$O(\mbox{depth})$-Competitive Algorithm for Online Multi-level Aggregation

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by  Niv Buchbinder

O
(depth)-Competitive Algorithm for Online Multi-level Aggregation
Niv Buchbinder∗ Moran Feldman† Joseph (Seffi) Naor‡ Ohad Talmon§
January 10, 2017
Abstract might be impossible or very costly. In general, various
7
Weconsideramulti-levelaggregationprobleminaweighted costfunctions andaggregationconstraintsgiverisetoa
1
0 rooted tree, studied recently by Bienkowski et al. [7]. In large family of interesting problems.
this problem requests arrive over time at the nodes of the
2 Scenarios where aggregation is beneficial are com-
tree, and each request specifies a deadline. A request is
n servedbysendingit totheroot beforeitsdeadlineatacost mon in supply-chain management, in which producing
a equal to the weight of the path from the node in which it (ordelivering)severaldemandstogethercanbecheaper
resides to the root. However, requests from different nodes
J thanproducingthemseparately[15,20,1,27]. Another
can be aggregated, and served together, so as to save on
8 cost. The cost of serving an aggregated set of requests is example is aggregating control packets in communica-
equal to the weight of the subtree spanning the nodes in tion networks (e.g., sensor networks) [16, 2, 11, 18]. In
] which the requests reside. Thus, the problem is to find a
S competitiveonlineaggregationalgorithmthatminimizesthe thissetting,transmittingtogetherseveralpacketsgoing
D total cost of the aggregated requests. This problem arises to a joint destination is, again, cheaper than transmit-
naturallyinmanyscenarios, includingmulticasting,supply- tingeachpacketseparately. Aggregationproblemswere
.
s chain management and sensor networks. It is also related
studiedextensivelyformanysettings,bothintheoffline
c tothewell studied TCP-acknowledgement problem and the
[ online joint replenishment problem. case,inwhichallrequestsareknowna-priori,andinthe
We present an online O(D)-competitive algorithm for online case in which requests arrive over time.
1 the problem, where D is the depth, or number of levels, of
v the aggregation tree. This result improves upon the D22D- We study averygeneralonline aggregationsetting,
6 competitive algorithm obtained recently by Bienkowski et considered recently by [7], and known as the Online
3 al. [7]. Multi-level Aggregation Problem with Deadlines. In this
9 problem requests arrive over time, and each request
1 Keywords: online algorithms, competitive analysis, aggre-
specifies a deadline for serving it. Thus, each request
0 gation of requests
is associated with a time interval in which it needs to
.
1
be served. At any point of time t, the online algorithm
0 1 Introduction
is only aware of requests whose arrival time is no later
7 Aggregationof tasks is a fundamental tool in optimiza-
1 than t.
tion, utilized in various areas. Suppose that there is a
: There is a tree with non-negativeedge costs rooted
v set of requests that need to be served, where serving
at a node r, and each request is assumed to reside at
i
X several requests together costs less than serving each some tree node. The cost of serving a single request is
request individually. There are aggregation constraints
r equaltothecostofthetreepathfromthenodewhereit
a on the request system that specify which sets of re-
residestotherootr. Requestswhoseintervalsintersect
quests can be jointly served, together with a function
in time can be aggregated and served together. The
that determines the cost of serving any aggregated set
aggregatecostofasetofrequestsisequaltothe costof
of requests. For example, serving all requests together
the subtree rooted at r and spanning the nodes where
these requests reside. The goal is to minimize the total
∗Department of Statistics and Operations Research, School service cost, i.e., the sum of the costs of the subtrees
of Mathematical Sciences, Tel Aviv university, Israel. Email:
used for serving the requests.
[email protected]. Researchissupported byISFgrant
The online multi-level aggregation problem with
1585/15andUS-IsraelBSFgrant2014414.
†Department of Mathematics and Computer Science, The deadlines was studied recently by Bienkowski et al. [7]
Open University of Israel. Email: [email protected]. Re- who designed a D22D-competitive algorithm for it,
searchsupportedinpartbyISFgrant1357/16. whereDdenotesthedepthofthetree. Wenotethatthis
‡Computer Science Department, Technion, Israel. Email:
bound is quite far fromthe best knownlower bound on
[email protected]. Research is supported by ISF grant
thecompetitivefactorfortheproblemwhichisonly2[9]
1585/15andUS-IsraelBSFgrant2014414.
§Computer Science Department, Technion, Israel. Email: (see Section 1.2 for more details). In the offline case,
[email protected]. Research is supported by ISF grant the problem is known to be NP-hard (and APX-hard)
1585/15andUS-IsraelBSFgrant2014414.
even when the aggregation tree is of depth 2 [1, 25, 8]. 3-decreasing trees to each tree of the resulting forest.
Currently,thebest(offline)approximationfactorknown A very simple argument shows that the algorithm
for the problem is 2 [6, 7]. obtained this way has a competitive factor of at most
O(D2). However, using a more involved analysis we
1.1 Our Results. Our main result is a competitive are able to show that the two above claims hold also
algorithmfortheOnlineMulti-levelAggregationProblem for general trees, which give us the promised O(D)-
with Deadlines. We have already mentioned that there competitiveness guarantee.
is a large gap between the lower bound of 2 on the
competitive ratio known and the current best upper 1.2 RelatedWork. Severalinterestingproblemsare
bound of D22D [7], which is exponential in the depth related to the Online Multi-level Aggregation Problem
D of the aggregation tree. We obtain a substantial with Deadlines. Suppose that each request accrues a
improvement over the upper bound of [7]. waiting cost over time, rather than having a strict
deadline. The cost of satisfying a set of requests is
Theorem 1.1. There exists an O(D)-competitive algo-
then defined to be the sum of both the service cost
rithmfortheOnlineMulti-levelAggregationProblemwith
and the sum of the waiting costs of the requests. For
Deadlines. thisproblemBienkowskietal.[7]designedanO(D42D)-
competitive algorithm. The best offline approximation
Our Techniques. Standard reductions allow us
for this variant is 2+ǫ [26] adapting ideas from [22].
to assume that the costs are associated with the tree
We note that a common choice for the waiting cost is
nodes rather than with the edges. To simplify the
a linear function over time. This model (with slight
presentation, we first present our result for a special
variations) was considered by Khanna et al. [19] who
kind of aggregation tree known as 3-decreasing trees.
achievedacompetitivefactorwhichislogarithmicinthe
Ina3-decreasingtree,aswetraverseanypathfromthe
weight of the aggregationtree.
root r to a leaf, the edge costs go down by a factor of
One of the earliest problems considered in the gen-
at least 3 at each step of the path. The idea of our
eralsetting of aggregationis the TCP-Acknowledgment
algorithm for 3-decreasing trees is simple and intuitive.
Problem. In this problem there is a single link over
Suppose thereis a requestthathasreachedits deadline
whichpacketsneedtobe acknowledgedby controlmes-
(and, thus, must be served). The algorithm recursively
sages. Any number of control messages can be aggre-
aggregates requests into a subtree (which eventually
gatedinto a single packetsentoverthe single link. The
will be served), starting from the root of the tree.
aggregationtree,thus,consistsofasingleedge. Theob-
The aggregation process has to balance between two
jective function is composed of two terms: the number
conflicting objectives. On the one hand, it is beneficial
of acknowledgments sent over the link and the sum of
to aggregate as many requests as possible, especially
the waiting times of the control messages. The optimal
those approaching their deadline. On the other hand,
competitive factors for the TCP-acknowledgmentprob-
it is necessary to bound the cost of the aggregate
lem are 2 for deterministic algorithms and e/(e−1) if
subtree,sothatitcanberelatedtotheoptimalsolution.
randomization is allowed [16, 18, 13]. Interestingly, the
We balance between the two by recursively providing
TCP-acknowledgmentproblemisequivalenttotheclas-
budgets to tree nodes that are added to the aggregate
sicalLotSizingProblem[16,27],whichhasbeenstudied
subtree. A budget provided to node u (that has joined
extensively by the operations research community.
thesubtree)canbeusedtoaggregateanadditional(yet
The aggregation problem for trees of depth 2 is
restricted) set of nodes into the subtree rooted at u.
known as the Joint Replenishment Problem in supply-
These nodes are then also given budgets, which can be
chainmanagement. The best competitive factor known
used to aggregate additional nodes (recursively).
forthis specialcaseis3withgeneralwaitingtimes,and
The analysis of the algorithm is performed by first
2 if requests have deadlines [12, 14, 9]. The best lower
showing that the total cost of each served subtree is
bound for general waiting times is 2.754 [9] improving
at most O(D ·c(r)), where D is the depth of the tree
upon an earlier bound of 2.64 by [14]. For the deadline
and c(r) is the cost of the root r. Then, we show
case the best lower bound is 2 [9]. The (offline) Joint
that the optimal cost of serving the remaining requests
Replenishment problem is known to be NP-hard (and
(including requests that at this point of time have not
APX-hard) [1, 25, 8]. The best approximation ratio
arrived yet) decreases by at least c(r). The two claims
currentlyknownforitis1.791[9],whichimprovesupon
together imply the desired bound.
previous results of [21, 24, 22, 23].
Extending our result to general trees is done via
Finally,whentheaggregationtreeisaninfinitehalf
a reduction that transforms a tree into a forest of 3-
line,Bienkowskietal.[7]gavea4-competitivealgorithm
decreasing trees. Then, we apply the algorithm for
forthedeadlineversionoftheproblem,andshowedthat c(T)= c(u).
u∈T
this is the best possible. For the more general problem P
2. Foreachnodeu,weassumec(u)isstrictlypositive.
with waiting costs, Bienkowski et al. [10] showed that
the competitive ratio is between 2 + φ ≈ 3.618 and Proving that these modifications are without loss
5, improving on an earlier upper bound of 8 for the of generality can be done in a rather standard way.
problem by Brito et al. [12]. It is also known that the First, we may assume that r(T) has only a single child.
optimal offline solution for this case can be computed One may observe that the transmission of a subtree
efficiently [10]. T containing multiple children of r(T) can always be
replaced with the transmission of multiple subtrees,
2 Problem Definition and Preliminaries
eachcontainingr(T)andthepartofT descendingfrom
An instance of the Online Multi-level Aggregation Prob- a single child of r(T). Moreover,this replacement does
lemwithDeadlines(OMAPD)isdefinedasatuple(T,I). not change the cost of the transmission. Hence, there
The first component of the tuple is a tree T rooted at is no loss in applying any algorithm independently to
some node r(T) with non-negative edge costs. We de- every subtree of T (more accurately, every instance of
notebyc(e)thecostofanedgeeofT,andbyD(T)the the algorithm faces an instance of OMAPD with a tree
depth of T, i.e., the maximum number of edges along containingr(T)andallthe nodesdescendinginT from
anypath fromthe rootofT to aleaf. The secondcom- asinglechildofr(T)). Next,weobservethatinOMAPD
ponent of the tuple (T,I) is a set of time intervals I, everyinterval associatedwith the rootr(T) can always
where each interval I = [a ,d ] ∈ I is associated with be serviced without any cost. Thus, we may assume
I I
a node w ∈ T, an arrival time a and a deadline d . that our instance does not contain any such intervals.
I I I
A solution for the problem is a sequence of subtrees We now transform T into a node-cost tree by moving
T ,T ,...,T ⊆ T rooted at r(T) that are transmitted the cost of every edge to its end point which is further
1 2 ℓ
at times t ,t ,...,t .1 The solution is feasible if, for away from the root, and then removing the root r(T)
1 2 ℓ
eachintervalI ∈I,thereexistsatreeT withtransmis- itself(whichistheonlynodeleftwithnocost). Finally,
i
siontimet ∈[a ,d ]containingthenodew . Moreover, if some node u has cost zero, we may merge it with
i I I I
wesaythatatreeofthesolutionhavingtheseproperties the next node on the path to the root. Thus, each
services I. The costof transmitting a tree T is denoted node has strictly positive cost. It is easy to see that
by c(T), c(e), where E[T ] is the set ofedges any solution for the original edge weighted instance
e∈E[Ti] i
of T . ThePcost of the solution is ℓ c(T ). can be transformed to a solution for the resulting node
i i=1 i
weighted instance with the same cost and vice-versa.
In the online setting the trePe T is known to the
Moreover, the depth of the node weighted instance
algorithm in advance, but the intervals of I arrive in
is always smaller than the depth of the original edge
an online fashion, i.e., each interval I ∈ I is revealed
weighted instance. From now on we abuse notation
to the online algorithm only upon its arrival time at
and refer to the variant of OMAPD obtained by our
a . The online algorithm can decide at every given
I
modifications simply as OMAPD.
time to transmit a subtree of T at this time, however,
Finally, a tree T is called 3-decreasing if, as we go
boththedecisiontotransmitthesubtreeandthechoice
along any path from the root r(T) to a leaf, the cost of
of the subtree to transmit must be done without any
each node is smaller than the cost of the previous node
knowledge about future intervals. The algorithm is
by a factor of at least 3. This definition is used in [7],
α-competitive if it produces a feasible solution whose
anditissimilartotheweightedversionofHSTsdefined
total cost is always at most α times the cost of the
in [3]. Note, however, that a 3-decreasing tree is not
optimal solution. In the online setting it is often useful
exactlyan(un-weighted)3-HST(asdefinedby[4,5,17])
to call intervals that have already been revealed to the
asthecostsoftheedgesalonganyroottoleafpathofa
algorithm,butwerenotservicedyet,bythetermactive.
3-decreasing tree decrease by factor of at least 3 rather
To simplify the presentation of our algorithm, we
thanbyexactly3. Moreover,differentroottoleafpaths
first modify the problem as described by the following
of a 3-decreasing tree might differ in their lengths.
two modifications.
1. We assume the tree T has node costs {c(u) | 3 Algorithm for 3-Decreasing Trees
u ∈ T} rather than edge costs. Consequently, the In this section we present and analyze an algorithm
transmissioncost ofa subtree T ⊆T now becomes for OMAPD on 3-decreasing trees. A generalization of
this algorithm for general trees is given in Section 4.
1Noticethatweabusenotationhereandtreattrees assets of Throughout the section we consider a fixed instance
nodes. Thisisdoneoccasionallythroughoutthepaper. (T,I) of OMAPD in which the tree T is a 3-decreasing
Algorithm 1: Transmission Tree Section
1 Initiate a transmission tree T ←{r}.
2 Let Q be a queue of tuples to be processed. Initially, Q contains only the tuple (r,cˆ(r)=2c(r)).
3 while Q6=∅ do
4 Dequeue the next tuple (u,cˆ(u)) from Q, and set Au ←∅.
5 while there are active intervals associated with nodes in Tu\T and c(Au)≤cˆ(u)/2 do
6 Let I be the interval with the earliest deadline among the active intervals in Tu\T (breaking ties
arbitrarily).
7 Add to Au all nodes on the path from wI to r that are not already in T.
8 For every v ∈Au, enqueue into Q the tuple v,c(v)· ccˆ(A(uu)) .
(cid:16) (cid:17)
9 Add the nodes of Au to T.
10 Transmit T.
tree. Since we have fixed the instance of OMAPD, we important to note that, when a node u chooses nodes
can use in this section a somewhat simplified notation. from its subtree T to add to the transmission tree, it
u
More specifically, we drop the parameter from the gives priority to satisfying intervals whose deadline is
notations r(T) and D(T) and use r and D to denote sooner (and thus are, intuitively, more “urgent”).
the root of T and its depth, respectively. We also need The following observation shows that each trans-
todefinesomenewnotation. GivenasetU ofnodeswe mission of Algorithm 1 makes progress, and thus, our
use c(U) to denote the totalcostof the nodes within it, algorithm, as a whole, terminates.
i.e., c(U) = c(u). Additionally, given a node u
u∈U Observation 3.1. Algorithm 1halts andthetransmis-
we denote byPTu the subtree of T rooted at u. sion tree T contains an active interval.
We remind the reader that an interval is said to
be active if and only if it has already appeared, but Proof. ObservethatLine6ofAlgorithm1isguaranteed
has not been serviced yet. Our algorithm transmits a to be reached at least once in every given execution of
tree whenever an active interval matures, i.e., reaches Algorithm 1. This line selects an active interval I, and
its deadline. More specifically, our algorithm invokes the nodes of the path from r to w are later added to
I
Algorithm 1 whenever an active interval matures, and T.
Algorithm1thenselectsasubtreeT ofT andtransmits
Wearenowreadytopresentthetwomainlemmata
it. Following the transmission of T, our algorithm
necessary for the analysis of the algorithm. The first of
returns to its idle state until another active interval
these lemmata bounds the cost paid by the algorithm
matures. We note that there might be multiple active
for every single transmission. The proof of this lemma
intervals that reach maturity at the same time. When
is deferred to Section 3.1.
this happens, the transmission of T might not service
all of these intervals, which might result in immediate Lemma 3.1. For every subtree T of T transmitted by
additional invocation of Algorithm 1. In other words, the algorithm, c(T)≤2(D+1)·c(r).
there might be a zero time gap between consecutive
The presentationof the other main lemma requires
executions of Algorithm 1.
some additionalnotation. Assume the online algorithm
Informally, Algorithm 1 starts the construction of
transmitsoverallℓtreesT ,...,T . Forevery0≤i≤ℓ,
the tree to be transmitted by assigning a budget of 1 ℓ
let I be the set of intervals that were not yet serviced
cˆ(r)=2c(r)tothe rootofthe tree. Thisbudgetisthen i
by the algorithm after it has made i transmissions.
used to recursively add new nodes to the transmission
Note thatI includes alsointervalsthatarriveafterthe
tree (and thus, serve the intervals residing in these i
transmission time of tree T , i.e., they were not active
nodes). Morespecifically,eachnodeuthatisassigneda i
yet when T was transmitted. By definition, T services
budgetusesittoaddtothetransmissiontreenewnodes i i
belonging to its subtree T . The total transmission alltheintervalsIi−1\Ii. Additionally,weuseOPT(I′)
u todenoteanoptimalsolutionforasetofintervalsI′ ⊆I
costof these newly added nodes is roughly equalto the
(more formally, OPT(I′) is an optimal solution for the
budget of u; and following the addition of these new
instance (T,I′) of OMAPD).
nodes u splits its budget proportionally among them
Our second main lemma can now be stated as
so that this process can be repeated recursively. It is
follows. We defer its proof to Section 3.2.
Lemma 3.2. OPT(Ii) ≤ OPT(Ii−1)−c(r) for every where the first inequality follows from Lemma 3.3,
1≤i≤ℓ. and the second inequality follows from the induction
hypothesis applied to u.
Analyzing our algorithm is now straightforward.
Theorem 3.1. There exists an O(D)-competitive algo- Note that for every node u such that Au 6= ∅, by
rithm for Online Multi-level Aggregation Problem with construction,
Deadlines on 3-decreasing trees.
cˆ(u)
cˆ(v)= c(v)· =cˆ(u).
Proof. By Lemma 3.1 the total cost suffered by our (cid:20) c(A )(cid:21)
algorithm is at most ℓ·[2(D+1)·c(r)]. On the other vX∈Au vX∈Au u
hand, Lemma 3.2 shows that the cost of the optimal Hence, the following observation can be proved by
solution OPT(I0) = OPT(I) is at least ℓ · c(r) + induction.
OPT(I )=ℓ·c(r).
ℓ
Observation 3.2. In each level of T, the sum of cˆ(u),
3.1 Proof of Lemma 3.1. In this section we prove taken over all nodes u, is at most cˆ(r)=2c(r).
Lemma 3.1. Let us first recall the lemma.
We now observe the following.
Lemma 3.1. For every subtree T of T transmitted by
the algorithm, c(T)≤2(D+1)·c(r). c(T)= c(u)≤ cˆ(u)≤2(D+1)·c(r) .
We begin the proof of the above lemma with the uX∈T uX∈T
following claim. The firstinequality followssince Lemma 3.4guarantees
Lemma 3.3. For every node u added to T, c(A ) ≤ that c(u) ≤ cˆ(u) for every node u ∈ T. The second
u
1(cˆ(u)+c(u)). inequalityfollowssincethereareatmostD+1possible
2 levels, and Observation 3.2 shows that the sum of cˆ(u)
Proof. Each time that Algorithm 1 adds nodes to Au, in each level is at most 2c(r).
it adds only nodes belonging to the path from u to w
I
of a single interval I. Since T is 3-decreasing, the total 3.2 Proof of Lemma 3.2. In this section we prove
cost of the nodes on such a path is upper bounded by Lemma 3.2. We begin by recalling the lemma.
D i ∞ i
1 ·c(u)≤ 1 ·c(u)= c(u)/3 = c(u) . Lemma 3.2. OPT(Ii) ≤ OPT(Ii−1)−c(r) for every
(cid:18)3(cid:19) (cid:18)3(cid:19) 1−1/3 2 1≤i≤ℓ.
Xi=1 Xi=1
On the other hand, Algorithm 1 stops adding nodes to To prove the lemma it suffices to construct a
Au once c(Au) exceeds cˆ(2u), thus completing the proof. solution S servicing all the intervals of Ii whose cost is
Next, define inductively the level of nodes in T as at most OPT(Ii−1)−c(r). The lemma will then follow
since, being an optimal solution, OPT(I ) is not more
follows. Theleveloftherootris0. Fornodev,suppose i
expensive then any other feasible solution for (T,I ).
thatnodeuaddedittoA ;thenthelevelofv isdefined i
u
Let us now construct the above mentioned solution
to be the level of u plus 1. Since the nodes of A are
u
S. Recall that T is the subtree of T transmitted by
all descendants of u, the level of each node u ∈ T is i
thealgorithmatitsi-thtransmission,andletusdenote
guaranteed to be at most D.
by t the time of this transmission. There are a few
i
Lemma 3.4. For every node u∈T, c(u)≤cˆ(u). assumptionsthatwecanmakeaboutOPT(Ii−1). First,
Proof. We prove the lemma by induction on the level we may assume that OPT(Ii−1) transmits at most
one subtree at every given time (otherwise, we may
of u. For the root r the lemma is immediate since
merge subtrees transmitted at the same time without
cˆ(u)=2c(u). Next,letv beanodeinA (addedbyu).
u increasing the total cost). Second, we may assume
Then,
that OPT(Ii−1) makes its first transmission at time ti
cˆ(u) because that is the time of the earliest deadline among
cˆ(v)=c(v)·
c(Au) the deadlines of the intervals of Ii−1 (specifically, the
cˆ(u) interval whose deadline triggered the i-th transmission
≥c(v)· 1(cˆ(u)+c(u)) of the algorithm is in Ii−1 and its deadline is ti).
2 Given these assumptions, there must be exactly one
cˆ(u)
≥c(v)· =c(v), transmission of OPT(Ii−1) at time ti. Let us denote
cˆ(u) the subtree transmitted at this time by T∗.
i
We obtainthe solutionS fromthe optimalsolution of any node, which contradicts the definition of u.
OPT(Ii−1) by applying the following two steps to the Hence, any node u which is reconstructed must belong
last solution. to T∗.
i
It remains to prove the second part of the lemma.
1. Remove T∗ from the solution.
i ThefactthatuwasreconstructedwhenI wasprocessed
2. Reconstruction step: Scan the intervals of I in means that prior to I’s processing the interval [ti,dI]
i
did not contain any transmission involving u. Since
a non-decreasingorder of their deadline. For every
the reconstruction step scans the intervals in a non-
such interval I we do the following to guarantee
decreasingorderoftheir deadlines,this means thatany
that it is serviced by S.
intervalI′ forwhichuisonthepathfromrtowI′ must
• Findthesubtreeofthecurrentsolutiontrans- have a deadline dI′ ≥ dI. Hence, when I′ is processed
mittedwithintherange[ti,dI]whichcontains by the reconstruction step, the range [ti,dI′] ⊇ [ti,dI]
the largest fraction of the path from r to wI already includes a subtree containing u, and thus, I′
(breaking ties arbitrarily), and add to this does not cause another reconstruction of the node u.
subtree the remaining part of this path. Note
thattheabovetreemightcontainallthepath, The last lemma shows that only a limited set of
inwhichcasethereisnoneedtoaddanything. nodes(namely,thenodesofT∗)mightbereconstructed.
i
Lemma 3.7 shows that even within this limited set
• Ifthecurrentsolutionmakesnotransmissions
there is a significant subset of nodes that are not
withintherange[t ,d ],thenintroduceintoit
i I
reconstructed. The following lemma proves a few
a new transmission transmitting the subtree
technical claims used later in the proof of Lemma 3.7.
consistingsolelyofthepathfromrtow . The
I
location of this new transmission within the
Lemma 3.6. Let u∈T be a node that is reconstructed.
range [t ,d ] can be chosen arbitrarily. i
i I Then, c(A )> cˆ(u) and A ⊆T∗.
u 2 u i
Observation 3.3. The solution S isafeasible solution
Proof. LetI be theintervalthatcausedthereconstruc-
for (T,I ).
i
tion of u. Since I belongs to I (intervals outside of I
i i
Proof. The construction of S guarantees that every arenotprocessedbythereconstructionstep),itmustbe
interval of Ii is serviced by S before its deadline. thatI isnotservicedbyTi. Ontheotherhand,thefact
thatI causedthe reconstructionofnodes means thatit
We are left to prove that the cost of S is at most isnotservicedbyanysubtreeofOPT(Ii−1)otherthan
OPT(Ii)−c(r). ItisclearthatStep1inthe procedure Ti∗, thus, it must be active at time ti.
for constructing S decreases the cost by c(T∗), thus, it The above observations imply that when Algo-
i
suffices to show that the second step increases the cost rithm 1 constructed T it left the inner loop growing
i
of S by at most c(T∗)−c(r). The restof this section is A while there were still active intervals associated
i u
devoted to proving this claim. with nodes of T \ T; which can only happen when
u
We say that a node u is reconstructed whenever it c(A ) > cˆ(u). Next, consider any node v ∈ A . Since
u 2 u
is added to some transmission of the solution during Algorithm 1 scans the intervals in a non-decreasing
Step 2 of the above procedure. deadline order while growing A , v must have been
u
Lemma 3.5. The reconstruction step obeys the follow- added to Au due to being on the path from r to wI′
ing claims. of some interval I′ having a deadline dI′ ≤dI. Assume
towards a contradiction that T∗ does not contain this
i
1. Only nodes of T∗ are reconstructed. path. Clearly, this assumption implies that OPT(Ii−1)
i servicesI′bysomesubtreetransmittedduringtherange
2. Each node of Ti∗ is reconstructed at most once. (ti,dI′]. Since any subtree servicing I′ must include u,
u is already present within the range (ti,dI′] ⊆ [ti,dI]
Proof. Consideranodeuwhichisreconstructed,andlet
when I is processed by the reconstruction step; which
I betheintervalwhoseprocessingbythereconstruction
contradictsthedefinitionofI astheintervalwhosepro-
step caused the first reconstruction of u. Thus, u is on
cessing caused the reconstruction of u.
the path from r to w . If u does not belong to T∗,
I i
then OPT(Ii−1) must service I by some transmission Lemma 3.7. There exists a set U of nodes obeying the
other than Ti∗ within the range (ti,dI]. Since this following properties.
transmission is not removed, the processing of I by the
reconstruction step could not cause the reconstruction 1. U ⊆T∗∩T .
i i
2. cˆ(u)=cˆ(r)=2c(r). an edge to the first node v on the path from u to r(T)
u∈U
P whose cost obeys c(v) ≥ 3c(u). If there is no such v,
3. cˆ(u)≤2c(u) for all u∈U. then u becomes the root of a new tree. Each node u in
the forest is now associated with a set B of all nodes
u
4. The nodes of U are not reconstructed.
on the path from u to v in the originaltree T (without
v, but including u itself). If v does not exist (i.e., u
Proof. We startwith a set U obeying all the properties
is a root in the new forest), then the associated set B
otherthanProperty4,andletitevolvewhilemaintain- u
is defined as the set of nodes on the path from u to
ingtheseproperties. TheevolutionofU endsassoonas
the root r(T) of the original tree (this time, including
it obeys also Property 4. More specifically, we initially
r(T)).
set U = {r}. Observe that this set indeed satisfies all
the properties other than Property4. If U also satisfies Observation 4.1. The forest F consists solely of 3-
Property4 then we are done. Otherwise,there mustbe decreasing trees.
a node u in the current set U which is reconstructed.
Proof. By definition, if u is a node of F which is not
Since we maintain U as a set obeying Property 1, u
a root of its tree, then the father v of u in F obeys
must belong to T . Hence, by Lemma 3.6, it must hold
i c(v)≥3c(u).
that c(A )≥ 1cˆ(u) and A ⊆T∗.
u 2 u i
AtthispointweevolveU byremovingufromitand Assume that F consists of m trees T1,T2,...,Tm.
adding the nodes of A instead. Since A ⊆ T ∩T∗, Notice that T and F have the same set of nodes, and
u u i i
Property 1 is preserved. Additionally, cˆ(v) = thus, each interval is naturally associated with a node
v∈Au
c(v)· cˆ(u) = cˆ(u), and thus,PProperty 2 is in one of the trees in the forest. Let Ii be the set of
v∈Auh c(Au)i intervals associated with the nodes of Ti.
Ppreservedaswell. Finally,Property3alsoremainsvalid
Ouralgorithmforgeneraltreesrunsanindependent
since Algorithm 1 sets cˆ(v) = c(v)· cˆ(u) ≤ 2c(v) for
c(Au) instance of the algorithm for 3-decreasing trees from
each node v ∈ Au (where the inequality holds since Section 3 on each tree Ti with its corresponding set of
Lemma 3.6 implies c(Au)≥ 12cˆ(u)). intervals Ii. For convenience, we denote the algorithm
We can now repeat the above evolution step as fromSection3byALGfromthispointon. Wheneveran
long as Property 4 is violated. However, this evolution instance of ALG choosesto transmit a subtree T ⊆Ti,
cannot continue forever since each step of it replaces a wetransmitinsteadthetree B . Itisusefultocall
u∈T u
single node with nodes appearing lower than it in T. T the virtual tree transmittSed by ALG, and u∈T Bu
Hence, the evolution presented is guaranteedto end up the concrete tree transmitted by ALG. ObseSrve that
eventually with a set U obeying Property 4 (as well as the concrete tree is alwaysa subtree ofthe originaltree
the three other properties). T, and thus, this is a description of a valid algorithm
for the original problem (OMAPD on general trees).
To conclude the proof of Lemma 3.2 we note that,
We next show that the total cost of the optimal
by combining all the properties of U in Lemma 3.7,
solutions for all the subtrees of the forest F is bounded
we get that there must exist a set U of node in T∗
i by the optimal cost of the original instance. Formally,
which are not reconstructed and have a total cost of let OPT(T′,I′) denote the costof the optimal solution
at least u∈Uc(u) ≥ 12 u∈Ucˆ(u) = c(r). Together for an instance (T′,I′) of OMAPD. Then,
rweictohntshtreuPccltaeidm,aonfdLeemvemnatPh3e.y5ctahnatoonnlylybneordeecsonosftTrui∗ctaerde Observation 4.2. OPT(T,I)≥ mi=1OPT(Ti,Ii).
once, we get that the increase in the cost of S during Proof. Consider the optimal solutPion for the instance
the reconstructionstepis atmostc(Ti∗)− u∈Uc(u)≤ (T,I) of OMAPD. Based on this optimal solution we
c(Ti∗)−c(r). P construct a solution for every one of the instances
{(Ti,Ii)}m asfollows. Whenevertheoptimalsolution
i=1
4 Algorithm for General Trees for (T,I) transmits a subtree T∗, the solution for the
In this section we show how the algorithm from Sec- instance(Ti,Ii)transmitsthesubtreeT∗∩Ti. Onecan
tion 3 can be modified to be O(D(T))-competitive also verifythatT∗∩Ti isindeedatreesincethesetofnodes
for OMAPD on general trees. We begin by describing a on the path connecting every node u∈Ti to r(Ti) is a
reduction, originallyshowedby [7], transforminga gen- subset of the set of nodes connecting u to r(T) in T.
eral tree into a forest of 3-decreasing trees. We complete the proof by observing that the costs
Given a tree T, we construct a forest F of 3- of the above solutions add up to exactly OPT(T,I),
decreasing trees as follows. The nodes of F are the and thus, the total cost of the optimal solutions for the
same as the nodes of T. We connect each node u with instances {(Ti,Ii)}m cannot exceed this value.
i=1
The above observation implies that in order to By Lemma 3.4, c(u) ≤ cˆ(u) for every node u ∈ T.
analyze our algorithm it is enough to relate the cost of Recallthatcˆ(u)isdefinedbyAlgorithm1onlyfornodes
the concretetreestransmittedby eachinstanceofALG u∈T. Wenowextendthedefinitionofcˆtoallthenodes
to the cost of the optimal solution for the instance of of T′ by setting cˆ(v) = cˆ(u) for every node v ∈ B .
u
OMAPD faced by this instance of ALG. Notice that Since, by definition, the nodes of B have costs of at
u
this is slightly different from what we do in Section 3 most 3c(u), we get the following corollary.
since in Section 3 we relate the cost of the virtual trees
transmitted by an instance of ALG to the cost of the Corollary 4.1. For every node u∈T′, c(u)≤3cˆ(u).
optimal solution for the instance of OMAPD faced by
thisinstanceofALG. Nevertheless,weshowintherest Next, define levels for the nodes of T′ using the
of this section that the arguments from Section 3 can following recursive procedure. The level of the root
be used, almost without change,to provealso the more r(Ti) is 0. Consider now a node u ∈ T which already
ambitiousgoalweneedtoprovehere. Morespecifically, has a level ℓu. Then, for every node v ∈Au, the nodes
we prove the following proposition. ofBv areassignedthelevelsℓu+1,ℓu+2,...,ℓu+|Bv|;
wherethelastlevelisassignedtovitself. Thedefinition
Proposition 4.1. For every 1 ≤ i ≤ m, the total cost ofA guaranteesthatallthenodesofB appearonthe
u v
of theconcretetreestransmittedby theinstanceof ALG pathfromu to v in T, andthus, the difference between
assigned to Ti is at most O(D(T))·OPT(Ti,Ii). the level of u and v is at most the difference between
their heights in T. Thus, no node of T′ is given a level
Clearly Theorem 1.1 follows from Observation 4.2 larger than D(T) by the above procedure.2
and Proposition 4.1. To prove Proposition 4.1 we Recallthatforeverynodeu∈T suchthatA 6=∅,
u
need to define some additional notation. Assume ALG by construction,
transmit ℓi trees when given (Ti,Ii). For every 0 ≤
j ≤ ℓi, let Ii be the set of intervals from Ii that were cˆ(u)
j cˆ(v)= c(v)· =cˆ(u).
not yet serviced by the algorithm after it has made i (cid:20) c(A )(cid:21)
transmissions. Note that Ii includes also intervals that vX∈Au vX∈Au u
j
arrive after the i-th transmission, i.e., they were not Let us now define for a node u ∈ T an additional set
activeyetwhenthis transmissionwasmade. Usingthis B = {w ∈ B | v ∈ A ,ℓ = ℓ +1}. In other words,
u v u w u
notation we can state the following lemma, which is a B isobtainedfromA byreplacingeverynodev ∈A
u u u
counterpart of Lemma 3.2, and also follows from it. with the single node w ∈ B whose level is larger by
v
1 than the level of u. Since cˆ(w) is identical for every
eLveemrym1a≤4.i1≤. OmPaTn(dIji1)≤≤j ≤OPℓiT. (Iji−1) − c(r(Ti)) for nwohdeeneovfeBrvBwue6=im∅m.edFioartealyngoedtealuso∈PTv′∈\BuTcˆ(wv)e=decˆfi(nue)
Bu = {w ∈ Bv | ℓw = ℓu + 1 ∧ ∃v∈Tu,w ∈ Bv}.
To proveProposition4.1wealsoneedthe following Informally, B contains the single node that belongs to
u
counterpart of Lemma 3.1. the same set B as u and has a level larger by 1 than
v
u. Again, the fact that cˆ(w) is identical for every node
Lemma 4.2. If the instance of ALG corresponding to
of a single set B guarantees cˆ(v)=cˆ(u).
(Ti,Ii) transmits a virtual tree T, then c B ≤ v v∈Bu
u∈T u P
u∈T c(Bu)≤6(D(T)+1)·c(r(Ti)). (cid:0)S (cid:1) Lemma 4.3. In each level 0 ≤ ℓ ≤ D(T), the sum
P of cˆ(u), taken over all nodes of level ℓ, is at most
One can verify that Lemmata 4.1 and 4.2 imply cˆ(r(Ti))=2c(r(Ti)).
Proposition 4.1 in the same way that Lemmata 3.1
and 3.2 imply Theorem 3.1. Thus, it remains to prove
Lemma 4.2. Recall thatALG generateseachone ofthe 2There is an alternative way to define this level assignment
virtualtreesittransmitsbyexecutingAlgorithm1,and whichmighthelptounderstandtheintuitionbehindit. Consider
consider the execution of Algorithm 1 which generated a tree TA defined as follows. The root of TA is r(Ti), and the
children of every node u ∈ TA are the nodes of Au. One can
the virtual tree T. verifythatTA hasexactlythesamesetofnodesasT. Moreover,
The sets {Bu}u∈T might not be disjoint. However, theheightofeverynodeinTA corresponds toitslevelaccording
for the purposes of the proofit is useful to assume they tothelevelassignmentusedinSection3. Ifwenowreplaceevery
are disjoint. In other words, if a node v appears in nodeu∈TA withapath consistingof thenodes ofBu inwhich
several sets out of {Bu}u∈T, we treat each one of its utreisehthaestlohweessatmneodseet,othfennodweesagsetTa′,naenwdttrheeehTeBig.hTtohfisevtiemrye,ntohdies
appearances as unique. Additionally, let us use the in TB corresponds to its level according to the level assignment
shorthand T′ = u∈T Bu. usedinthissection.
S
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[6] Luca Becchetti, Alberto Marchetti-Spaccamela, An-
u of a lower level than v. All this observations, taken
drea Vitaletti, Peter Korteweg, Martin Skutella, and
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LeenStougie. Latency-constrainedaggregation insen-
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[7] Marcin Bienkowski, Martin B¨ohm, Jaroslaw Byrka,
We now observe the following.
Marek Chrobak, Christoph Du¨rr, Luk´aˇs Folwarczny´,
Lukasz Jez, Jir´ı Sgall, Nguyen Kim Thang, and Pavel
′
c(T )= c(u)≤3· cˆ(u) Vesely´. Online algorithms for multi-level aggregation.
uX∈T′ uX∈T′ CoRR,abs/1507.02378, 2015. ToappearinESA2016.
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Neil B. Dobbs, Tomasz Nowicki, Maxim Sviridenko,
=6(D(T)+1)·c(r(Ti)) .
GrzegorzSwirszcz,andNealE.Young. Approximation
algorithms for the joint replenishment problem with
The first inequality follows since Corollary 4.1 guaran- deadlines. J. Scheduling, 18(6):545–560, 2015.
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ondinequality followssince there are at mostD(T)+1 Lukasz Jez, Dorian Nogneng, and Jir´ı Sgall. Bet-
possible levels, and Lemma 4.3 shows that the sum of ter approximation bounds for the joint replenishment
cˆ(u) in each level is at most 2c(r(Ti)). problem. In Proceedings of the Twenty-Fifth An-
nual ACM-SIAM Symposium on Discrete Algorithms,
SODA 2014, Portland, Oregon, USA, January 5-7,
5 Conclusions
2014, pages 42–54, 2014.
In this paper we have presented an O(D)-competitive
[10] Marcin Bienkowski, Jaroslaw Byrka, Marek Chrobak,
algorithmfortheOnlineMulti-levelAggregationProblem Lukasz Jez, Jir´ı Sgall, and Grzegorz Stachowiak. On-
with Deadlines. This result represents an exponential linecontrolmessage aggregation inchain networks. In
improvementoverthe previouslybestcompetitive ratio Algorithms and Data Structures - 13th International
of D22D given by [7]. Nevertheless, the competitive Symposium, WADS 2013, London, ON, Canada, Au-
ratioofouralgorithmisstillquitefarfromtheconstant gust 12-14, 2013. Proceedings, pages 133–145, 2013.
lowerboundsprovedby[9]and[7]. Narrowingthisgap, [11] Edward Bortnikov and Reuven Cohen. Schemes for
scheduling control messages by hierarchical protocols.
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stronger lower bounds, is an intriguing open problem
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that we leave open.
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