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132B_1_hw8_solutions

132B_1_hw8_solutions - EE132B Professor Izhak Rubin...

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Unformatted text preview: EE132B Professor Izhak Rubin Solution for Homework 8 Problem 1 a. GeonVGle q = 0.35 msg/slot i. p = 0.4 msglslot sz-q) 90— P) a =0.ll765 E[X}=—‘3— =o.1333 I—a E {Q} = E [XI -a = 0.015? a E [W]: 4(1-0) = 0.15636 slots ii. p = 0.8 msg/slol a = 0.70588 E IX] = 2.4 -EHM=L$4 E [W] = 2.82 slots iii. p = 10.0 msgfslot EEX1=EIQI=EfWJ=°° S lfiW>$=c{i:i] . l—p i. P(W>5)=wa104 ii. P(W>5)=0.1675 III. PAN-551:1 Problem 2 a. Let L denote the number of bits per message. The random variable I. has a geometric distribution with a mean of 500 bits so that P(L=k)=r{l-r)“, for k 21. where r = -——- . Let R denote the transmission-rate. A slot is set to be equal to 1' (sec). Assume that R -1." is a positive integer. Note that'R - 1' is the maximum number of bits which can be transmitted in a slot. Let 5 denote the number of slots for serving a message Then. we have he}. Using the expressiOn given above. we obtain P(S=k)=FUFL;-‘=k] I =F{k—1<%Sk] =P((k-1)-Rr<Lsk-Rr) = (1 _ r)(*~|}kr __ (I __ r)”: =[1-(1-rr1-[o-ritl‘". for k 21. Therefore. the service time S has the geOmett-ic distribution with parameter (1 _. r)". Set q =1 -(1- r)". Geot'nIGeom'l Eth1=500bits 15' = 0.01 second; E [Time between message arrivals] = E [T] a 4 siots i. ' R: 10.000bps P 1 0‘5 sgfslot = ——-'— = .5 In em q = I -(l- r)"'=0.1814 where r=—~l— 500 q < p. so system is unstable. Therefore. E {W} = °'° . P(W S 4) = 0 ii. R = 15.000 bps q = 0.2594 q > p. so system is smbie p(1- 0) 0(1 - P) a = =0.95166 iii. R = 20.000 bps. q = 0.32995 (stable) as = 0.677 E [W] = 6.35 P(W s 4) a 0.5687 Problem 3 E IL..I= 500 bits 1' = 0.01 seconds p = 0.25 mag/slot We must simultaneously meet E [W] s 4 and PM gs); 0.35 Restating these two conditions as formulas. we require a --—s4 ell-a) (i) and 6 (ii) {lug} 50.15 1-9 The only unknown in( i) and ( ii ) above is q (which alSo appears as a componentot'a . since 1 .. a = M . ) We don't know which of the two conditions, ( i ) or ( ii ). places a greater 4(1- p) restriction on q. Start by examining equation ( i ). (I --— S 4 (20 ~ 9) substituting a = Pu — 9') results in the condition ell - P) 933-93254 or 1qu —3q—120 _q(q — 0.25) This has one nonncgative solution: 4:} 2 0.36075. So. any value of q greater than 0.360% will satisfy the E{W]S 4 constraint. By substitution, we can show that q = 0.360% does not satisfy condition ( ii ) above. This means that condition ( ii ) places a greater restriction on q. and any value ofq that meets conditiou ( ii ) will also meet condition ( i ). 190 Ht) sill - :3) Turning attention to condition ( ii ). substituting a = Rcsuhs in the requircment: kil— S 0.03 ' ‘3 Starting at q = 0.36075. we tabulaac: 2' 0.36075 0.1209 0.38 0.093 0.39 I 0.0305 0.3905 0.08002 0.3906 0.0?9907 (less than 0.08) we choose q = 0.3906 1 mam) Usin = l -- l h “—- g q [ 500] Yeilds R = 24.739 bps Problem 4 MfM/l 1 A=-——=-——-=250 / EIT] 0.004 "”3 sec A Substituting p=— ivcs 400 3 -/I—S 3.333 400 —- 21 Soiving, we find fl 5 307.692msglscc . ...
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