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# Hw9s - ENGRI 1101 Engineering Applications of OR Fall 08...

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ENGRI 1101 Engineering Applications of OR Fall ’08 Homework 9 Homework 9 Solutions 1. (a) In this problem the initial feasible dictionary is: x 4 = 4 - x 1 - x 2 - 2 x 3 x 5 = 5 - 2 x 1 - 3 x 3 x 6 = 7 - 2 x 1 - x 2 - 3 x 3 z = 3 x 1 +2 x 2 +4 x 3 Now we want to increase the objective value, z . In this case, increasing any of x 1 , x 2 , x 3 would increase the value of z . We arbitrarily choose x 1 . Examining the upper bounds imposed by the constraints x 4 0 , x 5 0 , x 6 0, we find x 5 0 to be most stringent, which implies x 1 2 . 5. From the second equation in the previous dictionary we have x 1 = 2 . 5 - 1 . 5 x 3 - 0 . 5 x 5 and substituing it into the previous dictionary we arrive at a new dictionary with increased optimal value: x 4 = 1 . 5 - x 2 - 0 . 5 x 3 +0 . 5 x 5 x 1 = 2 . 5 - 1 . 5 x 3 - 0 . 5 x 5 x 6 = 2 - x 2 + x 5 z = 7 . 5 +2 x 2 - 0 . 5 x 3 - 1 . 5 x 5 Now we can only choose to increase x 2 to increase z . Of all the constraints, x 4 0 limits the increase of x 2 the most (making x 2 1 . 5). From the first equation in the previous dictionary we have x 2 = 1 . 5 - 0 . 5 x 3 - x 4 + 0 . 5 x 5 and substituting it into the previous dictionary we arrive at a new dictionary with increased optimal value: x 2 = 1 . 5 - 0 . 5 x 3 - x 4 +0 . 5 x 5 x 1 = 2 . 5 - 1 . 5 x 3 - 0 . 5 x 5 x 6 = 0 . 5 +0 . 5 x 3 + x 4 +0 . 5 x 5 z = 10 . 5 - 1 . 5 x 3 - 2 x 4 - 0 . 5 x 5

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