|x1||+ 2 x3||+ x4||< = 20|
|- 2 x1||+ x2||- x4||< = 15|
|6 x1||+ 2 x2||- 3 x3||< = 54|
|Inititial Tableau||Final Tableau|
|B = [ A4 , A6 , A2 ] =||[||
1 0 0 |
-1 1 1
0 0 2
|B-1 = [ A5* , A6* , A7* ] =||[||
1 0 0 |
1 1 -1/2
0 0 1/2
|A* = B-1 A||b* = B-1 b|
|c* = c - cB B-1 A||z* = z - cB B-1 b|
Changes in the Constant Column Vector
Changing the original column vector b into b' will affect b* and z* of the final tableau, but not c* and A*.
The modified b'* = B-1 b' can be calculated. If the entries remain nonnegative, since c* > = 0, the optimal solution to the modified problem will have the same optimal solution as the original problem, with values given by b'*; and z'* = z - cBb'*.
If some entries of b'* are negative, then to resolve this problem we use the Dual Simplex Algorithm on this new tableau.
Addition of a New Variable
Suppose that now we wish to add another variable in the formulation of the original problem. Let xn+1 be the new variable, with the objective coefficient cn+1 and the column vector of coefficients for the constraining equations An+1. Then the expanded, modified problem in canonical form is to minimize z' = c' X' subject to A' X' = b, X' > = 0, where
Addition of a Constraint
Suppose after solving the problem we wish to alter the original problem by the addition of a new constraint. Now it could be that X* satisfies this new constraint. If this is the case, X* is also optimal for the expanded problem, because clearly, by this addition of a constraint, we have not changed the objective function nor increased the set of feasible solutions to the system of constraints. On the other hand, if X* does not satisfy this new constraint, we must find a new optimal solution. Under certain circumstances, however, this problem may be resolved quite easily by creating a new canonical tableau ( the new costant column b' may contain some negative entries) from the final tableau solution to the original problem and the application of the Dual Simplex Algorithm.