Chapter 15

# Chapter 15 - Multicriteria Decision Problems 3 a Let x1 =...

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Multicriteria Decision Problems 3. a. Let x 1 = number of units of product 1 produced x 2 = number of units of product 2 produced Min P 1 ( d 1 + ) + P 1 ( d 1 - ) + P 1 ( d 2 + ) + P 1 ( d 2 - ) + P 2 ( d 3 - ) s.t. 1 x 1 + 1 x 2 - d 1 + + d 1 - = 350 Goal 1 2 x 1 + 5 x 2 - d 2 + + d 2 - = 1000 Goal 2 4 x 1 + 2 x 2 - d 3 + + d 3 - = 1300 Goal 3 x 1 , x 2 , d 1 + , d 1 - , d 2 + , d 2 - , d 3 - , d 3 + 0 b. In the graphical solution, point A provides the optimal solution. Note that with x 1 = 250 and x 2 = 100, this solution achieves goals 1 and 2, but underachieves goal 3 (profit) by \$100 since 4(250) + 2(100) = \$1200. 0 100 200 300 400 500 100 200 300 400 500 600 700 oal 1 Goal 2 A (250, 100) B (281.25, 87.5) oal 3 x 2 x 1 c. Max 4 x 1 + 2 x 2 s.t. 1 x 1 + 1 x 2 350 Dept. A 2 x 1 + 5 x 2 1000 Dept. B x 1 , x 2 0 The graphical solution indicates that there are four extreme points. The profit corresponding to each extreme point is as follows: Extreme Point Profit 1 4(0) + 2(0) = 0 2 4(350) + 2(0) = 1400 3 4(250) + 2(100) = 1200

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Chapter 15 - Multicriteria Decision Problems 3 a Let x1 =...

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