Problema de la m
* Modelo PL
Variables de decisión
X1 = cantidad a comprar de kg de la materia prima A
X2 = cantidad a comprar de kg de lamateria prima B
Función Objetivo: minimizar el costo.
minZ = 157X1 + 201X2
Restricciones
Ingrediente 1
3/20 X1 + 3/10 X2 ≥ 9/10
Ingrediente 2
3/20 X1 + 3/5 X2 ≥ 6/5
Ingrediente 3
7/10 X1 + 1/10 X2 ≥ 7/10
No Negatividad
X1, X2 ≥ 0
* Forma STD
Función Objetivo
0 Z – 157X1 – 201X2 – 0S1 – MA1 – 0S2 – MA2 – 0S3 – MA3 = 0
Ingrediente 1
1 3/20 X1+ 3/10 X2 – S1 + A1 = 9/10
Ingrediente 2
2 3/20 X1 + 3/5 X2 – S2 + A2 = 6/5
Ingrediente 3
3 7/10 X1 + 1/10 X2 – S3 + A3 = 7/10
V. B. | # | Z | X1 | X2 | S1 | A1 | S2 | A2 | S3 | A3 | 2º M |
|
Z | 0 | 1 | -157 | -201 | 0 | -M | 0 | -M | 0 | -M | 0 |
A1 | 1 | 0 | 3/20 | 3/10 | -1 | 1 | 0 | 0 | 0 | 0 | 9/10 |
A2 | 2 | 0 | 3/20 | 3/5 | 0 | 0 | -1 | 1 | 0 | 0 | 6/5 |A3 | 3 | 0 | 7/10 | 1/10 | 0 | 0 | 0 | 0 | -1 | 1 | 7/10 |
|
Z | 0 | 1 | M-157 | M-201 | -M | 0 | -M | 0 | -M | 0 | 14/5 M |
A1 | 1 | 0 | 3/20 | 3/10 | -1 | 1 | 0 | 0 | 0 | 0 | 9/10 |
A2 | 2 | 0 | 3/20 | 3/5 | 0 | 0 | -1 | 1 | 0 | 0 | 6/5 |
A3 | 3 | 0 | 7/10 | 1/10 | 0 | 0 | 0 | 0 | -1 | 1 | 7/10 |
|
Z | 0 | 1 | 3/10 M – 471/10 | 9/10 M – 1853/10 | -M | 0 | -M | 0 | -157 | -M + 157| 21/10 M + 1099/10 |
A1 | 1 | 0 | 9/200 | 57/200 | -1 | 1 | 0 | 0 | 3/20 | -3/20 | 159/200 |
A2 | 2 | 0 | 9/200 | 117/200 | 0 | 0 | -1 | 1 | 3/20 | -3/20 | 219/200 |
X1 | 3 | 0 | 7/10 | 1/10 | 0 | 0 | 0 | 0 | -1 | 1 | 7/10 |
|
Z | 0 | 1 | 3/13 M – 427/13 | 0 | -M | 0 | 7/13 M – 37060/117 | -20/13 M + 37060/117 | -3/13 M – 4270/ 39 | -10/13 M + 4270/39 | 27/65 M + 17813/39 |
A1 | 1 |0 | 3/130 | 0 | -1 | 1 | 19/39 | -19/39 | 1/13 | -1/13 | 17/65 |
X2 | 2 | 0 | 1/13 | 1 | 0 | 0 | -200/117 | 200/117 | 10/39 | -10/39 | 73/39 |
X1 | 3 | 0 | 9/13 | 0 | 0 | 0 | 20/117 | -20/117 | -40/39 | 40/39 | 20/39 |
|
Z | 0 | 1 | 39/190 M – 339/19 | 0 | 2/19 M – 37060/57 | -21/19 M + 37060/57 | 0 | -M | -6/19 M – 1130/19 | -13/19 M + 1130/19 | 12/95 M + 11909/19 |
S2 | 1 | 0 | 9/190| 0 | -39/19 | 39/19 | 1 | -1 | 3/19 | -3/19 | 51/95 |
X2 | 2 | 0 | 3/19 | 1 | -200/57 | 200/57 | 0 | 0 | 10/19 | -10/19 | 53/19 |
X1 | 3 | 0 | 13/19 | 0 | 20/57 | -20/57 | 0 | 0 | -20/19 | 20/19 | 8/19 |
|
Z | 0 | 1 | 0 | 0 | -25000/39 | -M + 25000/39 | 0 | -M | -1130/13 | -M + 1130/13 | 8291/13 |
S2 | 1 | 0 | 0 | 0 | -27/13 | 27/13 | 1 | -1 | 3/13 | -3/13 | 33/65 |
X2 | 2 | 0 | 0 |1 | -140/39 | 140/39 | 0 | 0 | 10/13 | -10/13 | 35/13 |
X1 | 3 | 0 | 1 | 0 | 20/39 | -20/39 | 0 | 0 | -20/13 | 20/13 | 8/13 |
Solución:
X1= 8/13
X2= 35/13
S2= 33/65
Z= 8291/13
Precios Sombra:
S1= -25000/39
S3= -1130/13
Recomendación:
INDUSTRIAS X debe comprar 8/13 kg de la materia prima A y 35/13 kg de la materia prima B para producir una caja de harina, con un costomínimo de $8291/13.
NRO |
| M - 157 | M - 201 | - M | 0 | - M | 0 | - M | 0 | 14/5 M |
- (M – 157) | 7/10 | 1/10 | 0 | 0 | 0 | 0 | - 1 | 1 | 7/10 |
| 3/10 M – 471/10 | 9/10 M – 1853/10 | - M | 0 | - M | 0 | - 157 | - M + 157 | 21/10 M + 1099/10 |
NR1 |
| 3/20 | 3/10 | - 1 | 1 | 0 | 0 | 0 | 0 | 9/10 |
- (3/20) | 7/10 | 1/10 | 0 | 0 | 0 | 0 | - 1 | 1 | 7/10 |
| 9/200 | 57/200 | - 1...
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