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BỘ CÔNG THƯƠNG TRƯỜNG ĐẠI HỌC CÔNG NGHIỆP TP. HCM Nguyễn Đức Phương Bài giảng Quy hoạch tuyến tính MSSV: ........................... Họ tên: ........................... TP. HCM – Ngày 27 tháng 6 năm 2011

Bài giảng Quy hoạch tuyến tínhdulieu.tailieuhoctap.vn/books/giao-duc-dai-cuong/kinh-te-hoc/file... · Mục lục Mục lục i 1 Giới thiệu quy hoạch tuyến tính

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  • B CNG THNGTRNG I HC CNG NGHIP TP. HCM

    Nguyn c Phng

    Bi gingQuy hoch tuyn tnh

    MSSV: . . . . . . . . . . . . . . . . . . . . . . . . . . .

    H tn: . . . . . . . . . . . . . . . . . . . . . . . . . . .

    TP. HCM Ngy 27 thng 6 nm 2011

  • Mc lc

    Mc lc i

    1 Gii thiu quy hoch tuyn tnh 1

    1.1 Mt s v d dn n bi ton quy hoch tuyn tnh . . . . . 1

    1.2 Cc dng ca bi ton quy hoch tuyn tnh . . . . . . . . . . 5

    1.2.1 Bi ton quy hoch tuyn tnh dng tng qut . . . . . 5

    1.2.2 Bi ton quy hoch tuyn tnh dng chun . . . . . . . 5

    1.2.3 Bi ton quy hoch tuyn tnh dng chnh tc . . . . . 6

    1.3 Quan h dng chun v chnh tc . . . . . . . . . . . . . . . . 8

    1.3.1 i chiu bt ng thc ca cc rng buc . . . . . . . 8

    1.3.2 Bin khng rng buc . . . . . . . . . . . . . . . . . . 9

    1.3.3 Quan h dng chun, chnh tc . . . . . . . . . . . . . 10

    1.4 Dng ma trn ca bi ton quy hoch . . . . . . . . . . . . . 13

    1.5 Phng n chp nhn c . . . . . . . . . . . . . . . . . . . 14

    1.6 ngha hnh hc ca bi ton quy hoch tuyn tnh . . . . . 16

    1.6.1 Phng php th . . . . . . . . . . . . . . . . . . . 16

    1.6.2 Tnh cht ca tp phng n chp nhn c . . . . . 17

    1.7 im cc bin . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

    1.8 Phng n c bn chp nhn c . . . . . . . . . . . . . . . 22

    1.8.1 Nghim c bn ca Ax D b . . . . . . . . . . . . . . . 23

    1.8.2 Thnh lp phng n cc bin . . . . . . . . . . . . . . 26

    1.8.3 Phng n cc bin v phng n ti u . . . . . . . . 30

  • MC LC ii

    1.9 Bi tp chng 1 . . . . . . . . . . . . . . . . . . . . . . . . . 31

    2 Phng php n hnh 33

    2.1 Phng php n hnh cho bi ton quy hoch dng chun . . 33

    2.1.1 Phng n cc bin ban u . . . . . . . . . . . . . . . 36

    2.1.2 Du hiu ti u . . . . . . . . . . . . . . . . . . . . . . 37

    2.1.3 Chn bin vo c s . . . . . . . . . . . . . . . . . . . 40

    2.1.4 Chn bin ra khi c s . . . . . . . . . . . . . . . . . 41

    2.1.5 Lp bng n hnh mi . . . . . . . . . . . . . . . . . . 42

    2.2 Thut ton n hnh cho bi ton min . . . . . . . . . . . . . 50

    2.3 Bi ton chnh tc khng c sn ma trn n v . . . . . . . . 52

    2.4 Bi tp chng 2 . . . . . . . . . . . . . . . . . . . . . . . . . 58

    3 L thuyt i ngu 64

    3.1 V d dn n bi ton i ngu . . . . . . . . . . . . . . . . 64

    3.1.1 Bi ton i ngu ca bi ton max . . . . . . . . . . . 66

    3.1.2 Bi ton i ngu ca bi ton min . . . . . . . . . . . 68

    3.2 Cc nh l v i ngu . . . . . . . . . . . . . . . . . . . . . 71

    3.3 Bi tp chng 3 . . . . . . . . . . . . . . . . . . . . . . . . . 79

    4 Bi ton vn ti 83

    4.1 Bi ton vn ti cn bng thu pht . . . . . . . . . . . . . . . 83

    4.2 Phng n cc bin ca bi ton vn ti . . . . . . . . . . . . 85

    4.3 Cc phng php thnh lp phng n cc bin . . . . . . . . 89

    4.3.1 Phng php cc ph thp nht . . . . . . . . . . . . 89

    4.3.2 Phng php gc Ty - Bc . . . . . . . . . . . . . . . 90

    4.3.3 Phng php Vogel (Fogel) . . . . . . . . . . . . . . . 90

    4.4 Thut ton th v gii bi ton vn ti . . . . . . . . . . . . . 92

    4.4.1 Thut ton quy khng cc ph chn . . . . . . . . . 92

    4.4.2 Xy dng phng n cc bin mi . . . . . . . . . . . 97

  • MC LC iii

    4.5 Mt s trng hp c bit ca bi ton vn ti . . . . . . . . 101

    4.5.1 Bi ton vn ti khng cn bng thu pht . . . . . . . 101

    4.5.2 Bi ton vn ti c cm . . . . . . . . . . . . . . . . 103

    4.6 Bi ton vn ti cc i cc ph . . . . . . . . . . . . . . . . 105

    4.7 Bi tp chng 4 . . . . . . . . . . . . . . . . . . . . . . . . . 106

    Ti liu tham kho 110

  • Chng 1

    Gii thiu quy hoch tuyn tnh

    1.1 Mt s v d dn n bi ton quy hoch tuyn tnh

    V d 1.1 (Bi ton lp k hoch sn xut). Mt tri ca cc khc g thnhcc tm vn. C hai loi vn: vn thnh phm v vn s dng trong xydng. Gi s, i vi:

    Vn thnh phm cn 2 gi ca v 5 gi bo 10m vn.

    Vn xy dng cn 3 gi ca v 3 gi bo 10m vn.

    My ca lm vic ti a 8 gi trong ngy, v my bo lm vic ti a 15 gitrong ngy. Nu li nhun ca 10m vn thnh phm l 120 (ngn ng), vli nhun ca 10m vn xy dng l 100 (ngn ng). Trong ngy, tri caphi ca bao nhiu vn mi loi li nhun ln nht?Gii.

  • 1.1 Mt s v d dn n bi ton quy hoch tuyn tnh 2

    V d 1.2 (Bi ton khu phn n). Chuyn gia dinh dng nh thnh lpmt thc n gm 2 loi thc phm chnh A v B. C mt (trm gram):

    Thc phm A cha 2 n v cht bo, 1 n v carbohydrate v 4 nv protein.

    Thc phm B cha 3 n v cht bo, 3 n v carbohydrate v 3 nv protein.

    Nu mt (trm gram) thc phm A gi 20 (ngn ng) v mt (trm gram)thc phm B gi 25 (ngn ng). Nh dinh dng mun thc n phi cungcp t nht 18 n v cht bo, 12 n v carbohydrate v 24 n v protein.Bao nhiu (trm gram) thc phm mi loi c gi nh nht nhng vncung cp dinh dng?

    Gii.

  • 1.1 Mt s v d dn n bi ton quy hoch tuyn tnh 3

    V d 1.3 (Bi ton vn ti). Mt nh sn xut c 2 nh my: Mt nhmy Vnh Phc v mt nh my Bnh Dng. C 3 kho hng phn phisn phm t H Ni, TP. HCM v Cn Th. Nh my Vnh phc; BnhDng, c kh nng cung cp ti a 100; 140 tn mi tun. Lng cu cacc kho H Ni, TP. HCM v Cn Th ln lt t 100; 60 v 80 tn trln. Chi ph vn chuyn (trm ngn) mi tn cho nh bng bn di. Hicn vn chuyn bao nhiu tn hng ha t nh sn xut n cc kho hng H Ni, TP. HCM v cn th chi ph nh nht nhng vn p ng nhu cu?

  • 1.1 Mt s v d dn n bi ton quy hoch tuyn tnh 4

    ``````````````````Trm pht

    Trm thu H Ni TP. HCM Cn thW1:100 W2:60 W3:80

    Vnh Phc-Q1: 100 5 7 9Bnh Dng-Q2:140 8 7 10

    Gii.

  • 1.2 Cc dng ca bi ton quy hoch tuyn tnh 5

    1.2 Cc dng ca bi ton quy hoch tuyn tnh

    1.2.1 Bi ton quy hoch tuyn tnh dng tng qut

    T cc v d mc 1.1, bi ton quy hoch tuyn tnh tng qut c phtbiu nh sau: Tm x1; x2; : : : ; xn sao cho

    z D c1x1 C c2x2 C C cnxn ! max .hay min/ (1.1)

    Vi cc rng buc8

  • 1.2 Cc dng ca bi ton quy hoch tuyn tnh 6

    1.2.3 Bi ton quy hoch tuyn tnh dng chnh tc

    Chng ta ni bi ton quy hoch tuyn tnh c dng chnh tc nu n cdng nh sau: Tm x1; x2; : : : ; n sao cho

    z D c1x1 C c2x2 C C cnxn ! max; .hay min/ (1.6)

    Vi cc rng buc8

  • 1.2 Cc dng ca bi ton quy hoch tuyn tnh 7

    Vi cc rng buc

    3x1 C 2x2 x3 C 2x5 D 4

    4x1 C 5x2 C 3x3 C 2x4 D 7

    x1 0; x2 0; x3 0; x4 0; x5 0

    e. z D 2x1 C 5x2 ! maxVi cc rng buc

    3x1 C 2x2 6

    2x1 C 9x2 8

    x1 0

    f. z D 2x1 C 3x2 ! minVi cc rng buc

    8

    0, u tin cng kt qu vi (1.27) ta cphng trnh (1.29), sau tr kt qu vi (1.27) ta c phng trnh(1.30).

    .x0

    1 C dc1/A0

    1 C .x0

    2 C dc2/A0

    2 C C .x0

    k C dck/A0

    k D b (1.29)

    .x0

    1 dc1/A0

    1 .x0

    2 C dc2/A0

    2 C C .x0

    k dck/A0

    k D b (1.30)

    By gi ta chn hai im trong Rn,

    v D .0; 0; : : : ; 0; x0

    1 C dc1; x0

    2 C dc2; : : : ; x0

    k C dck/

    vw D .0; 0; : : : ; 0; x

    0

    1 dc1; x0

    2 dc2; : : : ; x0

    k dck/

    Bi v d l hng s dng bt k, ta chn nh sau:

    0 < d < minj

    x0

    j

    jcj j; cj 0

  • 1.8 Phng n c bn chp nhn c 29

    Vi cch chn d nh trn, ta thy k thnh phn sau ca v; w l cc s dng.Mc khc, t (1.29) v (1.30) ta cng c v; w l phng n chp nhn c.Nhng ta li c

    x D1

    2vC

    1

    2w;

    tri vi gi thit ban u x l im cc bin. Vy gi s k ct cui ca A clp tuyn tnh l sai.

    H qu 1.15. S phng n cc bin ca tp phng n chp nhn cS D fxjAx D b; x 0g l hu hn.

    Chng minh. Bi v s h c m vct ct c lp tuyn tnh l hu hn, nntheo nh l 1.14 th s phng n cc bin ca S l hu hn.

    H qu 1.16. S thnh phn dng ca mt phng n cc bin ti a lm:

    Chng minh. Theo nh l 1.14, cc ct ca A tng ng vi cc thnh phndng ca phng n cc bin x 2 S l c lp tuyn tnh trong Rm. Nhngkhng th c nhiu hn m vct c lp tuyn tnh trong Rm: Do s thnhphn dng ca mt phng n cc bin ti a l m:

    nh l 1.17 (Tng ng gia phng n cc bin v phng n c bnchp nhn c). x l im cc bin ca S D fxjAx D b; x 0g khi v chkhi x l phng n c bn chp nhn c.

    V d 1.20. Tm tt c cc phng n cc bin ca

    z D 4x1 C 3x2 ! max

    Vi cc rng buc

    x1 C x2 C x3 D 4

    5x1 C 3x2 C x4 D 15

    xj 0; j D 1; : : : ; 4

    Gii.

  • 1.8 Phng n c bn chp nhn c 30

    1.8.3 Quan h gia phng n cc bin v phng n ti u

    nh l 1.18. Nu bi ton quy hoch tuyn tnh dng chnh tc c phngn ti u th s c mt phng n cc bin l phng n ti u.

    Nhn xt. Nh nh l 1.18, nu ta chng minh c bi ton quy hochtuyn tnh dng chnh tc c phng n ti u, th n s c phng n ccbin l phng n ti u. Trn y chng ta c th tm c tt c ccphng n cc bin (v s phng n cc bin l hu hn theo h qu). Do trong s cc phng n cc bin va ch ra, ln lt th tng phng nta c phng n ti u.

    Rng buc ca bi ton quy hoch tuyn tnh dng chnh tc Ax D b l mth m phng trnh tuyn tnh n n. nh l 1.12 v 1.14 cho ta mi quanh gia im cc bin ca tp cc phng n chp nhn c S D fxjAx Db; x 0g v s c lp tuyn tnh cc ct ca A:

    nh l 1.19. iu kin cn v bi ton quy hoch tuyn tnh dngchnh tc c phng n ti u l tp cc phng n khng rng v hm mctiu b chn trn (nu l bi ton max) hoc b chn di (nu l bi tonmin).

    V d 1.21. Gii bi ton quy hoch tuyn tnh

    z D 4x1 C 3x2 ! max

    Vi cc rng buc

    x1 C x2 C x3 D 4

    5x1 C 3x2 C x4 D 15

    xj 0; j D 1; : : : ; 4

    Gii. Bi ton quy hoch ny c cc phng n cc bin

  • 1.9 Bi tp chng 1 31

    Nghim c bn Phng n cc bin Gi tr hm mc tiux1 D .3=2I 5=2I 0I 0/ x1 D .3=2I 5=2I 0I 0/

    x2 D .3I 0I 1I 0/ x2 D .3I 0I 1I 0/

    x3 D .4I 0I 0I 5/

    x4 D .0I 5I 1I 0/

    x5 D .0I 4I 0I 3/ x5 D .0I 4I 0I 3/

    x6 D .0I 0I 4I 15/ x6 D .0I 0I 4I 15/

    1.9 Bi tp chng 1

    Bi tp 1.1. Bng phng php hnh hc gii bi ton quy hoch tuyntnh

    z D 4x1 C 3x2 ! min

    Vi cc rng buc8

    0 vi mi j th phng n cc bin hinthi l phng n ti u.

    Bc 2: Chn bin ra c s. Nu

    min

    bi

    aivI aiv > 0

    Dbr

    arv

    th bin xr l bin ra khi c s. Ngc li, nu khng tm c bin rakhi c s th bi ton khng c phng n ti u.

    Bc 3: Lp bng n hnh mi. Ta xc nh bin xv l bin vo vxr l bin ra khi c s. Ta lp bng n hnh mi

    Xc nh phn t trc arv:

    Chia dng cha phn t trc cho phn t trc.

    Cc phn t dng i ct j khc ca bng c tnh

    aij D

    aij aivarj arv

    =arv D

    aij arv arj aiv

    =arv

    V d 2.4. Gii li chi tit bi ton

    z D 4x1 C 3x2 ! max

    Vi cc rng buc

    x1 C x2 C x3 D 4

    5x1 C 3x2 C x4 D 15

    xj 0; j D 1; : : : ; 4

    c minh ha trong v d trn bng thut ton n hnh.

    Gii.

  • 2.1 Phng php n hnh cho bi ton quy hoch dng chun 45

    Nhn xt. Trong cc bng n hnh, ct cha z khng thay i qua cc bclp. Do , n gin ta s khng ghi ct z trong bng n hnh.

    V d 2.5. Gii bi ton quy hoch tuyn tnh

    z D 2x1 3x2 C x3 ! max

    Vi cc rng buc8

    d

    V d 2.12. Gi li bi ton quy hoch tuyn tnh v d 2.11

    z D 2x1 x2 2x3 ! max

    Vi cc rng buc

    x1 C x2 x3 D 1

    x1 C 2x2 C x3 D 2

    xj 0; j D 1; 2; 3

    Gii.

  • 2.3 Bi ton chnh tc khng c sn ma trn n v 56

  • 2.3 Bi ton chnh tc khng c sn ma trn n v 57

    V d 2.13. Gii bi ton quy hoch tuyn tnh

    z D 2x1 C 5x2 ! max

    Vi cc rng buc

    2x1 C 3x2 6

    x1 C 2x2 4

    x1; x2 0

    Gii.

  • 2.4 Bi tp chng 2 58

    2.4 Bi tp chng 2

    Bi tp 2.1. Gii cc bi ton quy hoch tuyn tnh:

    a. z D x1 2x2 C 2x3 ! minVi cc rng buc

    x1 C x2 C 4x4 D 6

    2x2 C x3 C 5x4 D 8

    xj 0; j D 1; : : : ; 4

    p n. Phng n ti u xT D .2I 4I 0I 0/ ; gi tr hm mc tiuz D 6

    b. z D 2x1 C 3x2 C x3 ! maxVi cc rng buc

    8