homework-2-solutions

homework-2-solutions - MthSc810 – Mathematical...

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Unformatted text preview: MthSc810 – Mathematical Programming Fall 2011. Solutions to Homework #2. Exercise 1 Solve the following problem using the graphical method: min x 1 + 4 x 2 s.t. x 1 + 2 x 2 ≤ 10 x 1 + x 2 ≥ 6 x 1 − x 2 ≤ 2 x 2 ≥ . Then write the problem in standard form. Solution. The standard form requires one slack for each constraint and x 1 , uncon- strained in sign, has to be rewritten as x + 1 − x − 1 . min x + 1 − x − 1 + 4 x 2 s.t. x + 1 − x − 1 + 2 x 2 + x 3 = 10 x + 1 − x − 1 + x 2 − x 4 = 6 x + 1 − x − 1 − x 2 + x 5 = 2 x + 1 , x − 1 , x 2 , x 3 , x 4 , x 5 ≥ . Exercise 2 Given two convex sets S 1 and S 2 in R n prove the following: – S 1 ∩ S 2 is convex; A.: Trivial: for any two vectors x , y ∈ S 1 ∩ S 2 and any λ ∈ [0 , 1], λ x +(1 − λ ) y ∈ S 1 and λ x +(1 − λ ) y ∈ S 2 for convexity of S 1 and S 2 , and hence λ x +(1 − λ ) y ∈ S 1 ∩ S 2 . – S 1 ⊕ S 2 = { x 1 + x 2 : x 1 ∈ S 1 , x 2 ∈ S 2 } is convex; A.: For any two vectors x , y ∈ S 1 ⊕ S 2 and any λ ∈ [0 , 1], there certainly exist x ′ ∈ S 1 and x ′′ ∈ S 2 such that x = x ′ + x ′′ , and there exist y ′ ∈ S 1 and y ′′ ∈ S 2 such that y = y ′ + y ′′ . Then for any λ ∈ [0 , 1], λ x + (1 − λ ) y = λ ( x ′ + x ′′ ) + (1 − λ )( y ′ + y ′′ ) = λ x ′ + (1 − λ ) y ′ + λ x ′′ + (1 − λ ) y ′′ = u + v , where u = λ x ′ + (1 − λ ) y ′ ∈ S 1 and v = λ x ′′ + (1 − λ ) y ′′ ∈ S 2 for convexity of S 1 and S 2 , hence u + v...
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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.

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homework-2-solutions - MthSc810 – Mathematical...

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