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Branch & bound módszer

Branch & bound módszer. A megoldandó feladat:. P(x) = 8x 1 + 5x 2  MAX x 1 + x 2 <= 6 9x 1 + 5x 2 <= 45 x 1 ,x 2 >= 0 x 1 ,x 2 - egész. Megoldás egészértékű feltételek nélkül:. X 1 = 3.75 X 2 = 2.25 Cfgv = 41.25. Korlátozás és Szétválasztás:. Bound =. 39.

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Branch & bound módszer

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  1. Branch & bound módszer

  2. A megoldandó feladat: P(x) = 8x1 + 5x2 MAX x1 + x2 <= 6 9x1 + 5x2 <= 45 x1,x2 >= 0 x1,x2 - egész

  3. Megoldás egészértékű feltételek nélkül: X1 = 3.75 X2 = 2.25 Cfgv = 41.25

  4. Korlátozás és Szétválasztás: Bound = 39 X1 <= 3 X1 >= 4 X1 = 3 x2 = 3 Cfgv = 39

  5. Korlátozás és Szétválasztás: Bound = 39 X1 <= 3 X1 >= 4 X1 = 3 x2 = 3 Cfgv = 39 X1 = 4 x2 = 1.8 Cfgv = 41 Bound<Cfgv

  6. Korlátozás és Szétválasztás: Bound = 39 X1 <= 3 X1 >= 4 X1 = 3 x2 = 3 Cfgv = 39 X1 = 4 x2 = 1.8 Cfgv = 41 X2 <= 1 X2 >= 2 X1 = 4.44 x2 = 1 Cfgv = 40.55

  7. X1 <= 3 X1 >= 4 X1 = 3 x2 = 3 Cfgv = 39 X1 = 4 x2 = 1.8 Cfgv = 41 X2 <= 1 X2 >= 2 X1 = 4.44 x2 = 1 Cfgv = 40.55 Korlátozás és Szétválasztás: Bound = 39 Bound < Cfgv

  8. Korlátozás és Szétválasztás: Bound = 39 X1 <= 3 X1 >= 4 X1 = 3 x2 = 3 Cfgv = 39 X1 = 4 x2 = 1.8 Cfgv = 41 X2 <= 1 X2 >= 2 X1 = 4.44 x2 = 1 Cfgv = 40.55 X1 <= 4 X1 >= 5 X1 = 4 x2 = 1 Cfgv = 37 Bound > Cfgv

  9. Korlátozás és Szétválasztás: Bound = 39 40 X1 <= 3 X1 >= 4 X1 = 3 x2 = 3 Cfgv = 39 X1 = 4 x2 = 1.8 Cfgv = 41 X2 <= 1 X2 >= 2 X1 = 4.44 x2 = 1 Cfgv = 40.55 X1 <= 4 X1 >= 5 X1 = 4 x2 = 1 Cfgv = 37 X1 = 5 x2 = 0 Cfgv = 40 Bound < Cfgv

  10. Korlátozás és Szétválasztás: Bound = 40 X1 <= 3 X1 >= 4 X1 = 3 x2 = 3 Cfgv = 39 X1 = 4 x2 = 1.8 Cfgv = 41 X2 <= 1 X2 >= 2 X1 = 4.44 x2 = 1 Cfgv = 40.55 S = Ǿ Nincs megoldása X1 <= 4 X1 >= 5 X1 = 4 x2 = 1 Cfgv = 37 X1 = 5 x2 = 0 Cfgv = 40

  11. A megoldás: X1 = 5 x2 = 0 Cfgv = 40

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