The following maximum problem is in standard form. Introduce slack variables and set up the initial simplex.
Maximize P = 2x1 + 4x2 + x3
Subject to the constraints:
x1 + x2 + x3 <= 12
x1 ≥ 0
x2 ≥ 0
x3 ≥ 0
x1 + x2 <= 6
2x1 - x2 + 3x3 <= 6
x1 + x2 + x3 + s1 = 12
x1 + x2 + s2 = 6
2x1 - x2 + 3x3 + s3 = 6
P = 2x1 + 4x2 + x3
Write that as this system:
1x1 + 1x2 + 1x3 + 1s1 + 0s2 + 0s3 + 0P = 12
1x1 + 1x2 + 0x3 + 0s1 + 1s2 + 0s3 + 0P = 6
2x1 - 1x2 + 3x3 + 0s1 + 0s2 + 1s3 + 0P = 6
-2x1 - 4x2 - 1x3 + 0s1 + 0s2 + 1s3 + 1P = 0
Then make this as the first tableau:
x1 x2 x3 s1 s2 s3 P
_____________________________
| 1 1 1 | 1 0 0 | 0 | 12 |
| 1 1 0 | 0 1 0 | 0 | 6 |
| 2 -1 3 | 0 0 1 | 0 | 6 |
|----------------------------|
|-2 -4 -1 | 0 0 1 | 1 | 0 |
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
The first step will be to select the most negative
indicator which is -4. so the pivot column is the
2nd column.
Divide each positive number above the -4 into the
corresponding element in the far right:
___ __
1)12 = 12, 1)6 = 6, 6 is smaller so pivot on the 1
in the 2nd row 2nd column.
Edwin