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Unformatted text preview: w 4 . Thus we pivot with x 1 as the entering variable and w 4 as the leaving variable. Add 2*row4 to the objective, 2*row4 to row1, 2*row4 to row2, and4*row4 to row3. The resulting dictionary is: 22 w 49 x 2 w 1 = 22 w 4 +9 x 2 w 2 = 1 +2 w 4 +7 x 2 w 3 = 1 +4 w 4 +19 x 2 x 1 = 1w 45 x 2 Now notice the the coﬃcients in the objective are all negative so we have an optimal solution: ( x 1 , x 2 )=(1,0) and Z=2. You should also include a plot of the feasible region a two dimensional plot of the problem constraints. An arrow should be drawn from the starting point (0,0) to the optimal point (1,0) which is reached after one pivot. 3...
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 Spring '08
 AlexandreW.d'Aspremont
 UCI race classifications, X1, largest positive coefficient

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