final-exam-solutions

# final-exam-solutions - MthSc 810 Mathematical Programming...

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MthSc 810 – Mathematical Programming Solutions to the final exam, Fall 2011. Problem 1 (5 pts.): 1. Prove that the set S = { ( x 1 , x 2 ) : x 1 x 2 1 , x 1 > 0 , x 2 > 0 } is convex. A. : There are several ways. First, note that x 1 x 2 1 x 2 1 x 1 1 x 1 x 2 0, a convex constraint given that 1 x 2 is convex for x 2 > 0. Another way is as follows: since we want to prove that x S, x ′′ S, λ x +(1 λ ) x ′′ S , we need to prove x 1 x 2 1 x ′′ 1 x ′′ 2 1 ( λx 1 +(1 λ ) x ′′ 1 )( λx 2 +(1 λ ) x ′′ 2 ) 1. To this purpose, let’s take the logarithm of both sides (which exists since S R 2 + ): ( λx 1 + (1 λ ) x ′′ 1 )( λx 2 + (1 λ ) x ′′ 2 ) 1 log( λx 1 + (1 λ ) x ′′ 1 ) + log( λx 2 + (1 λ ) x ′′ 2 ) 0 . log( λx 1 + (1 λ ) x ′′ 1 ) + log( λx 2 + (1 λ ) x ′′ 2 ) λ log x 1 + (1 λ ) log x ′′ 1 + λ log x 2 + (1 λ ) log x ′′ 2 = λ log( x 1 x 2 ) + (1 λ ) log( x ′′ 1 x ′′ 2 ) = 0 , the third inequality a consequence of the logarithm being a concave function. 2. Prove that the set G = { α R + : 1 min { αx : x 1 }} is convex. A. : Note that min { αx : x 1 } = α for any α 0, hence G = { α R + : α 1 } = [0 , 1], which is convex. Problem 2 (15 pts.): Consider the problem min 3 x 1 + x 2 s.t. x 1 x 2 x 3 = 1 2 x 1 x 2 + x 4 = 4 x 1 , x 2 , x 3 , x 4 0 . 1. The basis B = { 2 , 4 } is neither primal nor dual feasible. However, there is one pivoting operation that suffices to obtain a basis that is either primal or dual feasible. Carry out that pivoting operation and write the resulting dictionary (or tableau). A. : First dictionary: x 2 = 1 + x 1 x 3 x 4 = 3 x 1 x 3 z = 1 2 x 1 x 3 . Let’s try to obtain primal feasibility. To this purpose, x 3 can’t help because x 3 > 0 means x 2 < 1, so let x 1 enter the basis. For x 1 = 1 we get back primal feasibility while x 2 becomes nonbasic, and this would be the corre- sponding dictionary: x 1 = 1 + x 2 + x 3 x 4 = 2 x 2 2 x 3 z = 3 2 x 2 3 x 3 . 2. Consider again the first basis, and suppose you could change a coefficient of x 1 or x 3 in any of the two linear constraints. Does any such change make the solution primal feasible? Explain. Note: there is no need to try any change. A. : No. The initial solution is infeasible because B 1 b negationslash≥ 0, and a change in any coefficient of x 1 or x 3 would change N , not B .

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