Optimization using the Simplex Algorithm: A Second Pivoting Rule Example, Exams of Mathematics

Download Optimization using the Simplex Algorithm: A Second Pivoting Rule Example and more Exams Mathematics in PDF only on Docsity! 7 The simplex algorithm by example. 7.1 A second pivoting rule. Consider the primal problem 1 2 1 2 1 2 1 2 3 9 max 2 3 when 2 8 , 0 x x x x x x x x  + ≤+ + ≤ ≥ . The corresponding initial tableau is given by 1 3 1 0 9 2 1 0 1 8 2 3 0 0 0 . Since both 1x and 2x are non-basic variables, it is possible to pivot using either variable. It is reasonable to examine the current objective and choose the variable with the largest coefficient. In the example 2x has the larger coefficient. In terms of the tableau, the second pivoting rule selects the column with the largest value in the bottom row to the left of the vertical line. This choice leads to the ‘fastest’ increase in the objective. 7.2 The next tableau. Combine the two pivoting rules and pivot to get 1/ 3 1 1/ 3 0 3 5/ 3 0 1/3 1 5 1 0 1 0 9 − − − . The new solution is given by 1 20, 3x x= = . There is a positive coefficient in the ‘new’ objective 1 19 x y+ − . There is a need for a second pivot. 7.3 The final tableau. Apply the original pivoting rule and get the tableau 0 1 2/5 1/5 2 1 0 1/5 3/5 3 0 0 4/5 3/5 12 − − − − − . The ‘new’ objective is given by 4 35 51 212 y y− − with no hope of improvement. This must be the final tableau and the proposed solution is given by the feasible 1 2ˆ ˆ3, 2x x= = . To verify the solution observe that the dual problem is given by