Simplex
DIVISIÓN DE INGENIERÍA
DEPARTAMENTO DE INGENIERÍA INDUSTRIAL
LICENCIATURA EN INGENIERÍA INDUSTRIAL Y DE SISTEMAS
ADMINISTRACIÓN ESTRATÉGICA
METODO SIMPLEX
ACOSTA VILLAVICENCIO OSCAR OSWALDO
TAREA 3
Hermosillo, Sonora, a lunes 08 de septiembre de 2012
DECICIONES GERENCIALES
El gerente general de una empresa recibe las llamadas en su oficina de cincoclientes, los cuales quieren comprar un producto que ellos comercializan, cubetas para impermeabilizar. La siguiente información representa las formas en que él puede resolver la atención al servicio al cliente.
| | | Cliente | | | |
Sucursal | 1 | 2 | 3 | 4 | 5 | Oferta |
1 | | 25 | | 20 | | 18 | | 10 | | 11 | 200 |
2 | | 12 | | 22 | | 17 | | 09 | | 07 | 150 |
3 | | 14 | | 26 | | 24| | 21 | | 13 | 100 |
4 | | 15 | | 19 | | 16 | | 27 | | 30 | 250 |
Demanda | 100 | 200 | 50 | 150 | 200 | |
* Costo de transporte de la sucursal i al cliente j:
| x |
Resolución del problema mediante el método simplex.
Minimizar Z = 25x11 + 20x12 + 18x13 + 10x14 + 11x15 +
12x21 + 22x22 + 17x23 + 9x24 + 7x25 +
14x31 + 26x32 + 24x33 + 21x34 + 13x35 +
15x41 + 19x42 + 16x43+ 27x44 + 30x45
Sujeto a:
4 restricciones x11 + x12 + x13 + x14 + x15 ≤ 200
Oferta x21 + x22 + x23 + x24 + x25 ≤ 150
Sucursales x31 + x32 + x33 + x34 + x35 ≤ 100
x41 + x42 + x43 + x44 + x45 ≤ 250
5 restricciones x11 + x21 + x31 + x41 = 100
Demanda x12 + x22 + x32 + x42 = 200
Clientes x13 + x23 + x33 + x43 = 50
x44 + x34 + x24 + x14 = 150
x15 + x25 + x35 + x45 = 200Forma estándar
* Paso 1. Igualar la función objetivo a cero.
Minimizar Z = - 25x11 - 20x12 - 18x13 - 10x14 - 11x15 -
12x21 - 22x22 - 17x23 - 9x24 - 7x25 -
14x31 - 26x32 - 24x33 - 21x34 - 13x35 -
15x41 - 19x42 - 16x43 - 27x44 - 30x45 = 0
* Paso 2. Convertir las desigualdades en igualdades introduciendo una variable de holgura en las restricciones.
x11 + x12 + x13 + x14 + x15 + S1= 200
x21 + x22 + x23 + x24 + x25 + S2 = 150
x31 + x32 + x33 + x34 + x35 + S3 = 100
x41 + x42 + x43 + x44 + x45 + S4 = 250
x11 + x21 + x31 + x41 + A1 = 100
x12 + x22 + x32 + x42 + A2 = 200
x13 + x23 + x33 + x43 + A3 = 50
x44 + x34 + x24 + x14 + A4 = 150
x15 + x25 + x35 + x45 + A5 = 200
Xij ≥ 0
S4, S3, S2, S1 ≥ 0
A5, A4, A3, A2, A1 ≥ 0
INTERACCION 1:
* Paso 3. Escribir la tablainicial simplex
Básica | Z | X11 | X12 | X13 | X14 | X15 | X21 | X22 | X23 | X24 | X25 | X31 | X32 | X33 | X34 | X35 | X41 | X42 | X43 | X44 | X45 | S1 | S2 | S3 | S4 | A1 | A2 | A3 | A4 | A5 | Solución |
Z | 1 | -25 | -20 | -18 | -10 | -11 | -12 | -22 | -17 | -9 | -7 | -14 | -26 | -24 | -21 | -13 | -15 | -19 | -16 | -27 | -30 | 0 | 0 | 0 | 0 | -M | -M | -M | -M | -M | 0 |
S1 | 0 | 1 | 1 |1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 200 |
S2 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 150 |
S3 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 100 |
S4 | 0 | 0 | 0 |0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 250 |
A1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 100 |
A2 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 200 |
A3 | 0 | 0 | 0 | 1| 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 50 |
A4 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 150 |
A5 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 200 |
* Paso 4....
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