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Cost Changes


The first change we will consider is changing a cost value by tex2html_wrap_inline6034 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 tex2html_wrap_inline6034 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 >= 0

The optimal tableau to this problem (after adding tex2html_wrap_inline5636 and tex2html_wrap_inline5638 as slacks to place in standard form) is:


Suppose the cost for x is changed to tex2html_wrap_inline8470 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 tex2html_wrap_inline6034 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 tex2html_wrap_inline8486 . For any tex2html_wrap_inline6034 in that range, our previous basis (and variable values) is optimal. The objective changes to tex2html_wrap_inline8490 .

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 >= 0

Here, the optimal tableau is :


Now suppose we change the cost on w from 2.5 to tex2html_wrap_inline8496 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 tex2html_wrap_inline8500 , so tex2html_wrap_inline8502 . 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.

next up previous contents
Next: Right Hand Side Changes Up: Tableau Sensitivity Analysis Previous: Tableau Sensitivity Analysis

Michael A. Trick
Mon Aug 24 16:30:59 EDT 1998