Optic fiber

Electronic switch

the fiber serves as a transmission medium

Optical networks - 1st generation

1. Optical networks – 1. Optical networks – basic notionsbasic notions

Routing in the optical domainTwo 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 - 2nd generation

lightpaths

ADM

OADM

Data in electronic form

Data in electronic form

lightpaths

p1

p2

1 2( ) ( )w p w p

Valid coloring

Optical switch

lightpathlightpath

OADM (optical add-drop multiplexer)

No two inputs with the same wavelength should be routed on the same edge.

Electronic device at the endpoints of lightpaths

ADM (electronic add-drop multiplexer)

Where can we save?

an ADM can be shared by two lightpaths

2 ADMs 1 ADM

1 2 3

1 2 3

• 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

lightpaths - with grooming

Valid coloringg=2

Optical networksOptical networks

ADMs, OADMs, ADMs, OADMs, groominggrooming

Graph theoretical Graph theoretical modelmodel

Coloring and routingColoring and routing

12

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

2. Minimize number of 2. Minimize number of ADMsADMs

minADMminADM

Input:Input: a graph, a set of lightpaths, t>o. a graph, a set of lightpaths, t>o.

Output:Output: can the lightpath be colored such can the lightpath be colored such that #ADMs that #ADMs ≤ t ? t ?

The problem is easy on a path The problem is easy on a path networknetwork

k = 4

Reminder: coloring of an interval Reminder: coloring of an interval graphgraph

Go from left to right …Go from left to right …

2.1 minADM is NPC for a 2.1 minADM is NPC for a ringring

minADMminADM

Input:Input: a graph, a set of lightpaths, t>o. a graph, a set of lightpaths, t>o.

Output:Output: can the lightpath be colored such can the lightpath be colored such that #ADMs that #ADMs ≤ t ? t ?

Coloring of aColoring of a circular arc circular arc graphgraph

Coloring of aColoring of a circular arc circular arc graphgraph

Not always possible with max Not always possible with max loadload

Input:Input: circular arc graph G, k>o. circular arc graph G, k>o.

Output:Output: can the arcs be colored by can the arcs be colored by ≤ k k colors?colors?

Coloring of a Coloring of a circular arc circular arc graphgraph

Coloring of a circular arc graphColoring of a circular arc graph Input:Input: circular arc graph G, k>o. circular arc graph G, k>o.

Output:Output: can the arcs be colored with can the arcs be colored with ≤ k k colors?colors?

minADMminADM

Input:Input: a graph, a set of lightpaths, a graph, a set of lightpaths, t>o.t>o.

Output:Output: can the lightpath be can the lightpath be colored such that #ADMs colored such that #ADMs ≤ t ? t ?

G

Given an instance of the circular arc graph problem, construct an instance H of minADM:

Claim:Claim: can color G with ≤ k colors

iff can color H with ≤ k colors

iff can color H with #ADMs ≤ N.

G H

Assume a coloring with ≤ 3 colors …

Claim:Claim: can color H with ≤ 3 colors iff

can color H with #ADMs ≤ 13

Claim:Claim: can color with ≤ 3 colors iff

can color the lightpaths with ≤ 13 ADMs

Assume a coloring with ≤ 13 ADMs …

2.2 three basic 2.2 three basic observationsobservations

#ADMs = N + #chains

N lightpaths

cycleschain

s

Cycles are good, chains are bad

A. Structure of a solutionA. Structure of a solution

In the approximation algorithms there are two common techniques for saving ADMs:

Eliminate cycles of lightpaths

Find matchings of lightpaths

#ADMs = N + #chains

cost(S) = N + chains=13+6=19–Every path costscosts 1 ADM

cost(S) = 2N-savings=26-7=19–Every connection savessaves 1 ADM

N lightpaths

N=13

w/out grooming:

ALG 2N N OPT

ALG 2 OPT

N: # of lightpathsALG: #ADMs used by algorithmOPT: #ADMs used by an optimal solution

w/ grooming:

ALG 2N N/g OPT

ALG 2g OPT

B. The competitive ratioB. The competitive ratio

Lemma: Assume that a solution ALG saves y ADMs, and OPT saves x ADMs.

x 1 if y then cost(ALG) (2- )cost(S*).

k k³ £

C. A basic C. A basic lemmalemma

x if y then

23

cost(S)

f or

cost(S*)

exa

2

mple: ³

£

Optimal solution OPT saves x ADMsa solution ALG saves y ADMs x

yk

Recall :

cost(ALG) = 2N - y

cost(OPT) =2N - x

x N cost(OPT)

x 1 1cost(ALG)- cost(OPT) =x- y x - (1- )x (1- )cost(OPT)

k k k1

So: cost(ALG) (2- )cost(OPT)k

³

£ £

£ = £

£