This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 szq)
90— P) a =0.ll765 E[X}=—‘3— =o.1333
I—a E {Q} = E [XI a = 0.015? a E [W]:
4(10) = 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. PAN551:1
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{lr)“, for k 21. where r = —— . Let R denote the transmissionrate. 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((k1)Rr<LskRr)
= (1 _ r)(*~}kr __ (I __ r)”: =[1(1rr1[oritl‘". for k 21. Therefore. the service time S has the geOmettic 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
ella) (i)
and 6
(ii) {lug} 50.15
19 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)
93393254 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 . ...
View
Full Document
 Spring '09
 IzhakRubin

Click to edit the document details