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Unformatted text preview: CO250/CM340 INTRODUCTION TO OPTIMIZATION  HW 4 SOLUTIONS Exercise 1 (10 marks) . (a) No, since the columns { 1 , 2 , 5 } do not form an identity matrix (they are the right columns but the wrong order — by the definitions in this course, ‘canonical’ does not allow this). (b) No, since the columns { 1 , 5 , 6 } are not linearly independent, so the matrix they form is not invertible (it has a zero row). (c) Same as (a) — the columns { 2 , 5 , 6 } are still in the wrong order. (d) Yes — columns { 1 , 5 , 7 } give an identity matrix, and x 1 , x 5 and x 7 have zero coefficients in the objective function. (e) No — x 4 has nonzero coefficient in the objective function. Exercise 2 (20 marks) . (a) With B = { 1 , 4 } we have A B = 1 1 2 1 ! , A 1 B = 1 1 2 1 ! , A 1 B A N = 1 1 2 1 !  1 2 0 0 1 1 ! = 1 1 1 2 3 1 ! ; A 1 B b = 1 1 2 1 ! 1 1 ! = 2 3 ! . So the new constraints are 1 1 1 0 1 0 2 3 1 1 ! x = 2 3 ! . We use y = A T B c B = 1 2 1 1 ! 1 1 ! = 3 2 ! so the new objective function is y T b + ( c T y T A ) x = ( 3 , 2) 1 1 ! + ( (1 , 2 , , 1 , 3) (1 , 3 , 4 , 1 , 2) ) x = 5 + (0 , 5 , 4 , , 1) x. So altogether it is max 5 + (0 , 5 , 4 , , 1) x subject to 1 1 1 0 1 0 2 3 1 1 ! x = 2 3 ! x ≥ . 1 2 The basis is infeasible since the basic solution is ( 2 , , , 3 , 0) . (b) With B = { 3 , 5 } we have A B = 2 0 1 1 ! , A 1 B = 1 / 2 0 1 / 2 1 ! , A 1 B A N = 1 / 2 0 1 / 2 1 ! 1 1 1 2 1 ! = 1 / 2 1 / 2 1 / 2 3 / 2 1 / 2 1 / 2 ! ; A 1 B b = 1 / 2 0 1 / 2 1 ! 1 1 ! = 1 / 2 3 / 2 ! . So the new constraints are 1 / 2 1 / 2 1 1 / 2 0 3 / 2 1 / 2 1 / 2 1 ! x = 1 / 2 3 / 2 ! . We use y = A T B c B = 1 / 2 1 / 2 1 ! 3 ! = 3 / 2 3 ! so the new objective function is y T b + ( c T y T A ) x = ( 3 / 2 , 3) 1 1 ! + ( (1 , 2 , , 1 , 3) (9 / 2 , 3 / 2 , , 3 / 2 , 3) ) x = 9 / 2 + ( 7 / 2 , 7 / 2 , , 5 / 2 , 0) x. So altogether it is max 9 / 2 + ( 7 / 2 , 7 / 2 , , 5 / 2 , 0) x subject to 1 / 2 1 / 2 1 1 / 2 0 3 / 2 1 / 2 1 / 2 1 ! x = 1 / 2 3 / 2 ! x ≥ . The basis is infeasible since the basic solution is (0 , , 1 / 2 , , 3 / 2) . Exercise 3 (10 marks) . (a) If the variable to enter is x i then we need c i > in the objective function. Thus i = 1 , 4, 6 or 8, and of course all these are valid choices. (b) If 1 enters the basis, the ratios to consider are 2 / 1 and 4 / 2 . These are equal, so either the first or second basis elements may leave the basis, i.e. 2 or 5 may leave. (The new basis is { 1 , 5 , 7 } or { 1 , 2 , 7 } ). If 4 enters the basis, the ratios are ( , 4 / 1 , ) where means negative or 0. Hence only the second basis element, i.e. 5, may leave. 3 If 6 enters the basis, all the ratios are negative, so nothing may leave. (Explanation: this shows the LP is unbounded — the simplex algorithm comes to a halt here.) If 8 enters the basis, the ratios are (2 / 3 , , 1 / 1) . Hence only the first basis element, i.e. 2, may leave....
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 Spring '10
 GUENIN
 Linear Algebra, Determinant, objective function, LP

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