The Online Set Aggregation Problem

Rodrigo A. Carrasco; Kirk Pruhs; Cliff Stein; José Verschae

doi:10.1007/978-3-319-77404-6_19

Latin American Symposium on Theoretical Informatics

LATIN 2018: LATIN 2018: Theoretical Informatics pp 245-259 | Cite as

The Online Set Aggregation Problem

Authors
Authors and affiliations

Rodrigo A. CarrascoEmail author
Kirk Pruhs
Cliff Stein
José Verschae

Conference paper

First Online: 13 March 2018

Part of the Lecture Notes in Computer Science book series (LNCS, volume 10807)

Abstract

We introduce the online Set Aggregation Problem, which is a natural generalization of the Multi-Level Aggregation Problem, which in turn generalizes the TCP Acknowledgment Problem and the Joint Replenishment Problem. We give a deterministic online algorithm, and show that its competitive ratio is logarithmic in the number of requests. We also give a matching lower bound on the competitive ratio of any randomized online algorithm.

Keywords

Online algorithms Competitive analysis Set aggregation Multilevel aggregation

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A Detailed Proofs

In this section we detail some of the proofs from Sect. 3.

Lemma 1. There exists a proactive schedule whose objective value is at most twice optimal.

Proof

We show how to iteratively transform an arbitrary schedule \(\mathcal {Z}\) into a proactive schedule in such a way that the total cost at most doubles. Let t be the next time in \(\mathcal {Z}\) when there is an unserved set S, with the property that the waiting time for S, infinitesimally after t, is greater than the service cost for S. We then add the set S at time t to \(\mathcal {Z}\). The service cost of this set is at most the total waiting time of the requests that it serves. Thus the total service cost of the final schedule is at most the waiting time in the original \(\mathcal {Z}'\). Further the transformation can only decrease the total waiting cost, since the requests in S are being served no later than they were originally.

Lemma 2. The value \(\text {LB}^-(s, t, d)\) and \(\text {LB}^+(s, t, d)\) are monotone on the set of requests, that is, adding more requests between times s and t can not decrease either. Moreover, \(\text {LB}^-(s, t, d)\) and \(\text {LB}^+(s, t, d)\) are non-decreasing as a function of t and as a function of d, for any \(s\le t\le d\).

Proof

Any schedule that is proactive for the larger set of requests is also proactive for the smaller set of requests, because the waiting time of the requests serviced can not be more in the smaller set of requests than in the larger set of requests. The monotonicity on t follows directly from the monotonicity on the set of requests. The monotonicity on d is clear since for any \(d\le d'\), a proactive solution up to time \(d'\) is also proactive up to time d.

Lemma 3. Assume that RetrospectiveCover just computed a new deadline d[i] in line 7. Then process i will either reach a new milestone or be terminated by time d[i].

Proof

If d[i] is infinite, then this is obvious, so assume otherwise. If process i is terminated before time d[i] then this is obvious, so assume otherwise. If no requests arrive during the time interval (m[i], d[i]) then process i will reach a new milestone exactly at time d[i] by the definition of milestones. If requests arrive before d[i] then the claim follows by the monotonicity of \(\text {LB}^+\), see Lemma 2.

Lemma 5. It holds that \( \sum _{j=1}^{k-1} x_j \le x_0\).

Proof

First notice that \(\text {LB}^{-}(u_{-1},t,t)\le 2\text {LB}^{+}(u_{-1},u_0,u_0)=2x_0\), since otherwise process i would have hit a milestone within the interval \((u_{0},t)\). Then, it suffices to show that \( \sum _{j=0}^{k-1} x_j\le \text {LB}^{-}(u_{-1},t,t)\). To show this last bound, fix \(j\in \{0,\ldots ,k-1\}\) and consider a proactive schedule \(\mathcal {Z}\) for requests arriving in \((u_{-1},t)\). Within each interval \((u_{j-1},u_j)\), the solution is proactive when considering requests with release date in \((u_{j-1},u_j]\), and hence the serving cost of the requests served by \(\mathcal {Z}\) within this interval is at most \(\text {LB}^+(u_{j-1},u_{j-1},u_j)\). We remark that this is also true for \(j=k-1\) as \(u_{k-1}<t\). Taking \(\mathcal {Z}\) minimizing the service cost within \((u_{-1},t)\), we obtain the required bound, \( \sum _{j=0}^{k-1} x_j\le \text {LB}^{-}(u_{-1},t,t)\).

Lemma 6. For any set S serviced at time t in the schedule produced by the algorithm RetrospectiveCover, it will be the case that \( \mathop { W _t}(S) \le 2 \mathop { C }(S)\).

Proof

Assume that a process i hit a milestone at time t. We adopt the notation from Subsubsect. 3.3.1 and illustrated in Fig. 1. Additionally let \(U_j\) be the requests that are released in the time interval \((u_{j-1}, u_j)\) for \(j \in \{0,1,\ldots , k\}\).

We now prove that the sets serviced in line 5 have waiting time at most twice the service cost. Also, we show that after serving such sets, the set \(\cup _{h=k-j}^k U_h\) has no violated subset at time t, where \(j=a-\ell \). We show this by induction on \(j=a-\ell \). The base case is when \(j=0\) (\(\ell =a\)). There is no violated subset of \(U_k\) at time t since process \(i + k=a\) did not hit a milestone before time t. Thus, no set serviced in \(\text {LB}^-(s[a], t, t)\) is violated. Obviously, after servicing such sets, \(U_k\) still has no violated subset. Now let us show the two properties for an arbitrary j. By induction hypothesis, the set \(\cup _{h=k-(j-1)}^k U_h\) has no violated subset at time t. There is no violated subset of \(U_{k-j}\) at time t since the servicing of sets in \(\text {LB}^-(s[a- j], m[a-j], d[a-j])=\text {LB}^-(u_{k- j - 1}, u_{k-j}, d[a-j])\) at time \(m[a-j]\) guaranteed that \(U_{k-j}\) would not have any violated subsets until after time \(d[a-j]\) and \(t \le d[a-j]\). Thus no set serviced in \(\text {LB}^-(s[a-j], t, t)\) in line 5 has waiting time at most twice the service cost. Further since \(\text {LB}^-(s[a-j], t, t)=\text {LB}^-(u_{k-j-1}, t, t)\) is a proactive schedule, after serving sets in such solution the set \(\cup _{h=k-j}^k U_h\) has no violated subset at time t.

Finally the same argument can be applied to the sets serviced in line 8 in RetrospectiveCover.

Theorem 2. The RetrospectiveCover algorithm is \(O(\log |R|)\)-competitive.

Proof

Applying Lemma 4 to the original process, and noting that the service cost in the final invocation of line 8 is at most twice the previous service costs, one obtains that the service cost for the algorithm is \(O(\log |R|)\) times the lower bound. By Lemma 1 the lower bound is at most twice the optimal, and by Lemma 6 the waiting cost for requests serviced by the algorithm is at most the service cost of these requests. Finally the waiting cost of any unserviced requests is at most twice the service cost of the algorithm.

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Authors and Affiliations

Rodrigo A. Carrasco
- 1
Email authorView author's OrcID profile
Kirk Pruhs
- 2
Cliff Stein
- 3
José Verschae
- 4

1.Facultad de Ingeniería y CienciasUniversidad Adolfo IbáñezSantiagoChile
2.Department of Computer ScienceUniversity of PittsburghPittsburghUSA
3.Department of Industrial Engineering and Operations ResearchColumbia UniversityNew YorkUSA
4.Facultad de Matemáticas & Escuela de IngenieríaPontificia Universidad Católica de ChileSantiagoChile

The Online Set Aggregation Problem

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