método del cruce del arroyo trampolín

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Método Del Cruce Del Arroyo Trampolín

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MTODO DEL CRUCE DEL ARROYO (TRAMPOLN)

EJERCICIO N 01ORIGENDESTINOSOFERTA

ABCD

1100201115

212792025

301416185

DEMANDA515151045

ORIGENDESTINOSOFERTA

ABCD

1105

01000

201115

21275

915

205

25

301416185

5

DEMANDA515151045

MEN= 4101C = 20 9 + 7 0 = 181D = 11 0 + 7 20 = -2 2A = 12 10 + 0 7 = -53A = 0 18 + 20 7 + 0 - 10 = -153B = 14 7 + 20 18 = 93C = 16 9 + 20 18 = 9ORIGENDESTINOSOFERTA

ABCD

1100

01500

201115

21270

915

2010

25

301416180

5

DEMANDA515151045

Z = 335

1C = 20 0 + 7 9 = 181D = 11 0 + 7 20 = -2 2A = 12 0 + 18 20 = 103A = 14 18 + 20 7 = 93B = 16 18 + 20 8 = 9

ORIGENDESTINOSOFERTA

ABCD

11005

201110

15

212710

915

200

25

305

1416185

DEMANDA515151045

Z = 315

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