sol3 - 1 = 15 x 2 = 5 x 3 = 0#4 4.6 z x 1 x 2 s 1 s 2 RHS 1...

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AMS 341 Spring, 2011 Homework Set # 3 — Solution Notes 1). The standard form of the problem is: min z = 2 x 1 + x 2 s . t . 3 x 1 + 3 x 2 - e 1 = 6 x 1 + + s 2 = 5 x 1 - x 2 = 2 x 1 ,x 2 ,e 1 ,s 2 0 #1, 4.4: Basic Varibles BFS Corner Point x 1 ,x 2 ,s 1 x 1 = 40 ,x 2 = 40 ,s 1 = - 20 ,s 2 = s 3 = 0 infeasible x 1 ,x 2 ,s 2 x 1 = 40 ,x 2 = 20 ,s 2 = 20 ,s 1 = s 3 = 0 F x 1 ,x 2 ,s 3 x 1 = 20 ,x 2 = 60 ,s 3 = 20 ,s 1 = s 2 = 0 G x 1 ,s 1 ,s 2 x 1 = 40 ,s 1 = 20 ,s 2 = 40 ,s 3 = x 2 = 0 E x 1 ,s 1 ,s 3 x 1 = 80 ,s 1 = - 60 ,s 3 = - 40 ,s 2 = x 2 = 0 infeasible x 1 ,s 2 ,s 3 x 1 = 50 ,s 2 = 30 ,s 3 = - 10 ,s 1 = x 2 = 0 infeasible x 2 ,s 1 ,s 2 not possible infeasible x 2 ,s 1 ,s 3 x 2 = 80 ,s 1 = 20 ,s 3 = 40 ,s 2 = x 1 = 0 D x 2 ,s 2 ,s 3 x 2 = 100 ,s 2 = - 20 ,s 3 = 40 ,s 1 = x 1 = 0 infeasible s 1 ,s 2 ,s 3 s 1 = 100 ,s 2 = 80 ,s 3 = 40 ,x 1 = x 2 = 0 H #3, 4.5: z x 1 x 2 x 3 s 1 s 2 s 3 RHS 1 - 2 1 - 1 0 0 0 0 0 3 1 1 1 0 0 60 0 1 - 1 2 0 1 0 10 0 1 1 - 1 0 0 1 20 1 0 - 1 3 0 2 0 20 0 0 4 - 5 1 - 3 0 30 0 1 - 1 2 0 1 0 10 0 0 2 - 3 0 - 1 1 10 1 0 0 1 . 5 0 1 . 5 0 . 5 25 0 0 0 1 1 - 1 - 2 10 0 1 0 0 . 5 0 0 . 5 2 15 0 0 1 - 1 . 5 0 - 0 . 5 0 . 5 5 Optimal solution: z = 25 , x
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Unformatted text preview: 1 = 15 , x 2 = 5 , x 3 = 0 #4, 4.6: z x 1 x 2 s 1 s 2 RHS 1 3-8 4 2 1 12 2 3 1 6 x 1 enters the basis, there is a tie in the min ratio test,, so we can choose either s 1 or s 2 to leave the basis, we choose s 2 : z x 1 x 2 s 1 s 2 RHS 1-12 . 5-1 . 5-9-4 1-2 1 1 . 5 . 5 3...
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This note was uploaded on 05/28/2011 for the course AMS 341 taught by Professor Arkin,e during the Spring '08 term at SUNY Stony Brook.

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