x1 0 1 2 3 4 5 6 7 therefore

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Unformatted text preview: ............. ...... . ..... ... . .... ................ . . .... ... .............. . .. .... . ..... .......... −3.......... . ....−3 . .. .... ... ..... .................. .............. . . ..... ............. . ..... .... . ... .. . . .... ............................................................................................................ ..... x1 . . .. . 0 1 2 3 4 5 6 7 Therefore, only the first constraint is active, so we must have u2 = u3 = 0 in the dual feasibility condition − f (x) = −c − Cx = A u. Therefore, A u = u1 (1, 1) , so we are looking for x1 , x2 , and u1 which satisfy 1 = −c, 1 x1 + x2 = 4. Cx + u1 That is, 13 5 1 x1 67 5 13 1 x2 = 59 , 1 1 0 u1 4 which has solution x1 = 2.5, x2 = 1.5, and u1 = 27. Therefore, the optimality conditions are satisfied by 27 2.5 x= , u = 0 , 1.5 0 7 which implies that the optimal solution is x∗ = (2.5, 1.5) with f (x∗ ) = −182. (5.6) Note: this question refers to Example 1.3, not Example 1.2. Given µ = (µ1 , ....
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