EJERCICIO 1

PROTRAC, INC.
X1: excavadora maquinaria E-9
X2: madera maquinaria F-9

DEPARTAMENTO | E-9x-1 | F-9x-2 | RESTRICCIONES | A | 10 | 15 | 150 | B | 20 | 10 | 160 | UTILIDAD | $ 5000 | $ 4000 | |

MAXIMIZAR
Z= 500x1+400x2 SUJETO A
10x1+15x2 ≤ 150 X1-3x2 ≤ 0
20x1+10x2 ≤ 160 X1+x2 ≥ 5
30x1+10x2 ≥ 135 X1 ≥ 0 X2 ≥ +0

SOLUCION METODO GRAFICO
10x1+15x2 ≤ 150
10x1 + 5(0)= 150 10(0)+15x2=150
X1=150/10 x2=150/15
X1=15 x2=10
20x1+10x2 ≤ 160
20x1+10(0)= 160 20(0)+ 10x2=160
X1= 160/20 x2=160/10
X1=8 x2=16

30x1+10x2 ≤ 135
30x1+10(0)= 135 30(0)+10x2= 135
X1=135/30 x2=135/0
X1= 4.5 x2=13.5

X1+ x2 ≥ 5
X1=5 …ver más…

+ 16153.83 = 73589.63 FO 10 x1 + 9x2 c1 c2 1x1 + 0.70x2 = 7000 0.15x1 + 0.30x2 = 1400

C2/c1 = 0.70/ 1 Si c1 ≠ 0 0.70/1 ≤ c2/c1 ≤ 0.30/0.15 C2/c1 = 0.30/0.15 0.15/0.30 ≤ C1/C2 ≤ 1 /.70
0.15/0.30 ≤ C1/C2 ≤ 1 /.70 C1/c2 = 1 /0.70 Si c2 ≠ 0 C1/C2= 0.15/0.30

C1= 10: 0.70/1 ≤ C2/C1 ≤ 0.30/0.15 = 0.70/ 1 ≤ C2/10 ≤ 0.30/0.15 = 7/1 ≤ C2 ≤ 3/0.15 C2= 9: 0.15/0.30 ≤ C1/C2 ≤ 1/ .70 = 0.15/0.30 ≤ C 1/9 ≤ 1/0.70 = 1.35/0.30 ≤ c1 ≤ 9/ 070 20
20
12.855
12.855

7
7
4.5
4.5
FO 10 X1 + 9X2

C1 FIJO z min= 10(5743.58) + 7 (1794.87) = 69999.89 Z Max= 10 (5743.58) + 20(1794.87) = 93333.2 C2 FIJO z min = 4.5 (5743.58) + 9 (1794.87) = 41999.94 Z Max=