(jù)率，按照前面的Nyquist定理，最大數(shù)據(jù)率應(yīng)該是：2HlogN,但是答案沒有乘以2。）31. IEEE 802.16 supports four service classes.Which service class is the best choice for sending uncompressed video?（E）未壓縮的視頻有一個固定

30、的位速率。每幀都有與前一幀相同的點數(shù)量，因此，可能要準確計算需要的帶寬。最后，最好選用固定位速率服務(wù)。Chapter 51. Give two example computer applications for which connection-oriented serviceis appropriate. Now give two examples for which connectionless service is best.(E)文件傳送、遠程登錄和視頻點播需要面向連接的服務(wù)。另一方面，信用卡驗證和其他的銷售點終端、電子資金轉(zhuǎn)移，以及許多形式的遠程數(shù)據(jù)庫訪問生來具有無連接的性質(zhì)，在一個

31、方向上傳送查詢，在另一個方向上返回應(yīng)答。5. Consider the following design problem concerning implementation of virtual-circuit service. If virtual circuits are used internal to the subnet, each data packet must have a 3-byte header and each router must tie up 8 bytes of storage for circuit identification. If datagrams a

32、re used internally, 15-byte headers are needed but no router table space is required. Transmission capacity costs 1 cent per 106 bytes, per hop. Very fast router memory can be purchased for 1 cent per byte and is depreciated over two years, assuming a 40-hour business week. The statistically average

33、 session runs for 1000 sec, in which time 200 packets are transmitted. The mean packet requires four hops. Which implementation is cheaper, and by how much?（H）4 跳意味著引入了5個路由器。實現(xiàn)虛電路需要在1000 秒內(nèi)固定分配5*8=40 字節(jié)的存儲器。實現(xiàn)數(shù)據(jù)報需要比實現(xiàn)虛電路多傳送的頭信息的容量等于(15-3 )42009600 字節(jié)-跳段?，F(xiàn)在的問題就變成了40000 字節(jié)-秒的存儲器對比9600 字節(jié)-跳段的電路容量的開銷。如果

34、存儲器的使用期為兩年，即360085522= 1.5107秒，一個字節(jié)-秒的代價為1/( 1.5107) = 6.710-8 分，那么40000 字節(jié)-秒的代價為2.7 毫分。另一方面，1 個字節(jié)-跳段代價是10-6 分，9600 個字節(jié)-跳段的代價為10-6 9600=9.610-3分，即9.6 毫分，即在這1000 秒內(nèi)的時間內(nèi)便宜大約6.9 毫分。7. Consider the network of Fig. 5-7, but ignore the weights on the lines. Supposethat it uses flooding as the routing algo

35、rithm. If a packet sent by A to D has amaximum hop count of 3, list all the routes it will take. Also tell how many hopsworth of bandwidth it consumes.（E）所有的路由選擇如下： ABCD, ABCF, ABEF, ABEG, AGHD, AGHF, AGEB, andAGEF，所以總跳數(shù)為2410. If delays are recorded as 8-bit numbers in a 50-router network, and delay

36、vectors are exchanged twice a second, how much bandwidth per (full-duplex) line ischewed up by the distributed routing algorithm? Assume that each router has threelines to other routers.（E）路由表的長度等于8*50=400bit。該表每秒鐘在每條線路上發(fā)送2 次，因此400*2=800b/s，即在每條線路的每個方向上消耗的帶寬都是800 bps。17. In Fig. 5-20, do nodes H or

37、I ever broadcast on the lookup shown starting at A?（E）在d中，E，H，I接收到了廣播信息之后陰影節(jié)點是新的接收節(jié)點；箭頭顯示了可能的逆向路由路徑。H 收到分組A后，它廣播A；然而，I知道了如何到達I，所以I不廣播收到的分組。18. Suppose that node B in Fig. 5-20 has just rebooted and has no routinginformation in its tables. It suddenly needs a route to H. It sends out broadcasts withT

38、TL set to 1, 2, 3, and so on. How many rounds does it take to find a route?（E）從結(jié)點B到H 需要3跳，因此要花3圈來找到路由線路。24. Give an argument why the leaky bucket algorithm should allow just one packet per tick, independent of how large the packet is.（M）通常計算機能夠以很高的速率產(chǎn)生數(shù)據(jù)，網(wǎng)絡(luò)也可以用同樣的速率運行。然而，路由器卻只能在短時間內(nèi)以同樣高的速率處理數(shù)據(jù)。對于排在隊

39、列中的一個分組，不管它有多大，路由器必須做大約相同分量的工作。顯然，處理10 個100 字節(jié)長的分組所作的工作比處理1 個1000 字節(jié)長的分組要做的工作多得多。25. The byte-counting variant of the leaky bucket algorithm is used in a particularsystem. The rule is that one 1024-byte packet, or two 512-byte packets, etc., may besent on each tick. Give a serious restriction of thi

40、s system that was not mentioned inthe text.（E）不可以發(fā)送任何大于1024 字節(jié)的分組。28. Imagine a flow specification that has a maximum packet size of 1000 bytes, atoken bucket rate of 10 million bytes/sec, a token bucket size of 1 million bytes, anda maximum transmission rate of 50 million bytes/sec. How long can a

41、burst atmaximum speed last?（E）令最大突發(fā)時間長度為 t 秒。在極端情況下，漏桶在突發(fā)期間的開始是充滿的（1MB），這期間數(shù)據(jù)流入桶內(nèi)10 t MB，流出包含50 t MB，由等式1+10 t=50t，得到 t=1/40s，即25ms。因此，以最大速率突發(fā)傳送可維持25ms 的時間。32. Is fragmentation needed in concatenated virtual-circuit internets or only in datagram systems?（E）都需要分割功能。即使是在一個串接的虛電路網(wǎng)絡(luò)中，沿通路的某些網(wǎng)絡(luò)可能接受1024 字節(jié)

42、分組，而另一些網(wǎng)絡(luò)可能僅接受48字節(jié)分組，分割功能仍然是需要的。33. Tunneling through a concatenated virtual-circuit subnet is straightforward:the multiprotocol router at one end just sets up a virtual circuit to the other end andpasses packets through it. Can tunneling also be used in datagram subnets? If so,how?（E）可以。只需把分組封裝在屬于所

43、經(jīng)過的子網(wǎng)的數(shù)據(jù)報的載荷段中，并進行發(fā)送。34. Suppose that host A is connected to a router R 1, R 1 is connected to another router, R 2, and R 2 is connected to host B. Suppose that a TCP message that contains 900 bytes of data and 20 bytes of TCP header is passed to the IP code at host A for delivery to B. Show the Tot

44、al length, Identification, DF, MF, and Fragment offset fields of the IP header in each packet transmitted over the three links. Assume that link A-R1 can support a maximum frame size of 1024 bytes including a 14-byte frame header, link R1-R2 can support a maximum frame size of 512 bytes, including a

45、n 8-byte frame header, and link R2-B can support a maximum frame size of 512 bytes including a 12-byte frame header.（M）在I1 最初的IP數(shù)據(jù)報會被分割成兩個IP數(shù)據(jù)報，以后不會再分割了。鏈路A-R1：Length = 940; ID = x; DF = 0; MF = 0; Offset = 0鏈路R1-R2：(1) Length = 500; ID = x; DF = 0; MF = 1; Offset = 0(2) Length = 460; ID = x; DF = 0

46、; MF = 0; Offset = 60鏈路R2-B:(1) Length = 500; ID = x; DF = 0; MF = 1; Offset = 0(2) Length = 460; ID = x; DF = 0; MF = 0; Offset = 6036AnIPdatagramusingthestrictsourceroutingoptionhastobefragmented.Doyouthinktheoptioniscopiedintoeachfragment,orisitsufficienttojustputitinthefirstfragment?Explainyoura

47、nswer.Since the information is needed to route every fragment, the option must appear in every fragment.38Convert IP address whose hexadecimal notation is C22F1582 to dotted decimal notation?The address is 194.47.21.130.39AnetworkontheInternethasasubnetmaskof255.255.240.0.Whatisthemaximumnumberofhos

48、tsitcanhandle?The mask is 20 bits long, so the network part is 20 bits. The remaining 12 bits are for the host, so 4096 host addresses exist.40. A large number of consecutive IP address are available starting at 198.16.0.0.Suppose that four organizations, A, B, C, and D, request 4000, 2000, 4000, an

49、d 8000 addresses, respectively, and in that order. For each of these, give the first IP address assigned, the last IP address assigned, and the mask in the /s notation.（M）A：4000212 ；B：2000211 ；C：4000212 ；D：8000213 ；始地址，尾地址，和子網(wǎng)掩碼如下：A：198.16.0.0 子網(wǎng)寫作B：198.16.16.0 198.16.23.255 子網(wǎng)寫作C：198.16.32.0 198.16

50、.47.255 子網(wǎng)寫作43A router has the following (CIDR) entries in its routing table:Address/mask Next hop Interface 0 Interface 1 Router 1Default Router 2For each of the following IP addresses, what does the router do if a packet with that address arrives?The packets are routed as follows:(a) Interface 1(b

51、) Interface 0(c) Router 2(d) Router 1(e) Router 247Describe a way to reassemble IP fragments at the destination.In the general case, the problem is nontrivial. Fragments may arrive out of order and some may be missing. On a retransmission, the datagram may be fragmented in different-sized chunks. Fu

52、rthermore, the total size is not known until the last fragment arrives. Probably the only way to handle reassembly is to buffer all the pieces until the last fragment arrives and the size is known. Then build a buffer of the right size, and put the fragments into the buffer, maintaining a bit map wi

53、th 1 bit per 8 bytes to keep track of which bytes are present in the buffer. When all the bits in the bit map are 1,the datagram is complete.51IPv6uses16-byteaddresses. If ablockof 1 millionaddressesis allocated every picosecond, how long will theaddresseslast?With 16 bytes there are 2128 or 3.4 103

54、8 addresses. If we allocate them at a rate of 1018 per second, they will last for 1013 years. This number is 1000 times the age of the universe. Of course, the address space is not flat, so they are not allocated linearly, but this calculation shows that even with an allocation scheme that has an ef

55、ficiency of 1/1000 (0.1 percent), one will never run out.Chapter 65. Why does the maximum packet lifetime, T, have to be large enough to ensure that not only the packet but also its acknowledgements have vanished?（M）首先看三次握手過程是如何解決延遲的重復(fù)到達的分組所引起的問題的。正常情況下，當主機1 發(fā)出連接請求時，主機1 選擇一個序號x，并向主機2 發(fā)送一個包含該序號的請求TPD

56、U；接著，主機2 回應(yīng)一個接受連接的TPDU，確認x，并聲明自己所選用的初始序列號y；最后，主機1 在其發(fā)送的第一個數(shù)據(jù)TPDU中確認主機2 所選擇的初始序列號。當出現(xiàn)延遲的重復(fù)的控制TPDU 時，一個TPDU 是來自于一個已經(jīng)釋放的連接的延遲重復(fù)的連接請求（ CONNECTION REQUEST），該TPDU 在主機1 毫不知情的情況下到達主機2。主機2 通過向主機1 發(fā)送一個接受連接的TPDU（CONNECTION ACCEPTED）來響應(yīng)該TPDU，而該接受連接的TPDU 的真正目的是證實主機1 確實試圖建立一個新的連接。在這一點上，關(guān)鍵在于主機2 建議使用y 作為從主機2 到主機1 交

57、通的初始序列號，從而說明已經(jīng)不存在包含序列號為y 的TPDU，也不存在對y 的應(yīng)答分組。當?shù)诙€延遲的TPDU 到達主機2 時，z 被確認而不是y 被確認的事實告訴主機2 這是一個舊的重復(fù)的TPDU，因此廢止該連接過程。在這里。三次握手協(xié)議是成功的。最壞的情況是延遲的“連接請求”和對“連接被接收”的確認應(yīng)答都在網(wǎng)絡(luò)上存活?？梢栽O(shè)想，當?shù)? 個重復(fù)分組到達時，如果在網(wǎng)上還存在一個老的對序列號為y 的分組的確認應(yīng)答，顯然會破壞三次握手協(xié)議的正常工作，故障性的產(chǎn)生一條沒有人真正需要的連接，從而導(dǎo)致災(zāi)難性的后果。（The reason why T must be large enough to ensure that both the packets andacknowledgements have vanished is to deter duplicate packets from being introduced into the network.）6. Imagine that