Caso programación líneal

VARIABLES (9):
Kg de latas tomate entero grado A: X1 = Xea
Kg de latas tomate entero grado B: X2 = Xeb
Kg de latas tomate salsa grado A: X3 = Xsa
Kg de latas tomate salsa grado B: X4 = Xsb
Kg delatas tomate salsa grado C: X5 = Xsc
Kg de latas tomate pure grado A: X6 = Xpa
Kg de latas tomate pure grado B: X7 = Xpb
Kg de latas tomate pure grado C: X8 = Xpc
Kg de latas tomate pure grado D:X9 = Xpd

Se asume que no hay merma.
RESTRICCIONES (9):
Restricciones de oferta
X1 + X3 + X6 <= 2000
X2 + X4 + X7 <= 4000
Restricciones de Demanda
X5 + X8 <= 6000
X9 <= 1000X3+ X4 + X5 <= 3000
X6 + X7 + X8+ X9 <= 2000
Restricciones de calidad
9X1 + 6X2 = 7X1+ 7X2 2X1 – 1X2 = 0
9X3 + 6X4 + 5X5 = 5X3 + 5X4 + 5X5 4X3 + 1X4 = 0
9X6 + 6X7 + 5X8 + 2X9 = 2X6 + 2X7 +2X8 + 2X9 7X6 + 4X7 + 3X8 = 0
MAXIMIZACION DEL BENEFICIO
FO: MAX [(X1 + X2)*0.8 + (X3 + X4 + X5)*0.55 + (X6 + X7 + X8 + X9)*0.2] –
[(X1 + X2)*0.08 + (X3 + X4 + X5)*0.05 + (X6 + X7 + X8 +X9)*0.03] –
[(X1 + X3 + X6)*0.4 + (X2 + X4 + X7)*0.3 + (X5 + X8)*0.25 + (X9)*0.1]

FO: MAX [0.32(X1) + 0.42(X2) + 0.10(X3) + 0.20(X4) + 0.25(X5) – 0.23(X6) - 0.13 (X7) - 0.08(X8) + 0.07(X9)TOMATES
OPTIMAL SOLUTION - SUMMARY REPORT (Nonzero Variables)
Variable Value Cost
1 X1 E(A) 2000.0000 0.3200
2 X2 E(B) 4000.0000 0.4200
5 X5 S(C) 3000.0000 0.2500
9 X9 P(D)1000.0000 0.0700

Slack Variables
12 OFERTA C 3000.0000 0.0000
15 VTA PURE 1000.0000 0.0000

Objective Function Value = 3140

TOMATES
OPTIMAL SOLUTION - DETAILED REPORT
VariableValue Cost Red. Cost Status
1 X1 E(A) 2000.0000 0.3200 0.0000 Basic
2 X2 E(B) 4000.0000 0.4200 0.0000 Basic
3 X3 S(A) 0.0000 0.1000 0.0000 Basic
4 X4 S(B) 0.0000 0.2000-0.5925 Lower bound
5 X5 S(C) 3000.0000 0.2500 0.0000 Basic
6 X6 P(A) 0.0000 -0.2300 -0.0433 Lower bound
7 X7 P(B) 0.0000 -0.1300 -0.6033 Lower bound
8 X8 P(C) 0.0000 -0.0800...