1 / 15

Case 5.1 Old Dominion Energy

Solve gas transportation network problem for Old Dominion Energy to maximize profits and simplify demand requirements.

friend
Download Presentation

Case 5.1 Old Dominion Energy

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. Case 5.1 Old Dominion Energy Berg Bjarne Bong-Keun Jeong

  2. Background • Old Dominion Energy (ODE) – A gas trading company • Currently has 100,000 cf of gas in Katy • Customers in Joliet – 35,000 cf with $4.35 per thousands • Customers in Leidy – 60,000 cf with $4.63 per thousands

  3. Maumee 10 (.52, 25) Joliet 8 (.28, 15) (.53, 25) (.45, 20) Kiowa 6 (.33, 15) Leidy 11 (.47, 25) (.35, 25) (.47, 30) Lebanon 9 (.28, 20) Waha 2 Carthage 4 (.35, 15) (.48, 20) (.22, 15) (.51, 15) (.52, 15) (.42, 30) (.21, 10) (.30, 30) Perryville 7 Wharton 1 Katy 3 Henry 5 (.28, 20) (.39, 15) (.32, 20) • - (x, y) • x = the cost per thousand cubic feet (cf) of transporting gas along the arc • y = available transmission capacity of the arc in thousands • - Arcs in the network are bidirectional (either direction)

  4. MAXIMIZE PROFITS:: 4.35JK + 4.63EK- 0.21WA - 0.28WK - 0.21AW - 0.51AK - 0.35AI - 0.28KW - 0.51KA - 0.42KC - 0.39KH - 0.35IA - 0.28IC - 0.28IJ - 0.42CK - 0.28CI - 0.47CJ - 0.35CL - 0.3CP - 0.39HK - 0.32HP - 0.28JI - 0.47JC - 0.45JL - 0.52JP - 0.52JM - 0.35LC - 0.45LJ - 0.48LP - 0.33LM - 0.47LE - 0.3PC - 0.32PH - 0.52PJ - 0.48PL - 0.22PE - 0.52MJ - 0.33ML - 0.53ME - 0.47EL - 0.22EP - 0.53EM

  5. Simplifying of the problem A quick review of the network can simplify the problem. We have 100 (thousands) cubic feets of gas at Katy, but notice the obvious network constraints: 1) From Katy we can ship no more than 80 (25+15+30+15) 2) However, we cannot ship more that 70 past 3 'bottlenecks' (25+30+15).So we can either use 'brute force or we can simplify the demand requirements (max 70)

  6. Defining the network Simplified the problem and added 2 archs from Leidy and Joliet to Katy, but Joliet does not want more than 35 1) JK + EK <= 70 2) JK <= 35 3) EK - KH - KC - KA - KW <= 0 4) JK - KH - KC - KA - KW <= 0 5) AW + KW - WA - WK = 0 6) WA + KA + IA - AW - AK - AI = 0 7) WK + AK + CK + HK + LK + JK - KW - KA - KC - KH >= -100 8) AI + CI + JI - IA - IC - IJ = 0 9) KC + IC + JC + LC + PC - CK - CI - CJ - CL - CP = 0 10) KH + PH - HK - HP = 0 11) JK + JI + JC + JP + JL + JM - IJ - CJ - PJ - LJ - MJ = 0 12) CL + JL + PL + ML + EL - LC - LJ - LP - LM - LE = 0 13) CP + HP + JP + LP + EP - PC - PH - PJ - PL - PE = 0 14) JM + LM + EM - MJ - ML - ME = 0 15) EK + EL +EP + EM - LE - PE - ME =0 This is what causes the ‘flow to move’ Cannot send more than we have in Katy (not a problem anyway in this case, since this time, the network is the limitation) We cannot send more than we receive (input and output must match).

  7. Defining Network capacity 17) WA <= 10 18) WK <= 20 19) AW <= 10 20) AK <= 15 21) AI <= 25 22) KW <= 20 23) KA <= 15 24) KC <= 30 25) KH <= 15 26) IA <= 25 27) IC <= 20 28) IJ <= 15 29) CK <= 30 30) CI <= 20 31) CJ <= 25 32) CL <= 15 33) CP <= 30 34) HK <= 15 35) HP <= 20 36) JI <= 15 37) JC <= 25 38) JL <= 20 39) JP <= 15 40) JM <= 25 41) LC <= 15 42) LJ <= 20 43) LP <= 20 44) LM <= 15 45) LE <= 30 46) PC <= 30 47) PH <= 20 48) PJ <= 15 49) PL <= 20 50) PE <= 15 51) MJ <= 25 52) ML <= 15 53) ME <= 25 54) EL <= 30 55) EP <= 15 56) EM <= 25 Notice flow limitations both ways of the network pipes

  8. Solution Analysis - Sequence of events Solution Analysis - Costs and Profits

  9. Lindo Background Material

  10. Lindo Code LP O OTIMUM FOUND AT STEP 11 1) 231.5000 VARIABLE VALUE REDUCED COST JK 35.000000 0.000000 EK 35.000000 0.000000 WA 10.000000 0.000000 WK 0.000000 0.560000 AW 0.000000 0.440000 AK 0.000000 1.020000 AI 25.000000 0.000000 KW 10.000000 0.000000 KA 15.000000 0.000000 KC 30.000000 0.000000 KH 15.000000 0.000000 IA 0.000000 2.940000 IC 10.000000 0.000000 IJ 15.000000 0.000000 CK 0.000000 3.800000 CI 0.000000 0.560000 CJ 20.000000 0.000000 CL 15.000000 0.000000 CP 5.000000 0.000000 HK 0.000000 3.750000 HP 15.000000 0.000000 JI 0.000000 1.030000 JC 0.000000 0.940000 JL 0.000000 0.140000 JP 0.000000 0.690000 JM 0.000000 0.270000 LC 0.000000 1.130000 LJ 0.000000 0.760000 LP 0.000000 0.960000 LM 0.000000 0.390000 LE 20.000000 0.000000 PC 0.000000 0.600000 PH 0.000000 0.640000 PJ 0.000000 0.350000 PL 5.000000 0.000000 PE 15.000000 0.000000 MJ 0.000000 0.770000 ML 0.000000 0.270000 ME 0.000000 0.000000 EL 0.000000 0.940000 EP 0.000000 1.170000 EM 0.000000 1.060000 LK 0.000000 0.000000 MAX 4.35JK + 4.63EK- 0.21WA - 0.28WK - 0.21AW - 0.51AK - 0.35AI - 0.28KW - 0.51KA - 0.42KC - 0.39KH - 0.35IA - 0.28IC - 0.28IJ - 0.42CK - 0.28CI - 0.47CJ - 0.35CL - 0.3CP - 0.39HK - 0.32HP - 0.28JI - 0.47JC - 0.45JL - 0.52JP - 0.52JM - 0.35LC - 0.45LJ - 0.48LP - 0.33LM - 0.47LE - 0.3PC - 0.32PH - 0.52PJ - 0.48PL - 0.22PE - 0.52MJ - 0.33ML - 0.53ME - 0.47EL - 0.22EP - 0.53EM SUBJECT TO 1) JK + EK <= 70 2) JK <= 353) EK - KH - KC - KA - KW <= 0 4) JK - KH - KC - KA - KW <= 0 5) AW + KW - WA - WK = 0 6) WA + KA + IA - AW - AK - AI = 0 7) WK + AK + CK + HK + LK + JK - KW - KA - KC - KH >= -100 8) AI + CI + JI - IA - IC - IJ = 0 9) KC + IC + JC + LC + PC - CK - CI - CJ - CL - CP = 0 10) KH + PH - HK - HP = 0 11) JK + JI + JC + JP + JL + JM - IJ - CJ - PJ - LJ - MJ = 0 12) CL + JL + PL + ML + EL - LC - LJ - LP - LM - LE = 0 13) CP + HP + JP + LP + EP - PC - PH - PJ - PL - PE = 0 14) JM + LM + EM - MJ - ML - ME = 0 15) EK + EL +EP + EM - LE - PE - ME =0 17) WA <= 10 18) WK <= 20 19) AW <= 10 20) AK <= 15 21) AI <= 25 22) KW <= 20 23) KA <= 15 24) KC <= 30 25) KH <= 15 26) IA <= 25 27) IC <= 20 28) IJ <= 15 29) CK <= 30 30) CI <= 20 31) CJ <= 25 32) CL <= 15 33) CP <= 30 34) HK <= 15 35) HP <= 20 36) JI <= 15 37) JC <= 25 38) JL <= 20 39) JP <= 15 40) JM <= 25 41) LC <= 15 42) LJ <= 20 43) LP <= 20 44) LM <= 15 45) LE <= 30 46) PC <= 30 47) PH <= 20 48) PJ <= 15 49) PL <= 20 50) PE <= 15 51) MJ <= 25 52) ML <= 15 53) ME <= 25 54) EL <= 30 55) EP <= 15 56) EM <= 25 END

  11. Lindo Output - Sensitivity ROW SLACK OR SURPLUS DUAL PRICES 29) 30.000000 0.000000 30) 20.000000 0.000000 31) 5.000000 0.000000 32) 0.000000 0.430000 33) 25.000000 0.000000 34) 15.000000 0.000000 35) 5.000000 0.000000 36) 15.000000 0.000000 37) 25.000000 0.000000 38) 20.000000 0.000000 39) 15.000000 0.000000 40) 25.000000 0.000000 41) 15.000000 0.000000 42) 20.000000 0.000000 43) 20.000000 0.000000 44) 15.000000 0.000000 45) 10.000000 0.000000 46) 30.000000 0.000000 47) 20.000000 0.000000 48) 15.000000 0.000000 49) 15.000000 0.000000 50) 0.000000 0.730000 51) 25.000000 0.000000 52) 15.000000 0.000000 53) 25.000000 0.000000 54) 30.000000 0.000000 55) 15.000000 0.000000 56) 25.000000 0.000000 NO. ITERATIONS= 33 ROW SLACK OR SURPLUS DUAL PRICES 1) 0.000000 0.000000 2) 0.000000 0.500000 3) 35.000000 0.000000 4) 35.000000 0.000000 5) 0.000000 -0.280000 6) 0.000000 -0.510000 7) 65.000000 0.000000 8) 0.000000 -3.100000 9) 0.000000 -3.380000 10) 0.000000 -3.360000 11) 0.000000 3.850000 12) 0.000000 -4.160000 13) 0.000000 -3.680000 14) 0.000000 -4.100000 15) 0.000000 4.630000 17) 0.000000 0.020000 18) 20.000000 0.000000 19) 10.000000 0.000000 20) 15.000000 0.000000 21) 0.000000 2.240000 22) 10.000000 0.000000 23) 0.000000 0.000000 24) 0.000000 2.960000 25) 0.000000 2.970000 26) 25.000000 0.000000 27) 10.000000 0.000000 28) 0.000000 0.470000

  12. ExcelFormulation

  13. Maumee 10 (.52, 25) Joliet 8 (.28, 15) (.53, 25) (.45, 20) Kiowa 6 (.33, 15) Leidy 11 (.47, 25) (.35, 25) (.47, 30) Lebanon 9 (.28, 20) Waha 2 Carthage 4 (.35, 15) (.48, 20) (.22, 15) (.51, 15) (.52, 15) (.42, 30) (.21, 10) (.30, 30) Perryville 7 Wharton 1 Katy 3 Henry 5 (.28, 20) (.39, 15) (.32, 20) Maximal Flow Problem

  14. Excel formulation Max 4.35X83 + 4.63X113 – 0.21X12 – 0.28X13 - …… - 0.53X1110 S.T. X21 + X31 – X12 – X13 = 0 (Node 1 incoming and outgoing) …. X13 + X23 + X43 + X53 + X83 + X113 – X31 – X32 – X34 – X35 = 0 (Node 3 incoming and outgoing) …. X48 + X68 + X78 + X98 + X108 – X83 – X84 – X86 – X87 – X89 – X810 = 0 (Node 8 incoming and outgoing) …. X711 + X911 + X1011 – X113 – X117 – X119 – X1110 = 0 (Node 11 incoming and outgoing) 0 <= X12 <= 10 (Capacity constraint) 0 <= X13 <= 20 (Capacity constraint) ….. 0 <= X1011 <= 25 0 <= X83 <= 35 0 <= X113 <= 60

  15. Solution 35 Maumee 10 (.52, 25) Joliet 8 (.53, 25) 15 (.45, 20) Kiowa 6 (.33, 15) Leidy 11 20 25 20 Lebanon 9 10 Waha 2 Carthage 4 15 (.52, 15) 5 15 15 5 10 30 Perryville 7 Wharton 1 Katy 3 Henry 5 35 10 15 15

More Related