Note: constraint 4 uses the surplus variable in the initial (infeasible) solution MIN 4 X1 + 4 X2 + X3 SUBJECT TO 2) X1 + X2 + X3 <= 2 3) 2 X1 + X2 <= 3 4) 2 X1 + X2 + 3 X3 >= 3 END THE TABLEAU ROW (BASIS) X1 X2 X3 SLK 2 SLK 3 1 ART 4.000 4.000 1.000 0.000 0.000 2 SLK 2 1.000 1.000 1.000 1.000 0.000 3 SLK 3 2.000 1.000 0.000 0.000 1.000 4 SLK 4 -2.000 -1.000 -3.000 0.000 0.000 ART ART -2.000 -1.000 -3.000 0.000 0.000 ROW SLK 4 1 0.000 0.000 2 0.000 2.000 3 0.000 3.000 4 1.000 -3.000 ART 0.000 -3.000 X3 ENTERS AT VALUE 1.0000 IN ROW 4 OBJ. VALUE= -1.0000 THE TABLEAU ROW (BASIS) X1 X2 X3 SLK 2 SLK 3 1 ART 3.333 3.667 0.000 0.000 0.000 2 SLK 2 0.333 0.667 0.000 1.000 0.000 3 SLK 3 2.000 1.000 0.000 0.000 1.000 4 X3 0.667 0.333 1.000 0.000 0.000 ART ART 3.333 3.667 0.000 0.000 0.000 ROW SLK 4 1 0.333 -1.000 2 0.333 1.000 3 0.000 3.000 4 -0.333 1.000 ART 0.333 0.000 LP OPTIMUM FOUND AT STEP 1 OBJECTIVE FUNCTION VALUE 1) 1.000000 VARIABLE VALUE REDUCED COST X1 0.000000 3.333333 X2 0.000000 3.666667 X3 1.000000 0.000000 ROW SLACK OR SURPLUS DUAL PRICES 2) 1.000000 0.000000 3) 3.000000 0.000000 4) 0.000000 -0.333333 NO. ITERATIONS= 1 RANGES IN WHICH THE BASIS IS UNCHANGED: OBJ COEFFICIENT RANGES VARIABLE CURRENT ALLOWABLE ALLOWABLE COEF INCREASE DECREASE X1 4.000000 INFINITY 3.333333 X2 4.000000 INFINITY 3.666667 X3 1.000000 5.000000 1.000000 RIGHTHAND SIDE RANGES ROW CURRENT ALLOWABLE ALLOWABLE RHS INCREASE DECREASE 2 2.000000 INFINITY 1.000000 3 3.000000 INFINITY 3.000000 4 3.000000 3.000000 3.000000