Basic variable Coefficient of Z XI X2 X3 X4 X5 Right side 1/5 -1/3 1/5 2/5 (a) The optimal solution is x1=5/3, x2=0, X3=3, maxZ=17 minS=25y,+ 20y2 y1+3y2 (b)The dual problem 13y1+4y2≥1 5y1+5y2≥4 0 (c)The optimal solution for the dual problem from the final set of equations is yl=1/5y2=3/5,minS=17 (d) The coefficient of x2 in row O is 0,=CBB-A,? 0 55人3 so the previous optimal solution is not till optimal (e) The coefficient of new variable Xnew is 06=CBBA6-C6=231-2=>0, so the previous optimal solution is till 552 7. Solve the parameter programming as follow:(15 points Maxz=2x,+x2 ≤10+26 x1+x2 x2≤10+26 x1≥0, Solution: when 0=0, the optimal tableau is Basic variable Coefficient of X4 X5 Right side 0) 0 0 30 )0 0 0 10 2 0 When the right-side is b= 25-0 the final tableau is 10+265 Coefficient of : Basic variable Eq. Z X1 X2 X3 X4 X5 Right side Z (0) 1 0 2 0 1/5 3/5 17 X1 (1) 0 1 -1/3 0 1/3 -1/3 5/3 X3 (2) 0 0 1 1 -1/5 2/5 3 (a) The optimal solution is x1=5/3, x2=0, x3=3, maxZ=17 (b) The dual problem is ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ ≥ + ≥ + ≥ + ≥ = + , 0 5 5 4 3 4 1 6 3 3 . . min 25 20 1 2 1 2 1 2 1 2 1 2 y y y y y y y y st S y y (c) The optimal solution for the dual problem from the final set of equations is y1=1/5,y2=3/5,minS=17 (d) The coefficient of x2 in row 0 is 0 5 4 3 3 2 5 3 5 1 2 2 1 2 − = − < ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − = − σ CB B A C , so the previous optimal solution is not till optimal (e) The coefficient of new variable Xnew is 0 5 1 2 2 3 5 3 5 1 6 6 1 6 − = > ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − = − σ CB B A C , so the previous optimal solution is till optimal. 7.Solve the parameter programming as follow: (15 points) ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ ≥ ≥ ≤ + + ≤ − ≤ + = + 0, 0 10 2 25 10 2 . . 2 1 2 2 1 2 1 1 2 x x x x x x st Maxz x x θ θ θ Solution: when θ=0, the optimal tableau is Coefficient of : Basic variable Eq. Z X1 X2 X3 X4 X5 Right side Z (0) 1 0 0 2 0 1 30 X1 (1) 0 1 0 1 0 0 10 X4 (2) 0 0 0 -1 1 -1 5 X2 (3) 0 0 1 0 0 1 10 When the right-side is ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ + − + = θ θ θ 10 2 25 10 2 b ,the final tableau is