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Assignment4 management science

# Assignment4 management science - For a football –4/5 =...

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x2 300 250 200 150 A B 100 50 Z 0 50 100 150 C 200 300 350 x1 A: x1 = 0 x2 = 160 Z = 2,560 B: x1 = 128.5 x2 = 57.2 Z = 2,457.2 C: x1 = 167 x2 = 0 Z = 2,004 (a) A: 3(0) + 2(160) + s 1 = 500 s 1 = 180 4(0) + 5(160) + s 2 = 800 s 2 = 0 B: 3(128.5) + 2(57.2) + s 1 = 500 s 1 = 0 4(128.5) + 2(57.2) + s 2 = 800 s 2 = 0 C: 2(167) + 2(0) + s 1 = 500 s 1 = 0 4(167) + 5(0) + s 2 = 800 s 2 = 132 (b) Z = 12 x 1 + 16 x 2 and, x 2 = Z/16 – 12 x 1/16 The slope of the objective function, –12/16, would have to become steeper than the slope of the constraint line 4 x 1 + 5 x 2 = 800, for the solution to change. The profit, c 1, for a basketball that would change the solution point is, – 4/5 = – c 1/16 5 c 1 = 64 c 1 = 12.8 Since \$13 > 12.8 the solution point would change to B where x 1 = 128.5, x 2 = 57.2. The

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new Z value is \$2,585.70.
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Unformatted text preview: For a football, –4/5 = –12/ c 2 4 c 2 = 60 c 2 = 15 Thus, if the profit for a football decreased to \$15 or less, point B will also be optimal. The solution at B is x 1 = 128.5, x 2 = 57.2, and Z = \$2,400. (c) If the constraint line for rubber changes to 3 x 1 + 2 x 2 = 1,000, it moves outward, eliminating points B and C . However, since A is the optimal point, it will not change and the optimal solution remains the same, x 1 = 0, x 2 = 160, and Z = 2,560. There will be an increase in slack, s 1, to 680 lbs. If the constraint line for leather changes to 4 x 1 + 5 x 2 = 1,300, point A will move to a new location, x 1 = 0, x 2 = 250, Z = \$4,000....
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Assignment4 management science - For a football –4/5 =...

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