{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ECE302HW2Soln_Fa09

# ECE302HW2Soln_Fa09 - 1 Text problem 2.97(page 91(a We wish...

This preview shows pages 1–9. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1. Text, problem 2.97 (page 91) (a) We wish to ﬁnd the probability that the block is accepted, or equivalently, the probability that there are 1 or fewer errors. Note that there are 100 ways to receieve a single error (depending on which of the 100 bits were incorrect), but only one way to receive no errors, so that P [block accepted] I l P [no errors] + P [1 error] (1 — @100 +100(1— @9919 (099)100 + 100(0.99)99(0.01) 0.7358. (b) Let the probability of a block transmission failing be q. From part (a), we have that q = P [transmission fail] : 1 — P [block accepted] : 0.2642. If a block transmission fails, it is retransmitted, so that the num— ber, M, of retransmissions will consist of M - 1 failed transmission followed by 1 successful one. The probability of M transmissions is then qM‘1(1 — q) 2. Text, problem 3.18 (page 132) As in the hint, deﬁne I to be the index of the ﬁrst refresh message which is not lost. In order to ex— tend the reservation, we must send j messages before the 10 minute reservation expires, and have that I S j to guarantee an extension of reservation. Therefore, the probability that the reservation is extended within j sent messages is the probability PUSJ]=P]I=1]+P[I=2]+...+P[I=j]:ipﬂz‘], where p1[2] is the pmf of random variable I. We now need only ﬁnd the form for the pmf p] [i] = P [I = i]. We have that I 2 7L if and only if 2' — 1 refresh requests fail, followed by one that succeeds. If we let P [request fails] : p = 1/2, then P1<i>= PU = ii = M ~ p>=<1/2>2 1 Therefore Pl! 3 j] = 2291M: 2(1/2Y = viii/22);“ =1~<1/2)j. The problem asks to find j such that the probability of extension is greater than 0.99, or, in other words, ﬁnd 3' such that PUSjl > 0.99 :1—(1/2y' > 0.99 :>(1/2)j < 0.01 :>jln(1/2) < ln(0.01) =>j > ln(0.01) 1n(1/2) ::>j > 6.6439. Since we must get at least j = 7 requests in to guarantee 0.99 proba— bility of reservation, we should start sending refresh requests at least 70 seconds prior to expiration. 3‘ c4) MAXQI Raw/on Variable 41km *Httre. are, 1‘wa and (,ovvhkuoug [”11an @J-f LL) FIX<“] W—IOI FEW] = R W) — Excel—2) =— 73' C—w ﬂ it“ P[x€—l]= £0!) = [1:] PH at < WM] t PH < KS wk] = ?-x (-9.7!) _ fxC-I) = L07 vas')+ fI—°“”‘°/\/?r<“01 Ptx— 0} R v—Fxc—v-U-f’rwj Pliod’jar - _ .,_ - J f -SD ' o (D- l“ z/TDI/l :L g.» \- \‘_.U ‘ ”I; W' J, m _ ﬁr " 2 7° } W'JLIM { WWW] ”2»; > o I L) Pom—u): POM) FCX- ’)= R (4) IDCX’1):‘V(:][7Q(XMX “.7; = 0 PCX{"): FXQ ”3 I: AbeddX - (O{( f wr:J(xq):X) F£»0.IEX< 0) t: PCX=¢0 1) «r [)(-0K<7(<o)— [30“ o) W mam; ville/WI 0+ . 1 ‘L We» Hm. W 2 0 + ‘ ‘“ E £0 .3’ {7c 09‘4“ "’ T ﬁne MW“ ”f" Jr I. The. ,th52 ak- aif’ ~_ 0 +j V fxcxnlx = 7ix|+7;:~é Io :HL Cum/L #‘C M w of“ ?[ P4241024“) t FC0<X<03 f(o<x<o) PCX ) F. ‘ f+7lxt704x “0 2 AX) -0 :(j 0 e Fr‘035X<0 ): 1:; fwdx : (Ti—X11: 7A6— A W x}; 71m): m—ﬂ M2129 lad-)3 AH)? HT‘J‘TW’H 52/- I? #17; 2. I 2. “’9’ f((sx<é) =’Px('9+7x(*)= 72(4) + 645(4)”: 373’ {’c (< M) >7; Clﬁrxehfxa) WI) \- 71165231371 E» W!) ' O4 5 (A) 0: MAW): 32¢ij 1% : Fauna) RAM) :. Eﬁgg) 1(00’3) We) _ PCB/Icy - ( I’ M [/3 :> PCA/IBIC) = PMfc)P(g/C) @ WILL PCQNﬂ >9CW20 ~‘- P(S)P(¢}m FLSM) >PM) = Panm); L PM» PM) PUMA) > PM) ‘~ D=P(¢)I’CA)= o {DUI/194?): F645) 2 Foam) m4) 2 o 3;; Wm m Ida/arm. (@ “twat“? asfvﬁe (L) Ag “A 4) are, mmuud excl‘wz'ue ,Lm— ‘fkta am I‘AC’3Feno’enT" (65,2 [email protected];( Aﬁwrdﬂvd’fv IL} ‘ QM If are. .‘nolerendacr. and '14ch are, MW Mohave. (WW maPrL/mI/In) PrIAmlAu «,Mn.) ~~ PFCAJ/hnr‘ AA») = Fr (AnnAn~I)I’rCA"*’ A"”A"1:;;fr_£& “In" A A») =. frIAnnAn~.rl/m) ~~ MMHNV .32, y, 'k ”~10. I P;- (AolA’nﬂkn~—gﬂ’*n A»: ‘{I {3: CL)? CWT) 2C“ (“70) l / [Eur-am) Faisal a, i ' ' M..- W 1“ C" )r"o~P9""‘ -r-—7”/— (Z)r°<:~fj"= ‘ bl ‘3’ [Q v 32x0“) : 0‘ ECU) {szo t a! PX (X9!) .ohc A_ . ’ (W61 X 1‘5 (mﬁwoug 114/.) Eli} Ural; / (a Pr(as><{l¢)= Prcasxd») = PrM<><<LJsRm<X£U :fjf oddx X U E Pr (MX €19) — g; Pr (Jim) 2 2:” {you} F0[ 42.!) #1914064) RUJ 5an Pr (a 5% s L) 21K (a) TFXUoJ' MW 14}. x, .‘s dexcrexe, w wt 61 , Prune Hugo) ﬁCasJCéL) m‘a‘rh not“ 6W 54%)“ BK!) ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 9

ECE302HW2Soln_Fa09 - 1 Text problem 2.97(page 91(a We wish...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online