wykład michał pióro (profesor) andrzej mysłek (prawie doktor) Ćwiczenia (audytoryjne)

Michał Pióro

OPTYMALIZACJA SIECI TELEKOMUNIKACYJNYCH

Michał Pióro

Instytut Telekomunikacji Wydział Elektroniki i Technik Informacyjnych

Politechnika Warszawska

semestr letni 2003/2004

Michał Pióro

Wykład Michał Pióro (profesor) Andrzej Mysłek (prawie doktor)

Ćwiczenia (audytoryjne) Andrzej Mysłek ruszają w tygodniu nr 4 - 2 godziny tygodniowo

Projekt Mateusz Dzida (doktorant PW) Michał Zagożdżon (doktorant PW) rusza w tygodniu nr 6 - 2 godziny tygodniowo

Michał Pióro

Literatura podstawowa

M. Pióro and D. Medhi

Routing, Flow and Capacity Design in Communication and Computer Networks

Morgan Kaufmann Publishers (Elsevier), April 2004ISBN 0125571895

www.mkp.com ($54.95 – 20% off)

Michał Pióro

Warszawa

10

9

6

54

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12

2

8

11

13

Sieć szkieletowa (core/backbone network)- IP/OSPF- MPLS- IDN- SDH- WDM

Michał Pióro

Uncapacitated flow allocation problem

indices d=1,2,…,D demands p=1,2,…,Pd paths for flows realising demand d e=1,2,…,E links

constants hd volume of demand d

e unit (marginal) cost of link e

edp = 1 if e belongs to path p realising demand d, 0 otherwise

Michał Pióro

Uncapacitated flow allocation problem – LP formulation

variables xdp flow realizing demand d on path p

ye capacity of link e

objective minimize F = e eye

constraints

p xdp = hd d=1,2,…,D

d p edpxdp ye e=1,2,…,E all variables are continuous non-negative

Michał Pióro

Simple flow problem

given: capacities hd of all Layer 2 link d

- to be realised by means of flows in Layer 1

Layer 1:equipment

Layer 2:demand

link e with marginal cost ce and capacity ye

demand d with given volume hd

flow xd2 j xdj = hd demand d must be realised

flow through link e cannot exceed its capacity

flow xd1 nodes appearing only in Layer 1

Michał Pióro

3 = 1

5 = 1

4 = 1

Example - a solution

equipment

demands

cost of the network: C(y) = e eye = 85this is not an optimal solution - why?

flow x11 = 15

h1 = 15

h3 = 20

h2 = 10

1 = 22 = 1

flow x21 = 5

flow x22 = 5

flow x31 = 5flow x32 = 15

x11 = 15 = h1 demand 1 is realised

x21 + x22 = 10 = h2 demand 2 is realised

x31 + x32 = 20 = h3 demand 3 is realised

x21 = 5 = y1 load and capacity of link 1

x11 + x22 = 20 = y2 load and capacity of link 2

x22 + x32 = 20 = y3 load and capacity of link 3

x11 + x32 = 30 = y4 load and capacity of link 4

x31 = 5 = y5 load and capacity of link 5

Michał Pióro

3 = 1

5 = 1

4 = 1

Example - optimal solution

equipment

demand

cost of the network: F(y) = e eye = 70

flow x11 = 15

h1 = 15

h3 = 20

h2 = 10

1 = 22 = 1

flow x21 = z

flow x22 = 10 - z

flow x31 = 20flow x32 = 0

x11 = 15x21 = z , x22 = 10 - z ( 0 z 10 ) x31 = 20 , x32 = 0

y1 = zy2 = 25 - zy3 = 10 - zy4 = 15y5 = 20

The rule (SPAR): for each demand d realise the demanded volume hd

on its cheapest path(s)

Michał Pióro

Uncapacitated flow allocation problem - MIP formulation

variables xdp flow realising demand d on path p

ye capacity of link e

objective minimize F = e eye

constraints

p xdp = hd d=1,2,…,D

d p edpxdp Mye e=1,2,…,E all flow variables variables are non-negative and all capacity

variables are non-negative integers

Michał Pióro

Uncapacitated flow allocation problem - IP formulation

variables xdp flow realising demand d on path p

ye capacity of link e

objective minimise C = e eye

constraints

p xdp = hd d=1,2,…,D

d pedpxdp Mye e=1,2,…,E all variables are non-negative integers

Michał Pióro

Capacitated flow allocation problem

indices d=1,2,…,D demands p=1,2,…,Pd paths for flows realising demand d e=1,2,…,E links

constants hd volume of demand d

ce capacity of link e

edp = 1 if e belongs to path p realising demand d, 0 otherwise

Michał Pióro

Capacitated flow allocation problem – LP formulation

variables xdp flow realising demand d on path p

constraints p xdp = hd d=1,2,…,D

d p edpxdp ce e=1,2,…,E flow variables are continuous, non-negative

Michał Pióro

Capacitated flow allocation problem - IP formulation

variables xdp flow realising demand d on path p

constraints p xdp = hd d=1,2,…,D

d p edpxdp ce e=1,2,…,E flow variables are non-negative integers

Michał Pióro

Node-link formulation

indices d=1,2,…,D demands v,w=1,2,... ,V nodes

constants hd volume of demand d s(d), t(d) end-nodes of demand d A(v), B(v) sets of nodes “after” and “before” v cvw capacity of link (v,w)

so far we have been using link-path formulation

for directed graphs!

Michał Pióro

Node-link formulation

variables xdvw 0 flow of demand d on link (v,w)

constraints

= hd if v = s(d)

wA(v) xdvw - wB(v) xdwv = 0 if x s(d),t(d)

= - hd if x = t(d)

v=1,2,...,V d=1,2,…,D

d xdvw cvw v,w=1,2,…,V (v,w) is a link (arc)

Michał Pióro

Shortest Path Routing (IP/OSPF)

indices d=1,2,…,D demands p=1,2,…,Pd paths for flows realising demand d e=1,2,…,E links

constants hd volume of demand d

ce capacity of link e

edp = 1 if e belongs to path p realising demand d, 0 otherwise

Michał Pióro

Shortest Path Routing (IP/OSPF)

variables we weight (metric) of link e, w = (w1,w2,...,wE)

xdp(w) flow induced by metric system w on path (d,p)

constraints p xdp(w) = hd d=1,2,…,D

d p edpxdp(w) ce e=1,2,…,E w W

Michał Pióro

ECMP (Equal Cost Multi-Path) rule

Equal-split rule

ba

d

s

c e

t

Unfeasible paths

a

b

c

f

d

g

e

Michał Pióro

Flow allocation - single path allocation (non-bifurcated flows)

variables udp binary flow variable corresponding to demand d and path p

constraints

p udp = 1 d=1,2,…,D

d hd p edpudj = ye e=1,2,…,E u:s are binary