Metodo banquillo

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• Publicado : 25 de febrero de 2012

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BANQUILLO

Ejemplo No.1
Optimizar utilizando el método de banquillo. 1 2 A 300 10 B 200 16 300 C 14 D 0 Demanda 500 300 CT = 16100

3 14 10 6 0 100 150 250 8 14 10 0

4 12 19 16 0

250 200 450

Oferta 300 600 400 200

C12 -> C22-> C21 -> C11 -> C12 = 14 – 10 + 16 – 10 = 10 C13 -> C23 -> C21 -> C11 -> C13 = 8 – 14 + 16 – 10 = 0 C14 -> C34 -> C33 -> C23 -> C21 -> C11 -> C14 = 12 – 16 +10 – 14 + 16 – 10 = -2 C24 -> C34 -> C33 -> C23 -> C24 = 19 – 16 + 10 – 14 = -1 C31 -> C21 -> C23 -> C33 -> C31 = 14 – 16 + 14 – 10 = 2 C32 -> C22 -> C23 -> C33 -> C32 = 6 – 10 + 14 – 10 = 0 C41 -> C21 -> C23 -> C33 -> C34 -> C44 -> C41 = 0 – 16 + 14 – 10 + 16 – 0 = 4 C42 -> C22 -> C23 -> C33 -> C34 -> C44 -> C42 = 0 -10 + 14 – 10 + 16 – 0 = 10 C43 -> C33 -> C34 -> C44 -> C43 = 0 – 10 + 16 – 0 =6 1 300 (-) 200 (+) 2 10 16 14 0 300 14 10 6 0 3 100 (-) 150 (+) 250 3 14 10 6 0 4 8 (+) 14 10 250 (-) 0 200 450 4 8 100 14 10 150 0 200 450 12 19 16 0 12 19 16 0 Oferta 300 600 400 200

A B C D Demanda

500 1 200 300

300 2 10 16 14 0 300

A B C D Demanda

250 250

Oferta 300 600 400 200

500

300

C12 -> C22-> C21 -> C11 -> C12 = 14 – 10 + 16 – 10 = 10 C13 -> C33 -> C34 -> C14-> C13 = 8 – 10 + 16 – 12 = 2 C23 -> C21 -> C11 -> C14 -> C34 -> C33 -> C23 = 14 – 16 + 10 – 12 + 16 – 10 = 2 C24 -> C21 -> C11 -> C14 -> C24 = 19 – 16 + 10 – 12 = 1 C31 -> C11 -> C14 -> C34 -> C31 = 14 – 10 + 12 – 16 = 0 C32 -> C22 -> C21 -> C11 -> C14 -> C34 -> C32 = 6 – 10 + 16 – 10 + 12 - 16 = -2 C41 -> C11 -> C14 -> C44 -> C14 = 0 – 10 + 12 – 0 = 2 C42 -> C22 -> C21 -> C11 -> C14 -> C44 ->C42 = 0 -10 + 16 – 10 + 12 – 0 = 8 C43 -> C33 -> C34 -> C44 -> C43 = 0 – 10 + 16 – 0 = 6

A B C D Demanda

1 200 (-) 300 (+)

2 10 16 300 (-) 14 (+) 0 300 2 10 16 14 0 150 150 300 14 10 6 0 14 10 6 0

3

250 250 3

500 1 50 450

4 8 100(+) 14 10 150 (-) 0 200 450 4 8 250 14 10 0 200 450

12 19 16 0

Oferta 300 600 400 200

A B C D Demanda

250 250

12 19 16 0

Oferta 300600 400 200

500

C12 -> C22-> C21 -> C11 -> C12 = 14 – 10 + 16 – 10 = 10 C13 -> C33 -> C23 -> C22 -> C21 -> C11 -> C13 = 8 – 10 + 6 – 10 + 16 – 10 = 0 C23 -> C33 -> C32 -> C22 -> C23 = 14 – 10 + 6 – 10 = 0 C24 -> C21 -> C11 -> C14 -> C24 = 19 – 16 + 10 – 12 = 1 C31 -> C21 -> C22 -> C32 -> C31 = 14 – 16 + 10 – 6 = 2 C34 -> C32 -> C22 -> C21 -> C11 -> C14 -> C34 = 16 – 6 + 10 – 16 + 10 - 12 = 2C41 -> C11 -> C14 -> C44 -> C14 = 0 – 10 + 12 – 0 = 2 C42 -> C22 -> C21 -> C11 -> C14 -> C44 -> C42 = 0 -10 + 16 – 10 + 12 – 0 = 8 C43 -> C33 -> C32 -> C22 -> C21 -> C11 -> C14 -> C44 -> C43 = 0 – 10 + 6 –10 + 16 – 10 + 12 -0=4

Acá se termina el método, ya que no obtuvimos ningún valor negativo Procedemos a calcular el costo total de este nuevo tablero y obtenemos que el valor es CT = 15600. Sepuede apreciar que se logro optimizar, ya que el costo se redujo de 16100 a 15600. Por lo que podemos decir que este modelo de transporte es el más óptimo, ya que nos representa un menor costo

Ejemplo No.2 Optimizar utilizando multiplicadores. A 8 B 4 2 3 0 3 3 3 1 0 C 0 2 6 1 9 Demanda 8 5 6 1

1 2 3 4 Oferta CT = 65

5 6 2 0

8

3

C12 -> C13 -> C23 -> C22 -> C12 = 3 – 5 + 6 – 3 =1 C21 -> C11 -> C13 -> C23 -> C21 = 2 – 4 + 5 – 6 = -3 C31 -> C11 -> C13 -> C33 -> C31 = 3 – 4 + 5 – 2 = 2 C32 -> C22 -> C23 -> C33 -> C32 = 1 – 3 + 6 – 2 = 2 C41 -> C11 -> C13 -> C43 -> C41 = 0 – 4 + 5 – 0 = 1 C42 -> C22 -> C23 -> C43 -> C42 = 0 – 3 + 6 – 0 = 3

1 2 3 4 Oferta

A (-) 8 (+)

B 4 2 3 0 3

8

3

C 3 0(+) 3 2(-) 1 6 0 1 9

5 6 2 0

Demanda 8 5 6 1

1 2 3 4 OfertaA 6 2

B 4 2 3 0 3 3 3 1 0

C 2 6 1 9

5 6 2 0

Demanda 8 5 6 1

8

3

C12 -> C22 -> C21 -> C11 -> C12 = 3 – 3 + 2 – 4 = -2 C23 -> C21 -> C11 -> C13 -> C23 = 6 – 2 + 4 – 5 = 3 C31 -> C11 -> C13 -> C33 -> C31 = 3 – 4 + 5 – 2 = 2 C32 -> C22 -> C21 -> C11 -> C13 -> C32 = 1 – 3 + 2 – 4 + 5 – 2 = -1 C41 -> C11 -> C13 -> C43 -> C41 = 0 – 4 + 5 – 0 = 1 C42 -> C22 -> C21 -> C11 -> C13 ->...