1 / 121

Clearly

First lecture in Mathematics. Clearly. 1 + 1 = 2. But this is a complicated way to write this. Hereby we suggest some improvement. Step 1. and. and. 1 + 1 = 2. So can be simplifid as. which is clearly more intuitive. Step 2. and. So. 1 + 1 = 2.

tracen
Télécharger la présentation

Clearly

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related