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Deber # 5
1.
UNIVERSIDAD NACIONAL DE
CHIMBORAZO FACULTAD DE CIENCIAS POLÍTICAS Y ADMINISTRATIVAS CARRERA DE CONTABILIDAD Y AUDITORÍA DEBER # 05 NOMBRE: RUBÍ PARRA SEMESTRE: QUINTO “A” TEMA: PROBLEMAS DE MAXIMIZACIÓN POR EL MÉTODO SIMPLEX FECHA: 27 DE OCTUBRE DEL 2014 REALIZAR 2 EJERCICIOS DE MAXIMIZACIÓN 1. Z - 3 X1 - 2 X2 - 0 H1 - 0 H2 - 0 H3 = 0 2 X1 + 1 X2 +1 H1 = 18 2 X1 + 3 X2 + 1 H2 = 42 3 X1 + 1 X2 + 1 H3 = 24 X1, X2, H1, H2, H3 ≥ 0 VE: -3 VS: 8 PIVOT: 3 VE: -1 VS: 6 PIVOT: 1/3 Maximizar Z= 3x + 2y sujeto a: 2x + y 18 2x + 3y 42 3x + y 24 x 0 , y 0 VB Z X1 X2 H1 H2 H3 VALOR H1 0 2 1 1 0 0 18 H2 0 2 3 0 1 0 42 H3 0 3 1 0 0 1 24 Z 1 -3 -2 0 0 0 0 VB Z X1 X2 H1 H2 H3 VALOR H1 0 0 1 / 3 1 0 -2 / 3 2 H2 0 0 7 / 3 0 1 -2 / 3 26 X1 0 1 1 / 3 0 0 1 / 3 8 Z 1 0 -1 0 0 1 24
2.
La solución óptima
es Z = 33 X1 = 3 X2 = 12 H1= 0 H2= 0 H3= 3 VE: -1 VS: 3 PIVOT: 4 VB Z X1 X2 H1 H2 H3 VALOR X2 0 0 1 3 0 -2 6 H2 0 0 0 -7 1 4 12 X1 0 1 0 -1 0 1 6 Z 1 0 -2 -3 0 -1 30 VB Z X1 X2 H1 H2 H3 VALOR X2 0 0 1 -0.5 0.5 0 12 H3 0 0 0 -1.75 0.25 1 3 X1 0 1 0 0.75 -0.25 0 3 Z 1 0 0 1.25 0.25 0 33
3.
2. MAXIMIZAR: 3
X1 + 5 X2 + 2 X3 Z- 3 X1 + 5 X2 + 2 X3 + 0 X4 + 0 X5 + 0 X6 + 0 X7 3 X1 + 5 X2 + 6 X3 ≤ 34 4 X1 + 5 X2 + 9 X3 ≤ 12 2 X1 + 4 X2 + 6 X3 ≤ 23 3 X1 -7 X2 -3 X3 ≤ 12 3 X1 + 5 X2 + 6 X3 + 1 H1 = 34 4 X1 + 5 X2 + 9 X3 + 1 H2 = 12 2 X1 + 4 X2 + 6 X3 + 1 H3 = 23 3 X1 -7 X2 -3 X3 + 1 H4 = 12 X1, X2, X3 ≥ 0 X1, X2, X3, H1, H2, H3, H4 ≥ 0 VE: -5 VS: 2,4 PIVOT: 5 Base Z X1 X2 X3 H1 H2 H3 H4 VALOR H1 0 3 5 6 1 0 0 0 34 H2 0 4 5 9 0 1 0 0 12 H3 0 2 4 6 0 0 1 0 23 H4 0 3 -7 -3 0 0 0 1 12 Z 1 -3 -5 -2 0 0 0 0 0 Tabla 2 3 5 2 0 0 0 0 Base Z X1 X2 X3 H1 H2 H3 H4 VALOR H1 0 -1 0 -3 1 -1 0 0 22 X2 0 0.8 1 1.8 0 0.2 0 0 2.4 H3 0 -1.2 0 -1.2 0 -0.8 1 0 13.4 H4 0 8.6 0 9.6 0 1.4 0 1 28.8 Z 1 1 0 7 0 1 0 0 12 La solución óptima es Z = 12 X1 = 0 X2 = 2.4 X3 = 0 H1= 22 H2= 0 H3= 13,4 H4= 28,8
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