1. 1 Distributed Computing Optical networks: switching cost and traffic grooming Shmuel Zaks zaks@cs.technion.ac.il ©

2. 2 the fiber serves as a transmission medium Electronic switch Optic fiber Optical networks - 1 st generation

3. 3 Routing in the optical domain Two complementing technologies: - Wavelength Division Multiplexing (WDM): Transmission of data simultaneously at multiple wavelengths over same fiber - Optical switches: the output port is determined according to the input port and the wavelength Optical networks - 2 nd generation

4. 4 Wavelength Division Multiplexing (WDM) Directed: Symmetric: Undirected: Optic Fiber

5. 5 Optical Switches No two inputs with the same wavelength should be routed on the same edge.

6. 6 Lightpaths ADM Data in electronic form

7. 7 A virtual topology

8. 8 Lightpaths p1 p2 Valid coloring

9. 9 The Routing Problem Input : A graph G=(V,E) A set or sequence of node pairs (a i,b i ) Output: A set or sequence of paths p i =(a i, v 1, …, b i )

10. 10 The Load Given a graph G=(V,E) and a set P of paths on the graph, we define: for any edge e of the graph: the load on this edge l(e)=|P e | The (maximum, minimum, average) load on the network:

11. 11 Wavelength Assignment Problem (WLA) Input: A graph G=(V,E). A set or sequence of paths P. Output: A coloring w of the paths: Constraint:

12. 12 Routing and WLA (RLA/WRA) Input : A graph G=(V,E) A set or sequence of node pairs (a i,b i ) Output: A set or sequence of paths p i =(a i, v 1, …, b i ) A coloring w of the paths: Constraint:

13. 13 Cost Measure: # of colors For any legal coloring w of the paths:

14. 14 Electronic ADM

15. 15 Cost Measure: # of ADMs Each lightpath requires 2 ADM ’ s, one at each endpoint, as described before. A total of 2|P | ADM ’ s. But two paths p=(a, …,b) and p ’ =(b, …,c), such that w(p)=w(p ’ ) can share the ADM in their common endpoint b. This saves one ADM. For graphs with max degree at most 2 we define:

16. 16 Optimization Problems Goal: MINW: Minimize W. or MAXPC: Maximize |Domain(w)| under the constraint W<=W max. or MINADM: Minimize ADM.

17. 17 Static vs. Dynamic vs. Incremental Static: The input is a set (of pairs or paths), the algorithm calculates its output based on the input. Incremental (Online): The input is a sequence of input elements (pairs or paths). It is supplied to the algorithm one element at a time. The output corresponding to the input element is calculated w/o knowledge of the subsequent input elements. Dynamic: Similar to incremental The sequence may contain deletion requests for previous elements.

18. 18 WLA (A trivial lower bound) For any instance of the WLA problem: W>=L. Proof: Consider an edge e, such that L=l(e). There are L paths p 1, …, p |L| using e, because the paths are simple. Therefore :

19. 19 WLA (A trivial lower bound) For some instances W > L. L=2 W=3

20. 20 Static WLA in Line Graphs The GREEDY algorithm: // The set of integers for i = 1 to |V| do for each path p=(x,i) do for each path p=(i,x) do

21. 21 Static WLA in Line Graphs Correctness, obvious. Optimality: By induction, After node i is processed, the claim is correct, i.e. Where W(i) is the value of after node i is pocessed, and L(i) is the maximum load on the edges processed so far.

22. 22 Static WLA in Line Graphs Note: After a slight modification, the algorithm solves optimally the MINADM problem too: At each node, first use the colors added to at this step. It ’ s straigtforward to show that this: Does not harm the optimality w.r. to the MINW prb. Solves the MINADM problem optimally at each node.

23. 23 number of wavelengths Switching cost ADM

24. 24 W=2, ADM=8 W=3, ADM=7

25. 25 ring (Eilam, Moran, Zaks, 2002) reduction from coloring of circular arc graphs. NP-complete

26. 26 Coloring of Circular arc Graphs Consider: a ring H (the host graph) and A set of paths P in H. The graph G=(P,E) constructed as follows is a circular arc graph: There is an edge (p1,p2) in e if and only if p1 and p2 have a common edge in H. The problem of finding the chromatic number of a circular arc graph is NP-Hard [Tuc 75 ’ ]

27. 27 The reduction The min W problem is exactly the circular arc coloring problem. But we will show NP- hardness even of the special case L=L min. Given an instance C,P where C is the ring and P is the set of paths, we construct an instance C, P ’ (by adding paths to P) such that L min (P ’ )=L(P ’ )=L(P).

28. 28 The reduction (cont ’ d) Claim: P is L-colorable iff P ’ is L-colorable. Claim: ADM(P)=ADM(P ’ ). Therefore w.l.o.g. all the edges have the same load (L).

29. 29 |ADMs|=7=7+0 |ADMs|=9=6+3 |ADMs| = N + |chains| Basic observation N lightpaths cycles chains

30. 30 The reduction (cont ’ d) P ’ is L-colorable iff P ’ can be partitioned into L cycles iff ADM(P ’ )=|P ’ |.

31. 31 R  ALG  2R R  OPT  2R ALG  2 x OPT R: # of lightpaths ALG: # of ADMs used by the algorithm OPT: # of ADMs used by optimal solution Approximation algorithms

32. 32 3/2 - Calinescu, Wan, 2002 10/7+  - Shalom, Z., 2004 10/7 - Epstein, Levin, 2004 ALG  2 x OPT Approximation algorithms

33. 33 Algorithm PIM Preprocessing: While there is a cycle C in the instance 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

34. 34 Analysis of PIM Consider an optimal partition of the paths into chains and cycles. OPT=|P|+#Opt_Chains

35. 35 Analysis of PIM (cont ’ d) Let ’ s call the edges removed in the preprocessing phase “ blue ” (B), and the rest “ red ” (R).

36. 36 Analysis of PIM (cont ’ d) k=#Red_Chains ≤ #Opt_Chains + |B| = OPT - |P| + |B| = OPT - |R| There is a matching of size (|R|-k)/2 in the red chains. The algorithm will find this (or a better) matching. The number of chains will reduced to at most

37. 37 Analysis of PIM (cont ’ d) The total cost of the algorithm is at most ALG ≤ |P| + OPT/2 ≤ (3/2). OPT

38. 38 Lower Bound of PIM PIM (or any other algorithm doing “ oblivious ” cycle removal) can not do better than 3/2 – approximation [EL 04 ’ ]

39. 39 Lower Bound of PIM (cont ’ d) OPT consists of 3 cycles. (6 ADMs) ALG may break all of them (3+2*3=9 ADMs)

40. 40 Lower Bound of PIM The example just given is too simplistic. Actually it is easy to show that, given any cycle of two paths, there is an optimal solution containing it as a cycle. Therefore they can be removed greedily w/o affecting optimality. There is a far more complicated construction that uses unbounded rings showing that 3/2-  is impossible.

41. 41 Eilam, Moran, Z., 2002 EMZ <= OPT + 0.6 N Calinescu, Frieder, Wan, 2002 Algorithm MCC-WS: MCC-WS <= OPT + 0.6 N General Topology

42. 42 The Instance An instance of the problem is  = (G,P) where G is ANY graph P is a muti-set of simple paths on G. x u v w b a c d e

43. 43 The Shareability Graph G  The set of nodes is P There is an edge e=(p 1,p 2 ) labeled x if: p 1 and p 2 have a common endpoint p 1 and p 2 do not share an edge a b c de w v x x u x x

44. 44 The Conflict Graph The set of nodes is P There is an edge e=(p 1,p 2 ) labeled x if: p 1 and p 2 share an edge a b c de

45. 45 The Conflict Graph a b c de a b c de w v x x u x x The Shareability Graph x u v w b a c d e f f f

46. 46 A solution A Partition of P into feasible chains and cycles Feasible  1-colorable x u v w b a c d e

47. 47 The Sharing Graph G S of S Partition G  into paths and cycles. Each path or cycle satisfies: No pair of consecutive edges have same label Independent set of the conflict graph a b c de w v x x u x x a b c de

48. 48 Solution S : Partition Ga into paths and cycles. d i (S): # of nodes of degree i in S. E S : edges of G S P S : paths of S S*: An optimal solution

49. 49 36 lightpaths d 0 = 3 d 1 = 4 d 2 = 29

50. 50 Cost(S) = 2N-|E S | Every edge of G S is a “ saving ” Cost(S) = N + |P S | Every Path of S is a “ loss ”

51. 51 S – ALG, S* - OPT Solution S=chains + cycles |ADMs| = N + |chains|

52. 52 We define: and get

53. 53 36 lightpaths d 0 = 3 d 1 = 4 d 2 = 29 cost = N+5= 41

54. 54 An approximation algorithm PIM(l) G.Calinescu and P.-J.Wan[CW02]

55. 55 Algorithm Preprocessing: While there is a cycle C of length l or less in the instance 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

56. 56 Performance of PIM(l)

57. 57 The running time of the algorithm is exponential in l = O(1/  ) due to the preprocessing phase By removing the preprocessing phase (l=1) we obtain algorithm PIM(1) with performance guarantee of. This implies PIM(1) ≤ OPT + 2/3 N Algorithm PIM

58. 58 Flammini-Shalom-Zaks[FSZ09] A more refined analysis

59. 59 w/out preprocessing, l =1: improve: l =1 (no preprocessing)

60. 60 Sample Run of PIM(3) x u v w b a c d e

61. 61 Sample Run of PIM(3) a b c de w v x x u x x COST=5

62. 62 Sample Run of PIM(1) x u v w b a c d e

63. 63 Sample Run of PIM(1) a b c de w v x x u x x COST = 8 = OPT + 3 = = OPT + 0.6x5

64. 64 Lower Bound The performance of PIM(1) can be as bad as OPT + 0.6 N We will show that this is also an upper bound...

65. 65 Proof (upper bound) Orient the paths and cycles of in arbitrary directions. Let LAST be the set of nodes which are last elements of the paths according to this orientation. p p Edge of G S Edge of G S*

66. 66 Proof We will map each node p of D 0 (S) Either to a node p ’ of D 2 (S) Or to a node p ’’ of LAST Or to a some We show that this mapping is “ 1-to-1 ” : All the nodes p ’ are distinct All the nodes p ” are distinct All the subsets C are disjoint Recall: algorithm is doing IM

67. 67 The Mapping q 0 =p If q 0 is the last node of a path of S* then: p ’ =q 0 map p to p ’ return Otherwise, q 1 is the next node in q 0 ’ s path/cycle in S* q1q1 q 1 can not be in D 0 (S), otherwise the algorithm would add the edge (q 0,q 1 ) to the matching. If q 1 is in D 2 (S) then: p ” =q 1 map p to p ” return q2q2 Otherwise q 1 has exactly one neighbor q 2 in G S. Obviously q 2 is not in D 0 (S). If q 2 is in D 2 (S) then: p ” =q 2 map p to p ” return If q 2 is the last node of a path of S* then: p ’ =q 2 map p to p ’ return Otherwise, q 3 is the next node in q 2 ’ s path/cycle in S* q3q3 q 3 can not be in D 0 (S), otherwise the above path would be an augmenting path of the maximum matching found by the algorithm. As the graph is finite, this process continues until p is mapped, or we re-encounter a node. In this case: C = {q i,... } Map p to C return It is easy to see that |C| is odd. We also show that |C| > 3.

68. 68 All the possible edges between the two cycles are in the conflict graph The matching leaves only one isolated node, therefore maximum. There are no feasible cycles of length <= l. l +2 nodes l +1 nodes The matching can not be extended to a bigger solution, because of the conflict edges. l >1 (with preprocessing) - lower bound

69. 69 d 0 (S)=1 d 2 (S)=0 |P S* |=0 N=2 l +3 l +1 nodes l +2 nodes

70. 70 Simplifying Assumptions We will demonstrate the proof under the following simplifying (but worst case) assumptions: 1)d 0 (S*)=d 1 (S*)=0, i.e. the optimal solution consists of cycles only. 2)d 2 (S)=0 The preprocessing phase did not find any cycles. For each cycle C, |C| > l. l >1 (with preprocessing) - upper bound

71. 71 Note that the example in the lower bound satisfies the assumptions. Under these assumptions: = number of isolated nodes per node.

72. 72 [CFW02] : There is a matching M with one isolated node per odd cycle.  By our assumptions, each odd cycle has at least l +2 nodes.  At most N/( l +2) isolated nodes

73. 73 In the rest of the talk: we show how to prove that a larger matching exists, i.e. a matching having smaller d 0 (M).

74. 74 A chord of S Lemma: For every solution S, there is a solution S’ with no chords, such that cost(S’)=cost(S). Corollary: There is an optimal solution S* with no chords. Step 1: chordless

75. 75 No chords - proof p1p1 p6p6 p5p5 p4p4 p3p3 p2p2 b a d c b b a b c d p1p1 p2p2 p3p3 p4p4 p5p5 p6p6

76. 76 No chords - proof a b c d p1p1 p2p2 p3p3 p4p4 p5p5 p6p6 p1p1 p6p6 p5p5 p4p4 p3p3 p2p2 b a d c b

77. 77 Definition: Given a cycle C of the optimal solution, OUT(C) is the set of edges connecting it to other cycles. Lemma 1: Step 2 : edges between cycles

78. 78 Proof: Consider an edge : e e’e’ e ’’ C e could not be added to S by PIM( l ), therefore either e ’ or e ’’ should be in S. Let k be the number of edges like e. By the above argument we conclude

79. 79 Edges between cycles (Assumption 2) (No Chords) (Prev. Slide) The bound is tight 

80. 80 Input: A cycle C with some of the nodes colored, others white (=no- color). Output: A subset X of the nodes s.t.: The difference between two consecutive nodes is odd. The nodes in X have distinct colors. F( C ) … see example Step 3: Lemma 2 (cobinatorial)

81. 81 F= 4 N=11

82. 82 Invalid: even distance N=11

83. 83 F = 10 N=11

84. 84 A cycle with at most one color odd size: F = N even size : no solution

85. 85 Is this bound tight ? Lemma 2: For any instance of the problem, on a ring of size N, which is not of even size and one color, there is a solution with F ≥ N/3. In other words: after the process, # colored nodes ≤ 2N/3 # white nodes ≥ N/3 ≥ ½ # colored nodes

86. 86 Odd Cycle Step 4: putting it all together … Optimal solution – we assumed only cycles. Algorithm : no cycles, d 2 (M)=0. Use of lemma:

87. 87 Special case – two odd cycles Odd Cycle

88. 88 The Matching Consider an optimal solution (cycles) Def: An independent set of cycles is a set of cycles with no edges between them. I - maximum independent set of odd cycles. D - all the other odd cycles. E - even cycles.

89. 89 The Matching I Max ind. odd cycles D Other odd cycles I1I1 IDID I2I2 D1D1 D2D2 E Even cycles EDED E2E2 Comb. Lemma - even cycles Comb. lemma - odd cycles Matching of odd cycles even cycles with one color

90. 90 The Matching - Analysis I D I1I1 D2D2 E E2E2 I2I2 IDID EDED D1D1 The maximality of I implies: |D 1 |≤|I 2 |Even cycles, one color: |I D |≤|E D | At least 2 cycles per isolated node, therefore at least 2 l +3 (≥ 3/2 ( l +2)) nodes per isolated node.

91. 91 The Matching - Analysis I D I1I1 D2D2 E E2E2 ≥ l +2 nodes ≥ ( l +2)/3 “purple” nodes (Lemma 1) ≥ (1/2)( l +2)/3) white nodes (Lemma 2) ≥ (3/2)( l +2) nodes

92. 92 Minimizing # of ADMs – Gerstel, Lin, Sasaki, 1998 … Traffic grooming – Gerstel, Ramaswamy, Sasaki, 1998 Zhu, Mukherjee, 2003 … More References