HW_optional_soln

HW_optional_soln - Solution of Optional Assignment 6.10.1...

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Solution of Optional Assignment 6.10.1 a . min w = 600y 1 + 400y 2 + 500y 3 s.t. 4y 1 + y 2 + 3y 3 6 (1) 9y 1 + y 2 + 4y 3 10 (2) 7y 1 + 3y2 + 2y 3 9 (3) 10y 1 + 40y 2 + y 3 20 (4) y 1 , y 2 , y 3 , y 4 0 b . since s 3 >0, y 3 = 0. Since x 1 >0, (1) is binding. Since x 4 >0, (4) is binding. Setting y 3 = 0 and solving (1) and (4) simultaneously yields y 1 = 22/15 y 2 = 2/15. Thus the optimal dual solution is y 1 = 22/15, y 2 = 2/15, y 3 = 0, z = 2800/3. c . (40): s 3 >0 implies y 3 = 0. This is reasonable because if all available glass is not being used an additional ounce of glass will not increase z hence glass constraint should have 0 shadow price. (41): y 2 >0 implies s 2 = 0. Since y 2 >0, an additional minute of packaging time will increase z. Thus all packaging time that is currently available must be used (hence s 2 = 0). (42):e 2 = 9(22/15) + 2/15 - 10 = 50/15>0 implies x 2 = 0. Note that e 2 = [Cost of Resources Used to Make a Beer glass] -[Price of a Beer Glass]. Since e 2 >0, producing a beer glass would not be profitable so x 2 = 0 should hold. (43): x
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HW_optional_soln - Solution of Optional Assignment 6.10.1...

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