5. Consider the following problem(20 points) Maximize Z=2x,+7x2-3x3 3x,+4x3≤30 subject to x,+4x2-x3<10 x≥0,x2≥0,x3≥0 The corresponding final set of equations yielding the optimal solution is (0) +2x。=20 x2+5x3+x4-x5=20 x1+4x2-x3 10 Now you are to conduct sensitivity analysis by independently investigating each of the following changes in the original model. For each change, use the sensitivity analysis procedure to determine whether the previous optimal solution is till optimal (a) Change the right-hand sides to b2」[30 b)Change the coefficients of x3 to a13 (c)Introduce a new variable x6 with coefficients a (d)Introduce a new constraint 3x,+2x,+3x3 <25 (e)Change constraint 2 to x,+2x2+2x3 $35 Solution the final simplex tableau is Basic variable E Coefficient of X3 X4 X5 Right side 0 2 XI 1-120 (a)b- b= 30/x0, so the previous optimal solution is not till optimal (b) The new 3 In row a3=CB-A4-C3=(02 So the previous optimal solution is not till optimal4 5. Consider the following problem (20 points) ⎪ ⎩ ⎪ ⎨ ⎧ ≥ ≥ ≥ + − ≤ + + ≤ = + − 0, 0, 0 4 10 3 4 30 2 7 3 1 2 3 1 2 3 1 2 3 1 2 3 x x x x x x x x x subject to Maximize Z x x x The corresponding final set of equations yielding the optimal solution is (2) 4 10 (1) 5 20 (0) 2 20 1 2 3 5 2 3 4 5 2 3 5 + − + = − + + − = + + + = x x x x x x x x Z x x x Now you are to conduct sensitivity analysis by independently investigating each of the following changes in the original model. For each change, use the sensitivity analysis procedure to determine whether the previous optimal solution is till optimal. (a) Change the right-hand sides to ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 30 20 2 1 b b (b) Change the coefficients of x3 to ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 2 3 2 23 13 3 a a c (c) Introduce a new variable x6 with coefficients ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡− = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 2 1 3 21 11 6 a a c (d) Introduce a new constraint 3x1 + 2x2 + 3x3 ≤ 25 (e) Change constraint 2 to x1 + 2x2 + 2x3 ≤ 35 Solution: the final simplex tableau is Coefficient of : Basic variable Eq. Z X1 X2 X3 X4 X5 Right side Z (0) 1 0 1 1 0 2 20 X4 (1) 0 0 -1 5 1 -1 20 X1 (2) 0 1 4 -1 0 1 10 (a) 0 30 10 30 20 0 1 1 1 1 ≥⎥ ⎦ ⎤ ⎢ ⎣ ⎡− =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = − B b , so the previous optimal solution is not till optimal. (b) The new coefficient of x3 in row 0 is ( ) ( 2) 2 0 2 3 3 3 0 2 1 3 − − = − < ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = − = − σ CB B A C , So the previous optimal solution is not till optimal