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)nitro10x1+2x2 > 20 2)phos 6x1 + 6x2 > 36 3)potas x2 > 2

3. 5(cont’d) Nitr X1X2 010 20

4. 5(cont’d) Phos X1X2 0 6 60

5. 5(cont’d) Potas X1X2 0 2 Infinity0

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

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 Corner ptx1X2 A010 B15 C42

10. MIN 3x1+5x2 X1x23x1+5X2 0103(0)+5(10)=50 153(1)+5(5)=28 423(4)+5(2)=22= min cost

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 Objmax profit = 400x1+100x2 Constr (1)labor8x1+10x2 < 80 (2) wood2x1 + 6x2 < 36 (3) demand x1< 6

13. x2 0,8 10,0 0,6 18,06,0 feasible A B C D infeasible

14. Corner points X1X2 06A 4.34.6B 63.2C 60D

15. MAX 400x1+100x2 X1X2400x1+100x2 06600 4.34.62171 63.22720=max 602400

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 Wood2*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 X1X2400x1+500x2 063000 4.34.64000=max 63.24000=max 602400

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

21. Problem 24 X1=permanent X2=temp Objmin cost =64x1+42x2 Constr 1) claims16x1+12x2 > 450 2) computer station x1 + x2 < 40 3) defective.5x1 +1.4x2 < 25

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

23. Corner points x1x2 28.10A 20.110.7B 34.45.6C 400D

24. Min cost x1x264x1+42x2 28.101800 20.110.71736=min 34.45.62438 4002560

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

31. 25b: insensitive Problem 24: corner pt B optimal Problem 25b: corner pt B optimal same mix as 24

32. 26: Remove constraint from 24 Remove constraint 3 (defective)

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

34. Problem 26 x1x2Cost= 64x1+42x2 2801792 4002560 0371554=min 0401680

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