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. y = available transmission capacity of the arc in thousands
Wharton 1
Waha 2
Kiowa 6
Joliet 8
Maumee 10
Leidy 11
Carthage 4
Katy 3
Henry 5
Lebanon 9
Perryville 7
(.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)
(.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)

4. MAXIMIZE PROFITS::
4.35JK EK WA WK AW AK AI KW KA KC KH IA IC IJ CK CI CJ CL - 0.3CP HK HP JI JC JL JP JM LC LJ LP LM LE - 0.3PC PH PJ PL PE MJ ML ME EL EP EM

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 ( )
2) However, we cannot ship more that 70
past 3 'bottlenecks' ( ). So we can either use 'brute force or we can simplify the demand requirements (max 70)

6. Defining the network 1) JK + EK <= 70 2) JK <= 35
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 1) 231.5000 LP O OTIMUM FOUND AT STEP 111)
VARIABLE VALUE REDUCED COST JK EK WA WK AW AK AI KW KA KC KH IA IC IJ CK CI CJ CL CP HK HP JI JC JL JP JM LC LJ LP LM LE PC PH PJ PL PE MJ ML ME EL EP EM LK
MAX 4.35JK EK- 0.21WA WK AW AK AI KW KA KC KH IA IC IJ CK CI CJ CL - 0.3CP HK HP JI JC JL JP JM LC LJ LP LM LE - 0.3PC PH PJ PL PE MJ ML ME EL EP EM
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 1) 2) 3) 4) 5) 6) 7) 8) 9)
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ROW SLACK OR SURPLUS DUAL PRICES
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NO. ITERATIONS=

12. Excel Formulation

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

14. Excel formulation
Max 4.35X X113 – 0.21X12 – 0.28X13 - …… X1110
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 10 5 30
Perryville 7
Wharton 1
Katy 3
Henry 5 35 10 15 15