# fkey - MAT419 Linear Optimization May 1 2008 FINAL EXAM...

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Unformatted text preview: MAT419 Linear Optimization May 1, 2008 FINAL EXAM Name KEY Object of the Game: Solve 10 problems correctly! 0. Al’s refinery can buy two types of gasoline. Boosch Oil has available, at \$60 per barrel, 130000 barrels of 92 octane gasoline with vapor pressure 4 . 6 psi and sulfur content 0 . 58%. Chayni Oil has available, at \$70 per barrel, 140000 barrels of 85 octane gasoline with vapor pressure 6 . 5 psi and sulfur content 0 . 40%. Al needs to blend these two to produce 200000 barrels of a mixture with octane between 87 and 89, with vapor pressure at most 6 . 0 psi and sulfur content at most . 50%. Formulate a LOP to determine the proportions of each type he should use to minimize his cost. Mixing two different octanes yields a product with octane the weighted average of the two. Similarly for sulfur content and vapor pressure. Now, let B the amount of Boosch oil and C the amount of Chayni oil. Then, for the average vapor pressure, we have (46 B +65 C ) / ( B + C ) ≤ 60 , which implies 14 B − 5 C ≥ . With similar computations for the other constraints we get the following LOP. Min . w = 60 B + 70 C s . t . B + C = 200000 B ≤ 130000 C ≤ 140000 14 B − 5 C ≥ − 8 B + 10 C ≥ − 3 B + 4 C ≥ 5 B − 2 C ≥ & B , C ≥ 1. Solve the following LOP, including w ∗ , y ∗ , and x ∗ . Write the notation for each pivot along the way. Min . w = 12 y 1 + 12 y 2 + 2 y 3 − 5 y 4 s . t . − 4 y 1 − 7 y 2 − 8 y 3 + 4 y 4 ≥ − 7 9 y 1 − 7 y 2 + 13 y 3 + y 4 ≥ 10 12 y 1 + 6 y 2 − 12 y 3 + 12 y 4 ≥ 10 9 y 1 + 13 y 2 + 11 y 3 − 11 y 4 = − 7 y 2 , y 3 , y 4 ≥ After converting to standard maximum form, the initial tableau is F E 4 7 8 − 4 1 0 0 [0] 0 7 − 9 7 − 13 − 1 0 1 0 [0] 0 − 10 − 12 − 6 12 − 12 0 0 1 [0] 0 − 10 E − 9 − 13 − 11 11 0 0 0 [1] 0 7 12 12 2 − 5 0 0 0 [0] 1 . The sequence of pivots is as follows. Ph 0: 1 mapsto→ ∅ , Ph I: 4 mapsto→ 7 , 3 mapsto→ 6 , Ph II: 2 mapsto→ 5 . The optimal tableau is F E 0 19680 3360 960 − 320 [960] 23840 0 19680 3576 − 384 292 [1584] 37040 0 19680 5004 24 − 582 [2976] 61440 E 19680 − 3108 − 888 − 606 [ − 1872] − 19920 14844 24 7618 [22656] 19680 186080 , having optimum w ∗ = − 186080 / 19680 at y ∗ = ( − 19920 , 23840 , 37040 , 61440 T / 19680 , with certificate x ∗ = (14844 , 24 , 7618 , 22656) T / 19680 . 2. Consider the following LOP P ....
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## This note was uploaded on 02/24/2009 for the course MAT 18600 taught by Professor Hurlbert during the Spring '08 term at ASU.

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fkey - MAT419 Linear Optimization May 1 2008 FINAL EXAM...

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