The Simplex Method

Updated 15 February 2009

Main Steps of the Simplex Method1. 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.

2

Step 1

LP in Row-0 FormMaximize zs.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 LPMaximize 4.5 x1 + 4 x2s.t. 30 x1 + 12 x2 6000  10 x1 + 8 x2 2600

4 x1 + 8 x2 2000 x1, x2 0

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Steps 2 and 3

Initial BFS:BV = {z, x3, x4, x5}, NBV = {x1, x2}z = 0, x3 = 6,000, x4 = 2,600, x5 = 2,000x1 = x2 = 0

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Steps 4 and 5

x1 and x2 are eligible to enter the basis.

Select x1 to become a basic variable

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Step 6

• How much can we increase x1?

• Constraint in Row 1:30 x1 + 12 x2 + x3 = 6000

implies

x3 = 6000 - 30 x1 - 12 x2.

• x2 = 0 (it will stay non-basic)

• x3 0 forces x1 200.

6

Step 6

• How much can we increase x1?

• Constraint in Row 2:10 x1 + 8 x2 + x4 = 2600

implies

x4 = 2600 - 10 x1 - 8 x2

• x2 = 0 (it will stay non-basic)

• x4 0 forces x1 260.

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Step 6

• How much can we increase x1?

• Constraint in Row 3:4 x1 + 8 x2 + x5= 2000

implies

x5 = 2000 - 4 x1 - 8 x2

• x2 = 0 (it will stay non-basic)

• x5 0 forces x1 500.

8

Step 6• From constraint 1, we see that we can increase x1

up to 200, if simultaneously reduce x3 to zero.• From constraint 2, we see that we can increase x1

up to 260, if we simultaneously reduce x4 to zero.• From constraint 3, we see that we can increase x1

up to 500, if we simultaneously reduce x5 to zero.• Since x3 is the limiting variable, we make it non-

basic as x1 becomes basic.

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Step 6: Ratio Test for x1

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.

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Step 6: Ratio Test for x1

The minimum ratio occurs in Row 1.Thus, x3 leaves the basis when x1 enters.

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Step 7: Pivot x1 in and x3 out

Pivot on the x1 column of Row 1 to makex1 basic and x3 non-basic.

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First ERO: divide Row 1 by 30

Step 7: Pivot x1 in and x3 out

First ERO: divide Row 1 by 30

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

Step 7: Pivot x1 in and x3 outSecond ERO: Add –10 times Row 1 to Row 2

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

Step 7: Pivot x1 in and x3 outThird ERO: Add –4 times Row 1 to Row 3

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Fourth ERO: Add 4.5 times Row 1 to Row 0

Step 7: Pivot x1 in and x3 outFourth ERO: Add 4.5 times Row 1 to Row 0

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Steps 4 and 5

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

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Step 6: Ratio Test for x2

The minimum ratio occurs in Row 2.Thus, x4 leaves the basis when x2 enters.

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5.1874.612001200101333.04.600

1504600600013333.0400

5004.0200200000333.04.010

9000015.02.201

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Step 7: Pivot x2 in and x4 Out

BV = {z, x1, x2, x5}, NBV = {x3, x4}z = 1230, x1 = 140, x2 = 150, x5 = 240

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Steps 4 and 5

x3 is eligible to enter the basis

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Step 6: Ratio Test for x3

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6004.024024016.14.0000

18000833.0150150025.00833.0100

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15025.00833.0 432 xxx

432 25.00833.0150 xxx

If x3 enters the basis, then x2 will increase as well.

Step 6: Ratio Test for x3

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6004.024024016.14.0000

150025.00833.0100

21000667.014014001.00667.0010

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If x3 enters the basis, then x5 will leave the basis.

Step 7: Pivot x3 in and x5 out

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Steps 4 and 8

BV = {z, x1, x2, x3}, NBV = {x4, x5}z = 1250, x1 = 100, x2 = 200, x3 = 600This an optimal BFS.

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