計網(wǎng)第五次作業(yè)

1、精選優(yōu)質(zhì)文檔-傾情為你奉上精選優(yōu)質(zhì)文檔-傾情為你奉上專心-專注-專業(yè)專心-專注-專業(yè)精選優(yōu)質(zhì)文檔-傾情為你奉上專心-專注-專業(yè)Review Questions:1. What are some of the possible services that a link-layer protocol can offer to the network layer? Which of these link-layer services have corresponding services in IP? In TCP? 鏈路層協(xié)議提供給網(wǎng)絡(luò)層的服務(wù)有哪些？哪些是給IP的？哪些是給TCP的 ？Link

2、access, framing, reliable delivery between adjacent nodes, flow control, error detection, error correction, half-duplex and full-duplex.In IP: framing, error detection.In TCP: framing, reliable delivery between adjacent nodes, error detection, half-duplex and full-duplex.2. If all the links in the I

3、nternet were to provide reliable delivery service, would the TCP reliable delivery service be redundant? Why or why not?不會多余，因為TCP保證的是傳輸層的數(shù)據(jù)傳送，而link提供穩(wěn)定可靠傳輸保證鏈路層的傳輸穩(wěn)定，二者不完全重疊，所以TCP可靠傳輸也不會多余。3. In Section 5.3, we listed four desirable characteristics of a broadcast channel. Which of these characteris

4、tics does slotted ALOHA have? Which of these characteristics does token passing have?4. Suppose two nodes start to transmit at the same time a packet of length Lover a broadcast channel of rate R. Denote the propagation delay between the two nodes as dprop. Will there be a collision if dprop L / R?

5、Why or why not?因為L / R = 包傳遞的時間，如果dprop L / R，也就意味著A的信號跑到B的時候B的最后一個分組還沒離開B，所以發(fā)生碰撞。7. Suppose nodes A, B, and C each attach to the same broadcast LAN (through their adapters). If A sends thousands of IP datagrams to B with each encapsulating frame addressed to the MAC address of B, will Cs adapter pr

6、ocess these frames? If so, will Cs adapter pass the IP datagrams in these frames to the network layer C? How would your answers change if A sends frames with the MAC broadcast address?不會，C會拆封幀從而讀取報頭的MAC，因為每一個host的MAC都唯一，C讀取到數(shù)據(jù)報中的MAC和自己的不一樣就不會繼續(xù)拆封數(shù)據(jù)報，不會投遞給C。當(dāng)使用LAN口廣播地址的時候，C的適配器就會拆封幀，向C傳遞數(shù)據(jù)。8. How big

7、 is the MAC address space? The IPv4 address space? The IPv6 address space?248，232，2128.9. Why is an ARP query sent within a broadcast frame? Why is an ARP response sent within a frame with a specific destination MAC address?因為新加入網(wǎng)絡(luò)的主機(jī)是不知道路由器的IP的，自己也沒有IP，所以只能廣播才能得到IP。因為每個主機(jī)的MAC地址都是唯一的，而ARP建立轉(zhuǎn)發(fā)表的時候會帶上

8、MAC地址。12. In CSMA/CD, after the fifth collision, what is the probability that a node chooses K = 4? The result K = 4 corresponds to a delay of how many seconds on a 10 Mbps Ethernet?可能，因為第五次K的取值范圍是0-(25-1)即0-31。Bit time = 1 bit / R = 1 bit / 10 Mbps = 1 msec，K = 4，wait time = 4 * 512 * 1 msec = 2048

9、 msec。Problems:Suppose the information content of a packet is the bit pattern 1110 1100 1000 1010 and an even parity scheme is being used. What would the value of the field containing the parity bits be for the case of a two-dimensional parity scheme? Your answer should be such that a minimum-length

10、 checksum field is used.假設(shè)一個數(shù)據(jù)包的信息含量是XXX，使用偶校驗方案。采用二位奇偶校驗方案的字段包含的奇偶校驗位的字段的值是多少？答案要使用最小長度校驗。10100101001010010111000112. Suppose the information portion of a packet (D in Figure 5.4) contains 10 bytes consisting of the 8-bit unsigned binary ASCII representation of the integers 0 through 9. Compute the

11、 Internet checksum for this data.假設(shè)一個包的信息的一部分包括10bytes組成的8-bit無符號二進(jìn)制碼表示的整數(shù)0-9，計算該數(shù)據(jù)的網(wǎng)絡(luò)校驗。算校驗碼先把0-9加起來0000 00000000 00010000 00100000 00110000 01000000 01010000 01100000 01110000 10000000 10010001 01000001 1001取反可得校驗碼為1110 10111110 01103. Consider the previous problem, but instead of containing the b

12、inary of the numbers 0 through 9 suppose these 10 bytes contain. Compute the Internet checksum for this data.a. the binary representation of the numbers 1 through 10.0000 00010000 00100000 00110000 01000000 01010000 01100000 01110000 10000000 10010001 01000001 10110001 1110取反可得1110 01001110 0001b. t

13、he ASCII representation of the letters A through J (uppercase).0100 00010100 00100100 00110100 01000100 01010100 01100100 01110100 10000100 10010100 10100101 10000101 1111取反可得1010 01111010 0000c. the ASCII representation of the letters a through j (lowercase).小寫字母表示0110 00010110 00100110 00110110 01

14、000110 01010110 01100110 01110110 10000110 10010110 10101111 10011111 1101取反可得0000 01100000 00106. Consider the previous problem, but suppose that D has the valuea. 1001 0001.b. 1010 0011.c. 0101 0101.前一題題目：Consider the 7-bit generator, G=10011, and suppose that D has the value . What is the value o

15、f R？求余而已，記住不要做減法而是做與運算就好。a. R = 001b. R = 101c. R = 10112. Consider three LANs interconnected by two routers, as shown in Figure 5.38.a. Redraw the diagram to include adapters.重新畫圖b. Assign IP addresses to all of the interfaces. For Subnet 1 use addresses of the form 111.111.111.xxx; for Subnet 2 us

16、es addresses of the form 122.122.122.xxx; and for Subnet 3 use addresses of the form 133.133.133.xxx.所有的接口分配IP地址。c. Assign MAC addresses to all of the adapters.a.b.c如圖d. Consider sending an IP datagram from Host A to Host F. Suppose all of the ARP tables are up to date. Enumerate all the steps, as d

17、one for the single-router example in Section 5.4.2.1. host A發(fā)送一個數(shù)據(jù)報，通過轉(zhuǎn)發(fā)表查詢F的IP，向路由器1發(fā)送，其中destination IP為133.133.133.12，MAC未知，source IP為111.111.111.12，source MAC為aa-aa-aa-aa-aa-aa。2. 適配器更改destination的IP為111.111.111.12，MAC地址變?yōu)間g-gg-gg-gg-gg-gg3. 路由器1發(fā)現(xiàn)目標(biāo)IP和MAC不屬于子網(wǎng)1中任何host，屬于子網(wǎng)3（圖中忘了畫了，意會一下）。于是根據(jù)轉(zhuǎn)發(fā)表向

18、路由器2進(jìn)行轉(zhuǎn)發(fā)。Destination的IP為122.122.122.20，MAC為ii-ii-ii-ii-ii-ii。4. 路由器2收到了數(shù)據(jù)報，發(fā)現(xiàn)host F在自己的子網(wǎng)內(nèi)，于是修改destination的IP為133.133.133.12，MAC地址為ff-ff-ff-ff-ff-ff。修改source IP為133.133.133.20，MAC為jj-jj-jj-jj-jj-jj，然后向F發(fā)送數(shù)據(jù)報。5. F收到來自A的數(shù)據(jù)報。e. Repeat (d), now assuming that the ARP table in the sending host is empty (a

19、nd the other tables are up to date).發(fā)送方的ARP表為空，首先需要建立ARP表1. host A發(fā)送一個廣播，destination IP是255.255.255.255，MAC為空。Source IP為111.111.111.12，MAC為aa-aa-aa-aa-aa-aa2. 適配器收到了來自host A的數(shù)據(jù)報，更新自己的ARP表，同時發(fā)送一個ACK給host A，告訴host A自己的IP、MAC。3. host A建立ARP表4. 如d小問所答，開始進(jìn)行數(shù)據(jù)發(fā)送。14. Recall that with the CSMA/CD protocol,

20、the adapter waits K512 bit times after a collision, where K is drawn randomly. For K = 100, how long does the adapter wait until returning to Step 2 for a 10 Mbps Ethernet? For a 100 Mbps Ethernet?當(dāng)網(wǎng)速 = 10 Mbps時，bit time = 1 bit / 10 Mbpst = 100 * 512 * 1 / (106) = 5.12 msec當(dāng)網(wǎng)速 = 100 Mbps時，bit time

21、= 1 bit / 100 Mbpst = 100 * 512 * 1 / (107) = 0.512 msec15. Suppose nodes A and B are on the same 10 Mbps Ethernet bus, and the propagation delay between the two nodes is 225 bit times. Suppose A and B send frames at the same time, the frames collide, and then A and B choose different values of K in

22、 the CSMA/CD algorithm. Assuming no other nodes are active, can the retransmissions from A and B collide? For our purposes, it suffices to work out the following example. Suppose A and B begin transmission at t = 0 bit times. They both detect collisions at t = 225 bit times. They finish transmitting

23、 a jam signal at t = 225 + 48 = 273 bit times. Suppose KA = 0 and KB = 1. At what time does B schedule its retransmission? At what time does A begin transmission? (Note: The nodes must wait for an idle channel after returning to Step 2see protocol.) At what time does As signal reach B? Does B refrai

24、n from transmitting at its scheduled time?1. 因為A的K值 = 0，所以A從273 bit time開始檢測是否沖突2. 由于傳輸延遲的問題，B的最后一個bit要等到273 + 225 = 498 bit time才能傳到A，也就是說此時A檢測到?jīng)]有沖突。3. A傳輸前先等待96個bit time，即t = 498 + 96 = 594 bit time的時候A開始傳輸數(shù)據(jù)4. 因為propagation delay = 225 bit time，所以t = 594 +225 = 819 bit time的時候A傳輸完畢5. 因為KB = 1，當(dāng)B等

25、待1 * 512 * 1 = 512 bit time的時候，B可以重傳，此時B開始檢測信道是否空閑，此時t = 273 + 512 = 785 bit time6. 此時B檢測到信道忙，因此不能重傳，等待信道空閑7. 當(dāng)t = 819 bit time的時候信道空閑，此時B等待96 bit time，即t = 96+819= 915 bit time的時候B可以開始重傳16. Suppose nodes A and B are on the same 10 Mbps Ethernet bus, and the propagation delay between the two nodes i

26、s 225 bit times. Suppose node A begins transmitting a frame and, before it finishes, node B begins transmitting a frame. Can A finish transmitting before it detects that B has transmitted? Why or why not? If the answer is yes, then A incorrectly believes that its frame was successfully transmitted w

27、ithout a collision. Hint: Suppose at time t = 0 bit times, A begins transmitting a frame. In the worst case, A transmits a minimum-sized frame of 512 + 64 bit times. So A would finish transmitting the frame at t = 512 + 64 bit times. Thus, the answer is no, if Bs signal reaches A before bit time t =

28、 512 + 64 bits. In the worst case, when does Bs signal reach A?因為A最快傳輸時間t = 512 + 64 bit time，看了答案，微積分神馬的可以讓我shi了Let Y be a random variable denoting the number of slots until a success:P(Y = m) = (1 -)(m - 1)Where is the probability of a success.This is a geometric distribution, which has mean 1 /.

29、The number of consecutive wasted slots is X = Y 1 that x = E X = E Y -1 = (1 -) / = Np(1 - p)(N - 1)x = 1 Np * (1 - p)(N - 1) / Np * (1 - p)(N - 1)efficiency = k / (k + x) = k / k + 1 Np * (1 - p)(N - 1) / Np(1 - p)(N - 1)19. Suppose two nodes, A and B, are attached to opposite ends of a 900 m cable

30、, and that they each have one frame of 1,000 bits (including all headers and preambles) to send to each other. Both nodes attempt to transmit at time t = 0. Suppose there are four repeaters between A and B, each inserting a 20-bit delay. Assume the transmission rate is 10 Mbps, and CSMA/CD with back

31、off intervals of multiples of 512 bits is used. After the first collision, A draws K = 0 and B draws K = 1 in the exponential backoff protocol. Ignore the jam signal and the 96-bit time delay.a. What is the one-way propagation delay (including repeater delays) between A and B in seconds? Assume that

32、 the signal propagation speed is 2108 m/sec.傳輸時間t0 = 900 m / 2 * (108) m / sec = 4.5 secRepeater 傳輸延遲 t1 = 4 * 20 bit / 10 Mbps = 8 secT = t0 + t1 = 12.5 secb. At what time (in seconds) is As packet completely delivered at B? （為什么這里答案沒有96個bit time的等待？）1 bit time = 1 / 10 Mbps = 0.1 sec當(dāng)t = 12.5 sec

33、時，A、B檢測到?jīng)_突，停止傳輸。A的K值 = 0，于是A立即開始檢測信道是否空閑，當(dāng)B傳輸延遲結(jié)束后，信道空閑，此時t2 = 12.5 +12.5 = 25 secA等待兩個96 bit time，即 25 + 9.6 = 34.6 sec時，A開始重傳B在A的傳輸延遲：34.6 + 12.5 sec = 47.1 sec之后，檢測到A的第一個bitB等待的K值 = 1，即B在512 * 0.1 sec = 51.2 sec之后重傳，此時A已經(jīng)傳輸，信道忙，B偵聽信道，等待信道空閑A的傳輸時間tA = 1000 bit / 10 MbpsA傳輸完畢的時間T = 1000 bit / 10 Mb

34、ps + 47.1 sec= 147.1 secc. Now suppose that only A has a packet to send and that the repeaters are replaced with switches. Suppose that each switch has a 20-bit processing delay in addition to a store-and-forward delay. At what time, in seconds, is As packet delivered at B? （repeater和switch和router區(qū)別

35、在哪里？）t = 12.5 sec + 5 * 100 sec = 512.5sec21. Suppose now that the leftmost router in Figure 5.38 is replaced by a switch. Hosts A, B, C, and D and the right router are all star-connected into this switch. Give the source and destination MAC addresses in the frame encapsulating this IP datagram as the frame is transmitted (i) from A to the switch, (ii) from the switch to the