1 / 20

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

barney
Download Presentation

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

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related