This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Solution problem 1: (questions, comments, suggestions please to Steffen Rebennack, [email protected] ) As in the problem stated, we first transform the linear program into standard form. For this, we have to have equality constraints; hence we introduce the (positive) slack variables s 1 , s 2 and s 3 . They will be the initial basis giving us the following tableau: z x 1 x 2 x 3 s 1 s 2 s 3 RHS z 1311 s 1 1 1 1 6 s 2 1 1 1 12 s 3 1 1 2 1 20 We are maximizing, so we want a variable to enter the basis which has a neg ative entry in row 0 of our tableau. Here, we choose variable x 1 to enter the basis. The minimum ratio test gives us that the variable s 1 leaves the basis. The updated tableau reads then as follows z x 1 x 2 x 3 s 1 s 2 s 3 RHS z 1 21 3 18 x 1 1 1 1 6 s 2 1 1 1 12 s 3 21 1 14 Again, we choose a variable with negative coefficient in row 0. This time, it is the unique choice x 3 . The minimum ratio test give us the leaving variable which is in this case s 3 . After the pivoting step, we get z x 1 x 2 x 3 s 1 s 2 s 3 RHS z 1 2 2.5 0.5 25 x 1 1 1 1 6 s 2 1 0.5 10.5 5 x 3 10.5 0.5 7 As all the coefficients in row 0 are positive, we have found an optimal solution. Finally, the optimal solution is x 1 = 6 x 2 = 0 x 3 = 7 with the objective function value 25. 1 Solution problem 2: (questions, comments, suggestions please to Zeki Caner Taskin, [email protected] ) We first write the problem in tableu form. z x 1 x 2 x 3 x 4 RHS z 11 31 1 x 1 11 1 2 15 x 32 1 11 We first apply GaussJordan elimination to convert the x 1 column to identity....
View
Full
Document
This note was uploaded on 02/25/2010 for the course ESI 6314 taught by Professor Vladimirlboginski during the Fall '09 term at University of Florida.
 Fall '09
 VLADIMIRLBOGINSKI

Click to edit the document details