132B_1_HW3_Solutions

132B_1_HW3_Solutions - EEISEE Professor Iahak Rubin...

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Unformatted text preview: EEISEE Professor Iahak Rubin Solution for Homework No. 3 Set No. 5 Problem 1 Problem 1.1 If no collision occurs. it takes 1"; {sec} to transmit a massage. Problem 1.2 The channel is idle When all the stations at: idle. "JI'inIna1 we have T = mil-IN".-::'=l.'2,---.i'lu|"}r F{T}E} = Ffimin{fl:f=112,---,N}:~!] H 9mm a an H :1 2 33 u ll L'I Note that {I}. t' = 1,2. -- r. N} is a set of independent random variablot. Problem 1.3 No collision occurs il' all stations except Station i an: id]: during the mssaage transmission time. Note that the massage transmission time is % {sac}. Thus, 1. PH no collision” : - ll PHAH tbs oth or stations are idl: for more than M Fin {1'} 3" El} EE'ilr"'rN}l.iil M = H PU} 3' E} jEH.---.NIHH = moi—L ti.- R}- :Eh. -.N}t:-+ Problem 2 Problem 2.1 When we ignore the propagation delay, the time elapsed from the instant Station 1 receives the token to the instant Station 3 receives the token is equal to the total time the tolren is held by Station 1 and Station 2. Let 313 denote this time duration. Then1 313 = H] + H}. Noting that the random variables H. and H: are independentE we have P[313=n] = PIIHI+Hg=fl2| = F{H1+H:=anz=m}-F{Hg=m} m=fl : EFfH1=n—m]F1:Hl=m] m=|2l = Ethane." '" tl—vrlei" m=El = ll-v1ltl—vr}t?ElE]“ “=5 9'1 _ nl_[_E_]a+l -‘ Iii—EILHI-‘HJ'FI when: r1=l—ri and "i'!=l_P'.l- Problem 2.2 By the same method as in Problem 2.1, we can soIve this problem. Let 311 denote the time duration from the moment Station 1 receiver the token to the moment Station 1 receiver. it again. Let ffi,(n} denote the enrolls-titling.r mare function for the random variable Hg. 1' = 112,- - -,N, eueh that Iain] = F{H,- = 1:] =p§[l -- pd" for It = fl, 1,* Then1 we have N £1th- = n] lizll ffl'r'Efl} ‘5 ffleinj * ‘ "*IHHIZHJI P(3’|] = 11-] where e is the convolution operator. Problem 2.3 _The throughput measures the average amount of work per unit-time. As indicated in the hiritI the mount of work is equal to the total time for which atatione hold the token. Define a cycle to he the time elapeed from the instant the token is received by any station to the earlieet inatant it is received by the same station. Consider the amount of wort: during a cycle denoted as W. Then. the average amount of work during a cycle is E{W] : EiHr}, and the average ieogth of a cycle is E{W} + N - r. Thea, we have EEW} thmflhP“ = m — ' E(Hi) 2?; E(H.-)+N -r N L22; P- i=1 Zi=lqfi+Nlr ll Problem 2.4 The total time duration from the moment Station 1' releases the token to the moment. Station 1' receives it again is the sum of the token holding times of all stations except Station i and the propagation delay across the N links. Note that the propagation delay is a deterministic (non-random) and fixed value, and is equal to N ~1', so that the random variable R..- must be greater than or equal to N - 1'. Hence, we have P(R.- = n) = £1211,- + N—r = n) je{1.---.N}\{i) = H} = n—N-r) J‘E{1.---.N}\{i} fH;(n-N‘T) * "'*fH.-i("—N'T) *fH;+;(""N'T) * "JEAN-NT), forn=N-1',N-r+1,---. Fora:1,---,N-r-1,P(Rg=n)=0. Problem 3 - problem 3.1 The most critical difference between circuit-switching and packet-switching systems is that the circuit-switching system provides dedicated space, time and/or frequency resource to a user, regardless of the frequency and the efficiency of the resource utilization by the user. The packet-switching system does not dedicate resource to a user. It allocates resource to a user when the user needs such resource. The email system is a packet-switching system. Problem 3.2 There is no absolutely correct answer. Possible answers are: 1. reservation (demand-assigned TDMA) 2. reservation (demand—assigned TDMA) 3. TDMA 4. reservation (demand-assigned TDMA) Problem 4 Problem 4.1 (A,M), (E, K), (G,N), (F, 0), (D, P), (H, L), (BJ), (c, I). Problem 4.2 (11.1”), (E, K), (HO). HUHGEmHD h.“ 1 Eva." mun." aflflflun mafia. flanEmE Ftp firm n35." 3350: in; 3:3 rm BEE 3 "an Fun; n... gran—Em SEEK- Hrs? En “.39..” icing: $3.5 .E {Ewan Fm": H in? HEW- mind 39 man; wan SE “FF En LE 3553 15m E mum” HER? .Hwn 33.3.. E. nnwafiuwnmnu "main mu umm an may 3 :5". any 331." F 33333.“. E a 3:2,”... 3mm 3. m ER. 203 :55. H Li fin” H.355 mu Emian mun w REEF firming. F map—.5: 3E: Swami: E, Ema... m 3:. glam u. m3... ER gm 53.2.:an :EEEE: 3H .5 E ...E$:hfl-fl «than: H mum Frau“. Communication Network? (2“ Edition) Chapter 6 Solutions ' 3 . s : 87:68 x Ex {54 + s) :13440i‘aits 2?4 54 4 '.' ELK] : M : 0,75 1920034223 :2 2 = 4.43 x 10—3 0.7.? Therefore, the maximum throughput is 1 - 1 — 0.97. 1+ (2e+ 1)(4.43><10‘-) pH]th = 12. Can the Digital Sense Multiple Access protocol, which is used by CDPD, also be used on the digital carrier 0 TSM'? liyes, explain how. Solution: Digital Sense Multiple Access would require redesign of the GSM frame structure. In the downlink transmission broadcast from the base station, a version of the flag word would need to be inserted indicate whether the medium is busy. Furthermore since the frame structure in GSM effectively divides the frame into 8 channels, a separate Digital Sense Multiple Access protocol would need to be in operation for each channel. This in turn would require the presence of a flag word per channe in the downlink frames. 13. M terminals are attached by a dedicated pair of lines to a hub in a. star topology. The distance from each terminal to the hub is of meters, the speed of the transmission lines is R bits-iseeond, all frames are ot‘length 12500 bytes, and the signal propagates on the line at a speed of2.5 (103) metcrsisecond. For the four combinations ofth following parameters {o’= 25 meters or d = 2500 meters; R = 10 Mbps or R —- 10 Gbps}, compare the maximum network throughput achievable when the hub is implementing: Slotted ALOHA; CSMAECD. Solution: i. = 12500 x 8 bits, rpm, = d! (2.5 x 103 metersisee), a = rpm, at Values for a: R/d 2x25 2x2500 mow 2E-03_ 1.00E+10 2E-02 2E+00 Maximum Throughput for Slotted ALOHA: R/d 2x25 2x2500 1.er+07 0.367;er {9.3677879 1.00E+1U -. 0.3678T9 nearer-e Maximum throughput for CSMA—CD: {‘oaJm-unieurior.‘ .1'anl'oiittfl'm I':-'.|it'i't't1'l] Chapter 5 Solmioos are 2:95 EKESUG i.DGE+fl? - asseser asses 1.flGE+1fl 5 asssses oatsse 14. Consider the star—Lepnlugy nt-ntroriq in problem {all when The token-ring preteen] is used for medium access centre]. Assume singia-I'rame operation: eigitL—hit latency at each station. it? 125 stations. Assume a Free tel-:en ihrau bytes long. Solutions follow questions: M = 125 b = 3 bits LEM“ = 3 bytes Ill—frame = 'I.-"= as s is“ misee a. i"io-:l 111: effective I'rame transmission time Jet Liu: Your eeliitiiiifltitllifi Lifer and Ft. The distance from each terminal to the hub is of meters. so the total distance around the ring is trier" ares. Assuming singia frame transmission token rei'r'raertr'eo. then Km = token transmission time + frame transmission time + ring iatenezr I. ll f ‘5. [net-r; 4:_ ’_i'.i'rJ.IH-I:_ _|_ + _ R t 1.! R b. Assume that each station can transmit up to a maximum oHr {rallies-“token. Find The maximum network throughput for the four oases ol'rr' and it. The maximum throughput oeeurs I.Irhen all stations transmit it frames per tel-ten. After eornpleting tr transmission of it frames, eaeh station waits one ring lateney time and then transmits a free token ir the ring. 3H K k “X A! I'I'anirtl ] puma: n: — I, _'-I— ; — 3“: 11'; km,” '5' Emil”? [1+ T 1+ ___ ___ j: Hihirrrr Mk):- Ii'rrritur' where 3 2a" b 1" . r = M + E is the amount time spent in passing the toiten around the Hog. t ‘r J Calculation: i. o'=25m. R: 10 hist '1: = Xian-5 = m3. X||31|§|5= E-AHSEBI ’ s e 25 s ‘“ a”: (3+I2sati}s __ Mir :3: “Jr =ml1m5 113M iasste- Lflrl’fx I firm—['1' 'I win -"L1."'r:|1' =I : 3‘1 -" ...
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This note was uploaded on 11/11/2010 for the course EE 132B taught by Professor Izhakrubin during the Spring '09 term at UCLA.

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132B_1_HW3_Solutions - EEISEE Professor Iahak Rubin...

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