Original citation:
Goldberg, Leslie Ann, Jerrum, Mark, Kannan, Sampath and Paterson, Michael S. (2000)
A bound on the capacity of backoff and acknowledgement-based protocols. University of
Warwick. Department of Computer Science. (Department of Computer Science
Research Report). (Unpublished) CS-RR-365
Permanent WRAP url:
http://wrap.warwick.ac.uk/61127
Copyright and reuse:
The Warwick Research Archive Portal (WRAP) makes this work by researchers of the
University of Warwick available open access under the following conditions. Copyright ©
and all moral rights to the version of the paper presented here belong to the individual
author(s) and/or other copyright owners. To the extent reasonable and practicable the
material made available in WRAP has been checked for eligibility before being made
available.
Copies of full items can be used for personal research or study, educational, or
not-for-profit purposes without prior permission or charge. Provided that the authors, title and
full bibliographic details are credited, a hyperlink and/or URL is given for the original
metadata page and the content is not changed in any way.
A note on versions:
The version presented in WRAP is the published version or, version of record, and may
be cited as it appears here.For more information, please contact the WRAP Team at:
aknowledgement-based protools
Leslie Ann Goldberg Mark Jerrum Sampath Kannan
Mike Paterson
January 7, 2000
Abstrat
We studyontention-resolution protools for multiple-aess hannels. We show
thatevery bakoprotoolistransient ifthearrivalrate,,isatleast0:42and that
the apaity of every bako protool is at most 0:42. Thus, we show that bako
protools have (provably) smaller apaity than full-sensing protools. Finally, we
showthattheorrespondingresults,withthelargerarrivalboundof0:531, alsohold
forevery aknowledgement-based protool.
1 Introdution
Amultiple-aesshannel isabroadasthannelthatallowsmultipleuserstoommuniate
witheahotherbysendingmessages ontothe hannel. Iftwoormoreuserssimultaneously
send messages, thenthe messages interfere witheahother (ollide),andthe messages are
not transmitted suessfully. The hannel is not entrally ontrolled. Instead, the users
use a ontention-resolution protool to resolve ollisions. Thus, after a ollision, eah
user involved in the ollisionwaits a randomamount of time (whih is determined by the
protool) beforere-sending.
Following previous work on multiple-aess hannels, we work in a time-slotted model
in whih time is partitioned into disrete time steps. At the beginning of eah time step,
a random number of messages enter the system, eah of whih is assoiated with a new
user whih has noother messages tosend. The numberof messages that enter the system
is drawn from a Poisson distribution with mean . During eah time step, eah message
ResearhReport 365,Departmentof ComputerSiene,UniversityofWarwik,CoventryCV47AL,
UK. This work was partially supported by EPSRC grant GR/L60982, NSF Grant CCR9820885, and
hannel,then thismessageleavesthesystem. Otherwise, allof themessages remaininthe
system and the next time step is started.
The suess of a protool an be measured in several ways. Typially, one models the
exeution of the protool as a Markov hain. If the protool is good (for a given arrival
rate ), the orresponding Markov hain will be reurrent (with probability 1, it will
eventually return to the empty state in whih no messages are waiting.) Otherwise, the
hain is said to be transient (and we also say that a protool is transient). Note that
transiene is a very strong form of instability. Informally, if a protool is transient then
with probability 1 itis in \small"states (states with few baklogged messages) for only a
nite number of steps.
Anotherwayto measure the suess of aprotoolis tomeasure itsapaity. A protool
issaidtoahievethroughput if, whenitisrun withinputrate, theaveragesuess rate
is . The apaity of the protool[4℄ isthe maximum throughput that it ahieves.
The protools that we onsider in this paper are aknowledgement-based protools. In
the aknowledgement-based model, the only information that a user reeives about the
state of the system is the history of its own transmissions. An alternative model is the
full-sensing model, in whih every user listensto the hannel atevery step.
1
One partiularly simple and easy-to-implement lass of aknowledgement-based
proto-ols is the lass of bako protools. A bako protool is a sequene of probabilities
p 1
;p 2
;:::. If a message has sent unsuessfully i times before a time-step, then with
probability p
i
, it sends during the time-step. Otherwise, it does not send. Kelly and
MaPhee [12,13,16℄gave aformulafor theritialarrival rate,
,of abako protool,
whih is the minimum arrival rate for whih the expeted number of suessful
transmis-sions that the protoolmakes is nite.
2
Perhaps the best-known bako protool is the binary exponential bako protool in
whih p
i
= 2
i
. This protool is the basis of the Ethernet protool of Metalfe and
Boggs[17℄.
3
KellyandMaPheeshowedthattheritialarrivalrateofthisprotoolisln2.
Thus,if >ln2,thenbinaryexponentialbakoahievesonlyanitenumberofsuessful
transmissions (in expetation). Aldous [1℄ showed that the binary exponential bako
protoolis notagoodprotoolforany positivearrivalrate . In partiular,itis transient
andthe expetednumberofsuessfultransmissionsintsteps iso(t). MaPhee [16℄posed
the questionofwhether thereexists abako protoolwhihisreurrent forsomepositive
1
In pratie, it is possible to implement the full-sensing model when there is a single hannel, but
this beomesinreasingly diÆultinsituations wherethere aremultiple sharedhannels, suh asoptial
networks. Thus,aknowledgement-basedprotoolsaresometimes preferabletofull-sensingprotools. For
work onontention-resolutionin themultiple-hannel setting,see[6℄.
2
If >
, then the expeted number of suessesis nite, even if the protool runs forever. They
showedthat theritialarrivalrateis0iftheexpetednumberoftimesthatamessagesendsduring the
rsttstepsis!(logt). 3
Thereare severaldierenebetweenthe\real-life"Ethernetprotooland \pure"binaryexponential
In this paper, weshow that thereis nobako protoolwhih isreurrent for0:42.
(Thus, every bako protool is transient if 0:42.) Also, every bako protool has
apaity atmost0:42. Asfar asweknow, our resultisthe rst proofshowingthat bako
protools have smaller apaity than full-sensing protools. In partiular, Mosely and
Humblet [19℄ have disovered a full-sensing protool with apaity 0:48776.
4
Finally, we
show that no aknowledgement-based protool isreurrent for 0:530045.
1.1 Related work
Bakoprotoolsandaknowledgement-basedprotoolshavealsobeenstudiedinann-user
model, whih ombines ontention-resolution with queueing. In this model, it is assumed
that n users maintain queues of messages, and that new messages arrive at the tails of
the queues. At eah step, the users use ontention-resolutionprotools to try to send the
messagesattheheadsoftheirqueues. Itturnsoutthatthequeueshaveastabilisingeet,
sosome protools (suh as\polynomialbako")whihare unstable inour model[13℄ are
stable in the queueing model [11℄. We will not desribe queueing-modelresults here, but
the reader is referredto [2,8, 11, 21℄.
Muhworkhasgoneintodeterminingupperboundsontheapaitythatanbeahieved
by a full-sensing protool. The urrent best result is due to Tsybakov and Likhanov [23℄
who have shown that no protool an ahieve apaity higher than 0:568. (For more
information, see [4, 9, 18, 22℄.) In the full-sensing model, one typially assumes that
messages are born at real \times" whih are hosen uniformly from the unit interval.
Reently, Loher [14, 15℄ has shown that if a protool is required to respet these birth
times,inthe sensethatpakets mustbesuessfullydeliveredintheirbirth order,thenno
protoolan ahieve apaity higher than0:4906. Intuitively, the \rst-ome-rst-served"
restrition seems very strong, soit is somewhat surprising that the best known algorithm
withoutthe restrition(thatofVvedenskayaandPinsker)doesnotbeat thisupperbound.
The algorithmof Humblet and Mosely satises the rst-ome-rst-served restrition.
1.2 Improvements
Wehoose =0:42 inorder tomakethe proof ofLemma10(see the Appendix)as simple
as possible. The lemma seems to be true for down to about 0:41 and presumably the
parameters A and B ould be tweaked to get slightly smaller.
4
MoselyandHumblet'sprotoolisa\treeprotool"inthesenseofCapetanakis[3℄andTsybakovand
Mikhailov[24℄Forasimpleanalysisoftheprotool,see[25℄. VvedenskayaandPinskerhaveshownhowto
modifyMoselyandHumblet'sprotooltoahieveanimprovementintheapaity(intheseventhdeimal
AnirreduibleaperiodiMarkovhainX =fX 0
;X
1
;:::g(see[10℄)isreurrent ifitreturns
toitsstartstatewithprobability1. Thatis,itisreurrentifforsomestatei(andtherefore,
for all i), Prob[X
t
= ifor some t 1jX
0
=i℄=1. Otherwise, X is said tobe transient.
X is positive reurrent (or ergodi) if the expeted number of steps that it takes before
returningtoitsstart stateisnite. Weuse thefollowingtheorems whihwe takefrom[5℄.
Theorem 1 (Fayolle, Malyshev, Menshikov) An irreduible aperiodi time-homogeneous
Markov hain X with state spae is not positive reurrent if there is a funtion f with
domain and there are onstantsC and d suh that
1. there isa state x with f(x)>C, and a state x with f(x)C, and
2. E[f(X
1
) f(X
0
)j X
0
=x℄0 for all x with f(x)>C, and
3. E[jf(X
1
) f(X
0
)j jX
0
=x℄d for every state x.
Theorem 2 (Fayolle, Malyshev, Menshikov) An irreduible aperiodi time-homogeneous
Markov hain X with state spae is transient if there is a positive funtion f with
domain and there are positive onstants C, d and " suh that
1. there isa state x with f(x)>C, and a state x with f(x)C, and
2. E[f(X
1
) f(X
0
)j X
0
=x℄" for all x with f(x)>C, and
3. If jf(x) f(y)j>d then the probability of moving from x to y in a single move is0.
3 Stohasti Domination and Monotoniity
Suppose that X is aMarkov hain and that the (ountable) state spae of the hain is
a partial order with binary relation . If A and B are random variables taking states as
values, then B dominates A (written A B) if and only if there is a joint sample spae
for A and B in whih A B. We say that X is monotoni if for any states x x
0
, the
next state onditioned on starting at x
0
dominates the next state onditioned on starting
atx. (Formally,(X
1
jX
0
=x
0
) dominates (X
1
jX
0
=x).)
Whenanaknowledgement-basedprotoolisviewed asaMarkov hain, thestate isjust
the olletionof messages in the system. (Eah message is identied by the history of its
transmissions.) The state spae forms apartial order with respet to the subset inlusion
relation (for multisets). We say that a protool is deletion resilient [7℄ if its Markov
As we indiated earlier, we willgenerally assumethat the number of messages entering
the system at a given step is drawn from a Poisson proess with mean . However, it
will sometimes be useful to onsider other message-arrival distributions. If I and I
0 are
message-arrival distributions, we write I I
0
to indiate that the number of messages
generated under I isdominated by the number of messages generated under I
0 .
Observation 4 If aknowledgement-based protool P is reurrent under message-arrival
distribution I
0
and I I
0
then P is also reurrent under I.
Proof: Let X be the Markov hain orresponding to protool P with arrival
distribu-tion I with X
0
as the empty state. Let X
0
be the analogous Markov hain with arrival
distribution I
0
. We an now showby indution ont that X
0 t
dominates X
t
. 2
4 Bako protools
Inthissetion,wewillshowthatthereisnobakoprotoolwhihisreurrentfor0:42.
Our method will be to use the \drift theorems" in Setion 2. Let p
1 ;p
2
;::: be a bako
protool. Let = 0:42. Let X be the Markov hain whih desribes the behaviour of
the protoolwith arrival rate . First,we will onstrut a potentialfuntion (Lyapounov
funtion)f whihsatisesthe onditionsof Theorem1,that is,apotentialfuntionwhih
has a bounded positive drift. We will use Theorem 1 to onlude that the hain is not
positive reurrent. Next, we willonsider the behaviourof the protoolunder a trunated
arrival distribution and we will use Theorem 2 to show that the protool is transient.
Using Observation 4 (domination), we will onlude that the protool is also transient
with Poisson arrivalsat rate orhigher. Finally, we will show that the apaity of every
bako protoolis atmost 0:42.
Wewilluse the following tehnial lemma.
Lemma 5 Let 1t
i
d for i2[1;k℄ and
Q k i=1
t i
=. Then
P k i=1
(t i
1)(d 1)
log logd .
Proof: Let S =
P k i=1
t i
. S an be viewed as a funtion of k 1 of the t
i
's, for example
S =
P
k 1
i=1 t
i
+=
Q
k 1
i=1 t
i
. For i 2 f1;:::;k 1g, the derivative of S with respet to t
i is
1 =(t
i Q
k 1
j=1 t
j
). Thus, the derivativeispositiveif t
i >t
k
. Thus, S ismaximised(subjet
to ) by setting some t
i
's to 1, some t
i
's to d and at most one t
i
to some intermediate
value t2 [1;d). Given this, the maximum value of
P k i=1
(t i
1)is s(d 1)+t 1, where
= d
s
is true, sine
(d 1)
log
logd
s(d 1) (t 1) = (d 1) =d
s
+1
= (d 1) d
+1
(d 1) (1+ (d 1))+1
= 0:
2
Wenowdenesomeparametersofastatex. Letk(x)denotethenumberofmessages in
state x. Ifk(x)=0,then p(x)=r(x)=u(x)=0. Otherwise, letm
1
;:::;m k(x)
denotethe
messages in state x, with send probabilities
1
k(x)
. Let p(x) =
1
and let r(x)
denote the probability that at least one of m
2
;:::;m k(x)
sends onthe next step. Let u(x)
denote the probability that exatly one of m
2
;:::;m k(x)
sends on the next step. Clearly
u(x)r(x). If p(x)<r(x) then we use the following(tighter) upperbound foru(x).
Lemma 6 If p(x)<r(x) then u(x)
r(x) r(x)
2 =2
1 p(x)=2
.
Proof: Fix a state x. We will use k;p;r;::: to denote k(x);p(x);r(x);:::. Sine p <r,
we havek 2.
u=
k X
i=2
i
1
i k Y
i=2
(1
i
)=
k X
i=2 (t
i
1)(1 r);
where t
i
=1=(1
i
). Let d=1=(1 p),and note that 1t
i
d. By Lemma5
u(1 r)(d 1)
log( Q
k i=2
t i
)
logd
=(1 r)
p
1 p
log(1=(1 r))
logd
=(1 r)
p
1 p
( log(1 r))
( log(1 p))
:
Now we wish toshow that
(1 r)
p
1 p
( log(1 r))
( log(1 p))
r r
2 =2
1 p=2
;
i.e., that
(1 r)
( log(1 r))
r r
2 =2
(1 p)
( log(1 p))
p p
2 =2
:
This is true, sine the funtion (1 r)
( log (1 r))
r r
2 =2
is dereasing in r. To see this, note that
the derivative of this funtion with respet tor isy(r)=(r r
2 =2)
2
, where
y(r)=(1 r+r
2
=2)log(1 r)+(r r
2
=2) (1 r+r
2
=2)( r r
2
=2)+(r r
2
=2)= r
4 =4:
step starting instate x. Let
g(r;p)=e
[(1 r)p+(1 p)minfr;
r r
2 =2
1 p=2
g+(1 p)(1 r)℄:
Wenow have the following orollaryof Lemma 6.
Corollary 7 For any state x, S(x)g(r(x);p(x)).
Let s(x) denote the probability that at least one message in state x sends on the next
step. (Thus, if x isthe empty state, then s(x)=0.) LetA=0:9 and B =0:41. Forevery
probability , let ()=max(0; A+B). Forevery state x, let f(x)=k(x)+(s(x)).
The funtion f is the potential funtion alluded to earlier, whih plays a leading role in
Theorems 1 and 2. To a rst approximation, f(x) ounts the number of messages in the
state x, but the smallorretion term is ruial. Finally,let
h(r;p)= g(r;p) [1 e
(1 p)(1 r)(1+)℄(r+p rp)+e
p(1 r)(r):
Now we have the following.
Observation 8 For any state x, E[jf(X
1
) f(X
0
)j jX
0
=x℄1+B.
Lemma 9 For any state x, E[f(X
1
) f(X
0
)jX
0
=x℄h(r(x);p(x)).
Proof: The result follows from the following hain of inequalities, eah link of whih is
justied below.
E[f(X
1
) f(X
0
)jX
0
=x℄ = S(x)+E[(s(X
1
))jX
0
=x℄ (s(x))
g(r(x);p(x))+E[(s(X
1
))jX
0
=x℄ (s(x))
g(r(x);p(x))+e
(1 p(x))(1 r(x))(1+)(s(x))
+e
p(x)(1 r(x))(r(x)) (s(x))
= h(r(x);p(x)):
The rst inequality follows from Corollary 7. The seond omes from substituting exat
expressionsfor(s(X
1
))whenevertheformofX
1
allowsit,andusingthebound(s(X
1
))
0 elsewhere. If none of the existing messages send and there is at most one arrival, then
(s(X 1
))=(s(x)), givingthe thirdterm; if messagem
1
alonesends and thereare nonew
arrivals then (s(X
1
)) = (r(x)), giving the fourth term. The nal equality uses the fat
that s(x)=p(x)+r(x) p(x)r(x). 2
0
0.2
0.4
0.6
0.8
1
p
0
0.2
0.4
0.6
0.8
1
r
-0.1
-0.075
-0.05
-0.025
0
0
0.2
0.4
0.6
0.8
p
0
0.2
0.4
0.6
0.8
Figure1: h(r;p) overthe range r 2[0;1℄;p2[0;1℄.
Proof: Wedeferthe proofofthislemmatothe Appendixofthepaper. Figure1ontains
a (Mathematia-produed) plot of h(r;p) over the range r 2 [0;1℄;p 2 [0;1℄. The plot
suggests that h(r;p)is bounded belowzero.
Wenotehere thatour proof ofthe lemma(inthe Appendix)involvesevaluatingertain
polynomials at40;000 points,and wedid this using Mathematia. 2
Wenow have the following theorem.
Theorem 11 No bako protoolis positive reurrent when the arrival rate is =0:42.
Proof: This follows fromTheorem1,Observation 8andLemmas 9and10. The value C
in Theorem 1an betaken to be1 and the value d an be taken tobe1+B. 2
Now we wish toshow that every bako protool istransient for 0:42. One again,
x a bako protool p
1 ;p
2
;::: . Notie that our potential funtion f almost satises
the onditions in Theorem 2. The main problem is that there is no absolute bound on
the amount that f an hange in a single step, beause the arrivals are drawn from a
Poisson distribution. We get aroundthis problemby rst onsidering atrunated-Poisson
distribution, T
M;
, in whih the probability of r inputs is e
r
=r! (as for the Poisson
distribution) when r < M, but r = M for the remaining probability. By hoosing M
suÆiently large we an have E[T
M;
℄ arbitrarilylose to .
Lemma 12 Every bako protool is transient for the input distribution T
M;
when =
0:42 and
0
=E[T
M;
inthedenitionofh(r;p)(forLemmas9and10)mustbereplaedby 0
. Theorresponding
funtion h
0
satises h
0
(r;p)h(r;p) 0:001. ThusLemma 10 shows that h
0
(r;p) 0:002
for allr 2[0;1℄ and p2[0;1℄.
The potential funtion f(x) is dened as before, but under the trunated input
distri-bution we have the property required for Theorem 2. If jf(x) f(y)j>M +B then the
probability of movingfrom xto y ina single moveis 0.
The lemma follows from Theorem 2, where the values of C, ", and d an be taken to
be 1,0:002, and M +B, respetively. 2
Wenow have the following theorem.
Theorem 13 Every bako protool is transient under the Poisson distribution with
ar-rival rate 0:42.
Proof: The proof is immediate fromLemma 12and Observation 4. 2
Finally,we bound the apaity of every bako protool.
Theorem 14 The apaity of every bako protool is at most 0:42.
Proof: Let p
1 ;p
2
;::: be abako protool, let
00
0:42 bethe arrival rate and let =
0:42. ViewthearrivalsateahstepasPoisson()\ordinary"messages andPoisson(
00 )
\ghost" messages. We will show that the protool does not ahieve throughput
00
. Let
Y 0
;Y 1
;:::betheMarkovhaindesribingtheprotool. Letk(Y
t
)bethenumberofordinary
messages inthe systemaftert steps. Clearly,the expetednumberofsuesses inthe rst
t steps is at most
00
t E[k(Y
t
)℄. Now let X
1
;X
2
;::: be the Markov hain desribing
the evolution of the bako protool with arrival rate (with no ghost messages). By
deletion resiliene (Observation 3), E[k(Y
t
)℄ E[k(X
t
)℄. Now by Lemmas 9 and 10,
E[k(X t
)℄ E[f(X
t
)℄ B 0:003t B. Thus, the expeted number of suesses in
the rst t steps is at most (
00
0:003)t +B, whih is less than
00
t if t is suÆiently
large. (If X
0
is the empty state, then we donot require t to be suÆiently large, beause
E[f(X
t
)℄0:003t+B.) 2
5 Aknowledgement-based protools
We will prove that every aknowledgement-based protool is transient for all > 0:531;
see Theorem 20for apreise statement of this laim.
An aknowledgement-based protool an be viewed a system whih, at every step t,
bounds, we introdue the notion of a \genie", whih (in general) has more freedom in
making these deisions than a protool.
Sine we only onsider aknowledgement-based protools, the behaviour of eah new
message is independent of the other messages and of the state of the system until after
its rst send. This is why we ignore new messages until their rst send { for Poisson
arrivalsthis isequivalenttothe onvention thateahmessagesends atitsarrivaltime. As
a onsequene, we impose the limitation on a genie, that eah deision is independent of
the numberof arrivalsat that step.
A genie is a random variable over the natural numbers, dependent on the omplete
history (of arrivalsand sends of messages) up totime t 1,whih gives anatural number
representing the number of (old) messages to send at time t. It is lear that for every
aknowledgement-basedprotoolthere isa orrespondinggenie. However thereare genies
whih do not behave like any protool, e.g., a genie may give a umulative total number
of \sends" up totime t whih exeeds the atual number of arrivalsup to that time.
Weproveapreliminaryresultforsuh\unonstrained"genies,butthenweimposesome
onstraintsreeting properties of a given protoolin order toproveour nal results.
LetI(t);G(t) be the number of arrivalsand the genie's send value, respetively, atstep
t. It is onvenient to introdue some indiator variables to express various outomes at
the step under onsideration. Weuse i
0 ;i
1
forthe events ofno new arrival,orexatlyone
arrival,respetively,andg
0 ;g
1
fortheeventsofnosendandexatlyonesendfromthegenie.
The indiatorrandom variable S(t) for a suess at time t is given by S(t) =i
0 g
1
+i
1 g
0 .
Let In (t) =
P jt
I(j) and Out(t) =
P jt
S(j). Dene Baklog (t) = In(t) Out(t). Let
=
0
0:567 be the (unique) rootof =e
.
Lemma 15 For any genie and input rate >
0
, there exists ">0 suh that
Prob[Baklog(t)>"t for all tT℄!1as T !1:
Proof: Let3"= e
>0. Atanystept, S(t)isaBernoulli variablewithexpetation
0;e
;e
, aording as G(t) > 1;G(t) = 1;G(t) = 0, respetively, whih is dominated
by the Bernoulli variable with expetation e
. Therefore E[Out(t)℄ e
t, and also,
Prob[Out(t) e
t < "tfor all t T℄ ! 1 asT ! 1. (To see this note that, by a
Cherno bound, Prob[Out (t) e
t "t℄e Æt
for a positive onstantÆ. Thus,
Prob[9tT suh that Out(t) e
t "t℄
X
tT e
Æt ;
whihgoes to0 as T goes to1.)
Wealsohave E[In (t)℄=tand Prob [t In (t)"tfor allt T℄!1 asT !1,sine
Baklog (t) = In (t) Out(t)=( e
)t+(In(t) t)+(e
t Out(t))
= "t+("t+In (t) t)+("t+e
t Out(t));
the result follows. 2
Corollary 16 No aknowledgement-based protool is reurrent for >
0
or has apaity
greater than
0 .
Tostrengthen theaboveresult weintroduearestrited lassof genies. Wethinkofthe
messages whihhavefailedexatlyoneasbeingontainedinthe buket. (Moregenerally,
weouldonsideranarrayofbukets, wherethejthbuketontainsthose messageswhih
havefailedexatlyj times.) A 1-buketgenie, here alledsimplyabuketgenie, isagenie
whihsimulatesa given protoolfor the messages in the buket and is required to hoose
a send value whih is at least as great as the number of sends from the buket. For suh
onstrained genies,wean improvethe bound of Corollary16.
For the range of arrivalrates we onsider, an exellent strategy for a genie is to ensure
that atleast one message issent ateahstep. Of ourse abuket genie has torespet the
buket messages and isobligedsometimes tosend more thanone message(inevitably
fail-ing). Aneagergenie always sends atleastone message, but otherwise sends the minimum
number onsistent with itsonstraints.
An eager buket genie is easy to analyse, sine every arrival is bloked by the genie
and enters the buket. For any aknowledgement-based protool, let Eager denote the
orresponding eager buket genie.
Let =
1
0:531bethe (unique) root of =(1+)e
2 .
Lemma 17 Forany eagerbuketgenieand inputrate >
1
, there exists">0suhthat
Prob[Baklog(t)>"t for all tT℄!1as T !1:
Proof: Let r
i
be the probability that a message in the buket sends for the rst time
(and hene exits from the buket) i steps after its arrival. Assume
P 1 i=1
r i
=1, otherwise
thereis apositiveprobabilitythat the messageneverexitsfromthe buket, and theresult
follows trivially.
The generating funtion for the Poisson distribution with rate is e
(z 1)
(i.e., the
oeÆient of z
k
in this funtion gives the probability of exatly k arrivals;see, e.g., [10℄).
Considerthe sendsfromthe buketatstep t. SineEageralwaysbloksarrivingmessages,
is e . Some of these messages may send at step t; the generating funtion for the
number of sends is e
[(1 r
i )+r
i z 1℄
=e
r i
(z 1)
. Finally,the generating funtion for allsends
fromthe buket atstep t isthe onvolution of allthese funtions, i.e.,
t Y
i=1
exp (r i
(z 1))=exp
"
(z 1)
t X
i=1 r
i #
:
For any Æ > 0, we an hoose t suÆiently large so that
P t i=1
r i
> 1 Æ. The number
of sends fromthe buket at step t is distributed as Poisson(
0
), where (1 Æ) <
0
.
The number of new arrivals sending at step t is independently Poisson (). The only
situationin whih a messagesueeds under Eager is when there are no new arrivalsand
the number ofsends fromthe buket iszero orone. Thusthe suess probabilityatstep t
is e
e
0
(1+
0
). For suÆiently small Æ, we have
1 <
0
, and so e
0
(1+
0 ) <
e
1
(1+
1
)=e
1
1
<e
. Hene e
e
0
(1+
0
) 3 for suÆiently small. Thus
the suess event is dominatedby aBernoulli variable with expetation 3. Hene, as
in the previous lemma,
Prob[Baklog (t)>"tfor all tT℄!1 as T !1;
ompleting the proof. 2
LetAny beanarbitrary buket genieand letEager bethe eager buket geniebased on
the same buket parameters. Wemay ouplethe exeutions of Eager and Any sothat the
same arrival sequenes are presented to eah. It willbe lear that at any stage the set of
messages in Any's buket is a subset of those in Eager's buket. We may further ouple
the behaviour of the ommonsubset of messages.
Let =
2
0:659bethe (unique) root of =1 e
.
Lemma 18 For the oupled genies Any and Eager dened above, if Out
A
and Out
E are
the orresponding output funtions, we dene Out(t) = Out
E
(t) Out
A
(t). For any
2
and any " >0,
Prob[Out (t) "t for all tT℄!1 as T !1:
Proof: Let
0
;
1
;
be indiators for the events of the number of ommon messages
sending being 0, 1, or more than one, respetively. In addition, for the messages whih
are onlyinEager'sbuket, weuse thesimilarindiators e
0 ;e
1 ;e
. Leta
0 ;a
1
represent Any
not sending, orsending, additional messages respetively. (Note that Eager'sbehaviour is
fully determined.)
We write Z(t) for Out (t) Out(t 1), for t > 0, so Z represents the dierene in
E A
= i
0 g
E1
(t)+i
1 g E0 (t) i 0 g A1 (t) i 1 g A0 (t); where S E
(t) is the indiator random variable for a suess of Eager at time t and g
E1 (t)
is the event that Eager sends exatly one message during step t (and so on) as in the
paragraph before Lemma15. Thus,
Z(t)i
0 0 (a 0 (e 0 +e 1 ) a 1 e ) i 0 1 (e 1 +e ) i 1 0 a 0 :
Note that if the number of arrivals plus the number of ommon buket sends is more
than 1 then neithergenie an sueed. We alsoneed to keep trak of the number, B, of
extra messages in Eager's buket. At any step, at most one new suh extra message an
arrive;the indiatorfor thiseventis i
1
0 a
0
,i.e., there isasingle arrivaland nosends from
the ommon buket, soif Any doesnot send then this message sueeds but Eager's send
willausea failure. The number of\extra" messages leavingEager'sbuketatany step is
unbounded, given bya randomvariableweouldshowase =1e
1
+2e
2
+. However
e dominates e
1
+e
and it is suÆient to use the latter. The hange at one step in the
number of extra messages satises:
B(t) B(t 1)=i
1
0 a
0
e i
1 0 a 0 (e 1 +e ):
Nextwe deneY(t)=Z(t) (B(t) B(t 1)), for somepositiveonstant to be
hosen below. Notethat X(t)=
P t j=1
Y(j)=Out(t) B(t). We alsodene
Y 0
(t)=i
0 0 (a 0 (e 0 +e 1 ) a 1 e ) i 0 1 (e 1 +e ) i 1 0 a 0 (i 1 0 a 0 (e 1 +e )) and X 0 (t)= P t j=1 Y 0
(j). Note that Y(t)Y
0 (t).
We an identify ve (exhaustive) ases A,B,C,D,E depending on the values of the 's,
a'sande's,suhthatineahaseY
0
(t)dominatesagivenrandomvariabledependingonly
onI(t).
A.
: Y
0
(t)0;
B. ( 1 + 0 a 1 )(e 1 +e ): Y 0
(t) i
0 ; C. ( 1 + 0 a 1 )e 0 : Y 0
(t)0;
D. 0 a 0 (e 0 +e 1 ): Y 0
(t)i
0
(1+ )i
1 ; E. 0 a 0 e : Y 0
(t) (1+ )i
1 .
Forexample,theorretinterpretationofCaseBis\onditionedon(
1 + 0 a 1 )(e 1 +e )=
1, the value of Y
0
(t) is at least i
0
." Sine E[i
0
℄ = e
and E[i
1
℄ = e
, we have
E[Y 0
(t)℄0ineahase,provided thatmaxfe
;e
=(1 e
)g1= 1. There
exists suh an forany
2
; for suh wemay takethe value =e
, say.
Let F
t
be the -eld generated by the rst t steps of the oupled proess. Let
^
Y(t) =
Y 0
(t) E[Y
0
(t) j F
t 1
℄ and let
^ X(t) = P t i=1 ^
Y(t). The sequene
^ X(0);
^
martingale (see Denition 4.11 of [20℄) sine E[X(t) j F
t 1
℄ = X(t 1). Furthermore,
there is a positive onstant suh that j
^ X(t)
^
X(t 1)j . Thus, we an apply the
Hoeding-Azuma Inequality (see Theorem 4.16 of [20℄):
Theorem 19 (Hoeding, Azuma) Let X
0
;X
1
;::: be a martingale sequene suh that for
eah k
jX k
X
k 1
j
k ;
where
k
may depend upon k. Then, for all t0 and any >0,
Prob[jX t
X 0
j℄2exp
2
2 P
t k=1
2 k
!
:
In partiular, we an onlude that
Prob[ ^ X t
t℄2exp
2
t
2 2
!
:
Our hoie of above ensured that E[Y
0
(t) j F
t 1
℄ 0. Hene Y
0
(t)
^
Y(t) and
X 0
(t)
^
X(t). We observed earlier that X(t)X
0
(t). Thus, X(t)
^
X(t) sowe have
Prob[X t
t℄2exp
2
t
2 2
!
:
Sine 2exp
2 t 2
2
onverges, we dedue that
Prob[X(t) "tfor allt T℄!1 asT !1:
Sine Out(t)=X(t)+ B(t)X(t), for allt, we obtainthe requiredonlusion. 2
Finally,we an provethe main resultsof this setion.
Theorem 20 Let P be an aknowledgement-based protool. Let =
1
0:531 be the
(unique) root of =(1+)e
2
. Then
1. P is transient for arrival rates greater than
1 ;
2. P has apaity no greater than
1 .
Proof: Let be the arrival rate, and suppose >
1
. If >
0
0:567 then the
result follows from Lemma 15. Otherwise, we an assume that <
2
0:659. If E is
the eager genie derived from P, then the orresponding Baklogs satisfy Baklog
P
(t) =
Baklog E
(t)+Out(t). The resultsofLemmas17and 18showthat,for some">0,both
Prob[Baklog E
(t)>2"t for all tT℄ and Prob[Out (t) "tfor all tT℄ tend to1 as
[1℄ D.Aldous, Ultimateinstabilityofexponentialbak-oprotoolfor
aknowledgement-based transmission ontrol of random aess ommuniation hannels, IEEE Trans.
Inf. TheoryIT-33(2) (1987) 219{233.
[2℄ H.Al-Ammal,L.A.Goldberg andP.MaKenzie,BinaryExponentialBakoisstable
for high arrival rates, To appear in International Symposium on Theoretial Aspets
of Computer Siene 17 (2000).
[3℄ J.I. Capetanakis, Tree algorithms for paket broadast hannels, IEEE Trans. Inf.
Theory IT-25(5) (1979)505-515.
[4℄ A. Ephremides and B. Hajek, Information theory and ommuniation networks: an
unonsummated union, IEEE Trans. Inf.Theory 44(6) (1998)2416{2432.
[5℄ G. Fayolle,V.A. Malyshev, and M.V. Menshikov, Topis in the Construtive Theory
of Countable Markov Chains,(Cambridge University Press, 1995).
[6℄ L.A. Goldberg and P.D. MaKenzie, Analysisof PratialBako Protools for
Con-tentionResolutionwithMultipleServers, Journalof Computerand SystemsSienes,
58 (1999) 232{258.
[7℄ 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.u k/ les lie/ pub. html . (Extends apaperby the rst
twoauthors inPro.of the Symposium on Foundations of Computer Siene (IEEE)
1997andapaperbytheseondtwoauthorsinPro.of theSymposiumonFoundations
of Computer Siene (IEEE) 1995.)
[8℄ J.Goodman,A.G.Greenberg,N.MadrasandP.Marh,Stabilityofbinaryexponential
bako,J. of the ACM, 35(3) (1988)579{602.
[9℄ A.G. Greenberg, P. Flajoletand R. Ladner, Estimatingthe multipliitiesof onits
to speed their resolution in multiple aess hannels, J. of the ACM, 34(2) (1987)
289{325.
[10℄ G.R.GrimmetandD.R.Stirzaker, ProbabilityandRandomProesses,SeondEdition.
(Oxford University Press, 1992)
[11℄ J.Hastad,T.LeightonandB.Rogo,Analysisofbakoprotoolsformultipleaess
hannels, SIAM Journal on Computing25(4) (1996) 740-774.
[12℄ F.P. Kelly,Stohasti models of omputer ommuniation systems, J.R. Statist. So.
randomaess shemes,The Annalsof Probability,15(4) (1987) 1557{1568.
[14℄ U. Loher, EÆieny of rst-ome rst-served algorithms,Pro. ISIT(1998) p108.
[15℄ U.Loher,Information-theoretiandgenie-aidedanalysesofrandom-aessalgorithms,
PhD Thesis, Swiss Federal Institute of Tehnology, DISS ETH No. 12627, Zurih
(1998).
[16℄ I.M. MaPhee,On optimal strategies instohasti deisionproesses, D. PhilThesis,
University of Cambridge, (1987).
[17℄ R.M.Metalfeand D.R.Boggs,Ethernet: Distributedpaketswithing forloal
om-puter networks.Commun. ACM, 19 (1976)395{404.
[18℄ M.MolleandG.C.Polyzos,Conitresolutionalgorithmsandtheirperformane
anal-ysis,TehnialReportCS93-300,ComputerSystems ResearhInstitute,Universityof
Toronto,(1993).
[19℄ J. Mosely and P.A. Humblet, A lass of eÆient ontention resolution algorithms
for multiple aess hannels, IEEE Trans. on Communiations, COM-33(2) (1985)
145{151.
[20℄ R. Motwani and P.Raghavan, Randomized Algorithms(Cambridge University Press,
1995.)
[21℄ P. Raghavan and E. Upfal, Contention resolution with bounded delay, Pro. of the
ACM Symposium on the Theory of Computing 24 (1995) 229{237.
[22℄ R. RomandM. Sidi, Multipleaess protools: performane and analysis,
(Springer-Verlag, 1990).
[23℄ B.S. Tsybakov and N. B. Likhanov, Upper bound on the apaity of a random
multiple-aess system, Problemy Peredahi Informatsii, 23(3) (1987)64{78.
[24℄ B.S. Tsybakov and V. A. Mikhailov, Free synhronous paket aess in a broadast
hannel with feedbak, Probl. Information Transmission,14(4) (1978) 259{280.
[25℄ S. Verdu, Computationof the eÆieny of the Mosely-Humblet ontention resolution
algorithm: a simplemethod,Pro. of the IEEE 74(4) (1986) 613{614.
[26℄ N.S. Vvedenskaya and M.S. Pinsker, Nonoptimality of the part-and-try algorithm,
Abstr. Papers, Int. Workshop \Convolutional Codes; Multi-User Commun.," Sohi,
Letj(r;p)= h(r;p). Wewillshowthat for anyr 2[0;1℄and p2[0;1℄,j(r;p) 0:003.
Case 1: r+p rprB=A and pr:
In this ase we have
g(r;p) = e
((1 r)p+(1 p)r+(1 p)(1 r));
j(r;p) = g(r;p) :
Observe that
j(r;p)=e
1 X
i=0 1 X
j=0
i;j p
i r
j ;
where the oeÆients
i;j
are dened as follows.
0;0
= (1 e
)
1;0
= 1
0;1
= 1
1;1
= 2+:
NotethattheonlypositiveoeÆientsare
1;0
and
0;1
. Thus, ifp2[p
1 ;p
2
℄andr2[r
1 ;r
2 ℄,
then j(r;p) is atmost U(p
1 ;p
2 ;r
1 ;r
2
), whih we dene as
e
( 0;0
+
1;0 p
2
+
0;1 r
2
+
1;1 p
1 r
1 ):
Now we need only hek that for all r
1
2(B=A 0:01;1) and p
1
2 [r
1
;1) suh that p
1
and r
1
are multiples of 0:01, U(p
1 ;p
1
+0:01;r
1 ;r
1
+0:01) is at most 0:003. This is the
ase. (The highestvalue is U(0:45;0:46;0:45;0:46)= 0:00366228.)
Case 2: r+p rprB=A and p<r:
Now we have
g(r;p) = e
((1 r)p+(1 p)
r r
2 =2
1 p=2
+(1 p)(1 r))
j(r;p) = g(r;p) :
Observe that
(1 p=2)j(r;p)=e
2 X
i=0 2 X
j=0
i;j p
i r
i;j
0;0
= (1 e
)
1;0
= 1 3=2+e
=2 0;1 = 1 1;1
= 2+3=2
2;0
= 1=2+=2
0;2
= 1=2
2;1
= 1=2 =2
1;2 = 1=2 2;2 = 0:
Note that the only positive oeÆients are
1;0 , 0;1 , 2;1 and 1;2
. Thus, if p 2[p
1 ;p
2
℄ and
r2[r
1 ;r
2
℄, then j(r;p) isat most U(p
1 ;p 2 ;r 1 ;r 2
), whih we deneas
0;0 + 1;0 p 2 + 0;1 r 2 + 1;1 p 1 r 1 + 2;0 p 2 1 + 0;2 r 2 1 + 2;1 p 2 2 r 2 + 1;2 p 2 r 2 2 + 2;2 p 2 1 r 2 1
divided by e
(1 p
2 =2).
Now we need only hek that for all r
1
2 (B=A 0:005;1) and p
1
2[0;r
1
℄ suh that p
1
and r
1
are multiples of 0:005, U(p
1 ;p
1
+0:005;r
1 ;r
1
+0:005) is at most 0:003. This is
the ase. (The highest value is U(0:45;0:455;0:455;0:46)= 0:00479648.)
Case 3: r+p rpB=Ar and pr:
In this ase we have
g(r;p) = e
((1 r)p+(1 p)r+(1 p)(1 r)):
j(r;p) = g(r;p) ( Ar+B)e
(1 r)p:
Observe that
j(r;p)=e
2 X i=0 2 X j=0 i;j p i r j ;
where the oeÆients
i;j
are dened as follows.
0;0
= (1 e
)
1;0
= 1 B
0;1
= 1
1;1
2;0 0;2 = 0 2;1 = 0 1;2 = A 2;2 = 0:
NotethattheonlypositiveoeÆientsare
1;0
and
0;1
. Thus, ifp2[p
1 ;p
2
℄andr2[r
1 ;r
2 ℄,
then j(r;p) is atmost U(p
1 ;p 2 ;r 1 ;r 2
), whih we dene as
e ( 0;0 + 1;0 p 2 + 0;1 r 2 + 1;1 p 1 r 1 + 2;0 p 2 1 + 0;2 r 2 1 + 2;1 p 2 1 r 1 + 1;2 p 1 r 2 1 + 2;2 p 2 1 r 2 1 ):
Now we need only hek that for all p
1
2 [0;1)and r
1
2 [0;p
1
℄ suh that p
1
and r
1 are
multiples of 0:01, U(p
1 ;p
1
+0:01;r
1 ;r
1
+0:01) is at most 0:003. This is the ase. (The
highest value is U(0:44;0:45;0:44;0:45)= 0:00700507.)
Case 4: r+p rpB=Ar and p<r:
Now we have
g(r;p) = e
((1 r)p+(1 p)
r r
2 =2
1 p=2
+(1 p)(1 r))
j(r;p) = g(r;p) ( Ar+B)e
(1 r)p:
Observe that
j(r;p)=e
(1=2) P 2 i=0 P 2 j=0 i;j p i r j 1 p=2 ;
where the oeÆients
i;j
are dened as follows.
0;0
= 2(1 e
)
1;0
= 2 2B 3+e
0;1
= 2 2
1;1
= 4+2A+2B+3
2;0
= 1+B+
0;2
= 1
2;1
= 1 A B
1;2
= 1 2A
2;2
1;0 0;1 2;2 1 2
r2[r
1 ;r
2
℄, then j(r;p) isat most U(p
1 ;p 2 ;r 1 ;r 2
), whih we deneas
e (1=2) 0;0 + 1;0 p 2 + 0;1 r 2 + 1;1 p 1 r 1 + 2;0 p 2 1 + 0;2 r 2 1 + 2;1 p 2 1 r 1 + 1;2 p 1 r 2 1 + 2;2 p 2 2 r 2 2 1 p 2 =2 :
Now we need only hek that for all p
1
2[0;1)and r
1
2[p
1
;1) suh that p
1
and r
1 are
multiples of 0:01, U(p
1 ;p
1
+0:01;r
1 ;r
1
+0:01) is at most 0:003. This is the ase. (The
highest value is U(0:44;0:45;0:44;0:45)= 0:00337716.)
Case 5: B=Ar+p rpr and pr:
In this ase we have
g(r;p) = e
((1 r)p+(1 p)r+(1 p)(1 r)):
j(r;p) = g(r;p) +( A(r+p rp)+B)(1 (1 r)(1 p)e
(1+))
( Ar+B)e
(1 r)p:
Observe that
j(r;p)=e
2 X i=0 2 X j=0 i;j p i r j ;
where the oeÆients
i;j
are dened as follows.
0;0
= B+Be
+ B e
1;0
= 1+A Ae
+A+B
0;1
= 1+A+B Ae
+A+B
1;1
= 2 2A+Ae
+ 3A B
2;0
= A A
0;2
= A A
2;1
= 2A+2A
1;2
= A+2A
2;2
= A A:
Note that the only positive oeÆients are
1;0 , 0;1 , 2;1 and 1;2
. Thus, if p 2[p
1 ;p
2
℄ and
r2[r
1 ;r
2
℄, then j(r;p) isat most U(p
1 ;p 2 ;r 1 ;r 2
), whih we deneas
e 0;0 + 1;0 p 2 + 0;1 r 2 + 1;1 p 1 r 1 + 2;0 p 2 1 + 0;2 r 2 1 + 2;1 p 2 2 r 2 + 1;2 p 2 r 2 2 + 2;2 p 2 1 r 2 1 :
Now we need only hek that for all p
1
2 [0;1)and r
1
2 [0;p
1
℄ suh that p
1
and r
1 are
multiples of 0:01, U(p
1 ;p
1
+0:01;r
1 ;r
1
+0:01) is at most 0:003. This is the ase. (The
Now we have
g(r;p) = e
((1 r)p+(1 p)
r r
2 =2
1 p=2
+(1 p)(1 r))
j(r;p) = g(r;p) +( A(r+p rp)+B)(1 (1 r)(1 p)e
(1+))
( Ar+B)e
(1 r)p:
Observe that
(1 p=2)j(r;p)=e
3 X i=0 2 X j=0 i;j p i r j ;
where the oeÆients
i;j
are dened as follows.
0;0
= B+Be
+ B e
1;0
= 1+A+B=2 Ae
Be
=2 3=2+A+3B=2+e
=2
0;1
= 1+A+B Ae
+A+B
1;1
= 2 5A=2 B=2+3Ae
=2+3=2 7A=2 3B=2
2;0
= 1=2 3A=2+Ae
=2+=2 3A=2 B=2
0;2
= 1=2 A A
2;1
= 1=2+3A Ae
=2 =2+7A=2+B=2
1;2
= 1=2+3A=2+5A=2
2;2
= 3A=2 2A
3;0
= A=2+A=2
3;1
= A A
3;2
= A=2+A=2:
Note that the only positive oeÆients are
1;0 , 0;1 , 2;1 , 1;2 , 3;0 and 3;2
. Thus, if p 2
[p 1
;p 2
℄and r 2[r
1 ;r
2
℄,then j(r;p)is atmost U(p
1 ;p 2 ;r 1 ;r 2
), whihwe dene as
0;0 + 1;0 p 2 + 0;1 r 2 + 1;1 p 1 r 1 + 2;0 p 2 1 + 0;2 r 2 1 + 2;1 p 2 2 r 2 + 1;2 p 2 r 2 2 + 2;2 p 2 1 r 2 1 + 3;0 p 3 2 + 3;1 p 3 1 r 1 + 3;2 p 3 2 r 2 2
divided by e
(1 p
2 =2).
Now we need only hek that for all p
1
2[0;1)and r
1
2[p
1
;1) suh that p
1
and r
1 are
multiplesof0:005,U(p
1 ;p 1 +0:005;r 1 ;r 1
+0:005)isatmost 0:003. Thisisthe ase. (The
