1. 1. The Simplex Method for Problems in Standard Form 1

2. 2 The Simplex Method for Problems in Standard Form: 1. Introduce slack variables and state the problem in terms of a system of linear equations. 2. Construct the simplex tableau corresponding to the system.

3. 3 3. Determine if the left part of the bottom row contains negative entries. If none are present, the solution corresponding to the tableau yields a maximum and the problem is solved. 4. If the left part of the bottom row contains negative entries, construct a new simplex tableau.

4. 4 a) Choose the pivot column by inspecting the entries of the last row of the current tableau, excluding the right-hand entry. The pivot column is the one containing the most-negative of these entries.

5. 5 b)Choose the pivot element by computing ratios associated with the positive entries of the pivot column. The pivot element is the one corresponding to the smallest nonnegative ratio. c)Construct the new simplex tableau by pivoting around the selected element.

6. 6 5. Return to step 3. Steps 3 and 4 are repeated as many times as necessary to find a maximum.

7. 7

8.  Let u, v and w be the slack variables. The corresponding linear system is 8

9.  Set up the initial simplex tableau. 9 x y u v w M uvwMuvwM

10.  Determine if maximum has been reached. 10 At least one negative entry. Maximum has not been reached.

11.  Choose the pivot element 11 Most negative entry 72/2 = 36 Smallest positive ratio 96/6 = 16 18/1 = 18

12.  Pivot. 12 x y u v w M xvwMxvwM Group II variables

13.  Determine if maximum has been reached. 13 x y u v w M xvwMxvwM Group II variables At least one negative entry. Maximum has not been reached.

14.  Choose pivot. 14 pivot column 16/(1/2) = 32 2/(1/2) = 4 40/5 = 8 pivot row

15.  New tableau: 15 x y u v w M xywMxywM Group II variables No negative entries Solution: x = 14, y = 4 and Maximum = 1400

16.  The simplex method entails pivoting around entries in the simplex tableau until the bottom row contains no negative entries except perhaps the entry in the last column. The solution can be read off the final tableau by letting the variables heading columns with 0 entries in every row but the i th row take on the value in the i th row of the right-most column, and setting the other variables equal to 0. 16