Administracion de granja

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  • Publicado : 23 de julio de 2010
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The variables are:

X1 = existing cows (value of 35000/30 = 1167 equal to 1050 with 10% off)
X2 = new cows (value of 1500 equal to 1350 with 10% off)
X3 = existing hens (value of5000/2000 = 2.5 equal to 1.875 with 25% off)
X4 = new hens (value of 3 equal to 2.25 with 25% off)
X5 = acres of soybean
X6 = acres of corn
X7 = acres of wheat
X8 = work hours on neighboringfarms during the winter and spring
X9 = work hours on neighboring farms during the summer and fall

The constraints are:

850X1 + 85X2 + 4.25X3 + 4.25X4 (income fromlivestock)
+ 70X5 +60X6 + 40X7 (income from crops)
+ 5X8 + 5.5X9 (working on neighboring farms)
+ 1050X1 + 1350X2 +1.875 X3 + 2.25X4 (value of livestock after decreasing aging)
+ [20000 – (1500X2 + 3X4)] (remaining of investment)
- 400000(living expenses)

X1 = 850 + 1050 = 1900
X2 = 850 + 1350 – 1500 = 700
X3 = 4.25 + 1.875 = 6.125
X4 = 4.25+ 2.25 – 3 = 3.5

So the Objective Function is
Max Z =1900X1 + 700X2 + 6.125X3 + 3.5X4 + 70X5 + 60X6 + 40X7 + 5X8 + 5.5X9 - 20000

Subject to
ECOWS) X1 = 30
TCOWS) X1 + X2 <= 42
EHENS) X3 = 2000
THENS) X3 + X4 <= 5000
LSI) 1500X2 + 3X4 <=20000
ACRES) 3X1 + 3X2 + 0.05X3 + 0.05X4 + X5 + X6 + X7 <= 640
WH1) 60X1 + 60X2 + 0.3X3 + 0.3X4 + X5 + 0.9X6 + 0.6X7 + X8 <= 4000
WH2) 60X1 + 60X2 + 0.3X3 + 0.3X4 + 1.4X5 + 1.2X6 + 0.7X7 + X9<=4500

Run the problem and get the next optimal solution (assuming good weather conditions):

X1 = 30
X2 = 0
X3 = 2000
X4 = 0
X5 = 450
X6 = 0
X7 = 0
X8= 1150
X9 = 1470

Z = 114, 585 – 20,000 = $94,585

Then check Z and the constraints:

Z = 1900(30) + 700(0) + 6.125(2000) + 3.5(0) + 70(450) + 60(0) + 40(0) + 5(1150) + 5.5(1470) – 20000
Z=...
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