Unformatted text preview: ), and has objective value c t x ′ = c t xλc t c . Since c is not a zero vector, we must have that c t c > 0 (it is the norm of vector c ), and so c t x ′ < c t x . 3). (a). What are the extreme points and extreme directions of this LP? Extreme points are v 1 = (0 , 0) and v 2 = (1 , 1), there are no extreme directions (the region is a line segment, which is bounded!). (b). Write the equivalent LP in terms of the extreme points and extreme directions. c = (p 1 , p 2 ), cv 1 = 0, cv 2 =p 1 + p 2 max z = λ · 0 + (1λ )(p 1 + p 2 ) ≤ λ ≤ 1 (c). Now suppose that p 1 > p 2 . What is the optimal solution to the LP? (Make sure to state z as well as x 1 , x 2 .) In this casep 1 + p 2 < 0 so we see from part (b) that the optimal solution is λ = 1, z = 0, x 1 = x 2 = 0....
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 Fall '08
 Arkin,E
 Optimization, #, optimal solution, extreme points, Estie Arkin

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