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

OPTYMALIZACJA S IE CI TELEKOMUNIKACYJNYCH Michał Pióro Instytut Telekomunikacji Wydział Elektroniki i Technik Informacyjnych Politechnika Warszawska semestr letni 2003/2004. Wykład Michał Pióro (profesor) Andrzej Mysłek (prawie doktor) Ćwiczenia (audytoryjne) Andrzej Mysłek

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

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  1. OPTYMALIZACJA SIECI TELEKOMUNIKACYJNYCHMichał PióroInstytut Telekomunikacji Wydział Elektroniki i Technik InformacyjnychPolitechnika Warszawskasemestr letni 2003/2004

  2. 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

  3. Literatura podstawowa M. Pióro and D. Medhi Routing, Flow and Capacity Design in Communication and Computer Networks Morgan Kaufmann Publishers (Elsevier), April 2004 ISBN 0125571895 • www.mkp.com ($54.95 – 20% off)

  4. Sieć szkieletowa (core/backbone network) • IP/OSPF • MPLS • IDN • SDH • WDM 3 1 10 2 Warszawa 8 11 7 6 12 9 4 5

  5. 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

  6. Uncapacitated flow allocation problem – LP formulation • variables • xdp flow realizing demand d on path p • ye capacity of link e • objective minimize F = Se eye • constraints • Spxdp = hd d=1,2,…,D • Sd Spedpxdp ye e=1,2,…,E • all variables are continuous non-negative

  7. Sj xdj = hd demand d must be realised flow through link e cannot exceed its capacity Simple flow problem given: capacities hd of all Layer 2 link d - to be realised by means of flows in Layer 1 Layer 2: demand demand d with given volume hd Layer 1: equipment link ewith marginal cost ce and capacity ye flow xd2 flow xd1 nodes appearing only in Layer 1

  8. 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 Example - a solution demands x11= 15 = h1 demand 1 isrealised h1 = 15 x21 + x22= 10 = h2 demand 2 isrealised h2 = 10 x31 + x32= 20 = h3 demand 3 isrealised h3 = 20 flow x21 = 5 flow x11 = 15 2 = 1 1 = 2 4 = 1 flow x22 = 5 3 = 1 5 = 1 flow x31 = 5 cost of the network: C(y) = Se eye=85 this is not an optimal solution - why? flow x32 = 15 equipment

  9. Example - optimal solution demand The rule (SPAR): for each demand d realise the demanded volume hd on its cheapest path(s) h1 = 15 h2 = 10 h3 = 20 x11= 15 x21= z , x22= 10 - z ( 0 £ z £ 10 ) x31= 20, x32= 0 flow x21 = z flow x11 = 15 2 = 1 1 = 2 y1=z y2=25 - z y3=10 - z y4=15 y5=20 4 = 1 flow x22 = 10 - z 3 = 1 5 = 1 flow x31 = 20 cost of the network: F(y) = Se eye = 70 flow x32 = 0 equipment

  10. Uncapacitated flow allocation problem - MIP formulation • variables • xdp flow realising demand d on path p • ye capacity of link e • objective minimize F = Se eye • constraints • Spxdp = hd d=1,2,…,D • Sd Sp edpxdp£ Mye e=1,2,…,E • all flow variables variables are non-negative and all capacity variables are non-negative integers

  11. Uncapacitated flow allocation problem - IP formulation • variables • xdp flow realising demand d on path p • ye capacity of link e • objective minimise C = Se eye • constraints • Spxdp = hd d=1,2,…,D • Sd Spedpxdp£ Mye e=1,2,…,E • all variables are non-negative integers

  12. 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

  13. Capacitated flow allocation problem – LP formulation • variables • xdp flow realising demand d on path p • constraints • Spxdp = hd d=1,2,…,D • Sd Spedpxdp£ ce e=1,2,…,E • flow variables are continuous, non-negative

  14. Capacitated flow allocation problem - IP formulation • variables • xdp flow realising demand d on path p • constraints • Spxdp = hd d=1,2,…,D • Sd Spedpxdp£ ce e=1,2,…,E • flow variables are non-negative integers

  15. Node-link formulation so far we have been using link-path formulation for directed graphs! • 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 • cvwcapacity of link (v,w)

  16. Node-link formulation • variables • xdvw 0 flow of demand d on link (v,w) • constraints =hd if v = s(d) • SwA(v)xdvw - SwB(v)xdwv = 0 if x  s(d),t(d) = - hd if x = t(d) v=1,2,...,V d=1,2,…,D • Sd xdvw cvw v,w=1,2,…,V (v,w) is a link (arc)

  17. 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

  18. Shortest Path Routing (IP/OSPF) • variables • weweight (metric) of link e, w = (w1,w2,...,wE) • xdp(w) flow induced by metric system w on path (d,p) • constraints • Spxdp(w) = hd d=1,2,…,D • Sd Spedpxdp(w) ce e=1,2,…,E • w W

  19. s b a c e d a b t c Equal-split rule d e g f Unfeasible paths ECMP (Equal Cost Multi-Path) rule

  20. Flow allocation - single path allocation (non-bifurcated flows) • variables • udp binary flow variable corresponding to demand d and path p • constraints • Spudp = 1d=1,2,…,D • Sd hd Spedpudj= ye e=1,2,…,E • u:s are binary

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