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1 5 1! (Linear Programming, LP) LP OR LP ( ) ( ) 1.1

2 ( ) ? () 1.1 ( 3 ) j x j 10 j 1 10 j = 1,..., 10 x 1 + x x 10 = 900 [] x 1 120, x 2 120, x [] x 4 80, x 5 80, x 6 80 [] x 7 40, x 8 40, x 9 40, x []

3 x x x x x x x x x x 10 minimize 110x x x x x x x x x x 10 subject to x 1 + x x 10 = 900 x j 120, j = 1, 2, 3 x j 80, j = 4, 5, 6 x j 40, j = 7, 8, 9, 10 (1.1) () ( ) (1.1) (1.1) 1.1 (1.1) y 1 y 2 y 3

4 y 1 x 1 1.2y 1 0.8y 1 x 2 1.2y 1 0.8y 1 x 3 1.2y 1 0.8y 2 x 4 1.2y 2 0.8y 2 x 5 1.2y 2 0.8y 2 x 6 1.2y 2 0.8y 3 x 7 1.2y 3 0.8y 3 x 8 1.2y 3 0.8y 3 x 9 1.2y 3 0.8y 3 x y 3 y 1 = 3y 3 y 2 = 2y 3 minimize 110x x x x x x x 7 +88x x x 10 subject to x 1 + x x 10 = y 1 x 1 1.2y 1 0.8y 1 x 2 1.2y 1 0.8y 1 x 3 1.2y 1 0.8y 2 x 4 1.2y 2 0.8y 2 x 5 1.2y 2 0.8y 2 x 6 1.2y 2 0.8y 3 x 7 1.2y 3 0.8y 3 x 8 1.2y 3 0.8y 3 x 9 1.2y 3 0.8y 3 x y 3 y 1 = 3y 3 y 2 = 2y 3 (1.2) (1.2) (1.1) (1.1) (1.2) 1.2 ( ) x x x 1 1 1

5 (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 (1.3). a m1 x 1 + a m2 x a mn x n = b m x 1 0, x 2 0,..., x n 0 x 1, x 2,..., x n ( ) (decision variable) (b 1,..., b m ) (right hand side) 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 ) (1.3)) 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 (1.4) maximize subject to c x Ax = b x 0 (1.5) 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

6 10 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 (1.6). 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 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 (1.7) (1.8) 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 (1.8) (1.8) x 1 = x + 1 x 1 ( x + 1 0, x 1 0) 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 (1.9)

7 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 (diet problem)? g B1 C (kcal) (g) (g) (g) (g) (g) (g) (mg) (mg) (mg) (100g ) ( ) B1 C 1 (http://food.tokyo.jst.go.jp/)

8 g 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 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 (1.10) (1.10) x 1 = x 2 = 0.0 x 3 = 0.0 x 4 = 0.0 x 5 = 0.0 x 6 = x 7 = Stigler[8] Stigler 77 A B1 B2 C ( 1.4) No one recommends these diets for anyone, let alone everyone. Dantzig (simplex method) Dantzig 500 ([2]) ( ) 1 70(55)g B1 1.1(0.8)mg C 100(100)mg

9 Stigler 535 lb. 107 lb. 13 lb. 134 lb. 25 lb. 500 ( 1890L 3 ) (bran) 2 ( 900g) 4 (blackstrap molasses) 2 ( 900g) Dantzig Fletcher, Soden, Zinober[6] LP Dantzig 3

10 (g) 1( ) (200) ( / ) 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 3 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 x 1, x 2, x 3, x 4, x 5, y 1, y 2, y 3 0

11 [11] 5 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,..., A B C ( 15 ) []

12 16 1 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 6 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 (1.11) LP A C () (1.11) x 1 + x 2 + x 3 = 15 y 1 + y 2 + y 3 = 15 z 1 + z 2 + z 3 = A B C numeraire

13 K4 A 8% B 4% C 2% 4 1g g 1 A B C 0.03g 0.02g 0.01g 1.10 ( /g) A B C kg 1.5 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 kg kg ( ) kg ( 2 140kg )

14 K kg : 1kg : 1kg : 1kg : 1kg : 1kg : 1kg : 1kg : kg kg 0.3 kg kg 1500kg 3. 1kg 200 1kg 300 ( ) 4. 1kg

15 [ ] w t t (t = 1,..., 5) x t t (t = 2,..., 5) y t t (t = 2,..., 5) z t t (t = 1,..., 5) u t t (t = 1,..., 5) v t t (t = 1,..., 5)

16 20 1 [] min v 1 + v 2 + v 3 + v 4 + v 5 s.t w 1 u w t x t u t 0, t = 2,..., 5 u 1 v 1 5, v 1 12 u 2 v 2 4, v 2 12 u 3 v 3 4, v 3 12 u 4 v 4 3, v 4 5 u 5 v 5 2, v 5 5 w 1 z 1 = 0 w t + 0.9z t 1 10x t z t = 0, t = 2,..., 5 10x t 0.9z t 1 0, t = 2,..., 5 w 5 = 0, z 5 = 0 x 2 y 2 = 0 x t + y t 1 y t = 0, t = 3,..., 5 y w t 0, t = 1,..., 4 x t 0, t = 2,..., 5 y t 0, t = 2,..., 5 z t 0, t = 1,..., 5 u t 0, t = 1,..., 5 v t 0, t = 1,..., 5 (1.12) (u t ) (v t ) (w t ) (x t ) (y t ) (z t ) (1.12) 1.13 [ ] T L t (t = 1,..., T) T t (t = 1,..., T) B j (j = 1,..., n) ( ) c jt B j 1 p j t i t []

17 B j x j t v t min s.t. n j=1 p jx j n j=1 c jtx j + (1 + i t )v t 1 v t = L t, t = 1,..., T v 0 = 0 x j 0, j = 1,..., n v t 0, t = 1,..., T 1.14 Σc x jt j (1+i t )v t-1 v t L t 1.8 [] = 0:8 [ ] + 0:2 [ ] - 0:05 [] max x 1 + x 2 s.t. x 1 + 2x 2 3 x 1 0, x 2 0 (1.13)

18 22 1 x 1 + 2x 2 3 x 1 + 2x 2 0 (1.13) (x 1, x 2 ) (infeasible) (feasible) (feasible solution) (1.13) max x 1 + x 2 s.t. x 1 + 2x 2 3 x 1 0, x 2 0 (1.14) (1.14) (x 1, x 2 ) = (1, 1) (x 1, x 2 ) = (1, 1) = 2 1 t (1 + t, 1) (1 + t) = 1 t 3 (1 + t) + 1 = 2 + t t t 7 (unbounded) ^x d t ^x + td ^x ^x + td (1.14) max x 1 + x 2 s.t. x 1 + 2x 2 3 x 1 0, x 2 0 (1.15) (1.15) (1.14) d = (1, 0) (1.15) (0, 1.5) (infeasible) (feasible) (feasible solution) ( ) ( ) 7

19 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 (1.16) (1.16) ( x 1 x 2 ) S S x 1 x S x 2 A (0,3) E (3,3) (3.5,3) S B (4,2) C O D 0 (5,0) (6,0) F x 1 (c 1, c 2 ) ( )

20 S (1) x 2 (C1,C2) 0 x S (2) x 2 0 x 1 (C1,C2)

21 ( (1.17)) 1.19 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 (1.17) 1.19 x 2 (C1,C2) 0 x max x 1 - x 2 s. t. 2x 1-3x 2-3 x 1 - x x 1 ; x 2 0 [] x 1 -x 2 x 1 - x 2 = 3-3 (3; 3)

22 (1.16) (c 1, c 2 ) = (3, 2) (1.18) max 3x 1 + 2x 2 s. t. x 1 + x 2 + x 3 = 6 2x 1 + x 2 + x 4 = 10 (1.18) 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 (1.19) (1.19) 5 3 x 2 x 4 (1.19) 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 (1.20) (1.20) 2 x 3 = 0.5x x x x x 4 = 5 x 2 2x 3 +x 4 = 2 x 2 +x 5 = 3 (1.21) (1.18) z = 3x 1 + 2x 2 (1.21) 1 x 1 z = 3 (5 0.5x 2 0.5x 4 ) + 2x 2 = x 2 1.5x 4 (1.22) x 2 x 4 (1.21) (1.22) 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 (1.23)

23 (1.18) z z = 3x 1 + 2x 2 x 1 + x 2 + x 3 = 6 2x 1 + x 2 + x 4 = 10 (1.24) x 2 + x 5 = 3 (1.23) (1.24) 1.10 (1.24) (x 1 ; x 2 ; : : : ; x 5 ) (1.23) (1.23) (x 1 ; x 2 ; : : : ; x 5 ) (1.24) (1.23) z, x 1, x 3, x 5 x 2 x 4 x 2 x 4 z, x 1, x 3, x 5 8 x 1, x 3, x 5 (basic variable) x 2, x 4 (nonbasic variable) 0 (1.23) x 2 = x 4 = 0 x 1 = 5, x 3 = 1, x 5 = 3, z = 15 (1.24) (1.24) 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) (1.23) 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) 1.16 D 8 (dictionary)

24 28 1 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 (1.25) 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 1.11 x 1 ; : : : ; x [ ] S 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 (1.23) 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 (1.25) (1.23) ( ) x 2, x 4 0 (1.23) 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 (1.26)

25 t z = t t t x 1, x 3, x 5 t t = 2 x 3 = 0 (1.23) 3 x 3 = 1 0.5x x 4 (1.25) (1.23) x 2 (1.25) (1.23) x 3 (1.25) (1.25) (1.23) x 4 ( ) 0? 1.16? (1.23) (1.25) (1.25) z = 16 x 3 x 4 x 3 x 4 0 (1.25) (1.25) (^x 1,..., ^x 5 ) (1.18) (1.25) z = 3^x 1 + 2^x 2 = 16 ^x 3 ^x 4 9 ^x 3 0, ^x 4 0 z = 3^x 1 + 2^x (5; 0; 1; 0; 3) (0; 0; 6; 10; 3) (3; 3; 0; 1; 0)

26 30 1 (simplex method) 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 (1.27) (1.16) 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 (1.25)

27 (1.25) 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 ) 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 (1.28) 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 f (1.29)

28 32 1 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 (1.30) y 1 = 1 y 2 = 1 y 3 = 0 ( ) 16 (1.30) (shadow price) (1.30) (1.27) (1.30) (1.31) 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 (1.31) (1.32) (1.31) (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 (1.32)

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 15 20 10 25 35 20 18 30 12 4.1 7 min. z = 602.5x 1 + 305.0x 2 + 2

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 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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