[math-educare]_bai tap va huong dan cach giai bai toan quy hoach tuyen tinh

7/23/2019 [MATH-EDUCARE]_Bai Tap Va Huong Dan Cach Giai Bai Toan Quy Hoach Tuyen Tinh

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M A T H E D U C A R E . C O M

1

Chng 1BI TON QUY HOCH TUYN TNH

1.1 MT SV DVBI TON QUY HOCH TUYN TNH

V d1.Mt x nghip cn sn xut 3 loi bnh: bnh u xanh, bnh thp cm v bnhdo. Lng nguyn liu ng, u cho mt bnh mi loi, lng dtrnguyn liu, tinli cho mt bnh mi loi c cho trong bng sau:

Nguyn liu Bnh u xanh Bnh thp cm Bnh do Lng dtr

ng 0,04kg 0,06kg 0,05kg 500kg

u 0,07kg 0kg 0,02kg 300kg

Li 3000 2000 2500

Hy lp m hnh bi ton tm slng mi loi bnh cn sn xut sao cho khng bng

vnguyn liu m li t c cao nht.

GiiGi 1 2 3, ,x x x ln lt l sbnh u xanh, bnh thp cm, bnh do cn phi sn xut.

iu kin: 0jx , 1,2,3j = . Khi

1) Tin li thu c l: 1 2 3 1 2 3( ) ( , , ) 3 2 2,5f x f x x x x x x= = + + (ngn).

2) Lng ng c sdng l: 1 2 30,04 0,06 0,05x x x+ + (kg)

khng bng vnguyn liu th: 1 1 10,04 0,06 0,05 500x x x+ + .

3) Lng u c sdng l: 1 30,07 0,02x x+ (kg)

khng bng vnguyn liu th: 1 30,07 0,02 300x x+ .

Vy ta c m hnh bi ton

(1) 1 2 3 1 2 3( ) ( , , ) 3 2 2,5 axf x f x x x x x x m= = + +

(2) 1 1 10,04 0,06 0,05 500x x x+ +

1 30,07 0,02 300x x+

(3) 0jx , 1,2,3j = .

Ta ni y l bi ton quy hoch tuyn tnh 3 n tm max ca hm mc tiu.V d2.Gisyu cu ti thiu mi ngy vcc cht dinh dng m, ng, khongcho mt loi gia sc tng ng l 90g, 130g, 10g. Cho bit hm lng cc cht dinh dngtrn c trong 1g thc n A, B, C v gi mua 1kg thc n mi loi c cho trong bng sau:

Cht dinh dng A B C

m 0,1g 0,2g 0,3g

ng 0,3g 0,4g 0,2gKhong 0,02g 0,01g 0,03g

Gi mua 3000 4000 5000


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2

Hy lp m hnh ton hc ca bi ton xc nh khi lng thc n mi loi phi mua

tng s tin chi cho mua thc n t nht nhng p ng c nhu cu dinh dng mingy.

GiiGi 1 2 3, ,x x x ln lt l khi lng (g) thc n A, B, C cn mua.

iu kin: 0jx , 1,2,3j = . Khi

Tng khi lng cc cht dinh dng c trong thc n cn mua l

m: 1 2 30,1 0,2 0,3x x x+ + (g)

ng: 1 2 30,3 0,4 0,2x x x+ + (g)

Khong: 1 2 30,02 0,01 0,03x x x+ + (g)

p ng c nhu cu dinh dng ti thiu mi ngy th tng khi lng cc cht dinhdng c trong thc n cn mua khng thnhhn cc nhu cu ti thiu mi ngy vcc

cht dinh dng nn ta c cc iu kin:1 2 30,1 0,2 0,3 90x x x+ +

1 2 30,3 0,4 0,2 130x x x+ +

1 2 30,02 0,01 0,03 10x x x+ +

Tng stin phi chi mua sthc n trn l

1 2 33 4 5x x x+ + (ng)

Yu cu bi ton l stin chi cho mua thc n t nht nn ta c iu kin

1 2 33 4 5 minx x x+ +


(1) 1 2 3 1 2 3( ) ( , , ) 3 4 5 inf x f x x x x x x m= = + +

(2) 1 2 30,1 0,2 0,3 90x x x+ +

1 2 30,3 0,4 0,2 130x x x+ +

1 2 30,02 0,01 0,03 10x x x+ +

(3) 0j

x , 1,2,3j = .

V d3.(CHLH 2009) Mt cssn xut gdnh sn xut ba loi sn phm l bn,

ghv t. nh mc sdng lao ng, chi ph sn xut v gi bn mi sn phm mi loic tnh trong bng sau:

Cc yu t Bn Gh T

Lao ng (ngy cng) 2 1 3

Chi ph sn xut (ngn ng) 100 40 250

Gi bn (ngn ng) 260 120 600

Hy lp m hnh ton hc ca bi ton xc nh ssn phm mi loi cn phi sn xut

sao cho khng bng trong sn xut v tng doanh thu t c cao nht, bit rng csc s lao ng tng ng vi 500 ngy cng, s tin dnh cho chi ph sn xut l 40

triu ng v sbn, ghphi theo tl1/6.


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3

GiiGi 1 2 3, ,x x x ln lt l sbn, gh, tcn phi sn xut. Ta c cc iu kin: 1 2 3, , 0x x x .

Tng ngy cng v chi ph dnh sn xut l:

1 2 32 3x x x+ + (ngy cng)

1 2 3100 40 250x x x+ +

(ngn ng)khng bng trong sn xut ta c cc iu kin sau

1 2 32 3 500x x x+ +

1 2 3100 40 250 40000x x x+ +

Theo tlgia sbn v sghta c iu kin sau

1 26x x=

Tng doanh thu theo dkin l

1 2 3260 120 600x x x+ + (ngn ng)

tng doanh thu t c cao nht ta c iu kin

1 2 3260 120 600 axx x x m+ +

Nhvy, m hnh ton hc ca bi ton l

(1) 1 2 3260 120 600 axx x x m+ +

1 2 32 3 500x x x+ +

(2) 1 2 3100 40 250 40000x x x+ +

1 26x x=

(3) 1 2 3, , 0x x x .

1.2 PHN LOI DNG BI TON QUY HOCH TUYN TNH

1.2.1 Dng tng qut ca bi ton quy hoch tuyn tnhBi ton QHTT dngtng qutvi n n l bi ton c dng

(1) 1 1 2 2( ) max(min)n nf x c x c x c x= + + + L

(2) 1 1 2 2 , 1,2, ,i i in n ia x a x a x b i m

+ + + = =

L K

(3)

0

0 , 1,2, ,

jx j n

tuy y

=

K

Trong

(1) l hm mc tiu.

(2) l hrng buc chnh. (3) l rng buc du.

(2) v (3) c gi chung l hrng buc ca bi ton.


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4

Khi

Mi vector 1 2( , , , )nx x x x= K tha (2) v (3) c gi l mt phng n (PA)

ca bi ton.

Mi phng n x tha (1), ngha l ti hm mc tiu t gi tnhnht (lnnht) trn tp cc phng n c gi l mtphng n ti u (PATU) ca biton.

Gii mt bi ton QHTT l i tmmt phng n ti uca n hoc chra rngbi ton v nghim, ngha l bi ton khng c PATU.

1.2.2 Dng chnh tc ca bi ton QHTT


(2) 1 1 2 2 , 1,2, ,i i in n ia x a x a x b i m+ + + = =L K

(3) 0, 1,2, ,jx j n = K Nhn xt.Bi ton QHTTdng chnh tcl bi ton QHTT dng tng qut trong

Cc rng buc chnh u l phng trnh.

Cc n u khng m.V d.Bi ton sau c dng chnh tc

(1) 1 2 3 4( ) 2 4 6 minf x x x x x= +

(2)

1 2 4

1 2 3 4

1 2 3 4

4 12

12 3 3

6

x x x

x x x x

x x x x

+ =

+ + =

=

(3) 0, 1,2,3,4jx j =

1.2.3 Dng chun ca bi ton QHTTBi ton QHTTdng chunl bi ton QHTT dng chnh tc



(3) 0, 1,2, ,j

x j n = K

Trong Cc hstdo u khng m.

Trong ma trn hstdo c m vector ct n v: 1 2, , , me e eK .

1 1

1 0 0

0 1 0, , ,

0 0 1

me e e

= = =

KM M M

Khi : Cc n ng vi cc vector ct n vc gi l cc n cbn. Cthn ng vi

vector ct n v ke l n cbn thk.


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Mt phng n m cc n cbn u bng 0 c gi lphng n cbn.

Mt phng n c bn c m thnh phn dng c gi l khng suy bin.Ngc li mt phng n c bn c t hn m thnh phn dng c gi l suy

bin.V d.Xt bi ton QHTT sau

(1) 1 2 3 4( ) 2 4 6 minf x x x x x= +

(2)

1 4 5

1 3 6

1 2 3 4

12

12 3

6

x x x

x x x

x x x x

+ + =

+ + =

+ =

(3) 0, 1,2,3,4,5,6jx j =

Ta thy bi ton trn c dng chnh tc, hn naCc hstdo u khng m

Ma trn hsrng buc A l1 0 0 1 1 0

12 0 1 0 0 1

1 1 1 1 0 0

A

=

C cha y 3 vector ct n v 1e (ct 5), 2e (ct 6), 3e (ct 2).

Do bi ton c dng chun, trong

n cbn thnht l 5x

n cbn thhai l 6x n cbn thba l 2x

Nhn xt.Trong bi ton trn, khi cho n cbn thk bng hstdo thk, cn cc nkhng c bn bng 0, ngha l cho 2 6 2 1 3 415, 3, 6, 0, 0, 0x x x x x x= = = = = = ta c mt

phng n cbn ca bi ton (0,6,0,0,12,3)x = .

Phng n ny khng suy bin v c 3 thnh phn dng. Ta gi y l phng n c

bn ban u ca bi ton.Ch .Tng qut, trong bi ton QHTT dng chun bt k, khi cho n cbn thk bng

hstdo thk ( 1,2, ,k m= K ), cn cc n khng cbn bng 0, ta c mt phng ncbn ca bi ton. Ta gi y l phng n cbn ban u ca bi ton.

1.3 BIN I DNG BI TON QHTT

1.3.1 Dng tng qut vdng chnh tc

Ta c thbin i bi tondng tng qutvdng chnh tcbng cc bc sauBc 1.Kim tra hrng buc chnh

1)

Nu c rng buc chnh dng 1 1 2 2i i in n ia x a x a x b+ + + L th ta cng vo vtri rng

buc n ph n kx + , ngha l ta thay rng buc 1 1 2 2i i in n ia x a x a x b+ + + L trong bi

ton bng rng buc 1 1 2 2i i in n n k ia x a x a x x b++ + + + =L .


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2)Nu c rng buc chnh dng 1 1 2 2i i in n ia x a x a x b+ + + L th ta tr vo v tri rng


ton bng rng buc 1 1 2 2i i in n n k ia x a x a x x b++ + + =L .

Ch .Cc n phl cc n khng m v hsca cc n ph trong hm mc tiu l 0.

Bc 2.Kim tra iu kin du ca n s1) Nu c n 0jx th ta thc hin php i n s j jx x = vi 0jx .

2)

Nu c n jx c du ty th ta thc hin php i n s j j jx x x = vi

, 0j jx x .

Ch .Khi tm c PATU ca bi ton dng chnh tc ta chcn tnh gi trca cc nban u v bi cc n phth sc PATU ca bi ton dng tng qut cho.V d.Bin i bi ton sau vdng chnh tc

(1) 1 2 3 4( ) 2 4 6 minf x x x x x= +

(2)

1 2 3

1 3

1 2 3

4 6 5 50

7 30

2 3 5 25

x x x

x x

x x x

+

+

+ =

(3) 1 20, 0x x

GiiThm vo bi ton n ph 4 0x bin bt phng trnh 1 2 34 6 5 50x x x + vphng

trnh 1 2 3 44 6 5 50x x x x + + = .Thm vo bi ton n ph 5 0x bin bt phng trnh 1 37 30x x+ vphng trnh

1 3 57 30x x x+ = .

i bin 2 2x x = vi 2 0x .

i bin 3 3 3x x x= vi 3 3, 0x x .

Ta a bi ton vdng chnh tc

(1) 1 2 3 3( ) 3 2 2,5( ) maxf x x x x x = +

(2)

1 2 3 3 4

1 3 3 5

1 2 3 3

4 6 5( ) 50

7 ( ) 30

2 3 5( ) 25

x x x x x

x x x x

x x x x

+ + + = + =

=

(3) 1 2 3 3 4 50, 0, 0, 0, 0, 0x x x x x x

1.3.2 Dng chnh tc vdng chun

Tbi tondng chnh tcta c thxy dng bi tondng chunnhsau

1) Khi gp hstdo 0ib < ta i du hai vca rng buc thi.


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2) Khi ma trn hsrng buc A khng cha ct n vthk l ke , ta a vo n gi

0n kx + v cng thm n gi n kx + vo vtri phng trnh rng buc thk c

phng trnh rng buc mi: 1 1 2 2k k kn n n k k a x a x a x x b++ + + + =L .

3)

Hm mc tiu mrng ( )f x c xy dng thm mc tiu ban u nhsau

i vi bi ton min: ( ) ( ) ( gia)f x f x M an= + . i vi bi ton max: ( ) ( ) ( gia)f x f x M an= .

Trong M l i lng rt ln, ln hn bt k sno cho trc.

V d1.Bin i bi ton QHTT sau vdng chun(1) 1 2 3 4( ) 3 2 2 minf x x x x x= + + +

(2)

1 2 3

1 3 4

1 2 3

4 6 5 50

7 0

2 3 5 25

x x x

x x x

x x x

+ =

+ + =

+ =

(3) 0, 1, ,4jx j = K

GiiBi ton trn c dng chnh tc, trong vphi ca phng trnh rng buc thba l

-25 < 0. i du hai vca phng trnh ny ta c 1 2 32 3 5 25x x x + =

V (2) trthnh

1 2 3

1 3 4

1 2 3

4 6 5 50

7 0

2 3 5 25

x x x

x x x

x x x

+ =

+ + =

+ =

Ma trn hsca rng buc l

4 6 5 0 1 0

7 0 1 1 0 0

2 3 5 0 0 1

A

=

V A cn thiu 2 vector ct n vl 1e v 3e nn bi ton cha c dng chun.

Thm vo bi ton hai n gi 5 6, 0x x v xy dng bi ton mrng c dng chun nh

sau

(1) 1 2 3 4 5 6( ) 3 2 2 minf x x x x x Mx Mx= + + + + +

(2)

1 2 3 5

1 3 4

1 2 3 6

4 6 5 50

7 0

2 3 5 25

x x x x

x x x

x x x x

+ + =

+ + =

+ + =

(3) 0, 1, ,6jx j = K

V d2.Bin i bi ton QHTT sau vdng chun

(1) 1 2 3 4( ) 3 2 2 maxf x x x x x= + + +


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(2)

1 2 3

1 3 4

1 2 3

4 6 5 50

7 0

2 3 5 25

x x x

x x x

x x x

+ =

+ + =

+ =

(3) 0, 1, ,4jx j = K

Ta xy dng bi ton mrng dng chun nhsau(1) 1 2 3 4 5 6( ) 3 2 2 maxf x x x x x Mx Mx= + + +

(2)

1 2 3 5

1 3 4

1 2 3 6

4 6 5 50

7 0

2 3 5 25

x x x x

x x x

x x x x

+ + =

+ + =

+ + =

(3) 0, 1, ,6jx j = K

Ch .

n ph: Tng qut chuyn thnh chnh tc

n gi: Chnh tc chuyn thnh chun


(2)

1 3

3 4

1 2 3

9 15 50

6 2 120

3 5 45

x x

x x

x x x

+

+ =

+

(3) 0, 1, , 4jx j = K

GiiThm vo bi ton 2 n ph 5 6, 0x x ta c bi ton c dng chnh tc nhsau

(1) 1 2 3 4( ) 3 2 2 minf x x x x x= + + +

(2)

1 3 5

3 4

1 2 3 6

9 15 50

6 2 120

3 5 45

x x x

x x

x x x x

+ + =

+ =

+ =

(3) 0, 1, ,6jx j = K .

Bi ton trn cha c dng chun.Ta thy cc vphi ca hai phng trnh rng buc th2 v 3 u m nn bng cch i

du hai vca cc phng trnh ny ta c

(2)

1 3 5

3 4

1 2 3 6

9 15 50

6 2 120

3 5 45

x x x

x x

x x x x

+ + =

=

+ + =

Ma trn hsrng buc l

A =

0 9 15 0 1 0 0

0 0 6 2 0 0 11 3 5 0 0 1 0


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V A cn thiu mt vector ct l 2e nn bi ton cha c dng chun.

Thm vo rng buc chnh thhai n gi 7 0x ta c bi ton dng chun nhsau

(1) 1 2 3 4 7( ) 3 2 2 minf x x x x x Mx= + + + +

(2)

1 3 5

3 4 7

1 2 3 6

9 15 50

6 2 1203 5 45

x x x

x x xx x x x

+ + =

+ = + + =

(3) 0, 1, ,7jx j = K .

Ch .Quan hgia bi ton xut pht v bi ton mrngMi quan hgia bi ton xut pht (A) v bi ton mrng (B) nhsau

B v nghim suy ra A v nghim

B c nghim c hai trng hp:1) Nu mi n gica PATU bng 0 th bn gita c PATU ca A

2)

Nu c t nht mt n gi> 0 suy ra A khng c PATU

Chng 2PHNG PHP N HNH

2.1 PHNG PHP N HNH GII BI TON QUY HOCH TUYN TNHDNG CHUN2.1.1

Thut ton gii bi ton max

Bc lp thnht (bng n hnh thnht)1)

Lp bng n hnh xut phtVbng n hnh v ghi vo cc thnh phn sau ca bi ton dng chun

Dng 1. Ghi cc n ca bi ton (kcn ph)

Dng 2. Ghi cc hsca cc n trong hm mc tiu

Ct 2. Ghi cc n cbn ca bi ton theo thttn cbn thnht n n c

bn cui cng, ta gi ct ny lct n cbn.

Ct 1: Ghi tng ng cc hsca cc n cbn trong hm mc tiu, ta gi ctny lct hscbn.

Ct 3. Ghi cc shng tdo ca hrng buc chnh theo thtttrn xung di,ta gi ct ny lct phng n.

Ct 4. Ghi ma trn iu kin A ca bi ton.

Tnhhsc lngj ca cc n ( 1,2, , )jx j n= K v ghi tng ng vo dng di ct

4, vi j c tnh theo cng thc sau:

(cot1) ( ) ( )j j jA hsx = ( jhsx : hsca n jx trong hm mc tiu).

Ch .Nu jx l n cbn th 0j = .Tnh trs 0 (cot1) (cot3)f = v ghi di ct 3.

2)Xc nh phng n cbn xut pht


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Vi bng n hnh va lp c th phng n cbn xut pht 0x ca bi ton cxc nh nhsau: Cho cc n ct 2 nhn gi trtng ng ct 3, cc n cn li nhn

gi tr0. Trsca hm mc tiu ti phng n cbn xut pht 0x l 0 0( )f x f= .

3)

nh gi tnh ti u ca phng n cbn xut pht

Du hiu ti u. Nu h sc lng ca cc n u khng m, 0,j j thphng n cbn xut pht 0x l phng n ti u ca bi ton. Thut ton kt thc

vi kt lun: Bi ton c PATU l 0x v GTTU l 0( )f x .

Du hiu ca bi ton khng c PATU.Nu c n khng cbn kx c hsc

lng m v ct iu kin kA ca n c cc thnh phn u khng dng, 0k <

v 0;ika i th bi ton khng c phng n ti u. Thut ton kt thc vi kt lun:

Bi ton khng c PATU.

Nu khng xy ra chai trng hp trn th thut ton tip tc trong bc lp thhaiBc lp thhai(Bng n hnh thhai)

1) Tm n a voTrong tt c cc 0j < ta chn 0v < nh nht (ta nh du * cho 0v < nh nht

trong bng). Khi , vx l n m ta sa vo hn cbn. Ct vA c gi l ct

chyu.

2) Tm n a raThc hin php chia ln lt cc sca ct phng n cho cc sdng ca ct ch

yu v ghi cc thng s i vo ct cui cng.Xc nh min{ }r i = (Ta nh du * cho r nh nht trong bng). Khi rx l n

m ta a ra khi hn cbn. Dng c cha rx c gi ldng chyu. Sdng

nm trn dng chyu v ct chyu c gi lhschyu.Ch .Nu ct chyu chc mt sdng th sdng l hschyu, dng csdng l dng chyu, n nm trn dng chyu l n c a ra.

3)Lp bng n hnh thhai

Ct 2: Thay n a ra bng n a vo, cc n cbn cn li ginguyn. Dng c

n a vo gi ldng chun. Ct 1: Thay hsca n a ra bng hsca n a vo, cc hsca cc n c

bn cn li ginguyn.Cc thnh phn cn li c xc nh theo dng nhsau

Dng chun = Dng chyu chia cho hschyu.

Dng thi = Dng thi (c) aiv.dng chun. (aiv: snm trn giao ca dng i vct chyu).

Cc hsc lng v trsca hm mc tiu trong bng thhai c tnh v ghi

nhbng thnht.4)

Xc nh v nh gi phng n cbn thhai(nhbc lp thnht)


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2.1.2 Thut ton gii bi ton min

Gii tng tbi ton max vi ch sau

iu kin ti u: 0,j j

iu kin khng c PATU: 0k > v 0,ika i

n c chn a vo: n ng vi 0k > ln nht.

V d1.Gii bi ton QHTT sau

(1) 1 2 3 4 5( ) 2 5 4 6 m axf x x x x x x= +

(2)

1 2 4 5

2 3 4 5

2 5 6

6 2 9 32

2 3 30

3 36

x x x x

x x x x

x x x

+ =

+ + + =

+ + =

(3) 0, 1, ,6jx j = K .

GiiBi ton trn c dng chnh tc vi cc vphi ca cc phng trnh rng buc trong (2)

u khng m.Ma trn hsca rng buc

1 6 0 2 9 0

0 2 1 1 0 0

0 3 0 0 0 1

A

=

V A cha 3 ct n v 1e (ct 1), 2e (ct 3), 3e (ct 6) nn bi ton c dng chun

trong : n cbn thnht: 1x

n cbn thhai: 3x

n cbn thba: 6x

Ta gii bi ton bng phng php n hnh

0 2.32 4.30 184f = + = .

1 3 6 0 = = =

2 2.6 4.2 0.3 5 25 = + + + =

4 2.( 2) 4.1 0.0 1 1 = + + + =

5 2.( 9) 4.3 0.1 6 0 = + + + =

x1

2

x2

-5

x3

4

x4

-1

x5

-6

x6

0i

2

4

0

x1

x3

x6

32

30

36

1

0

0

6

2

3

0

1

0

-2

1

0

-9

(3)

1

0

0

1

184 0 -25 0 1 0 0


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Trong bng trn ta thy 0, 1, ,6j j = K nn bi ton c PATU l

0 (32,0,30,0,0,36)x = v 0 184f = .


(1) 1 2 3 4 5 6( ) 6 3 minf x x x x x x x= + + + +

(2)

1 2 4 6

1 3 6

1 4 5 6

15

2 2 9

4 2 3 2

x x x x

x x x

x x x x

+ + = + =

+ + =

(3) 0, 1, ,6j

x j = K .

GiiBi ton trn c dng chnh tc vi vphi ca phng trnh rng buc chnh thhai l 9.

i du hai vca phng trnh ny, ta a vbi ton sau

(1) 1 2 3 4 5 6( ) 6 3 7 minf x x x x x x x= + + + +

(2)

1 2 4 6

1 3 6

1 4 5 6

15

2 2 9

4 2 3 2

x x x x

x x x

x x x x

+ + =

+ =

+ + =

(3) 0, 1, ,6jx j = K .

Ma trn hsca rng buc

1 1 0 1 0 1

2 0 1 0 0 2

4 0 0 2 1 3

A

=


trong :

n cbn thnht: 2x

n cbn thhai: 3x

n cbn thba: 5x


x1

6

x2

1

x3

1

x4

3

x5

1

x6

-7i

1

1

1

x2

x3

x5

15

9

2

-1

-2

4

1

0

0

0

1

0

-1

0

2

0

0

1

(1)

-2

-3

15

26 -5 0 0 -2 0 3*

-7

1

1

x6

x3

x5

15

39

47

-1

-4

1

1

2

3

0

1

0

-1

-2

-1

0

0

1

1

0

0-19 -5 0 0 -2 0 0


13/58


13

Trong bng 1 ta thy tn ti 6 3 0 = > v trn ct tng ng c 13 1 0a = > ( 23 2a = v

23 3a = ) nn ta chn n a ra l 6x , n a vo l 2x , hschyu l 13 1a = . Sau

tnh cc dng bng 2 bng cng thc saudc= dcy, d2= d2+ 2dc, d3= d3+ 3dc.

Trong bng 2 ta thy tn ti 4 1 0 = >

m 4 0, 1,2,3ia i =

nn bi ton min ang xt vnghim.V d3.Gii bi ton QHTT sau

(1) 1 2 3 4 5( ) 2 6 4 2 3 m axf x x x x x x= + + +

(2)

1 2 3

2 3 4

2 5

2 4 52

4 2 60

3 36

x x x

x x x

x x

+ + =

+ + =

+ =

(3) 0, 1, ,5jx j = K .

GiiBi ton trn c dng chnh tc vi v phi ca phng trnh rng buc trong (2) ukhng m.

Ma trn hsrng buc l

1 2 4 0 0

0 4 2 1 0

0 3 0 0 1

A

=


trong :

n cbn thnht: 1x

n cbn thhai: 4x

n cbn thba: 5x


x1-2

x26

x34

x4-2

x53 i

-2

-2

3

x1

x4

x5

52

60

36

1

0

0

2

4

3

(4)

2

0

0

1

0

0

0

1

52/4*

60/2

-116 0 -9 -16* 0 0

4

-2

3

x3

x4

x5

13

34

36

1/4

-1/2

0

1/2

(3)

3

1

0

0

0

1

0

0

0

1

13.2

34/3*

36/3

92 4 -1* 0 0 0

4

6

3

x3

x2

x5

22/3

34/3

2

1/3

-1/6

1/2

0

1

0

1

0

0

-1/6

1/3

-1

0

0

1

310/3 23/6 0 0 1/3 0


14/58


14

Trong bng I ta thy tn ti cc 0j < : 2 39, 16 = = v trn mi ct tng ng c h

sdng. Ta chn 3 16 = m nhnht v n a vo l 3x , khi trn ct tng ng

c cc h sdng l 13 234, 2a a= = nn ta lp cc t s 1 252 / 4, 60 / 2 = = . Ta chn

1 5 2 / 4 = nhnht v n a ra l 1x , hschyu l 13 4a = . Sau bin i bng I

bng cc php bin i sau:dc= dcy/4 , d2= d2- 2dc, d3= d3.

Bin i tng tcho bng II.

Trong bng III ta thy 0, 1,2, ,5j j = K nn bi ton ang xt c PATU l

0 (0,34 / 3,22 / 3,0,2)x = vi 0( ) 310 / 3f x = .

2.2 PHNG PHP N HNH MRNG GII BI TON QUY HOCHTUYN TNH DNG CHNH TC

Thut ton n hnh mrng gii bi ton QHTT dng chnh tc tng tnhthut tonn hnh gii bi ton QHTT dng chun nhng c mt slu nhsau

1)Do hm mc tiu m rng l ( ) ( ) ( )f x f x angia= + i vi bi ton min v( ) ( ) ( )f x f x angia= i vi bi ton max, nn trong bng n hnh ct hsc

thc cc hsphthuc M. Khi dng cui cc hssc dng aM b+ , do ngi ta thng chia dng cui thnh hai dng nh: Dng trn ghi a v dng di ghi b.

2)V M l mt i lng dng rt ln, nn khi so snh cc shng aM + b v cM + d tac quy tc sau

a caM b cM d

b d

=+ = +

=

0

00

0

a

baM b

a

b

>

+ >

=

>

,a c

b daM b cM d

a c

b d

>

+ > +

=

>

3)Trong bng n hnh u tin cc n giu c trong n cbn. Mi khi mt n giba ra khi hn cbn th khng bao gi ta a n gi tr li na, v vy trong

bng n hnh ta c thbi cc ct ng vi cc n gi.V d1.Gii bi ton QHTT sau

f(x) = 6x1+ 3x2+ 2x3- 3x4 minx1+ x2+ x3- 2x4= 4

- x1+ x4 10


15/58


15

2x2+ x3 2x4= 12

xj 0 (j = 1,2,3,4)

Giia bi ton vdng chun:

f(x) = 6x1+ 3x2+ 2x3- 3x4+ Mx6+ Mx7 min

x1+ x2+ x3- 2x4+ x6= 4

- x1+ x4+ x5= 10

2x2+ x3 2x4+ x7= 12

xj 0 (j = 1,2,3,4,5,6,7)

Gii bi ton mrng bng phng php n hnh

PATU: x = (0, 8, 0, 2), f(x) = 18.

V d2.Gii bi ton quy hoch tuyn tnh sau:

f(x) = - 2x1- x2+ x3+ x4 maxx1+ x2+ 2x3- x4= 2

- x2- 7x3+ 3x4 3

- 3x3+ 2x4 7

xj 0, j = 1,2,3,4.


f(x) = - 2x1- x2+ x3+ x4 max

x1+ x

2+ 2x

3- x

4 = 2

- x2- 7x3+ 3x4+ x5= 3

- 3x3+ 2x4+ x6= 7

x1

6

x2

3

x3

2

x4

-3

x5

0

M0

M

x6x5

x7

410

12

1-1

0

(1)0

2

10

1

-21

-2

01

0

-6 -3 -2 3 0

1 3* 2 -4 0

3

0

M

x2

x5

x7

4

10

4

1

-1

-2

1

0

0

1

0

-1

-2

1

(2)

0

1

0

-3 0 1 -3 0

-2 0 -1 2* 0

3

0

-3

x2

x5

x4

8

8

2

-1

0

-1

1

0

0

0

1/2

-1/2

0

0

1

0

1

0

18 -6 0 -1/2 0 0


16/58


16

xj 0, j = 1,2,3,4,5,6.


.

PATU: x = (0, 0, 11, 20), f(x) = 31.

Chng 3

BI TON I NGU

3.1 nh ngha

Cho (P) l bi ton QHTT c dng chnh tc nhsau

1 1 2 2( ) max(min)n nf x c x c x c x= + + + L

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

0, 1,2, , .

n n

n n

m m mn n m

j

a x a x a x b

a x a x a x b

a x a x a x b

x j n

+ + + =

+ + + =

+ + + =

=

L

L

KKK

L

K

Tbi ton (P) ta lp c bi ton QHTT (D) nhsau v ta gi bi ton (D) l bi ton

i ngu ca bi ton (P)

1 1 2 2( ) min(max)m mf y b y b y b y= + + + L

11 1 21 2 1 1

12 1 22 2 1 2

1 1 2 2

( )

( )

( )

m m

m m

n n mn m n

a y a y a y c

a y a y a y c

a y a y a y c

+ + +

+ + +

+ + +

L

L

KKK

L

x1

-2

x2

-1

x3

1

x4

1

x5

0

x6

0-2

0

0

x1

x5

x6

2

3

7

1

0

0

1

-1

0

(2)

-7

-3

-1

3

2

0

1

0

0

0

1

0 -1 -5* 1 0 0

1

0

0

x3

x5

x6

1

9

8

1/2

7/2

3/2

1/2

5/2

3/2

1

0

0

-1/2

-1/2

(1/2)

0

1

0

0

0

1

5/2 3/2 0 -3/2* 0 01

0

1

x3

x4

x5

11

20

20

2

5

3

2

4

3

1

0

0

0

0

1

0

1

0

1

1

2

31 7 6 0 0 0 3


17/58


17

Ch .Bi ton (D) c lp tbi ton (P) theo nguyn tc sau1. Sn ca bi ton (D) bng srng buc chnh ca bi ton (P) v srng buc chnhca bi ton (D) bng sn ca bi ton (P).

2. Hsca n iy trong hm mc tiu ca bi ton (D) l shng tdo ib trong hrng

buc chnh ca bi ton (P).3.

Cc hsca cc n v hstdo trong rng buc chnh thj ca bi ton (D) l cc

hstng ng ca n jx trong hrng buc chnh v hm mc tiu ca bi ton (P).

4.

Nu (P) l bi ton max th (D) l bi ton min v hrng buc chnh ca bi ton (D)

l hbt phng trnh vi du . Nu (P) l bi ton min th (D) l bi ton max v hrng buc chnh ca bi ton (D) l hbt phng trnh vi du .

5. Cc n ca bi ton (D) u c du ty .

3.2 Cch lp bi ton i nguBi ton i ngu c lp trc tip theo quy tc sau, gi l quy tc i ngu

(P) (D)

1 1 2 2( ) maxn nf x c x c x c x= + + + L 1 1 2 2( ) minm mf y b y b y b y= + + + L

1 1 2 2i i in n ia x a x a x b

+ + + =

L

0

0iy

tuyy

0

0jxtuyy

1 1 2 2j j mj m ja y a y a y c

+ + +

=

L

V d.Tm bi ton i ngu ca bi ton saua)

(1)1 2 3 4

( ) 3 2 5 minf x x x x x= + +

(2)

1 2 3 4

1 3 4

1 2 3

4 6 5 5 50

7 30

2 3 5 25

x x x x

x x x

x x x

+

+ + =

+

(3) 1 20, 0x x .

b)

(1) 1 2 3( ) 2 8 m axf x x x x= +

(2)

1 2 3

1 2 3

1 2 3

7 4 2 28

3 3 10

2 3 15

x x x

x x x

x x x

+ +

+ =

+

(3) 1 2 3, , 0x x x .

Giia) Bi ton i ngu l


18/58


18

(1) 1 2 3( ) 50 30 25 maxg y y y y= +

(2)

1 2 3

1 3

1 2 3

1 2

4 7 2 3

6 3 2

5 5 5

5 1

y y y

y y

y y y

y y

+ +

+

+ = + =

(3) 1 0y , 2y ty , 3 0y .

b) Bi ton i ngu l

(1) 1 2 3( ) 28 10 15 ming y y y y= + +

(2)

1 2 3

1 2 3

1 2 3

7 3 2 2

4 3 1

2 3 8

y y y

y y y

y y y

+ +

+

+

(3) 1 0y , 3 0y .

3.3 Cp rng buc i ngu

Trong mt cp rng buc i ngu (P) v (D) nh trong nh ngha th ta c n cp rngbuc i ngu nhsau:

Trng hp 1. 1 1 2 2( ) maxn nf x c x c x c x= + + + L

1 1 2 20 , 1,2, ,j j j mj m jx a y a y a y c j n + + + =L K

Trng hp 2. 1 1 2 2( ) minn nf x c x c x c x= + + + L

1 1 2 2

0 , 1,2, ,j j j mj m j

x a y a y a y c j n + + + =L K

V d.Tm bi ton i ngu v chra cc cp rng buc i ngu ca cc bi ton QHTTsau:

a)

(1)1 2 3

( ) 2 3 minf x x x x= + +

(2)

1 2 3

1 2 3

1 2 3

1 2 3

2 3 2

4 3

2 6

3 5 7 8

x x x

x x x

x x x

x x x

+ =

+

+

(3) 1 2 30, 0, 0x x x .

b)

(1) 1 2 3 4( ) 2 3 4 m axf x x x x x= +

(2)

1 2 3 4

1 2 3 4

1 2 3 4

5 6 7 8 9

10 11 12 13 14

15 16 17 18 19

x x x x

x x x x

x x x x

+ + +

+ =

+ +

(3) 1 2 3, , 0x x x .

Giia) Bi ton i ngu


19/58


19

(1) 1 2 3 4( ) 2 3 6 8 m axg y y y y y= + +

(2)

1 2 3 4

1 2 3 4

1 2 3 4

2 3 1

3 7 2

4 2 7 3

y y y y

y y y y

y y y y

+ + +

+ +

+ + +

(3) 2 4, 0y y , 3 0y .Hrng buc chnh ca bi ton (P) c 3 bt phng trnh v bi ton (P) c 3 iu kin vdu ca n snn cp bi ton i ngu (P) v (D) c 6 cp rng buc i ngu

1 2 3 2

1 2 3 3

1 2 3 4

1 1 2 3 4

2 1 2 3 4

3 1 2 3 4

4 3 0

2 6 0

3 5 7 8 0

0 2 3 1

0 3 5 2

0 4 2 7 3

x x x y

x x x y

x x x y

x y y y y

x y y y y

x y y y y

+

+

+ + +

+ +

+ + +

b) Bi ton i ngu

(1) 1 2 3( ) 9 14 19 ming y y y y= +

(2)

1 2 3

1 2 3

1 2 3

1 2 3

5 10 15 1

6 11 16 2

7 12 17 3

8 13 18 4

y y y

y y y

y y y

y y y

+

+

+ + =

(3) 10y

, 2y

ty , 3 0y

.Hrng buc chnh ca bi ton (P) c hai bt phng trnh v bi ton (P) c ba iu kinvdu ca n snn cp bi ton i ngu (P) v (D) c 5 cp rng buc i ngu sau:

1 2 3 4 1

1 2 3 4 3

1 1 2 3

2 1 2 3

3 1 2 3

5 6 7 8 0

15 16 17 18 19 0

0 5 10 15 1

0 6 11 16 2

0 7 12 17 3

x x x x y

x x x x y

x y y y

x y y y

x y y y

+ + +

+ +

+

+

+

3.4 nh l i ngu

nh l lch b yu.iu kin cn v phng n 0x ca bi ton (P) v

phng n 0y ca bi ton (D) u l phng n ti u l trong cc cp rng buc i

ngu ca bi ton : Nu mt rng buc tha mn phng n vi du bt ng thc thcsth rng buc cn li phi tha mn phng n vi du bng.

ng dng.Nhnh l lch b yu, khi ta bit c mt phng n ti u ca mttrong hai bi ton ca cp bi ton i ngu th ta c thtm c tp phng n ti u ca

bi ton cn li. ng dng ny thng c sdng trong vic gii quyt cc vn cabi ton QHTT.

V d1. Cho bi ton quy hoch tuyn tnh sau


20/58


20

f(x) = x1+ 2x2+ 3x3+ 3x4 max

2x1+ x2+ x3+ 2x4 20

x1+ 2x2+ 3x3+ 4x4= 18

2x1+ x2+ 2x3+ x4 16

xj 0 (j = 1,2,3,4)

a.

Gii bi ton trn

b. Hy lp bi ton i ngu ca bi ton trn v gii bi ton i ngu .

Giia. a bi ton vdng chun:

f(x) = x1+ 2x2+ 3x3+ 3x4 Mx7 Mx8 max

2x1+ x2+ x3+ 2x4+ x5= 20

x1+ 2x2+ 3x3+ 4x4+ x7= 18

2x1+ x2+ 2x3+ x4 x6+ x8= 16

xj 0 (j = 1,2,3,4,5,6,7,8)Gii bi ton mrng bng phng php n hnh

.

PATU: x = (3, 0, 5, 0), f(x) = 18.

b. Bi ton i ngu

g(y) = 20y1+ 18y2+ 16y3 min

2y1+ y2+ 2y3 1

y1+ 2y2+ y3 2

y1+ 3y2+ 2y3 3

2y1+ 4y2+ y3 3

y1 0, y3 0.

x1

1

x2

2

x3

3

x4

3

x5

0

x6

0

0

-M

-M

x5

x7

x8

20

18

16

2

1

2

1

2

1

1

(3)

2

2

4

1

1

0

0

0

0

-1

-1 -2 -3 -3 0 0

-3 -3 -5* -5 0 1

0

3

-M

x5

x3

x8

14

6

4

5/3

1/3

(4/3)

1/3

2/3

-1/3

0

1

0

2/3

4/3

-5/3

1

0

0

0

0

-1

0 0 0 1 0 0

-4/3* 1/3 0 5/3 0 1

0

3

1

x5

x3

x1

9

5

3

0

0

1

3/4

3/4

-1/4

0

1

0

11/4

7/4

-5/4

1

0

0

5/4

1/4

-3/4

18 0 0 0 1 0 0


21/58


21

Theo nh l lch b yu, ta c hphng trnh ti u sau

=

=++

=++

0

323

122

1

321

321

y

yyy

yyy

Gii hphng trnh ta c PATU: y = (0, 1, 0) v g(y) = 18.V d2.Cho bi ton quy hoch tuyn tnh sau

f(x) = 6x1+ 3x2+ 2x3- 3x4 min

x1+ x2+ x3- 2x4= 4

- x1+ x4 10

2x2+ x3 2x4= 12

xj 0 (j = 1,2,3,4)

a.

Gii bi ton trnb. Lp bi ton i ngu ca bi ton trn v gii bi ton i ngu .

Giia. a bi ton vdng chun

f(x) = 6x1+ 3x2+ 2x3- 3x4+ Mx6+ Mx7 min

x1+ x2+ x3- 2x4+ x6= 4

- x1+ x4+ x5= 10

2x2+ x3 2x4+ x7= 12

xj 0 (j = 1,2,3,4,5,6,7)


x1

6

x2

3

x3

2

x4

-3

x5

0

M

0

M

x6

x5

x7

4

10

12

1

-1

0

(1)

0

2

1

0

1

-2

1

-2

0

1

0

-6 -3 -2 3 0

1 3* 2 -4 0

3

0

M

x2

x5

x7

4

10

4

1

-1

-2

1

0

0

1

0

-1

-2

1

(2)

0

1

0

-3 0 1 -3 0

-2 0 -1 2* 0

3

0

-3

x2

x5

x4

8

8

2

-1

0

-1

1

0

0

0

1/2

-1/2

0

0

1

0

1

0

18 -6 0 -1/2 0 0


22/58


22

PATU: x = (0, 8, 0, 2), f(x) = 18.

b. Bi ton i ngu

g(y) = 4y1+ 10y2+ 12y3 max

y1- y2 6

y1+ 2y3 3

y1+ y3 2

- 2y1+ y2 2y3 - 3

y2 0.

Theo nh l lch b yu, ta c hphng trnh ti u sau:

1 3

2

1 2 3

2 3

0

2 2 3

y y

y

y y y

+ =

= + =

Gii hphng trnh ta c PATU: y = (0, 0, 3/2) v g(y) = 18.V d3(CHLH 2009). Cho bi ton quy hoch tuyn tnh sau

1 2 3 4 5

1 2 3 4

2 3 4 5

1 2 3 4 5

3 4 5 ax

2 3 2 30

23

3 2 4 10

0; 1, 2,3, 4,5.j

x x x x x m

x x x x

x x x x

x x x x x

x j

+ +

+ + =

+ =

+ + +

=

a.

Hy gii bi ton trn bng phng php n hnh.b.

Hy lp bi ton i ngu ca bi ton trn v tm mt phng n ti u ca biton i ngu .

Giia.

Thm vo bi ton n ph 6x ri i du rng buc chnh thba, ta c bi ton

phsau:

1 2 3 4 5

1 2 3 4

2 3 4 5

1 2 3 4 5 6

3 4 5 ax

2 3 2 30

233 2 4 10

0; 1, 2, 3, 4, 5, 6.j

x x x x x m

x x x x

x x x xx x x x x x

x j

+ +

+ + =

+ = + + =

=

Ma trn iu kin

2 1 3 2 0 0 1 0

0 1 1 1 1 0 0 1

3 2 1 1 4 1 0 0

A

=

Thm vo bi ton hai n gi7 8

,x x ta c bi ton mrng sau


23/58


23

1 2 3 4 5 7 8

1 2 3 4 7

2 3 4 5 8

1 2 3 4 5 6

3 4 5 max

2 3 2 30

23

3 2 4 10

0; 1,2,3,4,5,6,7,8.j

x x x x x Mx Mx

x x x x x

x x x x x

x x x x x x

x j

+ +

+ + + =

+ + =

+ + =

=


x11

x2-3

x3-4

x41

x55

x60

i

-M

-M

0

x7

x8

x6

30

23

10

2

0

-3

1

1

2

-3

-1

-1

(2)

1

-1

0

-1

-4

0

0

1

15*

23

0 -1 3 4 -1 -5 0

-53 -2 -2 4 -3* 1 0

1

-M

0

x4

x8

x6

15

8

25

1

-1

-2

1/2

5/2

-3/2

(1/2)

-5/2

1

0

0

0

-1

-4

0

0

1

15 0 7/2 5/2* 0 -5 0

-8 1 -1/2 -1/2 0 1 0

1

-4

0

x4

x3

x6

39

16

65

-2

-2

-7

2

1

5

0

1

0

1

0

0

-3

-2

-9

0

0

1

-25 5 1 0 0 0 0

Trong bng n hnh thba ta thy hsc lng ca cc n u khng m nn biton mrng c nghim l:

0

0(0, 0,16, 39, 0, 65, 0, 0); 25x f= = .

Ta thy trong PATU ca bi ton mrng cc n giu nhn gi tr0 nn bi ton

cho gii c v c nghim nhsau:0

0(0, 0,16, 39, 0, 65, 0, 0); 25x f= =

b) Bi ton i ngu vi bi ton cho l:1 2 3

1 3

1 2 3

1 2 3

1 2 3

2 3

3

30 23 10 min

2 3 1

2 3

3 4

2 1

4 5

0

y y y

y y

y y y

y y y

y y y

y y

y

+

+

+

+

+ +

+

Do bi ton cho c PATU l 0 (0,0,16,39,0,65,0,0)x = nn ta c hphng trnh ti u

sau:


24/58


24

3 1

1 2 3 2

31 2 3

0 3

3 4 5

02 1

y y

y y y y

yy y y

= =

+ = =

=+ + =

Vy bi ton i ngu c mt PATU l 0 (3, 5,0)y = v GTTU l: 0( ) 25g y = .

BI TPPhn I1.1nui mt loi gia sc ngi ta sdng 3 loi thc n A1, A2, A3. Tl(%) cc cht dinh

dng D1, D2c trong cc loi thc n A1, A2, A3v gi 1kg mi loi nhsau:

Chtdinh dng

Loi thc nA1 A2 A3

D1

D2

30

20

20

20

20

30Gi mua 8000 6000 4000

Yu cu trong khu phn thc n ca loi gia sc ny l: cht dinh dng D1phi c t nht l70g v nhiu nht l 100g, cht dinh dng D2phi c t nht l 50g v nhiu nht l 80g. Hy lpm hnh ton hc ca bi ton xc nh khi lng thc n mi loi cn mua sao cho tng chi phthp nht v bo m cht lng theo yu cu.1.2 C hai loi thc n I v II cha 3 loi vitamin A, B, C. Hm lng vitamin trong min vthc n nhsau:

Loi thc n VitaminA B C

III

2

4

3

1

4

5

Gi mt n vthc n thI l 3, v II l 7. Mt khu phn n phi c ti thiu 5n vA, 4 n vB v 8 n vC. Tm mt cch n tt nht (t tin nht v dinhdng). Hy lp m hnh ton hc ca bi ton.1.3 Mt x nghip c khoch sn xut ba loi sn phm A1, A2, A3t3 loi nguyn liuN1, N2, N3c trlng tng ng l 50kg, 70kg v 100kg. nh mc tiu hao nguyn liu(kg/SP) v li nhun (ngn ng/SP) khi sn xut mt sn phm c cho trong bng sau

Nguyn liu Sn phmA1 A2 A3

N1N2

N3

0.20.1

0.1

0.10.2

0.3

0.10.1

0.0

Li nhun 8000 6000 4000Hy lp m hnh ton hc ca bi ton lp khoch sn xut ti u bit rng lng

sn phm A3chc thtiu thc ti a 400 sn phm.

1.4 nui mt loi gia sc trong 24h cn c khi lng ti thiu cc cht: Protit, Gluxit,khong tng ng l: 90, 130, 20 gram. Tlphn trm theo khi lng cc cht trn ctrong cc loi thc n A, B, C v gi mua 1kg thc n mi loi nhsau:

Cht Loi thc n


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25

dinh dng A B CProtit

Gluxit

Khong

10

30

2

20

40

1

30

20

3

Gi mua 3000 4000 5000

Hy lp m hnh ton hc ca bi ton xc nh khi lng thc n mi loi cnmua sao cho tng chi ph thp nht v bo m cht lng theo yu cu.2.5 C hai loi sn phm A, B c gia cng trn 3 my I, II, III. Thi gian gia cng miloi sn phm trn mi my cho bi bng:

LoiSP

My

I II III

AB

4

2

3

1

2

4

Thi gian cho php ca mi my I, II, II ln lt l 100, 300, 50 gi. Mt n vsn phm A li 6000 , B li 4000 .Vy cn phi sn xut bao nhiu sn phm mi loi li ti a. Hy lp m hnh

ton hc ca bi ton.1.6 Trong mt chu k sn xut, nh my sdng hai loi vt liu V1, V2sn xut 3 loisn phm S1, S2, S3. Lng vt liu Vidng sn xut mt n vsn phm Sjv gi bnmt n vsn phm Sjcho bi bng sau:

VL SP

S1 S2 S3

V1V2

42

26

53

Gi bn 12000 8000 14000

Bit rngsn vvt liu V1, V2nh my c ln lt l 10000 v 14000. Yu culp khoch sn xut ca nh my, xc nh slng sn phm mi loi cn sn xut saocho tng thu nhp ln nht.1.7 Gisyu cu ti thiu mi ngy vcc cht dinh dng m, ng, bo cho mtloi gia sc tng ng l 50g, 80g v 20g. Cho bit hm lng cc cht dinh dng trn ctrong 1g thc n A, B, C v gi mua 1 kg thc n mi loi trong bng sau:

ChtDinh dng

Loi thc nA B C

mng

Bo

0.1g

0.3g

0.05g

0.2g

0.1g

0.02g

0.2g

0.1g

0.01g

Gi mua 8000 6000 4000 Hy lp m hnh ton hc ca bi ton xc nh khi lng thc n mi loi phi

mua tng stin chi cho mua thc n t nht nhng p ng c nhu cu dinh dngmi ngy.1.8 Mt x nghip c khoch sn xut ba loi sn phm A, B, C t2 loi nguyn liu N1,N2c trlng tng ng l 50kg, 70kg. nh mc tiu hao nguyn liu (kg/SP) v linhun (ngn ng/SP) khi sn xut mt sn phm c cho trong bng sau

Nguyn Sn phm


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26

liu A B CN1

N2

0.2

0.1

0.1

0.1

0.1

0.2

Li nhun 5000 2000 6000

Hy lp m hnh ton hc ca bi ton lp khoch sn xut ti u bit rng lngsn phm B chc thtiu thc ti a 300 sn phm.

Phn II2.1 Cho bi ton quy hoch tuyn tnh sau

f(x) = 6x1+ 3x2+ 2x3- 3x4 min

x1+ x2+ x3- 2x4= 4- x1+ x4 10

2x2+ x3 2x4= 12

xj 0 (j = 1,2,3,4)

a.

Gii bi ton trnb.

Lp bi ton i ngu ca bi ton trn v gii bi ton i ngu .2.2 Cho bi ton quy hoch tuyn tnh sau:

f(x) = 2x1- 5x2+ 4x3+ x4 min

3x1+ x2+ 4x3- 6x4 20

x1+ x3- 2x4 6

3x1- x2+ 2x3 5x4= 24

xj 0 (j = 1,2,3,4)

a.


2.3 Cho bi ton quy hoch tuyn tnh sau:f(x) = x1+ 3x2- x3+ 3x4 minx1+ x2- 2x3+ x4 6

- x1+ x3 102x2- 3x3+ x4= 20

xj 0 (j = 1,2,3,4)

a. Gii bi ton trnb.

Lp bi ton i ngu ca bi ton trn v gii bi ton i ngu .2.4 Cho bi ton quy hoch tuyn tnh sau:

f(x) = - 2x1+ 3x2+ x3+ x4 4x5 max

3x1- 2x2+ x3 4x4+ 2x5= 97x1- 3x2- 7x4+ 5x5= 14

4x1- 2x2 4x4+ 3x5= 8

xj 0, j = 1,2,3,4,5.

a. Gii bi ton trnb.

Lp bi ton i ngu ca bi ton trn v hphng trnh ti u ca bi ton ingu .

2.5 Cho bi ton quy hoch tuyn tnh sau:

f(x) = - 3x1- 2x2- 3x3- 5x4 min

x1+ x2+ 2x3+ 2x4 182x1+ 2x2+ 3x3+ 4x4= 12

x1+ 2x2+ 2x3+ 3x4 11


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27

xj 0, j = 1,2,3,4.

a. Gii bi ton trnb. Lp bi ton i ngu ca bi ton trn v gii bi ton i ngu .

2.6 Cho bi ton quy hoch tuyn tnh sau:f(x) = - 2x1- x2+ x3+ x4 max

x1+ x2+ 2x3- x4= 2

- x2- 7x3+ 3x4 3- 3x3+ 2x4 7

xj 0, j = 1,2,3,4.

a. Gii bi ton trnb. Lp bi ton i ngu ca bi ton trn v gii bi ton i ngu .

2.7 Cho bi ton quy hoch tuyn tnh sau:f(x) = x1+ 2x2+ 3x3+ 3x4 max

2x1+ x2+ x3+ 2x4 20

x1+ 2x2+ 3x3+ 4x4= 18

2x1+ x2+ 2x3+ x4 16

xj 0 (j = 1,2,3,4)a.

Gii bi ton trnb. Hy lp bi ton i ngu ca bi ton trn v gii bi ton i ngu .

2.8 Cho bi ton quy hoch tuyn tnh sau:f(x) = 15x1+ 8x2+ 10x3 max

- 3x1+ 2x2+ 4x3 3

2x1- x2+ 2x3 4

- 4x1- 5x2+ 2x3 1

xj 0 (j = 1,2,3)

a.

Gii bi ton trnb.

Hy lp bi ton i ngu ca bi ton trn v gii bi ton i ngu .Hng dn v p sPhn I1.1 Gi xj, i = 1, 2, 3 ln lt l sgam thc n A1, A2, A3cn phi mua. Khi m hnhton hc ca bi ton l:

f(x) = 8x1+ 6x2+ 4x3 min

0.3x1+ 0.2x2+ 0.2x3 70

0.3x1+ 0.2x2+ 0.2x3 100

0.2x1+ 0.2x2+ 0.3x3 50

0.2x1+ 0.2x2+ 0.3x3 80x1, x2, x3 0

1.2 Gi x1, x2l lng thc n I v II trong mt khu phn. Khi m hnh ton hc cabi ton l:

f(x) = 3x1+ 7x2 min2x1+ 4x2 5

3x1+ x2 4

4x1+ 5x2 8

x1, x2 0

1.3 Gi x1, x2, x3ln lt l ssn phm A1, A2, A3cn sn xut. Khi m hnh ton hc

ca bi ton l:f(x) = 8x1+ 6x2+ 4x3 max

0.2x1+ 0.1x2+ 0.1x3 50


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28

0.1x1+ 0.2x2+ 0.1x3 70

0.1x1+ 0.3x2 100

x3 400

xj 0 (j = 1,2,3)

1.4 Gi xj, i = 1, 2, 3 ln lt l sgam thc n A, B, C cn mua. Khi m hnh ton hcca bi ton l:

f(x) = 3x1+ 4x2+ 5x3 min

0.1x1+ 0.2x2+ 0.3x3 90

0.3x1+ 0.4x2+ 0.2x3 130

0.02x1+ 0.01x2+ 0.03x3 20

xj 0 (j = 1,2,3)

1.5 Gi x1, x2l sn vsn phm loi A v B cn sn xut. Khi m hnh ton hc cabi ton l:

f(x) = 6000x1+ 4000x2 max

4x1+ 2x2 100

3x1+ x

2 300

2x1+ 4x2 50

x1, x2 0 .

1.6 Gi xj(j = 1,2,3) l sn vsn phm Sjcn sn xut. Khi m hnh ton hc cabi ton l:

f(x) = 12000x1+ 8000x2+ 14000x3 max

4x1+ 2x2+ 5x3 10000

2x1+ 6x2+ 3x3 14000

xj 0, j = 1,2,3.

1.7 Gi xj, i = 1, 2, 3 ln lt l sgam thc n A, B, C cn mua. Khi m hnh ton hc

ca bi ton l:f(x) = 8x1+ 6x2+ 4x3 min0.1x1+ 0.2x2+ 0.2x3 50

0.3x1+ 0.1x2+ 0.1x3 80

0.05x1+ 0.02x2+ 0.01x3 20

x1, x2, x3 0

1.8 Gi xj, i = 1, 2, 3 ln lt l ssn phm A, B, C cn sn xut. Khi m hnh tonhc ca bi ton l:

f(x) = 5x1+ 2x2+ 6x3 max

0.2x1+ 0.1x2+ 0.1x3 50

0.1x1+ 0.1x2+ 0.2x3

70x2 300x1, x2, x3 0

Phn II2.1 a. PATU: x = (0, 8, 0, 2), f(x) = 18.

b. PATU: y = (0, 0, 3/2) v g(y) = 18.

2.2 a. PATU: x = (18, 0, 0, 6), f(x) = 42.

b. PATU: y = (0, -13, 5) v g(y) = 42.

2.3 a. PATU: x = (0, 10, 0, 0), f(x) = 30.b. PATU: y = (0, 0, 3/2) v g(y) = 30.

2.4 a. PATU: x = (2, 0, 3, 0, 0), f(x) = -1.2.5 a. PATU: x = (0, 4, 0, 1), f(x) = -13.

b. PATU: y = (0, -2, 1) v g(y) = - 13.


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29

2.6 a. PATU: x = (0, 0, 11, 20), f(x) = 31.

b. PATU: y = (5, 0, 3) v g(y) = 31.

2.7 a. PATU: x = (3, 0, 5, 0), f(x) = 18.

b. PATU: y = (0, 1, 0) v g(y) = 18.

2.8 a. PATU: x = (1/5, 0, 9/10), f(x) = 12.

b. PATU: y = (7, 0, -9) v g(y) = 12.


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1

Chng 1BI TON QUY HOCH TUYN TNH

1.1 MT SV DVBI TON QUY HOCH TUYN TNH

V d1.Mt x nghip cn sn xut 3 loi bnh: bnh u xanh, bnh thp cm v bnhdo. Lng nguyn liu ng, u cho mt bnh mi loi, lng dtrnguyn liu, tinli cho mt bnh mi loi c cho trong bng sau:

Nguyn liu Bnh u xanh Bnh thp cm Bnh do Lng dtr

ng 0,04kg 0,06kg 0,05kg 500kg

u 0,07kg 0kg 0,02kg 300kg

Li 3000 2000 2500

Hy lp m hnh bi ton tm slng mi loi bnh cn sn xut sao cho khng bng

vnguyn liu m li t c cao nht.

GiiGi 1 2 3, ,x x x ln lt l sbnh u xanh, bnh thp cm, bnh do cn phi sn xut.

iu kin: 0jx , 1,2,3j = . Khi

1) Tin li thu c l: 1 2 3 1 2 3( ) ( , , ) 3 2 2,5f x f x x x x x x= = + + (ngn).

2) Lng ng c sdng l: 1 2 30,04 0,06 0,05x x x+ + (kg)

khng bng vnguyn liu th: 1 1 10,04 0,06 0,05 500x x x+ + .

3) Lng u c sdng l: 1 30,07 0,02x x+ (kg)

khng bng vnguyn liu th: 1 30,07 0,02 300x x+ .


(1) 1 2 3 1 2 3( ) ( , , ) 3 2 2,5 axf x f x x x x x x m= = + +

(2) 1 1 10,04 0,06 0,05 500x x x+ +

1 30,07 0,02 300x x+

(3) 0jx , 1,2,3j = .

Ta ni y l bi ton quy hoch tuyn tnh 3 n tm max ca hm mc tiu.V d2.Gisyu cu ti thiu mi ngy vcc cht dinh dng m, ng, khongcho mt loi gia sc tng ng l 90g, 130g, 10g. Cho bit hm lng cc cht dinh dngtrn c trong 1g thc n A, B, C v gi mua 1kg thc n mi loi c cho trong bng sau:

Cht dinh dng A B C

m 0,1g 0,2g 0,3g

ng 0,3g 0,4g 0,2gKhong 0,02g 0,01g 0,03g

Gi mua 3000 4000 5000


31/58


2

Hy lp m hnh ton hc ca bi ton xc nh khi lng thc n mi loi phi mua

tng s tin chi cho mua thc n t nht nhng p ng c nhu cu dinh dng mingy.

GiiGi 1 2 3, ,x x x ln lt l khi lng (g) thc n A, B, C cn mua.

iu kin: 0jx , 1,2,3j = . Khi

Tng khi lng cc cht dinh dng c trong thc n cn mua l

m: 1 2 30,1 0,2 0,3x x x+ + (g)

ng: 1 2 30,3 0,4 0,2x x x+ + (g)

Khong: 1 2 30,02 0,01 0,03x x x+ + (g)

p ng c nhu cu dinh dng ti thiu mi ngy th tng khi lng cc cht dinhdng c trong thc n cn mua khng thnhhn cc nhu cu ti thiu mi ngy vcc

cht dinh dng nn ta c cc iu kin:1 2 30,1 0,2 0,3 90x x x+ +

1 2 30,3 0,4 0,2 130x x x+ +

1 2 30,02 0,01 0,03 10x x x+ +

Tng stin phi chi mua sthc n trn l

1 2 33 4 5x x x+ + (ng)

Yu cu bi ton l stin chi cho mua thc n t nht nn ta c iu kin

1 2 33 4 5 minx x x+ +


(1) 1 2 3 1 2 3( ) ( , , ) 3 4 5 inf x f x x x x x x m= = + +

(2) 1 2 30,1 0,2 0,3 90x x x+ +

1 2 30,3 0,4 0,2 130x x x+ +

1 2 30,02 0,01 0,03 10x x x+ +

(3) 0j

x , 1,2,3j = .

V d3.(CHLH 2009) Mt cssn xut gdnh sn xut ba loi sn phm l bn,

ghv t. nh mc sdng lao ng, chi ph sn xut v gi bn mi sn phm mi loic tnh trong bng sau:

Cc yu t Bn Gh T

Lao ng (ngy cng) 2 1 3

Chi ph sn xut (ngn ng) 100 40 250

Gi bn (ngn ng) 260 120 600

Hy lp m hnh ton hc ca bi ton xc nh ssn phm mi loi cn phi sn xut

sao cho khng bng trong sn xut v tng doanh thu t c cao nht, bit rng csc s lao ng tng ng vi 500 ngy cng, s tin dnh cho chi ph sn xut l 40

triu ng v sbn, ghphi theo tl1/6.


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3

GiiGi 1 2 3, ,x x x ln lt l sbn, gh, tcn phi sn xut. Ta c cc iu kin: 1 2 3, , 0x x x .

Tng ngy cng v chi ph dnh sn xut l:

1 2 32 3x x x+ + (ngy cng)

1 2 3100 40 250x x x+ +

(ngn ng)khng bng trong sn xut ta c cc iu kin sau

1 2 32 3 500x x x+ +

1 2 3100 40 250 40000x x x+ +

Theo tlgia sbn v sghta c iu kin sau

1 26x x=

Tng doanh thu theo dkin l

1 2 3260 120 600x x x+ + (ngn ng)

tng doanh thu t c cao nht ta c iu kin

1 2 3260 120 600 axx x x m+ +

Nhvy, m hnh ton hc ca bi ton l

(1) 1 2 3260 120 600 axx x x m+ +

1 2 32 3 500x x x+ +

(2) 1 2 3100 40 250 40000x x x+ +

1 26x x=

(3) 1 2 3, , 0x x x .

1.2 PHN LOI DNG BI TON QUY HOCH TUYN TNH

1.2.1 Dng tng qut ca bi ton quy hoch tuyn tnhBi ton QHTT dngtng qutvi n n l bi ton c dng


(2) 1 1 2 2 , 1,2, ,i i in n ia x a x a x b i m

+ + + = =

L K

(3)

0

0 , 1,2, ,

jx j n

tuy y

=

K

Trong

(1) l hm mc tiu.

(2) l hrng buc chnh. (3) l rng buc du.

(2) v (3) c gi chung l hrng buc ca bi ton.


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4

Khi

Mi vector 1 2( , , , )nx x x x= K tha (2) v (3) c gi l mt phng n (PA)

ca bi ton.

Mi phng n x tha (1), ngha l ti hm mc tiu t gi tnhnht (lnnht) trn tp cc phng n c gi l mtphng n ti u (PATU) ca biton.

Gii mt bi ton QHTT l i tmmt phng n ti uca n hoc chra rngbi ton v nghim, ngha l bi ton khng c PATU.

1.2.2 Dng chnh tc ca bi ton QHTT



(3) 0, 1,2, ,jx j n = K Nhn xt.Bi ton QHTTdng chnh tcl bi ton QHTT dng tng qut trong

Cc rng buc chnh u l phng trnh.

Cc n u khng m.V d.Bi ton sau c dng chnh tc

(1) 1 2 3 4( ) 2 4 6 minf x x x x x= +

(2)

1 2 4

1 2 3 4

1 2 3 4

4 12

12 3 3

6

x x x

x x x x

x x x x

+ =

+ + =

=

(3) 0, 1,2,3,4jx j =

1.2.3 Dng chun ca bi ton QHTTBi ton QHTTdng chunl bi ton QHTT dng chnh tc



(3) 0, 1,2, ,j

x j n = K

Trong Cc hstdo u khng m.

Trong ma trn hstdo c m vector ct n v: 1 2, , , me e eK .

1 1

1 0 0

0 1 0, , ,

0 0 1

me e e

= = =

KM M M

Khi : Cc n ng vi cc vector ct n vc gi l cc n cbn. Cthn ng vi

vector ct n v ke l n cbn thk.


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5

Mt phng n m cc n cbn u bng 0 c gi lphng n cbn.

Mt phng n c bn c m thnh phn dng c gi l khng suy bin.Ngc li mt phng n c bn c t hn m thnh phn dng c gi l suy

bin.V d.Xt bi ton QHTT sau

(1) 1 2 3 4( ) 2 4 6 minf x x x x x= +

(2)

1 4 5

1 3 6

1 2 3 4

12

12 3

6

x x x

x x x

x x x x

+ + =

+ + =

+ =

(3) 0, 1,2,3,4,5,6jx j =

Ta thy bi ton trn c dng chnh tc, hn naCc hstdo u khng m

Ma trn hsrng buc A l1 0 0 1 1 0

12 0 1 0 0 1

1 1 1 1 0 0

A

=

C cha y 3 vector ct n v 1e (ct 5), 2e (ct 6), 3e (ct 2).

Do bi ton c dng chun, trong

n cbn thnht l 5x

n cbn thhai l 6x n cbn thba l 2x

Nhn xt.Trong bi ton trn, khi cho n cbn thk bng hstdo thk, cn cc nkhng c bn bng 0, ngha l cho 2 6 2 1 3 415, 3, 6, 0, 0, 0x x x x x x= = = = = = ta c mt

phng n cbn ca bi ton (0,6,0,0,12,3)x = .

Phng n ny khng suy bin v c 3 thnh phn dng. Ta gi y l phng n c

bn ban u ca bi ton.Ch .Tng qut, trong bi ton QHTT dng chun bt k, khi cho n cbn thk bng

hstdo thk ( 1,2, ,k m= K ), cn cc n khng cbn bng 0, ta c mt phng ncbn ca bi ton. Ta gi y l phng n cbn ban u ca bi ton.

1.3 BIN I DNG BI TON QHTT

1.3.1 Dng tng qut vdng chnh tc

Ta c thbin i bi tondng tng qutvdng chnh tcbng cc bc sauBc 1.Kim tra hrng buc chnh

1)

Nu c rng buc chnh dng 1 1 2 2i i in n ia x a x a x b+ + + L th ta cng vo vtri rng


ton bng rng buc 1 1 2 2i i in n n k ia x a x a x x b++ + + + =L .


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6

2)Nu c rng buc chnh dng 1 1 2 2i i in n ia x a x a x b+ + + L th ta tr vo v tri rng


ton bng rng buc 1 1 2 2i i in n n k ia x a x a x x b++ + + =L .

Ch .Cc n phl cc n khng m v hsca cc n ph trong hm mc tiu l 0.

Bc 2.Kim tra iu kin du ca n s1) Nu c n 0jx th ta thc hin php i n s j jx x = vi 0jx .

2)

Nu c n jx c du ty th ta thc hin php i n s j j jx x x = vi

, 0j jx x .

Ch .Khi tm c PATU ca bi ton dng chnh tc ta chcn tnh gi trca cc nban u v bi cc n phth sc PATU ca bi ton dng tng qut cho.V d.Bin i bi ton sau vdng chnh tc

(1) 1 2 3 4( ) 2 4 6 minf x x x x x= +

(2)

1 2 3

1 3

1 2 3

4 6 5 50

7 30

2 3 5 25

x x x

x x

x x x

+

+

+ =

(3) 1 20, 0x x

GiiThm vo bi ton n ph 4 0x bin bt phng trnh 1 2 34 6 5 50x x x + vphng

trnh 1 2 3 44 6 5 50x x x x + + = .Thm vo bi ton n ph 5 0x bin bt phng trnh 1 37 30x x+ vphng trnh

1 3 57 30x x x+ = .

i bin 2 2x x = vi 2 0x .

i bin 3 3 3x x x= vi 3 3, 0x x .

Ta a bi ton vdng chnh tc

(1) 1 2 3 3( ) 3 2 2,5( ) maxf x x x x x = +

(2)

1 2 3 3 4

1 3 3 5

1 2 3 3

4 6 5( ) 50

7 ( ) 30

2 3 5( ) 25

x x x x x

x x x x

x x x x

+ + + = + =

=

(3) 1 2 3 3 4 50, 0, 0, 0, 0, 0x x x x x x

1.3.2 Dng chnh tc vdng chun

Tbi tondng chnh tcta c thxy dng bi tondng chunnhsau

1) Khi gp hstdo 0ib < ta i du hai vca rng buc thi.


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7

2) Khi ma trn hsrng buc A khng cha ct n vthk l ke , ta a vo n gi

0n kx + v cng thm n gi n kx + vo vtri phng trnh rng buc thk c

phng trnh rng buc mi: 1 1 2 2k k kn n n k k a x a x a x x b++ + + + =L .

3)

Hm mc tiu mrng ( )f x c xy dng thm mc tiu ban u nhsau

i vi bi ton min: ( ) ( ) ( gia)f x f x M an= + . i vi bi ton max: ( ) ( ) ( gia)f x f x M an= .

Trong M l i lng rt ln, ln hn bt k sno cho trc.


(2)

1 2 3

1 3 4

1 2 3

4 6 5 50

7 0

2 3 5 25

x x x

x x x

x x x

+ =

+ + =

+ =

(3) 0, 1, ,4jx j = K

GiiBi ton trn c dng chnh tc, trong vphi ca phng trnh rng buc thba l

-25 < 0. i du hai vca phng trnh ny ta c 1 2 32 3 5 25x x x + =

V (2) trthnh

1 2 3

1 3 4

1 2 3

4 6 5 50

7 0

2 3 5 25

x x x

x x x

x x x

+ =

+ + =

+ =

Ma trn hsca rng buc l

4 6 5 0 1 0

7 0 1 1 0 0

2 3 5 0 0 1

A

=

V A cn thiu 2 vector ct n vl 1e v 3e nn bi ton cha c dng chun.

Thm vo bi ton hai n gi 5 6, 0x x v xy dng bi ton mrng c dng chun nh

sau

(1) 1 2 3 4 5 6( ) 3 2 2 minf x x x x x Mx Mx= + + + + +

(2)

1 2 3 5

1 3 4

1 2 3 6

4 6 5 50

7 0

2 3 5 25

x x x x

x x x

x x x x

+ + =

+ + =

+ + =

(3) 0, 1, ,6jx j = K

V d2.Bin i bi ton QHTT sau vdng chun

(1) 1 2 3 4( ) 3 2 2 maxf x x x x x= + + +


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8

(2)

1 2 3

1 3 4

1 2 3

4 6 5 50

7 0

2 3 5 25

x x x

x x x

x x x

+ =

+ + =

+ =

(3) 0, 1, ,4jx j = K

Ta xy dng bi ton mrng dng chun nhsau(1) 1 2 3 4 5 6( ) 3 2 2 maxf x x x x x Mx Mx= + + +

(2)

1 2 3 5

1 3 4

1 2 3 6

4 6 5 50

7 0

2 3 5 25

x x x x

x x x

x x x x

+ + =

+ + =

+ + =

(3) 0, 1, ,6jx j = K

Ch .

n ph: Tng qut chuyn thnh chnh tc

n gi: Chnh tc chuyn thnh chun


(2)

1 3

3 4

1 2 3

9 15 50

6 2 120

3 5 45

x x

x x

x x x

+

+ =

+

(3) 0, 1, , 4jx j = K

GiiThm vo bi ton 2 n ph 5 6, 0x x ta c bi ton c dng chnh tc nhsau

(1) 1 2 3 4( ) 3 2 2 minf x x x x x= + + +

(2)

1 3 5

3 4

1 2 3 6

9 15 50

6 2 120

3 5 45

x x x

x x

x x x x

+ + =

+ =

+ =

(3) 0, 1, ,6jx j = K .

Bi ton trn cha c dng chun.Ta thy cc vphi ca hai phng trnh rng buc th2 v 3 u m nn bng cch i

du hai vca cc phng trnh ny ta c

(2)

1 3 5

3 4

1 2 3 6

9 15 50

6 2 120

3 5 45

x x x

x x

x x x x

+ + =

=

+ + =

Ma trn hsrng buc l

A =

0 9 15 0 1 0 0

0 0 6 2 0 0 11 3 5 0 0 1 0


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9

V A cn thiu mt vector ct l 2e nn bi ton cha c dng chun.

Thm vo rng buc chnh thhai n gi 7 0x ta c bi ton dng chun nhsau

(1) 1 2 3 4 7( ) 3 2 2 minf x x x x x Mx= + + + +

(2)

1 3 5

3 4 7

1 2 3 6

9 15 50

6 2 1203 5 45

x x x

x x xx x x x

+ + =

+ = + + =

(3) 0, 1, ,7jx j = K .

Ch .Quan hgia bi ton xut pht v bi ton mrngMi quan hgia bi ton xut pht (A) v bi ton mrng (B) nhsau

B v nghim suy ra A v nghim

B c nghim c hai trng hp:1) Nu mi n gica PATU bng 0 th bn gita c PATU ca A

2)

Nu c t nht mt n gi> 0 suy ra A khng c PATU

Chng 2PHNG PHP N HNH

2.1 PHNG PHP N HNH GII BI TON QUY HOCH TUYN TNHDNG CHUN2.1.1

Thut ton gii bi ton max

Bc lp thnht (bng n hnh thnht)1)

Lp bng n hnh xut phtVbng n hnh v ghi vo cc thnh phn sau ca bi ton dng chun

Dng 1. Ghi cc n ca bi ton (kcn ph)

Dng 2. Ghi cc hsca cc n trong hm mc tiu

Ct 2. Ghi cc n cbn ca bi ton theo thttn cbn thnht n n c

bn cui cng, ta gi ct ny lct n cbn.

Ct 1: Ghi tng ng cc hsca cc n cbn trong hm mc tiu, ta gi ctny lct hscbn.

Ct 3. Ghi cc shng tdo ca hrng buc chnh theo thtttrn xung di,ta gi ct ny lct phng n.

Ct 4. Ghi ma trn iu kin A ca bi ton.

Tnhhsc lngj ca cc n ( 1,2, , )jx j n= K v ghi tng ng vo dng di ct

4, vi j c tnh theo cng thc sau:

(cot1) ( ) ( )j j jA hsx = ( jhsx : hsca n jx trong hm mc tiu).

Ch .Nu jx l n cbn th 0j = .Tnh trs 0 (cot1) (cot3)f = v ghi di ct 3.

2)Xc nh phng n cbn xut pht


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10

Vi bng n hnh va lp c th phng n cbn xut pht 0x ca bi ton cxc nh nhsau: Cho cc n ct 2 nhn gi trtng ng ct 3, cc n cn li nhn

gi tr0. Trsca hm mc tiu ti phng n cbn xut pht 0x l 0 0( )f x f= .

3)

nh gi tnh ti u ca phng n cbn xut pht

Du hiu ti u. Nu h sc lng ca cc n u khng m, 0,j j thphng n cbn xut pht 0x l phng n ti u ca bi ton. Thut ton kt thc

vi kt lun: Bi ton c PATU l 0x v GTTU l 0( )f x .

Du hiu ca bi ton khng c PATU.Nu c n khng cbn kx c hsc

lng m v ct iu kin kA ca n c cc thnh phn u khng dng, 0k <

v 0;ika i th bi ton khng c phng n ti u. Thut ton kt thc vi kt lun:

Bi ton khng c PATU.

Nu khng xy ra chai trng hp trn th thut ton tip tc trong bc lp thhaiBc lp thhai(Bng n hnh thhai)

1) Tm n a voTrong tt c cc 0j < ta chn 0v < nh nht (ta nh du * cho 0v < nh nht

trong bng). Khi , vx l n m ta sa vo hn cbn. Ct vA c gi l ct

chyu.

2) Tm n a raThc hin php chia ln lt cc sca ct phng n cho cc sdng ca ct ch

yu v ghi cc thng s i vo ct cui cng.Xc nh min{ }r i = (Ta nh du * cho r nh nht trong bng). Khi rx l n

m ta a ra khi hn cbn. Dng c cha rx c gi ldng chyu. Sdng

nm trn dng chyu v ct chyu c gi lhschyu.Ch .Nu ct chyu chc mt sdng th sdng l hschyu, dng csdng l dng chyu, n nm trn dng chyu l n c a ra.

3)Lp bng n hnh thhai

Ct 2: Thay n a ra bng n a vo, cc n cbn cn li ginguyn. Dng c

n a vo gi ldng chun. Ct 1: Thay hsca n a ra bng hsca n a vo, cc hsca cc n c

bn cn li ginguyn.Cc thnh phn cn li c xc nh theo dng nhsau

Dng chun = Dng chyu chia cho hschyu.

Dng thi = Dng thi (c) aiv.dng chun. (aiv: snm trn giao ca dng i vct chyu).

Cc hsc lng v trsca hm mc tiu trong bng thhai c tnh v ghi

nhbng thnht.4)

Xc nh v nh gi phng n cbn thhai(nhbc lp thnht)


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11

2.1.2 Thut ton gii bi ton min

Gii tng tbi ton max vi ch sau

iu kin ti u: 0,j j

iu kin khng c PATU: 0k > v 0,ika i

n c chn a vo: n ng vi 0k > ln nht.


(1) 1 2 3 4 5( ) 2 5 4 6 m axf x x x x x x= +

(2)

1 2 4 5

2 3 4 5

2 5 6

6 2 9 32

2 3 30

3 36

x x x x

x x x x

x x x

+ =

+ + + =

+ + =

(3) 0, 1, ,6jx j = K .

GiiBi ton trn c dng chnh tc vi cc vphi ca cc phng trnh rng buc trong (2)

u khng m.Ma trn hsca rng buc

1 6 0 2 9 0

0 2 1 1 0 0

0 3 0 0 0 1

A

=


trong : n cbn thnht: 1x

n cbn thhai: 3x

n cbn thba: 6x


0 2.32 4.30 184f = + = .

1 3 6 0 = = =

2 2.6 4.2 0.3 5 25 = + + + =

4 2.( 2) 4.1 0.0 1 1 = + + + =

5 2.( 9) 4.3 0.1 6 0 = + + + =

x1

2

x2

-5

x3

4

x4

-1

x5

-6

x6

0i

2

4

0

x1

x3

x6

32

30

36

1

0

0

6

2

3

0

1

0

-2

1

0

-9

(3)

1

0

0

1

184 0 -25 0 1 0 0


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12

Trong bng trn ta thy 0, 1, ,6j j = K nn bi ton c PATU l

0 (32,0,30,0,0,36)x = v 0 184f = .


(1) 1 2 3 4 5 6( ) 6 3 minf x x x x x x x= + + + +

(2)

1 2 4 6

1 3 6

1 4 5 6

15

2 2 9

4 2 3 2

x x x x

x x x

x x x x

+ + = + =

+ + =

(3) 0, 1, ,6j

x j = K .

GiiBi ton trn c dng chnh tc vi vphi ca phng trnh rng buc chnh thhai l 9.

i du hai vca phng trnh ny, ta a vbi ton sau

(1) 1 2 3 4 5 6( ) 6 3 7 minf x x x x x x x= + + + +

(2)

1 2 4 6

1 3 6

1 4 5 6

15

2 2 9

4 2 3 2

x x x x

x x x

x x x x

+ + =

+ =

+ + =

(3) 0, 1, ,6jx j = K .

Ma trn hsca rng buc

1 1 0 1 0 1

2 0 1 0 0 2

4 0 0 2 1 3

A

=


trong :

n cbn thnht: 2x

n cbn thhai: 3x

n cbn thba: 5x


x1

6

x2

1

x3

1

x4

3

x5

1

x6

-7i

1

1

1

x2

x3

x5

15

9

2

-1

-2

4

1

0

0

0

1

0

-1

0

2

0

0

1

(1)

-2

-3

15

26 -5 0 0 -2 0 3*

-7

1

1

x6

x3

x5

15

39

47

-1

-4

1

1

2

3

0

1

0

-1

-2

-1

0

0

1

1

0

0-19 -5 0 0 -2 0 0


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13

Trong bng 1 ta thy tn ti 6 3 0 = > v trn ct tng ng c 13 1 0a = > ( 23 2a = v

23 3a = ) nn ta chn n a ra l 6x , n a vo l 2x , hschyu l 13 1a = . Sau

tnh cc dng bng 2 bng cng thc saudc= dcy, d2= d2+ 2dc, d3= d3+ 3dc.

Trong bng 2 ta thy tn ti 4 1 0 = >

m 4 0, 1,2,3ia i =

nn bi ton min ang xt vnghim.V d3.Gii bi ton QHTT sau

(1) 1 2 3 4 5( ) 2 6 4 2 3 m axf x x x x x x= + + +

(2)

1 2 3

2 3 4

2 5

2 4 52

4 2 60

3 36

x x x

x x x

x x

+ + =

+ + =

+ =

(3) 0, 1, ,5jx j = K .

GiiBi ton trn c dng chnh tc vi v phi ca phng trnh rng buc trong (2) ukhng m.

Ma trn hsrng buc l

1 2 4 0 0

0 4 2 1 0

0 3 0 0 1

A

=


trong :

n cbn thnht: 1x

n cbn thhai: 4x

n cbn thba: 5x


x1-2

x26

x34

x4-2

x53 i

-2

-2

3

x1

x4

x5

52

60

36

1

0

0

2

4

3

(4)

2

0

0

1

0

0

0

1

52/4*

60/2

-116 0 -9 -16* 0 0

4

-2

3

x3

x4

x5

13

34

36

1/4

-1/2

0

1/2

(3)

3

1

0

0

0

1

0

0

0

1

13.2

34/3*

36/3

92 4 -1* 0 0 0

4

6

3

x3

x2

x5

22/3

34/3

2

1/3

-1/6

1/2

0

1

0

1

0

0

-1/6

1/3

-1

0

0

1

310/3 23/6 0 0 1/3 0


43/58


14

Trong bng I ta thy tn ti cc 0j < : 2 39, 16 = = v trn mi ct tng ng c h

sdng. Ta chn 3 16 = m nhnht v n a vo l 3x , khi trn ct tng ng

c cc h sdng l 13 234, 2a a= = nn ta lp cc t s 1 252 / 4, 60 / 2 = = . Ta chn

1 5 2 / 4 = nhnht v n a ra l 1x , hschyu l 13 4a = . Sau bin i bng I

bng cc php bin i sau:dc= dcy/4 , d2= d2- 2dc, d3= d3.

Bin i tng tcho bng II.

Trong bng III ta thy 0, 1,2, ,5j j = K nn bi ton ang xt c PATU l

0 (0,34 / 3,22 / 3,0,2)x = vi 0( ) 310 / 3f x = .

2.2 PHNG PHP N HNH MRNG GII BI TON QUY HOCHTUYN TNH DNG CHNH TC

Thut ton n hnh mrng gii bi ton QHTT dng chnh tc tng tnhthut tonn hnh gii bi ton QHTT dng chun nhng c mt slu nhsau

1)Do hm mc tiu m rng l ( ) ( ) ( )f x f x angia= + i vi bi ton min v( ) ( ) ( )f x f x angia= i vi bi ton max, nn trong bng n hnh ct hsc

thc cc hsphthuc M. Khi dng cui cc hssc dng aM b+ , do ngi ta thng chia dng cui thnh hai dng nh: Dng trn ghi a v dng di ghi b.

2)V M l mt i lng dng rt ln, nn khi so snh cc shng aM + b v cM + d tac quy tc sau

a caM b cM d

b d

=+ = +

=

0

00

0

a

baM b

a

b

>

+ >

=

>

,a c

b daM b cM d

a c

b d

>

+ > +

=

>

3)Trong bng n hnh u tin cc n giu c trong n cbn. Mi khi mt n giba ra khi hn cbn th khng bao gi ta a n gi tr li na, v vy trong

bng n hnh ta c thbi cc ct ng vi cc n gi.V d1.Gii bi ton QHTT sau

f(x) = 6x1+ 3x2+ 2x3- 3x4 minx1+ x2+ x3- 2x4= 4

- x1+ x4 10


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15

2x2+ x3 2x4= 12

xj 0 (j = 1,2,3,4)


f(x) = 6x1+ 3x2+ 2x3- 3x4+ Mx6+ Mx7 min

x1+ x2+ x3- 2x4+ x6= 4

- x1+ x4+ x5= 10

2x2+ x3 2x4+ x7= 12

xj 0 (j = 1,2,3,4,5,6,7)


PATU: x = (0, 8, 0, 2), f(x) = 18.

V d2.Gii bi ton quy hoch tuyn tnh sau:

f(x) = - 2x1- x2+ x3+ x4 maxx1+ x2+ 2x3- x4= 2

- x2- 7x3+ 3x4 3

- 3x3+ 2x4 7

xj 0, j = 1,2,3,4.


f(x) = - 2x1- x2+ x3+ x4 max

x1+ x

2+ 2x

3- x

4 = 2

- x2- 7x3+ 3x4+ x5= 3

- 3x3+ 2x4+ x6= 7

x1

6

x2

3

x3

2

x4

-3

x5

0

M0

M

x6x5

x7

410

12

1-1

0

(1)0

2

10

1

-21

-2

01

0

-6 -3 -2 3 0

1 3* 2 -4 0

3

0

M

x2

x5

x7

4

10

4

1

-1

-2

1

0

0

1

0

-1

-2

1

(2)

0

1

0

-3 0 1 -3 0

-2 0 -1 2* 0

3

0

-3

x2

x5

x4

8

8

2

-1

0

-1

1

0

0

0

1/2

-1/2

0

0

1

0

1

0

18 -6 0 -1/2 0 0


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16

xj 0, j = 1,2,3,4,5,6.


.

PATU: x = (0, 0, 11, 20), f(x) = 31.

Chng 3

BI TON I NGU

3.1 nh ngha

Cho (P) l bi ton QHTT c dng chnh tc nhsau

1 1 2 2( ) max(min)n nf x c x c x c x= + + + L

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

0, 1,2, , .

n n

n n

m m mn n m

j

a x a x a x b

a x a x a x b

a x a x a x b

x j n

+ + + =

+ + + =

+ + + =

=

L

L

KKK

L

K

Tbi ton (P) ta lp c bi ton QHTT (D) nhsau v ta gi bi ton (D) l bi ton

i ngu ca bi ton (P)

1 1 2 2( ) min(max)m mf y b y b y b y= + + + L

11 1 21 2 1 1

12 1 22 2 1 2

1 1 2 2

( )

( )

( )

m m

m m

n n mn m n

a y a y a y c

a y a y a y c

a y a y a y c

+ + +

+ + +

+ + +

L

L

KKK

L

x1

-2

x2

-1

x3

1

x4

1

x5

0

x6

0-2

0

0

x1

x5

x6

2

3

7

1

0

0

1

-1

0

(2)

-7

-3

-1

3

2

0

1

0

0

0

1

0 -1 -5* 1 0 0

1

0

0

x3

x5

x6

1

9

8

1/2

7/2

3/2

1/2

5/2

3/2

1

0

0

-1/2

-1/2

(1/2)

0

1

0

0

0

1

5/2 3/2 0 -3/2* 0 01

0

1

x3

x4

x5

11

20

20

2

5

3

2

4

3

1

0

0

0

0

1

0

1

0

1

1

2

31 7 6 0 0 0 3


46/58


17

Ch .Bi ton (D) c lp tbi ton (P) theo nguyn tc sau1. Sn ca bi ton (D) bng srng buc chnh ca bi ton (P) v srng buc chnhca bi ton (D) bng sn ca bi ton (P).

2. Hsca n iy trong hm mc tiu ca bi ton (D) l shng tdo ib trong hrng

buc chnh ca bi ton (P).3.

Cc hsca cc n v hstdo trong rng buc chnh thj ca bi ton (D) l cc

hstng ng ca n jx trong hrng buc chnh v hm mc tiu ca bi ton (P).

4.

Nu (P) l bi ton max th (D) l bi ton min v hrng buc chnh ca bi ton (D)

l hbt phng trnh vi du . Nu (P) l bi ton min th (D) l bi ton max v hrng buc chnh ca bi ton (D) l hbt phng trnh vi du .

5. Cc n ca bi ton (D) u c du ty .

3.2 Cch lp bi ton i nguBi ton i ngu c lp trc tip theo quy tc sau, gi l quy tc i ngu

(P) (D)

1 1 2 2( ) maxn nf x c x c x c x= + + + L 1 1 2 2( ) minm mf y b y b y b y= + + + L

1 1 2 2i i in n ia x a x a x b

+ + + =

L

0

0iy

tuyy

0

0jxtuyy

1 1 2 2j j mj m ja y a y a y c

+ + +

=

L

V d.Tm bi ton i ngu ca bi ton saua)

(1)1 2 3 4

( ) 3 2 5 minf x x x x x= + +

(2)

1 2 3 4

1 3 4

1 2 3

4 6 5 5 50

7 30

2 3 5 25

x x x x

x x x

x x x

+

+ + =

+

(3) 1 20, 0x x .

b)

(1) 1 2 3( ) 2 8 m axf x x x x= +

(2)

1 2 3

1 2 3

1 2 3

7 4 2 28

3 3 10

2 3 15

x x x

x x x

x x x

+ +

+ =

+

(3) 1 2 3, , 0x x x .

Giia) Bi ton i ngu l


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18

(1) 1 2 3( ) 50 30 25 maxg y y y y= +

(2)

1 2 3

1 3

1 2 3

1 2

4 7 2 3

6 3 2

5 5 5

5 1

y y y

y y

y y y

y y

+ +

+

+ = + =

(3) 1 0y , 2y ty , 3 0y .

b) Bi ton i ngu l

(1) 1 2 3( ) 28 10 15 ming y y y y= + +

(2)

1 2 3

1 2 3

1 2 3

7 3 2 2

4 3 1

2 3 8

y y y

y y y

y y y

+ +

+

+

(3) 1 0y , 3 0y .

3.3 Cp rng buc i ngu

Trong mt cp rng buc i ngu (P) v (D) nh trong nh ngha th ta c n cp rngbuc i ngu nhsau:

Trng hp 1. 1 1 2 2( ) maxn nf x c x c x c x= + + + L

1 1 2 20 , 1,2, ,j j j mj m jx a y a y a y c j n + + + =L K

Trng hp 2. 1 1 2 2( ) minn nf x c x c x c x= + + + L

1 1 2 2

0 , 1,2, ,j j j mj m j

x a y a y a y c j n + + + =L K

V d.Tm bi ton i ngu v chra cc cp rng buc i ngu ca cc bi ton QHTTsau:

a)

(1)1 2 3

( ) 2 3 minf x x x x= + +

(2)

1 2 3

1 2 3

1 2 3

1 2 3

2 3 2

4 3

2 6

3 5 7 8

x x x

x x x

x x x

x x x

+ =

+

+

(3) 1 2 30, 0, 0x x x .

b)

(1) 1 2 3 4( ) 2 3 4 m axf x x x x x= +

(2)

1 2 3 4

1 2 3 4

1 2 3 4

5 6 7 8 9

10 11 12 13 14

15 16 17 18 19

x x x x

x x x x

x x x x

+ + +

+ =

+ +

(3) 1 2 3, , 0x x x .

Giia) Bi ton i ngu


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19

(1) 1 2 3 4( ) 2 3 6 8 m axg y y y y y= + +

(2)

1 2 3 4

1 2 3 4

1 2 3 4

2 3 1

3 7 2

4 2 7 3

y y y y

y y y y

y y y y

+ + +

+ +

+ + +

(3) 2 4, 0y y , 3 0y .Hrng buc chnh ca bi ton (P) c 3 bt phng trnh v bi ton (P) c 3 iu kin vdu ca n snn cp bi ton i ngu (P) v (D) c 6 cp rng buc i ngu

1 2 3 2

1 2 3 3

1 2 3 4

1 1 2 3 4

2 1 2 3 4

3 1 2 3 4

4 3 0

2 6 0

3 5 7 8 0

0 2 3 1

0 3 5 2

0 4 2 7 3

x x x y

x x x y

x x x y

x y y y y

x y y y y

x y y y y

+

+

+ + +

+ +

+ + +

b) Bi ton i ngu

(1) 1 2 3( ) 9 14 19 ming y y y y= +

(2)

1 2 3

1 2 3

1 2 3

1 2 3

5 10 15 1

6 11 16 2

7 12 17 3

8 13 18 4

y y y

y y y

y y y

y y y

+

+

+ + =

(3) 10y

, 2y

ty , 3 0y

.Hrng buc chnh ca bi ton (P) c hai bt phng trnh v bi ton (P) c ba iu kinvdu ca n snn cp bi ton i ngu (P) v (D) c 5 cp rng buc i ngu sau:

1 2 3 4 1

1 2 3 4 3

1 1 2 3

2 1 2 3

3 1 2 3

5 6 7 8 0

15 16 17 18 19 0

0 5 10 15 1

0 6 11 16 2

0 7 12 17 3

x x x x y

x x x x y

x y y y

x y y y

x y y y

+ + +

+ +

+

+

+

3.4 nh l i ngu

nh l lch b yu.iu kin cn v phng n 0x ca bi ton (P) v

phng n 0y ca bi ton (D) u l phng n ti u l trong cc cp rng buc i

ngu ca bi ton : Nu mt rng buc tha mn phng n vi du bt ng thc thcsth rng buc cn li phi tha mn phng n vi du bng.

ng dng.Nhnh l lch b yu, khi ta bit c mt phng n ti u ca mttrong hai bi ton ca cp bi ton i ngu th ta c thtm c tp phng n ti u ca

bi ton cn li. ng dng ny thng c sdng trong vic gii quyt cc vn cabi ton QHTT.

V d1. Cho bi ton quy hoch tuyn tnh sau


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20

f(x) = x1+ 2x2+ 3x3+ 3x4 max

2x1+ x2+ x3+ 2x4 20

x1+ 2x2+ 3x3+ 4x4= 18

2x1+ x2+ 2x3+ x4 16

xj 0 (j = 1,2,3,4)

a.

Gii bi ton trn

b. Hy lp bi ton i ngu ca bi ton trn v gii bi ton i ngu .

Giia. a bi ton vdng chun:

f(x) = x1+ 2x2+ 3x3+ 3x4 Mx7 Mx8 max

2x1+ x2+ x3+ 2x4+ x5= 20

x1+ 2x2+ 3x3+ 4x4+ x7= 18

2x1+ x2+ 2x3+ x4 x6+ x8= 16

xj 0 (j = 1,2,3,4,5,6,7,8)Gii bi ton mrng bng phng php n hnh

.

PATU: x = (3, 0, 5, 0), f(x) = 18.

b. Bi ton i ngu

g(y) = 20y1+ 18y2+ 16y3 min

2y1+ y2+ 2y3 1

y1+ 2y2+ y3 2

y1+ 3y2+ 2y3 3

2y1+ 4y2+ y3 3

y1 0, y3 0.

x1

1

x2

2

x3

3

x4

3

x5

0

x6

0

0

-M

-M

x5

x7

x8

20

18

16

2

1

2

1

2

1

1

(3)

2

2

4

1

1

0

0

0

0

-1

-1 -2 -3 -3 0 0

-3 -3 -5* -5 0 1

0

3

-M

x5

x3

x8

14

6

4

5/3

1/3

(4/3)

1/3

2/3

-1/3

0

1

0

2/3

4/3

-5/3

1

0

0

0

0

-1

0 0 0 1 0 0

-4/3* 1/3 0 5/3 0 1

0

3

1

x5

x3

x1

9

5

3

0

0

1

3/4

3/4

-1/4

0

1

0

11/4

7/4

-5/4

1

0

0

5/4

1/4

-3/4

18 0 0 0 1 0 0


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21

Theo nh l lch b yu, ta c hphng trnh ti u sau

=

=++

=++

0

323

122

1

321

321

y

yyy

yyy

Gii hphng trnh ta c PATU: y = (0, 1, 0) v g(y) = 18.V d2.Cho bi ton quy hoch tuyn tnh sau

f(x) = 6x1+ 3x2+ 2x3- 3x4 min

x1+ x2+ x3- 2x4= 4

- x1+ x4 10

2x2+ x3 2x4= 12

xj 0 (j = 1,2,3,4)

a.


Giia. a bi ton vdng chun

f(x) = 6x1+ 3x2+ 2x3- 3x4+ Mx6+ Mx7 min

x1+ x2+ x3- 2x4+ x6= 4

- x1+ x4+ x5= 10

2x2+ x3 2x4+ x7= 12

xj 0 (j = 1,2,3,4,5,6,7)


x1

6

x2

3

x3

2

x4

-3

x5

0

M

0

M

x6

x5

x7

4

10

12

1

-1

0

(1)

0

2

1

0

1

-2

1

-2

0

1

0

-6 -3 -2 3 0

1 3* 2 -4 0

3

0

M

x2

x5

x7

4

10

4

1

-1

-2

1

0

0

1

0

-1

-2

1

(2)

0

1

0

-3 0 1 -3 0

-2 0 -1 2* 0

3

0

-3

x2

x5

x4

8

8

2

-1

0

-1

1

0

0

0

1/2

-1/2

0

0

1

0

1

0

18 -6 0 -1/2 0 0


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