1 Distributed Computing Optical networks: switching cost and traffic grooming Shmuel Zaks ©

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

3 Optical switch lightpath

4 A virtual topology

5 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

6 lightpaths p1 p2 Valid coloring

7 number of wavelengths

8 Switching cost ADM OADM

9 Electronic ADM

10 lightpath Optical ADM

11 ADM (add-drop multiplexer)

12 Switching cost

13 p1 p2 Valid coloring Switching cost: number of ADMs

14 W=2, ADM=8 W=3, ADM=7

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

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

17 In the approximation algorithms there are two common techniques for saving ADMs: Eliminate cycles of lightpaths Find matchings of lightpaths |ADMs| = N + |chains|

18 w/out grooming: 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

19 3/2 - Calinescu, Wan, /7+  - Shalom, Z., /7 - Epstein, Levin, 2004 ALG  2 x OPT Previous Work - ring

20 On-line algorithms when a request arrives:  if no endpoint common with others then assign a new color  if one endpoint in common with other(s) then assign same color  if two endpoints in common with others then assign one of the colors

21 Case a: 7/4=1.75

22 Case b: Case b1: 6/3 = 2 Case b2: 5/3 = ≤ r ≤ ≤ r ≤ 1.75 Show:

23 ring: 3/2 - Calinescu, Wan, /7+  - Shalom, Z., /7 - Epstein, Levin, 2004 ALG  2 OPT general topology: OPT+3/5 N - Eilam, Moran, Z Calinescu, Frieder, Wan, 2002 Approximation algorithms OPT+N α,α<1 – impossible ( Eilam, Moran, Z )

24 low capacity requests can be groomed into high capacity wavelengths (colors). colors can be assigned such that at most g lightpaths with the same color can share an edge g is the grooming factor Traffic grooming

25 W=2, ADM=8 W=1, ADM=7 g=2

26 W=3, ADM=10 W=2, ADM=8 g=2

27 R: # of lightpaths ALG: # of ADMs used by the algorithm OPT: # of ADMs used by optimal solution w/ grooming: R/g  ALG  2R R/g  OPT  2R ALG  2g x OPT

28 paths of length 2 paths of length 1 P - Star Networks: g=1

29 P - Star Networks: g=2 1. Create a new graph H having the same node set of G and an edge for any request in P between its endpoints. 2. Find cycles in H, assign a different color to each cycle. Remove the cycles and get H’.

30 P - Star Networks: g=2 3. Find paths between nodes having odd degrees in H’, assign a different color to each path.

31 The number of used ADM is exactly equal to the lower bound of needed ADM: P - Star Networks: g=2 node 0 nodes 1, …,n 0 1 n 2 i x i paths of length 2 y i paths of length 1

32 NPC - Star Networks, any fixed g ≥3 Sketch for g=3: 3-Exact Cover  Edge Partition into 3-regular graphs  Star grooming, g=3

33 3-Exact Cover: INPUT: set A of size 3n, and a collection S of subsets of A of size 3 each. QUESTION: are there n subsets in the collection S such that each element of A is in exactly one of these sets?

34 Edge Partition into 3-regular graphs: INPUT: undirected graph G = (V,E). QUESTION: can E be partitioned into subsets E 1,…,E m, each inducing a 3- regular subgraph G=(V t,E t ), t=1,…,m ?

35 NPC - Star Networks, any fixed g ≥3 3-Exact Cover  Edge Partition into 3-regular graphs  Star grooming, g=3

36 sets elements

37 3-Exact Cover  Edge Partition into 3-regular graphs  Star grooming, g=3

38 Claim: there exists a solution using at most 2|E|/3 ADMs iff the edges of G can be partitioned into 3-regular graphs G S

g=2  =2

40 Approximation algorithm (log g) Input: Graph G, set of lightpaths P, g > 0 Step 1 : Choose a parameter k = k(g). Step 2: Consider all subsets of P of size If a subset A is 1-colorable (i.e., any edge is used at most g times) then weight[A]=endpoints(A);

41 Algorithm (cont’d) Step 3: COVER  an approximation to the Minimum Weight Set Cover of S, using [Chvatal 79] Step 4: Convert COVER to a PARTITION Output: the coloring induced by PARTITION

42 Legal coloring For any fixed g, the number of subsets constructed in the first phase is

43 Analysis Legal coloring, B is 1-colorable  A is 1-colorable (  correctness). (and cost(A)  cost(B).)

44 for every set cover SC.

45 Lemma: There is a set cover SC, s.t.: for every set cover SC.

46 Conclusion: For k = g ln g :

47 Proof of Lemma Lemma: There is a set cover SC, s.t.:

48  Consider OPT  x - a color of OPT.  P x - the set of paths colored x.  endpoints(P x ) - the set of ADMs operating at wavelength x.  (assume |endpoints(P x )|= ) Partition endpoints(P x ) into sets of k consecutive nodes.

49 kk k k S 1 S 2 S m M=4 k=6

50 w/o the assumption we have:

51 undirected: Lemma: There is a set cover SC ’, s.t.: trees directed:

52 Eilam, Moran, Z., 2002 EMZ <= OPT N Calinescu, Frieder, Wan, 2002 Algorithm MCC-WS: MCC-WS <= OPT N Approximation algorithms for g=1

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

54 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

55 36 lightpaths d 0 = 3 d 1 = 4 d 2 = 29

56 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 ”

57

58 We define: and get

59 36 lightpaths d 0 = 3 d 1 = 4 d 2 = 29 cost = N+5= 41

60 G.Calinescu and P.-J.Wan[CW02] algorithm PIM( l ) eliminate short cycles, then find matchings An analysis of a basic algorithm …

61 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

62 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

63 The running time of is exponential in l (due to the preprocessing phase) The improvement is in the exponent of the running time to get same performance.

64 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

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

66 l >1 ( with preprocessing) improve:

67 l =1 (no preprocessing) - lower bound

68 OPT PIM(1) 5 8 ALG = 8 = OPT + 3 = OPT N N=5

69 OPT N …

70 l =1 (no preprocessing) - upper bound

71 D i (S): Set of Nodes of degree i in G S. d i (S): # of Nodes of degree i in G S. E S : The edges of G S : P S : Set of paths of solution S S*: An optimal solution

72 Proof 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*

73 Lemma 3.1 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” We define: It is easy to prove that: It remains to show that

74 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

75 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.

76 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 (no preprocessing) - lower bound

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

78 Simplifying Assumptions We will demonstrate the proof under the following simplifying 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 (no preprocessing) - upper bound

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

80 [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

81 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).

82 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

83 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

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

85 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

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

87 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)

88 F= 4 N=11

89 Invalid: even distance N=11

90 F = 10 N=11

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

92 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

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

94 Special case – two odd cycles Odd Cycle

95 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.

96 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

97 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.

98 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

99 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

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