From the above table the BF solution is 2 6 2 0 This is optimal since no im

# From the above table the bf solution is 2 6 2 0 this

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Ch 2: The Simplex Method MATH2230/MATH3205/Mark Lau 1. (Initialization) We choose ( x 1 , x 2 ) = (0 , 0) to be the initial CFP. 2. (Optimality test) This CFP is not optimal since z increases when either x 1 or x 2 increases, in order to satisfy the first constraint. 3. (Initial BF solution) Set up the initial tabular form z ( x 1 ) ( x 2 ) x 3 x 4 x 5 x 6 RHS 1 -3 -2 0 0 0 0 0 0 1 2 1 0 0 0 6 0 2 1 0 1 0 0 8 0 -1 1 0 0 1 0 1 0 0 1 0 0 0 1 2 and see that (0 , 0 , 6 , 8 , 1 , 2) is our first BF solution. 4. (Iteration 1, step 1) We choose x 1 as the entering variable, since z increases faster in the direction of x 1 . 5. (Iteration 1, step 2) We choose x 4 as the leaving variable, so that x 1 yields the greatest increase without leaving the feasible set. Note that x 5 cannot be the leaving variable since - x 1 + x 5 = 1, and hence x 1 would be negative if x 5 = 0. 6. (Iteration 1, step 3) We perform row operations on the simplex tableau in 3, in order to obtain a BF solution: z x 1 ( x 2 ) x 3 ( x 4 ) x 5 x 6 RHS 1 0 -1/2 0 3/2 0 0 12 0 1 1/2 0 1/2 0 0 4 0 0 3/2 1 -1/2 0 0 2 0 0 3/2 0 1/2 1 0 5 0 0 1 0 0 0 1 2 The BF solution found in the current iteration is (4 , 0 , 2 , 0 , 5 , 2). More iterations are need since the objective function value can be improved further by moving to one of the adjacent BF solutions. 7. (Iteration 2, step 1) Assign x 2 as the entering variable. Last updated: October 7, 2014 Page 10 of 18
Ch 2: The Simplex Method MATH2230/MATH3205/Mark Lau 8. (Iteration 2, step 2) Assign x 3 the leaving variable. 9. (Iteration 2, step 3) Perform row operations on the tabular form in Iteration 1 to get z x 1 x 2 ( x 3 ) ( x 4 ) x 5 x 6 RHS 1 0 0 1/3 3/2 0 0 38/3 0 1 0 -1/3 2/3 0 0 10/3 0 0 1 2/3 -1/3 0 0 4/3 0 0 0 -1 1 1 0 3 0 0 0 -2/3 1/3 0 1 2/3 The BF solution found in the current iteration is (10 / 3 , 4 / 3 , 0 , 0 , 3 , 2 / 3). The optimal solution is found with objective function value 38/3. Keep in mind that the algorithm stops when an optimal solution has been obtained. In problems having multiple optimal solutions, one can perform more iterations to obtain other optimal solutions. 2.1 Solving Minimization Problems The following minimization min c 1 x 1 + c 2 x 2 + · · · + c n x n subject to a set of constraints can be converted to a maximization, max - c 1 x 1 - c 2 x 2 - · · · - c n x n , over the same set of constraints. For example, the following minimization problem min x 1 ,x 2 R 0 . 4 x 1 + 0 . 5 x 2 s.t. 0 . 3 x 1 + 0 . 1 x 2 2 . 7 , 0 . 5 x 1 + 0 . 5 x 2 = 6 , 0 . 6 x 1 + 0 . 4 x 2 6 , x 1 , x 2 0 Last updated: October 7, 2014 Page 11 of 18
Ch 2: The Simplex Method