計算機網(wǎng)絡(luò)(第4版)-清華大學(xué)出版社-習(xí)題答案(中文版)

第 1 章 概述1. 答：狗能攜帶21千兆字節(jié)或者168千兆位的數(shù)據(jù)。18 公里/小時的速度等于0.005 公里/秒，走過x公里的時間為x / 0.005 = 200x秒， 產(chǎn)生的數(shù)據(jù)傳輸速度為168/200x Gbps或者840 /x Mbps。因此，與通信線路相比較，若x 32/x，或f max =x/32。36. 答：10-9 的漂移意味著109 秒中的1 秒，或1 秒中的10-9 秒。對于OC-1 速率，即51.840Mbps，取近似值50Mbps，大約一位持續(xù)20ns。這就說明每隔20 秒，時鐘就要偏離1位。這就說明，時鐘必須每隔10 秒或更頻繁地進行同步，才能保持不會偏離太大。37. 答：基本的SONET 幀是美125產(chǎn)生810 字節(jié)。由于SONET 是同步的，因此不論是否有數(shù)據(jù)，幀都被發(fā)送出去。每秒8000 幀與數(shù)字電話系統(tǒng)中使用的PCM 信道的采樣頻率完全一樣。810字節(jié)的SONET 幀通常用90列乘以9行的矩形來描述，每秒傳送51.84Mbps，即88108000bps。這就是基本的SONET 信道，它被稱作同步傳輸信號STS-1，所有的SONET 干線都是由多條STS-1構(gòu)成。每一幀的前3 列被留作系統(tǒng)管理信息使用，前3 行包含段開銷，后6 行包含線路開銷。剩下的87 列包含87988000bps。被稱作同步載荷信封的數(shù)據(jù)可以在任何位置開始。線路開銷的第一行包含指向第一字節(jié)的指針。同步載荷信封（SPE）的第一列是通路開銷。通路開銷不是嚴格的SONET 結(jié)構(gòu)，它在嵌入在載荷信封中。通路開銷端到端的流過網(wǎng)絡(luò)，因此把它與端到端的運載用戶信息的SPE 相關(guān)聯(lián)是有意義的。然而，它確實從可提供給端點用戶的50.112Mbps 中又減去1988000bps，即0.576Mbps，使之變成49.536Mbps 。OC-3相當于3個OC-1復(fù)用在一起，因此其用戶數(shù)據(jù)傳輸速率是49.546 3 148.608 Mbps。38. VT1.5 can accommodate 8000 frames/sec 3 columns 9 rows 8 bits =1.728 Mbps. It can be used to accommodate DS-1. VT2 can accommodate 8000 frames/sec 4 columns 9 rows 8 bits = 2.304 Mbps. It can be used to accommodate European CEPT-1 service. VT6 can accommodate 8000 frames/sec 12 columns 9 rows 8 bits = 6.912 Mbps. It can be used to accommodate DS-2 service.39. Message switching sends data units that can be arbitrarily long. Packet switching has a maximum packet size. Any message longer than that is split up into multiple packets.40. 答：當一條線路（例如OC-3）沒有被多路復(fù)用，而僅從一個源輸入數(shù)據(jù)時，字母c（表示conactenation，即串聯(lián)）被加到名字標識的后面，因此，OC-3 表示由3 條單獨的OC-1 線路復(fù)用成155.52Mbps，而OC-3c 表示來自單個源的155.52Mbps 的數(shù)據(jù)流。OC-3c 流中所包含的3 個OC-1 流按列交織編排，首先是流1 的第1 列，流2 的第1 列，流3 的第1 列，隨后是流1 的第2 列，流2 的第2 列，以此類推，最后形成270 列寬9 行高的幀。OC-3c 流中的用戶實際數(shù)據(jù)傳輸速率比OC-3 流的速率略高（149.760Mbps 和148.608Mbps），因為通路開銷僅在SPE 中出現(xiàn)一次，而不是當使用3 條單獨OC-1 流時出現(xiàn)的3 次。換句話說，OC-3c 中270 列中的260 列可用于用戶數(shù)據(jù)，而在OC-3 中僅能使用258列。更高層次的串聯(lián)幀（如OC-12c）也存在。OC-12c 幀有12*90=1080 列和9 行。其中段開銷和線路開銷占12*3=36 列，這樣同步載荷信封就有1080-36=1044 列。SPE 中僅1 列用于通路開銷，結(jié)果就是1043 列用于用戶數(shù)據(jù)。由于每列9 個字節(jié)，因此一個OC-12c 幀中用戶數(shù)據(jù)比特數(shù)是8 9104375096。每秒8000 幀，得到用戶數(shù)據(jù)速率750968000 =bps，即600.768Mbps。 所以，在一條OC-12c 連接中可提供的用戶帶寬是600.768Mbps。41. 答：The three networks have the following properties: 星型：最好為2，最差為2，平均為2；環(huán)型：最好為1，最差為n/2，平均為n/4如果考慮n 為奇偶數(shù)，則n 為奇數(shù)時，最壞為（n-1）/2，平均為（n+1）/4n 為偶數(shù)時，最壞為 n/2，平均為n2/4(n-1) 全連接：最好為1，最差為1，平均為1。42. 對于電路交換， t= s時電路建立起來；t s+ + x /d 時報文的最后一位發(fā)送完畢；t = s+ x/b+kd時報文到達目的地。而對于分組交換，最后一位在t=x/b 時發(fā)送完畢。為到達最終目的地，最后一個分組必須被中間的路由器重發(fā)k-1次，每次重發(fā)花時間p/ b，所以總的延遲為為了使分組交換比電路交換快，必須：所以：43. 答：所需要的分組總數(shù)是x /p ，因此總的數(shù)據(jù)加上頭信息交通量為(p+h)x/p位。源端發(fā)送這些位需要時間為(p+h )x / /pb中間的路由器重傳最后一個分組所花的總時間為(k-1)(p +h )/ b 因此我們得到的總的延遲為對該函數(shù)求p 的導(dǎo)數(shù)，得到令得到因為 p0，所以故時能使總的延遲最小。44. Each cell has six neighbors. If the central cell uses frequency group A, its six neighbors can use B, C, B, C, B, and C respectively. In other words, only 3 unique cells are needed. Consequently, each cell can have 280 frequencies.45. First, initial deployment simply placed cells in regions where there was high density of human or vehicle population. Once they were there, the operator often did not want to go to the trouble of moving them. Second, antennas are typically placed on tall buildings or mountains. Depending on the exact location of such structures, the area covered by a cell may be irregular due to obstacles near the transmitter. Third, some communities or property owners do not allow building a tower at a location where the center of a cell falls. In such cases, directional antennas are placed at a location not at the cell center.46. If we assume that each microcell is a circle 100 m in diameter, then each cell has an area of 2500. If we take the area of San Francisco, 1.2 108 m2 and divide it by the area of 1 microcell, we get 15,279 microcells. Of course, it is impossible to tile the plane with circles (and San Francisco is decidedly three-dimensional), but with 20,000 microcells we could probably do the job.47. Frequencies cannot be reused in adjacent cells, so when a user moves from one cell to another, a new frequency must be allocated for the call. If a user moves into a cell, all of whose frequencies are currently in use, the users call must be terminated.48. It is not caused directly by the need for backward compatibility. The 30 kHz channel was indeed a requirement, but the designers of D-AMPS did not have to stuff three users into it. They could have put two users in each channel, increasing the payload before error correction from 260 50= 13 kbps to 260 75 = 19.5 kbps. Thus, the quality loss was an intentional trade-off to put more users per cell and thus get away with bigger cells.49. D-AMPS uses 832 channels (in each direction) with three users sharing a single channel. This allows D-AMPS to support up to 2496 users simultaneously per cell. GSM uses 124 channels with eight users sharing a single channel. This allows GSM to support up to 992 users simultaneously. Both systems use about the same amount of spectrum (25 MHz in each direction).D-AMPS uses 30 KHz 892 = 26.76 MHz. GSM uses 200 KHz 124 =24.80 MHz. The difference can be mainly attributed to the better speech quality provided by GSM (13 Kbps per user) over D-AMPS (8 Kbps per user).50. The result is obtained by negating each of A, B, and C and then adding the three chip sequences. Alternatively the three can be added and then negated. The result is (+3 +1 +1 .1 .3 .1 .1 +1).51. By definitionIf T sends a 0 bit instead of 1 bit, its chip sequence is negated, with the i-th element becoming .Ti . Thus,52. When two elements match, their product is +1. When they do not match, their product is .1. To make the sum 0, there must be as many matches as mismatches. Thus, two chip sequences are orthogonal if exactly half of the corresponding elements match and exactly half do not match.53. Just compute the four normalized inner products:(.1 +1 .3 +1 .1 .3 +1 +1) d (.1 .1 .1 +1 +1 .1 +1 +1)/8 = 1(.1 +1 .3 +1 .1 .3 +1 +1) d (.1 .1 +1 .1 +1 +1 +1 .1)/8 = .1 (.1 +1 .3 +1 .1 .3 +1 +1) d (.1 +1 .1 +1 +1 +1 .1 .1)/8 = 0(.1 +1 .3 +1 .1 .3 +1 +1) d (.1 +1 .1 .1 .1 .1 +1 .1)/8 = 1The result is that A and D sent 1 bits, B sent a 0 bit, and C was silent.54. 答：可以，每部電話都能夠有自己到達端局的線路，但每路光纖都可以連接許多部電話。忽略語音壓縮，一部數(shù)字PCM電話需要64kbps 的帶寬。如果以64kbps 為單元來分割10Gbps，我們得到每路光纜串行 家?，F(xiàn)今的有線電視系統(tǒng)每根電纜串行數(shù)百家。55. 答：它既像TDM，也像FDM。100 個頻道中的每一個都分配有自己的頻帶（FDM），在每個頻道上又都有兩個邏輯流通過TDM 交織播放（節(jié)目和廣告交替使用頻道）。This example is the same as the AM radio example given in the text, but neither is a fantastic example of TDM because the alternation is irregular. 56. A 2-Mbps downstream bandwidth guarantee to each house implies at most 50 houses per coaxial cable. Thus, the cable company will need to split up the existing cable into 100 coaxial cables and connect each of them directly to a fiber node.57. The upstream bandwidth is 37 MHz. Using QPSK with 2 bits/Hz, we get 74 Mbps upstream. Downstream we have 200 MHz. Using QAM-64, this is 1200 Mbps. Using QAM-256, this is 1600 Mbps.58. Even if the downstream channel works at 27 Mbps, the user interface is nearly always 10-Mbps Ethernet. There is no way to get bits to the computer any faster than 10-Mbps under these circumstances. If the connection between the PC and cable modem is fast Ethernet, then the full 27 Mbps may be available. Usually, cable operators specify 10 Mbps Ethernet because they do not want one user sucking up the entire bandwidth.第 3 章 數(shù)據(jù)鏈路層1. 答：由于每一幀有0.8 的概率正確到達，整個信息正確到達的概率為 p=0.810=0.107。為使信息完整的到達接收方，發(fā)送一次成功的概率是p ，二次成功的概率是(1-p)p，三次成功的概率為(1-p )2 p，i 次成功的概率為(1-p)i-1 p，因此平均的發(fā)送次數(shù)等于：2. The solution is(a) (b) (c) 3. After stuffing, we get A B ESC ESC C ESC ESC ESC FLAG ESC FLAG D. 4. If you could always count on an endless stream of frames, one flag byte might be enough. But what if a frame ends (with a flag byte) and there are no new frames for 15 minutes. How will the receiver know that the next byte is actually the start of a new frame and not just noise on the line? The protocol is much simpler with starting and ending flag bytes.5. The output is .6. 答：可能。假定原來的正文包含位序列 作為數(shù)據(jù)。位填充之后，這個序列將變成。如果由于傳輸錯誤第二個0 丟失了，收到的位串又變成，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。7. 答：如果傳播延遲很長，例如在探測火星或金星的情況下，需要采用前向糾錯的方法。還有在某些軍事環(huán)境中，接收方不想暴露自己的地理位置，所以不宜發(fā)送反饋信號。如果錯誤率足夠的低，糾錯碼的冗余位串不是很長，又能夠糾正所有的錯誤，前向糾錯協(xié)議也可能是比較合理和簡單的。8. Making one change to any valid character cannot generate another valid character due to the nature of parity bit