If this modified problem were now to be minimized, the initial 

 matrix would be: 



Si = 



52 = 



53 = 



-Z = 



The bottom row indicates that the problem has been solved because 

 of the negative numbers, however, the matrix says this is accomplished 

 by having a negative amount of S2 in solution, which is not an accept- 

 able situation. The negative three in the last column must be removed. 

 It is here that this approach differs from the more commonly presented 

 techniques. Rather than becoming involved with an artificial variable, 

 its high negative value and additional computer storage space, the nega- 

 tive last column figure is removed mathematically with the following 

 rules: 



Rules for Removing Negative Resources ^ 



Neg-1. The row containing the negative number in the last column 

 becomes the pivot row. 



Neg-2. The pivot element must be positive. 



Neg-3. The pivot column is chosen arbitrarily if more than one 

 column satisfies rules Neg-1 and Neg-2. 



Neg— 4. If there are two or more negative numbers in the last col- 

 umn, then the pivot row is the row generating the largest 

 absolute value of Q. 



Once the pivot row and pivot column are determined, the pivoting 

 rules are followed as before. The first iteration for the above example 

 is shown here: 



Si = 



82 = 



83 = 

 -Z = 



S2 



X2 



KPo) 



Si = 

 Xi = 



83 = 

 — z = 



Old Matrix 



New Matrix 



^ These rules are derived in Appendix B. 



8 



