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Lecture 18: Robust Designs for WDM Routing and Provisioning. Jeff Kennington & Eli Olinick Southern Methodist University. Augustyn Ortynski & Gheorghe Spiride Nortel Networks. 1. LTE. TE. LTE. LTE. TE. LTE. LTE. TE. LTE. LTE. LTE. TE. …. …. …. …. …. …. TE. LTE. LTE. TE.
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Lecture 18: Robust Designs for WDM Routing and Provisioning Jeff Kennington & Eli Olinick Southern Methodist University Augustyn Ortynski & Gheorghe Spiride Nortel Networks 1
LTE TE LTE LTE TE LTE LTE TE LTE LTE LTE TE … … … … … … TE LTE LTE TE LTE LTE LTE TE LTE TE LTE TE LTE TE … … … … LTE LTE TE LTE TE LTE LTE LTE … … LTE LTE Regenerator Optical Amplifier Basic Building Block for the WDM Network 2
UnderprovisionedCase Dallas Atlanta Unmet demand Satisfied demand Excess capacity Perfect MatchCase LA Phoenix OverprovisionedCase Boston NYC 3
Regret UnderprovisionedCase OverprovisionedCase Underprovisioning Overprovisioning Perfect MatchCase 4
Example 2 Boston Chicago San Francisco 3 6 7 New York Los Angeles 5 1 4 Atlanta Dallas 5
Scenario 1 Scenario 2 Scenario 3 Robust Solution Figure 5. Solutions 10
Basic design model Minimize cx (equip. cost) Subject to Ax = b (structural const) Bx = r (demand const) 0<x<u (bounds) xj integer for some j (integrality) Integer Linear Program
Decision Variables Scenario s Model Robust Model Variable Description Variables Variables Type s p x continuous number of DS3s assigned to path x p p s continuous number of TEs assigned to node n ? ? n n s t contin uous number of TEs assigned to link e t e e s e a continuous number of optical amplifiers assigned to link a e e s e r continuous number of regens assigned to link r e e s f e integer number of fibers assigned to link f e e s c e integer number of channels assigned to link c e e s z e continuous number of DS3s assigned to link z e e (o,d) and positive infeasibility for demand + continuous - z ods scenario s negative infeasibility for demand (o,d) and - continuous - z ods scenario s
Constants Constant Value or Range Description s R 300 - 1500 traffic demand for pair (o,d) in scenario s in units of DS3s od TE 192 number of DS3s that each TE can accommodate M R 192 number of DS3s that each regen can accommodate M A 15,360 number of DS3s that each optical amplifier can accommodate M 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 C F 24 max number of fibers available on link e e max distance th at a signal can traverse without amplification, also R 80km called the reach max number of amplified spans above which signal regeneration is Q 5 required B 2 - 1106 the length of link e e
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 Test problems overview • Regional US network – DA problem • European multinational network – KL problem
60 29 37 24 4 1 50 14 3 65 48 40 18 17 5 53 51 49 66 25 69 68 28 22 56 63 26 43 16 11 21 10 9 7 20 6 55 44 58 12 54 59 35 41 42 45 33 8 47 62 46 30 64 36 67 52 32 39 38 61 13 57 31 2 19 34 27 23 DA Test Problem 16
10 Legend 1 Brussels 2 Copenhagen 3 Paris 4 Berlin 5 Athens 6 Dublin 7 Rome 8 Luxembourg 9 Amsterdam 10 Oslo 11 Lisbon 12 Madrid 13 Stockholm 14 Zurich 15 London 16 Zagreb 17 Prague 18 Vienna 13 2 6 4 9 15 1 8 3 17 14 18 16 7 11 12 European Problem 5 17
Scenario Prob. TEs Rs As CPU Seconds Equipment Cost 1 0.15 24,996 3962 563 0.5 1,848,000,000 2 0.20 39,456 6502 864 0.5 2,925,000,000 3 0.30 51,882 8074 1101 0.5 3,791,000,000 4 0.20 65,086 10,122 1355 0.6 4,742,000,000 5 0.15 76,848 12,447 1584 0.5 5,630,000,000 Expected — 51,749 8,208 1096 — 3,792,000,000 Value DA – method comparison
Equipment CPU Unrouted Scaled Budget Method TEs Rs As Cost Seconds Demand Regret Mean Value 51,800 8117 1081 3,780,000,000 0.7 15.5% 1.40 Stoch. Prog. 44,373 7446 918 3,273,000,00 0 1.8 20.4% 1.82 5,630,000,000 Worst Case 39,098 5495 757 2,773,000,000 4.6 27.2% 3.75 Robust Opt. 63,122 10,813 1425 4,734,000,000 2.7 5.2% 1.00 Mean Value 51,800 8117 1081 3,780,000,000 0.2 15.5% 1.11 Stoch. Prog. 44,373 7446 918 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 No Feasible Mean Value — — — 0.3 100% — Solution 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
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 — 23,837 15,545 1196 — 3,033,250,000 Value
CPU Unrouted Scaled Budget Method TEs Rs As Equip. Cost Seconds Demand Regret Mean Value 25,124 15,350 1221 3,094,700,000 1.1 15.4% 1.41 Stoch. Prog. 20,264 14,168 996 2,644,620,000 0.6 20.8% 1.94 4,554,610 ,000 Worst Case 17,977 11,812 872 2,279,830,000 1.1 27.8% 4.05 Robust Opt. 27,520 21,348 1614 3,890,840,000 1.0 5.6% 1.00 Mean Value 23,978 15,382 1198 3,028,460,000 0.5 15.4% 1.11 Stoch. Prog. KL – method comparison 20,264 14,168 996 2,644,620,000 0.2 20.8% 1.52 3,032,69 0,000 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 No Feasible Mean Value — — — ? 100% — Solution Stoch. Prog. 12,154 7456 666 1,537,150,000 2.7 42.7% 1.19 1,539,360,000 Wors t 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
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
Shared Dedicated No Protection P-Cycle 3 3 3 3 2 2 2 2 6 6 6 6 4 4 4 4 5 5 5 5 1 1 1 1 TE = 2 A = 11 R = 5 Cost = 6.00 TE = 8 A = 32 R = 20 Cost = 18.00 TE = 6 A = 32 R = 15 Cost = 17.50 TE = 6 A = 32 R = 15 Cost = 17.5 Example 1 Demand: (1,4) of 192 DS3s (1 ) 2 26
Shared Dedicated No Protection P-Cycle 3 3 3 3 2 2 2 2 6 6 6 6 4 4 4 4 5 5 5 5 1 1 1 1 TE = 18 A = 32 R = 45 Cost = 20.50 TE = 16 A = 32 R = 39 Cost = 19.92 TE = 18 A = 32 R = 43 Cost = 20.34 TE = 6 A = 20 R = 13 Cost = 11.34 Example 2 Demands: (1,4) 192 DS3s (1 ), (1,3) 384 DS3s ( 2 ) 2 2 2 2 2 2 2 2 2 2 2 2 27
Shared No Protection Dedicated P-Cycle 3 3 3 3 2 2 2 2 6 6 6 6 4 4 4 4 5 5 5 5 1 1 1 1 TE = 70 A = 47 R = 181 Cost = 66.48 TE = 34 A = 45 R = 91 Cost = 31.48 TE = 70 A = 88 R = 185 Cost = 62.30 TE = 72 A = 88 R = 194 Cost = 63.12 Example 3Demands: (1,4) of 1 , (1,3) of 2 , (2,5) of 4 4 4 2 4 2 2 4 2 4 2 2 4 4 2 4 4 4 2 2 2 4 6 2 2 4 2 2 6 2 2 4 28