midterm-2-solutions

midterm-2-solutions - MthSc 810 Mathematical Programming...

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MthSc 810 – Mathematical Programming Solutions of the second midterm exam Exercise 1 (25 pts.) Prove that if a polyhedron P = { x R n : A x b } does not contain an extreme point, then its recession cone is not { 0 } . Then consider the following polyhedron, where α > 0: P = { x R 2 + : αx 1 - x 2 ≥ - 1 , αx 1 - x 2 α } . 1. Find the extreme rays of P by applying the de±nition. 2. Prove that P is not contained in its recession cone. Solution. If a polyhedron P R n contains no extreme point, then it contains a line { x R n : x = ¯ x + λ d } , where ¯ x R n , d R n \ 0 , and λ R n . Therefore, d and - d are two extreme rays and the recession cone cannot be 0 . 1. C = { d R 2 + : αd 1 - d 2 0 , αd 1 - d 2 0 } = { d R 2 + : αd 1 - d 2 = 0 } . Hence C is the hal²ine { λ (1 , α ) , λ 0 } ⊂ R 2 + . 2. ¯ x = (0 , 1) P since - 1 ≥ - 1 and - 1 α for any α > 0, but ¯ x is not in C . Exercise 2 (25 pts.)
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midterm-2-solutions - MthSc 810 Mathematical Programming...

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