Wykład Michał Pióro (profesor) Andrzej Mysłek (prawie doktor) Ćwiczenia (audytoryjne)
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Transcript of 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
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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 = 15 = 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
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Unfeasible paths
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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