Case 5.1 Old Dominion Energy

Berg Bjarne

Bong-Keun Jeong

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

Wharton1

Waha2

Kiowa6

Joliet8

Maumee10

Leidy11

Carthage4

Katy3

Henry5

Lebanon9

Perryville7

(.21, 10)

(.35, 25)

(.28, 15)

(.52, 25)

(.53, 25)

(.33, 15)

(.47, 30)

(.22, 15)(.48, 20)

(.45, 20)

(.35, 15)

(.47, 25)

(.28, 20)

(.51, 15)

(.42, 30)

(.28, 20) (.39, 15)

(.30, 30)

(.32, 20)

(.52, 15)

- (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)

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

Simplifying of the problem

A quick review of the network can simplify the problem. For example, at the beginning of 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 and work on sensitivity analysis to reduce the requirements, or we can simplify the demand requirements (max 70)

Wharton (W)

Waha (A)

Katy (K)

Kiowa (I)

Carthage ( C)

Henry (H)

Joliet (J)

Lebanon (L)

Perryville (P)

Manumee (M)

Leidy (E)

TOTAL

Wharton (W) - 10 20 - - - - - - - - 30Waha (A) 10 - 15 25 - - - - - - - 50Katy (K) 20 15 - - 30 15 - - - - - 80Kiowa (I) - 25 - - 20 - 15 - - - - 60Carthage ( C) - - 30 20 - - 25 15 30 - - 120Henry (H) - - 15 - - - - - 20 - - 35Joliet (J) - - - 15 25 - - 20 15 25 - 100Lebanon (L) - - - - 15 - 20 - 20 15 30 100Perryville (P) - - - - 30 20 15 20 - - 15 100Manumee (M) - - - - - - 25 15 - - 25 65Leidy (E) - - - - - - - 30 15 25 - 70

MAX 70 can be sent 'north'

Capacity in 000 cubic ft. (cf)

Sending

Defining the network

Wharton (W)

Waha (A) Katy (K)

Kiowa (I)

Carthage ( C)

Henry (H)

Joliet (J)

Lebanon (L)

Perryville (P)

Manumee (M)

Leidy (E)

Wharton (W) WA WKWaha (A) AW AK AIKaty (K) KW KA KC KHKiowa (I) IA IC IJ

Carthage ( C) CK CI CJ CL CPHenry (H) HK HPJoliet (J) JI JC JL JP JM

Lebanon (L) LC LJ LP LM LEPerryville (P) PC PH PJ PL PEManumee (M) MJ ML ME

Leidy (E) EL EP EM

Sending Node

Receiving Node

1) JK + EK <= 702) JK <= 353) EK - KH - KC - KA - KW <= 04) JK - KH - KC - KA - KW <= 05) AW + KW - WA - WK = 06) WA + KA + IA - AW - AK - AI = 07) WK + AK + CK + HK + LK + JK - KW - KA - KC - KH >= -1008) AI + CI + JI - IA - IC - IJ = 09) KC + IC + JC + LC + PC - CK - CI - CJ - CL - CP = 010) KH + PH - HK - HP = 011) JK + JI + JC + JP + JL + JM - IJ - CJ - PJ - LJ - MJ = 012) CL + JL + PL + ML + EL - LC - LJ - LP - LM - LE = 013) CP + HP + JP + LP + EP - PC - PH - PJ - PL - PE = 014) JM + LM + EM - MJ - ML - ME = 015) EK + EL +EP + EM - LE - PE - ME =0

We cannot send more than we receive (input and

output must match).

Cannot send more than we have in Katy (not a problem

anyway in this case, since this time, the network is the

limitation)

Simplified the problem and added 2 archs from Leidy and Joliet to Katy, but Joliet does not want more than 35

Defining Network capacity

Notice flow limitations both ways of the network pipes

17) WA <= 1018) WK <= 2019) AW <= 1020) AK <= 1521) AI <= 2522) KW <= 2023) KA <= 1524) KC <= 3025) KH <= 1526) IA <= 2527) IC <= 2028) IJ <= 1529) CK <= 3030) CI <= 2031) CJ <= 25

32) CL <= 15 33) CP <= 30

34) HK <= 1535) HP <= 2036) JI <= 1537) JC <= 2538) JL <= 2039) JP <= 1540) JM <= 2541) LC <= 1542) LJ <= 2043) LP <= 2044) LM <= 1545) LE <= 3046) PC <= 30

47) PH <= 2048) PJ <= 1549) PL <= 2050) PE <= 1551) MJ <= 2552) ML <= 1553) ME <= 2554) EL <= 3055) EP <= 1556) EM <= 25

Wharton (W)

Waha (A)

Katy (K)

Kiowa (I)

Carthage ( C)

Henry (H)

Joliet (J) Lebanon (L)

Perryville (P)

Manumee (M)

Leidy (E)

Wharton (W) - 10 20 - - - - - - - -Waha (A) 10 - 15 25 - - - - - - -Katy (K) 20 15 - - 30 15 - - - - -Kiowa (I) - 25 - - 20 - 15 - - - -Carthage ( C) - - 30 20 - - 25 15 30 - -Henry (H) - - 15 - - - - - 20 - -Joliet (J) - - - 15 25 - - 20 15 25 -Lebanon (L) - - - - 15 - 20 - 20 15 30Perryville (P) - - - - 30 20 15 20 - - 15Manumee (M) - - - - - - 25 15 - - 25Leidy (E) - - - - - - - 30 15 25 -

Capacity in 000 cubic ft. (cf)

Sending

Solution Analysis - Sequence of eventsWharton

(W)Waha

(A)Katy (K)

Kiowa (I)

Carthage ( C)

Henry (H) Joliet (J) pay $4.35 need 35

Lebanon (L)

Perryville (P)

Manumee (M)

Leidy (E) pay $4.63 need 60

Wharton (W) (2) 10Waha (A) (3) 25Katy (K) -storage 100 (1) 10 (1) 15 (1) 30 (1) 15Kiowa (I) (4) 10 (4) 15Carthage ( C) (2) 20 (5)15 (2) 5Henry (H) (2) 15Joliet (J)Lebanon (L) (6) 20Perryville (P) (3)5 (3) 15Manumee (M)Leidy (E)

Wharton (W)

Waha (A)

Katy (K)

Kiowa (I)

Carthage ( C)

Henry (H)

Joliet (J) pay $4.35 need 35

Lebanon (L)

Perryville (P)

Manumee (M)

Leidy (E) pay $4.63 need 60

Wharton (W) - $ 2.10 - - - - - - - - -Waha (A) - - - $ 8.75 - - - - - - -Katy (K) -storage 100 $ 2.80 $ 7.65 - - $ 12.60 $ 5.85 - - - - -Kiowa (I) - - - - $ 2.80 - $ 4.20 - - - -Carthage ( C) - - - - - - $ 9.40 $ 5.25 $ 1.50 - -Henry (H) - - - - - - - - $ 4.80 - -Joliet (J) - - - - - - - - - - -Lebanon (L) - - - - - - - - - - $ 9.40 Perryville (P) - - - - - - - $ 2.40 - - $ 3.30 Manumee (M) - - - - - - - - - - -Leidy (E) - - - - - - - - - - -

Total costs =Total Income = (35*$4.35 + 35* 4.63)

82.80$ 314.30$

231.50$

Solution Analysis - Costs and Profits

Lindo Background Material

Lindo CodeMAX 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 TO1) JK + EK <= 702) JK <= 353) EK - KH - KC - KA - KW <= 04) JK - KH - KC - KA - KW <= 05) AW + KW - WA - WK = 06) WA + KA + IA - AW - AK - AI = 07) WK + AK + CK + HK + LK + JK - KW - KA - KC - KH >= -1008) AI + CI + JI - IA - IC - IJ = 09) KC + IC + JC + LC + PC - CK - CI - CJ - CL - CP = 010) KH + PH - HK - HP = 011) JK + JI + JC + JP + JL + JM - IJ - CJ - PJ - LJ - MJ = 012) CL + JL + PL + ML + EL - LC - LJ - LP - LM - LE = 013) CP + HP + JP + LP + EP - PC - PH - PJ - PL - PE = 014) JM + LM + EM - MJ - ML - ME = 015) EK + EL +EP + EM - LE - PE - ME =017) WA <= 1018) WK <= 2019) AW <= 1020) AK <= 1521) AI <= 2522) KW <= 2023) KA <= 1524) KC <= 3025) KH <= 1526) IA <= 2527) IC <= 2028) IJ <= 1529) CK <= 3030) CI <= 2031) CJ <= 2532) CL <= 1533) CP <= 3034) HK <= 1535) HP <= 2036) JI <= 1537) JC <= 2538) JL <= 2039) JP <= 1540) JM <= 2541) LC <= 1542) LJ <= 2043) LP <= 2044) LM <= 1545) LE <= 3046) PC <= 3047) PH <= 2048) PJ <= 1549) PL <= 2050) PE <= 1551) MJ <= 2552) ML <= 1553) ME <= 2554) EL <= 3055) EP <= 1556) EM <= 25END

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

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

Excel Formulation

Wharton1

Waha2

Kiowa6

Joliet8

Maumee10

Leidy11

Carthage4

Katy3

Henry5

Lebanon9

Perryville7

(.21, 10)

(.35, 25)

(.28, 15)

(.52, 25)

(.53, 25)

(.33, 15)

(.47, 30)

(.22, 15)(.48, 20)

(.45, 20)

(.35, 15)

(.47, 25)

(.28, 20)

(.51, 15)

(.42, 30)

(.28, 20) (.39, 15)

(.30, 30)

(.32, 20)

(.52, 15)

Maximal Flow Problem

Excel formulationMax 4.35X83 + 4.63X113 – 0.21X12 – 0.28X13 - …… - 0.53X1110S.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 <= 250 <= X83 <= 350 <= X113 <= 60

Solution

Wharton1

Waha2

Kiowa6

Joliet8

Maumee10

Leidy11

Carthage4

Katy3

Henry5

Lebanon9

Perryville7

10

25

15

(.52, 25)

(.53, 25)

(.33, 15)

20

155

(.45, 20)

15

20

10

15

30

10 15

5

15

(.52, 15)

35

35