Metodo de la gran M
Enviado por brenemn • 12 de Mayo de 2017 • Apuntes • 306 Palabras (2 Páginas) • 109 Visitas
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Minimizar igualar
Z= 4x1 + x2
3x1 + x2 = 3 3x1 + x2 = 3
4x1 + 3x2 >= 6 4x1 + 3x2 = 6
X1 + 2x2 <= 4 X1 + 2x2 = 4
3x1 + x2 + R1 = 3
4x1 + 3x2 + R2 – S1 = 6
X1 + 2x2 + S2 = 4+
Z= 4X1 + X2 + MR1 + MR2
Substitution en la formula
Z = 4x1 + x2 + M(3 - 3x1 – x2) + M (6- 4x1 – 3x2 + s1)
Se obtiene
Z = (4 – 7M)X1 – (1-4M)x2 + MS1 = 9M
Z | X1 | X2 | R1 | R2 | S1 | S2 | Coof. | |
F1 | 1 | -4+7M | -1+4M | 0 | 0 | -M | 0 | 9M |
F2 | 0 | 3 | 1 | 1 | 0 | 0 | 0 | 3 |
F3 | 0 | 4 | 3 | 0 | 1 | -1 | 0 | 6 |
F4 | 0 | 1 | 2 | 0 | 0 | 0 | 1 | 4 |
3/3 =1
6/4= 1.5
4/1=4
Operaciones
E.P.
0 | 1 | 1/3 | 1/3 | 0 | 0 | 0 | 1 |
1 | -4+7M | -1+4M | 0 | 0 | -M | 0 | 9M |
0 | 4-7M | (4-7M)/3 | (4-7M)/3 | 0 | 0 | 0 | 4-7M |
1 | 0 | (1+5M)/3 | (4-7M)/3 | 0 | -M | 0 | 4+2M |
0 | 4 | 3 | 0 | 1 | -1 | 0 | 6 |
0 | -4 | -4/3 | -4/3 | 0 | 0 | 0 | -4 |
0 | 0 | 5/3 | -4/3 | 1 | -1 | 0 | 2 |
0 | 1 | 2 | 0 | 0 | 0 | 1 | 4 |
0 | -1 | -1/3 | -1/3 | 0 | 0 | 0 | -1 |
0 | 0 | 5/3 | -1/3 | 0 | 0 | 1 | 3 |
Z | X1 | X2 | R1 | R2 | S1 | S2 | Coof. | |
F1 | 1 | 0 | (1+M)/3 | (4-7M)/3 | 0 | -M | 0 | 4+2M |
F2 | 0 | 1 | 1/3 | 0 | 0 | 0 | 1 | |
F3 | 0 | 0 | 5/3 | 1 | -1 | 0 | 2 | |
F4 | 0 | 0 | 5/3 | 0 | 0 | 1 | 3 |
E.P.
0 | 0 | 1 | -4/5 | 5/3 | -5/3 | 0 | 6/5 |
1 | 0 | (1+5M)/3 | (4-7M)/3 | 0 | -M | 0 | (4+2M) |
0 | 0 | (-1-M)/3 | 4+20M/18 | (-1+5M)/5 | (1-5M)/5 | 0 | -6-30M/15 |
1 | 0 | 0 | (16+15M) | (-1+5M)/5 | 1/5 | 0 | 18/5 |
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