Giải bài tập mạng máy tính Computer Networking A TopDown Approach 6th

Computer Networking: A Top-Down Approach Featuring the Internet, 5th Edition – Translate by K55CC  Page 1 Computer Networking - A Top-Down Approach Featuring the Internet, 5 th edition Solutions to Review Questions and Problems (Vietnamese version) Author : K55CC translate team.(K55CC-University of engineering and technology-Vietnam national university) This document belong to K55CC.All copies or sharing must be allowed by K55CC. Chapter1: Introduction-Review Question 1: S      thay th cho nhau. End system gt ni internet  v giao thc ngoi giao (diplomatic protocol) Gi s Alice, m i s ca quc gia A mun mi Bob, m i s c B,  i. Alice  ch  gi   cn gi Bob  n tho   n   ca    ". Thay    gi Bob  cho thy mt i gian. Bob  ng b  rng i  s c th  n m Alice  Bob tip tc gi p" qua li cho n khi h   i gian. Bob y ti s    tho thun, hy vng  c khi hoc sau khi th thu c  cho  hoc Alice hoc Bob lch s hy b s tham gia nu h   hp  n dch v t  M    c ni vs u vi cn dch v t  server.  truy cp. 1. Dial-up modem over telephone line: residential; 2. DSL over telephone line: residential or small office; 3. Cable to HFC: residential; 4. 100 Mbps switched Ethernet: company; 5. Wireless LAN: mobile; 6. Cellular mobile access (for example, WAP): mobile  truyn ci 1 t truyn mi user trong m truyc v  Computer Networking: A Top-Down Approach Featuring the Internet, 5th Edition – Translate by K55CC  Page 2  truys. VD vi X Mbps Ethernet ( X = 10, 100, 1,000 or 10,000), m truyn lien tc vi t X Mbps nu ch  i d liu. Nu user ho truyc vi t X Mbps. ng v  truy  truyng trc mng xo truyn qua si ng tr t khong t truyn c  c kiu chia s t? Dial up modems: up to 56 Kbps, bandwidth is dedicated; ISDN: up to 128 kbps, bandwidth is dedicated; ADSL: downstream channel is .5-8 Mbps, upstream channel is up to 1 Mbps, bandwidth is dedicated; HFC, downstream channel is 10-30 Mbps and upstream channel is usually less than a few Mbps, bandwidth is shared. FTTH: 2-10Mbps upload; 10-20 Mbps download; bandwidth is not shared.   truy c bin hi bit c Hi truy c bin. - M n 1 tr m truy cm c ng c kt ni Internet b - Mng truy cn r thc truy h t thn thoi, tr c qup dch v vin tp s truy cc kilomet t . n li ca mng chuyn mn so vi chuyn m Mng chuyn m m bng c a 2 m n c gi. Hu hng chuyn mn nay (bao gm c o s m b m ni.  s c chuyn t host gn host nhn. T truyn gia host g n l s switch s dng chuyn m   chuyn ting thi gian delay gim n g   tr khi truy  Ti thm t 0 , host gi bu truyn. Ti thm t 1 = L/R1, host gc truy  c nhn t khi truy i thm t 1  truyn host nhn ti thm t 1 . Ti thim t 2 = t 1  c truy c nhn ti host nhy, tng thi gian tr  Câu 15: Giả sử nhiều user chia sẻ 1 link 2Mbps. Và cũng giả sử rằng mỗi user truyền lien tục vs tốc độ 1Mbps khi truyền, nhưng mỗi user chỉ truyền 20% thời gian. Computer Networking: A Top-Down Approach Featuring the Internet, 5th Edition – Translate by K55CC  Page 3 a. Nếu dùng chuyển mạch điện thì có bao nhiêu user được sử dụng? b. Giả sử dùng chuyển mạch gói. Tại sao nếu có ít hơn hoặc bằng 2 user cùng truyền thì không phải đợi? và nếu nhiều hơn 2 user cùng truyền thì phải có hàng đợi? c. Tính xác suất để mỗi user được truyền. d. Giả sử có 3 user. Tính xác suất tại bất kì thời điểm nào, cả 3 user cùng truyền 1 lúc. Tính khoảng thời gian để hình thành hàng đợi. TL: a. c truyi user cn 1 nng truyn. b. Theo gi thit, mi user cn 1Mbps khi truyn, n cng truyi . c li, nn s 3Mbps, nhi ng h ng truyn. c. t = 0.2 d.   3 . (1-p) 3-3 = (0.2) 3 i t c ho tt c  truy   s     1 host ngu    ng c nh. Li  khong tr i? n tr gm: tr khi x  khi lan ta (propagation delay), tr khi truyn (transmission  i. Tr tr i, tt c nhng khong tr  nh. ng th truy ng truyn kho truy chuy thu ng tgian = d/s.  thu  chuyn R.  s host A mun gi 1 file ln  n B, t l  a. Gi s c nghn trong mng truyn file? b. Gi s ng 4 tric file bi gian xp x  truyn host B? c. i R2 gi TL: a. 500 kbps.  b. 4.000.000 byte = 4.000 KB = 32.000 Kbit. Tgian = 32000/500 = 64 s Computer Networking: A Top-Down Approach Featuring the Internet, 5th Edition – Translate by K55CC  Page 4 c. ng = 100 kbps. Tgian = 320 s. m v  thc hi xng hp mt hay nhiu nhim v c thc hin bi nhiu tc k? 5 nhim v u khin lui, dt kt ni. C xng hp 1 hay nhiu nhim v c thc hin bi nhiu tng. VD: kii c cung cp  nhiu tng.   t message  tng ng d   t segment  tng giao vn? Mt datagram  tng network? Mt frame  tt? -  -  . - a tng giao v. -  t router x  tt switch x  tt host x  t   tng 1, 2, 3 tng v  t 2 tng v  c 5 tng. Chapter1: : Introduction- Problem P5: Gi s t ni vi nhau b c  lan tn host B. a. Biu di tr lan ta (d prop  b. nh thi gian truyn c trans  c. B  tr x i. Vit biu th tr gim k d. Gi s Host A bu truyi thm t = 0. Ti thm t = d trans , bit cu c  e. Gi s d prop > d trans . Ti thm t = d trans   f. Gi s d prop < d trans . Ti thm t = d trans   g. Gi s s = 2,5.10 8  d prop = d trans TL: a. d prop = m/s (s)   tr lan ta (tht kho b. d trans = L/R (s)   tr truyn (thi gian t n khi bit cui i host). c. d A-B = d prop + d trans = m/s + L/R (s) d. Ti t = d trans , bit cua ri khi host A. e. Ti t = d trans , bit  ng truyc host B. f. Ti t = d trans n host B. Computer Networking: A Top-Down Approach Featuring the Internet, 5th Edition – Translate by K55CC  Page 5 g. d prop = d trans  m/s = L/R  m = Ls/R = 523158 (m) c gi gii gian thc t n host B qua 1 mng chuyn m A chuyn giu s  bytes. Ch ng duy nht t n B t truy tr lan t host A t n i gian k t c to (t u gc  n khi bit cuc gi host B)?  c truyt c c to y cn: 3 10.64 8.56 (s) = 7 (ms). Thi gian c truy 6 10.2 8.56 (s) = 224  s  tr lan t  tr n khi gi  s + 10 ms = 17,224 ms P12: Gi s ng truyn tc truyc phi x t ng truy tr  tin.  tr    1)    tr c (L/R + 2L/R + + (N-1).L/R) / N = L/(RN) * (1 + 2 + + (N-1)) = L/(RN) * N(N-1)/2 = LN(N-1)/(2RN) = (N-1)L/(2R) P24: Gi s c ni trc tip vs nhau bng t R = 2 Mbps. Tc  lan tng truy 8 m/s. a.  tr (R.d prop ). b. c g  n B. Gi s c truyc. S bit t c truyng truyn ti b c. Gi prop . d.  rng cng truyn b e. Vic t rng cng truyn. Cho bit t lan tc  truy ng truy TL: Computer Networking: A Top-Down Approach Featuring the Internet, 5th Edition – Translate by K55CC  Page 6 a. R.d prop = R. m/s = 2. 20.10 6 / (2,5.10 8 ) = 160.000 bits. b. S bit tc truyng truyn ti bm = R.d prop = 160.000 bits. c.  gi tr(d prop ) ca ng truy bit t  ng truyn d.  rng c ng truyn / (R.d prop ) = 20.10 6 / 160.000 = 125 m e.  rng ca 1 bit = m / (R.d prop ) = s/R. Chapter2: Application Layer -Review Question  1:  - The Web - HTTP; - File transfer (bittorrent) - FTP; - Remote login - Telnet; - Network News - NNTP; - E-mail - SMTP;  t gi - Network architecture t h thp v - Application architecture c hin ng d u khin hong c ng dng chia s file P2P. Bi nh server trong 1 giao tip gii sao? i giao dng dng chia s file P2P  g  s bn mut giao dch gii t cao, bn UDP hay TCP? N dn thit lng chuyn ch mt 1 RTT u tu s dng TCP bn mt  cho vic thit lp kt n u khi-of-band? FTP s dng 2 kt ni TCP song song. Mt kt n king hu chuyn giao 1 tp tint kt n chuyp tin . Bm c g kt ni vi t ng FTP  "out-of-band" .  t gia download-and-delete mode and the download-and-keep mode in POP3? Computer Networking: A Top-Down Approach Featuring the Internet, 5th Edition – Translate by K55CC  Page 7 - Vi download-and-delete y tin nhn t m n s b t ra mt v cn gi quyn t nhi) - Trong download-and-keep, tin nhn s  i s dng ln.  t tii li s dng ln, tt c   c chuy      mt  M       (alias) cho 1 hostname   host name of the mail server? c s d  a mail server va ch IP. Cverlay network N The edgesc c t  - t h thng g tt gi - a router. - The edges c. -  i   thn bia ch IP ca 1 hay nhiu node ca h th gode h s tr n ca h thng.  P2P cho 2 giao thc quan tr  ng dng quan trp vi ki a) File Distribution b) Instant Messaging c) Video Streaming d) Distributed Computing Chapter2: Application Layer -Problem  a. Mu t nh, client s gi 1 tin nhn 4 tin nhn phn hi? - Sai: gi 4 nhn 4 b.  gt n-  c. Vi mt kt n  c hin 2 request - Sai d. u ca giao th rng. Computer Networking: A Top-Down Approach Featuring the Internet, 5th Edition – Translate by K55CC  Page 8 :   - Application layer protocols: DNS and HTTP - Transport layer protocols: UDP for DNS; TCP for HTTP Gi s t web ba 1 trang web. Bn cn la ch IP cn phi l DNS bn mt RTT thi gian l 1 , RTT 2  RTT n ng text. Bn mt RTT 0   host ti server chi gian t n khi nhc ng. Tng th la ch  1  n ; a ch IP bn mt RTT 0  kt n 0 n gi ng. ng thi gian cn thi 0 + RTT 1  n ;   a. : Kng ta cn 8 ln thit ln gt th 0 = 16 RTT 0 ng th 0 + RTT 1  n + 16 RTT 0 b. t ni song song: 1 ln kt ni gi nhng cn 2 ln kt nt th 0 ng th 0 + RTT 1  n + 4 RTT 0 c.    0  ng th 0 + RTT 1  n + RTT 0 = 3 RTT 0 + RTT 1  n   s dc ging  k - SMTP kng m cha du chm. - HTTP qu  -  s dc ging SMTP    d    dng ASCII  s truy cp mail ca bn bng POP3 Computer Networking: A Top-Down Approach Featuring the Internet, 5th Edition – Translate by K55CC  Page 9 a. Gi s bnh da bc  ch   i: C: dele 1 C: retr 2   S: . C: dele 2 C: quit S: +OK POP3 server signing of b. Ch  down xong gi C: retr 2   S: . C: quit S: +OK POP3 server signing off u  ch  down xong gi ng hp  C: list S: 1 498 S: 2 912 S: . C: retr 1   S: . C: retr 2   S: . C: quit S: +OK POP3 server signing off : Suppose you can access the caches in the local DNS servers of your depart ment. Can you propose a way to roughly determine the Web servers (outside your department) that are most popular among the users in your department?   n ca DNS caches trong nh  Web server xut hi bin nh bi nu nhic gi  t hin trong DNS caches nhilocal DNS cache, y thi gian truy vn s ng hi gian truy vn l Computer Networking: A Top-Down Approach Featuring the Internet, 5th Edition – Translate by K55CC  Page 10 22: Consider distributing a file of F = 15 Gbits to N peers. The server has an upload rate of us = 30 Mbps, and each peer has a download rate of d i = 2 Mbps and an upload rate of u. For N = 10, 100, and 1,000 and u = 300 Kbps, 700 Kbps, and 2 Mbps, prepare a chart giving the minimum distribution time for each of the combinations of Nand u for both client-server distribution and P2P distribution.  u phi ti thiu phi client-server, ta s dc: Dcs = max {NF/us, F/dmin}  u phi ti thiu phi P2P, ta s dc:   F = 15 Gbits = 15 * 1024 Mbits u s = 30 Mbps d min = di = 2 Mbps Note, 300Kbps = 300/1024 Mbps 23. Consider distributing a file of F bits to N peers using a client-server architecture. Assume a fluid model where the server can simultaneously transmit to multiple peers, transmitting to each peer at different rates, as long as the combined rate does not exceed us a. Suppose that us/N <= dmin . Specify a distribution scheme that has a distribution time of NF/us. b. Suppose that us/ N ~ dmin . Specify a distribution scheme that has a distribution time of F/dmin . c. Conclude that the minimum distribution lime is in general given by max{NF/us, F/ dmin ). u phi file ti tng client, song song, v mt t  t   download ca t thit us/N <= dmin.   nhc file vi mt t  i client nhn t us/N, th mi client [...]... utilization (shown in the cross-country example), the designers of this application let the receiver keep sending back a number (more than two) of alternating ACK 0 and ACK 1 even if the corresponding data have not arrived at the receiver Would this application design increase the channel utilization? Why? Are there any potential problems with this approach? Explain Answer Yes This actually causes... 00101000 01110000 Đề cương mạng máy tính Page 27 Computer Networking: A Top-Down Approach Featuring the Internet, 5th Edition – Translate by K55CC P17: Giả sử gửi 2400 byte datagram vào trong một link, link này có MTU c a 700bytes Cho rằng datagram gốc đ ợc đ nh dấu với định danh số 422 Có bao nhiêu cờ đ ợc t o ra? Gi trị c a mỗi tr ờng kh c nhau trong datagram đ ợc t o ra có liên quan gì đến phân mảnh?... send a number of pipelined data into the channel Yes Here is one potential problem If data segments are lost in the channel, then the sender of rdt 3.0 won’t re-send those segments, unless there are some additional mechanism in the application to recover from loss Đề cương mạng máy tính Page 21 Computer Networking: A Top-Down Approach Featuring the Internet, 5th Edition – Translate by K55CC Chapter4:... lượng c a A gấp 2 lần c a B thì pA(1-pB) = 2pB(1-pA)  pA = 2 – (pA/pB) c) Thông lượng c a A là 2p(1-p)N-1, và các node khác là p(1-p)N-2(1-2p) Đề cương mạng máy tính Page 34 Computer Networking: A Top-Down Approach Featuring the Internet, 5th Edition – Translate by K55CC P11: Giả sử có 4 node ho t động là A, B,C,D đang c nh tranh truy cập vào kênh sử dụng ALOHA chia khe Giả sử mỗi node có vô h n packet... datagram có giới h n là 1500byte (bao gồm header) gi a host nguồn A và host đ ch B Th a nhận rằng có 20byte IP header Hỏi có bao nhiêu datagram cần đến để gửi 1 MP3 bao gồm 5 triệu byte ? giải th ch cho t nh to n c a b n Trả lời: 1 file MP3 = 5000000byte Th a nhận dữ liệu mang theo TCP segment, với mỗi TCP segment có 20byte c a header Mỗi datagram có thể mang 1500-40=1460 byte c a file MP3 Số datagram... Top-Down Approach Featuring the Internet, 5th Edition – Translate by K55CC a Kí hiệu EstimatedRTT(n) là ước lượng sau n sample EstimatedRTT(1) = SampleRTT1 EstimatedRTT(2) = α SampleRTT1 + (1-α) SampleRTT2 EstimatedRTT(3) = α SampleRTT1 + (1-α) α SampleRTT2 + (1-α).SampleRTT3 ] = α SampleRTT1 + (1-α).α SampleRTT2 + (1-α)2.SampleRTT3 EstimatedRTT(4) = α SampleRTT1 + (1-α).α SampleRTT2 + (1-α)2.α SampleRTT3... trên đ a chỉ IP c a packet (tầng 3) A link-layer switch muốn chuyển packet phải d a trên đ a chỉ MAC (tầng 2) R2: Hai chức năng quan trọng c a tầng network trong chuyển m ch gói là gì? Ba chức năng quan trọng c a tầng network trong một chuyển m ch ảo? Trả lời: Hai chức năng quan trọng c a tầng network trong datagram – network: chuyển và định tuyến Ba chức năng quan trọng c a tầng network trong virtual... cho cả hai R7: T i sao mỗi cổng vào trong vùng nhớ c a router có tốc độ cao (Discuss why each input port in a high-speed router stores a shadow copy of the forwarding table) With the shadow copy, the forwarding decision is made locally, at each input port, without invoking the centralized routing processor Such decentralized forwarding avoids creating a forwarding processing bottleneck at a single... t i sao gói tin trả lời ARP l i chỉ gửi đến 1 đ a chỉ MAC duy nhất ? V ửi ARP query thi t b a chỉ MAC c a thi t b gử ã t tr t rf a chỉ MAC c A P qu r ử n Khi gửi lạ n ph i gửi broadcast n a t tr l t R11 Với m ng ở Figure 5.19, router có 2 ARP modules, mỗi modules có bảng ARP c a nó Có thể hay không 1 đ a chỉ MAC xuất hiện ở cùng 1 bảng ARP K t v i mạ L 1 t p adapter gắ vs với m t rt 1 a chỉ MAC duy... TRR) P9: xem nh 1 datagram sử dụng 32 bit để đ nh đ a chỉ c a host Giả sử 1 router có 4 links từ 0 đến 3, và gói tin có thể chuyển qua link interface nh hình sau: Đề cương mạng máy tính Page 26 Computer Networking: A Top-Down Approach Featuring the Internet, 5th Edition – Translate by K55CC a Cung cấp 1 bảng chuyển sử dụng prefix matching dài nhất và chuy n gói tin qua link interface đúng b B n x c . Computer Networking: A Top-Down Approach Featuring the Internet, 5th Edition – Translate by K55CC  Page 1 Computer Networking - A Top-Down Approach. 1Mbps khi truyền, nhưng mỗi user chỉ truyền 20% thời gian. Computer Networking: A Top-Down Approach Featuring the Internet, 5th Edition – Translate by K55CC  Page 3 a byte = 4.000 KB = 32.000 Kbit. Tgian = 32000/500 = 64 s Computer Networking: A Top-Down Approach Featuring the Internet, 5th Edition – Translate by K55CC  Page 4 c.