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- わんど かいじ
- 5 years ago
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1 5 1! (Linear Programming, LP) LP OR LP ?
2 ( ) () () ( ) () () () () ( )
3 x 1 x 2 x 3 x 4 z z = 25.1x x x x 4 x 1 + x 2 + x 3 + x 4 = x x x x x x x x x x x x min. z = 25.1x x x x 4 s.t. x 1 + x 2 + x 3 + x 4 = x x x x x x x x x j 0, j = 1,..., 4 (1.1) \ x j 0, j = 1,..., 4 " x 1 0, x 2 0, x 3 0, x 4 0 () ( ) (1.1) 1 1 (linear programming problem) (linear programming)
4 8 1 (1.1) x 1 = x 2 = 0.0 x 3 = x 4 = z = (1.1) 2 3 x x 1 = x 2 = x 3 = 0.0 x 4 = z = x x 1 100, x 2 100, x 3 100, x min. z = 25.1x x x x 4 s.t. x 1 + x 2 + x 3 + x 4 = x x x x x x x x x j 100, j = 1,..., 4 (1.2) x 1 = x 2 = x 3 = x 4 = z = ( ) f(x) x X (1.3)
5 f(x) X x (1.3) X x f(x) ( ) x x (standard form) maximize c 1 x 1 + c 2 x c n x n subject to a 11 x 1 + a 12 x a 1n x n = b 1 a 21 x 1 + a 22 x a 2n x n = b 2. a m1 x 1 + a m2 x a mn x n = b m x 1 0, x 2 0,..., x n 0 (1.4) b 1,..., b m (right hand side) a ij (i = 1,..., m; j = 1,..., n) a 11 a 12 a 1n a 21 a 22 a 2n..... a m1 a m2 a mn x 1 0, x 2 0,..., x n 0 (nonnegativity constraint) (x 1,..., x n ) (feasible solution) (optimal solution) (1.4)) \ " maximize subject to n c j x j j=1 n a ij x j = b i, j=1 x j 0, i = 1,..., m j = 1,..., n (1.5)
6 10 1 maximize subject to c x Ax = b x 0 (1.6) x = x 1. x n c = c 1. c n b = b 1. b m A = a 11 a 12 a 1n a 21 a 22 a 2n..... a m1 a m2 a mn 2 maximize c 1 x 1 + c 2 x c n x n subject to a 11 x 1 + a 12 x a 1n x n b 1 a 21 x 1 + a 22 x a 2n x n b 2. a m1 x 1 + a m2 x a mn x n b m x 1 0, x 2 0,..., x n 0 x n+1, x n+2,..., x n+m (1.7) maximize c 1 x 1 + c 2 x c n x n subject to a 11 x 1 + a 12 x a 1n x n + x n+1 = b 1 a 21 x 1 + a 22 x a 2n x n + x n+2 = b 2. a m1 x 1 + a m2 x a mn x n + x n+m = b m x 1 0,..., x n 0, x n+1 0,..., x n+m 0 (1.8) (1.9) x 1 maximize c 1 x 1 + c 2 x c n x n subject to a 11 x 1 + a 12 x a 1n x n = b 1 a 21 x 1 + a 22 x a 2n x n = b 2. a m1 x 1 + a m2 x a mn x n = b m x 2 0,..., x n 0 (1.9) x 1 = x + 1 x 1 ( x + 1 0, x 1 0) 2 c = (c 1,..., c n )
7 (1.9) maximize c 1 x + 1 c 1x 1 + c 2x c n x n subject to a 11 x + 1 a 11x 1 + a 12x a 1n x n = b 1 a 21 x + 1 a 21x 1 + a 22x a 2n x n = b 2. a m1 x + 1 a m1x 1 + a m2x 2 + a mn x n = b m x + 1 0, x 1 0,..., x n 0 (1.10) 1.3 max 3x 1 + 2x 2 4x 3 s.t. x 1 + x 3 2 x 1 + 2x 2 + x 3 5 x 1 0, 4 x 2 5, x max x 1 + x 2 + x 3 s.t. 2x 1 x 2 + 3x 3 + x 4 = 6 x 1 4x 2 + 2x 3 + x 5 = 5 x 1, x 2, x 3, x 4, x Lancaster(1992) (diet problem)
8 g B1 C (kcal) (g) (g) (g) (g) (g) (g) (mg) (mg) (mg) (100g ) ( ) ? B1 C 10g 0.5mg 50mg j (j = 1,..., 7) x j (100g) x 1,..., x 7 B1 1.9x x x x x x x 7 C 0.08x x x x x x x x x x x x x x x x x x x x x 7 3 ( ( ) 1 70(55)g B1 1.1(0.8)mg C 100(100)mg
9 min 136.5x x x x x x x 7 subject to 1.9x x x x x x x x x x x x x x x x x x x x x 7 50 x j 0, j = 1,..., 7 (1.11) (1.11) x 1 = x 2 = 0.0 x 3 = 0.0 x 4 = 0.0 x 5 = 0.0 x 6 = x 7 = Stigler[7] Stigler 77 A B1 B2 C ( 1.5) No one recommends these diets for anyone, let alone everyone. 1.5 Stigler 535 lb. 107 lb. 13 lb. 134 lb. 25 lb. Dantzig (simplex method) Dantzig 500 ([2]) 500 ( 1890L 5 ) (bran) 2 ( 900g) 4 (blackstrap molasses) 2 ( 900g) Dantzig Dantzig 3
10 14 1 Fletcher, Soden, Zinober[5] \ " LP (g) 1( ) (200)
11 ( / ) A B C ( ) A B C [] 1 x 1 2 x 2 1 x 3 2 x 4 x 5 A B C y 1 y 2 y max 45x x x x x 5 s.t. 4.5x x x x x 5 y x x x x x 5 y x x x 5 y 3 0 2y 1 + 4y 2 + 8y y 1 150, y 2 100, y x 1, x 2, x 3, x 4, x 5, y 1, y 2, y 3 0 (1.12) [ ] (1.12) x 1 = 18.8 x 2 = 20 x 4 = 21.2 x 3 = x 5 = 0 y 1 = 150 y 2 = 100 y 3 = 100
12 16 1 [8] 7 100a [] x 1 (a) x 2 (a) x 3 (a) x 4 (a) [ ] max 29.8x x x x 4 s.t. x 1 + x 2 + x 3 + x 4 = x x x x x x x x j 0, j = 1,..., 4 x 1 = 4.14 x 2 = 0 x 3 = 9.62 x 4 =
13 ( ) (%) (%) % 4 40% = 2.30( ) j (1 ) x j (j = 1,..., 7) x x x x x x x x x 1
14 x 1 / x x x x x x x x x x x x x x ( x x x x x x x 7 )/ x x x x x x x maximize 0.232x x x x x x x 7 subject to x x x x x x x x x x x x x x x x x x x x x x 1 400, x 2 400, x 3 400, 92.53x 4 400, 97.09x 5 400, x 6 400, x x j 0 j = 1,..., 7 (1.13) ( 1 ) (x 1, x 2, x 3, x 4, x 5, x 6, x 7) = (0, 1.84, 3.82, 0, 0, 0, 2.78) A B C
15 ( 15 ) [] A x 1 x 2 x 3 B y 1 y 2 y 3 C z 1 z 2 z 3 A B C u A, u B, u C 8 u A = x x x 3 u B = y y y 3 u C = z z z 3 max u A + u B + u C s.t. u A x x 2 1 u B y y 2 1 u C z z 2 1 x 1 + y 1 + z 1 = 20 x 2 + y 2 + z 2 = 20 x 3 + y 3 + z 3 = 5 x j 0, j = 1, 2, 3 y j 0, j = 1, 2, 3 z j 0, j = 1, 2, x 3 = y 3 = z 3 = 0 (1.14) LP C B () (1.14) x 1 + x 2 + x 3 = 15 y 1 + y 2 + y 3 = 15 z 1 + z 2 + z 3 = (numeraire)
16 A B C min{u A, u B, u C } x, y, z min{x, y, z} x, y, z 1.6 K2 A 8% B 4% C 2% 4 1g g 1 A B C 0.03g 0.02g 0.01g 1.12 ( /g) A B C kg 1.7 A B C D P 1 d 1 P 2 d 2 p c 1 p c 2 p m 1 p m 2 P i (i = 1, 2) A B C D a A i a B i a C i a D i A B C D q A q B q C q D b kg 20kg 30kg 2 20% 10% 10% ( ) 1kg 5 6
17 kg kg ( ) kg ( 2 140kg ) K kg : 1kg : 1kg : 1kg : 1kg : 1kg : 1kg : 1kg : kg kg 0.3 kg kg 1500kg 3. 1kg 200 1kg 300 ( ) 4. 1kg
18 = : : = 500 : 600 : 750 = 1 : 1.2 :
19 x t t (t = 1,..., 5) y t t (t = 1,..., 5) u t t (t = 1,..., 5) v t t (t = 1,..., 5) w t t (t = 1,..., 5) 3 2 y y 2 3 x 3 3 y 3 y 3 = x y 2 (1.15) 3 3 x x 3 20 = 0.015x 3( ) 0.15x 3 u 3 + v 3 + w 3, u 3 1, v 3 2, w 3 2 (1.16) u 1 + u 2 + u 3 + u 4 + u v v v v v 5 (1.15)(1.16)(1.17) +1.5w w w w w 5 (1.17) min 5 u t + t= v t + t= w t t=1 s.t. y 1 = x 1 y 2 = x y 1 y 3 = x y 2 y 4 = x y 3 y 5 = x y 4 y x 1 u 1 + v 1 + w 1, 0.015x 2 u 2 + v 2 + w 2, 0.015x 3 u 3 + v 3 + w 3, 0.015x 4 u 4 + v 3 + w 4, 0.015x 5 u 5 + v 3 + w 5, u 1 4, u 2 5, u 3 1, u 4 3, u 5 1 v 1 2, v 2 2, v 3 2, v 4 2, v 5 2 w 1 4, w 2 4, w 3 2, w 4 2, w 5 2 x t 0, y t 0, u t 0, v t 0, w t 0, t = 1,..., 5 (1.18)
20 (u t ) (v t ) (w t ) (x t ) (y t ) (1.18) (u t ) (v t ) (w t ) (x t ) (y t ) (u t ) (v t ) (w t ) (x t ) (y t )
21 x 0 x t t (t = 1,..., 5) y t t (t = 1,..., 5) z t t (t = 1,..., 5) 2 2 x 2 5x 2 y 2 = y 1 + 5x x 2 z y 1 2 y 2 y 1 z 2 = z 1 10 x y 1 x 2 z 1 min x 0 s.t. z 1 = x 0 x 1 z 2 = z 1 10 x y 1 z 3 = z 2 x y 2 z 4 = z 3 40 x y 3 z 5 = z 4 20 x y 4 y 1 = 5x 1 y 2 = y 1 + 5x 2 y 3 = y 2 + 5x 3 y 4 = y 3 + 5x 4 y 5 = y 4 + 5x 5 y x 1 x 0, x 2 z 1, x 3 z 2, x 4 z 3, x 5 z 4 x t 0, y t 0, z t 0, t = 1,..., 5 (1.19) (1.19) x 1 = 13.1 x 2 = 0 x 3 = 6.3 x 4 = 0 x 5 = 0.6 y 1 = 65.3 y 2 = 65.3 y 3 = 97 y 4 = 97 y 5 = 100 z 1 = 0 z 2 = 6.3 z 3 = 16.3 z 4 = 0.6 z 5 = 4.3 x 0 = % (=100/10) (100 20) 0.01 = 10.8
22 % 1.11 [] = 0.8 [ ] [ ] 0.05 [] max x 1 + x 2 s.t. x 1 + 2x 2 3 x 1 0, x 2 0 (1.20) x 1 + 2x 2 3 x 1 + 2x 2 0 (1.20) (x 1, x 2 ) (infeasible) (feasible) (feasible solution) (1.20) max x 1 + x 2 s.t. x 1 + 2x 2 3 x 1 0, x 2 0 (1.21) (1.21) (x 1, x 2 ) = (1, 1) (x 1, x 2 ) = (1, 1) = 2 1 t (1 + t, 1) (1 + t) = 1 t 3
23 (1 + t) + 1 = 2 + t t t 10 (unbounded) ^x d t ^x + td ^x ^x + td (1.21) max x 1 + x 2 s.t. x 1 + 2x 2 3 x 1 0, x 2 0 (1.22) (1.22) (1.21) d = (1, 0) (1.22) (0, 1.5) (infeasible) (feasible) (feasible solution) ( ) ( ) LP (c 1, c 2 ) max c 1 x 1 + c 2 x 2 s. t. x 1 + x 2 6 2x 1 + x 2 10 x 2 3 x 1, x 2 0 (1.23) (1.23) ( x 1 x 2 ) S S x 1 {x
24 S x 2 A (0,3) (3,3) E (3.5,3) S B (4,2) C O D 0 (5,0) (6,0) F x 1 (c 1, c 2 ) \ " ( ) ( (1.24)) 1.21 max c 1 x 1 + c 2 x 2 s. t. x 1 + x 2 6 2x 1 + x 2 10 x 2 3 x 1, x 2 0 (1.24) 1.12 α max x 1 x 2 s. t. 2x 1 3x 2 3 αx 1 x 2 3α 3 x 1, x 2 0 α [] x 1 -x 2 αx 1 x 2 = 3α 3 α (3, 3)
25 S (1) x 2 (C1,C2) 0 x S (2) x 2 0 x 1 (C1,C2)
26 x 2 (C1,C2) 0 x (1.23) (c 1, c 2 ) = (3, 2) (1.25) max 3x 1 + 2x 2 s. t. x 1 + x 2 + x 3 = 6 2x 1 + x 2 + x 4 = 10 (1.25) x 2 + x 5 = 3 x 1, x 2, x 3, x 4, x x 1 + x 2 +x 3 = 6 2x 1 + x 2 +x 4 = 10 x 2 +x 5 = 3 (1.26) (1.26) 5 3 x 2 x 4 (1.26) 1 x 1 = x 2 x x 1 +x 2 +x 3 = 6 x 2 2x 3 +x 4 = 2 x 2 +x 5 = 3 (1.27)
27 (1.27) 2 x 3 = 0.5x x x x x 4 = 5 x 2 2x 3 +x 4 = 2 x 2 +x 5 = 3 (1.28) (1.25) z = 3x 1 + 2x 2 (1.28) 1 x 1 z = 3 (5 0.5x 2 0.5x 4 ) + 2x 2 = x 2 1.5x 4 (1.29) x 2 x 4 (1.28) (1.29) x 2, x 4 z = x 2 1.5x 4 x 1 = 5 0.5x 2 0.5x 4 x 3 = 1 0.5x x 4 x 5 = 3 x 2 (1.30) (1.25) z z = 3x 1 + 2x 2 x 1 + x 2 + x 3 = 6 2x 1 + x 2 + x 4 = 10 (1.31) x 2 + x 5 = 3 (1.30) (1.31) 1.13 (1.31) (x 1, x 2,..., x 5 ) (1.30) (1.30) (x 1, x 2,..., x 5 ) (1.31) (1.30) z, x 1, x 3, x 5 x 2 x 4 x 2 x 4 z, x 1, x 3, x 5 11 x 1, x 3, x 5 (basic variable) x 2, x 4 (nonbasic variable) 11 (dictionary)
28 (1.30) x 2 = x 4 = 0 x 1 = 5, x 3 = 1, x 5 = 3, z = 15 (1.31) (1.31) x 2 = x 4 = 0 x 1 + x 3 = 6 2x 1 + x 4 = 10 x 5 = 3 x 1 = 5, x 3 = 1, x 5 = 3 z = 3x 1 + 2x 2 z = 15 0 (basic solution) (feasible basic solution) (1.30) x 1 = 5, x 2 = 0, x 3 = 1, x 4 = 0, x 5 = 3 ( ) x 1 x 2 (x 1, x 2 ) = (5, 0) 1.18 D x 3 x 4 z = 16 x 3 x 4 x 1 = 4 +x 3 x 4 x 2 = 2 2x 3 +x 4 x 5 = 1 +2x 3 x 4 (1.32) x 1 = 4, x 2 = 2, x 3 = 0, x 4 = 0, x 5 = 1 x 1 -x 2 (x 1, x 2 ) = (4, 2) C 1.14 x 1,..., x [ ] S
29 D C z = x 2 1.5x 4 x 1 = 5 0.5x 2 0.5x 4 x 3 = 1 0.5x x 4 x 5 = 3 x 2 (1.30) z = 16 x 3 x 4 x 1 = 4 +x 3 x 4 x 2 = 2 2x 3 +x 4 x 5 = 1 +2x 3 x 4 (1.32) (1.30) ( ) x 2, x 4 0 (1.30) z = x 2 1.5x 4 x x 2 0 x 4 0 x 2 0 t z = t x 1 = 5 0.5t x 3 = 1 0.5t x 5 = 3 t (1.33) t z = t t t x 1, x 3, x 5 t t = 2 x 3 = 0 (1.30) 3 x 3 = 1 0.5x x 4 (1.32) (1.30) x 2 (1.32) (1.30) x 3 (1.32) (1.32) (1.30) x 4 ( ) 0? 1.18?
30 34 1 (1.30) (1.32) (1.32) z = 16 x 3 x 4 x 3 x 4 0 (1.32) (1.32) (^x 1,..., ^x 5 ) (1.25) (1.32) z = 3^x 1 + 2^x 2 = 16 ^x 3 ^x 4 12 ^x 3 0, ^x 4 0 z = 3^x 1 + 2^x (simplex method)
31 max 3x 1 + 2x 2 s. t. x 1 + x 2 6 2x 1 + x 2 10 x 2 3 x 1, x 2 0 (1.34) (1.23) z = 16 x 3 x 4 x 1 = 4 +x 3 x 4 x 2 = 2 2x 3 +x 4 x 5 = 1 +2x 3 x 4 (1.32) (1.32) z = 16 x 3 x 4 z = 3x 1 + 2x 2 z = 3x 1 + 2x 2 = 3(4 + x 3 x 4 ) + 2(2 2x 3 + x 4 ) 12 (5, 0, 1, 0, 3) (0, 0, 6, 10, 3) (3, 3, 0, 1, 0)
32 z = 3x x 2 = 3(4 + x 3 x 4 ) + 2.5(2 2x 3 + x 4 ) = 17 2x 3 0.5x ? z = 20 5x 3 +x 4 x 1 = 4 +x 3 x 4 x 2 = 2 2x 3 +x 4 x 5 = 1 +2x 3 x 4 (1.35) x 4 x 4 0 x 1, x 2, x 4 z = 21 3x 3 x 5 x 1 = 3 x 3 +x 5 x 2 = 3 x 5 x 4 = 1 +2x 3 x 5 (1.36) δ 2 + δ 2 + δ z = 3x 1 + (2 + δ)x 2 = 3(4 + x 3 x 4 ) + (2 + δ)(2 2x 3 + x 4 ) = δ + ( 1 2δ)x 3 + ( 1 + δ)x 4 (34 ) 0 δ 1 2δ δ δ
33 y 1 y 2 y 3 6y y 2 + 3y y 1 + y 2 + y 3 2 y 1 + y 2 + y 3 2 y 1 + 2y 2 3 min 6y y 2 + 3y 3 s. t. y 1 + 2y 2 3 y 1 + y 2 + y 3 2 y 1, y 2, y 3 0 (1.37) y 1 = 1 y 2 = 1 y 3 = 0 ( ) 16 (1.37) (shadow price) (1.37) (1.34) (1.37) (1.38) maximize c 1 x 1 + c 2 x c n x n subject to a 11 x 1 + a 12 x a 1n x n b 1 a 21 x 1 + a 22 x a 2n x n b 2. a m1 x 1 + a m2 x a mn x n b m x j 0, j = 1,..., n (1.38)
34 38 1 (1.39) (1.38) (dual problem) minimize b 1 y 1 + b 2 y b m y m subject to a 11 y 1 + a 21 y a m1 y m c 1 a 12 y 1 + a 22 y a m2 y m c 2. a 1n y 1 + a 2n y a mn y m c n y i 0, i = 1,..., m (1.39) y i (i = 1,..., m) (dual variable) (primal problem) i y i j x j (1.5.4 ) () max s.t. c x Ax b x 0 ( ) min s.t. b y A y c y 0 a 11 a 12 a 1n c 1 b 1 a 21 a 22 a 2n A =..... c = c 2. b = b 2. a m1 a m2 a mn c n b m (1.34) (1.37) 2. max 3x 1 x 2 + 4x 3 + 5x 4 s.t. x 1 + x 2 3x 3 4x 4 2 2x 1 4x 2 + x 3 2x 4 8 x 1, x 2, x 3, x f(x) f(x) 1 () (1.38) (1.38) (1.39)
35 (1.40) (1.38) maximize subject to n j=1 c jx j n j=1 a ijx j = b i, x j 0, i = 1,..., m j = 1,..., n (1.40) (1.40) minimize subject to m i=1 b iy i m i=1 a ijy i c j, j = 1,..., n (1.41) (1.41) y i (1.42) max 3x 1 + 2x 2 s. t. x 1 + x 2 + x 4 = 6 2x 1 + x 2 + x 5 = 10 x 2 + x 6 = 3 x 1, x 2, x 3, x 4, x 5 0 (1.42) (1.41) (1.42) min 6y y 2 + 3y 3 s. t. y 1 + 2y 2 3 y 1 + y 2 + y 3 2 y 1 0 y 2 0 y 3 0 (1.43) 3 (1.43) (1.37) (1.38) (1.39) (1.40) (1.41) ( ) x 1,..., x n y 1,..., y m c 1 x 1 + c 2 x c n x n b 1 y 1 + b 2 y b m y m (1.44)
36 ( ) () (1.44) ( n n m ) m n m c j x j a ij y i x j = a ij x j y i b i y i j=1 j=1 i=1 (x 1,..., x n ) (y 1,..., y m ) i=1 j=1 i=1 c 1 x 1 + c 2 x c n x n = b 1 y 1 + b 2 y b m y m (x 1,..., x n ) (y 1,..., y m ) 1.1 ( ) (1.38) (x 1,..., x n) (1.39) (y 1,..., y m) n m c j x j = b i y i j=1 [ ] ( ) i=1 n+m z = z + c j x j (1.45) x j c j = 0 c j 0 z = j=1 n c j x j (1.46) j=1 z (x 1,..., x n) z = n c j x j (1.47) j=1 y i = c n+i 0, i = 1,..., m (1.48) y i, i = 1,..., m
37 (1.45) z = z + z = z + = (z = z + j=1 n m c j x j + c n+i x n+i j=1 j=1 i=1 n m c j x j y i x n+i (1.49) i=1 n m c j x j y i (b i i=1 m b i y i ) + i=1 j=1 n a ij x j ) j=1 n m (c j + a ij y i )x j (1.50) (1.46) (1.50) n m c j x j = (z b i y i ) + j=1 i=1 j=1 i=1 n m (c j + a ij y i )x j i=1 c j + m b i y i = z i=1 m a ij y i = c j, i=1 j = 1,..., n c j 0 (D) (y 1,..., y m) [] 1.23
38 ( ) x = (x 1,..., x n) (1.38) y = (y 1,..., y m) (1.39) x y j = 1,..., n i = 1,..., m m a ij y i = c j x j = 0 (1.51) i=1 n a ij x j = b i y i = 0 (1.52) [ ] x y n c j x j j=1 j=1 j=1 n m m n m ( a ij y i )x j = ( a ij x j )y i b i y i (1.53) i=1 (1.51) (1.52) (1.53)) x y (1.53) (1.51) (1.52) i=1 j=1 i=1 [] x 1,..., x n, y 1,..., y m y i > 0 i () y i = 0 i 13 (1.37) 13 \ "
39 (1.34) y 1 y 2 y 3 (1.34) (x 1, x 2 ) = (4, 2) (1.37) (y 1, y 2, y 3 ) = (1, 1, 0) x 1 + x 2 = 6, 2x 1 + x 2 = 10 0 y 1 + 2y 2 = 3 y 1 + y 2 + y 3 = ( ) ( ) (1.38) (40 ) x n+i ^c n+i y i = ^c n+i 0 (1.34) (1.32) x 3 1 y 1 = 1 x 4 1 y 2 = 1 y 3 = b 1 b 2 b m. = b 1 + ɛ 1 b 2 + ɛ 2. b m + ɛ m ɛ i (i = 1,..., m) z z = c 1 x 1 + c 2 x c n x n = b 1 y 1 + b 2 y b m y m
40 44 1 (b 1 + ɛ 1, b 2 + ɛ 2,..., b n + ɛ n ) z z = (b 1 + ɛ 1 )y 1 + (b 2 + ɛ 2 )y (b m + ɛ m )y m z z = ɛ 1 y 1 + ɛ 2 y ɛ m y m (1.54) (1.34) z = 16 x 3 x 4 x 1 = 4 +x 3 x 4 x 2 = 2 2x 3 +x 4 x 5 = 1 +2x 3 x 4 (1.32) (x 1, x 2, x 3, x 4, x 5 ) = (4, 2, 0, 0, 1) (y 1, y 2, y 3 ) = (1, 1, 0) ɛ 3 (y 3 ) 0 (1.54) 0 y 1 1 ɛ 1 ɛ 1 ( ɛ 1 ) ɛ 1 = 0.2 x 1 x 2 x 5 z = 16.2 x 3 x 4 x 1 = 3.8 +x 3 x 4 x 2 = 2.4 2x 3 +x 4 x 5 = x 3 x 4 (1.55) (x 1, x 2, x 3, x 4, x 5 ) = (4, 2, 0, 0, 1) (3.8, 2.4, 0, 0, 0.6) ɛ 1 = 0.2 y 1 = 1 1 ɛ 1 1 ɛ 1 1 1
41 (0 ɛ 1 0.6) ɛ z x x x ɛ 1 0 x 1 x 2 x 5 x 2 x 1 x 5 ɛ 1 = 0.5 x 5 0 ɛ x 3 x 5 ɛ 1 = 0.5 b = (6.5, 10, 3) z = x 4 0.5x 5 x 1 = x x 4 x 2 = 3.5 x 4 x 3 = x x 4 (1.56) 1.25 ɛ (ɛ 1 0.5) ɛ z x x x ɛ 1 > 5 y 1 = 0 y 2 = 1.5 y 3 = b 1 0 b < b b 1 >
42 Obj. value Wood = max 3x 1 + 2x 2 + 8x 3 s. t. x 1 + x 2 + 3x 3 6 2x 1 + x 2 + 3x 3 10 x 2 + 2x 3 3 x 1, x 2, x 3 0 (1.57)
( ) ? () 1.1 ( 3 ) j x j 10 j 1 10 j = 1,..., 10 x 1 + x x 10 =
5 1! (Linear Programming, LP) LP OR LP 1.1 1.1.1 1. 2. 3. 4. 5. ( ) ( ) 1.1 6 1 1.1 ( ) 1 110 2 98 3 85 4 90 5 73 6 62 7 92 8 88 9 79 10 75 1.1.2 4? 900 40 80 120 () 1.1 ( 3 ) j x j 10 j 1 10 j = 1,...,
More information2 1/2 1/4 x 1 x 2 x 1, x 2 9 3x 1 + 2x 2 9 (1.1) 1/3 RDA 1 15 x /4 RDA 1 6 x /6 1 x 1 3 x 2 15 x (1.2) (1.3) (1.4) 1 2 (1.5) x 1
1 1 [1] 1.1 1.1. TS 9 1/3 RDA 1/4 RDA 1 1/2 1/4 50 65 3 2 1/15 RDA 2/15 RDA 1/6 RDA 1 1/6 1 1960 2 1/2 1/4 x 1 x 2 x 1, x 2 9 3x 1 + 2x 2 9 (1.1) 1/3 RDA 1 15 x 1 + 2 1/4 RDA 1 6 x 1 1 4 1 1/6 1 x 1 3
More information106 4 4.1 1 25.1 25.4 20.4 17.9 21.2 23.1 26.2 1 24 12 14 18 36 42 24 10 5 15 120 30 15 20 10 25 35 20 18 30 12 4.1 7 min. z = 602.5x 1 + 305.0x 2 + 2
105 4 0 1? 1 LP 0 1 4.1 4.1.1 (intger programming problem) 1 0.5 x 1 = 447.7 448 / / 2 1.1.2 1. 2. 1000 3. 40 4. 20 106 4 4.1 1 25.1 25.4 20.4 17.9 21.2 23.1 26.2 1 24 12 14 18 36 42 24 10 5 15 120 30
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