The first change we will consider is changing a cost value by 
in the original problem. We are given the original problem and an optimal
tableau. If we had done exactly the same calculations beginning with the
modified problem, we would have had the same final tableau except that
the corresponding cost entry would be
lower (this is because we never do anything except add or subtract scalar
multiples of Rows 1 through m to other rows; we never add or subtract
Row 0 to other rows). For example, take the problem
Max 3x+2y Subject to x+y <= 4 2x+y <= 6 x,y >= 0The optimal tableau to this problem (after adding
and 
as slacks to place in standard form) is:
Suppose the cost for x is changed to 
in the original formulation, from its previous value 3. After doing the
same operations as before, that is the same pivots, we would end up with
the tableau:
Now this is not the optimal tableau: it does not have a correct basis
(look at the column of x). But we can make it correct in form while
keeping the same basic variables by adding
times the last row to the cost row. This gives the tableau:
Note that this tableau has the same basic variables and the same variable values (except for z) that our previous solution had. Does this represent an optimal solution? It does only if the cost row is all non-negative. This is true only if
which holds for 
. For any
in that range, our previous basis (and variable values) is optimal. The
objective changes to 
.
In the previous example, we changed the cost of a basic variable. Let's go through another example. This example will show what happens when the cost of a nonbasic variable changes.
Max 3x+2y + 2.5w Subject to x+y +2w <= 4 2x+y +2w <= 6 x,y,w >= 0Here, the optimal tableau is :
Now suppose we change the cost on w from 2.5 to 
in the formulation. Doing the same calculations as before will result in
the tableau:
In this case, we already have a valid tableau. This will represent an
optimal solution if 
, so 
. As long as the objective coefficient of w is no more than 2.5+1.5=4
in the original formulation, our solution of x=2,y=2 will
remain optimal.
The value in the cost row in the simplex tableau is called the reduced cost. It is zero for a basic variable and, in an optimal tableau, it is non-negative for all other variables (for a maximization problem).
Summary: Changing objective function values in the original formulation will result in a changed cost row in the final tableau. It might be necessary to add a multiple of a row to the cost row to keep the form of the basis. The resulting analysis depends only on keeping the cost row non-negative.