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Unformatted text preview: Solution: (a) . The standard form is as follows max z = x 1 + 3 x 2 s.t. x 1 + 2 x 2 + s 1 = 4 3 x 1 + 2 x 2 + s 2 = 8 x 1 ,x 2 ,s 1 ,s 2 ≥ (b) . The basic solutions is listed in the following table Basic variables Basic solution Feasible? Objective Value ( x 1 ,x 2 ) (2,1) Yes 5 ( x 1 ,s 1 ) (8/3,4/3) Yes 8/3 ( x 1 ,s 2 ) (4,4) No( x 2 ,s 1 ) (4,4) No( x 2 ,s 2 ) (2,4) Yes 6 ( s 1 ,s 2 ) (4,8) Yes (c) . x 2 = 2, s 2 = 4, x 1 = s 1 = 0 is the optimal solution. (d) . The graph is as follows: ABCD is the feasible region, and the optimal solution is point B . 1 2 3 4 5 x 1 x 2 1 2 3 4 s 1 = 0 s 2 = 0 A B C D E F (e) . The two infeasible basic solutions are E and F . E : x 1 = 4 ,s 2 =4, F : x 2 = 4 ,s 1 =4. 2...
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 Spring '09
 Nedich
 Optimization, Standard form, min s.t.

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