ゲーム理論NEXT 線形計画問題第9回 -シンプレックス法4 人工変数-
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The document describes the formulation of an optimization problem to maximize the objective function f(x1, x2) = 10x1 + 20x2 subject to several constraints. It introduces additional variables x3, x4, x5, x6 to transform the constraints into equality constraints. The problem is then restated using only variables x1, x2, x5 by eliminating x3, x4, x6. Finally, a penalty term -Mx6 is added to the objective function to encourage x6 = 0 and further simplify the problem.
ゲーム理論NEXT 線形計画問題第9回 -シンプレックス法4 人工変数-
- 1. 9
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- 3. 1. (a)
- 5. m + n n n
- 6. m + n n n
- 7. m + n n n max 10x1 + 20x2 s . t . 3x1 + 2x2 ≤ 18 ⋯(1) 2x1 + 8x2 ≤ 52 ⋯(2) x1 + x2 ≥ 5 ⋯(3) x1 ≥ 0, x2 ≥ 0 x1 x2 9 6 26 6.5 (2, 6) 5 5
- 8. m + n n n max 10x1 + 20x2 s . t . 3x1 + 2x2 + x3 = 18 ⋯(1) 2x1 + 8x2 + x4 = 52 ⋯(2) x1 + x2 = 5 + x5 ⋯(3) x1 ≥ 0, x2 ≥ 0 x3 ≥ 0, x4 ≥ 0, x5 ≥ 0 x1 x2 9 6 26 6.5 (2, 6) 5 5
- 9. m + n n n max 10x1 + 20x2 s . t . 3x1 + 2x2 + x3 = 18 ⋯(1) 2x1 + 8x2 + x4 = 52 ⋯(2) x1 + x2 = 5 + x5 ⋯(3) x1 ≥ 0, x2 ≥ 0 x3 ≥ 0, x4 ≥ 0, x5 ≥ 0 x3, x4, x5 x1, x2 x3 = 18 − 3x1 − 2x2 x4 = 52 − 2x1 − 8x2 x5 = x1 + x2 − 5 f = 10x1 + 20x2 x1, x2 (x1, x2, x3, x4, x5) = (0, 0, 18, 52, − 5) x5
- 10. m + n n n max 10x1 + 20x2 s . t . 3x1 + 2x2 + x3 = 18 ⋯(1) 2x1 + 8x2 + x4 = 52 ⋯(2) x1 + x2 = 5 + x5 ⋯(3) x1 ≥ 0, x2 ≥ 0 x3 ≥ 0, x4 ≥ 0, x5 ≥ 0 x3, x4, x5 x1, x2 x3 = 18 − 3x1 − 2x2 x4 = 52 − 2x1 − 8x2 x5 = x1 + x2 − 5 f = 10x1 + 20x2 x1, x2 (x1, x2, x3, x4, x5) = (0, 0, 18, 52, − 5) x5 x5 = x1 + x2 − 5 + x6, x6 ≥ 0 x6 = 5 − x1 − x2 + x5
- 11. m + n n n max 10x1 + 20x2 s . t . 3x1 + 2x2 + x3 = 18 ⋯(1) 2x1 + 8x2 + x4 = 52 ⋯(2) x1 + x2 = 5 + x5 ⋯(3) x1 ≥ 0, x2 ≥ 0 x3 ≥ 0, x4 ≥ 0, x5 ≥ 0 x3, x4, x5 x1, x2 x3 = 18 − 3x1 − 2x2 x4 = 52 − 2x1 − 8x2 x5 = x1 + x2 − 5 f = 10x1 + 20x2 x1, x2 (x1, x2, x3, x4, x5) = (0, 0, 18, 52, − 5) x5 x5 = x1 + x2 − 5 + x6, x6 ≥ 0 x6 = 5 − x1 − x2 + x5 x6
- 12. m + n n n max 10x1 + 20x2 s . t . 3x1 + 2x2 + x3 = 18 ⋯(1) 2x1 + 8x2 + x4 = 52 ⋯(2) x1 + x2 = 5 + x5 − x6 ⋯(3′) x1 ≥ 0, x2 ≥ 0 x3 ≥ 0, x4 ≥ 0, x5 ≥ 0, x6 ≥ 0 x3, x4, x5 x1, x2 x3 = 18 − 3x1 − 2x2 x4 = 52 − 2x1 − 8x2 x5 = x1 + x2 − 5 f = 10x1 + 20x2 x1, x2 (x1, x2, x3, x4, x5) = (0, 0, 18, 52, − 5) x5 x5 = x1 + x2 − 5 + x6, x6 ≥ 0 x6 = 5 − x1 − x2 + x5 x6
- 13. m + n n n max 10x1 + 20x2 s . t . 3x1 + 2x2 + x3 = 18 ⋯(1) 2x1 + 8x2 + x4 = 52 ⋯(2) x1 + x2 = 5 + x5 − x6 ⋯(3′) x1 ≥ 0, x2 ≥ 0 x3 ≥ 0, x4 ≥ 0, x5 ≥ 0, x6 ≥ 0 x3, x4, x5 x1, x2 x3 = 18 − 3x1 − 2x2 x4 = 52 − 2x1 − 8x2 x5 = x1 + x2 − 5 f = 10x1 + 20x2 x1, x2 (x1, x2, x3, x4, x5) = (0, 0, 18, 52, − 5) x5 x5 = x1 + x2 − 5 + x6, x6 ≥ 0 x6 = 5 − x1 − x2 + x5 x6 x6 = 0 x6
- 14. m + n n n max 10x1 + 20x2 − Mx6 s . t . 3x1 + 2x2 + x3 = 18 ⋯(1) 2x1 + 8x2 + x4 = 52 ⋯(2) x1 + x2 = 5 + x5 − x6 ⋯(3′) x1 ≥ 0, x2 ≥ 0 x3 ≥ 0, x4 ≥ 0, x5 ≥ 0, x6 ≥ 0 x3, x4, x5 x1, x2 x3 = 18 − 3x1 − 2x2 x4 = 52 − 2x1 − 8x2 x5 = x1 + x2 − 5 f = 10x1 + 20x2 x1, x2 (x1, x2, x3, x4, x5) = (0, 0, 18, 52, − 5) x5 x5 = x1 + x2 − 5 + x6, x6 ≥ 0 x6 = 5 − x1 − x2 + x5 x6 x6 = 0 x6 x6 = 0 −Mx6 M > 0
- 15. m + n n n max 10x1 + 20x2 − Mx6 s . t . 3x1 + 2x2 + x3 = 18 ⋯(1) 2x1 + 8x2 + x4 = 52 ⋯(2) x1 + x2 = 5 + x5 − x6 ⋯(3′) x1 ≥ 0, x2 ≥ 0 x3 ≥ 0, x4 ≥ 0, x5 ≥ 0, x6 ≥ 0 x3, x4, x5 x1, x2 x3 = 18 − 3x1 − 2x2 x4 = 52 − 2x1 − 8x2 x5 = x1 + x2 − 5 f = 10x1 + 20x2 x1, x2 (x1, x2, x3, x4, x5) = (0, 0, 18, 52, − 5) x5 x5 = x1 + x2 − 5 + x6, x6 ≥ 0 x6 = 5 − x1 − x2 + x5 x6 x6 = 0 x6 x6 = 0 −Mx6 M > 0
- 16. m + n n n max 10x1 + 20x2 − Mx6 s . t . 3x1 + 2x2 + x3 = 18 ⋯(1) 2x1 + 8x2 + x4 = 52 ⋯(2) x1 + x2 = 5 + x5 − x6 ⋯(3′) x1 ≥ 0, x2 ≥ 0 x3 ≥ 0, x4 ≥ 0, x5 ≥ 0, x6 ≥ 0 x3, x4, x5 x1, x2 x3 = 18 − 3x1 − 2x2 x4 = 52 − 2x1 − 8x2 x5 = x1 + x2 − 5 f = 10x1 + 20x2 x1, x2 (x1, x2, x3, x4, x5) = (0, 0, 18, 52, − 5) x5 x5 = x1 + x2 − 5 + x6, x6 ≥ 0 x6 = 5 − x1 − x2 + x5 x6 x6 = 0 x6 x6 = 0 −Mx6 M > 0
- 17. m + n n n max 10x1 + 20x2 − Mx6 s . t . 3x1 + 2x2 + x3 = 18 ⋯(1) 2x1 + 8x2 + x4 = 52 ⋯(2) x1 + x2 = 5 + x5 − x6 ⋯(3′) x1 ≥ 0, x2 ≥ 0 x3 ≥ 0, x4 ≥ 0, x5 ≥ 0, x6 ≥ 0 x3, x4, x6 x1, x2, x5 x3 = 18 − 3x1 − 2x2 x4 = 52 − 2x1 − 8x2 x6 = 5 − x1 − x2 + x5 f′ = 10x1 + 20x2 − Mx6 = 10x1 + 20x2 − M(5 − x1 − x2 + x5) = (10 + M)x1 + (20 + M)x2 − 5M − Mx5 x1, x2, x5 (x1, x2, x3, x4, x5 x6) = (0, 0, 18, 52, 0, 5)
- 18. m + n n n max 10x1 + 20x2 − Mx6 s . t . 3x1 + 2x2 + x3 = 18 ⋯(1) 2x1 + 8x2 + x4 = 52 ⋯(2) x1 + x2 = 5 + x5 − x6 ⋯(3′) x1 ≥ 0, x2 ≥ 0 x3 ≥ 0, x4 ≥ 0, x5 ≥ 0, x6 ≥ 0 x3, x4, x6 x1, x2, x5 x3 = 18 − 3x1 − 2x2 x4 = 52 − 2x1 − 8x2 x6 = 5 − x1 − x2 + x5 f′ = 10x1 + 20x2 − Mx6 = 10x1 + 20x2 − M(5 − x1 − x2 + x5) = (10 + M)x1 + (20 + M)x2 − 5M − Mx5 x1, x2, x5 (x1, x2, x3, x4, x5 x6) = (0, 0, 18, 52, 0, 5) x1, x2
- 19. m + n n n max 10x1 + 20x2 − Mx6 s . t . 3x1 + 2x2 + x3 = 18 ⋯(1) 2x1 + 8x2 + x4 = 52 ⋯(2) x1 + x2 = 5 + x5 − x6 ⋯(3′) x1 ≥ 0, x2 ≥ 0 x3 ≥ 0, x4 ≥ 0, x5 ≥ 0, x6 ≥ 0 x3, x4, x6 x1, x2, x5 x3 = 18 − 3x1 − 2x2 x4 = 52 − 2x1 − 8x2 x6 = 5 − x1 − x2 + x5 f′ = 10x1 + 20x2 − Mx6 = 10x1 + 20x2 − M(5 − x1 − x2 + x5) = (10 + M)x1 + (20 + M)x2 − 5M − Mx5 x1, x2, x5 (x1, x2, x3, x4, x5 x6) = (0, 0, 18, 52, 0, 5) x1, x2 x6 x2
- 20. m + n n n max 10x1 + 20x2 − Mx6 s . t . 3x1 + 2x2 + x3 = 18 ⋯(1) 2x1 + 8x2 + x4 = 52 ⋯(2) x1 + x2 = 5 + x5 − x6 ⋯(3′) x1 ≥ 0, x2 ≥ 0 x3 ≥ 0, x4 ≥ 0, x5 ≥ 0, x6 ≥ 0 x2, x3, x4 x1, x5, x6 x2 = 5 − x1 + x5 − x6 x3 = 8 − x1 − 2x5 + 2x6 x4 = 12 + 6x1 − 8x5 + 8x6 f′ = 100 − 10x1 + 20x5 − (20 + M)x6 x1, x5, x6 (x1, x2, x3, x4, x5 x6) = (0, 5, 8, 12, 0, 0) x5 x4 x5
- 21. m + n n n max 10x1 + 20x2 − Mx6 s . t . 3x1 + 2x2 + x3 = 18 ⋯(1) 2x1 + 8x2 + x4 = 52 ⋯(2) x1 + x2 = 5 + x5 − x6 ⋯(3′) x1 ≥ 0, x2 ≥ 0 x3 ≥ 0, x4 ≥ 0, x5 ≥ 0, x6 ≥ 0 x2, x3, x5 x1, x4, x6 x2 = 13 2 − 1 4 x1 − 1 8 x4 x3 = 5 − 5 2 x1 + 1 4 x4 x5 = 3 2 + 3 4 x1 − 1 8 x4 + x6 f′ = 130 + 5x1 − 5 2 x4 − Mx6 x1, x4, x6 (x1, x2, x3, x4, x5 x6) = (0, 13 2 , 5, 0, 3 2 , 0) x1 x3 x1
- 22. m + n n n max 10x1 + 20x2 − Mx6 s . t . 3x1 + 2x2 + x3 = 18 ⋯(1) 2x1 + 8x2 + x4 = 52 ⋯(2) x1 + x2 = 5 + x5 − x6 ⋯(3′) x1 ≥ 0, x2 ≥ 0 x3 ≥ 0, x4 ≥ 0, x5 ≥ 0, x6 ≥ 0 x1, x2, x5 x3, x4, x6 x1 = 2 − 2 5 x3 + 1 10 x4 x2 = 6 + 1 10 x3 − 1 20 x4 x5 = 3 − 3 10 x3 − 1 20 x4 + x6 f′ = 130 − 2x3 − Mx6 x3, x4, x6 (x1, x2, x3, x4, x5 x6) = (2, 6, 0, 0, 3, 0)
- 23. m + n n n max 10x1 + 20x2 − Mx6 s . t . 3x1 + 2x2 + x3 = 18 ⋯(1) 2x1 + 8x2 + x4 = 52 ⋯(2) x1 + x2 = 5 + x5 − x6 ⋯(3′) x1 ≥ 0, x2 ≥ 0 x3 ≥ 0, x4 ≥ 0, x5 ≥ 0, x6 ≥ 0 x1, x2, x5 x3, x4, x6 x1 = 2 − 2 5 x3 + 1 10 x4 x2 = 6 + 1 10 x3 − 1 20 x4 x5 = 3 − 3 10 x3 − 1 20 x4 + x6 f′ = 130 − 2x3 − Mx6 x3, x4, x6 (x1, x2, x3, x4, x5 x6) = (2, 6, 0, 0, 3, 0) x6
- 25. See you next time 9