This last rule may seem rather complicated, but it will prove simple 

 if stepped through slowly. It may be thought of in terms of a rectangle. 

 The elements involved form a perfect rectangle within the matrix. The 

 element being changed is one corner of the rectangle, the pivot element 

 is the opposite corner, and the two elements that are to be multiplied 

 together complete the rectangle. In the example, the Si resource (24) 

 becomes 24, minus the value of 5 times minus 2 divided by minus 1, or. 



24 



[(5) (- 2)/(— 1)] = 24 — (+ 10) = 14 



Xi 



X2 KPo) 



S3 



X2 KPo) 



% 



—2 



new 



pivot 



row 



Old Matrix 



new pivot colamn 

 New Matrix 



It will be noticed that in the new matrix, the headings "Xj" and 

 "S3" have been switched, which can be visualized by the fact that they 

 have been "pivoted" on the element in the matrix common to both — 

 the pivot element. 



This new matrix indicates that the objective solution has been in- 

 creased to 15 by producing five units of Xj. There are 14 units of Si 

 and .5 units of S2 left over. The positive 2 in the bottom row, however, 

 indicates that the objective solution can be made even greater. A third 

 matrix is developed using the X2 column as the pivot column and the 

 S2 row as the pivot row. 



New matrices will continue to be developed as long as there is a 

 positive element in the bottom row (excepting the objective solution, 

 of course) . The solution of this example problem comes with the fourth 

 matrix. 



S3 = 

 X2 = 



Xi = 



z = 



Solution 



