Tutorial6sol - Tutorial 6: Outline of Solutions Q1. Solve...

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Unformatted text preview: Tutorial 6: Outline of Solutions Q1. Solve the following linear program by the big-M method min 3 x 1 + 2 x 2 + 3 x 3 s.t. 2 x 1 + x 2 + x 3 ≤ 2 3 x 1 + 4 x 2 + 2 x 3 ≥ 8 x 1 , x 2 , x 3 ≥ . Solution : Equivalent Standard LP: min 3 x 1 + 2 x 2 + 3 x 3 s.t. 2 x 1 + x 2 + x 3 + s 1 = 2 3 x 1 + 4 x 2 + 2 x 3- s 2 = 8 x 1 , x 2 , x 3 , s 1 , s 2 ≥ . The modified LP for large positive M is: min 3 x 1 + 2 x 2 + 3 x 3 + My s.t. 2 x 1 + x 2 + x 3 + s 1 = 2 3 x 1 + 4 x 2 + 2 x 3- s 2 + y = 8 x 1 , x 2 , x 3 , s 1 , s 2 , y ≥ . Start with x B = ( s 1 ,y ) , c B = (0 ,M ) , ¯ c i = c i- c B B- 1 A i . Thus to obtain the ¯ c-row: ¯ c-row = c-row- × ( s 1-row)- M × ( y-row) Basic x 1 x 2 x 3 s 1 s 2 y Solution c 3 2 3 M ¯ c 3- 3 M 2- 4 M 3- 2 M M- 8 M s 1 2 1 1 1 2 y 3 4 2- 1 1 8 ¯ c- 1 + 5 M 1 + 2 M- 2 + 4 M M- 4 x 2 2 1 1 1 2 y- 5- 2- 4- 1 1 Optimal tableau: x 1 = 0 ,x 2 = 2 ,x 3 = 0 with objective value 4. While the artificial variable y is still a basic variable, it has a value of 0. Hence it a optimal solution to the orginal linear program. NOTE : If x 2 is chosen as the entering variable, and y as the leaving variable, then we have the following optimal tableau: Basic x 1 x 2 x 3 s 1 s 2 y Solution ¯ c 3 / 2 2 1 / 2- 1 / 2 + M- 4 s 1 5 / 4 1 / 2 1 1 / 4- 1 / 4 x 2 3 / 4 1 1 / 2- 1 / 4 1 / 4 2 Again optimal tableau: x 1 = 0 ,x 2 = 2 ,x 3 = 0 with objective value 4, artificial variable is a nonbasic variable, but the surplus variable s 1 is basic with a zero value....
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Tutorial6sol - Tutorial 6: Outline of Solutions Q1. Solve...

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