EMIS 8374

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# EMIS 8374 - PowerPoint PPT Presentation

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

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.
• 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.
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
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

Example 1: Steps 4 and 5

x1 and x2 are eligible to enter the basis.

Select x1 to become a basic variable

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.
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.
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.
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.
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.

Example 1: Ratio Test

The minimum ratio occurs in row 1.

Thus, x3 leaves the basis when x1 enters.

Example 1: Steps 7 (Pivot)

Pivot on the x1 column of row 1 to make

x1 basic and x3 non-basic.

Example 1: Steps 7 (Pivot)

First ERO: divide row 1 by 30

Example 1: Steps 7 (Pivot)

Second ERO: Add –10 times row 1 to row 2

Example 1: Steps 7 (Pivot)

Third ERO: Add –4 times row 1 to row 3

Example 1: Steps 7 (Pivot)

Fourth ERO: Add 4.5 times row 1 to row 0

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.

Example 1: Ratio Test

The minimum ratio occurs in row 2.

Thus, x4 leaves the basis when x2 enters.

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.