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 must solve Phase 1, max { c T x | Ax = b, x 0 } where c T = [0 0 0 0 - 1 - 1 - 1] A = - 1 - 1 8 1 1 0 0 - 6 1 1 - 4 0 1 0 1 - 8 1 4 0 0 1 , b = 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: 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 . 1
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Iteration 2: Since c j 0 j N , the current solution is optimal. A T B y = c B y = 1 / 4 - 1 - 1 Since c j 0 j N , the current solution is optimal. The vector y satisfies the conditions of Farkas’ Lemma (compute for yourself) and hence acts as a certificate of infeasibility. 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 : We need to put the LP into standard equality form (using v 1 - v 2 to replace the free variable x 3 ): max z = 5 x 1 - 2 x 2 + v 1 - v 2 s.t. x 1 + 4 x 2 + v 1 - v 2 + s 1 = 6 2 x 1 + x 2 + 3 v 1 - 3 v 2 - s 2 = 2 x i , v i , s i 0 Now to formulate the auxiliary problem we use s 1 as the auxiliary variable for the first constraint and add auxiliary variable u 2 for the second constraint. The auxiliary problem is: max w = - s 1 - u 2 s.t. x 1 + 4 x 2 + v 1 - v 2 + s 1 + = 6 2 x 1 + x 2 + 3 v 1 - 3 v 2 - s 2 + u 2 = 2 x i , v i , s i , u i 0 The first tableau, using the basis { s 1 ,u 2 } : w - 3 x 1 - 5 x 2 - 4 v 1 + 4 v 2 + s 2 = - 8 x 1 + 4 x 2 + v 1 - v 2 + s 1 = 6
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asst6sol - C&O 350 Linear Optimization – Winter 2009...

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