1. Linear Programming (LP) Decision Variables Objective (MIN or MAX) Constraints Graphical Solution

2. History of LP World War II shortages Limited resources Research at RAND in Santa Monica Examples: limited number of machines Limited number of skilled workers Budget limits Time restrictions (deadlines) Raw materials

3. LP Requirements Single objective: MAX or MIN Objective must be linear function Linear constraints

4. Linear Programming I. Profit maximization example II. Cost minimization example

5. I. Profit MAX Example Source: Render and Stair, Quantitative Analysis for Management, Ch 2 Furniture factory Decision variables: X1 = number of tables to make X2 = number of chairs to make

6. Objective: MAX profit Each table: $ 7 profit Each chair: $ 5 profit Total profit = 7X1 + 5X2

7. Carpenter Constraint Labor Constraint 240 hours available per week Each table requires 4 hours from carpenter Each chair requires 3 hours from carpenter 4X1 + 3X2 < 240

8. Painter Constraint 100 hours available per week Each table requires 2 hours from painter Each chair requires 1 hour from painter 2X1 + X2 < 100

9. Non-negativity constraints X1 > 0 X2 > 0 Can’t have negative production

10. Graphical Solution Non-negativity constraints imply positive (northeast) quadrant

11. X1 X2 0,0

12. Plot carpenter constraint Temporarily convert to equation 4X1 + 3X2 = 240 Intercept on X1 axis: X2 =0 4X1 + 3(0) = 240 4X1 = 240 X1= 240/4 = 60 Coordinate (60,0)

13. X1 X2 0,0 60,0

14. Plot carpenter constraint Temporarily convert to equation 4X1 + 3X2 = 240 Intercept on X2 axis: X1 =0 4(0) + 3X2 = 240 3X2 = 240 X2= 240/3 = 80 Coordinate (0,80)

15. X1=tables X2=chairs 0,0 (0,80). (60,0).

16. X1=tables X2=chairs 0,0 (0,80). (60,0).

17. Convert back to inequality 4X1 + 3X2 < 240

18. X1=tables X2=chairs 0,0 (0,80). (60,0).

19. Plot painter constraint Equation: 2X1 + X2 = 100

20. 2X1 + X2 = 100 X1X2 0100 100/2 = 500

21. X1=tables X2=chairs 0,0 (0,80). (60,0). 0,100 50,0

22. Feasible Region Decision: how many tables and chairs to make Feasible allocation: satisfies all constraints

23. MAXIMUM PROFIT Must be on boundary If not on boundary, could increase profit by making more tables or chairs Must be feasible “Corner Point”

24. X1=tables X2=chairs 0,0 (0,80). (60,0). 0,100 50,0 NOT FEASIBLE CORNER POINTS 3

25. 3 RD CORNER POINT Intersection of 2 constraints Temporarily convert to equations (1) 4X1 + 3X2 = 240 (2) 2X1 + X2 = 100 Solve 2 equations in 2 unknowns (2)*3implies 6X1 + 3X2 = 300 Subtract (1) - 4X1 - 3X2 = -240 2X1 = 60 X1 = 30

26. Substitute into equation (1) 4X1 + 3X2 = 240 4(30) + 3X2 = 240 120 + 3X2 = 240 3X2 = 240 – 120 = 120 X2 = 120/3 = 40 3 rd corner point: (30,40)

27. X1=tables X2=chairs 0,0 (0,80). (60,0). 0,100 50,0 NOT FEASIBLE (30,40)

28. MAXIMUM PROFIT X1=tablesX2=chairsInterpretProfit= 7X1+5X2 500Make tables only 7(50)+0= $ 350 080Make chairs only 7(0)+5(80) = $ 400 3040Mix of tables, chairs 7(30)+5(40) = $ 410 =MAX

29. Exam Format Make 30 tables and 40 chairs for $410 profit

30. II. Cost minimization example Diet problem Decision variables: number of pounds of brand #1 and brand #2 to buy to prepare processed food Objective Function: MINIMIZE cost Each pound of brand #1 costs 2 cents, pound of brand #2 costs 3 cents Objective: MIN 2X1 + 3X2

31. CONSTRAINTS Each pound of brand #1 has 5 ounces of ingredient A, 4 ounces of ingredient B, and 0.5 ounces of ingr C Each pound of brand #2 has 10 ounces of ingr A and 3 ounces of ingr B We need at least 90 ounces of ingr A, 48 ounces of ingr B, and 1.5 ounces of ingr C

32. CONSTRAINTS (A) 5X1 + 10X2 > 90 (B) 4X1 + 3X2 > 48 (C) 0.5X1 > 1.5 (D) X1 > 0 (E) X2 > 0

33. X1 X2

34. (A) 5X1+10X2>90 X1X2INTERCEPT 090/10=9(0,9) 90/5=180(18,0)

35. X1 X2 0,9 18,0 A

36. (B) 4X1+3X2>48 X1X2 048/3=16 48/4=120

37. X1 X2 0,9 18,0 A 0,16 12,0 B

38. (C) 0.5X1 > 1.5 X1X2 01.5/0 undefined, so no intercept on vertical axis 1.5/.5= 30

39. (C) Must be vertical line.5X1 = 1.5 X1= 3

40. X1 X2 0,9 18,0 A 0,16 12,0 B C

41. X1 X2 0,9 18,0 A 0,16 12,0 B C FEASIBLE REGION UNBOUNDED

42. CORNER POINTS Only 1 intercept feasible: (18,0) Solve 2 equations in 2 unknowns: B and C A and B

43. B and C B: 4X1 + 3X2 = 48 C: X1 = 3 Substitute X1=3 into B B: 4(3) + 3X2 = 48 12 + 3X2 = 48 3X2 = 36 X2 = 12

44. X1 X2 18,0 A B C FEASIBLE REGION UNBOUNDED 3,12

45. A and B A: 5X1 + 10X2 =90 B: 4X1 + 3X2 = 48 (A)(4): 20X1+ 40X2 = 360 (B)(5): 20X1 + 15X2 = 240 Subtract: 25X2 = 120 X2 = 4.8 Substitute5X1 + 10(4.8) = 90 X1 = 8.4

46. X1 X2 18,0 A B C FEASIBLE REGION UNBOUNDED 3,12 8.4,4.8

47. MINIMIZE COST X1= BRAND #1 X2= BRAND #2 COST= 2X1+3X2 180BRAND#1 ONLY 2(18)+3(0) = 36 CENTS 3122(3)+3(12) =42 CENTS 8.44.82(8.4)+ 3(4.8)=31= MIN

48. EXAM FORMAT BUY 8.4 POUNDS OF BRAND #1 AND 4.8 POUNDS OF BRAND #2 AT COST OF 31 CENTS

49. COMPUTER OUTPUT If computer output says “no feasible solution”, no feasible region Reason #1: unrealistic constraints Reason #2: computer input error

50. No feasible region

51. MULTIPLE OPTIMA If 2 corner points have same objective function value, ok to pick midway point

52. X2 0,0 $ 100.. $ 80 NOT FEASIBLE $ 100 Any point between 2 $100 points is MAX X1