【正文】 ALOHA，發(fā)送可以立即開(kāi)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。44 Ten thousand airline reservation stations are peting for the use of a single slotted ALOHA channel. The average station makes 18 requests/hour. A slot is 125 181。sec. What is the approximate total channel load?每個(gè)終端每200（=3600/18）秒做一次請(qǐng)求，總共有10 000 個(gè)終端，因此，總的負(fù)載是200 秒做10000 次請(qǐng)求。平均每秒鐘50 次請(qǐng)求。每秒鐘8000 個(gè)時(shí)隙，所以平均每個(gè)時(shí)隙的發(fā)送次數(shù)為50/8000=1/45 A large population of ALOHA users manages to generate 50 requests/sec, including both originals and retransmissions. Time is slotted in units of 40 msec. (a) What is the chance of success on the first attempt? (b) What is the probability of exactly k collisions and then a success? (c) What is the expected number of transmission attempts needed?一大群ALOHA用戶每秒鐘產(chǎn)生50個(gè)請(qǐng)求，包括原始的請(qǐng)求和重傳的請(qǐng)求。時(shí)槽單位為40ms。(a)首次發(fā)送成功的幾率是多少？(b)恰好k次沖突之后成功的概率是多少? (c)所需傳送次數(shù)的期望值是多少？答：（a）在任一幀時(shí)間內(nèi)生成k 幀的概率服從泊松分布生成0 幀的概率為eG對(duì)于純的ALOHA，發(fā)送一幀的沖突危險(xiǎn)區(qū)為兩個(gè)幀時(shí)，在兩幀內(nèi)無(wú)其他幀發(fā)送的概率是eGe –G=e2G對(duì)于分隙的ALOHA，由于沖突危險(xiǎn)區(qū)減少為原來(lái)的一半，任一幀時(shí)內(nèi)無(wú)其他幀發(fā)送的概率是eG ?，F(xiàn)在時(shí)隙長(zhǎng)度為40ms，即每秒25 個(gè)時(shí)隙，產(chǎn)生50 次請(qǐng)求，所以每個(gè)時(shí)隙產(chǎn)生兩個(gè)請(qǐng)求，G=2。因此，首次嘗試的成功率是：e2 = 1/ e2 （b） （c）嘗試k 次才能發(fā)送成功的概率（即前k1 次沖突，第k 次才成功）為：那么每幀傳送次數(shù)的數(shù)學(xué)期望為46 Measurements of a slotted ALOHA channel with an infinite number of users show that 10 percent of the slots are idle. (a) What is the channel load, G? (b) What is the throughput? (c) Is the channel underloaded or overloaded?對(duì)一個(gè)無(wú)限用戶的分槽ALOHA信道的測(cè)試表明，10%的時(shí)槽是空閑的。(a)信道載荷G是多少?(b)吞吐量是多少?（c)該信道是載荷不足，還是過(guò)載了?答：（a）從泊松定律得到p0＝e –G ，因此G＝ lnp0= ＝ （b） S＝G e G ， G ＝，e G= S==（c）因?yàn)槊慨?dāng)G1 時(shí)，信道總是過(guò)載的，因此在這里信道是過(guò)載的。48 How long does a station, s, have to wait in the worst case before it can start transmitting its frame over a LAN that uses (a) the basic bitmap protocol? (b) Mok and Ward39。s protocol with permuting virtual station numbers?如果一個(gè)LAN使用了下列協(xié)議，請(qǐng)問(wèn)在最差情況下，一個(gè)站s在開(kāi)始傳送幀之前必須要等到多長(zhǎng)時(shí)間？（a）基本的位圖協(xié)議（b）改變虛擬站序列號(hào)的MokWard協(xié)議(a) 最壞的情況是：所有站都想發(fā)送，其中s是編號(hào)最小的站點(diǎn)。等待時(shí)間是N位爭(zhēng)用周期+(N1) d 位幀傳輸，一共是N+(N1) d位時(shí)間； (b) (b) 最壞的情況是：所有站都有幀要傳輸，s具有其中最小的虛擬站編號(hào)。因此，s在其它N1個(gè)站各發(fā)送了一個(gè)幀之后將獲得傳輸機(jī)會(huì)，以及每個(gè)大小為log2 N 的N個(gè)爭(zhēng)用周期。等待時(shí)間是(N+1)d+Nlog2 Nbits.？？？410 Sixteen stations, numbered 1 through 16, are contending for the use of a shared channel by using the adaptive tree walk protocol. If all the stations whose addresses are prime numbers suddenly bee ready at once, how many bit slots are needed to resolve the contention?16個(gè)站的編號(hào)從1到16，它們正在競(jìng)爭(zhēng)使用一個(gè)使用了可適應(yīng)樹(shù)徑協(xié)議的共享信道。如果地址編號(hào)為素?cái)?shù)的所有站突然間全部要發(fā)送幀，請(qǐng)問(wèn)需要多少位時(shí)槽才能解決競(jìng)爭(zhēng)？答：在自適應(yīng)樹(shù)遍歷協(xié)議中，可以把站點(diǎn)組織成二叉樹(shù)（見(jiàn)圖）的形式。在一次成功的傳輸之后，在第一個(gè)競(jìng)爭(zhēng)時(shí)隙中，全部站都可以試圖獲得信道，如果僅其中之一需用信道，則發(fā)送沖突，則第二時(shí)隙內(nèi)只有那些位于節(jié)點(diǎn)B 以下的站（0 到7）可以參加競(jìng)爭(zhēng)。如其中之一獲得信道，本幀后的時(shí)隙留給站點(diǎn)C 以下的站；如果B 點(diǎn)下面有兩個(gè)或更多的站希望發(fā)送，在第二時(shí)隙內(nèi)會(huì)發(fā)生沖突，于是第三時(shí)隙內(nèi)由D 節(jié)點(diǎn)以下各站來(lái)競(jìng)爭(zhēng)信道。本題中，站11 和13 要發(fā)送，需要13 個(gè)時(shí)隙，每個(gè)時(shí)隙內(nèi)參加競(jìng)爭(zhēng)的站的列表如下：第一時(shí)隙：113第二時(shí)隙：7第三時(shí)隙：3第四時(shí)隙：空閑第五時(shí)隙：3第六時(shí)隙：2第七時(shí)隙：3第八時(shí)隙：7第九時(shí)隙：5第十時(shí)隙：7第十一時(shí)隙：113第十二時(shí)隙：11第十三時(shí)隙：13414 Six stations, A through F, municate using the MACA protocol. Is it possible that two transmissions take place simultaneously? Explain your answer.是的，想像它們排成一條直線，每個(gè)站只能到達(dá)其最近的鄰居。當(dāng)E向F發(fā)送時(shí)A也能向B發(fā)送。421 Consider building a CSMA/CD network running at 1 Gbps over a 1km cable with no repeaters. The signal speed in the cable is 200,000 km/sec. What is the minimum frame size?考慮在一條1km長(zhǎng)的電纜（無(wú)中繼器）上建立一個(gè)1Gbps速率的CSMA/CD網(wǎng)絡(luò)。信號(hào)在電纜中的速度為200000km/s。請(qǐng)問(wèn)最小的幀長(zhǎng)度為多少？答：對(duì)于1km 電纜，單程傳播時(shí)間為1/200000 =5106 s，即5，來(lái)回路程傳播時(shí)間為2t =10。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。以1Gb/s 速率工作，10可以發(fā)送的比特?cái)?shù)等于：因此，最小幀是10 000 bit 或1250 字節(jié)長(zhǎng)。422 An IP packet 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?一個(gè)通過(guò)以太網(wǎng)傳送到IP分組有60字節(jié)長(zhǎng)，其中包括所有的頭部。如果沒(méi)有使用LLC的話，則以太網(wǎng)幀中需要填補(bǔ)字節(jié)碼？如果需要的話，請(qǐng)問(wèn)需要填補(bǔ)多少字節(jié)？最小的以太幀是64bytes，包括了以太幀頭部的二者地址、類型/長(zhǎng)度域、校驗(yàn)和。因?yàn)轭^部域占用18 bytes 報(bào)文是60 bytes，總的幀長(zhǎng)度是78 bytes, 已經(jīng)超過(guò)了64byte 的最小限制。因此，不需要填補(bǔ)。423 Ethernet 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 64byte minimum frame size but can get the bits out ten times faster. How is it possible to maintain the same minimum frame size?快速以太網(wǎng)的最大線路長(zhǎng)度是以太網(wǎng)的1/10 。424 Some books quote the maximum size of an Ethernet frame as 1518 bytes instead of 1500 bytes. Are they wrong? Explain your answer.有效載荷是1500 bytes, 但將目的地址、源地址、類型/長(zhǎng)度和校驗(yàn)和域都計(jì)算進(jìn)去的話，總和就是1518.437 Consider the interconnected LANs showns in Fig. 444. Assume that hosts a and b are on LAN 1, c is on LAN 2, and d is on LAN 8. Initially, hash tables in all bridges are empty and the spanning tree shown in Fig 444(b) is used. Show how the hash tables of different bridges change after each of the following events happen in sequence, first (a) then (b) and so on. (a) a sends to d. (b) c sends to a. (c) d sends to c. (d) d moves to LAN 6. (e) d sends to a.。假設(shè)主機(jī)a和b在LAN1上，主機(jī)c在LAN2上，主機(jī)d在LAN8上。剛開(kāi)始的時(shí)候，所有的網(wǎng)橋內(nèi)部的散列表都是空的。在下面給出的每個(gè)事件依次發(fā)生以后，不同網(wǎng)橋的散列表將如何變化。（a)a向d發(fā)送幀(b)c向a發(fā)送幀(c)d向c發(fā)送幀(d)d移動(dòng)到LAN6上(e)d向a發(fā)送幀第一個(gè)幀會(huì)被每個(gè)網(wǎng)橋轉(zhuǎn)發(fā)。這次傳輸后，每個(gè)網(wǎng)橋的散列表會(huì)得到一個(gè)帶有適當(dāng)端口的目的地為a的項(xiàng)目。例如D的散列表會(huì)有一個(gè)向在LAN2上的目的地為a轉(zhuǎn)發(fā)幀的項(xiàng)目。第二個(gè)信息會(huì)被網(wǎng)橋B, D和A看到。這些網(wǎng)橋會(huì)在它們的散列表中添加一個(gè)目的地為c的新項(xiàng)目。例如，網(wǎng)橋D的散列表現(xiàn)在會(huì)有另一個(gè)向在LAN2上的目的地為c轉(zhuǎn)發(fā)幀的項(xiàng)目。第三個(gè)信息會(huì)被網(wǎng)橋H, D, A和B看到。這些網(wǎng)橋會(huì)在它們的散列表中添加一個(gè)目的地為d的新項(xiàng)目。第五條信息會(huì)被網(wǎng)橋E, C, B, D和A看到。網(wǎng)橋E和C會(huì)在他們的散列表添加一個(gè)目的地為d的新項(xiàng)目，與此同時(shí)，網(wǎng)橋D, B和A 將會(huì)更新它們對(duì)應(yīng)目的地d的散列表項(xiàng)目。438 One consequence of using a spanning tree to forward frames in an extended LAN is that some bridges may not participate at all in forwarding frames. Identify three such bridges in Fig. 444. Is there any reason for keeping these bridges, even though they are not used for forwarding?在一個(gè)擴(kuò)展的LAN中使用生成樹(shù)來(lái)轉(zhuǎn)發(fā)幀的一個(gè)結(jié)果是，有的網(wǎng)橋可能根本不參與幀的轉(zhuǎn)發(fā)過(guò)程。既然這些網(wǎng)橋沒(méi)有被用于轉(zhuǎn)發(fā)幀，那么是否有理由要保留這些網(wǎng)橋呢？網(wǎng)橋 G, I 和 J 沒(méi)有被用來(lái)轉(zhuǎn)發(fā)任何幀。在一個(gè)擴(kuò)展的LAN中具有回路的主要原因是增加可靠性。如果當(dāng)前生成樹(shù)中的任何網(wǎng)橋出了故障，（動(dòng)態(tài)）生成樹(shù)算法重構(gòu)一個(gè)新的生成樹(shù)，其中可能包括一個(gè)或更多不屬于先前生成樹(shù)部分的網(wǎng)橋。442 Briefly describe the difference between storeandforward and cutthrough switches.存儲(chǔ)轉(zhuǎn)發(fā)型交換機(jī)完整存儲(chǔ)輸入的每個(gè)幀，然后檢查并轉(zhuǎn)發(fā)。直通型交換機(jī)在輸入幀沒(méi)有全部到達(dá)之前就開(kāi)始轉(zhuǎn)發(fā)。一得到目的地址，轉(zhuǎn)發(fā)就開(kāi)始了。443 Storeandforward switches have an advantage over cutthrough switches with re