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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 = milIN".::'=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 = llv1ltl—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 enrollstitling.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 unittime. 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 (nonrandom) 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—Nr)
J‘E{1..N}\{i}
fH;(nN‘T) * "'*fH.i("—N'T) *fH;+;(""N'T) * "JEANNT), forn=N1',Nr+1,. Fora:1,,Nr1,P(Rg=n)=0. Problem 3  problem 3.1 The most critical difference between circuitswitching and packetswitching systems is that
the circuitswitching 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 packetswitching 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 packetswitching system. Problem 3.2 There is no absolutely correct answer. Possible answers are:
1. reservation (demandassigned TDMA)
2. reservation (demand—assigned TDMA)
3. TDMA
4. reservation (demandassigned 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: $3.5 .E {Ewan Fm": H in? HEW mind 39 man; wan SE “FF En LE 3553 15m E
mum” HER? .Hwn 33.3.. E. nnwaﬁuwnmnu "main mu umm an may 3 :5". any 331." F 33333.“. E a 3:2,”... 3mm 3. m ER. 203 :55. H Li ﬁn” H.355 mu Emian mun w REEF ﬁrming. F map—.5: 3E: Swami: E,
Ema... m 3:. glam u. m3... ER gm 53.2.:an :EEEE: 3H .5 E ...E$:hﬂﬂ «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 bitsiseeond, 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 2E03_
1.00E+10 2E02 2E+00 Maximum Throughput for Slotted ALOHA: R/d
2x25 2x2500 1.er+07 0.367;er {9.3677879
1.00E+1U . 0.3678T9 nearere Maximum throughput for CSMA—CD: {‘oaJmunieurior.‘ .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 ntntroriq in problem {all when The tokenring preteen] is used for medium access
centre]. Assume singiaI'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. [netr; 4:_ ’_i'.i'rJ.IHI:_ __ + _ 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 telten. 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: = Xian5 = m3. X31§5= EAHSEBI
’ 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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 Spring '09
 IzhakRubin
 Stations of the Cross, base station, Channel access method, eaeh station

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