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assign8.sol - CO 350 Assignment 8 – Winter 2010 Solutions...

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Unformatted text preview: CO 350 Assignment 8 – Winter 2010 Solutions Question # Max. marks Part marks 1 10 1 2 1 2 1 3 2 8 4 4 3 22 1 1 2 2 3 2 2 3 3 3 Total 40 1. (Exercise 1 of Section 10.4 of the course notes.) (a) We solve A B x * B = b to get x * B = [3 , 4 , 2] T . So x * = [3 , 4 , 2 , , , 0] T . It is feasible. (b) We solve A T B y = c B to get y = [- 3 , 2 , 1] T . So ¯ c = c- A T y = [0 , , , ,- 5 ,- 6] T . Since ¯ c ≤ ,x * is optimal. (c) We compute ¯ c new 4 = c new 4- A T 4 y = θ . So x * remains optimal if and only if ¯ c new 4 = θ ≤ 0. (d) If θ = 3, then ¯ c new 4 > 0 and so x 4 enters. We then solve A B d = A 4 to get d = [- 1 , 1 , 2] T . The minimum ratio t = min(- , 4 / 1 , 2 / 2) = 1. So x 3 leaves. (e) We compute ¯ c 7 = c 7- A T 7 y = θ + 1. So B remains an optimal basis if and only if ¯ c 7 ≤ 0, if and only if θ ≤ - 1. (f) Note that x 1 is basic. We solve A T B y new = c new B to get y new = [- 3- θ, 2- 2 θ, 1 + 2 θ ] T . Then we compute ¯ c new N = c N- A T N y new = [ θ,- 5- 2 θ,- 6 + 2 θ ] T . Now, ¯ c new ≤ 0 if and only if θ ≤ 0 and θ ≥ - 5 / 2. So x * stays optimal if and only if- 5 / 2 ≤ θ ≤ 0. 2. (Exercise 4 of Section 10.4. of the course notes.) (a) The new constraint, with slack variable x 8 added to it, is x 1 + x 3 + x 8 = 5 , x 8 ≥ . Eliminating the variables x 1 and x 3 from the above using the x 1-row and x 3-row gives- 2 x 2- 3 x 5- x 6- 2 x 7 + x 8 =- 1 . The initial basis in the augmented system is B = { 1 , 3 , 4 , 8 } . B is not an optimal basis because the added constraint has a negative right-hand side. However, B is a dual feasible basis. So we can apply the dual simplex method with x 8 as the leaving variable. Now, min ¯ a rj < ¯ c j ¯ a rj = min { 4 2 , 10 3 , 5 1 , 3 2 } = 3 2 . So we pivot on (8 , 7). The new basis is B = { 1 , 3 , 4 , 7 } . The tableau for this basis is z + x 2 + 11 2 x 5 + 7 2 x 6 + 3 2 x 8 = 33 2 x 1 +2 x 2 + 5 2 x 5 + 3 2 x 6- 1 2 x 8 = 3 2- 2 x 2 + x 3- 5 2 x 5- 3 2 x 6 + 3 2 x 8 = 7 2 5 x 2 + x 4 + 11 2 x 5 + 7 2 x 6- 1 2 x 8 = 13 2 x 2 + 3 2 x 5 + 1 2 x 6 + x 7- 1 2 x 8 = 1 2 This tableau is optimal. So the modified linear programming problem reaches an optimal value of 33 2 at x * = [ 3 2 , , 7 2 , 13 2 ] T .....
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This note was uploaded on 05/28/2011 for the course CO 250 taught by Professor Guenin during the Fall '10 term at Waterloo.

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assign8.sol - CO 350 Assignment 8 – Winter 2010 Solutions...

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