sol3 - , x 2 = 80. #3, 4.6: z x 1 x 2 s 1 s 2 RHS 1-2 5 3 8...

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AMS 341 Spring, 2010 Homework Set # 3 — Solution Notes #3, 4.1: The standard form of the problem is: min z = 3 x 1 + x 2 s . t . x 1 - e 1 = 3 x 1 + x 2 + s 2 = 4 2 x 1 - x 2 = 3 x 1 , x 2 , e 1 , s 2 0 #3, 4.4: Basic Varibles BFS Corner Point x 1 , x 2 x 1 = 150 , x 2 = 100 , s 1 = s 2 = 0 (150 , 100) x 1 , s 1 x 1 = 200 , s 1 = 150 , x 2 = s 2 = 0 (200 , 0) x 1 , s 2 x 1 = 350 , s 2 = - 300 , x 2 = s 1 = 0 Infeasible x 2 , s 1 x 2 = 400 , s 1 = - 450 , x 1 = s 2 = 0 Infeasible x 2 , s 2 x 2 = 175 , s 2 = 225 , x 1 = s 1 = 0 (0 , 175) s 1 , s 2 s 1 = 350 , s 2 = 400 , x 1 = x 2 = 0 (0 , 0) #5, 4.5: z x 1 x 2 s 1 s 2 s 3 RHS 1 - 1 - 1 0 0 0 0 0 4 1 1 0 0 100 0 1 1 0 1 0 80 0 1 0 0 0 1 40 1 0 0 0 1 0 80 0 3 0 1 - 1 0 20 0 1 1 0 1 0 80 0 1 0 0 0 1 40 Optimal solution: z = 80 , x 1 = 0
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Unformatted text preview: , x 2 = 80. #3, 4.6: z x 1 x 2 s 1 s 2 RHS 1-2 5 3 8 1 12 2 3 1 6 x 2 enters the basis, s 1 wins the min ratio test,, so it leaves the basis. z x 1 x 2 s 1 s 2 RHS 1-31 / 8-5 / 8-7 . 5 3 / 8 1 1 / 8 1 . 5 7 / 8-3 / 8 1 1 . 5...
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This note was uploaded on 05/03/2010 for the course AMS 341 taught by Professor Arkin,e during the Spring '08 term at SUNY Stony Brook.

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