1. The Simplex Method Updated 15 February 2009

2. Main Steps of the Simplex Method 1.Put the problem in Row-Zero Form. 2.Construct the Simplex tableau. 3.Obtain an initial basic feasible solution (BFS). 4.If the current BFS is optimal then go to step 9. 5.Choose a non-basic variable to enter the basis. 6.Use the ratio test to determine which basic variable must leave the basis. 7.Perform the pivot operation on the appropriate element of the tableau. 8.Go to Step 4. 9.Stop. 1

3. Step 1 LP in Row-0 Form Maximize z s.t. z - 4.5 x 1 - 4 x 2 = 0 30 x 1 + 12 x 2 + x 3 = 6000 10 x 1 + 8 x 2 + x 4 = 2600 4 x 1 + 8 x 2 + x 5 = 2000 x 1, x 2, x 3, x 4, x 5  0 Original LP Maximize 4.5 x 1 + 4 x2 s.t. 30 x 1 + 12 x 2  6000 10 x 1 + 8 x 2  2600 4 x 1 + 8 x 2  2000 x 1, x 2  0 2

4. Steps 2 and 3 Initial BFS: BV = {z, x 3, x 4, x 5 }, NBV = {x 1, x 2 } z = 0, x 3 = 6,000, x 4 = 2,600, x 5 = 2,000 x 1 = x 2 = 0 3

5. Steps 4 and 5 x 1 and x 2 are eligible to enter the basis. Select x 1 to become a basic variable 4

6. Step 6 How much can we increase x 1 ? Constraint in Row 1: 30 x 1 + 12 x 2 + x 3 = 6000 implies x 3 = 6000 - 30 x 1 - 12 x 2. x 2 = 0 (it will stay non-basic) x 3  0 forces x 1  200. 5

7. Step 6 How much can we increase x 1 ? Constraint in Row 2: 10 x 1 + 8 x 2 + x 4 = 2600 implies x 4 = 2600 - 10 x 1 - 8 x 2 x 2 = 0 (it will stay non-basic) x 4  0 forces x 1  260. 6

8. Step 6 How much can we increase x 1 ? Constraint in Row 3: 4 x 1 + 8 x 2 + x 5 = 2000 implies x 5 = 2000 - 4 x 1 - 8 x 2 x 2 = 0 (it will stay non-basic) x 5  0 forces x 1  500. 7

9. Step 6 From constraint 1, we see that we can increase x 1 up to 200, if simultaneously reduce x 3 to zero. From constraint 2, we see that we can increase x 1 up to 260, if we simultaneously reduce x 4 to zero. From constraint 3, we see that we can increase x 1 up to 500, if we simultaneously reduce x 5 to zero. Since x 3 is the limiting variable, we make it non- basic as x 1 becomes basic. 8

10. Step 6: Ratio Test for x 1 Row 1: 30 x 1 + 12 x 2 + x 3 = 6000 => 30 x 1 + x 3 = 6000 => x 1  6000/30 = 200. Row 2: 10 x 1 + 8 x 2 + x 4 = 2600 => 10 x 1 + x 4 = 2600 => x 1  2600/10 = 260. Row 3: 4 x 1 + 8 x 2 + x 5 = 2000 => 4 x 1 + x 5 = 2000 => x 1  2000/4 = 500. 9

11. Step 6: Ratio Test for x 1 The minimum ratio occurs in Row 1. Thus, x 3 leaves the basis when x 1 enters. 10

12. Step 7: Pivot x 1 in and x 3 out Pivot on the x 1 column of Row 1 to make x 1 basic and x 3 non-basic. 11 First ERO: divide Row 1 by 30

13. Step 7: Pivot x 1 in and x 3 out First ERO: divide Row 1 by 30 12 Second ERO: Add –10 times Row 1 to Row 2

14. Step 7: Pivot x 1 in and x 3 out Second ERO: Add –10 times Row 1 to Row 2 13 Third ERO: Add –4 times Row 1 to Row 3

15. Step 7: Pivot x 1 in and x 3 out Third ERO: Add –4 times Row 1 to Row 3 14 Fourth ERO: Add 4.5 times Row 1 to Row 0

16. Step 7: Pivot x 1 in and x 3 out Fourth ERO: Add 4.5 times Row 1 to Row 0 15

17. Steps 4 and 5 BV = {z, x 1, x 4, x 5 }, NBV = {x 2, x 3 } z = 900, x 1 = 200, x 4 = 600, x 5 = 1200 Increasing x 2 may lead to an increase in z. 16

18. Step 6: Ratio Test for x 2 The minimum ratio occurs in Row 2. Thus, x 4 leaves the basis when x 2 enters. 17

19. Step 7: Pivot x 2 in and x 4 Out BV = {z, x 1, x 2, x 5 }, NBV = {x 3, x 4 } z = 1230, x 1 = 140, x 2 = 150, x 5 = 240 18

20. Steps 4 and 5 x 3 is eligible to enter the basis 19

21. Step 6: Ratio Test for x 3 20 If x 3 enters the basis, then x 2 will increase as well.

22. Step 6: Ratio Test for x 3 21 If x 3 enters the basis, then x 5 will leave the basis.

23. Step 7: Pivot x 3 in and x 5 out 22

24. Steps 4 and 8 BV = {z, x 1, x 2, x 3 }, NBV = {x 4, x 5 } z = 1250, x 1 = 100, x 2 = 200, x 3 = 600 This an optimal BFS. 23