maxZ=50x,+40x。+30x 0.02x+0.03xa+0.05x≤40 s0.05x,+0.02xn+0.04x≤40 0 3. Consider a preemptive goal programming problem with three priority levels, just one goal for each priority level, and just two activities to contribute toward these goals, as summarized in the following table Unit contribution activity Priority Level Goal First priority Second priority 2211 Third priority Use the goal programming technique to formulate one complete linear programming model for this problem. (10 points) Solution we set the activities level are xix min Z,=d min z=d.+d min z,=d The linear goal programming model is 2x,+d1-d1=20 d2-d2+=1 2x1+x2+d3-d3=4 4. Use the simplex method to solve the following problem (15 points Maximize Z=4x,+3x2+6x3 subject to 2x,+2x,+3x,<40 x1≥0,x2≥0,x3≥0 Solution: the process of simplex method iteration described as follows Coefficient of Basic variable (4 X5 Right side X4 (1)0 X5 (2) 0 2 2 Z 00100 60 l/3 (2) 363301002 ⎪ ⎩ ⎪ ⎨ ⎧ ≥ + + ≤ + + ≤ = + + , , 0 0.05 0.02 0.04 40 0.02 0.03 0.05 40 . . max 50 40 30 A B C A B c A B c A B C x x x x x x x x x st Z x x x 3. Consider a preemptive goal programming problem with three priority levels, just one goal for each priority level, and just two activities to contribute toward these goals, as summarized in the following table: Unit contribution activity Priority Level 1 2 Goal First priority 1 2 ≤20 Second priority 1 1 =15 Third priority 2 1 ≥40 Use the goal programming technique to formulate one complete linear programming model for this problem. (10 points) Solution: we set the activities level are x1,x2 The linear goal programming model is ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ + + − = + + − = + + − = = = + = − + − + − + − + − + 2 40 15 2 20 . . min min min 1 2 3 3 1 2 2 2 1 2 1 1 3 3 2 2 2 1 1 x x d d x x d d x x d d st Z d Z d d Z d 4. Use the simplex method to solve the following problem. (15 points) ⎪ ⎩ ⎪ ⎨ ⎧ ≥ ≥ ≥ + + ≤ + + ≤ = + + 0, 0, 0 2 2 3 40 3 3 30 4 3 6 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 Solution: the process of simplex method iteration described as follows: Coefficient of : Basic variable Eq. Z X1 X2 X3 X4 X5 Right side Z (0) 1 -4 -3 -6 0 0 0 X4 (1) 0 3 1 3 1 0 30 X5 (2) 0 2 2 3 0 1 40 Z (0) 1 2 -1 0 2 0 60 X3 (1) 0 1 1/3 1 1/3 0 10 X5 (2) 0 -1 1 0 -1 1 10 Z (0) 1 1 0 0 1 1 70