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LP Examples: another Max and a min

Dr. Ron Lembke. LP Examples: another Max and a min. Example 2. mp3 - 4 min electronics - 2 min assembly DVD - 3 min electronics - 1 min assembly Min available : 240 (elect) 100 ( assy ) Profit / unit: mp3 $7, DVD $5 X 1 = number of mp3 players to make

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LP Examples: another Max and a min

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  1. Dr. Ron Lembke LP Examples: another Max and a min

  2. Example 2 • mp3 - 4 min electronics - 2 min assembly • DVD - 3 min electronics - 1 min assembly • Min available: 240 (elect) 100 (assy) • Profit / unit: mp3 $7, DVD $5 X1 = number of mp3 players to make X2 = number of DVD players to make

  3. Standard Form Max 7x1 + 5x2 s.t. 4x1 + 3x2 <= 240 2x1 + 1x2 <= 100 x1 >= 0 x2 >= 0 electronics assembly

  4. 100 80 60 40 20 0 0 20 40 60 80 Graphical Solution X2 mp3 X1 DVD players

  5. 100 80 60 40 20 0 0 20 40 60 80 Graphical Solution X1 = 0, X2 = 80 4x1+3x2 <= 240 x1 =0, x2 =80 x2 =0, x1 =60 X2 mp3 X1 = 60, X2 = 0 Electronics Constraint X1 DVD players

  6. 100 80 60 40 20 0 0 20 40 60 80 Graphical Solution X1 = 0, X2 = 100 2x1+1x2 <= 100 x1 =0, x2 =100 x2 =0, x1 =50 Assembly Constraint X2 mp3 X1 = 50, X2 = 0 X1 DVD players

  7. 100 80 60 40 20 0 0 20 40 60 80 Graphical Solution Assembly Constraint Feasible Region – Satisfies all constraints X2 mp3 Electronics Constraint X1 DVD players

  8. 100 80 60 40 20 0 0 20 40 60 80 Isoprofit Lnes Isoprofit Line: X2 mp3 $7X1 + $5X2 = $210 (0, 42) (30,0) X1 DVD players

  9. 100 80 60 40 20 0 0 20 40 60 80 Isoprofit Lines X2 $280 mp3 $210 X1 DVD players

  10. 100 80 60 40 20 0 0 20 40 60 80 Isoprofit Lines $350 X2 $280 mp3 $210 X1 DVD players

  11. 100 80 60 40 20 0 0 20 40 60 80 Isoprofit Lines (0, 82) $7X1 + $5X2 = $410 X2 mp3 (58.6, 0) X1 DVD players

  12. Mathematical Solution • Obviously, graphical solution is slow • We can prove that an optimal solution always exists at the intersection of constraints. • Why not just go directly to the places where the constraints intersect?

  13. 100 80 60 40 20 0 0 20 40 60 80 Constraint Intersections X1 = 0 and 4X1 + 3X2 <= 240 So X2 = 80 (0, 80) X2 mp3 4X1 + 3X2 <= 240 (0, 0) X1 DVD players

  14. 100 80 60 40 20 0 0 20 40 60 80 Constraint Intersections (0, 80) X2 mp3 X2 = 0 and 2X1 + 1X2 <= 100 So X1 = 50 (0, 0) (50, 0) X1 DVD players

  15. 100 80 60 40 20 0 0 20 40 60 80 Constraint Intersections 4X1+ 3X2= 240 2X1 + 1X2= 100 – multiply by -2 4X1+ 3X2= 240 -4X1 -2X2= -200 add rows together (0, 80) 0X1+ 1X2= 40 X2 = 40 substitute into #2 X2 2X1+ 40 = 100 So X1 = 30 mp3 (0, 0) (50, 0) X1 DVD players

  16. 100 80 60 40 20 0 0 20 40 60 80 Constraint Intersections Find profits of each point. Substitute into $7X1 + $5X2 (0, 80) $400 X2 mp3 (30,40) $410 (50, 0) $350 (0, 0) $0 X1 DVD players

  17. Do we have to do this? • Obviously, this is not much fun: slow and tedious • Yes, you have to know how to do this to solve a two-variable problem. • We won’t solve every problem this way.

  18. Constraint Intersections • Start at (0,0), or some other easy feasible point. • Find a profitable direction to go along an edge • Go until you hit a corner, find profits of point. • If new is better, repeat, otherwise, stop. 100 Good news: Excel can do this for us. Using the Simplex Algorithm 80 60 mp3 X2 40 20 0 0 20 40 60 80 X1 DVD players

  19. Minimization Example Min8x1 + 12x2 s.t.5x1 + 2x2≥20 4x1 + 3x2 ≥ 24 x2 ≥ 2 x1 , x2 ≥ 0

  20. Minimization Example 5x1 + 2x2=20 Min8x1 + 12x2 s.t. 5x1 + 2x2≥20 4x1 + 3x2 ≥ 24 x2 ≥ 2 x1 , x2 ≥ 0 If x1=0, 2x2=20, x2=10 (0,10) If x2=0, 5x1=20, x1=4 (4,0) 4x1 + 3x2=24 If x1=0, 3x2=24, x2=8 (0,8) If x2=0, 4x1=24, x1=6 (6,0) x2= 2 If x1=0, x2=2 No matter what x1 is, x2=2

  21. 10 8 6 4 2 0 0 2 468 Graphical Solution 5x1 + 2x2=20 X2 4x1 + 3x2 =24 x2=2 X1

  22. 10 8 6 4 2 0 0 2 468 15x1+6x2 = 60 8x1 +6x2 = 48 [5x1+ 2x2 =20]*3 [4x1+3x2=24]*2 (0,10) - 7x1 = 12 5x1 + 2x2=20 x1 = 12/7= 1.71 5x1+2x2=20 5*1.71 + 2x2 =20 2x2= 11.45 x2 = 5.725 (1.71,5.73) X2 4x1+3x2 =24 (1.71,5.73) x2 =2 X1

  23. 10 8 6 4 2 0 0 2 468 4x1+3*2 =24 4x1=18 x1=18/4 = 4.5 (4.5,2) 4x1+3x2 =24 x2 =2 (0,10) 5x1 + 2x2=20 X2 4x1+3x2 =24 (1.71,5.73) x2 =2 (4.5,2) X1

  24. 10 8 6 4 2 0 0 2 468 Lowest Cost (0,10) Z=8x1 +12x2 8*0 + 12*10 = 120 5x1 + 2x2=20 Z=8x1 +12x2 8*1.71 + 12*5.73 = 82.44 X2 4x1+3x2 =24 (1.71,5.73) Z=8x1 +12x2 8*4.5+ 12*2 = 60 x2 =2 (4.5,2) X1

  25. IsoCost Lines X2 Z=8x1 +12x2 Try 8*12 = 96 x1=0 12x2=96, x2=8 x2=0 8x1=96, x1=12 10 5x1 + 2x2=20 8 6 4x1 + 3x2 =24 4 x2=2 2 0 0 2 46810 12 X1

  26. Summary • Method for solving a two-variable problem graphically • Find end points of each constraint • Draw constraints • Figure out which intersections are interesting • Use algebra to solve for intersection pts • Find profits (or costs) of intersections • Choose the best one • Iso-profit (or Iso-Cost) lines can help find the most interesting points

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