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8. Sample3_Solution - ADM 2302 Problem 1 Part a Midterm...

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ADM 2302 Midterm Exam Winter 04 Problem 1 Part a) Let xij = number of ads of type i (1=N, 2=T, 3=R) to be produce at MRj (j: 1. 2, 3, 4) Minimize 16x11 + 10x12 + 12x13 + 12x14 + 26x21 + 20x22 + 30x23 + 21x24 + 22x31 + 15x32 + 23x33 + 14x34 subject to : x11 + x12 + x13 + x14 = 30 x21 + x22 + x23 + x24 = 15 x31 + x32 + x33 + x34 = 25 x11 + x21 + x31 <= 20 x12 + x22 + x32 <= 25 x13 + x23 + x33 <= 15 x14 + x24 + x34 <= 20 All variables 0 Part b) The network presentation can be used, or just a paragraph stating: The types of ads can be considered to be sources. The four market research firms can be considered to be destinations. (OR VICE VERSA). In either interpretation, we have supplies and demands, and the xij’s in the LP represent the shipment amounts for these supplies and demands. Problem 2: X1 = the number of households with children < 16 and are premium subscribers. (c1 = 5+4+3) X2 = the number of households with children <16 and are not premium subscribers. (c2 = 5+4) X3 = the number of households without children and are premium subscribers. (c3 = 5+3) X4 = the number of households without children and are not premium subscribers. (c4= 5) Min Cost = 12X1 + 9X2 + 8X3 + 5X4 Subject to X1 + X2 + X3 + X4 > 50 X1 + X3 >= 0.5(X1+X2+X3+X$) .5X1 - .5X2 + .5X3 - .5X4 > 0 X1+X2 >= 0.6 (X1+X2+X3+X4) .4X1 + .4X2 - .6X3 - .6X4 >
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