Service commands: RETR, STOR, STOU, APPE, ALLO, REST, RNFR, RNTO, ABOR, DELE, RMD, MRD, PWD, LIST, NLST, SITE, SYST, STAT, HELP, NOOP.

Transcription

1 Chapter 2 Prblems Prblem a) F b) T c) F d) F e) F Prblem 2 Access cntrl cmmands: USER, PASS, ACT, CWD, CDUP, SMT, REI, QUIT Transfer parameter cmmands: PORT, PASV, TYPE STRU, MODE Service cmmands: RETR, STOR, STOU, APPE, ALLO, REST, RFR, RTO, ABOR, DELE, RMD, MRD, PWD, LIST, LST, SITE, SYST, STAT, HELP, OOP Prblem 3 Applicatin layer prtcls: DS and HTTP Transprt layer prtcls: UDP fr DS; TCP fr HTTP Prblem 4 a) The dcument request was The Hst : field indicates the server's name and /cs453/indexhtml indicates the file name b) The brwser is running HTTP versin, as indicated just befre the first <cr><lf> pair c) The brwser is requesting a persistent cnnectin, as indicated by the Cnnectin: keep-alive d) This is a trick questin This infrmatin is nt cntained in an HTTP message anywhere S there is n way t tell this frm lking at the exchange f HTTP messages alne One wuld need infrmatin frm the IP datagrams (that carried the TCP segment that carried the HTTP GET request) t answer this questin e) Mzilla/50 The brwser type infrmatin is needed by the server t send different versins f the same bject t different types f brwsers

2 Prblem 5 a) The status cde f 200 and the phrase OK indicate that the server was able t lcate the dcument successfully The reply was prvided n Tuesday, 07 Mar :39:45 Greenwich Mean Time b) The dcument indexhtml was last mdified n Saturday 0 Dec :27:46 GMT c) There are 3874 bytes in the dcument being returned d) The first five bytes f the returned dcument are : <!dc The server agreed t a persistent cnnectin, as indicated by the Cnnectin: Keep-Alive field Prblem 7 The ttal amunt f time t get the IP address is RTT RTT2 Once the IP address is knwn, RTT O elapses t set up the TCP cnnectin and anther elapses t request and receive the small bject The ttal respnse time is 2 RTT RTT RTT2 RTT O Prblem 8 a) b) c) RTT 2RTT RTTn 2RTT 8 8RTT RTT RTT 2RTT RTTn 2RTT 2 6RTT RTT RTT RTT 2RTT RTT 3 RTT RTT n Prblem a) Yes, because Bb has mre cnnectins, he can get a larger share f the link bandwidth b) Yes, Bb still needs t perfrm parallel dwnlads; therwise he will get less bandwidth than the ther fur users

3 Prblem 3 The MAIL FROM: in SMTP is a message frm the SMTP client that identifies the sender f the mail message t the SMTP server The Frm: n the mail message itself is OT an SMTP message, but rather is just a line in the bdy f the mail message Prblem 22 Fr calculating the minimum distributin time fr client-server distributin, we use the fllwing frmula: Dcs = max {F/us, F/dmin} Similarly, fr calculating the minimum distributin time fr P2P distributin, we use the fllwing frmula: D P2 P max{f/u s, F/d min, F/(u Where, F = 5 Gbits = 5 * 024 Mbits us = 30 Mbps dmin = di = 2 Mbps s i ui)} te, 300Kbps = 300/024 Mbps Client Server Kbps u 700 Kbps Mbps Peer t Peer Kbps u 700 Kbps Mbps Prblem 23 a) Cnsider a distributin scheme in which the server sends the file t each client, in parallel, at a rate f a rate f us/ te that this rate is less than each f the client s dwnlad rate, since

4 by assumptin us/ dmin Thus each client can als receive at rate us/ Since each client receives at rate us/, the time fr each client t receive the entire file is F/( us/) = F/ us Since all the clients receive the file in F/ us, the verall distributin time is als F/ us b) Cnsider a distributin scheme in which the server sends the file t each client, in parallel, at a rate f dmin te that the aggregate rate, dmin, is less than the server s link rate us, since by assumptin us/ dmin Since each client receives at rate dmin, the time fr each client t receive the entire file is F/ dmin Since all the clients receive the file in this time, the verall distributin time is als F/ dmin c) Frm Sectin 26 we knw that DCS max {F/us, F/dmin} (Equatin ) Suppse that us/ dmin Then frm Equatin we have DCS F/us But frm (a) we have DCS F/us Cmbining these tw gives: DCS = F/us when us/ dmin (Equatin 2) We can similarly shw that: DCS =F/dmin when us/ dmin (Equatin 3) Cmbining Equatin 2 and Equatin 3 gives the desired result Prblem 24 a) Define u = u + u2 + + u By assumptin us <= (us + u)/ Equatin Divide the file int parts, with the i th part having size (ui/u)f The server transmits the i th part t peer i at rate ri = (ui/u)us te that r + r2 + + r = us, s that the aggregate server rate des nt exceed the link rate f the server Als have each peer i frward the bits it receives t each f the - peers at rate ri The aggregate frwarding rate by peer i is (-)ri We have (-)ri = (-)(usui)/u <= ui, where the last inequality fllws frm Equatin Thus the aggregate frwarding rate f peer i is less than its link rate ui In this distributin scheme, peer i receives bits at an aggregate rate f ri r ji j u s Thus each peer receives the file in F/us

5 b) Again define u = u + u2 + + u By assumptin us >= (us + u)/ Equatin 2 Let ri = ui/(-) and r+ = (us u/(-))/ In this distributin scheme, the file is brken int + parts The server sends bits frm the i th part t the i th peer (i =,, ) at rate ri Each peer i frwards the bits arriving at rate ri t each f the ther - peers Additinally, the server sends bits frm the (+) st part at rate r+ t each f the peers The peers d nt frward the bits frm the (+) st part The aggregate send rate f the server is r+ + r + r+ = u/(-) + us u/(-) = us Thus, the server s send rate des nt exceed its link rate The aggregate send rate f peer i is (-)ri = ui Thus, each peer s send rate des nt exceed its link rate In this distributin scheme, peer i receives bits at an aggregate rate f ri r r j u /( ) ( us u /( )) / ( us u) / ji Thus each peer receives the file in F/(us+u) (Fr simplicity, we neglected t specify the size f the file part fr i =,, + We nw prvide that here Let Δ = (us+u)/ be the distributin time Fr i =,,, the i th file part is Fi = ri Δ bits The (+) st file part is F+ = r+ Δ bits It is straightfrward t shw that F+ + F+ = F) c) The slutin t this part is similar t that f 7 (c) We knw frm sectin 26 that D P2 P max{f/u s, F/(u s u)} Cmbining this with a) and b) gives the desired result Prblem 25 There are ndes in the verlay netwrk There are (-)/2 edges Prblem 33 Yes, yu can cnfigure many brwsers t pen multiple simultaneus cnnectins t a Web site The advantage is that yu will yu ptentially dwnlad the file faster The disadvantage is that yu may be hgging the bandwidth, thereby significantly slwing dwn the dwnlads f ther users wh are sharing the same physical links