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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.

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emis 8374

EMIS 8374

LP Review: The Ratio Test

main steps of the simplex method
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.
example 1 step 1
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
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
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
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 61
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 62
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 63
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
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
Example 1: Ratio Test

The minimum ratio occurs in row 1.

Thus, x3 leaves the basis when x1 enters.

example 1 steps 7 pivot
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 pivot1
Example 1: Steps 7 (Pivot)

First ERO: divide row 1 by 30

example 1 steps 7 pivot2
Example 1: Steps 7 (Pivot)

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

example 1 steps 7 pivot3
Example 1: Steps 7 (Pivot)

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

example 1 steps 7 pivot4
Example 1: Steps 7 (Pivot)

Fourth ERO: Add 4.5 times row 1 to row 0

example 1 steps 4
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 test1
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
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.