Data Migration

Placing Regenerators in Optical networks to satisfy Multiple Sets requestsShahar Chen1 G. Mertzios I. Sau M. Shalom S. ZaksMotivation2In modern networks:Optical signals must be amplified in order for them to be carried far.Amplifiers introduce noise into the signal.Regenerators must be allocated in order to keep the Signal-to-Noise Ratio above threshold.The goal:There might be several possible traffic configurations in the network.We would like to place the minimum number of regenerators so all possible traffic configurations are satisfied.

Definitions and Problem Formulation3Lightpath a simple path in an undirected graph .Length of number of edges contains.Lightpaths are symmetric.Given an integer ,a lightpath is d-satisfied if there are no d consecutive internal vertices in without a regenerator.A set of lightpaths is d-satisfied if each of its lightpaths isd-satisfied.

Definitions and Problem Formulation4Given sets of lightpaths with we denote .An assignment of regenerators is aFunctionwhere if and only if a regenerator is used at vertex by lightpath .

Examples5Easy cases: There is only one solution

In this Lecture6Hardness results for general graphsPreface complexity and approximation algorithmsp=2, d=2Reduction to any (d,p)Approximation algorithmsPolynomial-time solvable casesEdge instancesBounded loadBounded number of regenerators per vertexComplexity and Approximation Algorithms7APX class of all NP-hard optimization problems that can be approximated within a constant factor.Examples Minimum Vertex Cover, Max-SAT

Polynomial Time Approximation Scheme (PTAS) Subclass of APXContains all the problems that can be approximated in polynomial time within a ratio for any fixed .The time complexity of the scheme is polynomial in the input size, but may be exponential in .Examples Euclidean TSP, Minimum Makespan Scheduling

Complexity and Approximation Algorithms8Clearly .If then .

In order to prove that (d,p)-TR does not admit , we will see a reduction from a problem in to TR.Minimum Vertex Cover (VC), in particular in cubic graphs, is a problem in .Proposition I

Hardness Results9Lemma (Thomassen)The edges of any cubic graph can be 2-coloredIn polynomial time.Each monochromatic connected component is a path of length at most 5.

Observation in such coloring each vertex appearsexactly once as an endpoint of a path, andexactly once as an internal vertex of another path

Insert 2 examples910Proof of PropositionLet be a cubic graph as an instance of VC.We set , so is also cubic.Let be the partition of given by the mentioned 2-coloring.We define 2 sets of lightpaths such thateach path in will correspond to a lightpath in .each path in will correspond to a lightpath in .Hardness Results Proof

1011More precisely, Let be a path with endpoints and :Let be a neighbor of in . Let be a neighbor of in such that .The lightpath in associated with is

The length of each light path is at most 7Hardness Results Proof

Show L1, L21112We can assume that no regenerator is placed at the endpoint of a lightpath.Each vertex of appears as an internal vertex in exactly 2 lightpathsOne inOne in So for any vertex in .Hardness Results Proof

Show L1, L21213Claim:

Let be a vertex cover of Let us place regenerators in at each vertex belonging to .Solution cost .The solution is feasible as at least one endpoint of each internal edge of the lightpath contains a regenerator.Therefore, Hardness Results Proof

VC to TR

Show L1, L21314Claim:

Suppose we are given a solution to with regenerators.Recall .The set of internal edges of the lightpaths in is equal to .The set of internal edges of the lightpaths in is equal to .So the regenerators places at the endpoints of the internal edges of the lightpaths constitute a VC of , of size .Therefore, Hardness Results Proof

TR to VC

Show L1, L214Hardness Results from (2,2) to (d,p)15d>2 there is a reduction from a cubic graph to a graph where,Each edge in is divided into about edges.Each lightpath has length of at most .Like in the previous case,

p>2 trivialTheorem I

Insert 2 examples15Approximation Algorithms16Altough (d,p)-TR does not admit PTAS, it has polynomial time constant-factor approximation algorithms:Theorem II:

The latter ratio is obtained by the following 2 algorithms:A simple -approximation algorithm for (d,p)-TR.An -approximation algorithm that reduces (d,p)-TR to Minimum Set Cover Problem.

Algorithm 117Anlysis:Clearly,

And also,

Therefore, - for each set of lightpaths in find optimal solution of with and .- unite all solutions.

Example18

Algorithm 219We can rephrase the (d,p)-TR problem as follows:Each lightpath has an orientation.Each regenerator covers at most d edges (the ones that come right after it in the lightpath).The first d edges in every lightpath does not have to be covered.Each regenerator can be assigned to no more than one lightpath in each configuration.So in every solution each regenerator covers at most edges.

Algorithm 220So, we can present (d,p)-TR as a Minimum Set Cover:Each set is made ofA vertex From each , a choice of at most one lightpath that goes through

At most sets for each vertex v.

At most total number of sets.Polynomial in input size since p is fixed.

Each set in the cover corresponds to a regenerator in the original problem.

We now apply an algorithm for MSC that achieves an approximation ratio of where k is the maximum size of a set.Therefore, we get a solution which is at most times larger than OPT.

Example21

The case of a path22One of the most important topologies in real networks.

Two polynomial-time solvable casesAll lightpaths share the first edge.Bounded load the number of lightpaths crossing each edge is bounded by an appropriate function of the size of the instance.

Shared Edge Algorithm23Let be a lightpath with All the lightpaths share the edge .

Claim: There is an optimal solution using regeneratorsonly at vertices d,2d,3d,

Notations: the number of lightpaths in using edge . .

Proof

Shared Edge Algorithm & Analysis24Feasibilityby definition of , there are enough regenerators to d-satisfy each set of lightpaths in any .Optimalitywe can assume that regenerators are only placed at vertices which are multiples of d.At least regenerators are needed at such vertex.

Shared Edge26In case the shared edge is not the first, the algorithm is not optimal.Example:

Bounded Load27Polynomial time algorithm if the load is Uses dynamic programmingNotations: the subset of lightpaths in crossing edge .By definition, . the set of lightpaths crossing edge . Every vector assigns a value in to each lightpath crossing . denotes for each lightpath the distance to the closest regenerator on the left of used by .for every , Let denote the projection of to index set . the entry at index of .

Bounded Load Algorithm28Matrix dimensions:Entry for each vertex and vector Matrix entries 2 values in each entry: minimum cost covering all edges left of by regenerators, such that for each lightpath , the rightmost regenerator of is at vertex . a vector in achieving cost of .Optimal Cost:

Bounded Load Algorithm29InitializationFor each we set, General stepFor each and vertex we define

Bounded Load Algorithm30Filling up the matrix:

Constructing a solution:Backtracking and gathering for each .Assign a regenerator at vertex to lightpath if .

Bounded Load Running Time31 Minimum calculation stepsEach step calculates maximum among p valuesSo each entry is calculated in steps overall.After initialization, we fill up at most entries.So the total number of steps is .Given a constant , if Then running time is boundedby the polynomial .

32Technologically, there might be a bound on the number of regenerators per vertex.Let us define a related problem:p=1The number of regenerators in each vertex is bounded by kFormally:

The k-location ProblemThe k-location Problem33The new constraint now casts 2 questions:Feasibility

Optimality (in case exists a feasible solution)Let us focus on the case of (2,1)-RL.

2 or 3 examples33(2,1)-RL Feasibility34Done by reduction to an instance of 2-SAT problem.Define variable corresponds to .Construct a boolean expression with the following clauses that capture exactly the constraints of (2,1)-RL:Proposition:

2 or 3 examples34(2,1)-RL Optimality35ProofDone by reduction from the VC problemLet be an instance of VC.A (2,1)-RL instance is constructed as follows:

Vertices for each vertex , we add vertices to vertices with even indices vertices with odd indicesProposition:

2 or 3 examples35(2,1)-RL Optimality36Lightpaths Long lightpaths for each , we add a lightpath through vertices

Short lightpaths for each we add a lightpath of length 1. Its endpoints are a vertex and a vertex where, are both even indices have not yet been assigned as endpoints of a short lightpathFinally, we add an additional vertex and edge to each endpoint of all lightpaths.

(2,1)-RL Optimality37

Notice that:There is exactly one way to 2-satisfy each long lightpath optimally (even indices).Any other solution needs at least regenerators.

Claim:There is a VC with cardinality at most if and only if there is a solution of with cost at most .

Proof of Claim VCRL38Let be a vertex cover of of size .For each we putRegenerators in all the odd vertices of ( )A regenerator in every even vertex (which is an endpoint of a short lightpath).For each we putRegenerators in all the even vertices of ( )If a short lightpath has 2 regenerators, we remove one of them arbitrarily

Proof of Claim VCRL39Correctness:Every lightpath is 2-satisfied.At each vertex there is at most one regenerator.

Analysis:The solution uses regenerators for the short lightpaths. regenerators for the long lightpaths.

Therefore, its cost is regenerators.

Proof of Claim RLVC40Let be a solution to the instance using at most regenerators.

Each short lightpath has a regenerator in at least one endpoint.Therefore, has at least regenerators satisfying the short lightpaths.

Let be a set of vertices of such that, only if there is a vertex containing one of these regenerators.return as the vertex cover.

Proof of Claim RLVC41Correctness: is a vertex cover of since each edge of corresponds to a short lightpath in , and one of the endpoints contains a regenerator.

Analysis:Consider a lightpath with : cannot be covered with regenerators, since at least one of the even vertices cannot be used.So needs at least regenerators.

Consider a lightpath where : needs at least regenerators.

Therefore, uses at least regenerators.

Questions?42