bai giai quy hoach tuyen tinh chuong 1

... t h ịj (m) [t j (m)] t k ịj (m) [t j (m)−t ij ] D ij (c) [t j (m)−t i (s)−t ij ] y 1 2 0 2 2 2 0 0 y 2 4 0 4 7 7 3 3 y 3 3 0 3 3 3 0 0 y 4 5 2 7 7 7 2 0 y 5 4 2 6 9 9 5 3 y 6 6 2 8 10 10 4 2 y 7 3 7 10 10 10 7 0 y 8 11 3 13 14 14 3 0 y 9 4 10 14 14 14 10 0 y 10 5 6 11 14 14 ... đường Găng (y 1 , y 4 , y 7 , y 9 ) và (y 3 , y 8 ) 1 2 2 0 4 6 3 9 0 0 0 0 3 7 7 0 2 3 3 0 5 10 10 0 6 14 14 0 y 1 2 y 2 7 y 3 3 y 8 11 y 7 3 y 9 4 y 6 6 y 5 4 y 10 5 Hình 10 y 4 5 c. ... đường Găng chỉ còn là 12 . 1 2 2 0 4 6 1 7 0 0 0 0 3 5 5 0 2 3 3 0 5 8 8 0 6 12 12 0 y 1 2 y 2 4 y 3 3 y 8 9 y 7 3 y 9 4 y 6 6 y 5 4 y 10 5 Hình 11 y 4 3 Thời gian Quý 1/ 08 Quý 2/08 Quý 3/08...

... với n ẩn số: a 11 x 1 + a 12 x 2 +… +a 1n x n = b 1 a 21 x 1 + a 22 x 2 +… +a 2n x n = b 1 (*) …………………………… a m1 x 1 + a m2 x 2 +… +a 1m x n = b 1 với x 1 , x 2 ,…, x n là ... phương trình B là: a 11 a 12 … a 1n a 11 a 12 … a 1n a 21 a 22 … a 2n a 21 a 22 … a 2n A = …………… X = …………… a n1 a n2 … a nn a n1 a n2 … a nn B = b 1 b 2 … b n Vậy ... mi ,1= và nj ,1= Bài toán được mô tả theo bảng sau: S 1 S 2 … S j … S n Dự trữ NVL1 a 11 a 12 … a 1j … a 1n b 1 NVL2 a 21 a 22 … a 2j … a 2n b 2 … … … … … … … … 15 Một...

... X_9_2_3 *1 + X_9_2_4 *1 + X_9_2_6 *1 = 597.64 X _10 _2 _1* 1 + X _10 _2_2 *1 + X _10 _2_3 *1 + X _10 _2_4 *1 + X _10 _2_5 *1 + X _10 _2_6 *1 = 11 63 .15 Các ràng buộc tương quan tỷ lệ X_3 _1_ 1 *1 + X_4 _1_ 1 *1 + X_5_2 _1* 1 ... X_6_2 _1* 1 + X_7_2 _1* 1 + X_8_2 _1* 1 + X_9_2 _1* 1 + X _10 _2 _1* 1 ≤ 3499.74 X_3 _1_ 2 *1 + X_4_2_2 *1 + X_5 _1_ 2 *1 + X_6_2_2 *1 + X_7_2_2 *1 + X_8_2_2 *1 + X_9_2_2 *1 + X _10 _2_2 *1 ≤ 3499.74 X_2 _1_ 3 *1 + X_3_2_3 *1 ... nhập Z1 = X _1_ 2_5 *10 9 41 + X _1_ 2_6*6239 + X_2 _1_ 3*27746 + X_3 _1_ 1 *14 0 51 + X_3 _1_ 2 *11 914 + X_3_2_3*27746 + X_3_2_4 *12 854 + X_3_2_5 *10 9 41 + X_3_2_6*6239 + X_4 _1_ 1 *14 0 51 + X_4_2_2 *11 914 + X_4_2_3*27746...

... 1 1 1/ 3 -1/ 3 0 0 1/ 3 0 0 A7 M 1 0 2/3 1/ 3 -1 0 -1/ 3 1 0 A8 M 4 0 3 1 0 -1 -1 0 1 (M) 0 11 /3 4/3 -1 -1 -7/3 0 0 0 -14 -5 0 0 5 0 0 A1 15 5/9 1 0 -4/9 0 1/ 9 4/9 0 -1/ 9 A7 M 1/ 9 0 0 1/ 9 -1 2/9 ... A 4 . 000-532-3f(x) 0 1 0 1 0 0 0 0 1 0 -1 4 3 1 1 1 2 0 4 3 3 0 0 -1 A5 A6 A4 x6x5x4x3x2x1PaHscs 00 -11 -4-2 0 -10 -5 01- 7f(x) 0 1 0 1/ 3 -1/ 3 -1/ 3 0 0 1 0 -1 4 1 0 0 1/ 3 5/3 -1/ 3 4/3 5/3 5/3 -4 0 -1 A2 A6 A4 -3/5-4/50-22/500-8f(x) -1/ 5 3/5 1/ 5 2/5 -1/ 5 -2/5 0 0 1 ... bảng sau: 00 01- 320f(x) 0 0 1 0 1 0 1 0 0 1 -2 1 -5 2 0 1 3 4 15 20 10 0 0 0 A4 A5 A6 x6x5x4x3x2x1PaHscs 000 -13 -2 -1/ 20 01/ 2-30-5f(x) -1/ 4 -3/4 1/ 4 0 1 0 1 0 0 3/4 -11 /4 1/ 4 -5 2 0 0 0 1 25/2 25/2 5/2 0 0 -2...

... 1/ 9 1 -1/ 45 0 8/45 0 -1/ 15 A5 0 2/3 0 20/3 0 -1/ 3 1 0 A3 30 1/ 9 0 17 /45 1 -1/ 45 0 2 /15 0 -12 7/9 0 26/9 0 8/3 A1 20 17 /15 0 1 0 0 53/300 1/ 300 -1/ 15 A2 25 1/ 10 0 1 0 -1/ 20 3/20 0 A3 30 11 /15 0 ... 1 1 1/ 3 -1/ 3 0 0 1/ 3 0 0 A7 M 1 0 2/3 1/ 3 -1 0 -1/ 3 1 0 A8 M 4 0 3 1 0 -1 -1 0 1 (M) 0 11 /3 4/3 -1 -1 -7/3 0 0 0 -14 -5 0 0 5 0 0 A1 15 5/9 1 0 -4/9 0 1/ 9 4/9 0 -1/ 9 A7 M 1/ 9 0 0 1/ 9 -1 2/9 ... ñược. 15 19 0 0 0 M M M Co So CJ Ph.An A1 A2 A3 A4 A5 A6 A7 A8 A6 M 3 3 1 -1 0 0 1 0 0 A7 M 2 1 1 0 -1 0 0 1 0 A8 M 7 3 4 0 0 -1 0 0 1 (M) 7 6 -1 -1 -1 0 0 0 -15 -19 0 0 0 0 0 0 A1 15 1 1...

... u −u T 0  e k 1 1  +  0 k 1 k  =  Me k 1 + u −u T e k 1 + k  . Dễ thấy Me k 1 + u = e k 1 (từ định nghĩa của u) và −u T e k 1 + k = −(e k 1 − Me k 1 ) T e k 1 + k = −e T k 1 e k 1 + k = 1 (e T k 1 Me k 1 = ... τ. Thật vậy, với θ = 1/ (3 √ n), ta có θ 2 n (1 θ) 2 ≤ 1 4 . Do đó nếu δ(x, t) ≤ τ = 1/ √ 2 thì từ Bổ đề 6 .1 suy ra δ(x + , t + ) 2 ≤ 1 4 + 1 4 = 1 2 hay δ (x + , t + ) ≤ 1/ √ 2 = τ . Như vậy, ... −q, z ≥ 0} 51 Mệnh đề 4 .1. (x + , s(x, t)) chấp nhận được (chặt) khi và chỉ khi −e ≤ x 1 ∆x ≤ e (−e < x 1 ∆x < e). Điều này tương đương với x 1 ∆x ∞ ≤ 1 (x 1 ∆x ∞ < 1) . Trong...

... sau: (min)max)( 1    n i ii xcxf           ); ,1; ,1( 0,0 ) ,1; ,1( 1 nmmknibx mnjmkbxax ki k mn j jkmjk (min)max)( 1    n i ii xcxf           ); ,1; ,1( 0,0 ) ,1; ,1( 1 nmmknibx mnjmkbxax ki k mn j jkmjk (min)max)( 1    n i ii xcxf           ); ,1; ,1( 0,0 ) ,1; ,1( 1 nmmknibx mnjmkbxax ki k mn j jkmjk (min)max)( 1    n i ii xcxf           ); ,1; ,1( 0,0 ) ,1; ,1( 1 nmmknibx mnjmkbxax ki k mn j jkmjk (min)max)( 1    n i ii xcxf           ); ,1; ,1( 0,0 ) ,1; ,1( 1 nmmknibx mnjmkbxax ki k mn j jkmjk (min)max)( 1    n i ii xcxf           ); ,1; ,1( 0,0 ) ,1; ,1( 1 nmmknibx mnjmkbxax ki k mn j jkmjk (min)max)( 1    n i ii xcxf           ); ,1; ,1( 0,0 ) ,1; ,1( 1 nmmknibx mnjmkbxax ki k mn j jkmjk (min)max)( 1    n i ii xcxf           ); ,1; ,1( 0,0 ) ,1; ,1( 1 nmmknibx mnjmkbxax ki k mn j jkmjk ... sau: (min)max)( 1    n i ii xcxf           ); ,1; ,1( 0,0 ) ,1; ,1( 1 nmmknibx mnjmkbxax ki k mn j jkmjk (min)max)( 1    n i ii xcxf           ); ,1; ,1( 0,0 ) ,1; ,1( 1 nmmknibx mnjmkbxax ki k mn j jkmjk (min)max)( 1    n i ii xcxf           ); ,1; ,1( 0,0 ) ,1; ,1( 1 nmmknibx mnjmkbxax ki k mn j jkmjk (min)max)( 1    n i ii xcxf           ); ,1; ,1( 0,0 ) ,1; ,1( 1 nmmknibx mnjmkbxax ki k mn j jkmjk (min)max)( 1    n i ii xcxf           ); ,1; ,1( 0,0 ) ,1; ,1( 1 nmmknibx mnjmkbxax ki k mn j jkmjk (min)max)( 1    n i ii xcxf           ); ,1; ,1( 0,0 ) ,1; ,1( 1 nmmknibx mnjmkbxax ki k mn j jkmjk (min)max)( 1    n i ii xcxf           ); ,1; ,1( 0,0 ) ,1; ,1( 1 nmmknibx mnjmkbxax ki k mn j jkmjk (min)max)( 1    n i ii xcxf           ); ,1; ,1( 0,0 ) ,1; ,1( 1 nmmknibx mnjmkbxax ki k mn j jkmjk ... sau: (min)max)( 1    n i ii xcxf           ); ,1; ,1( 0,0 ) ,1; ,1( 1 nmmknibx mnjmkbxax ki k mn j jkmjk (min)max)( 1    n i ii xcxf           ); ,1; ,1( 0,0 ) ,1; ,1( 1 nmmknibx mnjmkbxax ki k mn j jkmjk (min)max)( 1    n i ii xcxf           ); ,1; ,1( 0,0 ) ,1; ,1( 1 nmmknibx mnjmkbxax ki k mn j jkmjk (min)max)( 1    n i ii xcxf           ); ,1; ,1( 0,0 ) ,1; ,1( 1 nmmknibx mnjmkbxax ki k mn j jkmjk (min)max)( 1    n i ii xcxf           ); ,1; ,1( 0,0 ) ,1; ,1( 1 nmmknibx mnjmkbxax ki k mn j jkmjk (min)max)( 1    n i ii xcxf           ); ,1; ,1( 0,0 ) ,1; ,1( 1 nmmknibx mnjmkbxax ki k mn j jkmjk (min)max)( 1    n i ii xcxf           ); ,1; ,1( 0,0 ) ,1; ,1( 1 nmmknibx mnjmkbxax ki k mn j jkmjk (min)max)( 1    n i ii xcxf           ); ,1; ,1( 0,0 ) ,1; ,1( 1 nmmknibx mnjmkbxax ki k mn j jkmjk ...

... xxxxxf            )5 ,1( 0 8032 404 40 011 10 543 21 543 21 543 21 ix xxxxx xxxxx xxxxx i min78)( 43 21      xxxxxf            )5 ,1( 0 8032 404 40 011 10 543 21 543 21 543 21 ix xxxxx xxxxx xxxxx i SV ... xxxxxf            )5 ,1( 0 8032 404 40 011 10 543 21 543 21 543 21 ix xxxxx xxxxx xxxxx i min78)( 43 21      xxxxxf            )5 ,1( 0 8032 404 40 011 10 543 21 543 21 543 21 ix xxxxx xxxxx xxxxx i min78)( 43 21      ... TOÁN QUY HOCH TUYN TÍNH BÀI 4. GII BTQHTT BNG PP HÌNH HC Vd2: Gii BTQHTT sau bng PP hình hc: min78)( 43 21      xxxxxf            )5 ,1( 0 8032 404 40 011 10 543 21 543 21 543 21 ix xxxxx xxxxx xxxxx i min78)( 43 21     ...

... 2 211 2222 212 1 11 212 111 nix bxaxaxa bxaxaxa bxaxaxa i mnmnmm nn nn (I) (min)max )( 2 211  nn xcxcxcxf              ) ,1( 0 2 211 2222 212 1 11 212 111 nix bxaxaxa bxaxaxa bxaxaxa i mnmnmm nn nn (min)max ... 01 21 22 212 12 111                  mnmmmm nmm nmm aaa aaa aaa A 1 00 0 10 0 01 21 22 212 12 111                  mnmmmm nmm nmm aaa aaa aaa A 1 00 0 10 0 01 21 22 212 12 111                  mnmmmm nmm nmm aaa aaa aaa A ... 3)               13 020 011 30 010 21 A               13 020 011 30 010 21 A               13 020 011 30 010 21 A 16  CHNG I- BÀI TOÁN QUY HOCH TUYN TÍNH BÀI 3....

... PACB                         14 29 ,0, 612 1 , 3 2 6 5 , 6 7 12 29   X 0         0, 12 1 , 6 5 , 12 29 0 x 2         2, 12 5 , 6 13 , 12 1 * x                         14 29 ,0, 612 1 , 3 2 6 5 , 6 7 12 29   X 0         ... 2, 12 5 , 6 13 , 12 1 * x                         14 29 ,0, 612 1 , 3 2 6 5 , 6 7 12 29   X 0         0, 12 1 , 6 5 , 12 29 0 x 2         2, 12 5 , 6 13 , 12 1 * x                         14 29 ,0, 612 1 , 3 2 6 5 , 6 7 12 29   X 0         ... BN                         14 29 ,0, 612 1 , 3 2 6 5 , 6 7 12 29   X 0         0, 12 1 , 6 5 , 12 29 0 x 2         2, 12 5 , 6 13 , 12 1 * x LÀ PA NHNG KHÔNG LÀ PACB                         14 29 ,0, 612 1 , 3 2 6 5 , 6 7 12 29   X 0        ...

... toán:                     5 ,1, 0,0,,,, 2680 012 2843 875002 412 1 814 15 48004,28 ,15 ,12 ,15 ,1 154008,22,22,35,22,2 max4,34 ,18 ,25 ,12 ,1) ( 543 21 543 21 543 21 543 21 543 21 543 21 inguyenxnguyenxxxxx xxxxx xxxxx xxxxx xxxxx xxxxxxf i                     5 ,1, 0,0,,,, 2680 012 2843 875002 412 1 814 15 48004,28 ,15 ,12 ,15 ,1 154008,22,22,35,22,2 max4,34 ,18 ,25 ,12 ,1) ( 543 21 543 21 543 21 543 21 543 21 543 21 inguyenxnguyenxxxxx xxxxx xxxxx xxxxx xxxxx xxxxxxf i                     5 ,1, 0,0,,,, 2680 012 2843 875002 412 1 814 15 48004,28 ,15 ,12 ,15 ,1 154008,22,22,35,22,2 max4,34 ,18 ,25 ,12 ,1) ( 543 21 543 21 543 21 543 21 543 21 543 21 inguyenxnguyenxxxxx xxxxx xxxxx xxxxx xxxxx xxxxxxf i                     5 ,1, 0,0,,,, 2680 012 2843 875002 412 1 814 15 48004,28 ,15 ,12 ,15 ,1 154008,22,22,35,22,2 max4,34 ,18 ,25 ,12 ,1) ( 543 21 543 21 543 21 543 21 543 21 543 21 inguyenxnguyenxxxxx xxxxx xxxxx xxxxx xxxxx xxxxxxf i 8  CHNG ... toán:                     5 ,1, 0,0,,,, 2680 012 2843 875002 412 1 814 15 48004,28 ,15 ,12 ,15 ,1 154008,22,22,35,22,2 max4,34 ,18 ,25 ,12 ,1) ( 543 21 543 21 543 21 543 21 543 21 543 21 inguyenxnguyenxxxxx xxxxx xxxxx xxxxx xxxxx xxxxxxf i                     5 ,1, 0,0,,,, 2680 012 2843 875002 412 1 814 15 48004,28 ,15 ,12 ,15 ,1 154008,22,22,35,22,2 max4,34 ,18 ,25 ,12 ,1) ( 543 21 543 21 543 21 543 21 543 21 543 21 inguyenxnguyenxxxxx xxxxx xxxxx xxxxx xxxxx xxxxxxf i                     5 ,1, 0,0,,,, 2680 012 2843 875002 412 1 814 15 48004,28 ,15 ,12 ,15 ,1 154008,22,22,35,22,2 max4,34 ,18 ,25 ,12 ,1) ( 543 21 543 21 543 21 543 21 543 21 543 21 inguyenxnguyenxxxxx xxxxx xxxxx xxxxx xxxxx xxxxxxf i                     5 ,1, 0,0,,,, 2680 012 2843 875002 412 1 814 15 48004,28 ,15 ,12 ,15 ,1 154008,22,22,35,22,2 max4,34 ,18 ,25 ,12 ,1) ( 543 21 543 21 543 21 543 21 543 21 543 21 inguyenxnguyenxxxxx xxxxx xxxxx xxxxx xxxxx xxxxxxf i 8  CHNG ... toán:                     5 ,1, 0,0,,,, 2680 012 2843 875002 412 1 814 15 48004,28 ,15 ,12 ,15 ,1 154008,22,22,35,22,2 max4,34 ,18 ,25 ,12 ,1) ( 543 21 543 21 543 21 543 21 543 21 543 21 inguyenxnguyenxxxxx xxxxx xxxxx xxxxx xxxxx xxxxxxf i                     5 ,1, 0,0,,,, 2680 012 2843 875002 412 1 814 15 48004,28 ,15 ,12 ,15 ,1 154008,22,22,35,22,2 max4,34 ,18 ,25 ,12 ,1) ( 543 21 543 21 543 21 543 21 543 21 543 21 inguyenxnguyenxxxxx xxxxx xxxxx xxxxx xxxxx xxxxxxf i                     5 ,1, 0,0,,,, 2680 012 2843 875002 412 1 814 15 48004,28 ,15 ,12 ,15 ,1 154008,22,22,35,22,2 max4,34 ,18 ,25 ,12 ,1) ( 543 21 543 21 543 21 543 21 543 21 543 21 inguyenxnguyenxxxxx xxxxx xxxxx xxxxx xxxxx xxxxxxf i                     5 ,1, 0,0,,,, 2680 012 2843 875002 412 1 814 15 48004,28 ,15 ,12 ,15 ,1 154008,22,22,35,22,2 max4,34 ,18 ,25 ,12 ,1) ( 543 21 543 21 543 21 543 21 543 21 543 21 inguyenxnguyenxxxxx xxxxx xxxxx xxxxx xxxxx xxxxxxf i 8  CHNG...

... : 1 2 1 2 j m j j mj A x A x A x A= + + + 1 2 ( ; ; ; ) j j j mj x x x x= 1 1 2 1 2 1 11 21 1. 0. (1, 0)A x A x A A A x= + = + ⇒ = 2 1 2 1 2 2 12 22 0. 1. (0 ,1) A x A x A A A x= + = + ⇒ = 3 1 ... ưu. 0, 1, j j n∆ ≥ ∀ = -5 -4 0 0 2 Cơ sở Hệ số cj Ph. án x 1 x 2 x 3 x 4 x 5 A 1 A 2 A 3 -5 -4 0 10 12 15 1 0 0 0 1 0 0 0 1 2 1 3 1 3 1 f(x) -98 0 0 0 -14 -19 Bài toán ... các véctơ x j : 1 1 3 1 3 1 11 31 1. 0. (1, 0)A x A x A A A x= + = + ⇒ = 2 1 3 1 3 2 12 32 2 1. ( 2, 1) A x A x A A A x= + = − − ⇒ = − − 3 1 2 1 2 3 13 23 0. 1. (0 ,1) A x A x A A A x= + = + ⇒ = , j j...

... ữ ữ = ữ ÷  ÷   1 2 1 2 n n Ax x A x A x A= + + + 1 1 2 2 ( ) n n f x c x c x c x= + + + với các ràng buộc: i1 1 i2 2 in 1 i1 1 i2 2 in 2 i1 1 i2 2 in 3 1 2 3 ; (1) ; (2) ; (3) 0 ... phương án tối ưu của bài toán QHTT. Ví dụ 2: 1 2 3 4 1 2 5 1 3 4 5 1 2 3 5 1 2 3 4 5 1 2 3 4 1 5 2 3 4 ( ) 4 max 2 6 1 4 4 2 16 2 4 5 5 17 9 5 2 11 ; 0, ; , 0. f x x x x x x x x x x x x x x x ... , ( ) 1, 1, i m ij j n A a = = = m nì 11 12 1 21 22 2 1 2 n n m m mn a a a a a a A a a a ữ ữ = ữ ữ 1 1 2 2 , n m x b x b x b x b ữ ữ ữ ữ = = ữ ữ ữ ữ 1 2 j j j mj a a A a ...

... -1 -1 -3 (-2) 1 0 0 A 6 -3 -1 -4 -2 -1 0 1 0 -4 -6 -5 -3 0 0 3 A 4 5/2 1/ 2 1/ 2 3/2 1 -1/ 2 0 0 A 6 ( -1/ 2) -1/ 2 -7/2 ( -1/ 2) 0 -1/ 2 1 15/2 -5/2 -9/2 -1/ 2 0 -3/2 0 3 A 4 1 -1 -10 0 1 -2 3 5 A 3 1 ... x 0 1 10 8 sở x 1 x 2 x 3 8 A 1 2/3 0 1 1 1 A 1 -1/ 3 1 ( -1) 0 0 -3 0 8 A 3 1/ 3 1 0 1 10 A 2 1/ 3 -1 1 0 6 -3 0 0 Ở bước 2, do x 0 ≥ 0 nên suy ra x =  0, 1 3 , 1 3  là phương án tối ưu. 53 Quy ... A 2 16 /7 0 0 0 1 -2/7 2/7 0 -1 A 1 22/7 0 0 1 0 6/7 1/ 7 0 0 A 0 -16 /7 1 1 0 0 9/7 -2/7 0 0 A 5 -4/7 0 0 0 0 ( -10 /7) 3/7 0 0 0 0 -9/7 -5/7 0 -2 A 2 12 /5 0 0 0 1 0 1/ 5 -1/ 5 -1 A 1 14/5 0 0 1 0...

... x J 1 1 x 1 x 2 x 3 x 4 x 5 x g 1 x g 3 1 x g 1 28 2 2 0 1 0 1 0 0 x 5 31 1 5 3 −2 1 0 0 1 x g 3 16 [2] −2 2 1 0 0 1 P 44 4 0 2 2 0 0 0 1 x g 1 0 x 5 x 1 P 0 8 1 1 1 1/ 2 0 0 0 0 0 1 1 0 12 4 ... )        ∈≤≥ ∈= →= ∑ ∑ ∑ = = = 2 n 1j ijij 1i n 1j jij n 1j jj Ii bxa Ii bxa (max)min xcxf c J J x J 1 1 x 1 x 2 x 3 x 4 x 5 x g 1 x g 3 1 x g 1 28 2 2 0 1 0 1 0 0 x 5 31 1 5 3 − 2 1 0 0 1 x g 3 16 2 −2 2 1 0 0 1 P 44 ... − 5 0 0 x 1 x 2 x 3 x 4 x 5 x 6 − 2 x 1 8 1 2 − 3 1 0 0 0 x 5 18 0 5 − 5 − 3 1 0 0 x 6 12 0 1 − 4 [2] 0 1 f(x) − 16 0 2 − 2 3 0 0 − 2 x 1 2 1 3/2 − 1 0 0 − 1/ 2 0 x 5 36 0 13 /2 − 11 0 1 3/2 − 5...