1. www.cs.technion.ac.il/~reuven 1 Approximation Algorithms for Bandwidth and Storage Allocation Reuven Bar-Yehuda Joint work with Michael Beder, Yuval Cohen and Dror Rawitz Technion IIT Slides and paper at: http://www.cs.technion.ac.il/~reuven

2. www.cs.technion.ac.il/~reuven 2 The Local-Ratio Technique: Basic definitions The Local-Ratio Technique: Basic definitions Given a profit [penalty] vector p. Maximize[Minimize] p·x Subject to:feasibility constraints F(x) x is r-approximation if F(x) and p·x  [  ] r · p·x* An algorithm is r-approximation if for any p, F it returns an r-approximation

3. www.cs.technion.ac.il/~reuven 3 The Local-Ratio Theorem: The Local-Ratio Theorem: x is an r-approximation with respect to p 1 x is an r-approximation with respect to p- p 1  x is an r-approximation with respect to p Proof: ( For maximization) p 1 · x  r × p 1 * p 2 · x  r × p 2 *  p · x  r × ( p 1 *+ p 2 *)  r × ( p 1 + p 2 )*

4. www.cs.technion.ac.il/~reuven 4 Special case: Optimization is 1-approximation Special case: Optimization is 1-approximation x is an optimum with respect to p 1 x is an optimum with respect to p- p 1 x is an optimum with respect to p

5. www.cs.technion.ac.il/~reuven 5 A Local-Ratio Schema for Maximization[Minimization] problems: A Local-Ratio Schema for Maximization[Minimization] problems: Algorithm r-ApproxMax[Min]( Set, p ) If Set = Φ then return Φ ; If  I  Set p(I)  0 then return r-ApproxMax( Set-{I}, p ) ; [ If  I  Set p(I)=0 then return {I}  r-ApproxMin( Set-{I}, p ) ; ] Define “good” p 1 ; REC = r-ApproxMax[Min]( S, p- p 1 ) ; If REC is not an r-approximation w.r.t. p 1 then “fix it”; return REC;

6. www.cs.technion.ac.il/~reuven 6 The Local-Ratio Theorem: Applications Applications to some optimization algorithms (r = 1): ( MST) Minimum Spanning Tree (Kruskal) MST ( SHORTEST-PATH) s-t Shortest Path (Dijkstra) SHORTEST-PATH (LONGEST-PATH) s-t DAG Longest Path (Can be done with dynamic programming)(LONGEST-PATH) (INTERVAL-IS) Independents-Set in Interval Graphs Usually done with dynamic programming)(INTERVAL-IS) (LONG-SEQ) Longest (weighted) monotone subsequence (Can be done with dynamic programming)(LONG-SEQ) ( MIN_CUT) Minimum Capacity s,t Cut (e.g. Ford, Dinitz) MIN_CUT Applications to some 2-Approximation algorithms: (r = 2) ( VC) Minimum Vertex Cover (Bar-Yehuda and Even) VC ( FVS) Vertex Feedback Set (Becker and Geiger) FVS ( GSF) Generalized Steiner Forest (Williamson, Goemans, Mihail, and Vazirani) GSF ( Min 2SAT) Minimum Two-Satisfibility (Gusfield and Pitt) Min 2SAT ( 2VIP) Two Variable Integer Programming (Bar-Yehuda and Rawitz) 2VIP ( PVC) Partial Vertex Cover (Bar-Yehuda) PVC ( GVC) Generalized Vertex Cover (Bar-Yehuda and Rawitz) GVC Applications to some other Approximations: ( SC) Minimum Set Cover (Bar-Yehuda and Even) SC ( PSC) Partial Set Cover (Bar-Yehuda) PSC ( MSP) Maximum Set Packing (Arkin and Hasin) MSP Applications Resource Allocation and Scheduling : ….

7. www.cs.technion.ac.il/~reuven 7 Maximum Independent Set in Interval Graphs Maximum Independent Set in Interval Graphs time Maximize s.t. For each instance I: For each time t:

8. www.cs.technion.ac.il/~reuven 8 Maximum Independent Set in Interval Graphs: How to select P 1 to get optimization? Maximum Independent Set in Interval Graphs: How to select P 1 to get optimization? Î time Let Î be an interval that ends first; 1 if I in conflict with Î For all intervals I define: p 1 (I) = 0 else For every feasible x: p 1 ·x  1 Every Î- maximal is optimal. For every Î- maximal x: p 1 ·x  1 P1=1P1=1 P1=1P1=1 P1=1P1=1 P1=1P1=1 P1=0P1=0 P1=0P1=0 P1=0P1=0 P1=0P1=0 P1=0P1=0

9. www.cs.technion.ac.il/~reuven 9 Maximum Independent Set in Interval Graphs: An Optimization Algorithm Maximum Independent Set in Interval Graphs: An Optimization Algorithm Activity9 Activity8 Activity7 Activity6 Activity5 Activity4 Activity3 Activity2 Activity1 Î time Algorithm MaxIS( S, p ) If S = Φ then return Φ ; If  I  S p(I)  0 then return MaxIS( S - {I}, p); Let Î  S that ends first;  I  S define: p 1 (I) = p(Î)  (I in conflict with Î) ; IS = MaxIS( S, p- p 1 ) ; If IS is Î- maximal then return IS else return IS  {Î}; P1=0P1=0 P1=0P1=0 P1=0P1=0 P1=0P1=0 P1=0P1=0 P 1 =P (Î )

10. www.cs.technion.ac.il/~reuven 10 Maximum Independent Set in Interval Graphs: Running Example Maximum Independent Set in Interval Graphs: Running Example P(I 1 ) = 5 -5 P(I 4 ) = 9 -5 -4 P(I 3 ) = 5 -5 P(I 2 ) = 3 -5 P(I 6 ) = 6 -4 -2 P(I 5 ) = 3 -4 -5 -4 -2

11. www.cs.technion.ac.il/~reuven 11 banwith I w(I) s(I) e(I) time Maximize s.t. For each instance I: For each time t: Bandwidth Allocation Problem (BAP)

12. www.cs.technion.ac.il/~reuven 12 1987 Arkin and Silverberg: requests have same duration. 2000 Philips Uma Wein: 6 2001 Bar-Noy Bar-Yehuda Freund Naor Schieber: 3 2002 Chen Hassin Tzur: DP for restricted case. 2002 Calinescu, Chakrabarti, Karloff, Rabani:  2 Randomized Also dynamic programming 2003 Chekuri, Maydlarz, Shepherd: Extentions (e.g. trees) 2005 This talk:  2+1/(e-1)  2.582 Bandwidth Allocation Problem (BAP)

13. www.cs.technion.ac.il/~reuven 13 Î I w(I) s(I) e(I) time 1 if I = Î For all intervals I define: p 1 (I) = 2*w(I) if I is in conflict with Î 0 else For every feasible x: p 1 ·x  1+2*1 Every Î- maximal is 1/3-approximation For every Î- maximal x: p 1 ·x  1 Bandwidth Allocation for w  1/2 How to select P 1 to get 1/3-approximation? Bandwidth Allocation for w  1/2 How to select P 1 to get 1/3-approximation?

14. www.cs.technion.ac.il/~reuven 14 Î I w(I) s(I) e(I) time 1 if I = Î For all intervals I define: p 1 (I) = w(I)/ Ŵ if I is in conflict with Î 0 else Where Ŵ =1- w( Î )  feasible x: p 1 ·x  max{ 1 + Ŵ / Ŵ, 1 / Ŵ } Î - maximal is 1/2-approximation  Î- maximal x: p 1 ·x  min{ 1, Ŵ / Ŵ } Bandwidth Allocation for w  1/2 … A better P 1 to get 1/2-approximation? Bandwidth Allocation for w  1/2 … A better P 1 to get 1/2-approximation?

15. www.cs.technion.ac.il/~reuven 15 Bandwidth Allocation: 1/3-approximation in O(nlogn) time w > ½ w > ½ w > ½ w > ½ w > ½ w > ½ w > ½ w > ½ w > ½ Algorithm: 1. GRAY = Find 1-approximation for gray (w>1/2) intervals; 2. COLORED = Find 1/2-approximation for colored intervals 3. Return the one with the larger profit Analysis: If GRAY*  33%OPT then GRAY  1*(33%OPT)=33%OPT else COLORED*  66%OPT thus COLORED  1/2(66%OPT)=33%OPT

16. www.cs.technion.ac.il/~reuven 16 Bandwidth Allocation: What if we get R-approx for w >  and r-approx for w ≤  ? w >  w >  w >  w >  w >  w >  w >  w >  w >  Algorithm: 1. GRAY = Find R-approximation for gray (w>  ) intervals; 2. COLORED = Find r-approximation for colored intervals 3. Return the one with the larger profit Analysis: If GRAY *  r/(R+r)×OPT then GRAY  R× r/(R+r)×OPT else COLORED *  R/(R+r)×OPT thus COLORED  r× R/(R+r)×OPT

17. www.cs.technion.ac.il/~reuven 17 Banwidth Allocation: 1-approx for w >  Algorithm: 2002 Chen Hassin Tzur 2002 Calinescu, Chakrabarti, Karloff, Rabani Dynamic Programming  R = 1

18. www.cs.technion.ac.il/~reuven 18 Î 1/k I w(I) s(I) e(I) time 1-1/k if I = Î For all intervals I define: p 1 (I) = k × w(I) if I is in conflict with Î 0 else For every 1-feasible x: p 1 ·x  k*1 +1 1/k-Î - maximal is  k+3-apx For every 1/k- Î- maximal x: p 1 ·x  1-1/k Bandwidth Allocation:  1-1/e-apx for w ≤  Let 1/k 2 ≥  Bandwidth Allocation:  1-1/e-apx for w ≤  Let 1/k 2 ≥  1/k 2

19. www.cs.technion.ac.il/~reuven 19 Bandwidth Allocation: 1-1/e-approx for w ≤  Algorithm: Iteratively: Loose 1/k bandwidth and gain 1/(k+3) fraction from the residual opt ≥ Gain 1 ≥ (1/(k+3)) ×OPT ≤ Left 1 ≤ (1-1/(k+3)) ×OPT ≥ Gain 2 ≥ (1/(k+3)) ×Left 1 ≤ Left 2 ≤ (1-1/(k+3)) 2 ×OPT ≤ Left 3 ≤ (1-1/(k+3)) 3 ×OPT ≤ Left k ≤ (1-1/(k+3)) k ×OPT ≥ Gain ≥ OPT- Left k >  (1-1/e)×OPT 

20. www.cs.technion.ac.il/~reuven 20 Storage Allocation Problem (SAP)

21. www.cs.technion.ac.il/~reuven 21 2000 Philips, Uma, Wein: 35 2000 Leonardi, Marchetti-Spaccamela, Vitaletti: 12 2001 Bar-Noy, Bar-Yehuda, Freund, Naor, Schieber: 7 2002 Chen, Hassin, Tzur: DP 2005 This talk:  2+1/(e-1)  2.582 Storage Allocation Problem (SAP)

22. www.cs.technion.ac.il/~reuven 22 Storage Allocation Problem (SAP) for w>  =  Let k =  1/  + Gravity  O(n k ) possible values for H H   Dynamic programming in polynomial time for constant 

23. www.cs.technion.ac.il/~reuven 23 For w ≤  R-apx for BAP   R-apx for SAP Dynamic Storage Allocation (DSA)  Find Min bandwidth with “perfect” SAP allocation Run BAP (w ≤  ) If Load ≤ 1/3 // Load = Max total height at time t use Gergov 99 to get DSA in bandwidth ≤ 1 Else use 2003 Buchsbaum, Karloff, Kenyon, Reingold and Thorup to get DSA in bandwidth ≤ 1 +O(  1/7 ) Corollary: for w ≤  :  1-apx randomized  1.582-apx (deterministic and combinatorial) Corollary: 2 randomized and 2.582 deterministic

24. www.cs.technion.ac.il/~reuven 24 Bandwidth Allocation on trees. 2003 Chekuri, Maydlarz, Shepherd: Input: Capacitated tree, where each edge has an integral capacity, and a set of paths, each of which has an integral demand and a profit. Solution: A set of paths such that the demands of the paths can be simultaneously routed without violating edge capacities. Results:  4-approx for the case of unit demands.  48-approx for the case where max(demand) ≤ min(capacity) Our results:  We consider the uniform capacities bounded degree case Note: in this case max(demand) ≤ min(capacity)  (2+1/(e 1/2 -1)+  )-approx  3.54 -aprox  (2+1/(e 1/d -1)+  )-approx when we are given a set of trees of degree d (instead of paths).

25. www.cs.technion.ac.il/~reuven 25 Bandwidth Allocation of uniform capacities general demands trees on trees.

26. www.cs.technion.ac.il/~reuven 26 Bounded degress trees: Bandwidth Allocation: 1-approx for w >  Dynamic Programming  R = 1

27. www.cs.technion.ac.il/~reuven 27 Bandwidth Allocation of uniform capacities general demands trees on trees. 1-1/k if I = Î For all intervals I define: p 1 (I) = k×w(I) if I is in conflict with Î 0 else For every 1-feasible x: p 1 ·x  k*d +1 1/k-Î-maximal is  d(k+1)-apx For every 1/k-Î-maximal x: p 1 ·x  1-1/k Let 1/k 2 ≥ 

28. www.cs.technion.ac.il/~reuven 28 Algorithm: Iteratively: Loose 1/k bandwidth and gain 1/(dk+d) fraction from the residual opt ≥ Gain 1 ≥ (1/(dk+d)) ×OPT ≤ Left 1 ≤ (1-1/(dk+d)) ×OPT ≥ Gain 2 ≥ (1/(dk+d)) ×Left 1 ≤ Left 2 ≤ (1-1/(dk+d)) 2 ×OPT ≤ Left 3 ≤ (1-1/(dk+d)) 3 ×OPT ≤ Left k ≤ (1-1/(dk+d)) k ×OPT ≥ Gain ≥ OPT- Left k >  (1-e 1/d )×OPT  Bandwidth Allocation:  -apx for w ≤  Bandwidth Allocation:  -apx for w ≤ 