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The Simplex Method. Updated 15 February 2009. Main Steps of the Simplex Method. Put the problem in Row-Zero Form . Construct the Simplex tableau . Obtain an initial basic feasible solution (BFS). If the current BFS is optimal then go to step 9. - PowerPoint PPT Presentation
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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.
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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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5
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2000100840
26000108100
600000112300
000045.41
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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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000045.41
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Step 6
• How much can we increase x1?• Constraint in Row 1:
30 x1 + 12 x2 + x3 = 6000 impliesx3 = 6000 - 30 x1 - 12 x2.
• x2 = 0 (it will stay non-basic) • x3 0 forces x1 200.
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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.
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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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500420002000100840
26010260026000108100
200306000600000112300
000045.41
ratiobasic
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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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000045.41
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First ERO: divide Row 1 by 30
Step 7: Pivot x1 in and x3 outFirst ERO: divide Row 1 by 30
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2000100840
26000108100
200000333.04.010
000045.41
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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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2000100840
600013333.0400
200000333.04.010
000045.41
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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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5
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1200101333.04.600
600013333.0400
200000333.04.010
000045.41
basic
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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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5
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1200101333.04.600
600013333.0400
200000333.04.010
9000015.02.201
basic
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bxxxxxz
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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5
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1200101333.04.600
600013333.0400
200000333.04.010
9000015.02.201
basic
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bxxxxxz
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
ratiobasic
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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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24016.14.0000
150025.00833.0100
14001.00667.0010
1230055.00333.0001
basic
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Steps 4 and 5
x3 is eligible to enter the basis
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24016.14.0000
150025.00833.0100
14001.00667.0010
1230055.00333.0001
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Step 6: Ratio Test for x3
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6004.024024016.14.0000
18000833.0150150025.00833.0100
21000667.014014001.00667.0010
1230055.00333.0001
ratiobasic
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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
1230055.00333.0001
ratiobasic
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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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6005.241000
2002083.00833.00100
1001667.01667.00010
12500833.04167.00001
basic
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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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6005.241000
2002083.00833.00100
1001667.01667.00010
12500833.04167.00001
basic
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