asst6sol - C&O 350 Linear Optimization Winter 2009...

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Question # Max. marks Part marks 1 8 2 10 3 14 10 2 2 4 15 3 3 3 3 3 5 8 3 1 1 3 6 5 7 Bonus (20) (0 6 4 6 4) 8 No Credit 9 No Credit Total 60 1. 8 marks: 4 marks for auxiliary LP and initial tableau, distributed as: 1 for the artificial variables of auxiliary LP 1 for rest of auxiliary LP 1 for the w -row of initial tableau 1 for the BFS in initial tableau 2 for pivots and optimal tableau 2 for ” y ” (as in farkas lemma) and checking infeasibility Exercise 7.7.2 from the course notes. Solution : We are given the system of equations Ax = b , where A = 1 1 - 8 - 1 - 6 1 1 - 4 1 - 8 1 4 ,b = - 1 1 2 We begin with Phase 1 by formulating the problem, max { c T x | Dx = d, x 0 } where D = - 1 - 1 8 1 1 0 0 - 6 1 1 - 4 0 1 0 1 - 8 1 4 0 0 1 , d = 1 1 2 , and c T = ± 0 0 0 0 - 1 - 1 - 1 ² The initial basis is B = { 5 , 6 , 7 } and the corresponding basic variables are [ x 5 x 6 x 7 ] T = [1 1 2] T . The initial tableau is: w + 6 x 1 + 8 x 2 - 10 x 3 - x 4 = - 4 - x 1 - x 2 + 8 x 3 + x 4 + x 5 = 1 - 6 x 1 + x 2 + x 3 - 4 x 4 + x 6 = 1 x 1 - 8 x 2 + x 3 + 4 x 4 + x 7 = 2 . Iteration 1: We take x 3 as the entering variable, and calculate t = min { 1 / 8 , 1 / 1 , 2 / 1 } = 1 / 8. So x 5 leaves, and our new basis is { 3 , 6 , 7 } . The updated tableau is: 1
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w + 19 / 4 x 1 + 27 / 4 x 2 + 1 / 4 x 4 + 5 / 4 x 5 = - 11 / 4 - 1 / 8 x 1 - 1 / 8 x 2 + x 3 + 1 / 8 x 4 + 1 / 8 x 5 = 1 / 8 - 47 / 8 x 1 + 9 / 8 x 2 - 33 / 8 x 4 - 1 / 8 x 5 + x 6 = 7 / 8 9 / 8 x 1 - 63 / 8 x 2 + 31 / 8 x 4 - 1 / 8 x 5 + x 7 = 15 / 8 . Iteration 2: Since c j 0 j N , the current solution is optimal. We now proceed to calculate a certificate of infeasibility for the original problem. First we calculate the vector y as, D T B y = c B y = 1 / 4 - 1 - 1 . However, b 1 < 0, so in order to recover a certificate for the original system, we multiply y 1 by - 1 to obtain z = - 1 / 4 - 1 - 1 . We calculate z T A = ± 5 - 1 / 4 7 - 1 / 4 0 1 / 4 ² 0, and z T b = 1 / 4 - 3 < 0. Thus z , together with Farkas’ Lemma shows that the system Ax = b is infeasible. 2. 10 marks: 6 marks distributed same as first 6 marks in Q1 2 marks for initial tableau of phase 2 2 marks for final tableau of phase 2 Exercise 7.7.4 from the course notes, which is reproduced here: Solve the following LP problem by the two-phase simplex method: ( P ) max z = 5 x 1 - 2 x 2 + x 3 subject to x 1 + 4 x 2 + x 3 6 2 x 1 + x 2 + 3 x 3 2 x 1 , x 2 0 , x 3 free Hint : Observe that x 3 is a free variable and the simplex method (the version studied in CO350) applies only to LPs in standard equality form. Solution
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asst6sol - C&amp;O 350 Linear Optimization Winter 2009...

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