1. Robust Designs for WDM Routing and Provisioning Jeff Kennington & Eli Olinick Southern Methodist University Augustyn Ortynski & Gheorghe Spiride Nortel Networks 1

2. LTE …… …… TE …… Basic Building Block for the WDM Network Optical Amplifier Regenerator 2

3. Unmet demand Satisfied demand Excess capacity Underprovisioned Case DallasAtlanta Perfect Match Case LAPhoenix Overprovisioned Case BostonNYC 3

4. Overprovisioned Case Underprovisioned Case Perfect Match Case UnderprovisioningOverprovisioning Regret 4

5. 7 4 3 1 2 5 6 Atlanta San Francisco Los Angeles Dallas Chicago Boston New York Example 5

6. ATLBOSCHIDALLANYSF ATL--1126128735401368- BOS-1609--322- CHI-1448-12873540 DAL-2253-2816 LA--644 NY-- SF- Distance Matrix (KM) 6

7. ATLBOSCHIDALLANYSF ATL-50003700880030009500- BOS--86007100-7400 CHI----- DAL---- LA--- NY-5400 SF- Scenario #1 Demand Matrix (DS3s) 7

8. ATLBOSCHIDALLANYSF ATL-1500013700188001300019500- BOS--1860017100-17400 CHI----- DAL---- LA--- NY-15400 SF- Scenario #2 Demand Matrix (DS3s) 8

9. ATLBOSCHIDALLANYSF ATL-2500023700288002300029500- BOS--2860027100-27400 CHI----- DAL---- LA--- NY-25400 SF- Scenario #3 Demand Matrix (DS3s) 9

10. Scenario 2Scenario 1 Scenario 3Robust Solution Figure 5. Solutions 10

11. Basic design model Minimizecx(equip. cost) Subject to Ax=b(structural const) Bx=r(demand const) 0 < x < u(bounds) x j integer for some j(integrality) Integer Linear Program

12. Decision Variables Scenarios Model Variables Robust Model Variables Variable Type Description x s p x p continuous number of DS3s assigned to pathp ? s n ? n continuous number of TEs assigned to noden t s e t e continuous number of TEs assigned to linke a s e a e continuous number of optical amplifiers assigned to linke r s e r e continuous number of regens assigned to linke f s e f e integer number of fibers assigned to linke c s e c e integer number of channels assigned to linke z s e z e continuous number of DS3s assigned to linke - z + ods continuous positive infeasibility for demand(o,d) and scenarios - z - ods continuous negative infeasibility for demand(o,d) and scenarios

13. Constants Constant Value or Range Description R s od 300-1500 traffic demand for pair(o,d) in scenario s in units of DS3s M TE 192 number of DS3s that eachTE can accommodate M R 192 number of DS3s that eachregen can accommodate M A 15,360 number of DS3s that eachoptical amplifier can accommodate C TE 50,000 unit cost for an TE C R 80,000 unit cost for a regen C A 500,000 unit cost for an optical amplifier F e 24 max number of fibers available on linke R 80km max distance that a signal can traversewithout amplification, also called the reach Q 5 max number of amplified spans above which signal regeneration is required B e 2-1106 the length of linke

14. Routing for scenario s

15. Robust model

16. Robust model (cont.)

17. Mean-Value model

18. Stochastic Programming Model

19. Worst Case Model

20. Regional US network – DA problem European multinational network – KL problem Test problems overview Source Total Nodes 67 Total Links 107 Total Demand Pairs 200 Number of Paths/Demand 4 Total Demand Scenarios 5 Source Total Nodes 18 Total Links 35 Total Demand Pairs 100 Number of Paths/Demand 4 Total Demand Scenarios 5

21. DA Test Problem 16

22. 1112 3 15 4 6 16 1 10 9 2 8 13 1418 7 17 5 Legend 1Brussels 2Copenhagen 3Paris 4Berlin 5Athens 6Dublin 7Rome 8Luxembourg 9Amsterdam 10Oslo 11Lisbon 12Madrid 13Stockholm 14Zurich 15London 16Zagreb 17Prague 18Vienna European Problem 17

23. DA – method comparison

24. DA – results 3,273,000,000 0.6 20.4% 1.44 3,787,000,000 Worst Case 39,098 5495 757 2,773,000,000 2.1 27.2% 2.95 Robust Opt. 52,159 8108 1061 3,787,000,000 4.5 12.6% 1.00 Mean Value — — — No Feasible Solution 0.3 100% — Stoch. Prog. 25,583 3696 515 1,832,000,000 3.9 42.3% 1.15 1,848,000,000 Worst Case 27,180 2960 505 1,848,000,000 6.6 42.3% 1.51 Robust Opt. 25,856 3575 539 1,848,000,000 5.6 43.3% 1.00

25. KL – individual scenarios Scenario Prob. TEs Rs As CPU Seconds Equipment Cost 1 0.15 12,767 7275 638 0.3 1,539,356,770 2 0.20 17,493 11,691 958 0.3 2,288,919,583 3 0.30 24,020 15,783 1178 0.3 3,052,619,167 4 0.20 29,295 19,196 1455 0.2 3,727,940,417 5 0.15 35,732 23,606 1760 0.3 4,554,614,375 Expected Value — 23,837 15,545 1196 — 3,033,250,000

26. KL – method comparison 20,264 14,168 996 2,644,620,000 0.2 20.8% 1.52 3,032,690,000 Worst Case 17,977 11,812 872 2,279,830,000 0.4 27.9% 3.20 Robust Opt. 23,967 15,548 1181 3,032,690,000 200.0 13.1% 1.00 Mean Value — — — No Feasible Solution ? 100% — Stoch. Prog. 12,154 7456 666 1,537,150,000 2.7 42.7% 1.19 1,539,360,000 Worst Case 12,782 7222 645 1,539,360,000 1.9 44.4% 1.71 Robust Opt. 13,562 7172 575 1,539,360,000 5.6 43.3% 1.00

27. Network Protection Dedicated Protection – 1 + 1 Protection P-Cycle Protection – Grover Stamatelakis (restoration speed of bi-directional rings at the cost of shared protection) Shared Protection – Path Restoration 25

28. 2 6 15 4 3 No Protection TE = 2 A = 11 R = 5 Cost = 6.00 2 2 6 15 4 3 P-Cycle TE = 8 A = 32 R = 20 Cost = 18.00 2 6 15 4 3 Dedicated TE = 6 A = 32 R = 15 Cost = 17.5 2 6 15 4 3 Shared TE = 6 A = 32 R = 15 Cost = 17.50 Example 1 Demand: (1,4) of 192 DS3s (1 ) 26

29. 2 6 15 4 3 No Protection TE = 6 A = 20 R = 13 Cost = 11.34 2 2 6 15 4 3 P-Cycle TE = 18 A = 32 R = 43 Cost = 20.34 2 6 15 4 3 Dedicated TE = 18 A = 32 R = 45 Cost = 20.50 2 6 15 4 3 Shared TE = 16 A = 32 R = 39 Cost = 19.92 Example 2 Demands: (1,4) 192 DS3s (1 ), (1,3) 384 DS3s ( 2 ) 2 2 2 2 2 2 2 2 2 2 2 27

30. 2 6 15 4 3 No Protection TE = 34 A = 45 R = 91 Cost = 31.48 2 2 6 15 4 3 P-Cycle TE = 70 A = 88 R = 185 Cost = 62.30 2 6 15 4 3 Dedicated TE = 70 A = 47 R = 181 Cost = 66.48 2 6 15 4 3 Shared TE = 72 A = 88 R = 194 Cost = 63.12 Example 3 Demands: (1,4) of 1, (1,3) of 2, (2,5) of 4 2 2 2 4 4 4 4 2 2 2 4 4 4 4 6 6 2 2 2 2 2 2 2 2 2 4 4 4 4 4 28

31. Protection Type (1,4) = 1 (1,3) = 2 (1,4) = 1 (1,3) = 2 (2,5) = 4 None6.00 (working) 11.34 (working) 31.48 (working) 1+117.5020.5066.48 P-Cycle18.00 (same working) 20.34 (same working) 62.30 (new working) Shared17.5019.9263.12 29