Transcript Wavelength Division Multiplexing (WDM) in Optical Networks:

Wavelength Division Multiplexing
(WDM) in Optical Networks:
Modeled as a Graph Coloring Problem
By Joshua Schoenly
Outline
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Background
Definitions
Graph Theoretical Approach
Special instances
Special Cases: Various networks
Background
• Three transmission
passbands in the near-IR
region of the light
spectrum.
• Each of band has 25,000
GHz (106 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
space
Optical networks – Current Usage
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Video Network Conferencing
Real –Time Medical Imaging
Scientific Visualization
High-Speed Supercomputing
Use of Fiber Optics in Networks
• 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 conversion
• Third generation – All optical network. The only
conversions occur at the transmitters / receivers. Higher
bandwidth and little noise.
Transferring Information
Transferring Information, cont.
Wavelength Division Multiplexing
• Partitions 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 time
• Consists of nodes interconnected by fiber optic
links (hmmm…seems like we can model this
with a graph theoretic approach)
Some Problems
• Two signals of the same wavelength cannot
travel down the same optical fiber (in the same
direction)
• Bandwidth is typically only around 30-40 optical
wavelengths
• Current optical fibers can only hold 2-20
wavelengths per fiber
• A transmitter/receiver cannot modulate a signal
(i.e. its wavelength cannot change)
Why not model this as a graph problem?
D
G
B
I
A
E
C
F
H
J
• Each vertex (node) is a receiver/transmitter
• 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
• 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.
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.
• 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.
• A routing R for an instance I is a set of dipaths:
R  {P  x, y  |  x, y   I }
Examples
D
G
B
I
A
E
C
F
H
J
• P(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 requests
• The routing R consists of all dipaths {P(A, J), P(D, F), and
P(F, A)} in the instance I
The Conflict Graph of G
D
B
5
I
3
A
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G
6
E
2
1
C
F
H
J
The optical network G
with a routing R
containing all colored
paths in this particular
instance I
3
6
The conflict graph of G
associated with the
routing R in G at a
particular instance I.
2
1
4
5
Solving our Network Problems
• 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).
• 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 instance
3
6
2
1
4
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c(G) = 3 due to K3 in the
conflict graph
Another Definition, the load of G
The conflict graph of G
The graph network G
The load p(G) of G is the
maximum number of paths
which share the same
directed arc.
So
c G   3
p G   2
Lemma: c(G) ≥ p(G) for any instance I in any network G
Special Instances
• The all-to-all instance IA (called gossiping) is
when each vertex makes a request with every
other vertex in a network G.
• A one-to-all instance I0 is when one vertex makes
a request with every other vertex.
All-to-all gossiping instance IA
One-to-all gossiping instance I0
Special Instance: One-to-all instance
Theorem 1: Let G be any digraph. Then c(G) = p(G)
for any one-to-all instance in G.
The network G
The conflict graph of G
Because of K3 in the conflict graph, the chromatic number is 3.
The middle arc as the highest load which is also three. Thus,
c(G) = p(G) = 3.
Special Case: Tree Network
• 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.
Theorem 2: Let T be a tree and symmetric digraph. Then c(T) = p(T),
for the all-to-all gossiping instance IA. 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).
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 p(T) = c(T) = 3.
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
Class Problem
Find the minimum number of wavelengths needed in
the instance of this network.
Answer to Class Problem
Since the chromatic number of the conflict graph is 3,
the number of wavelengths needed is three.
Special Case: Star network
Theorem 3: Let T be a symmetric tree. Then for all
instances I, p(T) = c(T) if and only if T is a subdivided star
Example: Star Network
The star network T
The conflict graph of T
Given some instance in the left network T, we find that p(T) = c(T) = 2.
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,
2


 2
N
c G   p G   
  4  
Example: Ring Network
The ring network G where N = 3
The conflict graph of G
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,
  N 2  2    32  2     9  2    2 2   1
  4     4     4    