EMIS 8374 The Ratio Test 1 2 3

EMIS 8374 The Ratio Test

1. 2. 3. 4. 5. 6. Main Steps of the Simplex Method Put the problem in row-0 form. Construct the simplex tableau. Obtain an initial BFS. If the current BFS is optimal then goto step 9. Choose a non-basic variable to enter the basis. 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. Goto Step 4. 9. Stop. 1

Example 1: Step 1 Original LP Maximize 4. 5 x 1 + 4 x 2 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 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 2

Example 1: 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

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

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

Example 1: Step 6 • How much can we increase x 1? • Constraint in Row 2: 10 x 1 + 8 x 2 + x 4 = 2600 => x 4 = 2600 - 10 x 1 - 8 x 2 • x 2 = 0 (it will stay non-basic) • x 4 0 • Thus x 1 260. 6

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

Example 1: Step 6 • From constraint 1, we see that we can increase x 1 up to 200, but we must reduce x 3 to zero to satisfy the constraints. • From constraint 2, we see that we can increase x 1 up to 260, but we must also reduce x 4 to zero to satisfy the constraints. • From constraint 3, we see that we can increase x 1 up to 500, but we must reduce x 5 to zero to satisfy the constraints. • Since x 3 is the limiting variable, we make it nonbasic as x 1 becomes basic. 8

Step 6: The Ratio Test 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

Example 1: Ratio Test The minimum ratio occurs in row 1. Thus, x 3 leaves the basis when x 1 enters. 10

Example 1: Step 7 (Pivot) Pivot on the x 1 column of row 1 to make x 1 basic and x 3 non-basic. 11

Example 1: Step 7 (Pivot) First ERO: divide row 1 by 30 12

Example 1: Step 7 (Pivot) Second ERO: Add – 10 times row 1 to row 2 13

Example 1: Step 7 (Pivot) Third ERO: Add – 4 times row 1 to row 3 14

Example 1: Step 7 (Pivot) Fourth ERO: Add 4. 5 times row 1 to row 0 15

Example 1: Step 4 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

Example 1: Ratio Test The minimum ratio occurs in row 2. Thus, x 4 leaves the basis when x 2 enters. 17

Example 1: Pivoting x 2 into the basis 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. 18