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Answers to set 3

Answers to set 3. Set 4 due Thurs March 14 Memo C-2 due Tues March 19 Next exam Thurs April 4 Tutor Mon 4-6PM, Thurs 2-4PM, BB 4120. Problem 5. X1=ingred 1 X2=ingred 2 Objective: MIN cost = 3x1+ 5x2 Constraints 1)nitro 10x1+2x2 > 20 2)phos 6x1 + 6x2 > 36

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Answers to set 3

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  1. Answers to set 3 • Set 4 due Thurs March 14 • Memo C-2 due Tues March 19 • Next exam Thurs April 4 • Tutor Mon 4-6PM, Thurs 2-4PM, BB 4120

  2. Problem 5 • X1=ingred 1 • X2=ingred 2 • Objective: MIN cost = 3x1+ 5x2 • Constraints • 1)nitro 10x1+2x2 > 20 • 2)phos 6x1 + 6x2 > 36 • 3)potas x2 > 2

  3. 5(cont’d)

  4. 5(cont’d)

  5. 5(cont’d)

  6. x2 A 0,10 feasible infeas 0,6 B (1) (2) C (3) 0,2 infeasible infeasible 2,0 6,0

  7. Corner pt B: nitro, phos • 10x1+2x2=20 • 6x1+6x2=36 • X1=1 • X2=5

  8. Corner pt C: Phos,potas • 6x1+6x2=36 • X2=2 • X1=4

  9. Feasible corner points

  10. MIN 3x1+5x2

  11. Exam format • Buy 4 pounds of ingredient #1 and 2 pounds of ingred #2 for $22 cost

  12. Problem 6 • X1=chairs • X2=tables • Obj max profit = 400x1+100x2 • Constr • (1)labor 8x1+10x2 < 80 • (2) wood 2x1 + 6x2 < 36 • (3) demand x1 < 6

  13. x2 0,8 infeasible 0,6 A B feasible infeasible C 10,0 infeasible 6,0 18,0 D

  14. Corner points

  15. MAX 400x1+100x2

  16. Exam Format • Make 6 chairs and 3.2 tables per day for $2720 profit

  17. Problem 7 • Plug optimal solution to 6 into labor constraint: 8*6+10*3.2=80=avail, so no labor slack • Wood 2*6+6*3.2=31.2<36 avail, so wood slack = 36-31.2=4.8

  18. Problem 8 • Re-do 6 with new table profit = $500 • NEW objective function = 400x1+500x2

  19. MAX 400x1+500x2

  20. Exam Format • Sensitive to change since two corner points are now equally optimal

  21. Problem 24 • X1=permanent • X2=temp • Obj min cost =64x1+42x2 • Constr • 1) claims 16x1+12x2 > 450 • 2) computer station x1 + x2 < 40 • 3) defective .5x1 +1.4x2 < 25

  22. x2 0,40 0,37 (1) (2) Infeasible 0,18 B (3) C infeasible A 28,0 50,0 D 40,0

  23. Corner points

  24. Min cost

  25. Exam format • Hire 20.12 permanent operators and 10.67 temporary operators for cost of $1736 • Real world: probably round up to 21 and 11 in anticipation of sick leave, attrition, etc

  26. Problem 25a • Sensitivity on previous problem • change objective function coefficient • a: change x1 coefficient

  27. New cost = 54x1+42x2

  28. New min =1519 • X1>0 only • previous problem: x1>0 and x2>0 • different corner pt • no more temps • solution sensitive to change

  29. 25b: change x2 coefficient • Min cost = 64x1 + 36x2

  30. 25b: insensitive • Problem 24: corner pt B optimal • Problem 25b: corner pt B optimal • same mix as 24

  31. 26: Remove constraint from 24 • Remove constraint 3 (defective)

  32. () x2 3 0,40 0,37 Infeasible (1) (2) Infeasible 0,18 B C infeasible A 28,0 50,0 D 40,0

  33. Problem 26

  34. Problem 26: sensitive • Problem 24: permanent and temp • Problem 26: temp only • New corner point

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