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# Tutorial9sol - Tutorial 9 Outline of Solutions Q1 Consider...

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Tutorial 9: Outline of Solutions Q1. Consider the following linear programming problem. min 5 x 1 - 5 x 2 - 13 x 3 s.t. - x 1 + x 2 + 3 x 3 20 12 x 1 + 4 x 2 + 10 x 3 90 x 1 , x 2 , x 3 0 Let x 4 and x 5 denote the slack variables for the respective constraints. The optimal tableau is given below: Basic x 1 x 2 x 3 x 4 x 5 Solution ¯ c 0 0 2 5 0 100 x 2 - 1 1 3 1 0 20 x 5 16 0 - 2 - 4 1 10 You are expected to conduct sensitivity analysis by independently investigating each of the following changes in the original model. For each change, use the sensitivity analysis procedure to obtain the new optimal solution and the new optimal objective value. (a) Change the right-hand side of the first constraint to b 1 = 30. (b) Change the right-hand sides to b 1 b 2 = 10 100 . (c) Change the coefficient of x 3 in the objective function to c 3 = 8. (d) Change the coefficient of x 2 in the objective function to c 2 = - 3. (d) Change the coefficients of x 1 to c 1 a 11 a 21 = - 2 0 5 . Solution : (a) x B = B - 1 30 90 ! = 30 - 30 ! 6≥ 0 not feasible. And - c 0 B B - 1 b = 150. Apply dual simplex algorithm: Basic x 1 x 2 x 3 x 4 x 5 Solution ¯ c 0 0 2 5 0 150 x 2 - 1 1 3 1 0 30 x 5 16 0 - 2 - 4 1 - 30 ¯ c 16 0 0 1 1 120 x 2 23 1 0 - 5 3 2 - 15 x 3 - 8 0 1 2 - 1 2 15 ¯ c 103 5 1 5 0 0 13 10 117 x 4 - 23 5 - 1 5 0 1 - 3 10 3 x 3 6 5 2 5 1 0 - 1 10 9 1

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(b) b 1 b 2 ! = 10 100 ! . x B = B - 1 10 100 ! = 10 60 ! 0 is feasible. Optimal solution is x = (0 , 10 , 0 , 0 , 60) 0 , with optimal objective value - 50. (c) Changing c 3 may affect the optimality conditions. ( x 3 nonbasic variable) Check ¯ c 3 : ¯ c 3 = c 3 - c 0 B B - 1 A 3 = 8 - ( - 5 , 0) 3 - 2 = 23 > 0 . Optimality condition satisfied. The current basis matrix is optimal. Same optimal solution. (d) x 2 is a basic variable. Check reduced costs for all nonbasic variables. New c B = ( - 3 , 0). c 1 , ¯ c 3 , ¯ c 4 ) = ( c 1 , c 3 , c 4 ) - c 0 B B - 1 ( A 1 , A 3 , A 4 ) = (5 , - 13 , 0) - ( - 3 , 0) 1 0 - 4 1 - 1 3 1 12 10 0 = (5 , - 13 , 0) - ( - 3 , 0) - 1 3 1 16 - 2 - 4 = (2 , - 4 , 3) Re-optimization Basic x 1 x 2 x 3 x 4 x 5 Solution ¯ c 2 0 - 4 3 0 60 x 2 - 1 1 3 1 0 20 x 5 16 0 - 2 - 4 1 10 ¯ c 2 3 4 3 0 4 1 3 0 260 3 x 3 - 1 3 1 3 1 1 3 0 20 3 x 5 15 1 3 2 3 0 - 3 1 3 1 70 3 The optimal solution is x * = (0 , 0 , 20 3 , 0 , 70 3 ) 0 , with the optimal objective value - 260 3 . (e) x 1 non-basic variable. Check ¯ c 1 : ¯ c 1 = c 1 - c 0 B B - 1 A 1 = - 2 - ( - 5 , 0) 1 0 - 4 1 0 5 = - 2 6≥ 0 Optimality condition not satisfied. Basic x 1 x 2 x 3 x 4 x 5 Solution ¯ c - 2 0 2 5 0 100 x 2 0 1 3 1 0 20 x 5 5 0 - 2 - 4 1 10 ¯ c 0 0 6 5 17 5 2 5 104 x 2 0 1 3 1 0 20 x 1 1 0 - 2 5 - 4 5 1 5 2 2
Thus optimal solution is: x = (2 , 20 , 0 , 0 , 0) 0 , with optimal objective value - 104.

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