a10sol

# a10sol - C O350 L INEAR O PTIMIZATION S OLUTIONS 10...

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CO350 LINEAR OPTIMIZATION - SOLUTIONS 10 Exercise 10.4.2 (a) Solve - 3 1 1 1 1 - 2 1 0 0 y 1 y 2 y 3 = 2 - 1 0 y 1 y 2 y 3 = 0 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 0 which implies R 2 . (c) - 3 1 1 1 1 0 1 - 2 0 d 1 d 2 d 3 = 2 1 1 d 1 d 2 d 3 = 1 0 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 , 0 , 0 , 1) T . (d) Solve

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## This note was uploaded on 01/05/2012 for the course CO 350 taught by Professor S.furino,b.guenin during the Spring '07 term at Waterloo.

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a10sol - C O350 L INEAR O PTIMIZATION S OLUTIONS 10...

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