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Wavelength Division Multiplexing (WDM) in Optical Networks: Modeled as a Graph Coloring Problem By Joshua Schoenly.

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Presentation on theme: "Wavelength Division Multiplexing (WDM) in Optical Networks: Modeled as a Graph Coloring Problem By Joshua Schoenly."— Presentation transcript:

1 Wavelength Division Multiplexing (WDM) in Optical Networks: Modeled as a Graph Coloring Problem By Joshua Schoenly

2 Outline 1.Background 2.Definitions 3.Graph Theoretical Approach 4.Special instances 5.Special Cases: Various networks

3 Background Three transmission passbands in the near-IR region of the light spectrum.Three transmission passbands in the near-IR region of the light spectrum. Each of band has 25,000 GHz (10 6 s -1 ) of capacity  equivalent to peak hour of phone calls in the U.S.Each of band has 25,000 GHz (10 6 s -1 ) of capacity  equivalent to peak hour of phone calls in the U.S. Capacity is also equivalent to 1000 times the entire RF spectrum (radio waves) in free spaceCapacity is also equivalent to 1000 times the entire RF spectrum (radio waves) in free space

4 Optical networks – Current Usage Video Network ConferencingVideo Network Conferencing Real –Time Medical ImagingReal –Time Medical Imaging Scientific VisualizationScientific Visualization High-Speed SupercomputingHigh-Speed Supercomputing

5 Use of Fiber Optics in Networks First generation – All copper networks. Contained a lot of noise (electron-electron interactions)First generation – All copper networks. Contained a lot of noise (electron-electron interactions) Second generation – Partially fiber/copper. Better signal to noise ratio. A lot of optical – electrical / electrical – optical conversionSecond generation – Partially fiber/copper. Better signal to noise ratio. A lot of optical – electrical / electrical – optical conversion Third generation – All optical network. The only conversions occur at the transmitters / receivers. Higher bandwidth and little noise.Third generation – All optical network. The only conversions occur at the transmitters / receivers. Higher bandwidth and little noise.

6 Transferring Information

7 Transferring Information, cont.

8 Wavelength Division Multiplexing Partitions the optical bandwidth into a large number of channelsPartitions the optical bandwidth into a large number of channels Allows multiple data streams to be transferred along the same piece of fiber at the same timeAllows multiple data streams to be transferred along the same piece of fiber at the same time Consists of nodes interconnected by fiber optic links (hmmm…seems like we can model this with a graph theoretic approach)Consists of nodes interconnected by fiber optic links (hmmm…seems like we can model this with a graph theoretic approach)

9 Some Problems Two signals of the same wavelength cannot travel down the same optical fiber (in the same direction)Two signals of the same wavelength cannot travel down the same optical fiber (in the same direction) Bandwidth is typically only around optical wavelengthsBandwidth is typically only around optical wavelengths Current optical fibers can only hold 2-20 wavelengths per fiberCurrent optical fibers can only hold 2-20 wavelengths per fiber A transmitter/receiver cannot modulate a signal (i.e. its wavelength cannot change)A transmitter/receiver cannot modulate a signal (i.e. its wavelength cannot change)

10 Why not model this as a graph problem? Each vertex (node) is a receiver/transmitter Each vertex (node) is a receiver/transmitter Each edge/arc is a piece of optical fiber connecting two nodes Each edge/arc is a piece of optical fiber connecting two nodes Colored arrows starting at one node and ending at another are directed paths representing signals of a certain wavelength being transmitted from one node to another Colored arrows starting at one node and ending at another are directed paths representing signals of a certain wavelength being transmitted from one node to another The graph is considered to be a digraph since edge {a, b} is different from edge {b, a}. Thus edges will be considered as directed arcs. The graph is considered to be a digraph since edge {a, b} is different from edge {b, a}. Thus edges will be considered as directed arcs. A F I GDJ H C B E

11 Definitions P(x, y) is a dipath in a network graph G from node x to y. That is, it is a path of interconnecting nodes starting at x and ending at y.P(x, y) is a dipath in a network graph G from node x to y. That is, it is a path of interconnecting nodes starting at x and ending at y. A request is an ordered pair of nodes (x, y) in G. This serves as a signal sent from x to y.A request is an ordered pair of nodes (x, y) in G. This serves as a signal sent from x to y. An instance I is a collection of requests.An instance I is a collection of requests. A routing R for an instance I is a set of dipaths:A routing R for an instance I is a set of dipaths:

12 Examples P(A, J) is a dipath from A to J which also represents a request from A to JP(A, J) is a dipath from A to J which also represents a request from A to J P(A, J), P(D, F), and P(F, A) are an instance I of requestsP(A, J), P(D, F), and P(F, A) are an instance I of requests The routing R consists of all dipaths {P(A, J), P(D, F), and P(F, A)} in the instance IThe routing R consists of all dipaths {P(A, J), P(D, F), and P(F, A)} in the instance I A F I GDJ H C B E

13 The Conflict Graph of G A F I GDJ H C B E The optical network G with a routing R containing all colored paths in this particular instance I The conflict graph of G associated with the routing R in G at a particular instance I

14 Solving our Network Problems We know that two signals of the same wavelength cannot travel down the same optical fiber in the same direction.We know that two signals of the same wavelength cannot travel down the same optical fiber in the same direction. Let the chromatic number of the conflict graph be c(G).Let the chromatic number of the conflict graph be c(G). Finding c(G) means we have properly colored the conflict graph and so no vertex is adjacent to another vertex of the same color. Thus, c(G) optical wavelengths are needed in this instanceFinding c(G) means we have properly colored the conflict graph and so no vertex is adjacent to another vertex of the same color. Thus, c(G) optical wavelengths are needed in this instance c(G)= 3 due to K 3 in the conflict graph c(G) = 3 due to K 3 in the conflict graph

15 Another Definition, the load of G The graph network G The conflict graph of G The load  (G) of G is the maximum number of paths which share the same directed arc. Lemma: c(G) ≥  (G) for any instance I in any network G

16 Special Instances The all-to-all instance I A (called gossiping) is when each vertex makes a request with every other vertex in a network G.The all-to-all instance I A (called gossiping) is when each vertex makes a request with every other vertex in a network G. A one-to-all instance I 0 is when one vertex makes a request with every other vertex.A one-to-all instance I 0 is when one vertex makes a request with every other vertex. All-to-all gossiping instance I A One-to-all gossiping instance I 0

17 Special Instance: One-to-all instance Theorem 1: Let G be any digraph. Then c(G) =  (G) for any one-to-all instance in G. The network G The conflict graph of G Because of K 3 in the conflict graph, the chromatic number is 3. The middle arc as the highest load which is also three. Thus, c(G) =  (G) = 3.

18 Special Case: Tree Network Let the graph network be a tree T where vertices and arcs were defined previously.Let the graph network be a tree T where vertices and arcs were defined previously. Remember: Every path in T is unique. Thus there is only one path from x to y in T. Therefore for n vertices, there are n(n-1) different paths.Remember: Every path in T is unique. Thus there is only one path from x to y in T. Therefore for n vertices, there are n(n-1) different paths. Theorem 2: Let T be a tree and symmetric digraph. Then c(T) =  (T), for the all-to-all gossiping instance I A. That is the minimum number of wavelengths needed is equal to the maximum load on an arc (often called the “forwarding index” in the all-to-all instance).

19 Example: All-to-all gossiping in a Tree The network T The conflict graph of T Looking in T, we see that the load on every arc is 3. Thus, by coloring the conflict graph of T, we find the chromatic number is also equal to 3. Thus  (T) = c(T) = 3.

20 An interesting observation The max load of an arc in a tree is simply the clique number of the conflict graph. Saying that the clique number equals the chromatic number implies that the conflict graph is perfect (i.e. the graph and its complement does not contain an odd circuit). The graph network G The conflict graph of G

21 Class Problem Find the minimum number of wavelengths needed in the instance of this network.

22 Answer to Class Problem Since the chromatic number of the conflict graph is 3, the number of wavelengths needed is three.

23 Special Case: Star network Theorem 3: Let T be a symmetric tree. Then for all instances I,  (T) = c(T) if and only if T is a subdivided star

24 Example: Star Network Given some instance in the left network T, we find that  (T) = c(T) = 2. The star network T The conflict graph of T

25 Special case: Ring Network A ring network G where the number of vertices N = 10. Theorem 4: For the all-to-all instance in the ring network G with nodes N,

26 Example: Ring Network We note that the load on each arc is 1 and the chromatic number of the conflict graph is also 1. Substituting N = 3 into the equation in theorem 4, The ring network G where N = 3 The conflict graph of G


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