優(yōu)質(zhì)計(jì)算機(jī)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(jì)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(huì)影響高層或低層的協(xié)議13. What is the principal difference between connectionless communication and connection-oriented communication?(E)主要的區(qū)別有兩條。其一：面向連接通信分為三個(gè)階段，第一是建立連接，在此階段，發(fā)出一個(gè)建立連接的請(qǐng)求。 第二階段，只有在連接成功建立之后

2、，保持連接狀態(tài)，才能開始數(shù)據(jù)傳輸。第三階段，當(dāng)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(jìn)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.An swer: (a) Data link l

3、ayer. (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(bào)服務(wù)。25. When a file is transferred between two computers, two acknowledgement strategies are possible. In the first one, the file is chopped up into packets, which are individually acknowledged by the

4、 receiver, 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(duì)每一個(gè)分組逐一進(jìn)行確認(rèn)較好，此時(shí)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(cuò)的情況下，僅在整個(gè)文件傳送的結(jié)尾發(fā)送一次確認(rèn)，而減少了

5、確認(rèn)的次數(shù)，節(jié)省了帶寬；不過，即使有單個(gè)分組丟失，也需要重傳整個(gè)文件。(課堂練習(xí))若待發(fā)送數(shù)據(jù)為：1010001101,現(xiàn)要計(jì)算CRC校驗(yàn)和。如果我們選G=1101011) 請(qǐng)給出對(duì)應(yīng)的生成多項(xiàng)式G(X)。2) 請(qǐng)給出實(shí)際發(fā)送出去的數(shù)據(jù)(比特流)。除1.1.0 mi io loadriipiaoooo111011110101T I I I11101Q I11111010X1001X0101r110 D10fl Hl.工缶*-左余數(shù)循環(huán)冗余竝益的原理說明補(bǔ)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(bào)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(shí)延為d秒，數(shù)據(jù)率為b

6、(bit/s)。在電路交換時(shí)電路的建立時(shí)間為 s秒。在分組交換時(shí)分組長(zhǎng)度為p (bit)，且各結(jié)點(diǎn)的排隊(duì)等待時(shí)間可忽略不計(jì)。問在怎樣的條件下，分組交換的時(shí)延比電路交換要?。侩娐方粨Q時(shí)延：s+x/b+kd分組交換時(shí)延：x/b+kd+(k-1)p/bx/b+kd+(k-1)p/b < s+x/b+kds > (k-1)p/b*但前提是：x>>p,或分組數(shù)大于鏈路數(shù).補(bǔ)充題2：在上題的分組交換網(wǎng)中，設(shè)報(bào)文和分組長(zhǎng)度分別為x和(p+h)(bit)，其中p為分組的數(shù)據(jù)部分的長(zhǎng)度，而h為每個(gè)分組所帶的控制信息固定長(zhǎng)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(b

7、it/s)，但傳播時(shí)延和結(jié)點(diǎn)的排隊(duì)時(shí)延均可忽略不計(jì)。若打算使總的時(shí)延為最小，問分組的數(shù)據(jù)部分長(zhǎng)度p應(yīng)該取多大？分組數(shù)：f -發(fā)送的數(shù)據(jù)量：2；鼻總時(shí)延"卜+制¥心1)-也嚴(yán)求D對(duì)P的導(dǎo)數(shù)，令其為0： 0 = 0p - yjxh/(k -1)(2010考研)在下圖所示的采用“存儲(chǔ)-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，分組大小為1000 B，其中分組頭大小為20 B。若主機(jī)H1向主機(jī)H2發(fā)送 一個(gè)大小為980 000 B的文件，則在不考慮分組拆裝時(shí)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(shí)間？b=100Mbps; x=9

8、80 000 B p=(1000-20) B; h=20B;k=3nrD=80.16 msec分組長(zhǎng)度 L=1000B=980B+20B分組數(shù) N=980000/980=1000發(fā)送 1 個(gè)分組的時(shí)間 Ttran=(1000x8"(100x106) =8x10-5 secTtotal=N x Ttra n+2 x Ttra n= 80.16 msec某局域網(wǎng)采用CSMA/CD協(xié)議實(shí)現(xiàn)介質(zhì)訪問控制，數(shù)據(jù)傳輸速率為10Mbps，主機(jī)甲和主機(jī)乙之間的距離為2km，信號(hào)傳播速度是200 000km/s。請(qǐng)回答下列問題，要求說明理由或?qū)?出計(jì)算過程。（ 1） 若主機(jī)甲和主機(jī)乙發(fā)送數(shù)據(jù)時(shí)發(fā)生沖突

9、，則從開始發(fā)送數(shù)據(jù)時(shí)刻起，到兩臺(tái)主機(jī)均 檢測(cè)到?jīng)_突時(shí)刻止，最短需經(jīng)過多長(zhǎng)時(shí)間？最長(zhǎng)需以過多長(zhǎng)時(shí)間？（假設(shè)主機(jī)甲和主機(jī)乙 發(fā)送數(shù)據(jù)過程中，其他主機(jī)不發(fā)送數(shù)據(jù)）（ 2） 若網(wǎng)絡(luò)不存在任何沖突與差錯(cuò)主同甲總是以標(biāo)準(zhǔn)的最長(zhǎng)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(jī)乙成功收到一個(gè)數(shù)據(jù)幀后立即發(fā)送下一個(gè)數(shù)據(jù)幀。此時(shí)主機(jī)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(dāng)甲乙同時(shí)向?qū)Ψ桨l(fā)送數(shù)據(jù)時(shí)，兩臺(tái)主機(jī)均檢測(cè)到?jīng)_突所需時(shí)間最短： 1km/200000km/s x 2=1 x 10A（-5）s當(dāng)一方發(fā)送的數(shù)據(jù)馬上要到達(dá)另一方時(shí)， 另一方開始發(fā)送數(shù)據(jù)， 兩臺(tái)主機(jī)均檢測(cè)到?jīng)_突 所需時(shí)間最長(zhǎng)：2k

10、m/200000km/s x 2=2 x 10A（-5）s（2）發(fā)送一幀所需時(shí)間： 1518B/10Mbps=1.2144ms 數(shù)據(jù)傳播時(shí)間： 2km/200 000km/s=1 x 10A（-5）s=0.01ms 有效的數(shù)據(jù)傳輸速率 =10Mbpsx 1.2144ms/1.2244ms=9.92MbpsChapter 35. A bit string, 0111101111101111110, needs to be transmitted at the data link layer.What is the string actually transmitted after bit stu

11、ffing? （ E） 輸出： 011110111110011111010.6. When bit stuffing is used, is it possible for the loss, insertion, or modification of a 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

12、，這個(gè)序列將變成 011111010。如果由于傳輸錯(cuò)誤第二個(gè)0 丟失了，收到的位串又變成01111110，被接收方看成是幀尾。然后接收方在該串的前面尋找檢驗(yàn)和，并對(duì)它進(jìn)行驗(yàn)證。如果檢驗(yàn)和是 16 位， 那么被錯(cuò)誤的看成是檢驗(yàn)和的 16 位的內(nèi)容碰巧經(jīng)驗(yàn)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(huì)導(dǎo)致不正確的幀被接收。 顯然， 檢驗(yàn)和段越長(zhǎng)，傳輸錯(cuò)誤不被發(fā)現(xiàn) 的概率會(huì)越低，但該概率永遠(yuǎn)不等于零。15. A bit stream 10011101 is transmitted using the standard CRC method described in the te

13、xt. The generator polynomial is x3 + 1. Show the actual bit string transmitted. Suppose the third bit from the left is inverted during transmission. Show that this error is detected at the receiver's end.A: The frame is 10011101. The generator is 1001. The message after appending three zeros is

14、10011101000. The remainder on dividing 10011101000 by 1001 is 100. So, the actual bit string transmitted is 10011101100. The receivedbit stream with an errorin the third bit from the left is 10111101100. Dividing this by 1001 produces a remainder 100, which is different from zero. Thus, the receiver

15、 detects the error and can ask for a retransmission.16. Data link protocols almost always put the CRC in a trailer rather than in a header. Why?（E）CR（是在發(fā)送期間進(jìn)行計(jì)算的。一旦把最后一位數(shù)據(jù)送上外出線路，就立即把CR（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(gè)幀先檢查一遍來 計(jì)算CRC這樣每個(gè)字節(jié)都要處理兩遍，第一遍是為了計(jì)算檢驗(yàn)碼，第二遍是為了發(fā)送。把CR（放在尾部就可以把處理時(shí)間減半。17. A chan

16、nel has a bit rate of 4 kbps and a propagation delay of 20 msec. For what range of frame sizes does stop-and-wait give an efficiency of at least 50 percent?（ E）當(dāng)發(fā)送一幀的時(shí)間等于信道的傳播延遲的2倍時(shí)，信道的利用率為50%或者說，當(dāng)發(fā)送一幀的時(shí)間等于來回路程的傳播延遲時(shí)，效率將是50%而在幀長(zhǎng)滿足發(fā)送時(shí)間大于延遲的兩倍時(shí)，效率將會(huì)高于50%現(xiàn)在發(fā)送速率為4Mb/s，發(fā)送一位需要0.25微秒。(20，I0'5 2) ('

17、0.25 10-1)160000 ki只有在幀長(zhǎng)不小于160kb時(shí)，停等協(xié)議的效率才會(huì)至少達(dá)到50%18. A 3000-km-long T1 trunk is used to transmit 64-byte frames using protocol 5f the propagation speed is 6 卩 sec/km, how many bits should the sequence numbers be? （M） 為了有效運(yùn)行，序列空間（實(shí)際上就是發(fā)送窗口大?。┍仨氉銐虻拇螅栽试S發(fā)送方在收到第一個(gè)確認(rèn)應(yīng)答之前可以不斷發(fā)送。信號(hào)在線路上的傳播時(shí)間為6000?lSfl（X>

18、;f 抑庚齦 發(fā)建的黑禎從幵鯨覽送起.】畀3血 岳克全測(cè)達(dá)按收方”琥認(rèn)應(yīng)葆叉葩 了悝少的發(fā)理時(shí)風(fēng)I憩略平計(jì)）和回程前也皿這樣，刖臨一耀的時(shí)冊(cè)是北少36 3W31=1O也It星說.為充需娃吐管道.績(jī)耍垂抄】0禎.因it庠列號(hào)為了皿的覽方是口的庫廈*T1的載據(jù)堪宰根關(guān)亡的特構(gòu)聞國(guó)理仃好J-24）/193'1 5+UU44Gib 回葩善秦卬最（i?3d> nS3*i441.5J6mbps31. Consider an error-free 64-kbps satellite channel used to send 512-byte data frames in one directi

19、on, with very short acknowledgements coming back the other way. What is the maximum throughput for window sizes of 1, 7, 15, and 127? The earth-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(shí)間為64ms,我們需要604/64=9個(gè)幀才能保持通道不空。對(duì)于窗口值1，每604ms發(fā)送4096位，吞吐率為4096/0.604=

20、6.8kb/s。對(duì)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(duì)于窗口值超過9 (包括15、127),吞吐率達(dá)到最大值，即 64kb/s。32. A 100-km-long cable runs at the T1 data rate. The propagation speed in the cable 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

21、的電纜將會(huì)在0.5ms內(nèi)填滿。T1速率125微秒傳送一個(gè)193位的幀，0.5ms可以傳送4個(gè)T1幀，即193*4=772bitChapter 43. Consider the delay of pure ALOHA versus slotted ALOHA at low load. Which one is less? Explain your answer. ( E)對(duì)于純的ALOHA發(fā)送可以立即開始。對(duì)于分隙的ALOHA它必須等待下一個(gè)時(shí)隙。這樣，平均會(huì)引入半個(gè)時(shí)隙的延遲。因此，純ALOHA的延遲比較小。16. What is the baud rate of the standard 1

22、0-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(gè)信號(hào)周期。標(biāo)準(zhǔn)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20MBaud19. 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 o

23、verhead. The first bit slot after a successful transmission is reserved for the receiver to capture the channel in order to send a 32-bit acknowledgement frame. What is the effective data rate, excluding overhead, assuming that there are no collisions?(M)依題知一公里的在銅纜中單程傳播時(shí)間為1/200000 =5X 10-6 s=5 usee

24、，往返傳播時(shí)間為2t =10 usee ，一次完整的傳輸分為 6步： 發(fā)送者偵聽銅纜時(shí)間為10usee,若線路可用發(fā)送數(shù)據(jù)幀傳輸時(shí)間為 256 bits / 10Mbps = 25.6 usee數(shù)據(jù)幀最后一位到達(dá)時(shí)的傳播延遲時(shí)間為5.0usee接收者偵聽銅纜時(shí)間為10 usee，若線路可用接收者發(fā)送確認(rèn)幀用時(shí) 3.2 usee確認(rèn)幀最后一位到達(dá)時(shí)的傳播延遲時(shí)間為5.0 usee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

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

26、t to be transmitted 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(zhǎng)度域，以及校驗(yàn)和。由于幀頭域占用18 bytes，并且分組是60 bytes ,總幀長(zhǎng)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. Ethern

27、et frames must be at least 64 bytes long to ensure that the transmitter is still going in the event of a collision at the far end of the cable. Fast Ethernet has the same 64-byte minimum frame size but can get the bits out ten times faster. How is it possible to maintain the same minimum frame size?

28、(E)將快速以太網(wǎng)的電纜長(zhǎng)度至為以太網(wǎng)的1/10即可。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(zhǎng)度域以及校驗(yàn)和域也算上，幀總長(zhǎng)就為 1518 bytes26. How many frames per second can gigabit Ethernet handle? Think ca

29、refully and take into account 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*106frames/sec，然而，這只是在充滿最小的幀時(shí)是這樣，如果沒有充滿幀，填充短幀至4096 bits ，這時(shí)每秒處理的幀的最大數(shù)量為244,140 bytes，對(duì)于最大的幀長(zhǎng)12,144 bits，每秒處理的幀的最大數(shù)量為82,345 frames/sec.28

30、. In Fig. 4-27, four stations, A, B, C, and D, are shown. Which of the last two stations do you think is closest to A and why?(E).A_S|如DCT* | AChB I 二 _ D|卜恣Tirek站C最接近A。因?yàn)镃最先聽到A發(fā)出的RTS并且通過插入一個(gè)NAVt號(hào)作為回應(yīng)。D對(duì)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 frames back-t

31、o-back over a radio channel with a bit error rate of 10-7. How many frames per second will be damaged on average?( E)一幀是64bytes=512 bits，位出錯(cuò)率為p=10-7，所有512位正確到達(dá)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(gè)數(shù)乘一下，大約每秒鐘有一幀被破壞。30. An 802.16 network has a channel w

32、idth of 20 MH 乙 How many bits/sec can be sent to a subscriber station?（ E）這取決于離子站有多遠(yuǎn)。如果子站就在附近，那么使用QAM-64可得帶寬120Mbps；中等距離 時(shí)，使用QAM-16可得帶寬80 Mbps;遠(yuǎn)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 t

33、he best choice for sending uncompressed video? (E)未壓縮的視頻有一個(gè)固定的位速率。每幀都有與前一幀相同的點(diǎn)數(shù)量，因此，可能要準(zhǔn)確計(jì) 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 service is appropriate. Now give two examples for which connectionless service is best.(E) 文件傳送、遠(yuǎn)程登錄和視頻點(diǎn)播需

34、要面向連接的服務(wù)。另一方面，信用卡驗(yàn)證和其他的銷售點(diǎn)終端、電子資金轉(zhuǎn)移，以及許多形式的遠(yuǎn)程數(shù)據(jù)庫訪問生來具有無連接的性質(zhì)，在一個(gè)方向上傳送查詢，在另一個(gè)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

35、up 8 bytes of storage for circuit identification. If datagrams are 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 ye

36、ars, assuming a 40-hour business week. The statistically average 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(gè)路由器。實(shí)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(chǔ)器。 實(shí)現(xiàn)數(shù)據(jù)報(bào)需要比實(shí)現(xiàn)虛電路多傳送的頭信息的容量等于(15

37、-3 ) X 4X 200 = 9600字節(jié)-跳段?，F(xiàn)在的問題就變成了 40000字節(jié)-秒的存儲(chǔ)器對(duì)比9600字節(jié)-跳段的電路容量的開銷。如果 存儲(chǔ)器的使用期為兩年，即3600X 8X 5 X 52X 2= 1.5 X 107秒，一個(gè)字節(jié)-秒的代價(jià)為1/( 1.5 X 107) = 6.7 X 10-8分，那么40000字節(jié)-秒的代價(jià)為2.7毫分。另一方面，1個(gè) 字節(jié)-跳段代價(jià)是10-6分，9600個(gè)字節(jié)-跳段的代價(jià)為10-6 X 9600=9.6 X 10-3分，即9.6 毫分，即在這1000秒內(nèi)的時(shí)間內(nèi)便宜大約 6.9毫分。7. Consider the network of Fig. 5

38、-7, but ignore the weights on the lines. Suppose that it uses flooding as the routing algorithm. If a packet sent by A to D has a maximum hop count of 3, list all the routes it will take. Also tell how many hops worth of bandwidth it consumes. ( E)所有的路由選擇如下：ABCD, ABCF, ABEF, ABEG, AGHD, AGHF, AGEB,

39、andAGEF所以總跳數(shù)為24 10. If delays are recorded as 8-bit numbers in a 50-router network, and delay vectors are exchanged twice a second, how much bandwidth per (full-duplex) line is chewed up by the distributed routing algorithm? Assume that each router has three lines to other routers. ( E)路由表的長(zhǎng)度等于8*50=

40、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(gè)方向上消耗的帶寬都是800 bps。17. In Fig. 5-20, do nodes H or I ever broadcast on the lookup shown starting at A? ( E)%1TK；LA r b用比V淖I伸？神*在d中，E，H, I接收到了廣播信息之后陰影節(jié)點(diǎn)是新的接收節(jié)點(diǎn)；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(dá)I，所以I不廣播收到的分組。18. Suppose that node B in Fig. 5-20 has

41、 just rebooted and has no routing information in its tables. It suddenly needs a route to H. It sends out broadcasts with TTL set to 1, 2, 3, and so on. How many rounds does it take to find a route? （ E） 從結(jié)點(diǎn)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 o

42、ne packet per tick,independent of how large the packet is.（ M ）通常計(jì)算機(jī)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(yùn)行。 然而， 路由器卻只能在短時(shí)間內(nèi)以同樣高的速率處理數(shù)據(jù)。 器必須做大約相同分量的工作。顯然，處理 1000 字節(jié)長(zhǎng)的分組要做的工作多得多。對(duì)于排在隊(duì)列中的一個(gè)分組， 不管它有多大，10 個(gè) 100 字節(jié)長(zhǎng)的分組所作的工作比處理路由1 個(gè)25. The byte-counting variant of the leaky bucket algorithm is used in a particular sys

43、tem. The rule is that one 1024-byte packet, or two 512-byte packets, etc., may be sent on each tick. Give a serious restriction of this system that was not mentioned in the 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, a token bucket ra

44、te of 10 million bytes/sec, a token bucket size of 1 million bytes, and a maximum transmission rate of 50 million bytes/sec. How long can a burst at maximum speed last?（ E）令最大突發(fā)時(shí)間長(zhǎng)度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) 10A t MB，流出包含50A t MB，由等式1+10A t=50 t，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(shí)間

45、。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(gè)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)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

46、 sets up a virtual circuit to the other end and passes 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經(jīng)過的子網(wǎng)的數(shù)據(jù)報(bào)的載荷段中，并進(jìn)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

47、. 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 Total length, Identification, DF, MF, and Fragment offset fields of the IP header in each packet transmitted over the three links. Assume that link A

48、-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 an 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11最初的IP數(shù)據(jù)報(bào)會(huì)被分割成兩個(gè)IP數(shù)據(jù)報(bào)

49、，以后不會(huì)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; 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 =6036 An IP datagram using

50、the strict source routing option has to be fragmented. Do you thi nk the option iscopied into each fragment, or is it sufficient to just put it in the first fragment ? Explain your answer.Since the information is needed to route every fragment, the option must appear in every fragment.38Convert IP a

51、ddress whose hexadecimal notation is C22F1582 to dotted decimal notation? The address is 194.47.21.130.39 A network on the Internet has a subnet mask of 255.255.240.0. What is the maximum n umber ofhosts it can handle?The mask is 20 bits long, so the network part is 20 bits. The remaining 12 bits ar

52、e 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, and 8000 addresses, respectively, and in that order. For each of these, give the first IP address

53、assigned, the last IP address assigned, and the mask in the w.x.y.z/s notation.( M )A:4000c212 ; B: 2000c211 ;C:4000212 ;D:8000生13;始地址，尾地址，和子網(wǎng)掩碼如下：A: 198.16.0.0- 198.16.15.255 子網(wǎng)寫作 198.16.0.0/20B: 198.16.16.0- 198.16.23.255子網(wǎng)寫作 198.16.16.0/21C: 198.16.32.0- 198.16.47.255子網(wǎng)寫作 198.16.32.0/2043 A route

54、r has the following (CIDR) entries in its routing table:Address/maskNext hop135.46.56.0/22Interface 0135.46.60.0/22Interface 1192.53.40.0/23Router 1DefaultRouter 2For each of the following IP addresses, what does the router do if a packet with that address arrives?(a) 135.46.63.10(b) 135.46.57.14(c)

55、 135.46.52.2(d) 192.53.40.7(e) 192.53.56.7The packets are routed as follows:(a) Interface 1(b) Interface 0(c) Router 2(d) Router 1(e) Router 247 Describe 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

56、 be missing. On a retransmission, the datagram may be fragmented in different-sized chunks. Furthermore, 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 bui

57、ld a buffer of the right size, and put the fragments into the buffer, maintaining a bit map with 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.51 IPv6 uses 16-byte addresses. If a block of 1 millionaddresses

58、is allocated every picosecond, how long will the addresseslast?With 16 bytes there are 2128 or 3.4 x 1038 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