(c)The coefficient of new variable in row O is =C4-C=02(2-=70,50 previous optimal solution till optil (d) Input the optimal solution x1=10, x2=0, x3=0 into the new constraint 3x,+2x,+3x3≤25,3×10+2×0+3×02 the previous optimal solution is not till optimal (e) If change constraint 2 to x, +2x2+2x, <35, the final tableau is changed into Basic variable Coefficient of X2 5 Right side (0) 0 0 So the previous optimal solution is till optimal 6. Solve the following parametric linear programming problem (15 points) Maxz()=(10-46)x1+(4-6)x2+(7+6)x 3x,+x,+2x2≤7 s12x1+x2+3x3≤5 x120,x20,x320 when 0=0, the final simplex tableau is Basic variable E Coefficient of: X4 X5 Right side 0 0 24 (1 0 Solution: when 0 >0, the simplex table is Basic variable Coefficient of XI X3 X5 Right side Z 03-202-202+0 (1)0 1 1 When O<e<1, x1=2, X2=l is the optimal solution When 0>1, the simplex tableau is Basic variable Eq Coefficient of ght side -40 2)0 3 When 1<0<5/4, X4=2, x2=5 is the optimal solution When 0>5/4, the simplex tableau is5 (c) The coefficient of new variable in row 0 is ( ) ( 3) 7 0 2 1 6 6 0 2 1 6 − − = > ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = − = − σ CB B A C , So the previous optimal solution is till optimal. (d) Input the optimal solution x1=10,x2=0,x3=0 into the new constraint: 3x1 + 2x2 + 3x3 ≤ 25,3×10 + 2× 0 + 3× 0 ≤ 25 So the previous optimal solution is not till optimal. (e) If change constraint 2 to x1 + 2x2 + 2x3 ≤ 35 , the final tableau is changed into Coefficient of : Basic variable Eq. Z X1 X2 X3 X4 X5 Right side Z (0) 1 0 1 5 0 2 20 X4 (1) 0 0 1 2 1 -1 -5 X1 (2) 0 1 2 2 0 1 35 So the previous optimal solution is till optimal. 6. Solve the following parametric linear programming problem (15 points) ⎪ ⎩ ⎪ ⎨ ⎧ ≥ ≥ ≥ + + ≤ + + ≤ = − + − + + 0, 0, 0 2 3 5 3 2 7 . . ( ) (10 4 ) (4 ) (7 ) 1 2 3 1 2 3 1 2 3 1 2 3 x x x x x x x x x st MaxZ θ θ x θ x θ x when θ=0, the final simplex tableau is Coefficient of : Basic variable Eq. Z X1 X2 X3 X4 X5 Right side Z (0) 1 0 0 3 2 2 24 X1 (1) 0 1 0 -1 1 -1 2 X2 (2) 0 0 1 5 -2 3 1 Solution: when θ>0, the simplex table is Coefficient of : Basic variable Eq. Z X1 X2 X3 X4 X5 Right side Z (0) 1 0 0 3-2θ 2-2θ 2+θ X1 (1) 0 1 0 -1 1 -1 2 X2 (2) 0 0 1 5 -2 3 1 When 0<θ≤1, x1=2,x2=1 is the optimal solution. Whenθ>1, the simplex tableau is Coefficient of : Basic variable Eq. Z X1 X2 X3 X4 X5 Right side Z (0) 1 2θ-2 0 5-4θ 0 4-θ X4 (1) 0 1 0 -1 1 -1 2 X2 (2) 0 2 1 3 0 1 5 When 1<θ≤5/4, x4=2,x2=5 is the optimal solution. Whenθ>5/4, the simplex tableau is