(完整版)計(jì)算機(jī)網(wǎng)絡(luò)課后作業(yè)以及答案(中英文對(duì)照)

1、Chapter11-11.What are two reasons for using layered protocols?（ 請(qǐng)說(shuō)出使用分層協(xié)議的兩個(gè)理由）答：通過(guò)協(xié)議分層可以把設(shè)計(jì)問(wèn)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。1-13. What is the principal difference between connectionless communication and connection-oriented communication?（ 在無(wú)連接通信和面向連接的通信兩者之間，最主要的區(qū)別是什么？）答：主要的區(qū)別有兩條。其一：面向連接通

2、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。只有在連接成功建立之后，才能開(kāi)始數(shù)據(jù)傳輸，這是第二階段。接著，當(dāng)數(shù)據(jù)傳輸完畢，必須釋放連接。而無(wú)連接通信沒(méi)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(wú)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。1-20. A system has an n-layer protocol hierarchy. Applications generate messages of length M bytes. At each of the layers, an h-byte header is added.

3、 What fraction of the network bandwidth is filled with headers?（ 一個(gè)系統(tǒng)有n 層協(xié)議的層次結(jié)構(gòu)。應(yīng)用程序產(chǎn)生的消息的長(zhǎng)度為M 字節(jié)。在每一層上需要加上一個(gè)h 字節(jié)的頭。請(qǐng)問(wèn)， 這些頭需要占用多少比例的網(wǎng)絡(luò)帶寬）答： hn/（hn+m）*100%1-28. An image is 1024 x 768 pixels with 3 bytes/pixel. Assume the imageis uncompressed. Howlong does it take to transmit it over a 56-kbps modem

4、 channel? Over a 1-Mbps cable modem?Over a 10-Mbps Ethernet? Over 100-Mbps Ethernet?（ 一幅圖像的分辨率為1024 x 768 像素，每個(gè)像素用3 字節(jié)來(lái)表示。假設(shè)該圖像沒(méi)有被壓縮。請(qǐng)問(wèn)，通過(guò)56kbps 的調(diào)制解調(diào)器信道來(lái)傳輸這幅圖像需要多長(zhǎng)時(shí)間？通過(guò)1Mbps的電纜調(diào)制解調(diào)器呢？通過(guò) 10Mbps的以太網(wǎng)呢？通過(guò)100Mbps的以太網(wǎng)呢？）答： The image is 1024*768*3 bytes or 2359296 bytes.This is 18874368 bit.At 56,000 bits

5、/sec, it takes about 337.042 sec. At 1,000,000 bits/sec,it takes about 18.874 sec. At 10,000,000 bits/sec, it takes about 1.887sec. At 100,000,000 bits/sec, it takes about 0.189 sec.Chapter22-2. A noiseless 4-kHz channel is sampled every 1 msec. Whatis the maximum data rate?（一條無(wú)噪聲4kHz信道按照每1ms一次進(jìn)行采樣，

6、請(qǐng)問(wèn)最大數(shù)據(jù)傳輸率是多少？ ）答：無(wú)噪聲信道最大數(shù)據(jù)傳輸率公式：最大數(shù)據(jù)傳輸率=2Hlog2V b/s。因此最大數(shù)據(jù)傳輸率決定于每次采樣所產(chǎn)生的比特?cái)?shù)，如果每次采樣產(chǎn)生16bits ,那么數(shù)據(jù)傳輸率可達(dá)128kbps；如果每次采樣產(chǎn)生1024bits ,那么可達(dá)8.2Mbps。 注意這是對(duì)無(wú)噪聲信道而言的，實(shí)際信道總是有噪聲的，其最大數(shù)據(jù)傳輸率由 香農(nóng)定律給出。2-4. If a binary signal is sent over a 3-kHz channel whose signal-to-noise ratio is 20 dB, what is the maximum achieva

7、ble data rate?(如果在一條3kHz的信道上發(fā)送一個(gè)二進(jìn)制信號(hào)，該信道的信噪比為20dB,則最大可達(dá)到的數(shù)據(jù)傳輸率為多少？)答：信噪比為20 dB即S/N =100由于log 2101=6.658,由香農(nóng)定理，該信道 的信道容量為 310g 2(1+100)=19.98kbps 。又根據(jù)乃奎斯特定理，發(fā)送二進(jìn)制信號(hào)的3kHz信道的最大數(shù)據(jù)傳輸速率為2*3*log 22=6kbps。所以可以取得的最大數(shù)據(jù)傳輸速率為6kbps。2-5. What signal-to-noise ratio is needed to put a T1 carrier on a 50-kHz line?(

8、在50kHz的線(xiàn)路上使用仃線(xiàn)路需要多大的信噪比？)答：為發(fā)送T1信號(hào)，我們需要J/log2(l + ) = 1.544 X10J/6H 二 50000=231-1 N101og10(231 - 1) = 93 dB所以，在50kHz線(xiàn)路上使用T1載波需要93dB的信噪比。2-34. A signal is transmitted digitally over a 4-kHz noiseless channelwith one sample every 125 仙 sec. How many bits per second are actually sent for each of these

9、encoding methods?(a) CCITT 2.048 Mbps standard.(b) DPCM with a 4-bit relative signal value.(c) Delta modulation.(一個(gè)信號(hào)在4kHz的無(wú)噪聲信道上以數(shù)字方式進(jìn)行傳輸，沒(méi)125us采樣一次。請(qǐng)問(wèn)，按照以下的編碼方法，每秒鐘實(shí)際發(fā)送多少位？(a) CCITT 2.048Mbps標(biāo)準(zhǔn)。(b)有4位相對(duì)信號(hào)值的DPCM(c)增量調(diào)制。)答：a. CCITT 2.048Mbps標(biāo)準(zhǔn)用32個(gè)8位數(shù)據(jù)樣本組成一個(gè)125的基本幀， 30個(gè)信道用于傳信息，2個(gè)信道用于傳控制信號(hào)。在每一個(gè) 4kHz信道

10、上發(fā)送的數(shù)據(jù)率就是8*8000=64kbps。b.差分脈碼調(diào)制（DPCM是一種壓縮傳輸信息量的方法，它發(fā)送的不是每 一次抽樣的二進(jìn)制編碼值，而是兩次抽樣的差值的二進(jìn)制編碼?，F(xiàn)在相對(duì)差值 是4位，所以對(duì)應(yīng)每個(gè)4kHz信道實(shí)際發(fā)送的比特速率為 4*8000=32bps。c.增量調(diào)制的基本思想是：當(dāng)抽樣時(shí)問(wèn)間隔s t很短時(shí)，模擬數(shù)據(jù)在兩次抽樣之間的變化很小，可以選擇一個(gè)合適的量化值作為階距。把兩次抽樣的差 別近似為不是增加一個(gè)就是減少一個(gè)。這樣只需用1bit二進(jìn)制信息就可以表示 一次抽樣結(jié)果，而不會(huì)引入很大誤差。因此，此時(shí)對(duì)應(yīng)每個(gè)4kHz信道實(shí)際發(fā)送的數(shù)據(jù)速率為1*8000=8kHzo2-43. S

11、uppose that x bits of user data are to be transmitted over a k-hop path in a packet-switched network as a series of packets, each containing p data bits and h header bits,with x >> p+h.The bit rate of the lines is b bps and the propagation delay is negligible.What value ofp minimizesthe total

12、delay?（假定x位用戶(hù)數(shù)據(jù)將以一系列分組形式，在一個(gè)分組交換網(wǎng)絡(luò)中沿著一條 共有k跳的路徑向前傳播，每個(gè)分組包含 p位數(shù)據(jù)和h位的頭，這里x>>p+ho 線(xiàn)路的傳輸率為b bps,傳播延遲忽略不計(jì)。請(qǐng)問(wèn)，什么樣的p值使總延遲最??？） 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(shí)間為（p+h）*x/pb 中間的路由器重傳最后一個(gè)分組所花的總時(shí)間為（k-1）*（p+h）/ b因此我們得到的總的延遲為pbb對(duì)該函數(shù)求p的導(dǎo)數(shù)，得到p-（p + h）x -Ip1 b b令得到p-因?yàn)閜>0,所以故時(shí)能使總的延遲最小2

13、-53. A CDMAreceiver gets the following chips: (-1 +1 -3 +1 -1 -3 +1 +1). Assuming the chip sequences defined in Fig. 2-45(b), which stations transmitted, and which bits did each one send?(一個(gè)CDM破收器得到了下面的時(shí)間片：(-1 +1-3 +1-1-3 +1 +1)。假設(shè)時(shí)間片序列如圖2.45 (b)中所定義，請(qǐng)問(wèn)哪些移動(dòng)站傳輸了數(shù)據(jù)？每個(gè)站發(fā) 送了什么位?)答： Just compute the four

14、 normalized inner products:1+1 3+1 1 3+1+11+1 3+1 1 3+1+11+1 3+1 1 3+1+11+1 3+1 1 3+1+1The result is that Ad( 1 1 1+1+1 1+1+1 ) /8=1 d( 1 1+1 1+1+1+1 1 ) /8=1 d( 1+1 1+1+1+1 1 1 ) /8=0 d( 1+1 1 1 1 1+1 1 ) /8=1and D sent 1 bits, B sent a 0 bit, and C was silent.Chapter33-3.The following data fragmen

15、t occurs in the middle of a data stream for which the byte-stuffing algorithm described in the text is used:A B ESCC ESC FLAG FLAG D. What is the output after stuffing?(數(shù)據(jù)片斷(A B ESC C ESC FLAG FLAG D出現(xiàn)在一個(gè)數(shù)據(jù)流的中間，而成 幀方式采用的是本章介紹的字節(jié)填充算法，請(qǐng)問(wèn)經(jīng)過(guò)填充之后的輸出時(shí)什么？)答： After stuffing, we get A B ESC ESC C ESC ESC ESC

16、 FLAG ESC FLAG D.3-15. A bit stream 10011101 is transmitted using the standard CRC method described in the text. The generator polynomial is x3 + 1. Showthe actual bit string transmitted. Suppose the third bit from the left is inverted during transmission. Show that this error is detected at the rec

17、eiver's end.(利用本章中介紹的標(biāo)準(zhǔn)多項(xiàng)式CRCT法來(lái)傳輸位流1001101。生成器多項(xiàng)式為x3+1o請(qǐng)給出實(shí)際被傳輸?shù)奈挥?。假設(shè)在傳輸過(guò)程中左邊第三位變反了。請(qǐng) 證明，這個(gè)錯(cuò)誤可以在接收端被檢測(cè)出來(lái)。)答： The frame is 10011101. The generator is 1001. The message after appending three zeros is 10011101000. The remainder on dividing 10011101000 by 1001 is 100. So, the actual bit stringTrans

18、mitted is 10011101100. The received bit stream with an error in the third bit from the left is 10111101100.Dividing this by 1001 produces a remainder 100, which is different from zero. Thus, the receiver detects the error and can ask for a retransmission.3-18. A 3000-km-long T1 trunk is used to tran

19、smit 64-byte frames usingprotocol 5. If the propagation speed is 6 仙 sec/km, how many bits should the sequence numbers be?(一條3000公里長(zhǎng)的T1 骨干線(xiàn)路被用來(lái)傳輸64字節(jié)的幀，兩端使用了協(xié)議5。如果傳輸速度為6us/公里，則序列號(hào)應(yīng)該有多少位？)答：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?，以允許發(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(xiàn)路上的傳播時(shí)間為6X3000=18000us,即18ms 在T1速率，發(fā)送64字節(jié)的數(shù)據(jù)幀需花的時(shí) 問(wèn)

20、：64X 8+（1.536 X106）= 0.33us。所以，發(fā)送的第一幀從開(kāi)始發(fā)送起，18.33ms 后完全到達(dá)接收方。確認(rèn)應(yīng)答又花了很少的發(fā)送時(shí)間（忽略不計(jì)）和回程的18ms。這樣，加在一起的時(shí)間是36.33ms。發(fā)送方應(yīng)該 有足夠大的窗口，從而能夠連續(xù)發(fā)送36.33ms。 36. 33/0.33=110 也就是說(shuō)，為充滿(mǎn)線(xiàn)路管道，需要至少110幀，因此序列號(hào)為7 位。3-22. In protocol 6, when a data frame arrives, a check is made to seeif the sequence number differs from the on

21、e expected and no_nak is true.If both conditions hold, a NAK is sent. Otherwise, the auxiliary timeris started. Suppose that the else clause were omitted. Would this change affect the protocol's correctness?（ 在協(xié)議 6 中，當(dāng)一個(gè)數(shù)據(jù)幀到達(dá)的時(shí)候，需要執(zhí)行一個(gè)檢查，看它的序列號(hào)是否與期望的序列號(hào)不同，并且no_nak 為真。如果這兩個(gè)條件都成立，則發(fā)送一個(gè)NAK否則的話(huà)，啟用輔助

22、定時(shí)器。假定 else子句被省略掉。這種改變會(huì)影響協(xié)議的正確性嗎？）答：可能導(dǎo)致死鎖。假定有一組幀正確到達(dá)，并被接收。然后，接收方會(huì)向前移動(dòng)窗口?，F(xiàn)在假定所有的確認(rèn)幀都丟失了，發(fā)送方最終會(huì)產(chǎn)生超時(shí)事件，并且再次發(fā)送第一幀，接收方將發(fā)送一個(gè)NAK然后NONA啾置成偽。假定NAK也丟失了。那么從這個(gè)時(shí)候開(kāi)始，發(fā)送方會(huì)不斷發(fā)送已經(jīng)被接收方接受了的幀。接收方只是忽略這些幀，但由于 NONA偽，所以不會(huì)再發(fā)送 NAK從而產(chǎn)生死鎖。如 果設(shè)置輔助計(jì)數(shù)器（實(shí)現(xiàn)“ else”子句），超時(shí)后重發(fā)NAK終究會(huì)使雙方重新 獲得同步。3-31. Consider an error-free 64-kbps satellite channel used to send512-byte data frames in one direction, with very short acknowledgementscoming back the other way. Whatis the maximumthroughput for window sizes of 1, 7, 15, and 127? The earth-satellite propagation time is 270 msec.（考慮在一個(gè)無(wú)錯(cuò)誤的64kbps 衛(wèi)星信道上單向發(fā)送512 字節(jié)的數(shù)據(jù)幀