In this section, we will use the dual simplex method. Let's see the following Linear Programming Problem (LPP).
Minimize z = 80x1 + 100x2
subject to
80x1 + 60x2 ≥
1500
20x1 + 90x2 ≥
1200
x1, x2 ≥ 0
Solution.
Minimize z = 80x1 + 100x2
Multiplying the constraints by -1 on both sides
-80x1 - 60x2 ≤
-1500
-20x1 - 90x2 ≤
-1200
After adding slack variables, the constraints are
-80x1 - 60x2 + x3 = -1500
-20x1 - 90x2 + x4 = -1200
Where x3 and x4 are slack variables.
Substituting x1 = 0, x2 = 0, z = 0
This gives the solution values as
x3 = -1500, x4 = -1200
The initial simplex table is exhibited below.
cj | 80 | 100 | 0 | 0 | ||
---|---|---|---|---|---|---|
cB | Basic variables B |
x1 | x2 | x3 | x4 | Solution values b (=XB) |
0 | x3 | -80 | -60 | 1 | 0 | -1500 |
0 | x4 | -20 | -90 | 0 | 1 | -1200 |
zj-cj | -80 | -100 | 0 | 0 |
Key Row :
Min { XB1, XB2} = Min ( -1500, -1200) = -1500
So x3 row is the key row.
Key Column:
Min | z1 - c1 |
, | z2 - c2 |
Min | -80 |
, | -100 |
Min(1, 5/3) = 1
So column under x1 becomes the key column.
Pivot element = - 80.
Therefore, x3 departs & x1 enters.
Table 2
cj | 80 | 100 | 0 | 0 | ||
---|---|---|---|---|---|---|
cB | Basic variables B |
x1 | x2 | x3 | x4 | Solution values b (=XB) |
80 | x1 | 1 | 3/4 | -1/80 | 0 | 75/4 |
0 | x4 | 0 | -75 | -1/4 | 1 | -825 |
zj-cj | 0 | -40 | -80 | 0 |
Key row = x4 row as it has the only negative value under
the XB column.
Key column = x2 column.
Pivot element = -75.
x4 departs & x2 enters.
Table 3
cj | 80 | 100 | 0 | 0 | ||
---|---|---|---|---|---|---|
cB | Basic variables B |
x1 | x2 | x3 | x4 | Solution values b (=XB) |
80 | x1 | 1 | 0 | -3/200 | 1/100 | 21/2 |
100 | x2 | 0 | 1 | 1/300 | -1/75 | 11 |
zj-cj | 0 | 0 | -13/15 | -8/15 |
The values for x1 & x2 are 21/2 & 11
respectively.
The associated objective function value is
z = 80 X 21/2 + 100 X 11 = 1940.