The Leaving Variable

Example: Product Planning

We need to find out which solution variable will first be driven to zero as units of x2 are introduced into the solution. Since variable x1 will remain an outside variable with value zero, we can write the constraint equations as:

If x3 were to be the leaving variable (and hence become zero), we could solve the above equations for x2 and x4, obtaining x2 = 8, x4 = 8. This is a feasible solution.

If x4 were to be the leaving variable (and hence become zero), solving the equations for x2 and x3 gives x2 = 16, x3 = -24. This is an infeasible solution, since it would require negative amounts of x3. In other words, as units of x2 are added, at the point x2 = 8, the solution variable x3 would be driven to zero; this is the first solution variable driven to zero (x4, the other solution variable, is still positive at that point), so x3 is the leaving variable.

The same result can be obtained using the table and a more mechanical rule. Divide each amount in the "Solution values" column by the amount in the comparable row of the entering xj column using the formula:

For x3 row: 24/3 = 8

For x4 row: 16/1 = 16

The smallest number obtained by this computation gives the maximum number of units of x2 - 8, which may be injected into the solution without driving some solution variable negative. If any of the amounts calculated by dividing are negative, they should be eliminated from consideration; otherwise, the smallest amount, as computed above, determines the leaving variable. In this case, the x3 row has the smallest number (8), and x3 is the leaving variable.. The x3 row is marked with an arrow :
Note what we have done so far in the simplex procedure. First, we identified a variable that will increase profitability and indicated that it is to be included in the next solution. Then, we guaranteed that the next solution would be a basic feasible one by identifying a variable to be removed from the solution that will keep all variable values nonnegative. The following step is to determine the new solution and construct the second simplex table. In the intersection of the entering variable column and the leaving variable line lies the pivot element.