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LPSolutions

# LPSolutions - Solutions LP Formulation and Sensitivity...

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Solutions LP Formulation and Sensitivity Analysis Quality Assurance X1 = pounds of bean 1, X2 = pounds of bean 2, X3 = pounds of bean 3 The objective is to minimize the total cost of the blend. Min 0.50x1 + 0.70x2 +0.45X3 S.T. (75x1+85x2+60x1)/(x1+x2+x3)>=75 or; 10x2 -15x3 >= 0 Aroma (86x1 +88x2+75x3)/(x1+x2+x3) >= 80 or; 6x1+ 8x - 5x3 >= 0 Taste x1 <= 500 Bean1 x2 <= 600 Bean2 x3 <= 400 Bean3 x1 + x2 + x3 = 1000 pounds xl, x2, x3 >=0 Optimal solution: x1 = 500, x2 = 300, X3 = 200 Total cost at the optimal solution = \$550. To answer the rest of the question we assume the problem was solved on Excel. b. The total cost is \$550. The total amount of blend produced is 1000lb. Therefore, the cost per pound is \$550/1000 = S0.55 c. We can use the optimal values of x1=500, x2 = 300, X3 = 200 to calculate the aroma and taste rating: Aroma rating = [75(500)+85(300)+60(200)]/(1000)=75 Taste rating = [86(500) + 88(300) + 75(200)]/1000= 84.4 d. If additional coffee is to be produced, the total amount of 1000lb is changing. Thus, we need to focus on the right hand side of the last constraint in the model above. The shadow price of this constraint is \$.60. Extra coffee can be produced at an additional cost of \$0.60 per pound, and this is true up to 500lb additional amount.

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