sol9 - CO350 LINEAR OPTIMIZATION - SOLUTION HW9 Exercise...

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Unformatted text preview: CO350 LINEAR OPTIMIZATION - SOLUTION HW9 Exercise 10.4.2 (a) Solve - 3 1 1 1 1- 2 1 y 1 y 2 y 3 = 2- 1 y 1 y 2 y 3 = 1 1 Next we have c 4 = c 4- A T 4 y =- 1- (0 , 1 , 0)(0 , 1 , 1) T = 0 c 5 = c 5- A T 5 y = 1- (2 , 1 , 1)(0 , 1 , 1) T =- 1 Since both values are not positive, x * is optimal. (b) With the change we have c 5 = c 5- A T 5 y = R- (2 , 1 , 1)(0 , 1 , 1) T = R- 2 For optimality we need c 5 which implies R 2 . (c) - 3 1 1 1 1 1- 2 0 d 1 d 2 d 3 = 2 1 1 d 1 d 2 d 3 = 1 5 t = min(3 / 1 ,- , 5 / 5) = 1 hence the leaving variable is x 3 . For the new solution we have x 5 := 1 , x 2 := 2- 1 0 = 2 and x 1 := 3- 1 1 = 2 , or x * = (2 , 1 , , , 1) T . (d) Solve - 3 1 1 1 1- 2 1 y 1 y 2 y 3 = 2 R y 1 y 2 y 3 = (4 + R ) / 3 (2- R ) / 3 We have c 4 = c 4- A T 4 y =- 1- (0 , 1 , 0)(0 , (4 + R ) / 3 , (2- R ) / 3) T = (- 7- R ) /...
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This note was uploaded on 02/21/2010 for the course CO 350 taught by Professor S.furino,b.guenin during the Winter '07 term at Waterloo.

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sol9 - CO350 LINEAR OPTIMIZATION - SOLUTION HW9 Exercise...

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