Clearly

1. First lecture in Mathematics Clearly 1 + 1 = 2 But this is a complicated way to write this. Hereby we suggest some improvement. Bertinoro, 5/5/08

2. Step1 and and Bertinoro, 5/5/08

3. 1 + 1 = 2 So can be simplifid as which is clearly more intuitive Bertinoro, 5/5/08

4. Step2 and Bertinoro, 5/5/08

5. So 1 + 1 = 2 can be further simplifid as Bertinoro, 5/5/08

6. Step2 and hence Bertinoro, 5/5/08

7. 1 + 1 = 2 So eventually gets its final form: Now, decide for yourself: 1. which of the forms is the simplest. 2. which of the forms will make your career faster? 3. which of the forms will mostly impress your boss/boyfriend/girlfriend/mother? Bertinoro, 5/5/08

8. 1.2. Cost functions We are given lightpaths we want to get a coloring S such that cost(S) is minimal. What is cost(S) ? Bertinoro, 5/5/08

9. 1.2. Cost functions 1. number of wavelengths #colors = 4 Bertinoro, 5/5/08

10. 2. Switching cost OADM ADM #ADMs + #OADMs Bertinoro, 5/5/08

11. ADM OADM #ADMs + #OADMs = 12 + 8 Bertinoro, 5/5/08

12. but number of OADMs is fixed, so … Bertinoro, 5/5/08

13. 2. number of ADMs #ADMs = 12 Bertinoro, 5/5/08

14. #ADMs=12 #ADMs=9 Bertinoro, 5/5/08

15. Trade-off between #colors and #ADMs #ADMs=12 #colors=4 #ADMs=9 #colors=3 Bertinoro, 5/5/08

16. #ADMs=8 #colors=2 #ADMs=7 #colors=3 Bertinoro, 5/5/08

17. g=2 With grooming #ADMs=9 #ADMs=8 Bertinoro, 5/5/08

18. 1.3. Problems in this talk Minimize the number of ADMs with and without grooming • Complexity • special networks, general networks • Approximation algorithms • on-line Bertinoro, 5/5/08

19. Traffic Grooming Ring Flammini, Moscardeli, Gianpierro, Shalom, Z. 2005/6/7 On-line Shalom, Wong, Zaks, 2007 Bertinoro, 5/5/08

20. 2.1 approximation ratio N: # of lightpaths ALG: #ADMs used by the algorithm OPT: #ADMs used by an optimalsolution ALG  2N N  OPT ALG  2 xOPT Bertinoro, 5/5/08

21. R: # of lightpaths ALG: #ADMs used by the algorithm OPT: #ADMs used by anoptimalsolution w/out grooming: N  ALG  2N N  OPT  2N ALG  2 x OPT w/ grooming: N/g  ALG  2N N/g  OPT  2N ALG  2g x OPT Bertinoro, 5/5/08

22. 2.2 Basic relation Cycles are good, chains are bad N lightpaths cycles chains #ADMs = N + #chains Bertinoro, 5/5/08

23. #ADMs = N + #chains In the approximation algorithms there are two common techniques for saving ADMs: Eliminate cycles of lightpaths Find matchings of lightpaths Bertinoro, 5/5/08

24. 2.3 Note Min ADM problem: (cost=#ADMs) Connections are good, chains are bad N lightpaths cost(S) = N + chains=13+6=19 • Every path costs 1 ADM cost(S) = 2N-savings=26-7=19 • Every connection saves 1 ADM N=13 Bertinoro, 5/5/08

25. 2.4 a basic lemma Assume that an optimal solution S* savesx ADMs, and a solution S savesy ADMs Bertinoro, 5/5/08

26. Optimal solution S* savesx ADMs a solution S savesy ADMs Bertinoro, 5/5/08

27. 2.5 notaton D0(S) – lightpath not sharing any ADM D1(S) – lightpath sharing ONE ADM D2(S) – lightpath sharing BOTH ADMs N=25 lightpaths d0(S) = 2 d1(S) = 4 d2(S) = 19 d0(S) +d1(S)+d2(S)= 25 = N Bertinoro, 5/5/08

28. 2.6 a basic tool Solution S=chains + cycles #ADMs = N + #chains S – ALG, S* - OPT Bertinoro, 5/5/08

29. We define: and get Bertinoro, 5/5/08

30. 2.7 example D0(S) – lightpath not sharing any ADM D1(S) – lightpath sharing ONE ADM D2(S) – lightpath sharing BOTH ADMs N=25 lightpaths d0(S) = 2 d1(S) = 4 #ADMs=29 d2(S) = 19 d0(S) +d1(S)+d2(S)= 25 = N Bertinoro, 5/5/08

31. N=25 suppose #chains(S*)=2, cost(S*)=25+2=27 d0(S) = 2 d2(S) = 19 Bertinoro, 5/5/08

32. 3.3 minADM with grooming is NP-complete for a star path of length 1 path of length 2 Bertinoro, 5/5/08

33. Star, g=1 ( trivial ) path of length 1 path of length 2 Bertinoro, 5/5/08

34. Star, g=2 The number of used ADM is exactly equal to the lower bound of needed ADM: 2 1 xi paths of length 2 yi paths of length 1 0 i n nodes 1,…,n node 0 Bertinoro, 5/5/08

35. Star, g≥3 - NP-complete Sketch for g=3: 3-Exact Cover  Edge Partition into 3-regular graphs  Star grooming, g=3 Bertinoro, 5/5/08

36. 3-Exact Cover: Input: set A of size 3n, and a collection S of subsets of A of size 3 each. Output: are there n subsets in the collection S that cover A? Bertinoro, 5/5/08

37. Edge Partition into 3-regular graphs: Input: undirected graph G = (V,E). Output: can E be partitioned into subsets E1,…,Em , each inducing a 3-regular subgraph G=(Vt,Et), t=1,…,m ? Bertinoro, 5/5/08

38. 3-Exact Cover  Edge Partition into 3-regular graphs  Star grooming, g=3 Bertinoro, 5/5/08

39. sets elements Bertinoro, 5/5/08

40. 3-Exact Cover  Edge Partition into 3-regular graphs  Star grooming, g=3 Bertinoro, 5/5/08

41. 4 S 1 4 G Claim: there exists a solution using at most 2|E|/3 ADMs iff the edges of G can be partitioned into 3-regular graphs. 3 1 0 3 2 2 Bertinoro, 5/5/08

42. 4 5 5 4 3 1 1 0 3 2 2 =2 g=2 Bertinoro, 5/5/08

43. 4.1 basic algorithm eliminate short cycles, then find matchings • Preprocessing: • While there is a cycle C of length ≤ l do: • Remove (the lightpaths of) C from the instance • Processing: • Designate each lightpath as a chain • Do • Build the matching graph M of the chains • Find a maximum matching MM of M • Combine chains according to MM • Until M has no edges Bertinoro, 5/5/08

44. The running time of the algorithm is exponential in l due to the preprocessing phase • By removing the preprocessing phase (l=1) we obtain algorithm PIM(1) • Recall: Bertinoro, 5/5/08

45. algorithm PIM(l) Bertinoro, 5/5/08

46. Need to prove: We will show: Bertinoro, 5/5/08

47. Take S*: in the example : 38 lightpaths, 5 chains, 3 cycles Bertinoro, 5/5/08

48. preprocessing stage of S: eliminate cycles of The lightpaths that we used are colored red: in the example : 10 lightpaths are used to form cycles in S Bertinoro, 5/5/08

49. So, after preprocessing stage of S: in S* we have the following lightpaths: Bertinoro, 5/5/08

50. Now the algorithm is doing MM,MM,MM,… We show that already after the first MM we are ok. Bertinoro, 5/5/08