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Linear Programming (LP)

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

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Linear Programming (LP)

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

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

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

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

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

  20. 2X1 + X2 = 100

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

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

  25. 3RD 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 3rd corner point: (30,40)

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

  28. MAXIMUM PROFIT

  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. X2 X1

  34. (A) 5X1+10X2>90

  35. X2 0,9 A X1 18,0

  36. (B) 4X1+3X2>48

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

  38. (C) 0.5X1 > 1.5

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

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

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

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

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

  47. MINIMIZE COST

  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

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