________________________________ LINDO OUTPUT FOR TV PROBLEM ________________________________ MAX 15 X1 + 45 X2 SUBJECT TO 2) X2 <= 50 3) X1 + 1.6 X2 <= 240 4) 0.5 X1 + 2 X2 <= 162 END THE TABLEAU ROW (BASIS) X1 X2 SLK 2 SLK 3 SLK 4 1 ART -15.000 -45.000 0.000 0.000 0.000 0.000 2 SLK 2 0.000 1.000 1.000 0.000 0.000 50.000 3 SLK 3 1.000 1.600 0.000 1.000 0.000 240.000 4 SLK 4 0.500 2.000 0.000 0.000 1.000 162.000 ART ART -15.000 -45.000 0.000 0.000 0.000 0.000 X2 ENTERS AT VALUE 50.000 IN ROW 2 OBJ. VALUE= 2250.0 THE TABLEAU ROW (BASIS) X1 X2 SLK 2 SLK 3 SLK 4 1 ART -15.000 0.000 45.000 0.000 0.000 2250.000 2 X2 0.000 1.000 1.000 0.000 0.000 50.000 3 SLK 3 1.000 0.000 -1.600 1.000 0.000 160.000 4 SLK 4 0.500 0.000 -2.000 0.000 1.000 62.000 X1 ENTERS AT VALUE 124.00 IN ROW 4 OBJ. VALUE= 4110.0 THE TABLEAU ROW (BASIS) X1 X2 SLK 2 SLK 3 SLK 4 1 ART 0.000 0.000 -15.000 0.000 30.000 4110.000 2 X2 0.000 1.000 1.000 0.000 0.000 50.000 3 SLK 3 0.000 0.000 2.400 1.000 -2.000 36.000 4 X1 1.000 0.000 -4.000 0.000 2.000 124.000 SLK 2 ENTERS AT VALUE 15.000 IN ROW 3 OBJ. VALUE= 4335.0 THE TABLEAU ROW (BASIS) X1 X2 SLK 2 SLK 3 SLK 4 1 ART 0.000 0.000 0.000 6.250 17.500 4335.000 2 X2 0.000 1.000 0.000 -0.417 0.833 35.000 3 SLK 2 0.000 0.000 1.000 0.417 -0.833 15.000 4 X1 1.000 0.000 0.000 1.667 -1.333 184.000 LP OPTIMUM FOUND AT STEP 4 OBJECTIVE FUNCTION VALUE 1) 4335.000 VARIABLE VALUE REDUCED COST X1 184.000000 0.000000 X2 35.000000 0.000000 ROW SLACK OR SURPLUS DUAL PRICES 2) 15.000000 0.000000 3) 0.000000 6.250000 4) 0.000000 17.500000 NO. ITERATIONS= 4 RANGES IN WHICH THE BASIS IS UNCHANGED: OBJ COEFFICIENT RANGES VARIABLE CURRENT ALLOWABLE ALLOWABLE COEF INCREASE DECREASE X1 15.000000 13.125000 3.750000 X2 45.000000 15.000000 21.000000 RIGHTHAND SIDE RANGES ROW CURRENT ALLOWABLE ALLOWABLE RHS INCREASE DECREASE 2 50.000000 INFINITY 15.000000 3 240.000000 84.000000 36.000000 4 162.000000 18.000000 42.000000 0 0.000 0.000