IE3. Section 3.3: Problem -3 Problem P.68 Max z 4x +x s.t. 8x +x 6 ---- C 5x +x ---- C xx 0 ---- C3 X 8 Feasible Region 6 C 4 C CASE The LP has alternative or multiple optimal solutions: To extreme points are optimal and the LP ill have an ininite number o optimal solutions. Z4 A Any points along line AB are the optimal solutions. C3 B.4 X Problem 3 P.68 Max z -x +3x s.t. x -x 4 ---- C x +x 4 ---- C xx 0 ---- C3 X Feasible Region C Z0 C CASE 4 The LP is unbounded: There are points in the easible region ith the arbitrarily large z-value. C3 4 X
IE3. Section 3.3: Problems 5-6. Make sure that you justiy your ansers. Problem 5: True An unbounded easible region is a necessary condition or a LP to be unbounded. Problem 6: False It is possible or a LP ith unbounded easible region to have optimal solution. See an example LP belo: 3. Construct an improving direction rom the gradient o each objective unction at the point indicated. x gives the direction o the largest increase improving i max problem. - x gives the direction o the largest decrease improving i min problem. a. Max 3 - + 4 at 05 0 3 4 3 At 05 the improving direction is 05 3-0. b. Min 4 + 3-5 4 at 7-5 40 4 3 At 7- the improving direction is - 7- -40-5. c. Min + - at 3 + At 3 the improving direction is - 3 -*3+-3-83. Isocost Line An optimal Solution unbounded easible region
IE3 3 d. Max 3-4 +6 at 5 3 46 At 5 the improving direction is 5 3* -46 86. 4. Consider the mathematical program Max 4z +7z Objective Function s.t. z +z 9 Constraint C 0 z 4 Constraint C 0 z 3 Constraint 3 C3 a Sho that directions Δz 0 and Δz -4 are improving directions or this model at every z. For max problem the objective unction is increased i x.δx>0. z z From objective unction e get z 47. z z The direction Δz 0 is an improving directions or this model at every z since z. Δz 47. 0 8 > 0. The direction Δz -4 is an improving directions or this model at every z since z. Δz 47. -4-8+80 > 0. b Beginning at z 0 00 execute the Improving Search algorithm on the model see PoerPoint slides or algorithm. Limit yoursel to the to directions in part a and continue until neither is improving and easible. Improving search algorithm rom PPT slide: 0 Choose initial solution I no improving direction Δz STOP Find improving direction Δz rom a. e have improving direction Δz 0 and Δz -4. 3 Choose the largest step size λ t+ that remains easible and improves perormance 4 Update z t+ z t + λ t+ Δz t+ Let tt+ and return to step At t0 initial solution is z 0 00. We have to improving direction rom a. hich are Δz and Δz. Let start improving ith Δz Find the largest step size λ We have z z 0 + λ Δz 00+ λ.0 λ 0 Substitute z into C CC3 to ind λ C: z +z 9 λ +04λ 9 λ 9/4 C: 0 z 4 0 λ 4 0 λ C3: 0 z 3 0 0 3 λ R To remain easible and improves perormance the largest λ is hence z 40. Since e cannot improve more perormance rom z using Δz e start moving in Δz direction.
IE3 4 At t ne initial solution is z 40. Start improving ith Δz Find the largest step size λ We have z z + λ Δz 40+ λ.-4 4-λ 4λ Substitute z into C CC3 to ind λ C: z +z 9 4-λ + 4λ 8 9 λ R C: 0 z 4 0 4-λ 4 0 λ C3: 0 z 3 0 4λ 3 0 λ 3/4 To remain easible and improves perormance the largest λ is 3/4 hence z 5/3. Since e cannot improve more perormance rom z using Δz e try improving ith Δz direction again. At t ne initial solution is z 5/3. Start improving ith Δz 3 hich is equal to Δz Find the largest step size λ 3 We have z 3 z + λ 3 Δz 3 5/3+ λ 3.0 5/+λ 3 3 Substitute z 3 into C CC3 to ind λ 3 C: z +z 9 5/+λ 3 +3 9 λ /4 C: 0 z 4 0 5/+λ 3 4-5/4 λ 3 3/4 C3: 0 z 3 0 3 3 λ 3 R To remain easible and improves perormance the largest λ 3 is /4 hence z 3 33. Since e cannot improve more perormance rom z 3 using Δz 3 or Δz e try improving ith Δz direction again. At t3 ne initial solution is z 3 33. Start improving ith Δz 4 hich is equal to Δz Find the largest step size λ 4 We have z 4 z 3 + λ 4 Δz 4 33+ λ 4.-4 3-λ 4 3+4λ 4 Substitute z 4 into C CC3 to ind λ 4 C: z +z 9 3-λ 4 +3+4λ 4 9 9 λ 4 R C: 0 z 4 0 3-λ 4 4 -/ λ 4 3/ C3: 0 z 3 0 3+4λ 4 3-3/4 λ 4 0 To remain easible and improves perormance the largest λ 4 is 0 hich mean e cannot improve rom z 3 or no improving direction Δz STOP searching And e get the optimal solution at z33 ith the objective value o 33. IF e start the search algorithm ith direction Δz instead o Δz the solution may dierent. At t0 initial solution is z 0 00. We have to improving direction rom a. hich are Δz and Δz. Let start improving ith Δz Find the largest step size λ We have z z 0 + λ Δz 00+ λ.-4 -λ 4λ Substitute z into C CC3 to ind λ C: z +z 9 -λ+4λ0 9 λ R C: 0 z 4 0 -λ 4 - λ 0 C3: 0 z 3 0 4λ 3 0 λ 3/4 To remain easible and improves perormance the largest λ is 0 hich mean e cannot improve rom z 0 or no improving direction Δz STOP searching We might think that the optimal solution is at z00 ith the objective value o 0. Hoever ith the previous search method start searching ith Δz direction e can get the better solution. We can see that using dierent search methods may result in the dierent solutions.
IE3 5 c Sho in a to-dimensional plot the easible space and objective unction contours o the models. Then plot the path o your search in part b. z Feasible Region C C C3 z z 3 z 0 z Z8 z Z4
IE3 6 5. Section 3.8: 4 Problem P.9 Notation: X ij Amount o input j use to produce product i lb Product I here i or orange and or orange juice Input J here j or grade 9 orange or grade 6 orange Decision Variable: X X X X LP Objective Function: Max Proit 0.3X +X + 0.45X +X Constraints Quality requirements Avg grade in bags o orange Avg grade in cartons o orange juice Input Availability Constraints: grade 9 orange grade 6 orange 9X +6X >7X +X or X -X >0 9X +6X >8X +X or X -X >0 X +X <00000 X +X <0000 Sign Constraints: X >0 X >0 X >0 X >0 Equation
IE3 7 Problem 4 P.94 Notation: X ij Amount o input j use to produce product i Product I here i or Regular Gasoline and or Premium Gasoline Input J here j or Reormate or FCG 3 or ISO 4 or POL 5 or MTB 6 or BUT Decision Variable: X X X 3 X 4 X 5 X 6 X X X 3 X 4 X 5 X 6 LP Objective Function: Max Proit 9.49X +X +X 3 +X 4 +X 5 +X 6 + 3.43X +X +X 3 +X 4 +X 5 +X 6 -.75X +X +X 3 +X 4 +X 5 +X 6 -.75X +X +X 3 +X 4 +X 5 +X 6 Constraints: Equation Demand Constraints: Regular Gasoline Premium Gasoline Ration Constraints: FCG in Regular Gasoline FCG in Premium Gasoline Attribute requirements RON in Regular Gasoline RON in Premium Gasoline RVP in Regular Gasoline RVP in Premium Gasoline ASTM70 in Regular Gasoline ASTM70 in Premium Gasoline ASTM30 in Regular Gasoline ASTM30 in Premium Gasoline Input Availability Constraints: Reormate FCG ISO POL MTB X +X +X 3 +X 4 +X 5 +X 6 >9800 X +X +X 3 +X 4 +X 5 +X 6 >30000 X <0.38X +X +X 3 +X 4 +X 5 +X 6 or 0.6X -0.38X -0.38X 3-0.38X 4-0.38X 5-0.38X 6 <0 X <0.38 X +X +X 3 +X 4 +X 5 +X 6 or -0.38X +0.6X -0.38X 3-0.38X 4-0.38X 5-0.38X 6 <0 98.9X +93.X +86.X 3 +97X 4 +7X 5 +98X 6 > 90X +X +X 3 +X 4 +X 5 +X 6 or 8.9X +3.X -3.9X 3 +7X 4 +7X 5 +8X 6 >0 98.9X +93.X +86.X 3 +97X 4 +7X 5 +98X 6 > 96X +X +X 3 +X 4 +X 5 +X 6 or.9x -.8X -9.9X 3 +X 4 +X 5 +X 6 >0 7.66X +9.78X +9.5X 3 +4.5X 4 +3.45X 5 +66.99X 6.8X +X +X 3 +X 4 +X 5 +X 6 or -3.5X -.40X +8.34X 3-6.67X 4-7.73X 5 +45.8X 6 0 7.66X +9.78X +9.5X 3 +4.5X 4 +3.45X 5 +66.99X 6.8X +X +X 3 +X 4 +X 5 +X 6 or -3.5X -.40X +8.34X 3-6.67X 4-7.73X 5 +45.8X 6 0-5X +57X +07X 3 +7X 4 +98X 5 +30X 6 > 0X +X +X 3 +X 4 +X 5 +X 6 or -5X +47X +97X 3-3X 4 +88X 5 +0X 6 >0-5X +57X +07X 3 +7X 4 +98X 5 +30X 6 > 0X +X +X 3 +X 4 +X 5 +X 6 or -5X +47X +97X 3-3X 4 +88X 5 +0X 6 >0 46X +03X +00X 3 +73X 4 +00X 5 +00X 6 > 50X +X +X 3 +X 4 +X 5 +X 6 or -4X +53X +50X 3 +3X 4 +50X 5 +50X 6 >0 46X +03X +00X 3 +73X 4 +00X 5 +00X 6 >50X +X +X 3 +X 4 +X 5 +X 6 or -4X +53X +50X 3 +3X 4 +50X 5 +50X 6 >0 X +X <557 X +X <5434 X 3 +X 3 <6709 X 4 +X 4 <90 X 5 +X 5 <748 Sign Constraints: X >0 X >0 X 3 >0 X 4 >0 X 5 >0 X 6 >0 X >0 X >0 X 3 >0 X 4 >0 X 5 >0 X 6 >0
IE3 8 6. Section 3.9: 8 Problem P.97-98 X + X hr o skilled labors X + X units Uninished table 3X hr o skilled labors X units Uninished table sold X units Finished table sold 40X +X +30X 3 +X 4 board t. Wood X 3 units Uninished chair sold X 3 + X 4 hr o skilled labors X 3 + X 4 units Uninished chair X 4 hr o skilled labors X 4 units Finished chair sold Notation: X i Amount o product i sold Product I here i or Uninished table or Finished table 3 or Uninished chair and 4 or Finished chair Decision Variable: X X X 3 X 4 LP Objective Function: Max Proit 70X +40X +60X 3 +0X 4 40X +X 30X 3 +X 4 30X +00X +30X 3 +80X 4 Constraints: Equation Input Availability Constraints: Wood Skilled labor 40X +X +30X 3 +X 4 <40000 X +X +X 3 +X 4 +3X +X 4 <6000 or X +5X +X 3 +4X 4 <6000 Sign Constraints: X >0 X >0 X 3 >0 X 4 >0
IE3 9 Problem 8 P.98 Purchase R Produce X Product Reprocess X -Y Produce X Sold Y Produce X 4 Product 4 Sold Y Sold Y 4 Ra Material Product Reprocess X -Y Produce X 5 Product 5 Sold Y 5 Destroy X 5 -Y 5 Produce X 3 Product 3 Notation: X i Amount o product i produced Y i Amount o product i sold R Amount o ra material purchased Product I here i 3 4 5 6 Sold Y 3 Produce X 6 Product 6 Sold Y 6 Destroy X 6 -Y 6 Decision Variable: X X X 3 X 4 X 5 X 6 Y Y Y 3 Y 4 Y 5 Y 6 R LP Objective Function: Max Proit 7Y +6Y +4Y 3 +3Y 4 +0Y 5 +35Y 6 4X +4X +X 3 +X 4 +5X 5 +5X 6 4X 5 Y 5 3X 6 Y 6 6R Constraints: Amount Sold Constraints: Product Product Product 3 Product 4 Product 5 Product 6 Balance Constraints: Y <00 Y < X Y <300Y < X Y 3 < X 3 Y 4 < X 4 Y 5 <000 Y 5 < X 5 Y 6 <800 Y 6 < X 6 R X /4 X / X 3 X -Y X 4 X -Y X 5 /0.8 X 6 /0.3 Equation Attribute requirements Ra Material R<3000 Sign Constraints: X >0 X >0 X 3 >0 X 4 >0 X 5 >0 X 6 >0 Y >0 Y >0 Y 3 >0 Y 4 >0 Y 5 >0 Y 6 >0 R>0