homework-2-solutions

# homework-2-solutions - MthSc810 Mathematical Programming...

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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 0 . 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 0 . 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 S 1 S 2 . The proof is complete.

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S 1 S 2 = { x 1 x 2 : x 1 S 1 , x 2 S 2 } is convex.
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