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mt.f09.sol

# mt.f09.sol - C&O 350 Linear Optimization Fall 2009...

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C&O 350 Linear Optimization – Fall 2009 – Solutions to Mid-term Exam 1. 6 marks Solution: x 2 is free. Introduce u 2 , v 2 0 and substitute x 2 by u 2 - v 2 . Thus, the given LP in SIF is: max x 1 + u 2 - v 2 + x 3 + x 4 s. t. 2 x 1 - u 2 + v 2 + 4 x 3 - 10 - u 2 + v 2 + 5 x 3 - x 4 20 - x 1 - 3 x 3 + 4 x 4 - 30 x 1 + 3 x 3 - 4 x 4 30 x 1 , u 2 , v 2 , x 3 , x 4 0 2. 13 marks = 4+3+3+3 Solution: (a) The dual of given LP is min 15 y 1 + 10 y 2 s. t. - 3 y 1 + y 2 1 y 1 - 11 y 2 - 1 y 1 , y 2 0 (b) Yes. ˆ x 0 and satisfies the two constraints of (P). - 3(10 + 0 15 10 - 0 10 (c) No. ˆ y violates the second constraint in the dual of (P) (1(0) - 11(1) 6≥ - 1). (d) By part (b), (P) is not infeasible. Therefore, by the FTLP, (P) is either unbounded or has an optimal solution. 3. 16 marks = 2+2+8+4 Solution: (a) B ⊆ { 1 , . . . , n } is a basis if | B | = m and { A j : j B } is linearly independent. (b) x * is basic solution determined by a basis B if Ax * = b and x * j = 0 , j 6∈ B . (c) Notice that { 1 , 3 , 5 } and { 2 , 3 , 5 } are bases, which give us basic solutions x 1 := (3 , 0 , 3 , 0 , 9) T and x 2 := (0 , 6 , 12 , 0 , 6) T respectively. Since x 1 , x 2

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