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# S1 - C&O 350 Linear Optimization Fall 2009 Solutions to...

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C&O 350 Linear Optimization – Fall 2009 – Solutions to Assignment 1 Question # Max. marks Part marks 1 6 2 2 2 2 8 2 2 2 2 3 10 4 5 5 12 3 3 3 3 Total 41 1. (Linear Algebra Review) 6 marks = [2+2+2] Solution: (a) S is a linearly independent set if and only if for all scalars c 1 , . . . , c n , the equation n i =1 c i v ( i ) = 0 implies that c 1 = c 2 = · · · = c n = 0 . (b) Since the set is linearly dependent, there exist scalars c 1 , . . . , c n that are not all zeros such that n i =1 c i v ( i ) = 0. Pick k ∈ { 1 , . . . , n } such that c k 6 = 0. Then, we have - c k v k = i ∈{ 1 ,...,n } ,i 6 = k c i v ( i ) . Thus, v k can be written as - i ∈{ 1 ,...,n } ,i 6 = k c i c k v ( i ) , a linear combinations of other vectors in S . (c) The given set is not a subspace of R 2 . Observe that both the vectors v = [ - 1 , 1] T and w = [1 , 1] T are in the given set, but v + w = [0 , 1] T is not in the set. 2. (Linear Algebra Review) [8 marks = 2+2+2+2] Solution: (a) TRUE. rank( A ) < n implies that the column vectors of A are linearly dependent. Therefore, the zero vector can be written as a nontrivial linear combination of the column vectors of A . Hence, there exist scalars c 1 , . . . , c n that are not all zeros such that Ac = 0, where c denotes the column vector ( c 1 , . . . , c n ) T . (b) FALSE. Notice that a system Ax = b may have a unique solution if rank( A ) = n . Therefore, we can construct a counter example that has rank( A ) = n and n < m . One such example is: A = 1 3 and b = 1 3 . Then the unique solution to Ax = b is x = 1. (c) FALSE. In general, if we start with a feasible system of linear inequalities, and then we change the right-hand-side, then the resulting system may be infeasible. A counter example: A = 1 - 1 , b = 2 0 and b 0 = 0 - 2 . Although Ax b has a solution (any x [0 , 2] would do), Ax b 0 implies that x 0 and - x ≤ - 2, which is obviously infeasible. (d) TRUE. Since A is square and Ax = b has a unique solution, we know that A must be non-singular, and hence so is A T . Thus, ( A T ) - 1 exists. Then A T x = b 0 ⇐⇒ ( A T ) - 1 ( A T x ) = ( A T ) - 1 ( b 0 ) ⇐⇒ x = ( A T ) - 1 b 0 , and so x = ( A T ) - 1 b 0 is the unique solution to A T x = b 0 . 3. (LP Problem formulation) 1

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[10 marks] 3 for correct definition of decision variables 2 for objective function (1 for correct data, 1 for “min”) 4 for the constraints 1 for non-negativity constraints Solution: Decision Variables: For i = 1 , 2 , 3 , 4 , 5, we define the decision variables x i
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S1 - C&O 350 Linear Optimization Fall 2009 Solutions to...

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