Constructing a Final Tableau from an an Initial Tableau and an Optimal Solution
Consider the following linear Linear Programming Problem:
Minimize z = -30 x1 - 6 x2 + 5 x3 - 18 x4
subject to:
| x1 | | + 2 x3 | + x4 |
< = 20 |
| - 2 x1 | + x2 | | - x4 | < = 15 |
| 6 x1 | + 2 x2 | - 3 x3 | | < = 54 |
x1, x2, x3, x4 > = 0
An initial canonical tableau of the problem is as follows:
|
|
|
|
b |
|
|
| 1 |
0 |
2 |
1 |
| -2 |
1 |
0 |
-1 |
| 6 |
2 |
-3 | 0 |
|
|
|
| C |
|
| 0 |
Suppose an optimal solution X* = ( 0 , 27 , 0 , 20 , 0 , 8 , 0 )t to the problem is
known. Then a final tableau may be constructed in the following way.
Step 1.
Notice that the basic variables are:
x2* = 27 , x4* = 20 and x6* = 8.
By using A2 , A4 and A6 we construct the matrix
| B = [ A2 , A4 , A6 ] =
| [ |
0 1 0
1 -1 1
2 0 0 | ] |
|
Step 2.
Now, we find the inverse of the matrix B.
| B-1 = | [ |
0 0 1/2 1 0 0 1 1 -1/2 | ] |
Step 3.
We use B-1 to obtain A* = B-1 A and
b* = B-1 b.
| [ |
0 0 1/2
1 0 0
1 1 -1/2 | ] |
|
[ |
1 0   2   1 1 0 0
-2 1 0 -1 0 1 0
6 2 -3 0 0 0 1 | | | 20 15 54 |
] = [ |
3 1 -3/2 0 0 0 1/2
1 0 2
1 1 0 0
-4 0 7/2 0 1 1 -1/2
| | | 27 20 8 | ] |
Step 4.
Let CB=( C2 , C4 , C6 ), then
C* = C- CB A* and z* = z - CB b. Thus the last row of our final tableau is:
[ -30 -6 5 -18 0 0 0 | 0 = z ] -
( -6 , -18 , 0 ) [ A* | b* ] = [ 6 0 32 0 18 0 3 | 522 = z*]
According to the Duality Theorem, Y* = ( 18 , 0 , 3 )t is an optimal solution to the Dual problem.
Here is our final tableau:
|
|
|
|
b* |
|
|
| 3 |
1 |
-3/2 |
0 |
| 1 |
0 |
2 |
1 |
| -4 |
0 |
7/2 |
0 |
|
|
|
| C |
|
| 522 |
Note
1. By rearranging the columns of the matrix B, we obtain a different B that gives
us a different final tableau.
2. Suppose instead of X*, the vector Y* = ( 18 , 0 , 3 )t, a solution to the dual of this problem is known, then by the Complementary Slackness Theorem, X* may be found.