ISYE 3133 Modeling2Sol

ISYE 3133 Modeling2Sol - Mi >=0 i=1,,3...

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ISyE 3133 C, Engineering Optimization Solutions to Modeling Handout #2 The solutions presented here might only include the variable description and LP formulation. Remember that you should also include the description of the objective and constraints in the Quiz. 1. In pages 86—91 of the Book 2. Let Mi = Tons of coal shipped from Mine i and Xij = Tons of coal shipped from Mine i to Customer j. MIN 50 M1 + 55 M2 + 62 M3 + 4 X11 + 6 X12 + 8 X13 + 12 X14 + 9 X21 + 6 X22 + 7 X23 + 11 X24 + 8 X31 + 12 X32 + 3 X33 + 5 X34 SUBJECT TO M1 <= 120 M2 <= 100 M3 <= 140 X11 + X21 + X31 = 80 X12 + X22 + X32 = 70 X13 + X23 + X33 = 60 X14 + X24 + X34 = 40 0.03 M1 + 0.01 M2 - 0.01 M3 <= 0 0.01 M1 - 0.01 M3 <= 0 M1 - X11 - X12 - X13 - X14 = 0 M2 - X21 - X22 - X23 - X24 = 0 M3 - X31 - X32 - X33 - X34 = 0
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Unformatted text preview: Mi &gt;=0 i=1,,3 Xij&gt;=0 i=1,,3 j=1,,4 3. Let (all variables are in ounces) Ing. 1 = Sugar, Ing. 2 = Nuts, Ing. 3 = Chocolate, Candy 1 = Slugger and Candy 2 = Easy Out Let x ij = Ounces of Ing. i used to make candy j. The appropriate LP is max z = 25(x 12 + x 22 + x 32 ) + 20(x 11 + x 21 + x 31 ) s.t. x 11 + x 12 100 (Sugar Const.) x 21 + x 22 20 (Nuts Constraint) x 31 + x 32 30 (Chocolate Const.) .8x 22-.2x 12-.2x 32 0 .9x 21-.1x 11-.1x 31 0 .9x 31-.1x 11-.1x 21 0 x ij 0 i=1,,3 j=1,,2 4. xij = barrels of oil i used to make product j (j = 1 is gasoline and j = 2 is heating oil) ai = dollars spent advertising product i max z = 25(x11 + x21) + 20(x12 + x22) - a1 - a2 s.t. x11 + x21 = 5a1 x12 + x22 = 10a2 2x11 - 3x21 0 x11 + x12 5,000 x21 + x22 10,000 4x12 - x22 0 xij 0 i=1,2 j=1,2...
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