In linear programming, unbounded solution means 16 (aha 1) Degenerate solution Infinite solutions Infeasible solution Unique solution o None of the above o (1 ) The variable added to the LHS of greater than or equal to ( 2 ) constraint to convert it into .17 equality is called (1 surplus variable o artificial variable o slack variable O additional variable surplus variable & artificial variable In the simplex algorithm if there exists an artificial variable as a basic variable in the optimal and 18 feasible table, then we have (1 unbounded solution o (1 ) Jnfeasible solution multiple solutions unbounded solution infeasible solution multiple solutions non of the above Based on the following model 4 Maximize Z = 4X1+2X2+3X3+5X4 s.t 2X1+ 3X2+ 4X3+ 2X4 = 300 8X1 + X2+ X3+ 5X4 = 200 X1 >= 0; X2 >= 0; X3>= 0; X4>=0 We can solve this problem by only 1 (1 ) Simplex method Big MO Two phase Both B and C None of the above Based on the following model.5 Maximize Z = 4X1+2X2+3X3+5X4 st 2X1+ 3X2+ 4X3+ 2X4 = 300 8X1+ X2+ X3+5X4 = 200 X1 >= 0; X2 >=0;X3>= 0; X4>=0 repared the constraints to be ready to solve the model (convert) 1) (1 ) 2X1+3x2+ 4X3+ 2X4+ R1 = 300 & 8X1+ X2+X3+ 5X4-S=200 O 2X1+ 3X2 + 4X3+ 2X4 -S+ R1 = 300 & 8X1+ X2+ X3+ 5X4+R2= 200 O 2X1+3x2+ 4x3+ 2X4 + R1= 300 & 8X1+ X2+ X3+ 5X4 +R2= 2000 2X1+ 3X2+ 4X3+ 2X4 -S+ R1= 300 & 8X1+ X2+ X3+ 5X4 - XS1= 200 O None of the above O Based on the following model.6 Maximize Z = 4X1-2X2+3X3+5X4 st 2X1+ 3X2+ 4X3+ 2X4 = 300 8X1 + X2 X3-5X4 = 200 X1 >= 0; X2 >= 0; X3>= 0; X4>=0 **The optimal solution is (1 (1 365.00 233.05 322 22 106.45 :Based on the following model.7 Maximize Z = 4X1+2X2+3X3+5X4 st 2X1 + 3X2+ 4X3+ 2X4 = 300 8X1 + X2+ X3+ 5X4 = 200 X1 >= 0; X2 >= 0; X3>= 0; X4>=0 : The optimal value of decision variables is 1) (1 ) X1 = 0: X2= 61.5; X3= 24: X4= 25.54 O X1 = 27.05: X2= 0: X3= 0; X4= 31 O X1 = 24 X2=27.78: X3 = 0; X4= 0 O X1 =0: X2=0; X3=61.11: X4= 27.78 None of the above Based on question 6.10 The dual prices of A and B and their feasibility ranges (1 (1 ) CP/C_M 21 s2 O C_P/CM 31 22 O C_M/C_P 31 1/2 O C_P/CM = 0; X2 >= 0; X3>= 0; X4>=0 : The optimal value of decision variables is 1) (1 ) X1 = 0: X2= 61.5; X3= 24: X4= 25.54 O X1 = 27.05: X2= 0: X3= 0; X4= 31 O X1 = 24 X2=27.78: X3 = 0; X4= 0 O X1 =0: X2=0; X3=61.11: X4= 27.78 None of the above Based on the following model.5 Maximize Z = 4X1+2X2+3X3+5X4 st 2X1+ 3X2+ 4X3+ 2X4 = 300 8X1+ X2+ X3+5X4 = 200 X1 >= 0; X2 >=0;X3>= 0; X4>=0 repared the constraints to be ready to solve the model (convert) 1) (1 ) 2X1+3x2+ 4X3+ 2X4+ R1 = 300 & 8X1+ X2+X3+ 5X4-S=200 O 2X1+ 3X2 + 4X3+ 2X4 -S+ R1 = 300 & 8X1+ X2+ X3+ 5X4+R2= 200 O 2X1+3x2+ 4x3+ 2X4 + R1= 300 & 8X1+ X2+ X3+ 5X4 +R2= 2000 2X1+ 3X2+ 4X3+ 2X4 -S+ R1= 300 & 8X1+ X2+ X3+ 5X4 - XS1= 200 O None of the above O Based on the following model.6 Maximize Z = 4X1-2X2+3X3+5X4 st 2X1+ 3X2+ 4X3+ 2X4 = 300 8X1 + X2 X3-5X4 = 200 X1 >= 0; X2 >= 0; X3>= 0; X4>=0 **The optimal solution is (1 (1 365.00 233.05 322 22 106.45