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EMIS 8374

EMIS 8374. LP Review: The Ratio Test. 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.

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EMIS 8374

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  1. EMIS 8374 LP Review: The Ratio Test

  2. 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. • Perform the pivot operation on the appropriate element of the tableau. • Goto Step 4. • Step 9. Stop.

  3. LP in Row-0 Form Maximize z s.t. z - 4.5 x1 - 4 x2 = 0 30 x1 + 12 x2 + x3 = 6000  10 x1 + 8 x2 + x4 = 2600 4 x1 + 8 x2 + x5 = 2000 x1, x2, x3, x4, x5  0 Original LP Maximize 4.5 x1 + 4 x2 s.t. 30 x1 + 12 x2  6000  10 x1 + 8 x2  2600 4 x1 + 8 x2  2000 x1, x2  0 Example 1: Step 1

  4. Example 1: Steps 2 and 3 Initial BFS: BV = {z, x3, x4, x5}, NBV = {x1, x2} z = 0, x3 = 6,000, x4 = 2,600, x5 = 2,000 x1 = x2 = 0

  5. Example 1: Steps 4 and 5 x1 and x2 are eligible to enter the basis. Select x1 to become a basic variable

  6. Example 1: Step 6 • How much can we increase x1? • Constraint in Row 1: 30 x1 + 12 x2 + x3 = 6000 => x3 = 6000 - 30 x1 - 12 x2. • x2 = 0 (it will stay non-basic) • x3  0 • Thus x1  200.

  7. Example 1: Step 6 • How much can we increase x1? • Constraint in Row 2: 10 x1 + 8 x2 + x4 = 2600 => x4 = 2600 - 10 x1 - 8 x2 • x2 = 0 (it will stay non-basic) • x4  0 • Thus x1  260.

  8. Example 1: Step 6 • How much can we increase x1? • Constraint in Row 3: 4 x1 + 8 x2 + x5= 2000 => x5 = 2000 - 4 x1 - 8 x2 • x2 = 0 (it will stay non-basic) • x5  0 • Thus x1  500.

  9. Example 1: Step 6 • From constraint 1, we see that we can increase x1 up to 200, but we must reduce x3 to zero to satisfy the constraints. • From constraint 2, we see that we can increase x1 up to 260, but we must also reduce x4 to zero to satisfy the constraints. • From constraint 3, we see that we can increase x1 up to 500, but we must reduce x5 to zero to satisfy the constraints. • Since x3 is the limiting variable, we make it non-basic as x1 becomes basic.

  10. Step : The Ratio Test Row 1: 30 x1 + 12 x2 + x3 = 6000 => 30 x1 + x3 = 6000 => x1  6000/30 = 200. Row 2: 10 x1 + 8 x2 + x4 = 2600 => 10 x1 + x4 = 2600 => x1  2600/10 = 260. Row 3: 4 x1 + 8 x2 + x5= 2000 => 4 x1 + x5 = 2000 => x1  2000/4 = 500.

  11. Example 1: Ratio Test The minimum ratio occurs in row 1. Thus, x3 leaves the basis when x1 enters.

  12. Example 1: Steps 7 (Pivot) Pivot on the x1 column of row 1 to make x1 basic and x3 non-basic.

  13. Example 1: Steps 7 (Pivot) First ERO: divide row 1 by 30

  14. Example 1: Steps 7 (Pivot) Second ERO: Add –10 times row 1 to row 2

  15. Example 1: Steps 7 (Pivot) Third ERO: Add –4 times row 1 to row 3

  16. Example 1: Steps 7 (Pivot) Fourth ERO: Add 4.5 times row 1 to row 0

  17. Example 1: Steps 4 BV = {z, x1, x4, x5}, NBV = {x2, x3} z = 900, x1 = 200, x4 = 600, x5 = 1200 Increasing x2 may lead to an increase in z.

  18. Example 1: Ratio Test The minimum ratio occurs in row 2. Thus, x4 leaves the basis when x2 enters.

  19. Example 1: Pivoting x2 into the basis BV = {z, x1, x2, x3}, NBV = {x4, x5} z = 1250, x1 = 100, x2 = 200, x3 = 600 This an optimal BFS.

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