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Unformatted text preview: CO 350 Assignment 6 – Winter 2010 Solutions Question # Max. marks Part marks 1 8 3 1 2 2 2 6 3 5 2 3 4 5 5 8 2 1 2 3 6 8 2 2 2 2 Total 40 1. Consider the LP problem maximize x 1 + x 2 ( P ) subject to 2 x 1 x 2 ≤  4 x 1 x 2 ≤  1 2 x 1 + x 2 ≤ 2 x 1 , x 2 ≥ (a) Formulate an auxiliary LP problem ( A ) for finding a feasible solution of ( P ). Use slack variables instead of artificial variables wherever possible (see Section 7.3 of the course notes). Clearly indicate the artificial variables and the slack variables. (b) Determine an initial feasible tableau for ( A ). (c) Solve ( A ) with the simplex method. Use the largest coefficient rule when choosing entering variables. Use of other rules will be assigned a grade of zero for this part. (d) Use your solution from part (c) to find a feasible tableau for ( P ). (You do not need to solve ( P )). Solution: (a) We add slack variables x 3 ,x 4 ,x 5 to the first, second and third constraints, respectively. We also multiply by 1 those constraints with negative righthand side. We get maximize z = x 1 + x 2 subject to 2 x 1 + x 2 x 3 = 4 x 1 + x 2 x 4 = 1 2 x 1 + x 2 + x 5 = 2 x 1 , x 2 , x 3 , x 4 , x 5 ≥ Then we add artificial variables x 6 and x 7 to the first and second constraints, respectively. The auxiliary problem is max w = x 6 x 7 ( A ) s.t. 2 x 1 + x 2 x 3 + x 6 = 4 x 1 + x 2 x 4 + x 7 = 1 2 x 1 + x 2 + x 5 = 2 x 1 , x 2 , x 3 , x 4 , x 5 , x 6 , x 7 ≥ (b) An initial feasible basis for ( A ) is B = { 5 , 6 , 7 } . We subtract the x 6row and the x 7row from the wrow ( w + x 6 + x 7 = 0) and obtain the following initial feasible tableau: w x 1 2 x 2 + x 3 + x 4 = 5 2 x 1 + x 2 x 3 + x 6 = 4 x 1 + x 2 x 4 + x 7 = 1 2 x 1 + x 2 + x 5 = 2 1 (c) c 2 = 2 > 0 is largest, so x 2 enters. t = min { 4 , 1 , 2 } = 1, so x 7 leaves. We pivot on (7 , 2) and obtain the following tableau for B = { 2 , 5 , 6 } : w 3 x 1 + x 3 x 4 + 2 x 7 = 3 3 x 1 x 3 + x 4 + x 6 x 7 = 3 x 1 + x 2 x 4 + x 7 = 1 x 1 + x 4 + x 5 x 7 = 1 Since x 7 is artificial, we can remove x 7 from the tableau: w 3 x 1 + x 3 x 4 = 3 3 x 1 x 3 + x 4 + x 6 = 3 x 1 + x 2 x 4 = 1 x 1 + x 4 + x 5 = 1 Now x 1 enters and x 6 leaves. We pivot on (6 , 1) and get the tableau for B = { 1 , 2 , 5 } : w + x 6 = 0 x 1 1 3 x 3 + 1 3 x 4 + 1 3 x 6 = 1 x 2 1 3 x 3 2 3 x 4 + 1 3 x 6 = 2 1 3 x 3 + 4 3 x 4 + x 5 + 1 3 x 6...
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 Fall '10
 GUENIN
 Optimization, Negative and nonnegative numbers, BMW Sports Activity Series, x4

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