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Unformatted text preview: MthSc 810 – Mathematical Programming Solutions of Midterm Exam I, Fall 2011. Exercise 1 (25 pts.) Are the following sets convex? Explain why or why not. 1. S 1 = { x ∈ R n : x = α u + β v ∀ α ∈ R , β ∈ R } , for two fixed u and v in R n . 2. S 2 = { x ∈ R n : f ( x )+ αg ( x ) ≤ } , where f is affine, g is convex, and α ∈ R . 3. S 3 = { x ∈ R n : A x = b , x ≤ } , with A ∈ R m × n and b ∈ R m . 4. S 4 = { x ∈ R n : x = x + P u , u ∈ R p ,  u  2 ≤ ǫ } , where x ∈ R n , P ∈ R n × p , and ǫ ∈ R are given, and for any vector a ∈ R n ,  a  2 = radicalbig ∑ n i =1 a 2 i . Solution. 1. Short answer: the affine hull of any k vectors is convex. Long version: S 1 is convex. Any two x ′ and x ′′ admit α ′ , β ′ , α ′′ , β ′′ such that x ′ = α ′ u + β ′ v and x ′′ = α ′′ u + β ′′ v . Therefore, for λ ∈ [0 , 1], λ x ′ + (1 − λ ) x ′′ = λ ( α ′ u + β ′ v ) + (1 − λ )( α ′′ u + β ′′ v ) = ( λα ′ + (1 − λ ) α ′′ ) u + ( λβ ′ + (1 − λ ) β ′′ ) v = α ′′′ u + β ′′′ v ∈ S 1 since α ′′′ , β ′′′ ∈ R . 2. S 2 is certainly convex for α ≥ 0: f is affine hence convex, and f + αg is the sum of two convex functions and hence convex. Consequently, the constraint f ( x ) + αg ( x ) ≤ 0 is convex and so is S 2 . For α < 0, S 2 may be nonconvex....
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This note was uploaded on 03/14/2012 for the course MTHSC 810 taught by Professor Staff during the Fall '08 term at Clemson.
 Fall '08
 Staff
 Math, Sets

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