ゲーム理論NEXT 線形計画問題第10回 -シンプレックス法5 人工変数続き-
0 likes104 views
The document describes the optimization of a linear function subject to linear constraints. It involves defining the objective function and constraints, substituting variables to eliminate parameters, and determining the optimal solution satisfies all constraints. The optimal solution is found to be x1=0, x2=5, x3=0, x4=0, x5=12.
ゲーム理論NEXT 線形計画問題第10回 -シンプレックス法5 人工変数続き-
- 1. 10
- 2. 1.
- 3. 1. (a)
- 5. m + n n n
- 6. m + n n n max 10x1 + 20x2 s . t . 2x1 + 8x2 ≥ 52 ⋯(1) x1 + x2 ≤ 5 ⋯(2) x1 ≥ 0, x2 ≥ 0 x1 x2 9 6 26 6.5 5 5
- 7. m + n n n max 10x1 + 20x2 s . t . 2x1 + 8x2 = 52 + x3 ⋯(1) x1 + x2 + x4 = 5 ⋯(2) x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0
- 8. m + n n n max 10x1 + 20x2 − Mx5 s . t . 2x1 + 8x2 + x3 − x5 = 52 ⋯(1) x1 + x2 + x4 = 5 ⋯(2) x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0,x5 ≥ 0 x3, x4 x1, x2 x3 = 2x1 + 8x2 − 52 x4 = 5 − x1 − x2 f = 10x1 + 20x2 x1, x2 (x1, x2, x3, x4) = (0, 0, − 52, 5) x3 x3 = 2x1 + 8x2 − 52 + x5, x5 ≥ 0 x5 = 52 − 2x1 − 8x2 + x3
- 9. m + n n n max 10x1 + 20x2 − Mx5 s . t . 2x1 + 8x2 + x3 − x5 = 52 ⋯(1) x1 + x2 + x4 = 5 ⋯(2) x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0,x5 ≥ 0 x4, x5 x1, x2, x3 x4 = 5 − x1 − x2 x5 = 52 − 2x1 − 8x2 + x3 f = 10x1 + 20x2 − M(52 − 2x1 − 8x2 + x3) = (10 + 2M)x1 + (20 + 8M)x2 − 52M − Mx3 x1, x2, x3 (x1, x2, x3, x4, x5) = (0, 0, 0, 5, 52) x1, x2
- 10. m + n n n max 10x1 + 20x2 − Mx5 s . t . 2x1 + 8x2 + x3 − x5 = 52 ⋯(1) x1 + x2 + x4 = 5 ⋯(2) x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0,x5 ≥ 0 x4, x5 x1, x2, x3 x4 = 5 − x1 − x2 x5 = 52 − 2x1 − 8x2 + x3 f = (10 + 2M)x1 + (20 + 8M)x2 − 52M − Mx3 x1, x2, x3 (x1, x2, x3, x4, x5) = (0, 0, 0, 5, 52) x1, x2 x2 x4
- 11. m + n n n max 10x1 + 20x2 − Mx5 s . t . 2x1 + 8x2 + x3 − x5 = 52 ⋯(1) x1 + x2 + x4 = 5 ⋯(2) x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0,x5 ≥ 0 x2, x5 x1, x3, x4 x2 = 5 − x1 − x4 x5 = 12 + 6x1 + 8x4 + x3 f = 100 − (10 + 6M)x1 − (20 + 8M)x4 − Mx3 − 12M x1, x3, x4 (x1, x2, x3, x4, x5) = (0, 5, 0, 0, 12) x1, x3, x4
- 12. m + n n n max 10x1 + 20x2 − Mx5 s . t . 2x1 + 8x2 + x3 − x5 = 52 ⋯(1) x1 + x2 + x4 = 5 ⋯(2) x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0,x5 ≥ 0 x2, x5 x1, x3, x4 x2 = 5 − x1 − x4 x5 = 12 + 6x1 + 8x4 + x3 f = 100 − (10 + 6M)x1 − (20 + 8M)x4 − Mx3 − 12M x1, x3, x4 (x1, x2, x3, x4, x5) = (0, 5, 0, 0, 12) x1, x3, x4 x5 = 12 > 0
- 13. m + n n n max 10x1 + 20x2 − Mx5 s . t . 2x1 + 8x2 + x3 − x5 = 52 ⋯(1) x1 + x2 + x4 = 5 ⋯(2) x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0,x5 ≥ 0 x2, x5 x1, x3, x4 x2 = 5 − x1 − x4 x5 = 12 + 6x1 + 8x4 + x3 f = 100 − (10 + 6M)x1 − (20 + 8M)x4 − Mx3 − 12M x1, x3, x4 (x1, x2, x3, x4, x5) = (0, 5, 0, 0, 12) x1, x3, x4 x5 = 12 > 0
- 15. See you next time 10