In the new matrix the negative three has been eliminated, however, a 

 negative one has appeared in the S3 row. For the next pivoting process, 

 the S;} row will therefore become the pivot row, and the X2 column will 

 become the pivot column since it contains the only positive element 

 in the S.-? row. The third and final table. 



Si = 



Xi = 



Xo = 



z = 



indicates that the solution is minimized at 19 ( — Z = — 19, therefore Z 

 =: 19 I . All negative values in the last column have been eliminated and 

 the negative nund3ers in the bottom row indicate that no further min- 

 imizing can be accomplished. If, after eliminating all the last column 

 negative values, positive numbers exist in the bottom row, the matrix is 

 further iterated by choosing the pivot column and the pivot row in the 

 manner described on page 3. 



