The Simplex Method
Updated 15 February 2009

1. 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.
1

Original LP
Maximize 4.5 s.t. 30 x
1 x
1
+ 12 x
+ 4 x 2
2

6000
10
4 x x x
1
1
1
+ 8
+ 8
, x x
2 x
2
2

2600

2000

0
Step 1
LP in Row-0 Form
Maximize z s.t. z - 4.5 x
1
30 x
1
+ 12
- 4 x
2 x
2
+
= 0 x
3
= 6000
10
4 x x
1
1
+ 8
+ 8 x x
1
, x
2
, x
3 x
2
2
, x
+ x
4
+ x
5
4
, x
5
= 2600
= 2000

0
2

Steps 2 and 3 z
1 x
1

4 .
5 x
2

4 x
3
0 x
4
0 x
5
0
0 30 12 1 0 b
0
0 6000 basic
0
0
10
4
8
8
0
0
1
0
0
1
2600
2000 x
3 x
4 x
5 z
Initial BFS:
BV = { z , x
3
, x
4
, x
5
}, NBV = { x
1
, x
2
} z x
1
= 0,
= x
2 x
3
= 6,000,
= 0 x
4
= 2,600, x
5
= 2,000
3

Steps 4 and 5 z
1 x
1

4 .
5 x
2

4 x
3
0 x
4
0 x
5
0
0 30 12 1 0 b
0
0 6000 basic
0
0
10
4
8
8
0
0
1
0
0
1
2600
2000 x
3 x
4 x
5 z x
1 and x
2 are eligible to enter the basis.
Select x
1 to become a basic variable
4

Step 6
• How much can we increase x
1
?
• Constraint in Row 1:
30 x
1
+ 12 x
2
+ x
3
= 6000 implies x
3
• x
2
• x
3
= 6000 - 30 x
1

0 forces x
1
- 12 x
2
.
= 0 (it will stay non-basic)

200.
5

Step 6
• How much can we increase x
1
?
• Constraint in Row 2:
10 x
1
+ 8 x
2
+ x
4
= 2600 implies x
4
• x
2
• x
4
= 2600 - 10 x
1

0 forces x
1
- 8 x
2
= 0 (it will stay non-basic)

260.
6

Step 6
•
How much can we increase x
1
?
•
Constraint in Row 3:
4 x
1
+ 8 x
2
+ x
5
= 2000 implies x
5
• x
2
• x
5
= 2000 - 4 x
1
= 0 (it will stay non-basic)

0 forces
- 8 x
2 x
1

500.
7

Step 6
• From constraint 1, we see that we can increase x
1 up to 200, if simultaneously reduce x
3 to zero.
• From constraint 2, we see that we can increase x
1 up to 260, if we simultaneously reduce x
4 to zero.
• From constraint 3, we see that we can increase x
1 up to 500, if we simultaneously reduce x
5 to zero.
• Since x
3 is the limiting variable, we make it nonbasic as x
1 becomes basic.
8

Step 6: Ratio Test for x
1
Row 1:
30 x
1
30 x
1
+ 12
+ x
3 x
2
+ x
3
= 6000 =>
= 6000 => x
1

6000/30 = 200.
Row 2:
10 x
1
10 x
1
+ 8 x
+ x
4
2
+ x
4
= 2600 =>
= 2600 => x
1

2600/10 = 260.
Row 3:
4 x
1
4 x
1
+ 8 x
+ x
5
2
+ x
5
= 2000 =>
= 2000 => x
1

2000/4 = 500.
9

Step 6: Ratio Test for x
1 z
1 x
1

4 .
5 x
2

4 x
3
0 x
4
0 x
5
0
0 30 12 1 0 b
0
0 6000 basic
0
0
10
4
8
8
0
0
1
0
0
1
2600
2000 ratio z x
3 x
4 x
5
6000
30

200
2600
10

260
2000
4

500
The minimum ratio occurs in Row 1.
Thus, x
3 leaves the basis when x
1 enters.
10

Step 7: Pivot x
1 in and x
3 out z
1 x
1

4 .
5 x
2

4 x
3
0 x
4
0 x
5
0
0 30 12 1 0 b
0
0 6000 basic
0
0
10
4
8
8
0
0
1
0
0
1
2600
2000 x
3 x
4 x
5 z
Pivot on the x
1 x
1 basic and x
3 column of Row 1 to make non-basic.
First ERO: divide Row 1 by 30
11

Step 7: Pivot x
1 in and x
3 out
First ERO: divide Row 1 by 30 z
1 x
1

4 .
5 x
2

4
0 1 0 .
4 0 .
0333 x
3
0 x
4
0 x
5
0
0 0
0
0
10
4
8
8
0
0
1
0
0
1
0
200 b basic z
2600
2000 x
1 x
4 x
5
Second ERO: Add –10 times Row 1 to Row 2
12

Step 7: Pivot x
1 in and x
3 out
Second ERO: Add –10 times Row 1 to Row 2 z
1 x
1

4 .
5 x
2

4
0
0
1
0
0 .
4
4
0 .
0333

0 .
3333 x
3
0 x
4
0 x
5
0
0
1
0
0
0 4 8 0 0 1 b basic
0
200
600
2000 z x
1 x
4 x
5
Third ERO: Add –4 times Row 1 to Row 3
13

Step 7: Pivot x
1 in and x
3 out
Third ERO: Add –4 times Row 1 to Row 3 z
1 x
1

4 .
5 x
2

4
0
0
0
1
0
0
0
6
.
4
4
.
4
0 .
0333

0 .
3333

0 .
1333 x
3
0 x
4
0 x
5
0
0
1
0
0
0
1 b
0
200
600
1200 basic x
1 x
4 x
5 z
Fourth ERO: Add 4.5 times Row 1 to Row 0
14

Step 7: Pivot x
1 in and x
3 out
Fourth ERO: Add 4.5 times Row 1 to Row 0 z
1 x
1
0 x
2

2 .
2
0
0
0
1
0
0
0 x
3
.
15
0 .
4
4
6 .
4
0 .
0333

0 .
3333

0 .
1333 x
4
0
0
1
0 x
5
0
0
0
1 b
900
200
600
1200 basic z x
1 x
4 x
5
15

Steps 4 and 5 z
1 x
1
0 x
2

2 .
2
0
0
0
1
0
0
0 x
3
.
15
0 .
4
4
6 .
4
0 .
0333

0 .
3333

0 .
1333 x
4
0
0
1
0 x
5
0
0
0
1 b
900
200
600
1200 basic z x
1 x
4 x
5
BV = { z , x
1
, x
4
, x
5
}, NBV = {x
2
, x
3
} z = 900, x
1
Increasing
= 200, x
4 x
2
= 600, x
5
= 1200 may lead to an increase in z .
16

Step 6: Ratio Test for x
2 z
1
0
0
0 x
0
1
1
0
0

2
0
6
.
.
x
2
2
4
4
.
4
0 x
3
.
15
0 .
0333

0 .
3333

0 .
1333 x
4
0
0
1
0 x
5
0
0
0
1 b
900
200
600
1200 basic ratio z x x x
1
4
5
200
0 .
4

500
600
4

150
1200
6 .
4

187 .
5
The minimum ratio occurs in Row 2.
Thus, x
4 leaves the basis when x
2 enters.
17

Step 7: Pivot x
2 in and x
4
Out z
1
0
0
0 x
0
1
0
0
1 x
2
0
0
1
0 x
3

0 .
0333
0 .
0667

0 .
0833
0 .
4
0 .
55

0 .
1
0 .
x
4
25

1 .
6 x
5
0
0
0
1 b
1230
140
150
240 basic z x
1 x
2 x
5
BV = { z , x
1
, x
2
, x
5
}, NBV = { x
3
, x
4
} z = 1230, x
1
= 140, x
2
= 150, x
5
= 240
18

Steps 4 and 5 z
1
0
0
0 x
0
1
0
0
1 x
2
0
0
1
0 x
3

0 .
0333
0 .
0667

0 .
0833
0 .
4
0 .
55

0 .
1
0 .
x
4
25

1 .
6 x
5
0
0
0
1 b
1230
140
150
240 basic z x
1 x
2 x
5 x
3 is eligible to enter the basis
19

Step 6: Ratio Test for x
3 z
1
0
0
0 x
0
1
1
0
0 x
2
0 x
3

0 .
0333 0 .
x
4
55 x
5
0 b
1230 basic z ratio
0
1
0 x x

2
2
0
0 .
.
0667
0833


0 .
4

0

.
0
1
.
1
25
.
6
0
0
1
140
150
240 x x x
1
2
5
0 .
0833 x
3
150


0 .
25
0 .
0833 x
3 x
4


150
0 .
25 x
4
140
0 .
0667

2100
150

0 .
0833
 
1800
240
0 .
4

600
If x
3 enters the basis, then x
2 will increase as well.
20

Step 6: Ratio Test for x
3 z
1
0
0
0 x
0
1
1
0
0 x
2
0
0
1
0 x
3

0 .
0333
0 .
0667

0 .
0833
0 .
4
0 .
55

0 .
1
0 .
x
4
25

1 .
6 x
5
0
0
0
1 b
1230
140
150
240 basic ratio z x
1
140
0 .
0667

2100 x
2 x
5
240
0 .
4

600
If x
3 enters the basis, then x
5 will leave the basis.
21

Step 7: Pivot x
3 in and x
5 out z
1 x
1
0
0
0
0
1
0
0 x
2
0
0
1
0 x
3 x
4 x
5 b
0 0 .
4167
0
0
1
0 .
1667

0 .
0833

4
0 .
0833 1250

0 .
1667 100
0 .
2083
2 .
5
200
600 basic x z
3 x
1 x
2
22

Steps 4 and 8 z
1 x
1
0
0
0
0
1
0
0 x
2
0
0
1
0 x
3 x
4 x
5 b
0 0 .
4167
0
0
1
0 .
1667

0 .
0833

4
0 .
0833 1250

0 .
1667 100
0 .
2083
2 .
5
200
600 basic x z
3 x
1 x
2
BV = { z , x
1
, x
2
, x
3
}, NBV = { x
4
, x
5
} z = 1250, x
1
= 100, x
This an optimal BFS.
2
= 200, x
3
= 600
23