sol2 - ), and has objective value c t x = c t x-c t c ....

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AMS 540 / MBA 540 (Fall, 2010) Estie Arkin Homework Set # 2: Solution notes (1). (a) Let S be a feasible set, x 1 , x 2 S , 0 λ 1 then: A ( λx 1 + (1 - λ ) x 2 ) = λAx 1 + (1 - λ ) Ax 2 = λb + (1 - λ ) b = b Also: λ 0 and x 1 0 imply λx 1 0, and 1 - λ 0 and x 2 0 imply (1 - λ ) x 2 0, so λx 1 +(1 - λ ) x 2 0. Thus λx 1 + (1 - λ ) x 2 S . (b) Let S be the set of all optimal solutions, x 1 , x 2 S , 0 λ 1 then by part (a) λx 1 + (1 - λ ) x 2 S . Also c ( λx 1 + (1 - λ ) x 2 ) = λcx 1 + (1 - λ ) cx 2 = λz + (1 - λ ) z = z , where z = min { cx | x S } , therefore λx 1 + (1 - λ ) x 2 S . 2). Intuitively, since x 0 is stirctly interior to the feasible region, we can move in any direction we want from x 0 some small but positive distance, while remaining in the feasible region. In particuler if we go in a direction for which the objective function improves, we will get a feasible point with a better objective value, and thus x 0 is not an optimal solution. To make this more precise, consider x = x 0 - λc for λ > 0 but small. x is also feasible (for small enough λ
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Unformatted text preview: ), and has objective value c t x = c t x-c 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 = 0 + (1- )(-p 1 + p 2 ) 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 case-p 1 + p 2 < 0 so we see from part (b) that the optimal solution is = 1, z = 0, x 1 = x 2 = 0....
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