z -3XE -2XI - 0S1 - 0S2 - 0S3- 0S4 = 0 (0)
XE +2XI + S1 = 6 (1)
2XE + XI + S2 = 8 (2)
-XE + XI + S3 = 1 (3)
XI + S4 = 2 (4)
XE , XI,, S1,, S2,, S3,, S4 ³ 0
Example 1: Reddy-Mikks Problem
FStep 0.- Since we have four equations representing the constraints we can have at most four linearly independent variables to get the solution of the system of equations.
FWe set XE , XI, to zero (non-basic variables) and solve for S1,, S2,, S3,, S4 (basic variables). Notice that we easily get a feasible solution. In particular a basic feasible solution.
F
z -3XE -2XI - 0S1 - 0S2 - 0S3- 0S4
XE +2XI + S1 = 6 (1)
2XE + XI + S2 = 8 (2)
-XE + XI + S3 = 1 (3)
XI + S4= 2 (4)
XE , XI,, S1,, S2,, S3,, S4 ³ 0