Traffic and routing

Network Queueing Model

• Packets are buffered in egress queues waiting for serialization on line• Link capacity is C bps• Average packet length is P bits• Service is modelled as independent with exponential distribution• Service rate is ¹ = P/C• Ingress and egress flows are homogeneous Poisson

Queue and Delay Calculations

• Each link is modelled as and M/M/1 queue• Average queue length Q=½/(1-½) =¸/(¹ -¸)• ½ = ¸/¹ <1• Average waiting time (delay): T=1/(¹ -¸)• Little: ¤ T = Q (also for the entire network)• Q= i ¸i /(¹i - ¸i) (for entire network)

• T=1/¤ i ¸i /(¹i - ¸i) (average delay for entire network)

Delay optimal routing

• Given a set of edge inflows ¤sd

• s,d: source, destination• ¤ =

s,d ¤sd

• A delay optimal route mimizes T under flow constraints and given inflows.

• Nonlinear cost• Difficult non-convex problem.

Afine (linear cost)• For a given link ”i” the cost of a flow is:

bi + ai ¸i

• Overall cost: C= i bi + ai sd ¸sdi = B + i ai sd ¸sdi

= B + sdi ai ¸sdi

• ¸sdi : is the flow from s to d routed over i

• Minimization of C w.r.t. {¸sdi}

• min{C}=B + min i ai sd ¸sdi

=B + min sd i ai ¸sdi

=B + sd min i ai ¸sdi

Afine cost• min{C} = B + sd min i ai ¸sdi

• min i ai ¸sdi for fixed s,d s.t. {¸sdi} routes ¸sd from s to d

¸sd ¸sd ¸sd

¸sd ½¸sd

¸sd ¸sd

¸sd

½¸sd

½¸sd

(1-½) ¸sd

a1 a2a3 a4

a5 a6

a10a11

a12

a9a8

a7

Afine cost

• C= B + ¸sd(a1 + a2 + a3 + a10 + a11

+ a12 ) + ½¸sd (a4 + a5

+ a6) + (1-½) ¸sd (a7 + a8 + a9)

• Assume (a4 + a5 + a6) < (a7 + a8 + a9)

• Then• C’ = B + ¸sd(a1 + a2 + a3 + a10 + a

11 + a12 )

+¸sd (a4 + a5 + a6) < C

• Routing everything through a4,a5,a6 is better• Optimal routes are single path• Optimal routes are optimal paths where link costs are {ai}

¸sd ¸sd ¸sd

¸sd ½¸sd

¸sd ¸sd

¸sd

½¸sd

½¸sd

(1-½) ¸sd

a1 a2a3 a4

a5 a6

a10a11

a12

a9a8

a7

Afine approximation of delay

• T=1/¤ i ¸i /(¹i - ¸i)

• ¸i =sd ¸sdi

• T({¸sdi}’) = T({ ¸sdi })+ sdi (¸sdi’ -¸ sdi) T/¸sdi

• T/¸sdi = T/¸i = ai =1/¤ ¹i /(¹i - ¸i)2

• T({¸sdi}’) = T({ ¸sdi })+ sdi (¸sdi’ -¸ sdi) T/¸sdi

= T({ ¸sdi })+ sdi (¸sdi’ -¸ sdi) a_i = T({ ¸sdi })- sdi ¸ sdi a_i +sdi ¸sdi’ a_i = B +sdi ¸sdi’ a_i

Small Flow Devitation

• Let ¸sdi and ¸sdi’ both be a flows routing {¤sd}• Then for 0<a<1,

¸sdi’’ = a¸sdi +(1-a)¸sdi’ is also a flow routing {¤sd} (The set of flows is convex)

• ¸sdi’’ may not comply to capacity contraints, but if ¸sdi does we can find a ¼ 1 so that ¸sdi’’ also does

The Flow Deviation Method

• Fratta, Gerla and Kleinrock (1973)• Given a flow {¸sdi} • For each s,d take a portion of the flow and re-route• Re-routing will be along shortest paths w.r.t. link

costs:T/¸sdi = ai =1/¤ ¹i /(¹i - ¸i)2

• Since cost is approximately affine for small deviations, the new flow will have lower cost

The Flow Deviation Method

• Fratta, Gerla and Kleinrock (1973)• Let old flow be: {¸sdi} and new flow: {¸sdi’}• Find a: 0<a<1, mimimizing

T({¸sdi’’}) = T({a¸sdi +(1-a)¸sdi’})• Fratta, et. al. suggest:

The portion rerouted is the portion previously routed along highest cost route w.r.t {ai}

Miniproject

• Suggest a network flow problem for which the Flow Deviation Method is meaningfully illustrated.

• Write a program implementing the Flow Deviation Method for the suggested network.

• Initial routes could be shortest paths (hop-count)

Wavelength Division MultiplexingWDM

Routing and wavelength assignment

• Since wavelengths are not mixed, traffic sources do not interfere.

• Queueing is limited to edge nodes.• Routing delay is purely propagation –

proportional to hop count.• Routes may be assigned initially and wavelengths

assigned afterwoods.• Routes may be assigned as shortest path and in

traffic load order

Wavelength assignment• Given routes R={Ri} previously assigned.• Create the auxiliary graph G• V(G): routes in R • (Ri,Rj) 2 E(G) iff Ri share fibre with Rj • Assign colours to vertices (routes), such that no neighbouring vertices

have identical colours.• Classical graph colouring problem.• The problem:

is it possible i n colours ?is generally NP-complete

• The 4-colour problem for planar graphs is solved.(colouring of countries on map)

Wavelength assignment• Sequential policy (colours are used in order):

given a vertice order v1,..,vn

1) select a colour for v1

2) j=23) If (all assigned colours are in the neighbours of vj )

assign new colour to vj

elseAssign vj an already assigned colour not within its neighbours.

4) j=j+15) if (j<=n)

goto 3

Wavelength assignment

• Smallest number of colours Â(G): chromatic number of G

• Th: Some sequential assignment uses Â(G) colours• Proof: Given a Â(G) assignment

Define a colour orderingOrder vertices in colour order v1,..,vn

defines a sequential assignment• There are n! (!!!!) different orderings.

Operational sequential policy

• Â(G) · deg(G)+1 = max {deg(Ri)} +1• If deg(G)+1 colours assigned, there is allways 1

non-neighbouring color assigned and no new colour is needed

• Heuristic: assign colours to highest degree nodes first.

Operational sequential policy

• Ordering: deg(v1) ¸ deg(v2) .. ¸ deg(vn)• Â(G) · max 0 <= i <= n min{i,1+ deg(vi)}• Proof:

i

deg(vi)+1

i

At step i we consider the subgraph v1,..,vi

As long as i · deg(vi)+1 assign i colours – one to each verticeLet i* be the smallest i so that

i ¸ deg(vi)+1Then i* ¸ deg(vi)+1 for all i ¸ i* For i ¸ i* we need no new colours

Alternate order

• Given a vertice order v1,..,vn

• Number of colours needed:· max 1<= i <= n (1+degi(vi))

• degj(vi): the degree of vi in the subgraph (v1,..,vj), where i · j

• Proof: We assign colours in in order v1,..,vn. When assigning colour to vi, you will never need more colours than 1+degi(vi)

• Note: degi(vi) · deg(vi) 8 i

Alternate order

1) select minimum degree vertice vn

2) j=n-13) Select vj so that degj(vj) is minimum (degj(vj): degree in G-(vn,..,vj+1))4) j=j-15) if (j ¸ 1) goto 3)

• The order obtained mimimizes degj(vj) for all j

Miniproject• Suggest a network flow problem for which the colouring schemes

may be meaningfully illustrated• Write a program that:

- Finds shortest path routes for s,d pairs- Constructs the auxiliary graph- Estimates the number of colours needed for colouring in degree order- Performs the colouring in degree order- Finds the alternate order- Estimate number of colours needed in alternate order - Performs colouring in alternate order

Linear programming (ILP)

Number of colours

All lines have at most Fmax colours

Flow equationspr colour --Wavelength continuity

Number of paths required from s to d

Number of LP over link i,j between s and d, at colour w

At most 1 LP over link i,j at colour w

Free variables:

ILP example

• ¤1 = ¤2 = ¤3 = 1

• F i,wj 2 {0,1} (s=j 2 {1..3}, d fixed)

• w,j F i,wj · F

• w ¸ j,w = ¤j

• j F i,wj · 1

• j=1: s=1=j, d=3: F 3,w

1 - F 1,w1 = - ¸ 1,w

s=2, d=1=j: F 3,w2 - F 1,w

2 = ¸ 2,w

s=3, d=2: F 3,w3 - F 1,w

3 = 0

F1 F2F3

¤1

¤2 ¤3

1

2

3

YALMIP• %YALMIP program for exercise in Network Performance• • %A number of integer F matrices - one for each colour• F=3; %number of colours• F1=intvar(3,3);• F2=intvar(3,3);• F3=intvar(3,3);• • %lambdas• lambda=binvar(3,F);• • %Lambdas;• L=ones(3,1);• • %problem formulation• P1=[lambda*ones(F,1)==L];• P2=[(F1+F2+F3)*ones(3,1) <= F*ones(3,1)]• P3=[F1*ones(3,1)<=ones(3,1),F2*ones(3,1)<=ones(3,1),F3*ones(3,1)<=ones(3,1)];• P=[P1 P2 P3];• S=solvesdp(P);

Miniproject

• Encode flow constraints in the YALMIP program for the example network

Miniproject

• Write a summary on the paper:Wavelength-Routed Optical Networks: Linear Formulation, Resource Budgeting Tradeoffs, and a Reconfiguration Study

by: Dhritiman Banerjee and Biswanath Mukherjee

• Objective ?• Method ?• Results ?