Original citation:
Al-Ammal, H., Goldberg, Leslie Ann and MacKenzie, P. D. (1999) Binary exponential
backoff is stable for high arrival rates. University of Warwick. Department of Computer
Science. (Department of Computer Science Research Report). (Unpublished)
CS-RR-359
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Hesham Al-Ammal
Leslie AnnGoldberg y
PhilMaKenzie z
July 28,1999
Abstrat
Goodman, Greenberg, Madras and Marh gave a lower bound of n
(logn)
for the maximumarriavalrateforwhihthen-userbinaryexponentialbakoprotoolisstable. Thus, they showed that the protool is stable as long as the arrival rate is at most n
(logn)
. We improvethelowerbound, showingthat the protool isstable for arrival ratesuptoO(n
:9 ).
Classiation of topi: Algorithmsand datastrutures (distributedalgorithms)
1 Introdution
A multiple-aess hannel isa broadast hannelthat allows multipleusers to ommuniate with eah other by sending messagesonto thehannel. If two ormore userssimultaneously send messages, then the messages interfere with eah other (ollide), and the messages are nottransmittedsuessfully. Thehannelisnotentrallyontrolled. Instead,theusersusea ontention-resolutionprotool toresolve ollisions. Thus,afteraollision, eahuserinvolved in theollisionwaitsarandom amount of time(whihisdetermined bytheprotool)before re-sending. Perhaps thebest-known ontention-resolution protool is the Ethernet protool of Metalfe and Boggs [9 ℄. The Ethernet protool is based on the following simple binary exponential bako protool. Time is divided into disrete units alled steps. If the i'th user hasa message to sendduring a given step, then it sendsthis message with probability 2
bi
, where b i
denotes the number of ollisions that this message has already had. With probability1 2
b i
, user i does not send during the step. The Ethernet protool is based on binary exponential bako, but some modiations are made to make it easier to build. See [6,9℄ fordetails.
Hastad, Leightonand Rogo [6 ℄ have studied theperformane ofthe binary exponential bako protool inthe following natural model. The system onsists of n users. Eah user maintainsa queueofmessages thatitwishestosend. Atthebeginningofthet'thtimestep,
heshamds.warwik.a.uk, Department of Computer Siene, University of Warwik, Coventry, CV47AL,UnitedKingdom.
y
leslieds.warwik.a.uk, http://www.ds.warwik.a.uk/le slie /, Department ofComputer Si-ene,UniversityofWarwik,Coventry,CV47AL,UnitedKingdom. Thisworkwaspartiallysupportedbythe EPSRCgrantGR/L6098andbyESPRITProjetRAND-II(Projet21726).
z
i message at the head of its queue has ollided is denoted b
i
(t). At the beginning of the t'th step, eah queue reeives 0 or 1 new messages. In partiular, a new message is added to theendofeah queue independentlywithprobability=n,where isthearrival rate ofthe system. After the new messages are added to the queues, eah user makes an independent deisionaboutwhetherornotto sendthemessageat theheadofitsqueue, usingthebinary exponential bako protool. (If the message at the head of the i'th queue has never been sent before then b
i
=0, soit is now sent. Otherwise, b i
=b i
(t), soit is sent independently with probability2
b i
(t)
.) If exatlyone message issent (sothere areno ollisions),then this message is delivered suessfully, and it leaves its queue. Otherwise, the messages that are sent ollideandno messagesare deliveredsuessfully.
Sine the arrivals are modelled by a stohasti proess, the evolution of the whole sys-tem over time an be viewed as a Markov hain in whih the state just before step t is X(t) =((q
1
(t);:::;q n
(t));(b 1
(t);:::;b n
(t))) and the next state is X(t+1). One measure of theperformaneofthesystemistheexpetationoftherandomvariableT
ret
,whihisthe num-berofstepsrequiredforthesystemtoreturnto thestartstateX(0)=((0;:::;0);(0;:::;0)). Hastadetal. [6 ℄provedthatifthearrivalrate istoo high,thenthesystemisunstable,inthe sensethat theexpetedreurrene timeisinnite.
Theorem1 (Hastad, Leighton, and Rogo) Supposethatforsomepositive, 1 2
+. Suppose that n is suÆientlylarge (asa funtion of ). Then E[T
ret ℄=1.
Ontheotherhand,Goodman,Greenberg,MadrasandMarh[5℄showedthatifthearrival rate is suÆientlylow, thenthesystemis stable.
Theorem 2 (Goodman, Greenberg, Madras and Marh) Thereisapositiveonstant suh that E[T
ret
℄is nite for the n-usersystem, provided that < 1 n
logn .
While Goodman, Greenberg, Madras, and Marh's result is the only known stability resultforthenitely-many-usersbinary-exponential-bakoprotool,theirupperbound(<
1 n
logn
)isverysmall. Inthispaper,wenarrowthegapbetweenthetworesults. Inpartiular, we prove thefollowingtheorem.
Theorem 3 There is a positive onstant suh that, as long as n is suÆiently large and <
1 n :9
then E[T ret
℄is nite for the n-usersystem.
The point of Theorem3 is to showthat n-userBinary Exponential Bako is stable for arrivalrateswhihgrowfasterasymptotiallythan1=n. That is,thepurposeof theresultis to showthat, forpositiveonstants and ,<
1 n
1
guaranteesstability. We have hosen =:1 foronreteness. We believe that thesame methodsould be used forslightlylarger valuesof,butaninteresting(anddiÆult)questionraisedbythisworkiswhetherthesame resultwouldbe truefor =1. That is,is there a onstant suhthat then-usersystem is stablewhenever<
1 ?
We nowsummarizesome other relatedwork. We startbyobserving thattheresultsin The-orem 1and 2 an be extendedto more general models. Forexample,the resultof Goodman et al. an be extended to a more general model of stohasti arrivals in whih theexpeted number of arrivals at user iat time t (onditioned on all events up to time t) is a quantity,
i , and
P i
i
is required to be equal to . The result of Hastad et al. an be extended to small values of n, provided that >:568+1=(4n 2). The instability result of Hastad et al.impliesthat, whenissuÆientlylarge,theexpeted average waitingtimeofmessagesis innite.
Next, we mention that the binary exponential bako protool is known to be unstable intheinnitely-many-usersPoisson-arrivalsmodel. KellyandMaPhee[7 , 8℄showedthisfor <ln2 and Aldous[1℄showed thatit holdsforallpositive .
Finally, we mention that, while the goal of this paper is to understand the binary-exponential bakoprotool, on whihEthernet is based,there are other aknowledgement-based protools whih are known to be stable inthe same model forlarger arrivalrates. In partiular, Hastad et al. have shownthat polynomial-bako protools are stable aslong as < 1. The expeted waiting time of messages is high in polynomial-bakoprotools, but Raghavan and Upfal [10 ℄ have given a protool that is stable for < 1=10, in whih the expeted waiting time of every message is O(logn), providedthat the usersare given a rea-sonably good estimate of logn. Finally, Goldberg, MaKenzie, Paterson and Srinivasan [4 ℄ have given a protool that is stable for < 1=e, in whih the expeted average message waiting-time isO(1), providedthat theusersaregiven an upperboundon n.
We onludebyobservingthatthetehniqueofGoldbergand MaKenzie[3℄ anbeused to extend Theorem 3 so that it applies to a non-geometri version of binary-exponential bako, whih is loser to theversion used inthe Ethernet. (Instead of deidingwhether to sendoneahstepindependentlywithprobability2
b i
,theusersimplyhoosesthenumberof stepsto wait before sending uniformlyat random from [1;:::;2
b i
℄.) The ideasare the same as those used in the proof that follows, but the details are messier. Our result an also be extendedalong thelinesof [6 ℄ to show that, when issuÆientlylow, theexpeted average messagewaitingtimeis nite.
3 The stability proof
In order to prove Theorem 3, let = 1
0 n
:9
, where 0
. We will now dene the relevant potential funtion. Letf(X(t)) be thefollowingfuntionof thestate just before step t.
f(X(t))=n 1:8
n X
i=1 q
i (t)+
n X
i=1 2
b i
(t) :
riodi Markov hain X with a ountable state spae A is positive reurrent i there exists a positivefuntionf(),2A,a number >0, a positiveinteger-valued funtionk(), 2A, anda nite setC A, suh that the following inequalitieshold.
E[f(X(t+k(X(t)))) f(X(t))jX(t)=℄ k(); 62C (1) E[f(X(t+k(X(t))))jX(t)=)℄ < 1; 2C : (2)
We use the following notation, where = 3. For a state X(t), let m(X(t)) denote the numberofusersiwithq
i
(t)>0andb i
(t)<lg+lgn,andletm 0
(X(t))denotethenumberof usersiwithq
i
(t)>0andb i
(t)<:8lgn+1. Wewilltaketo be 1 2=and C tobetheset onsisting of the single state ((0;:::;0);(0;:::;0)). We dene k(((0;:::;0);(0;:::;0))) =1, so Equation 2 is satised. For every state 62 C, we will dene k() in suh a way that Equation1 isalso satised. We give thedetailsinthree ases.
3.1 Case 1: m 0
(X(t))=0 and m(X(t))<n :8
.
Forevery state suh thatm 0
()=0 and m() <n :8
we dene k()=1. We wish to show that, if 6= ((0;:::;0);(0;:::;0)) and X(t)= , then E[f(X(t+1) f(X(t))℄ . Our general approah is the same as the approah used in the proof of Lemma 5.7 of [6 ℄. For onveniene,weusemasshorthandform(X(t))andweuse`to denotethenumberofusersi withq
i
(t)>0. Without lossofgenerality,we assumethat theseareusers 1;:::;`. We usep i to denote the probability that user i sends on step t. (So p
i = 2
b i
(t)
if i 2 [1;:::;`℄ and p
i
= =n otherwise.) We let T denote Q
n i=1
(1 p i
) and we let S denote P n i=1 p i 1 p i . Note that theexpeted numberofsuessesat steptis ST. LetI
a;i;t
bethe0=1 indiatorrandom variablewhihis1ithereisanarrivalatuseriduringsteptandletI
s;i;t
bethe0=1indiator random variable whih is1 iuser isueeds insendinga message at stept. Then
E[f(X(t+1) f(X(t))℄=n 1:8 n X i=1 ( E[I a;i;t ℄ E[I s;i;t ℄)+ n X i=1 E[2 b i (t+1) ℄ 2 b i (t) ; = n 1:8 n 1:8 ST + n X i=1 2 bi(t) i (2 bi(t) 1) i ; (3) = n 1:8 n 1:8 ST + n X i=1 2 b i (t) p i (1 T 1 p i ) (2 b i (t) 1)p i T 1 p i ; = n 1:8 n 1:8 ST + ` X i=1 (1 T 1 p i )+ n X i=`+1 n (1 T 1 p i
) `T;
= n
1:8
n
1:8
ST +` `T +
(n `) n T 0 ` X i=1 1 1 p i + n X i=`+1 p i 1 p i 1 A ; = n 1:8 n 1:8
ST +` `T +
(n `)
n
ST `T;
= n 1:8 +`+ (n `) n T((n 1:8
i i
probabilitythat user isendssuessfully at step t. We now ndlower bounds forS and T. First,
S =
n X
i=1 p
i
1 p
i
= ` X
i=1 2
b i
(t)
1 2
bi(t) !
+
(n `)
n
m X
i=1
1
n 1
+
(n `)
n
= m
n 1
+
(n `)
n
: (5)
Next,
T =
n Y
i=1 (1 p
i )
(1
1 2n :8
) m
(1 1 n
) ` m
(1 n )
n `
1
m 2n
:8
` m
n
(n `)
n
(6)
CombiningEquations4,5 and6, we getthefollowingequation.
E[f(X(t+1) f(X(t))℄n 1:8
+`+
(n `)
n
(7)
1 m 2n
:8
` m
n
(n `)
n
(n 1:8
+1)
m n 1
+
(n `)
n
+2`
:
We will let g(m;`) be the quantity in Equation 7 plus and we will show that g(m;`) is negative forall valuesof 0 m <n
:8
and all ` m. In partiular, for every xed positive value ofm, we willshowthat
1. g(m;m)isnegative,
2. g(m;n) isnegative,and
3.
2 ` 2
g(m;`)>0. (g(m;`)isonave upasafuntionof`forthexedvalueofm sog(m;`) is negativeforall `2[m;n℄.)
Wewillhandletheasem=0similarlyexeptthatm=`=0orrespondstothestartstate, so we willreplaeItem 1with thefollowingform=0.
1'. g(0;1) is negative.
3.2 Case 2: m(X(t)) n or m(X(t))>n .
For every state suh that m() n :8
or m 0
() > n :4
, we will dene an integer k (whih dependsupon ) and we will show that, if X(t)=, then E[f(X(t+k) f(X(t))℄ k, where =1 2=.
Foronveniene,wewillusemasshorthandform(X(t))andm 0
asshorthandform 0
(X(t)). If mn
:8
thenwe willdener=m,W =m 1=4
dlgre2 8
,A=W,b=lg+lgnand v=n. Otherwise, we will dene r = m
0
, W = dlgre2 8
, A = 0, b = :8lgn+1, and v = 2dn :8
e. In either ase, we will dene k =4(r+v)dlgre. Let be the set of steps ft;:::;t+k 1g and letS betherandomvariablewhihdenotesthenumberofsuessesthat thesystemhas during. Let pdenote Pr(S W). Thenwehave
E[f(X(t+k) f(X(t))℄ n 1:8
k n
1:8
E[S℄+ n X i=1 t+k X t 0 =t+1 E[2 b i (t 0 ) 2 b i (t 0 1) ℄ n 1:8 k n 1:8
Wp+kn
k;
wherethenal inequalityholdsaslongasp2 13
and nissuÆientlybig. Thus, itsuÆes to nda positive lower bound for p whih is independent of n. We do this with plenty to spare. Inpartiular,weshowthatp1 510
5 .
We startwith atehnial lemma,whihdesribesthebehaviourof asingle user.
Lemma 5 Let j be a positive integer, and let Æ be a positive integer whih is at least 2. Suppose that q
i
(t)>0. Then, with probability at least 1
dlgje j
Æ=(2ln2)
, either user i sueeds in steps [t;:::;t+Æjdlgje 1℄, or b
i
(t+Æjdlgje) dlgje.
Proof: Suppose that user i is running in an externally-jammed hannel (so every send results in a ollision). Let X
z
denote the number of steps t 0
2 [t;:::;t+dÆjlg (j)e℄ with b
i (t
0
) = z. We laim that Pr(X z
> Ædlgje2 z 1
) < j
Æ=(2ln2)
. This proves the lemma sine
P dlgje 1 z=0
Ædlgje2 z 1
Æjdlgje. To prove the laim, note that X 0
1, so Pr(X 0
> Ædlgje2
1
)=0<j
Æ=(2ln2)
. Forz>0,note that
Pr(X z
>Ædlgje2 z 1
)(1 2 z ) Ædlgje2 z 1 <j Æ=(2ln2) : 2
Next,wedenesomeevents. Wewillshowthattheeventsarelikelyto our,and,ifthey doour,thenSislikelytobeatleastW. Thiswillallowustoonludethatp1 510
5 , whihwillnishCase2. We startbydeningB =dWe+dAe, k
0
=4rdlgre,k 00
=4BdlgBe and
0
=ft;:::;t+k 0
1g. Let 0
(i) be the set of all t 0
2 suh that b i
(t 0
) =0 and either (1)q
i (t
0
)>0 or(2)there isan arrivalat user iat t 0
. Let 2
bethe setof all t 0
2 suh that jf(t
00 ;i) jt
00 2
0
(i) and t 00
<t 0
gj B. Finally, let 1
be the set of all t 0 2 0 2 suh that,forsome i,
0 (i)\[t
0 k
00 +1;t
0
℄6=;. We an nowdene theevents E1{E4.
E1. Thereareat mostA arrivals during.
E2. Every station with q i
(t) > 0 and b i
(t) < b either sends suessfully during 0
or has b
i (t+k
0
E3. Every stationwith q i
(t)>0 and b i
(t)<b hasb i
(t)b+lg (r)=2+3forall t 2.
E4. Forall t 0
2 0
(i) and allt 00
>t 0
suh that t 00
2
1
2 ,b
i (t
0
)dlgBe.
Next, weshowthat E1{E4arelikelyto our.
Lemma 6 If n issuÆiently large then Pr(E1)10 5
.
Proof: The expeted numberof arrivalsin is k. Ifmn :8
,thenA=m 1=4
dlgre2 8
2k. By a Cherno bound, the probability that there are this many arrivals is at most e
k=3
10
5
. Otherwise, A =0 and k = o(1). Thus, Pr(E1) ( 1 =n) nk
1 k
1 10 5
. 2
Lemma 7 If n issuÆiently large then Pr(E2)10 5
.
Proof: Apply Lemma 5 to eah of the r users with Æ = 4 and j = r. Then Pr(E2 ) r
dlgre r
2=(ln2) 10
5
. 2
Lemma8 If n issuÆiently large then Pr(E3)10 5
.
Proof: Let y= l
dlgre 4
m
. Note that 2y lgr
2
+3. Supposethat user i hasb i
(t 0
)>b+2y. Then this user sent when its bako ounter was db+ze for all z 2 fy;:::;2y 1g. The probabilityof suh a sendon anypartiular step isat most
1 2
b 2
y
. Thus, theprobabilitythat itmakesall y ofthe sendsis at most
k y
!
1 2
b 2
y
y
ke 2
b y2
y
y 10
5 =r:
Thus,theprobabilitythatanyoftherusersobtainssuhabigbakoounterisatmost10 5
. 2
Lemma9 If n issuÆiently large then Pr(E4)10 5
.
Proof: We an applyLemma5 separately to eah of the(upto B) pairs(t 0
;i) withÆ =4 and j =B. The probabilitythat thereis a failureisat most
BdlgBe B
2=(ln2) 10
5
. 2
We now wish to show that Pr(S < W jE1^E2^E3^E4)10 5
. We begin withthe followinglemma.
Lemma10 Given any xed sequeneof states X(t);:::;X(t+z) whih does notviolate E2 orE4, and satises t+z2
0
1
2 , q
i
(t+z)>0, andb i
(t+z)b+lg (r)=2+3, the probability that useri sueeds at step t+z isat least
1 2
10 2
b r
1=2 .
Proof: The onditionsinthelemma implythefollowing.
There areno usersj withb j
(t+z)<dlgBe.
There areat mostB usersj withb j
j
There areat mostm+B usersj with b j
(t+z)<lg+lgn.
Thus, theprobabilitythat user isueedsis at least
2
(b+lg(r)=2+3)
1 1 B
B
1
1 r
r
1
1 2
b
m r
1 1 n
n m B
1 2
b r
1=2 2
3 1 4 1 4 1 4
1
n m B
n
1 2
10 2
b r
1=2 :
2
Corollary 11 Given any xed sequene of states X(t);:::;X(t+z) whih does not violate E2, E3, or E4, and satises t+z2
0
1
2
, the probability that some user sueeds at step t+z is atleast
r B 2
10 2
b r
1=2
1 2
13 n
:6 .
Proof: Sine t+z62 2
,at leastr B of theusersiwithq i
(t)>0 and b i
(t)<b have not sueededbefore stept+z. SineE3 holds,eah ofthese hasb
i
(t+z)b+lg (r)=2+3. For alliand i
0
,theevent that useri sueedsat step t+z isdisjointwith theevent that useri 0
sueeds at step t+z. 2
Lemma 12 If n is suÆientlylarge then Pr(S <W jE1^E2^E3^E4)10 5
:
Proof: IfE1 issatisedthen 2
doesnotstartuntiltherehave beenat leastW suesses.
Sine j
0
1
j k k 0
Bk 00
vdlgre=2, Corollary 11 shows that the probability of having fewer than W suesses is at mostthe probabilityof havingfewer thanW suesses in vdlgre=2 Bernoulli trials with suess probability
1 2
13 n
:6
. Sine W is at most half of the expeted numberof suesses, a Cherno bound shows that the probabilityof having fewer than W suessesis at mostexp(
vdlgre 2
17 n
:6 )10
5
. 2
WeonludeCase2byobservingthatpisatleast1 Pr(E1 ) Pr(E2 ) Pr(E3) Pr(E4 ) Pr(S <W jE1^E2^E3^E4). ByLemmas 6, 7,8,9,and 12,thisisat least1 510
5 .
3.3 Case 3: 0<m 0
(X(t))n :4
and m(X(t))<n :8
.
Foreverystatesuhthat0<m 0
()n :4
andm()<n :8
,wewilldenek =32m 0
()dlgm 0
()e+ dn
:8
e. Wewillshowthat,ifX(t)=,thenE[f(X(t+k) f(X(t))℄ k. Oneagain,wewill usemasshorthandform(X(T))andm
0
asshorthandform 0
(X(t)). Let =ft;:::;t+k 1g, let S be the number of suesses that the system has in . Let p denote Pr(S 1). As in Case2, E[f(X(t+k) f(X(t))℄ n
1:8
k n
1:8
p+kn, and this is at most k as long asp>9. Thus, we willnishbyndingapositive lower boundforp whihis independent ofn.
Sine m 0
> 0, there is a user suh that b
(t) < :8lgn+1. Let k 0
= 32m 0
dlgm 0
e and
0
=ft;:::;t+k 0
1g. Wewillnowdene some events, asinCase2.
i i 0 hasb
i (t+k
0
)dlgm 0
e.
E3. b
(t 0
)<:8lgn+7 forall t 0
2.
Lemma 13 If n is suÆientlylarge then Pr(E1 )10 5
.
Proof: Asintheproofof Lemma6,
Pr(E1)
1 n
nk
1 k1 10
5 :
2
Lemma 14 Pr(E2)10 5
.
Proof: We use lemma5 withÆ =32 and j=m 0
to get
Pr(E3 )m 0
dlgm 0
e (m
0 )
16=ln(2) 10
5
: (8)
2
Lemma15 If n is suÆientlylarge then Pr(E3 )10 5
.
Proof: Let y = 6, and suppose that user sends with bako b
= d:8lgn+re for r2f1;:::;6g. The probabilityofthishappening is
Pr(E3 ) k 6
! 6 Y
r=1 2
d:8lgne r
ke 6
6
1 n
:8
2 P
6 r=1
r
2en :8 6n
:8 2
3
6
10
5 :
2
Lemma 16 Given any xed sequeneof states X(t);:::;X(t+z) whih does not violate E1, E2, or E3 suh that t+z2
0
and there are no suesses during steps[t;:::;t+z 1℄, the probability that user sueeds at step t+z isat least
1 2
12 n
:8 .
Proof: The onditionsinthestatement of thelemma implythefollowing.
q
(t+z)>0 and b
(t+z)<:8lgn+7.
There areno usersj withb j
(t+z)<dlgm 0
e.
There areat mostm 0
usersj with b j
j
There willbe no arrivals onstep t+z.
The probabilityof suess foruser is at least
2
(:8lgn+7)
1 1 m 0
m
0 1
1 2n :8
m m
0 1
1 n
n m
1 2
7 n
:8 1 4 1 4 1 2
1 2
12 n
:8 :
2
Lemma 17 If n is suÆientlylarge then Pr(S <1jE1^E2^E3)e 1=2
12 .
Proof: Lemma16 impliesthattheprobabilityofhavingno suessesisat mostthe prob-abilityof havingnosuesses inj
0
jBernoullitrials,eah withsuess probability 1 2
12 n
:8 .
Sine j
0 jn
:8
,thisprobabilityis at most
1 1 2
12 n
:8
n :8
e 1=2
12 :
2
WeonludeCase3byobservingthatpisatleast1 Pr (E1) Pr (E2) Pr (E3) Pr (S < 1jE1^E 2^E 3). ByLemmas13,14,15,and17,thisisatleast1 310
5 e
1=2 12
:0002:
Referenes
[1℄ D. Aldous, Ultimate instability of exponential bak-o protool for aknowledgement-based transmission ontrol of randomaess ommuniation hannels,IEEE Trans. Inf. TheoryIT-33(2) (1987) 219{233.
[2℄ G. Fayolle, V.A. Malyshev and M.V. Menshikov, Topis in the Construtive Theory of Countable Markov Chains, (Cambridge Univ.Press, 1995)
[3℄ L.A.GoldbergandP.D.MaKenzie, Analysisofpratialbakoprotoolsforontention resolution with multiple servers, Journal of Computer and Systems Sienes, 58 (1999) 232{258.
[4℄ L.A. Goldberg, P.D. MaKenzie, M. Paterson and A. Srinivasan, Con-tention resolution with onstant expeted delay, Pre-print (1999) available at http://www.ds.warwik.a.uk/leslie/pub.html. (Extends a paper by the rsttwo authorsinPro.of the Symposium on Foundations of Computer Siene(IEEE) 1997 and a paperby theseond two authors inPro. of the Symposium on Foundations of Computer Siene(IEEE) 1995.)
hannels,SIAM Journal on Computing 25(4)(1996) 740-774.
[7℄ F.P. Kelly, Stohasti models of omputer ommuniation systems, J.R. Statist. So. B 47(3)(1985) 379{395.
[8℄ F.P. Kelly and I.M. MaPhee, The number of pakets transmitted by ollision detet randomaessshemes,The Annals of Probability, 15(4) (1987) 1557{1568.
[9℄ R.M.MetalfeandD.R.Boggs,Ethernet: Distributedpaketswithingforloalomputer networks. Commun. ACM,19(1976) 395{404.
1. g(m;m) is negative: g(m;m)2 0 n 1:9 (n 1)( 0 n 1:9
1) isequal tothe following.
2m 4m
2
+2n+6mn+2mn+6m 2
n+2 0 m 2 n 1:1 +2 0 n 1:9 +2 0 m 2 n 1:9 2n 2 2n 2 8mn 2 0 mn 2:1 3 0 m 2 n 2:1 +2n 2:8 +2mn 2:8 +2m 2 n 2:8 2 0 n 2:9 2 0 mn 2:9 +2 0 mn 2:9 4 0 m 2 n 2:9 +2n 3 0 2 m 2 n 3 + 0 mn 3:1 2 0 m 2 n 3:7 2n 3:8 2n 3:8 2 0 2 n 3:8 4 mn 3:8 0 mn 3:9 +4 0 mn 3:9 0 m 2 n 3:9 +2 02 m 2 n 4 +2 0 mn 4:7 +2 0 mn 4:7 +2n 4:8 +2 02 n 4:8 2 0 2 mn 4:8 + 0 2 m 2 n 4:8 + 0 mn 4:9 2 0 2 mn 5:6
Thedominanttermis 2 0
2 mn
5:6
. Notethatthereisapositiveterm( 0 2 m 2 n 4:8 )whih ouldbehalfthisbigifmisasbigasn
:8
(theupperboundforCase1),butallotherterms are asymptotiallysmaller.
2. g(m;n) is negative: g(m;n)2 0
n(n 1) is equalto thefollowing.
2 0 m 2 + 0 m 2 n :2 2 0
n+6
0 mn 2 0 mn 2 0 mn 1:2 2 0 m 2 n 1:8 2n 1:9 4 0 n 2 +2 0 n 2 +2 0 2 n 2 4 0 mn 2 + 0 m 2 n 2 +2 0 2 mn 2:2 +2 0 mn 2:8 2 0 mn 2:8 +2 2 n 2:9 + 4 0 n 3 2 0 2 n 3
Sine > 2, the term 2 0 2 n 3 dominates +4 0 n 3
. For the same reason, the term 2
0 mn
2:8
dominates the two terms +2 0 mn 2:8 and + 0 m 2 n 2
. The other terms are asymptotiallysmaller.
3.
2 ` 2
g(m;`)>0:
2 `
2
g(m;`)=2 1 n n 2 (n 1:8 +1) n ! :
1'. g(0;1) is negative: g(0;1) 0 n 1:9 ( 0 n 1:9
1) isequalto the following.
+ 4 3
0 n
:9
5n+n
1:8 + 0 n 1:9 0 m 1=9 0 n 1:9 + n 2 0 n 2:7 +2 0 2 n 2:8 3n 2:8 +2 0 n 2:9 + 0 n 3:7 + 0 n 3:7 +n 3:8 0 2 n 3:8 + 0 2 n 3:8 Sine 0
(1 )(1 ) >1,the term 0
2
(1 )n
3:8
dominates the term+n 3:8
