Wavelength Division Multiplexing Wavelength Division Multiplexing (WDM) in Optical Networks:(WDM) in Optical Networks:

Modeled as a Graph Coloring ProblemModeled as a Graph Coloring Problem

By Joshua SchoenlyBy Joshua Schoenly

OutlineOutline

1.1. BackgroundBackground

2.2. DefinitionsDefinitions

3.3. Graph Theoretical ApproachGraph Theoretical Approach

4.4. Special instancesSpecial instances

5.5. Special Cases: Various networksSpecial Cases: Various networks

BackgroundBackground

• Three transmission Three transmission passbands in the near-IR passbands in the near-IR region of the light region of the light spectrum. spectrum.

• Each of band has 25,000 Each of band has 25,000 GHz (10GHz (106 6 ss-1-1) of capacity ) of capacity equivalent to peak equivalent to peak hour of phone calls in the hour of phone calls in the U.S.U.S.

• Capacity is also Capacity is also equivalent to 1000 times equivalent to 1000 times the entire RF spectrum the entire RF spectrum (radio waves) in free (radio waves) in free spacespace

Optical networks – Current Usage Optical networks – Current Usage

• Video Network Conferencing Video Network Conferencing • Real –Time Medical ImagingReal –Time Medical Imaging• Scientific VisualizationScientific Visualization• High-Speed SupercomputingHigh-Speed Supercomputing

Use of Fiber Optics in NetworksUse of Fiber Optics in Networks

• First generation – All copper networks. Contained a lot of First generation – All copper networks. Contained a lot of noise (electron-electron interactions)noise (electron-electron interactions)

• Second generation – Partially fiber/copper. Better signal Second generation – Partially fiber/copper. Better signal to noise ratio. A lot of optical – electrical / electrical –to noise ratio. A lot of optical – electrical / electrical –optical conversionoptical conversion

• Third generation – All optical network. The only Third generation – All optical network. The only conversions occur at the transmitters / receivers. Higher conversions occur at the transmitters / receivers. Higher bandwidth and little noise.bandwidth and little noise.

Transferring InformationTransferring Information

Transferring Information, cont.Transferring Information, cont.

Wavelength Division MultiplexingWavelength Division Multiplexing

• Partitions the optical bandwidth into a large Partitions the optical bandwidth into a large number of channelsnumber of channels

• Allows multiple data streams to be transferred Allows multiple data streams to be transferred along the same piece of fiber at the same timealong the same piece of fiber at the same time

• Consists of nodes interconnected by fiber optic Consists of nodes interconnected by fiber optic links (hmmm…seems like we can model this links (hmmm…seems like we can model this with a graph theoretic approach)with a graph theoretic approach)

Some ProblemsSome Problems

• Two signals of the same wavelength cannot Two signals of the same wavelength cannot travel down the same optical fiber (in the same travel down the same optical fiber (in the same direction)direction)

• Bandwidth is typically only around 30-40 optical Bandwidth is typically only around 30-40 optical wavelengthswavelengths

• Current optical fibers can only hold 2-20 Current optical fibers can only hold 2-20 wavelengths per fiberwavelengths per fiber

• A transmitter/receiver cannot modulate a signal A transmitter/receiver cannot modulate a signal (i.e. its wavelength cannot change)(i.e. its wavelength cannot change)

Why not model this as a graph problem?Why not model this as a graph problem?

• Each vertex (node) is a receiver/transmitterEach vertex (node) is a receiver/transmitter

• Each edge/arc is a piece of optical fiber connecting two nodesEach edge/arc is a piece of optical fiber connecting two nodes

• Colored arrows starting at one node and ending at another are directed Colored arrows starting at one node and ending at another are directed paths representing signals of a certain wavelength being transmitted from one paths representing signals of a certain wavelength being transmitted from one node to anothernode to another

• The graph is considered to be a digraph since edge {a, b} is different from 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.edge {b, a}. Thus edges will be considered as directed arcs.

AA

FF

IIGG

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JJHHCC

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DefinitionsDefinitions

• PP((xx, , yy) is a dipath in a network graph ) is a dipath in a network graph GG from node from node xx to to yy. That is, it is a path of interconnecting . That is, it is a path of interconnecting nodes starting at nodes starting at xx and ending at and ending at yy..

• A request is an ordered pair of nodes (A request is an ordered pair of nodes (xx, , yy) in ) in GG. . This serves as a signal sent from This serves as a signal sent from xx to to yy..

• An instance An instance II is a collection of requests. is a collection of requests.• A routing A routing RR for an instance for an instance II is a set of dipaths: is a set of dipaths:

{ , | , }R P x y x y I

ExamplesExamples

• PP((AA, , JJ)) is a dipath from is a dipath from AA to to JJ which also represents a which also represents a request from request from AA to to JJ

• PP((AA, , JJ)), , PP((DD, , FF)), and , and PP((FF, , AA)) are an instance are an instance II of requests of requests• The routing The routing RR consists of all dipaths { consists of all dipaths {PP((AA, , JJ)), , PP((DD, , FF)), and , and

PP((FF, , AA))} in the instance } in the instance II

AA

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IIGG

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JJHHCC

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The Conflict Graph of The Conflict Graph of GG

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IIGG

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JJHHCC

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The optical network The optical network G G with a routing with a routing RR

containing all colored containing all colored paths in this particular paths in this particular

instance instance II

The conflict graph of The conflict graph of GG associated with the associated with the routing routing RR in in GG at a at a

particular instance particular instance II..

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33

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Solving our Network ProblemsSolving our Network Problems• We know that two signals of the same wavelength cannot We know that two signals of the same wavelength cannot

travel down the same optical fiber in the same direction.travel down the same optical fiber in the same direction.• Let the chromatic number of the conflict graph be Let the chromatic number of the conflict graph be cc((GG). ). • Finding Finding cc((GG) means we have properly colored the conflict ) means we have properly colored the conflict

graph and so no vertex is adjacent to another vertex of the graph and so no vertex is adjacent to another vertex of the same color. Thus, same color. Thus, cc((GG) optical wavelengths are needed in ) optical wavelengths are needed in this instancethis instance

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cc((GG)) = = 3 due to 3 due to KK33 in the in the

conflict graphconflict graph

Another Definition, the load of Another Definition, the load of GG

The graph network The graph network GG The conflict graph of The conflict graph of GG

The load The load ((GG) of ) of GG is the is the maximum number of paths maximum number of paths

which share the same which share the same directed arc.directed arc.

So

3

2

c G

G

Lemma: c(G) ≥ (G) for any instance I in any network G

Special InstancesSpecial Instances

• The all-to-all instance The all-to-all instance IIAA (called gossiping) is (called gossiping) is

when each vertex makes a request with every when each vertex makes a request with every other vertex in a network other vertex in a network GG..

• A one-to-all instance A one-to-all instance II00 is when one vertex makes is when one vertex makes

a request with every other vertex.a request with every other vertex.

All-to-all gossiping instance All-to-all gossiping instance IIAA One-to-all gossiping instance One-to-all gossiping instance II00

Special Instance: One-to-all instanceSpecial Instance: One-to-all instance

Theorem 1: Let Theorem 1: Let GG be any digraph. Then be any digraph. Then cc((GG) = ) = ((GG) ) for any one-to-all instance in for any one-to-all instance in GG. .

The network The network GG The conflict graph of The conflict graph of GG

Because of Because of KK33 in the conflict graph, the chromatic number is 3. in the conflict graph, the chromatic number is 3.

The middle arc as the highest load which is also three. Thus, The middle arc as the highest load which is also three. Thus, cc((GG) = ) = ((GG) = 3.) = 3.

Special Case: Tree NetworkSpecial Case: Tree Network

• Let the graph network be a tree Let the graph network be a tree TT where vertices where vertices and arcs were defined previously.and arcs were defined previously.

• Remember: Every path in Remember: Every path in TT is unique. Thus is unique. Thus there is only one path from x to y in there is only one path from x to y in TT. Therefore . Therefore for for nn vertices, there are vertices, there are nn((nn-1) different paths.-1) different paths.

Theorem 2: Let Theorem 2: Let TT be a tree and symmetric digraph. Then be a tree and symmetric digraph. Then cc((TT) = ) = ((TT), ), for the all-to-all gossiping instance for the all-to-all gossiping instance IIAA. That is the minimum number of . That is the minimum number of

wavelengths needed is equal to the maximum load on an arc (often wavelengths needed is equal to the maximum load on an arc (often called the “forwarding index” in the all-to-all instance).called the “forwarding index” in the all-to-all instance).

Example: All-to-all gossiping in a TreeExample: All-to-all gossiping in a Tree

The network The network TT The conflict graph of The conflict graph of TT

Looking in Looking in TT, we see that the load on every arc is 3. Thus, by , we see that the load on every arc is 3. Thus, by coloring the conflict graph of coloring the conflict graph of TT, we find the chromatic number is , we find the chromatic number is

also equal to 3. Thus also equal to 3. Thus ((TT) = ) = cc((TT) = 3.) = 3.

An interesting observationAn interesting observation

The max load of an arc in a tree is simply the clique The max load of an arc in a tree is simply the clique number of the conflict graph. Saying that the clique number of the conflict graph. Saying that the clique number equals the chromatic number implies that number equals the chromatic number implies that the conflict graph is perfect (i.e. the graph and its the conflict graph is perfect (i.e. the graph and its

complement does not contain an odd circuit).complement does not contain an odd circuit).

The graph network The graph network GG The conflict graph of The conflict graph of GG

Class ProblemClass Problem

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

Answer to Class ProblemAnswer to Class Problem

Since the chromatic number of the conflict graph is 3, Since the chromatic number of the conflict graph is 3, the number of wavelengths needed is three.the number of wavelengths needed is three.

Special Case: Star networkSpecial Case: Star network

Theorem 3: Let Theorem 3: Let TT be a symmetric tree. Then for all be a symmetric tree. Then for all instances I, instances I, ((TT) = ) = cc((TT) if and only if ) if and only if TT is a subdivided star is a subdivided star

Example: Star NetworkExample: Star Network

Given some instance in the left network Given some instance in the left network TT, we find that , we find that ((TT) = ) = cc((TT) = 2.) = 2.

The star network The star network TT The conflict graph of The conflict graph of TT

Special case: Ring NetworkSpecial case: Ring Network

A ring network A ring network GG where the number of vertices where the number of vertices NN = 10. = 10.

Theorem 4: For the all-to-all instance in the ring network Theorem 4: For the all-to-all instance in the ring network GG with nodes with nodes NN,,

224

Nc G G

Example: Ring NetworkExample: Ring Network

We note that the load on each arc is 1 and the chromatic number of the We note that the load on each arc is 1 and the chromatic number of the conflict graph is also 1. Substituting conflict graph is also 1. Substituting NN = 3 into the equation in theorem 4, = 3 into the equation in theorem 4,

2 23 92 2 2 2 2 14 4 4N

The ring network The ring network GG where where NN = 3 = 3 The conflict graph of The conflict graph of GG