Introduction to Online Algorithms

Transcription

1 Introduction to Online Algorithms Naveen Sivadasan Indian Institute of Technology Hyderabad

2 Online Computation In an online setting, the complete input is not known in advance. Input is a request sequence that is revealed gradually over time. Can be viewed as a request-answer game between the algorithm and an adversary. Algorithm has to perform well under the lack of information on future requests. Applications in scheduling, data structures, OS, networking etc.

3 Online Makespan Scheduling Givenmidentical machines. That is, processing time for a job is same across all machines. Consider a sequence of requestsσ = j 1,j 2,j 3,,j n of lengthn. Letj i denote the processing time of jobi. Each jobj i has to be assigned to exactly one machine. Once a job is assigned to a machine, it remains there. Objective is to minimize the completion time of the last finishing job (makespan).

4 Example

5 Example

6 Example

7 Example

8 Example

9 Example

10 Example

11 Example

12 Example

13 Example

14 Example

15 Example

16 Better Schedule

17 Competitive Ratio Compare the performance of the algorithm against offline optimal strategy. Letσ = σ(1),σ(2),σ(3),...,σ(t) denote atlength sequence. Requestσ(i) is revealed to the algorithm in roundi. LetA(σ) denote the cost incurred by the algorithmafor servingσ (Make span in prev. example) LetOPT(σ) denote the optimal cost incurred if the completeσ is known in advance. A is said to bec competitive ifa(σ) c OPT(σ)+afor any sequence σ. (Hereais some fixed constant)

18 Back to Makespan Scheduling Consider the following greedy approach : Schedule the new job to the machine having least total processing time. (Graham s list scheduling) The scheduling given in the previous example follows this approach. How competitive is this approach?

19 Competitive ratio of greedy Consider any request sequenceσ = j 1,j 2,...,j n. Focus on the makespan machine. Letw be the last job andr be the remaining total processing time. HenceA(σ) = r +w. Whenw was assigned greedily, all other machines also have load at leastr. Hencem r+w j 1 +j j n Observe thatopt(σ) is at least the average load and also the size of any one job. That is, m r+w m OPT(σ) and alsow OPT(σ). Putting together,a(σ) = r +w (2 1 m )OPT(σ).

20 Self-organizing lists Consider a listlofnelements{a 1,a 2,...,a n }. Cost of accessing an elementxinlisrank(x). Algorithm is allowed to reorganize the list using transposition of adjacent elements. Cost of a single transposition is1. Input is an online sequenceσ = x 1,x 2,x 3, of elements inl. Objective is to minimize the total cost of servingσ.

21 Self-organizing lists : Move To Front (MTF) algorithm Under standard worst case analysis, any algorithm would incur a cost of σ nif each request is to access the last element of the list. This would not allow us to compare algorithms. Let analyze a simple, practical algorithm under online setting. MTF (Move to Front): When an element is accessed, move it to front of the list. Hence accessing an elementxwould incur a cost at most2 rank(x).

22 Amortized analysis - Potential functions An aggregate cost analysis technique using potential functions. Consider a request sequenceσ of lengths. LetC i denote the cost of MTF to serveith request. LetC(σ) = s i=1 C i. Define a potential functionφ i that maps the state of the list afterirounds to a non negative real number. We will defineφsuch thatφ 0 = 0. Amortized cost of the algorithm in stepidenoted byĉi = C i +Φ i Φ i 1. Cost of servingσ is C(σ) = s i=1 (Ĉi +Φ i 1 Φ i ) = Φ 0 Φ s + s i=1ĉi s i=1ĉi

23 MTF - Amortized analysis We will bound cost of servingith request, sayx. Lets compare the list configurationsl i 1 andl i 1 of MTF and OPT respectively before serving ith request. Letr = rank(x) inl i 1 andr berank(x) inl i 1. HenceC i 2r. LetC i denote the OPT cost for servingith request. LetC i = r +t i, wheret i is the number of transpositions done by OPT. DefineΦ i 1 to be twice the number of inversions inl i 1 with respect to L i 1. That isφ i 1 is the no. of ordered pairs of elements inl i 1 that appear in opposite order inl i 1. ClearlyΦ i 1 0 andφ 0 = 0. (Both start with same list)

24 We now boundφ i Φ i 1 MTF - Amortized analysis r A C A B r x x B D C D L i 1 L i 1 x A B C D L i MTF step creates A new inversions and destroys B existing inversions. Note thatr = A + B +1andr = A + C +1. New inversions due tot i transpositions of OPT are at mostt i. Hence the potential difference Φ is, Φ i Φ i 1 2( A B +t i ) 4 A +2+2t i 2r 4C i 2r

25 MTF - Amortized analysis HenceĈi = C i +Φ i Φ i 1 2r +Φ i Φ i 1 4C i. SinceC(σ) s i=1ĉi, we obtainc(σ) 4C (σ). Thus MTF is4competitive.

26 The Metrical Task Systems Framework Start ALG[S] = 6+4, opt[s] = 2+4. c.ratio(s) =

27 The Metrical Task Systems Framework Start S = τ 1 ALG[S] = 6+4, opt[s] = 2+4. c.ratio(s) =

28 The Metrical Task Systems Framework Start S = τ 1 τ 2 ALG[S] = 6+4, opt[s] = 2+4. c.ratio(s) =

29 The Metrical Task Systems Framework Start S = τ 1 τ 2 τ 3 ALG[S] = 6+4, opt[s] = 2+4. c.ratio(s) =

30 The Metrical Task Systems Framework Start S = τ 1 τ 2 τ 3 τ 4 ALG[S] = 6+4, opt[s] = 2+4. c.ratio(s) =

31 Metrical Task Systems (MTS) Lower Bound Theorem 1 On anynstate metric space and for any deterministic algorithm the c.ratio is at least 2n 1. That is, a bad adversarial instance for the specified graph and the specified algorithm. There is an algorithm called work function algorithm that matches this lower bound.

32 Metrical Task Systems (MTS) Lower Bound Fix any deterministic algorithm A. Consider2n 1algorithmsB = {B 1,B 2,...,B 2n 1 } such that the following invariant is always maintained. One alg fromb occupy the same node asaand the rest of the nodes are occupied by exactly2algs fromb. IfAmakes a transition to vertexv fromuin a roundi, then one of the two algs fromv moves tou. Thus invariant is maintained. Letv i denote the node wherearesides afterirounds. The adversarial inputσ is such that in roundt, processing cost at nodev t 1 isand0everywhere else.

33 Lets = σ. Metrical Task Systems (MTS) Lower Bound IfAmakes totalk transitions to serveσ thena(σ) = (s k)+t, where T is the total travel cost. LetB(σ) denote the sum total of cost of all algs inb, which is 2n 1 i=1 B i (σ) Note thatb(σ) = (s k)+t +2k = A(σ)+2k. Also,OPT(σ) 1 2n 1 B(σ). HenceOPT(σ) 1 2n 1 (A(σ)+2k) 1 2n 1 A(σ)(1+2). That is,alg(σ)/opt(σ) 2n 1.

34 k-server problem There arek machines/servers that can move around in annnode weighted graph (which is a metric) No two servers reside in the same node. Each requestσ(i) is a node in the graph where the request should be served.

35 2 server σ =

36 2 server σ = 2

37 2 server σ = 2,1. A(σ) = total travel cost for servingσ.

38 k-server problem There arek machines/servers that can move around in annnode weighted graph (which is a metric) No two servers reside in the same node. Each requestσ(i) is a node in the graph where the request should be served. Algorithm can send any of thek servers to serve the request. Cost incurred in a step is the distance the chosen server has to travel to serve the request. Total cost on a request sequence is the sum of the travel cost in each round. Generalization of problems such as paging problem.

39 k-server problem Actively researched area to bound the competitive ratio on arbitrary metric and on special cases. It is known that the best possible competitive ratio lies betweenk and2k 1 for any arbitrary metric. It is still open whether the competitive ratio of the problem is exactlyk. It is conjectured so.

40 References [1] Allan Borodin and Ran El-Yaniv, Online Computation and Competitive Analysis, Cambridge Univ. Press, Cambridge, [2] S. Albers, Online algorithms: A survey, Mathematical Programming, 97:3-26, [3] J. Sgall, On-line scheduling - A survey, Online Algorithms: The State of the Art, LNCS. 1442, pages , Springer, [4] Elias Koutsoupias, The k-server problem, Survey, 2009.