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# Generalized Network Flow (GNF) Problem - PowerPoint PPT Presentation

Generalized Network Flow (GNF) Problem. Each arc ( i , j ) has a multiplier  ij If 1 unit of flow leaves node i on arc ( i , j ), then  ij will arrive node j . When  ij < 1 the arc is said to be lossy. When  ij > 1 the arc is said to be gainy.

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## PowerPoint Slideshow about ' Generalized Network Flow (GNF) Problem' - dareh

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Presentation Transcript

• Each arc (i, j) has a multiplier ij

• If 1 unit of flow leaves node i on arc (i, j), then ij will arrive node j.

• When ij< 1 the arc is said to be lossy.

• When ij> 1 the arc is said to be gainy.

• cij, ij and uij apply to the amount of flow leaving node i.

Note: the flows are usually not integral

in GNFP

• Three types of paper plus fresh wood

• Minimize use of fresh wood subject to:

ij = 0.85

ij =0.90

cij=1

ij =0.80

Formulation as GNFP: Transportation Subproblem

1a

1b

2a

2b

F

3a

3b

1a

1b

-3475

4000

2a

2b

F

1600

-1223

?

3a

3b

1000

-2260

• Add arc (F, F) with multiplier FF .

• Flow Out = xF1b + xF2b + xF3b + xFF

• Flow In = FFxFF

• Out – In = xF1b + xF2b + xF3b + (1-FF)xFF

• Let bF = 0 and FF = 2.

• 0 = xF1b + xF2b + xF3b + (-1)xFF

• xFF= xF1b + xF2b + xF3b

• Add arc (1a, 1a) with multiplier 1a1a.

• Flow Out = x1a1a + x1a1b + x1a2b

• Flow In = 1a1ax1a1a

• Out – In = x1a1b + x1a2b + (1-1a1a)x1a1a

• Let b1a = 4000 and 1a1a = 0.5.

• x1a1b + x1a2b + (0.5)x1a1a= 4000

• Unused supply of wood type 1 = x1a1a

 = 0.5

1a

1b

 = 2

2a

2b

F

 = 0.5

3a

3b

 = 0.5