[math-educare]_giáo trình quy hoạch tuyến tính_le duc thang

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Gio trnh quy hoch tuyn tnh

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Gio trnh quy hoch tuyn tnh

Bin tp bi:thang leduc

Cc tc gi:thang leduc

Phin bn trc tuyn:http://voer.edu.vn/c/78021439

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MC LC

1. Thng tin v tc gi2. Gii thiu bi ton quy hoch tuyn tnh3. Quy hoch tuyn tnh tng qut v chnh tc4. c im ca cc tp hp cc phng n5. L thuyt c bn v quy hoch tuyn tnh-Mt s v d m u6. Du hiu ti u7. Gii thut n hnh c bn8. Phng php bin gi ci bin9. Quy hoch tuyn tnh suy bin10. Khi nim v i ngu11. Gii thut i ngu12. ng dng quy hoch tuyn tnh-M u13. Bi ton vn ti14. Bi ton dng trn mng15. Quy hoch tuyn tnh16. cng17. Bi tp tng hpTham gia ng gp

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Thng tin v tc gi

Thng tin v tc gi gio trnh:

H v tn: L c Thng Sinh nm: C quan cng tc: B mn H thng Thng tin v Ton ng dng, Khoa Cng

ngh Thng tin & Truyn thng, i hc Cn Th. i ch e-mail: [email protected]

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Gii thiu bi ton quy hoch tuyn tnhC th tm nh ngha quy hoch tuyn tnh l lnh vc ton hc nghin cu cc bi tonti u m hm mc tiu (vn c quan tm) v cc rng buc (iu kin ca biton) u l hm v cc phng trnh hoc bt phng trnh tuyn tnh. y ch l mtnh ngha m h, bi ton quy hoch tuyn tnh s c xc nh r rng hn thngqua cc v d .

cc bc nghin cu v ng dng mt bi ton quy hoch tuyn tnh in hnh l nhsau :

a- xc nh vn cn gii quyt, thu thp d liu.

b- lp m hnh ton hc.

c- xy dng cc thut ton gii bi ton m hnh hoc bng ngn ng thun licho vic lp trnh cho my tnh

d- tnh ton th v iu chnh m hnh nu cn.

e- p dng gii cc bi ton thc t.

Bi ton vn u t

ngi ta cn c mt lng (ti thiu) cht dinh dng i=1,2,..,m do cc thc nj=1,2,...,n cung cp. gi s :

aij l s lng cht dinh dng loi i c trong 1 n v thc n loi j

(i=1,2,...,m) v (j=1,2,..., n)

bi l nhu cu ti thiu v loi dinh dng i

cj l gi mua mt n v thc n loi j

vn t ra l phi mua cc loi thc n nh th no tng chi ph b ra t nht mvn p ng c yu cu v dinh dng. vn c gii quyt theo m hnh sau y:

gi xj 0 (j= 1,2,...,n) l s lng thc n th j cn mua .

tng chi ph cho vic mua thc n l :

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v chi ph b ra mua thc n phi l thp nht nn yu cu cn c tha mn l :

lng dinh dng i thu c t thc n 1 l : ai1x1 (i=1m)

lng dinh dng i thu c t thc n 2 l : ai2x2

.........................................................

lng dinh dng i thu c t thc n n l : ainxn

vy lng dinh dng th i thu c t cc loi thc n l :

ai1x1+ai2x2+...+ainxn (i=1m)

v lng dinh dng th i thu c phi tha yu cu bi v dinh dng loi nn ta crng buc sau :

ai1x1+ai2x2+...+ainxn bi (i=1m)

khi theo yu cu ca bi ton ta c m hnh ton sau y :

Bi ton lp k hoch sn xut

t m loi nguyn liu hin c ngi ta mun sn xut n loi sn phm

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gi s :

aij l lng nguyn liu loi i dng sn xut 1 sn phm loi j

(i=1,2,...,m) v (j=1,2,..., n)

bi l s lng nguyn liu loi i hin c

cj l li nhun thu c t vic bn mt n v sn phm loi j

vn t ra l phi sn xut mi loi sn phm l bao nhiu sao cho tng li nhun thuc t vic bn cc sn phm ln nht trong iu kin nguyn liu hin c.

gi xj 0 l s lng sn phm th j s sn xut (j=1,2,...,n)

tng li nhun thu c t vic bn cc sn phm l :

v yu cu li nhun thu c cao nht nn ta cn c :

lng nguyn liu th i=1m dng sn xut sn phm th 1 l ai1x1

lng nguyn liu th i=1m dng sn xut sn phm th 2 l ai2x2

...............................................

lng nguyn liu th i=1m dng sn xut sn phm th n l ainxn

vy lng nguyn liu th i dng sn xut l cc sn phm l

ai1x1+ai2x2+...+ainxn

v lng nguyn liu th i=1m dng sn xut cc loi sn phm khng th vtqu lng c cung cp l bi nn :

ai1x1+ai2x2+...+ainxn bi (i=1,2,...,m)

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vy theo yu cu ca bi ton ta c m hnh sau y :

Bi ton vn ti

ngi ta cn vn chuyn hng ho t m kho n n ca hng bn l. lng hng ho kho i l si (i=1,2,...,m) v nhu cu hng ho ca ca hng j l dj (j=1,2,...,n). cc vnchuyn mt n v hng ho t kho i n ca hng j l cij 0 ng.

gi s rng tng hng ho c cc kho v tng nhu cu hng ho cc ca hng l bngnhau, tc l :

bi ton t ra l lp k hoch vn chuyn tin cc l nh nht, vi iu kin l mica hng u nhn hng v mi kho u trao ht hng.

gi xij 0 l lng hng ho phi vn chuyn t kho i n ca hng j. cc vn chuynchuyn hng ho i n tt c cc kho j l :

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Quy hoch tuyn tnh tng qut v chnh tcQuy hoch tuyn tnh tng qut

Tng qut nhng bi ton quy hoch tuyn tnh c th trn, mt bi ton quy hochtuyn tnh l mt m hnh ton tm cc tiu (min) hoc cc i (max) ca hm mc tiutuyn tnh vi cc rng buc l bt ng thc v ng thc tuyn tnh. Dng tng qutca mt bi ton quy hoch tuyn tnh l :

Trong :

? (I) Hm mc tiu

L mt t hp tuyn tnh ca cc bin s, biu th mt i lng no m ta cn phiquan tm ca bi ton.

? (II) Cc rng buc ca bi ton

L cc phng trnh hoc bt phng trnh tuyn tnh n bin s, sinh ra t iu kin cabi ton.

? (III) Cc cc hn ch v du ca cc bin s

Ngi ta cng thng trnh by bi ton quy hoch tuyn tnh di dng ma trn nhsau :

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Gi ai (i=1m) l dng th i ca ma trn A, ta c :

Ngi ta gi :

- A l ma trn h s cc rng buc.

- c l vect chi ph (cT l chuyn v ca c)

- b l vect gii hn cc rng buc.

Quy hoch tuyn tnh dng chnh tc

Bi ton quy hoch tuyn tnh chnh tc l bi ton quy hoch tuyn tnh m trong cc rng buc ch c du = v cc bin s u khng m.

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Ngi ta c th bin i bi ton quy hoch tuyn tnh dng tng qut thnh bi tonquy hoch tuyn tnh dng chnh tc nh cc quy tc sau y :

- Nu gp rng buc i c dng th ngi ta cng thm vo v tri ca rng buc mtbin ph xn+i 0 c du = .

- Nu gp rng buc i c dng th ngi ta tr vo v tri ca rng buc mt bin phxn+i 0 c du = .

Cc bin ph ch l nhng i lng gip ta bin cc rng buc dng bt ng thc thnhng thc, n phi khng nh hng g n hm mc tiu nn khng xut hin tronghm mc tiu.

- Nu bin xj 0 th ta t xj = -xj vi xj 0 ri thay vo bi ton.

- Nu bin xj l tu th ta t

ri thay vo bi ton.

- Trong trng hp trong s cc rng buc c dng m v phi ca dng l gi tr mth i du c hai v c v phi l mt gi tr khng m.

Da vo cc php bin i trn m ngi ta c th ni rng b i ton quy hoch tuyntnh chnh tc l bi ton quy hoch tuyn tnh m trong cc rng buc ch c du =, v phi v cc bin s u khng m.

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V d :

Bin i bi ton quy hoch tuyn tnh sau y v dng chnh tc :

Bng cc thay th :

ta c :

hay :

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Phng n

Xt bi ton quy hoch tuyn tnh chnh tc :

(P)

x=[x1 x2 ... xn] T l mt phng n ca (P) khi v ch khi Ax = b. x=[x1 x2 ... xn] T l mt phng n kh thi ca (P) khi v ch khi Ax = b v x

0 .

Mt phng n ti u ca (P) l mt phng n kh thi ca (P) m gi tr ca hm mctiu tng ng t min/max.

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c im ca cc tp hp cc phng nKhi nim li v cc tnh cht

T hp li

- Cho m im xi trong khng gian Rn . im x c gi l t hp li ca cc im xi

nu :

- Khi x l t hp li ca hai im x1, x2 ngi ta thng vit :

x=x1+(1-)x2 (01)

Nu 0

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Tp hp rng v tp hp ch c mt phn t c xem l tp hp li.

nh l

Giao ca mt s bt k cc tp hp li l mt tp hp li.

nh l

Nu S l mt tp hp li th S cha mi t hp li ca mt h im bt k trong S.

im cc bin ca mt tp hp li

im x trong tp li S Rn c gi l im cc bin nu khng th biu din c xdi dng t hp li tht s ca hai im phn bit ca S.

4- a din li v tp li a din

a din li

Tp hp S tt c cc t hp ca cc im x1, x2,....,xm cho trc c gi l a din lisinh ra bi cc im .

a din li l mt tp hp li.

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Trong a din li ngi ta c th loi b dn cc im l t hp ca cc im cn li.Khi ngi ta thu c mt h cc im, gi s l y1, y2,...,yp (pm) . Cc im nychnh l cc im cc bin ca a din li, chng sinh ra a din li .

S im cc bin ca a din li l hu hn.

Siu phng - Na khng gian

A=[aij]m.n l ma trn cp m.n

Ai (i=1,2,...,m) l hng th i ca A

Siu phng trong Rn l tp cc im x=[x1,x2,.....,xn]T tha

Ai x = bi

Na khng gian trong Rn l tp cc im x=[x1,x2,.....,xn]T tha

Ai x bi

Siu phng v na khng gian u l cc tp hp li.

Tp li a din

Giao ca mt s hu hn cc na khng gian trong Rn c gi l tp li a din.

Tp li a din l mt tp hp li.

Nu tp li a din khng rng v gii ni th l mt a din li

c im ca tp hp cc phng n

nh l

Tp hp cc phng n ca mt quy hoch tuyn tnh l mt tp li a din.

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Nu tp hp li a din ny khng rng v gii ni th l mt a din li, s imcc bin ca n l hu hn.

nh l

Tp hp cc phng n ti u ca mt quy hoch tuyn tnh l mt tp li.

Xt quy hoch tuyn tnh chnh tc

Gi s A=[aij]m.n c cp m.n, m n, rang(A)=m .

Gi Aj (j=1,2,...,n) ct th j ca ma trn A, quy hoch tuyn tnh chnh tc trn c thvit :

Gi S={x=[x1,x2,...,xn]T 0 / x1A1+ x2A2+...+ xnAn=b} l tp cc phng n ca biton.

S l mt phng n khc 0.

nh l

iu kin cn v x0 l phng n cc bin ( im cc bin ca S) l cc ct Aj

ng vi xj0>0 l c lp tuyn tnh.

H qu

S phng n cc bin ca mt quy hoch tuyn tnh chnh tc l hu hn. S thnhphn > 0 ca mt phng n cc bin ti a l bng m.

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Khi s thnh phn > 0 ca mt phng n cc bin bng ng m th phng n cgi l mt phng n c s.

nh l

Nu tp cc phng n ca mt quy hoch tuyn tnh chnh tc khng rng th quy hochtuyn tnh c t nht mt phng n cc bin.

B

Nu

x l mt phng n ti u ca quy hoch tuyn tnh.

x1, x2 l cc phng n ca quy hoch tuyn tnh.

x l t hp li thc s ca x1, x2

th x1, x2 cng l phng n ti u ca quy hoch tuyn tnh.

nh l

Nu quy hoch tuyn tnh chnh tc c phng n ti u th th s c t nht mt phngn cc bin l phng n ti u.

V d : xt quy hoch tuyn tnh chnh tc

Vi h A1 A2 ta tnh c

Vi h A1 A3 ta tnh c

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Vi h A2 A3 ta tnh c

V cc thnh phn ca phng n cc bin l > 0 nn ta chi xt x2 v x3 . Khi :

z(x2)=2.1+3.0=2

z(x3)=2.0+3.1/3=1

Vy x2 = [ 1 0 1 ]Tl mt phng n ti u.

nh l

iu kin cn v mt quy hoch tuyn tnh c phng n ti u l tp cc phngn khng rng v hm mc tiu b chn.

nh l

Nu tp cc phng n ca mt quy hoch tuyn tnh khng rng v l mt a din lith quy hoch tuyn tnh s c t nht mt phng n cc bin l phng n ti u.

Phng php hnh hc

T nhng kt qu trn ngi ta c cch gii mt quy hoch tuyn tnh hai bin bngphng php hnh hc thng qua v d sau :

V d : xt quy hoch tuyn tnh

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A,B,C,D,O l cc im cc bin. Gi tr hm mc tiu ti l :

z(A)=3.6+2.0=18

z(B)=3.4+2.5=22

z(C)=3.2+2.6=18

z(D)=3.0+2.8=8

z(O)=3.0+2.0=0

Phng n ti u ca bi ton t c ti B

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L thuyt c bn v quy hoch tuyn tnh-Mt s v d m uXt bi ton quy hoch tuyn tnh :

a bi ton v dng chnh tc bng cch a vo cc bin ph w1, w2, w3 0 ( lmcho cc rng buc bt ng thc thnh ng thc ) . Ta c :

Thc hin vic chuyn v ta c bi ton ban u nh sau :

(I)

Mt phng n kh thi xut pht ( cha l phng n ti u ) ca bi ton l :

x1 = x2 = x3 = 0

w1=5 w2=11 w3 = 8

Gi tr tng ng ca hm mc tiu l z(x) = 0

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Ngi ta s ci tin phng n xut pht ny c mt phng n mi tt hn, nlm cho gi tr ca hm mc tiu gim xung. Ngi ta lm nh sau :

V h s ca x1 trong hm mc tiu l m v c gi tr tuyt i ln nht nn nu tngx1 t bng 0 ln mt gi tr dng ( cng ln cng tt ) v ng thi vn gi x2 v x3bng 0 th gi tr ca hm ca hm mc tiu s gim xung. Khi cc bin v trica bi ton (I) s b thay i theo nhng phi tho 0 . S thay i ca chng khngnh hng n s thay i ca hm mc tiu. Thc hin tng trn ta c :

Suy ra :

(dng 1 c chn)

Ngi ta chn x1 = 52 nn nhn c mt phng n tt hn c xc nh nh sau :

Gi tr tng ng ca hm mc tiu l

Bc tip theo l bin i bi ton (I) thnh mt bi ton tng ng bng cch tdng 1 ( dng c chn ) tnh x1 theo cc bin cn li v th gi tr nhn c vo ccdng cn li, ta c :

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(II)

Thc hin tng t nh trn, ngi ta tng x3 t bng 0 ln mt gi tr dng cho phpv ng thi vn gi x2 v w1 bng 0 th gi tr ca hm ca hm mc tiu s gimxung. Khi cc bin v tri ca bi ton (II) s b thay i theo nhng phi tho 0 . Ta c :

( dng 3 c chn )

Khi ngi ta chn x3=1 nn thu c mt phng n tt hn c xc nh nh sau:

Gi tr tng ng ca hm mc tiu l z(x)=-13

Bc tip theo l bin i bi ton (II) thnh mt bi ton tng ng bng cch tdng 3 ( dng c chn ) tnh x3 theo cc bin cn li v th gi tr nhn c vo ccdng cn li, ta c :

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(III)

n y v khng c h s no ca hm mc tiu l m nn khng th lm gim gi trca hm mc tiu theo cch nh trn na. Phng n thu c bc sau cng chnhl phng n ti u ca bi ton.

i vi bi ton max, thay cho vic lm tng bin c h s m trong hm mc tiu ngita lm tng bin c h s dng cho n khi cc h s trong hm mc tiu hon ton m.

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Du hiu ti uMa trn c s - Phng n c s - Suy bin

Xt bi ton quy hoch tuyn tnh chnh tc

Ma trn c s

Ngi ta gi c s ca bi ton quy hoch tuyn tnh chnh tc (P) l mi ma trn Bkhng suy bin (c ma trn nghch o) mxm trch ra t m ct ca ma trn rng buc A.Cc ct cn li c gi l ma trn ngoi c s, k hiu l N .

Phng n c s - Phng n c s kh thi

B l mt c s ca bi ton (P).

Khi , bng cch hon v cc ct ca A ngi ta c th lun lun t A di dng :

Phng n c s

Ngi ta gi mt phng n c s tng ng vi c s B l mt phng n c bit,nhn c bng cch cho :

xN = 0

Khi xB c xc nh mt cch duy nht bng cch gii h phng trnh tuyn tnhbng phng php Cramer :

BxB = b xB = B-1b

Phng n c s kh thi

Mt phng n c s l phng n c s kh thi nu :

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xB = B-1b 0

C s tng ng vi mt phng n kh thi c gi l c s kh thi .

V d : xt bi ton quy hoch tuyn tnh dng chnh tc :

Cc ct x5 x6 x3 to thnh mt ma trn c s . Cc bin tng ng c gi l cc bin(trong) c s .

Cc ct x1 x2 x4 to thnh mt ma trn ngoi c s. Cc bin tng ng c gi l ccbin ngoi c s.

Mt phng n c s kh thi ca bi ton l :

x1 x2 x3 x4 x5 x6

0 0 28 0 20 10

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Suy bin

Mt phng n c s kh thi c gi l suy bin nu xB = B-1b 0 c nhng thnhphn bng 0. S suy bin l mt hin tng thng xy ra trong mt s bi ton nh biton vn ti, dng d liu, ng i ngn nht....... y l hin tng kh phc tp (cnhiu cch gii quyt s c xt sau). V vy trong nhng phn tip theo ta gi s rngphng n c s kh thi l khng suy bin, tc l xB = B-1b > 0 ( dng thc s ) .

Du hiu ti u

Theo trn, khi mt bi ton quy hoch tuyn tnh c phng n ti u th tn ti mt cs kh thi (ti u) B* , tc l phng n c s x* tng ng vi B* l phng n ti u.

Vn by gi l xc nh mt th tc tm B*. Chng ta s thy rng th tc csuy ra mt cch trc tip t vic chng minh du hiu ti u sau y.

nh l 4 (du hiu ti u)


Ngi ta thng gi :

cN l chi ph ngoi c s

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cB l chi ph c s

cNT

l chi ph trt gim

cBTB 1Nl lng gia gim chi ph

Chng minh (cho bi ton max)

iu kin

Gi s x* l mt phng n c s kh thi vi ma trn c s B v tho

cNT

= cNT cB

TB 1N 0

th cn chng minh x* l phng n ti u, ngha l chng minh rng vi mi phngn bt k ca bi ton ta lun c :

z(x) z(x*)

Xt mt phng n kh thi x bt k , x tho :

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Tnh gi tr hm mc tiu i vi phng n x ta c :

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Vy x* l phng n ti u.

iu kin cn

Gi s

l phng n ti u vi ma trn c s B, cn chng minh rng : cNT

= cNT cB

TB 1 N0 .

( cN l vect c n-m thnh phn)

Ta s chng minh iu ny bng phn chng.

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Gi s rng tn ti mt thnh phn cs ca cN m cs > 0. Da vo cs ngi ta xy dngmt vect x nh sau :

Trong >0 v Is l mt vect c (n-m) thnh phn bng 0, tr thnh phn th s bng1 . Vy

(*)

Do B-1b 0 nn ngi ta c th chn >0 nh xB > 0

Vy x c chn nh trn s tho :

x 0 (3)

Ta kim chng x tha rng buc ca bi ton bng cch tnh :

T (3) v (4) cho thy x l mt phng n kh thi ca bi ton

By gi ta ch ra mu thun bng so snh gi tr hm mc tiu ti x v x* . Ta c :

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Vy x* khng phi l phng n ti u nn mu thun vi gi thit .

Ch

Qua vic chng minh nh l du hiu ti u ta thy rng t mt phng n c s khthi cha ti u c th tm c cc phng n kh thi cng lc cng tt hn nh lp linhiu ln cng thc (*). Vn c t l i lng c chn nh th no nhanhchng nhn c phng n ti u.

B


vi B l mt c s kh thi no v x0 l phng n c s tng ng, tc l

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v z(x0) = cBTB 1b

Xt

.

Nu tn ti mt bin ngoi c s xs sao cho cs>0 vi csl thnh phn th s ca cN th :

a- Hoc l ngi ta c th lm tng mt cch v hn gi tr ca xs m khng i ra khitp hp cc phng n kh thi, v trong trng hp ny phng n ti u ca bi tonkhng gii ni.

b- Hoc l ngi ta c th xc nh mt c s kh thi khc l B c phng n c s khthi x tng ng vi n l tt hn , tc l :

z(x0) < z( x)

Chng minh

Trong qu trnh chng minh nh l du hiu ti u ta c phng n mi c xc nhnh sau :

Hai trng hp c th xy ra nh sau :

a- Trng hp Ns 0

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Trong trng hp ny xs c th nhn mt gi tr ln tu m vn m bo xB 0, nghal x lun lun tho 0 . Khi nh bit gi tr hm mc tiu tng ng l

vi cs c th ln v hn th gi tr ca hm mc tiu l khng gii ni.

b- Trng hp tn ti i=1m sao cho Nis > 0

( Nis > 0 l thnh phn th i ca Ns)

Trong trng hp ny gi tr ca >0 m xs c th nhn khng th tng v hn v phim bo xB>0. Gi tr ln nht ca m xs c th nhn c xc nh nh sau :

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Ghi ch :

Trong trng hp bi ton khng suy bin, nu c xc nh mt cch duy nht thphng n mi x c ng m thnh phn khc 0. Tht vy :

- Bin xs ang bng 0 trong phng n x0 tr thnh dng tht s v xs =

- Bin xr ang dng tht s by gi nhn gi tr :

Vy phng n mi x l mt phng n c s. N tng ng vi c s B c suy rat B bng cch thay th ct r bng ct s.

Ngi ta ni rng hai c s B v B l k nhau, chung tng ng vi nhng im ccbin k nhau trong tp hp li S cc phng n kh thi ca bi ton.

CU HI CHNG 1

1- Trnh by cc bc nghin cu mt quy hoch tuyn tnh.

2- nh ngha quy hoch tuyn tnh chnh tc.

3- Trnh by khi nim v phng n ca mt quy hoch tuyn tnh.

4- Trnh by c s l thuyt ca phng php hnh hc gii mt quy hoch tuyn tnhhai bin.

BI TP CHNG 1

1- Mt nh my cn thp c th sn xut hai loi sn phm : thp tm v thp cun.Nu ch sn xut mt loi sn phm th nh my ch c th sn xut 200 tn thp tmhoc 140 tn thp cun trong mt gi . Li nhun thu c khi bn mt tn thp tm l25USD, mt tn thp cun l 30USD. Nh my lm vic 40 gi trong mt tun v thtrng tiu th ti a l 6000 tn thp tm v 4000 tn thp cun .

Vn t ra l nh my cn sn xut mi loi sn phm l bao nhiu trong mt tun t li nhun cao nht. Hy trnh by bi ton quy hoch tuyn tnh cho vn trn.

2- C 3 ngi cng phi i mt qung ng di 10km m ch c mt chic xe p mtch ngi. Tc i b ca ngi th nht l 4km/h, ngi th hai l 2km/h, ngi th

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ba l 2km/h. Tc i xe p ca ngi th nht l 16km/h, ngi th hai l 12km/h,ngi th ba l 12km/h.

Vn t ra l lm sao thi gian ngi cui cng n ch l ngn nht. Hy trnhby bi ton quy hoch tuyn tnh cho vn trn.

3- Mt nh my sn xut ba loi tht : b, ln v cu vi lng sn xut mi ngy l 480tn tht b, 400 tn tht ln, 230 tn tht cu. Mi loi u c th bn c dng tihoc nu chn. Tng lng cc loi tht c th nu chn bn l 420 tn trong gi v250 tn ngoi gi. Li nhun thu c t vic bn mt tn mi loi tht c cho trongbng sau y :

Hy trnh by bi ton quy hoch tuyn tnh nh my sn xut t li nhun cao nht.

4- Mt xng mc lm bn v gh. Mt cng nhn lm xong mt ci bn phi mt 2gi, mt ci gh phi mt 30 pht. Khch hng thng mua nhiu nht l 4 gh kmtheo 1 bn do t l sn xut gia gh v bn nhiu nht l 4:1. Gi bn mt ci bn l135USD, mt ci gh l 50USD. Hy trnh by bi ton quy hoch tuyn tnh xngmc sn xut t doanh thu cao nht, bit rng xng c 4 cng nhn u lm vic 8 gimi ngy.

5- Mt nh my sn xut hai kiu m. Thi gian lm ra mt ci m kiu th nhtnhiu gp 2 ln thi gian lm ra mt ci kiu th hai. Nu sn xut ton kiu m th haith nh my lm c 500 ci mi ngy. Hng ngy, th trng tiu th nhiu nht l150 ci m kiu th nht v 200 ci kiu th hai. Tin li khi bn mt ci m kiu thnht l 8USD, mt ci m th hai l 5USD. Hy trnh by bi ton quy hoch tuyn tnh nh my sn xut t li nhun cao nht.

6- Trong hai tun mt con g mi c 12 trng hoc p c 4 trng n ra g con.Sau 8 tun th bn tt c g con v trng vi gi 0,6USD mt g v 0,1USD mt trng.Hy trnh by bi ton quy hoch tuyn tnh b tr 100 g mi trng hoc p trngsao cho doanh thu l nhiu nht.

7- Gii nhng bi ton quy hoch tuyn tnh sau y bng phng php hnh hc :

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Gii thut n hnh c bnChng ny trnh by mt phng php gii bi ton quy hoch tuyn tnh lphng php n hnh. Phng php n hnh c George Bernard Dantzig a ranm 1947 cng lc vi vic ng khai sinh ra quy hoch tuyn tnh. y l mt phngphp thc s c hiu qu gii nhng bi ton quy hoch tuyn tnh c ln trong thct. Vi cch nhn hin i tng ca phng php n hnh rt n gin. C nhiucch tip cn phng php n hnh, chng ny trnh by mt trong cc cch .

C s xy dng gii thut n hnh c bn

Xt bi ton quy hoch tuyn tnh chnh tc :

Gi s rng B0 l mt c s kh thi xut pht ca bi ton ( khng nht thit l m ctu tin ca ma trn A ) . Thut ton n hnh c bn c xy dng da trn cc bcsau :

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V mt hnh hc, gii thut ny c hiu nh l mt qu trnh duyt qua cc im ccbin ca a din li S cc phng n kh thi ca bi ton.

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V mt i s, gii thut ny c hiu nh l mt qu trnh xc nh mt chui cc matrn c s k B0 B1 B2 ......... m cc phng n c s tng ng x0 x1 x2........ l ngycng tt hn, tc l :

z(x0) < z(x1) < z(x2) .............

Ch :

Nu c s ban u B0 chnh l m ct u tin ca ma trn A th trong gii thut trn tchnh l r .

nh l v s hi t

Vi gi thit bi ton khng suy bin, gii thut n hnh trn y s hi t v phngn ti u sau mt s hu hn ln lp.

Bng s thng k ngi thy rng ni chung gii thut n hnh s hi t vi s ln lpt nht phi l t m n 3m ( m l s rng buc ) .

Gii thut n hnh c bn


Gi s rng sau khi hon v cc ct trong A ta chn c ma trn c s B tho s phnhoch sau y :

Gii thut n hnh c bn c thc hin nh sau :

a- Tnh ma trn nghch o B-1

b- Tnh cc tham s :

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. Phng n c s kh thi tt hn

. Gi tr hm mc tiu z(x) = cBTxB

. Ma trn__N= B-1N

c- Xt du hiu ti u :

d- Xc nh ch s ca phn t pivot trong ma trn N

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e- Thc hin cc hon v :

. Ct th s trong ma trn N vi ct th r trong ma trn B

. Phn t th s trong cNT vi phn t th r trong cB

T

. Bin xs trong xNT vi bin xr trong xB

T

f- Quay v (a)

V d : Tm phng n ti u cho bi ton quy hoch tuyn tnh chnh tc sau y bnggii thut n hnh c bn

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Ln lp1

a- Tnh ma trn nghch o B-1

b- Tnh cc tham s

. Phng n c s kh thi tt hn :

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c- Xt du hiu ti u :

Chuyn sang bc d

d- Xc nh ch s ca pivot

. Xc nh ch s ct pivot s :

cs = max{ck > 0 cN} = max{2 , 1} = 2 = __c 1Vy s=1

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Ma trn ct s=1 trong ma trn N l

11

1

righ

[][]

N1 =

. Xc nh ch s dng pivot r :

Vy r = 1

e- Hon v

. Ct th s=1 trong ma trn N v ct th r=1 trong ma trn B

. Phn t th s=1 trong cNT vi phn t th r=1 trong cB

T

. Bin th s=1 trong xNT vi bin th r=1 trong xB

T

f- Quay v bc a

Ln lp 2

a. Tnh ma trn nghch o B-1

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b- Tnh cc tham s


. Gi tr hm mc tiu :

. Tnh ma trn :

c- Xt du hiu ti u :

Chuyn sang bc d

d- Xc nh ch s ca pivot

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. Xc nh ch s ct pivot s :

e- Hon v

. Ct th s=2 trong ma trn N v ct th r=2 trong ma trn B

. Phn t th s=2 trong cNT vi phn t th r=2 trong cB

T

. Bin th s=2 trong xNT vi bin th r=2 trong xB

T

f- Quay v bc a

Ln lp 3

a. Tnh ma trn nghch o B-1

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b- Tnh cc tham s


. Gi tr hm mc tiu :

. Tnh ma trn :

c- Xt du hiu ti u :

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: dng

Vy phng n ti u s l :

Gi tr hm mc tiu l z(x) = 9 vi x1 = 4 v x2 = 1

Ch trong trng hp suy bin

Trong trng hp bi ton suy bin, ngha l br = 0, ta c :

xs =br

ars= 0

cho nn gi tr ca hm mc tiu khng thay i khi thay i c s, v :

z(x) = z(x) + csxs = z(x)

Vy th, c th sau mt s ln thay i c s li quay tr v c s gp v lp nh vymt cch v hn. Ngi ta c nhiu cch khc phc hin tng ny bng cch xotrn mt cht cc d liu ca bi ton, s dng th tc t vng, quy tc chn pivot trnh b kh.

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Phng php bin gi ci binBi ton ci bin

Ci bin bi ton quy hoch tuyn tnh

Ngi ta c th bin i mt bi ton quy hoch tuyn tnh chnh tc thnh dng chunbng cch cng mt cch ph hp vo v tri ca rng buc i mt bin gi xn+i 0 lm xut hin ma trn n v. V cc bin gi ci bin c nh hng n hm mc tiunn cng s c s ci bin hm mc tiu.

Vy, ngi ta c th bin i bi ton quy hoch tuyn tnh tng qut, gi l bi tonxut pht, thnh bi ton dng chun, gi l bi ton ci bin (m rng)

V d :

Bin i bi ton quy hoch tuyn tnh sau y thnh dng chun

Bi ton xut pht c cc bin, ma trn rng buc v chi ph :

Bng cch thm bin gi x5, x6 ln lt vo rng buc 2 v 3 . Ta c bi ton cibin :

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z'(x) l hm mc tiu ci bin s c gii thch trong phn tip theo.

Cc bin, ma trn rng buc cc h s v chi ph ca bi ton ci bin l

Quan h gia bi ton xut pht v bi ton ci bin

Ngi ta kim chng rng :

- Nu xT = [x1x2...xn] l phng n (ti u) ca bi ton xut pht th xT

= [x1x2...xn0 0...0]l phng n (ti u) ca bi ton ci bin tng ng.

Vy nu bi ton ci bin khng c phng n ti u th bi ton xut pht cng skhng c phng n ti u.

- Nu xT

= [x1x2...xn0 0...0] l phng n ti u ca bi ton ci bin th xT = [x1x2...xn] lphng n ti u ca bi ton xut pht

- Nu bi ton ci bin c mt phng n ti u m trong c t nht mt bin gi cgi tr dng th bi ton xut pht khng c phng n ti u.

- Nu bi ton ci bin (dng chun) c phng n ti u th cng s phng n c sti u.

V d

1- Xt bi ton :

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Bi ton ci bin khng c phng n ti u nn bi ton xut pht cng khng cphng n ti u .

2- Xt bi ton :

Phng n ti u ca bi ton ci bin :

Phng n ti u ca bi ton xut pht :

3- Xt bi ton :

Phng n ti u ca bi ton ci bin :

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Bi ton xut pht khng c phng n ti u .

Hai phng php bin gi ci bin thng dng l phng php hai pha v phng phpM v cng ln .

Phng php hai pha

Pha 1

Tm phng n ti u cho bi ton ci bin vi hm mc tiu ci bin l :

min (tng tt c bin gi ci bin)

Pha 2

Tm phng n ti u cho bi ton xut pht vi phng n c s kh thi xut pht lphng n ti u tm c pha 1. pha 2 ny cc bin gi ci bin b loi ra khi matrn cc h s rng buc, v vect chi ph c cp nht li, do du hiu ti u cngc cp nht li

y l phng php thun li cho vic lp trnh ng dng gii thut n hnh ci tin.

V d : Xt bi ton quy hoch tuyn tnh

a bi ton v dng chnh tc bng cch thm bin ph x4 , x5 ta c

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Pha 1

Thm bin gi (ci bin ) x6 0 vo rng buc th hai c ma trn n v . Khi bi ton ci bin c dng :

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Pha 2

Loi b bin gi ci bin x6 0

Khi to

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Kt qu ca bi ton cho :

. Phng n ti u

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. Gi tr hm mc tiu z(x)=z(x3)= 8

Phng php M v cng ln

Phng php M v cng ln ( M l s v cng ln ) tng t nh phng php hai pha,ngoi tr pha 1 hm mc tiu ci bin c dng sau y cho bi ton max/min

max [z(x) - M*( tng cc bin gi ci bin) ]

min [z(x) + M*( tng cc bin gi ci bin) ]

Bng phng php ny, trong qu trnh ti u, cc bin gi ci bin s c loi dnra khi ma trn c s : tt c u bng 0. Nu trong qu trnh tm phng n ti u mkhng loi b c cc bin gi ci bin ra khi c s th bi ton v nghim.

So vi phng php hai pha th phng php ny trnh c vic phi cp nht li dliu cho bi ton gc nhng khng tin li bng trong lp trnh ng dng.

V d : Xt bi ton tng t nh trn

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Tm phng n ti u cho bi ton ci bin ny bng phng php n hnh ci tin

Khi to

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Do x6 = 0 (v ngoi c s) nn b loi ra khi bng v ta tip tc tm phng n ti ucho bi ton gc cho c phng n c s kh thi c khi to nh sau :

Cc bc tip theo c thc hin ging nh phng php hai pha.

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Quy hoch tuyn tnh suy binKhi thc hin thut ton n hnh trng hp bt thng c th xy ra l khi xc nhbin ra th tn ti t s, tc

biaik

= 0 l tn ti bi=0, hay khng c t s no dng tht s.Ngi ta xem y l trng hp suy bin. Khi mt bng n hnh ri vo tnh trng suybin th c th gy kh khn m cng c th khng khi ta tip tc thc hin thut tonn hnh.

Cc v d v quy hoch tuyn tnh suy bin

V d 1 : xt quy hoch tuyn tnh :

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y l trng hp suy bin, bin vo l x2, n c tng ln n mc vn tha nhngiu kin v du ca cc bin trong c s x3, x3, x5 . l :

Nh vy x2 c th ln ty nn hm mc tiu khng b gii ni. Vy bi ton khng cphng n ti u. Trng hp ny bng n hnh khng c t s no dng tht s xc nh bin ra.


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y l bng n hnh ti u.


a bi ton v dng chun :

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y l bng n hnh ti u

V d 4 : xt quy hoch tuyn tnh

vi ma trn h s

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c cha ma trn n v . p dng phng php n hnh ci tin

x2 vo , x6 ra

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x6 vo , x4 ra

Bng n hnh hin thi ging vi bng n hnh xut pht : y l hin tng xoayvng .

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X l trng hp suy bin

Theo cc v d trn, trong trng hp quy hoch tuyn tnh suy bin th sau mt s lnlp c th phng n nhn c vn nh c m khng c s thay i no, c th phngn nhn c tt hn, c th phng n nhn c l mt phng n nhn trc ri v t c xoay vng mi. Do nu khng c bin php phng nga th thut tonn hnh s c th ko di v tn.

Khi thc hin thut ton n hnh th hin tng suy bin xy ra khi c s tnh c khln nhau lm cho tn ti bi no bng 0. Trong trng hp ny c th c nhiu bintha iu kin ca bin ra. Gp trng hp ny cn phi la chn bin ra sao cho trnhc hin tng xoay vng.

Ngi ta thng dng phng php nhiu lon, phng php t vng trnh s tnhc kh ln nhau ny. Trong thc tin tnh ton ngi ta ra mt quy tc x l khn gin, gi l quy tc Bland, khi dng gii thut n hnh gii cc quy hoch tuyntnh suy bin, l :

Vi xk l bin vo , bin ra xr c chn l bin c ch s nh nht tha iu kin chnbin ra :

p dng quy tc Bland ta thy :

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Bin ra c th l x1 hay x2 . Chn x1

Bin ra l x2

Bin ra c th l x4 hay x5 . Chn x4

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Bin ra l x5

Bin ra l x3

n y khng cn hin tng suy bin.

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Bin vo l x7

CU HI CHNG 2

1- Trnh by c s l thuyt ca thut ton n hnh c bn.

2- nh ngha quy hoch tuyn chun.

3- Trnh by cc bc lp bng n hnh theo php ton trn dng .

4- Ci bin mt quy hoch tuyn tnh tng qut nh th no ? . Cch gii quy hochtuyn tnh ci bin v quy hoch tuyn tnh gc.

BI TP CHNG 2

1- Tm phng n ti u ca bi ton sau y bng phng php n hnh c bn

2- Tm phng n ti u ca bi ton sau bng phng php n hnh ci tin

a) max z = 5x1 + 3x2

2x1 + 2x2 80

x1 30

x1, x2 0

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b) max z = x1 + 2x2

2x1 + 3x2 7

x1 - x2 1

x1 0, x2 0

c) max z = 5x1 + 3x2 + x3

2x1 + 3x2 - x3 4

3x1 - x2 + 2x3 2

x1 + x2 + 3x3 5

x1 0, x2 0, x3 0

3- Tm phng n ti u ca cc bi ton sau bng phng php bin gi ci bin.

a) max z = 3x1 - x2

2x1 + x2 100

x1 10

x2 0

b) min w = 3x1 + x2

x1 + x2 3

2x1 5

x1, x2 0

c) max z = 3x1 + x2 - 3x3

x1 + 2x2 - x3 = 2

-10x2 + 5x3 = 5

-3x2 + 2 x3 = 4

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xi 0, i = 13

d

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Khi nim v i ngui ngu l mt khi nim c bn ca vic gii bi ton quy hoch tuyn tnh v lthuyt i ngu dn n mt kt qu c tm quan trng v mt l thuyt v c mt thchnh.

i ngu ca quy hoch tuyn tnh dng chnh tc

Xt mt bi ton quy hoch dng chnh tc:

Gi s rng x* l phng n ti u cn tm ca bi ton v x0 l mt phng n ca biton th mt cn trn ca gi tr mc tiu ti u c xc nh v :

cTx* cTx0

Tuy cha tm c phng n ti u x* nhng nu bit thm c mt cn di ca gitr mc tiu ti u th ta gii hn c phn no gi tr mc tiu ti u. Ngi ta clng cn di ny theo cch nh sau :

Vi mi vect xT = [x1 x2 ... xn] 0 thuc Rn cha tho rng buc ca bi ton, tc l

b Ax 0

ngi ta ni lng bi ton trn thnh bi ton ni lng :

yT = [ y1 y2 ... ym] tu Rm

Gi g(y) l gi tr mc tiu ti u ca bi ton ni lng, ta c :

g(y) = min { cTx + yT(b - Ax) } (x 0)

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cTx + yT(b - Ax)

Trong trng hp x l phng n ca bi ton ban u, tc l :

b - Ax = 0

th

g(y) cTx

Vy g(y) l mt cn di ca gi tr mc tiu bt k nn cng l cn di ca gi trmc tiu ti u.

Mt cch t nhin l ngi ta quan tm n bi ton tm cn di ln nht, l :

max g(y)

y tu Rm

Bi ton ny c gi l bi ton i ngu ca bi ton ban u. Trong phn sau ngita s chng minh gi tr mc tiu ti u ca bi ton i ngu bng vi gi tr mc tiuti u ca bi ton gc ban u.

Ngi ta a bi ton i ngu v dng d s dng bng cch tnh nh sau :

g(y) = min { cTx+yT(b - Ax) } (x 0)

= min { cTx + yTb - yTAx } (x 0)

= min { yTb + (cT - yTA)x } (x 0)

= yTb + min { (cT - yTA)x } (x 0)

Ta thy :

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nh ngha i ngu trong trng hp quy hoch tng qut

Trong trng hp quy hoch tuyn tnh tng qut, nhng quy tc sau y c p dng xy dng bi ton i ngu :

- Hm mc tiu i ngu :

. max min

- Bin i ngu :

. Mi rng buc mt bin i ngu

- Chi ph i ngu v gii hn rng buc :

. Chi ph i ngu gii hn rng buc

- Ma trn rng buc i ngu :

. Ma trn chuyn v

- Chiu ca rng buc v du ca bin :

. Rng buc trong bi ton max c du th bin i ngu trong bi ton min c du 0( tri chiu )

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. Rng buc trong bi ton max c du = th bin i ngu trong bi ton min c du ty.

. Rng buc trong bi ton max c du th bin i ngu trong bi ton min c du 0( tri chiu )

. Bin ca bi ton max c du 0 th rng buc i ngu trong bi ton min c du (cng chiu )

. Bin ca bi ton max c du ty th rng buc i ngu trong bi ton min c du =

.

. Bin ca bi ton max c du 0 th rng buc trong bi ton i ngu min c du (cng chiu )

Xt cc rng buc dng ma trn ca mt bi ton quy hoch tuyn tnh tng qut nhsau :

V d

a- Hai bi ton sau y l i ngu :

b- Hai bi ton sau y l i ngu :

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i vi cp bi ton i ngu (P) v (D) ch xy ra mt trong ba trng hp sau :

- C hai bi ton u khng c phng n ti u .

- C hai bi ton u c phng n, lc chng u c phng n ti u v gitr hm mc tiu i vi hai phng n ti u l bng nhau.

- Mt trong hai bi ton khng c phng n, cn bi ton kia th c phng n,khi bi ton c phng n khng c phng n ti u.

Cc nh l v s i ngu

nh l 1 ( i ngu yu )

Xt hai bi ton i ngu :

Nu x l phng n ca bi ton (P)

y l phng n ca bi ton (D)

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th z(x) w(y)

ngha l gi tr hm mc tiu ca bi ton max khng vt qu gi tr hm mc tiu cabi ton i ngu min trn cc phng n bt k ca mi bi ton .

Chng minh

x l phng n ca (P) nn : Ax = b

yTAx = y

Tb = bTy = w(y)

y l phng n ca (D) nn : ATy c

yTA cT

yTAx cTx = z(x)

Vy z(x) w(y)

nh l ny c pht biu v chng minh cho hai bi ton i ngu trong trng hptng qut .

nh l 2


x l phng n kh thi ca bi ton (P)

y l phng n kh thi ca bi ton (D)

Nu z(x) = w(y) th x, y ln lt l phng n ti u tng ng ca (P v (D).

Chng minh

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- Nu x khng l phng n ti u ca bi ton (P) th tn ti mt phng nx sao cho :

z(x) < z(x)

w(y) < z(x) : iu ny mu thun vi nh l 1.

- Nu y khng l phng n ti u ca bi ton (D) th tn ti mt phng ny sao cho :

w(y) < w(y)

w(y) < z(x) : iu ny mu thun vi nh l 1.

Vy x v y ln lt l phng n ti u ca (P) v (D).

nh l 3


Nu x* l phng n ti u ca bi ton (P) i vi c s B th phng n tiu y* ca bi ton (D) c tnh bi cng thc :

y

T = cBTB 1

Chng minh

Do x* l phng n ti u ca (P) vi c s B nn tho du hiu ti u

cT cBT.B 1A 0

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cBT.B 1A cT

y

TA cT

y* l mt phng n ca (D)

Mt khc x* c tnh bi cng thc :

v gi tr mc tiu ti u ca (P) l :

z(x*) = cTx* = cBTxB

Ta c :

Theo nh l 2 th y* l phng n ti u ca (D).

nh l ny cho php tm phng n ti u ca bi ton quy hoch tuyntnh i ngu t bi ton gc. Trong :

- cBT c xc nh trong bng n hnh ti u ca (P).

- B-1 gm m ct tng ng vi m ct ca ma trn c s ban u ly t bngn hnh ti u ca bi ton gc.

nh l 4 ( s i ngu)

Xt hai bi ton i ngu

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- Nu (P) v (D) u c phng n kh thi th chng c phng n ti u vgi tr ca hm mc tiu tng ng l bng nhau.

- Nu mt trong hai bi ton c phng n ti u khng gii ni th bi toncn li khng c phng n kh thi.

Chng minh

- y l kt qu ca nh l 3 .

- Gi s rng phng n ti u ca (D) khng gii ni, tc l tn ti mtphng n kh thi y ca (D) sao cho w(y)= bTy nh tu . iu ny cng cngha l : vi mi M>0 ln tu lun tm c mt phng n kh thi yca(D) sao cho :

bTy M

Nu (P) c phng n kh thi l x th theo nh l 1 ta c :

iu ny dn n mu thun

nh l 5 (tnh b sung )

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Chng minh

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Gii thut i nguXt hai bi ton i ngu :

Chng ta s xt xem gii thut n hnh c bn bit trong chng trc c p dngnh th no i vi bi ton i ngu.

Gi s rng B l mt c s ca bi ton (P) tho :

y = cBTB 1 v NTy cN

Nu B cng l mt c s kh thi ca bi ton gc, tc l

, th (theo nh l i ngu) y, x ln lt l phng n ti u ca bi ton i ngu v

bi ton gc. Nu khng th

xBxN

righ

[]

x =

khng l phng n ca bi ton gc v

xB = b = B 1b khng th 0.

tin vic trnh by ta xt (m=3 , n=5) :

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v bi ton i ngu

(D)

Ngi ta a (D) v dng chnh tc bng cch thm cc bin ph y4 y5, y6, y7, y8 0.Chng khng nh hng n hm mc tiu.

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Cc d liu ca (D) c trnh by trong bng sau :

Gi s rng m ct u tin ca A l mt c s B ca (P) th hai bng trn c trnh byrt gn nh sau :

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a bi ton i ngu v dng chun ngi ta nhn (bn tri) bng (D) vi bng sauy :

Khi ngi ta c bng kt qu c dng :

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Bng ny cho ta mt quy hoch tuyn tnh dng chun vi ma trn n v (c s) tngng vi cc ct y1 y2 y3 y7 y8 .

p dng gii thut n hnh c bn vo kt qu ny cho ta quy tc i c s nh sau :

Tnh : b = B 1b 0

a- Nu b 0 th gii thut kt thc, khi :

y = cBTB 1 l phng n ti u ca bi ton i ngu .

xBxN

righ

b0

righ

[]

x =

l phng n ti u ca bi ton gc .

b- Nu tn ti r sao cho br b,br < 0 th xy ra mt trong hai trng hp sau :

- Nu trong dng r ca N c thnh phn < 0 th ngi ta tnh :

Nh vy : i vi bi ton i ngu th bin yr i vo c s v bin ys ra khi c s,trong khi i vi bi ton gc th bin xs i vo c s v bin xr ra khi c s.

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- Nu mi thnh phn trong dng r ca N u > 0 th phng n ti u ca bi ton ingu l khng gii ni, iu ny (theo nh l i ngu) dn n bi ton gc khng cphng n.

Ta c th chn bi ton (D) hoc (P) gii tm phng n ti u bng phng phpn hnh, t suy ra phng n ti u ca bi ton cn li theo kt qu trn. Trong vd ny ta chn bi ton (D) gii v c cha sn ma trn n v.

Gii bi ton (D) bng phng php n hnh ci tin ta c :

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Gii thut dng v tho du hiu ti u ca bi ton min.

Phng n ti u ca bi ton (D) l :

CU HI CHNG 3

1- Bn hiu nh th no v khi nim i ngu ?

2- Quy hoch tuyn tnh i ngu ca mt quy hoach tuyn tnh chnh tc c dng nhth no ?

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3- Bn hy nu ra cc quy tc i ngu. Cho v d .

4- Gi tr hm mc tiu ca hai quy hoch tuyn tnh i ngu th nh th no ? . Chngminh

BI TP CHNG 3

1- Xt bi ton quy hoch tuyn tnh

max z = 7x1 + 5x2

2x1 + 3x2 19

(P) 2x1 + x2 13

3x2 15

3x1 18

x1 , x2 0

a- Tm bi ton i ngu (D) t bi ton (P)

b- Tm phng n ti u cho bi ton (P)

c- T bng n hnh ti u ca (P). Hy tm phng n ti u cho bi ton (D)


min w= x1 + x2

x1 - 2x3 + x4 = 2

(D) x2 - x3 + 2x4 = 1

x3 - x4 + x5 = 5

xi 0, i = 15

a- Tm bi ton i ngu ca bi ton (D)

b- Tm phng n ti u ca bi ton (D)

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c- T bng n hnh ti u ca bi ton (D). Hy tm phng n ti u cho bi ton ingu cu a.


min w = -2x1 - x4

x1 + x2 + 5x3 = 20

(D) x2 + 2x4 5

x1 + x2 - x3 8

xi ty (i=1 4)

Tm bi ton i ngu (P) ca bi ton (D). T bi ton (P) hy ch ra rng (P) khngtn ti phng n ti u do (D) cng tn ti phng n ti u.

4- Cho bi ton quy hoch tuyn tnh

(D)

1- Tm bi ton i ngu ca bi ton cho.

2- Gii bi ton cho ri suy ra kt qu ca bi ton i ngu.

5- Cho bi ton quy hoch tuyn tnh

(D)

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a- Tm bi ton i ngu ca bi ton cho.

b- Gii bi ton i ngu ri suy ra kt qu ca bi ton cho.

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ng dng quy hoch tuyn tnh-M uTrong chng ny, chng ta s tm hiu s lc mt s khi nim v phng php cbn trong l thuyt tr v mt s bi ton thc t m ngi ta s a v bi ton quyhoch tuyn tnh gii .

Trong thc t hay gp tnh hung l phi chn mt quyt nh (bp bnh) dophi i mt vi mt i th thng minh v c quyn li i lp vi ta : v dtrong cc tr chi tranh chp, trong qun s, trong vn ng tranh c....

Nghin cu vic chn quyt nh trong nhng trng hp i khng ny ctn gi l l thuyt tr chi. y ngi chn quyt nh v i th u cgi l ngi chi. Mi ngi chi c mt tp hp cc hnh ng la chnc gi l chin lc.

Chng ta xt mt trng hp n gin l tr chi hai ngi : phn thngs l ci c ca mt ngi v chnh l ci mt ca ngi kia.

Gii mt tr chi ngha l tm chin lc tt nht cho mi ngi chi. Haingi chi thng c k hiu l A v B, chin lc tng ng ca mingi c k hiu l :

A : i (i=1m)

B : j (j=1n)

Gii thng ng vi chin lc (i,j) ca hai ngi c k hiu l aij v cvit thnh mt bng nh sau :

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i vi A :

- Nu A i nc 1 (dng 1) th A s :

. Thng 1 im nu B i nc 1 (thng)

. Thng 0 im nu B i nc 2 (ho)

. Thng -2 im nu B i nc 3 (thua)

. Thng 1 im nu B i nc 4 (thng)

Nhng trng hp cn li l tng t .

i vi B :

- Nu B i nc 2 (ct 2) th B s :

. Thua 0 im nu A i nc 1

. Thua 2 im nu A i nc 2

. Thua -1 im nu A i nc 3

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Nhng trng hp cn li l tng t .

Nghim ti u ca tr chi, c khi gi tt l nghim, l b chin lc (i*,j*) ctnh cht l nu mt ngi ly chin lc khc cn ngi kia vn gi nguynth phn thng cho ngi i khc s b thit hi. Gii tr chi c ngha ltm nghim ti u.

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Bi ton vn tiM u

Bi ton vn ti l bi ton quan trng nht trong cc bi ton quy hoch tuyn tnh.Ngi ta tng kt rng 85% cc bi ton quy hoch tuyn tnh gp trong ng dng l biton vn ti hoc m rng ca n. Thut ng bi ton vn ti thng c hiu l biton vn chuyn sao cho cc ph nh nht.

Cc khi nim c bn

Bi ton vn ti c m t nh l mt bi ton v dng d liu gm tp hp cc nt Nc chia thnh hai phn ri nhau : cc nt ngun S v cc nt ch D, tc l :

i vi bi ton vn ti ngi ta thng k hiu

si S l ngun pht nt i(i=1m)

dj D l nhu cu thu ca nt j (j=1n)

Trong trng hp cc ngun pht khng chuyn ht sang cc nt cu v nhu cuth bi ton vn ti c gi l bi ton vn ti m. C th a mt bi ton vn ti m

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v mt bi ton vn ti (ng) bng cch thm vo mt nt cu gi th (n+1) vi nhucu c xc nh nh sau :

Bi ton vn ti cn bng thu pht

Thit lp bi ton

C m ni A1, A2,....,Am cung cp mt loi hng vi khi lng tng ng l a1,a2,....,am. Hng c cung cp cho n ni B1, B2,...., Bn vi khi lng tiu th tngng l b1, b2,....,bn.

Cc ph chuyn ch mt n v hng t im pht Ai n im thu Bj l cij .

Hy lp k hoch vn chuyn t mi im pht n mi im thu bao nhiu hng :

- Cc im pht u pht ht hng

- Cc im thu u nhn hng

- Tng cc ph phi tr l t nht

Gi xij l lng hng chuyn t im pht Ai n im thu Bj , xij 0 .

V tng lng hng pht i t mi im pht Ai n mi im thu Bj bng lng hngpht t Ai nn :

xi1 + xi2 + ....+xin = ai(i = 1,2,...,m)

V tng lng hng thu c ti mi im thu Bj t mi im pht Ai bng lng hngcn thu ti Bj nn :

x1j + x2j + ....+xmj = bji(j = 1,2,..., n)

tng cc ph l t nht cn phi c :

Vi cc phn tch trn ta c m hnh ca bi ton nh sau :

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Phng n - Phng n ti u

Mt ma trn X=[xij]m.n tha (2) v (3) c gi l phng n, tha thm (1) c gi lphng n ti u.

Dng bng ca bi ton vn ti

C th gii bi ton vn ti theo cch ca quy hoch tuyn tnh. Tuy nhin do tnh chtc bit ca bi ton vn ti nn ngi ta ngh ra mt thut ton hiu qu hn. Trctin ngi ta trnh by bi ton vn ti di dng bng nh sau :

Trong bng mi hng m t mt im pht, mi ct m t mt im thu, mi m tmt tuyn ng i t mt im pht ti mt im thu.

Dy chuyn - Chu trnh

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Mt dy cc ca bng m hai lin tip nm trong cng mt hng hoc mt ct, ba lin tip khng cng nm trn mt hng hoc mt ct c gi l mt dy chuyn. Tathy rng hai lin nhau trong mt dy chuyn c ch s hng hoc ch s ct bng nhau

chn - loi

Gi s ma trn X=[xij]m.n (i=1,2,...,m) (j=1,2,...,n) l mt phng n ca bi ton vnti.

Nhng trong bng tng ng vi xij >0 c gi l chn, nhng cn li c gil loi.

Phng n c bn

Mt phng n m cc chn khng to thnh mt chu trnh c gi l phng n cbn.

Mt phng n c m+n-1 chn c gi l khng suy bin, c t hn m+n-1 chnc gi l suy bin. Trong trng hp suy bin ngi ta chn b sung vo phng nc bn mt s loi c lng hng bng 0 phng n c bn tr thnh khng suybin

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Gii bi ton vn ti

Xt bi ton vn ti c s lng pht, s lng thu v ma trn cc ph dng bng nhsau :

80 20 60

50 5 4 1

40 3 2 6

70 7 9 11

LP PHNG N C BN BAN U

Phng n c bn ban u c xc nh bng cch u tin phn phi nhiu nht vo c cc ph nh nht (r,s) ( gi l chn). Khi : nu im pht r pht ht hng thxa hng r ca bng v s lng cn thu ti im s ch cn l bs-ar ; nu im thu s nhn hng th xa ct s ca bng v s lng pht cn li ti im pht r l ar-bs

Bng mi thu c c kch thc gim i. Tip tc phn phi nh trn cho n khi hthng.

Cc chn trong qu trnh phn phi, s khng cha chu trnh, l mt phng n cbn. Nu phng n c bn suy bin, cha m+n-1 , th b sung thm mt s " chn 0 "

p dng vo bi ton ang xt :

1- Phn vo (1,3) 50 . Hng (1) b xa . Ct (3) cn thu 60-50=10

80 20 10

0 5 4 1 50

40 3 2 6

70 7 9 11

2- Phn vo (2,2) 20 . Ct (2) b xa . Hng (2) cn pht 40-20=20

80 0 10

0 5 4 1 50

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20 3 2 20 6

70 7 9 11

3- Phn vo (2,1) 20 . Hng (2) b xa . Ct (1) cn thu 80-20=60

60 0 10

0 5 4 1 50

0 3 20 2 20 6

70 7 9 11

4- Phn vo (3,1) 60 . Ct (1) b xa . Hng (3) cn pht 70-60=10

0 0 10

0 5 4 1 50

0 3 20 2 20 6

10 7 60 9 11

5- Phn vo (3,3) 10. Ht hng.

0 0 0

0 5 4 1 50

0 3 20 2 20 6

0 7 60 9 11 10

c 5 c chn, chng to thnh mt phng n c bn khng suy bin v s bng vi m+n-1=3+3-1.

THUT TON "QUY 0 CC PH CC CHN"

nh l

Nu cng vo hng i v ct j ca ma trn cc ph C=[cij] mt s ty ri v sj th biton vn ti mi vi ma trn cc ph mi C'=[c'ij=cij+ri+sj] th phng n ti u cabi ton ny cng l phng n ti u ca bi ton kia v ngc li.

Thut ton "Quy 0 cc ph cc chn" gm ba giai on.

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Giai on 1 : Quy 0 cc ph cc chn

Sau khi xc nh c phng n c bn c m+n-1 chn, ngi ta cng vo mi hngi v mi ct j ca ma trn cc ph C=[cij] mt s ri v sj sao cho ma trn cc ph miC' ti cc chn tha c'ij=cij+ri+sj=0.

Tip tc v d trn ta thy :

5 4 1 50 r1=6

3 20 2 20 6 r2=0

7 60 9 11 10 r3=-4

s1=-3 s2=-2 s3=-7

Cc gi tr cng vo phi tha h phng trnh :

1 + r1 + s3 = 0

3 + r2 + s1 = 0

2 + r2 + s2 = 0

7 + r3 + s1 = 0

11+r3 + s3 = 0

{{{{

Chn r2=0 , gii h ta c kt qu trn

Ma trn cc ph mi thu c l :

8 8 0 50

0 20 0 20 -1

0 60 3 0 10

Giai on 2 : Kim tra tnh ti u

Sau khi quy 0 cc ph cc chn nu : cc loi u c cc ph 0 th phng nang xt l ti u, ngc li th chuyn sang giai on 3

Trong v d ny ta chuyn sang giai on 3.

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Giai on 3 : Xy dng phng n mi tt hn

1- Tm a vo.

a vo l loi (i*,j*) c cc ph nh nht v tr thnh chn

Trong v d ny l (2,3).

2- Tm chu trnh iu chnh.

Chu trnh iu chnh c tm bng cch b sung (i*,j*) vo m+n-1 chn ban u,khi s xut hin mt chu trnh duy nht, gi l chu trnh iu chnh V .

Trong v d ny chu trnh iu chnh l :

V : (2,3) (3,3) (3,1) (2,1) (2,3)

3- Phn chn l cho chu trnh iu chnh.

nh s th t cc trong chu trnh iu chnh V bt u t (i*,j*). Khi chu trnhiu chnh V c phn thnh hai lp :

VC : cc c s th t chn.

VL : cc c s th t l.

4- Tm a ra v lng iu chnh.

Trong s cc c th t chn chn (r,s) c phn phi t hng nht lm a ra, trthnh loi. Lng hng xrs a ra gi l lng iu chnh.

Trong v d ny a ra l (3,3), lng iu chnh l 10.

5- Lp phng n mi.

Phng n mi c c bng cch thm hoc bt lng iu chnh trn chu trnh iuchnh nh sau :

c th t chn b bt i lng iu chnh.

c th t l c cng thm lng iu chnh.

ngoi chu trnh iu chnh khng thay i

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Trong v d ny ta thy nhng trong chu trnh iu chnh c s thay i nh sau :

(2,3) c thm 10 tr thnh 10

(3,3) b bt 10 tr thnh 0

(3,1) c thm 10 tr thnh 70

(2,1) b bt 10 nn tr thnh 10

Khi phng n mi l :

8 8 0 50

0 10 0 20 -1 10

0 70 3 0

Quay v giai on 1.

Giai on 1 : Quy 0 cc ph chn

8 8 0 50 r1=-1

0 10 0 20 -1 10 r2=0

0 70 3 0 r3=0

s1=0 s2=0 s3=1

Ma trn cc ph mi l :

7 7 0 50

0 10 0 20 0 10

0 70 3 1

Giai on 2 : Kim tra tnh ti u

y l phng n ti u

80 20 60

50 5 4 1 50

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40 3 10 2 20 6 10

70 7 70 9 11

Vi cc ph l :

1.50+3.10+2.20+6.10+7.70=670

Khi s dng phng n ban u

80 20 60

50 5 4 1 50

40 3 20 2 20 6

70 7 60 9 11 10

th cc ph l :

1.50+3.20+2.20+7.60+11.10=680

Cc bi ton c a v bi ton vn ti

C nhiu bi ton thc t c tnh cht khng phi l vn ti nhng c m hnh tonhc l bi ton vn ti. Mt s bi ton nh vy l :

a- Bi ton b nhim

Gi s tp hp S gm m ngi v tp hp D gm n cng vic (chc v). Cc ph cavic b nhim ngi iS vo vic jD l cij (i=1m , j=1n). Bi ton t ra l tmcch chia mi ngi ng mt vic sao cho cc ph b nhim l nh nht.

Ngi ta t bin (bin trn dng) nh sau :

th bi ton tr thnh :

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V mi ngi nhn ng 1 vic nn :

V mi vic ch giao cho mt ngi nn :

y l bi ton vn ti nhng c thm yu cu l cc bin xij ch ly gi tr 0 hoc 1.

Bi ton b nhim cng c khi c gi l bi ton chn (Choice Problem). Nhiu biton thc t a dng c m hnh ton hc l bi ton b nhim, chng hn nh bi tonphn b ho lc vo mc tiu cn tiu dit.

b- Bi ton vn ti vi cung t hn cu

Xt mt bi ton mt bi ton vn ti vi S l tp hp m nt cung v D l tp hp n ntcu m tng ngun cung nh hn tng nhu cu, tc l

Trong trng hp ny tt nhin khng th p ng nhu cu dj cho mi nt j=1ncho nn rng buc c dng bt ng thc thay v l ng thc. Vy :

Ngi ta thng a bi ton ny v bi ton vn ti (ng) theo mt trong hai trnghp sau y :

1.Trng hp th nht l c tnh n s thit hi bng tin khi thiu mt n v hngho nt cu j l rj (j=1n)

Lc ny ngi ta a thm vo mt nt cung gi (m+1) vi ngun cung l

v cc ph tng ng l

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c(m+1) j = rj (j=1n)

Khi ta nhn c mt bi ton vn ti (ng)

2.Trng hp th hai l khng tnh n s thit hi do thiu hng nt cu

Lc ny ta cng a v bi ton vn ti (ng) nh trn, nhng v khng tnh n sthit hi nn mc tiu s l

Ghi ch :

Vi bi ton vn ti m, ngun chuyn khng ht sang cc nhu cu, ngi ta c th tnhthm cc ph lu kho mi ngun cho mi n v hng l ci (n+1) (i=1m) . Honton tng t nh trn, khi a bi ton ny v bi ton vn ti (ng) bng cch thmvo nt cu gi (n+1) th hm mc tiu tr thnh

Nh vy ta ch cn xt bi ton vn ti (ng)

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c- Bi ton vn ti c ng cm

y l bi ton vn ti nhng khng phi mi ngun u c cung ni vi mi ch.ngha l c ng cm. Cch a v bi ton vn ti l dng phng php M-ln, tc lphng php pht nh sau :

Gi E l tp cc cung khng cm, tc l cc cung (i,j), iS, jD v bi ton c thmiu kin

xij=0 vi (i,j)E

ta a bi ton c cc yu cu

(*)

v bi ton vn ti bng cch t cc vn chuyn mi nh sau :

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cijnu(i,j) E

M nu(i,j) E

cij = {

y M l mt s rt ln, c coi l s ln hn mi s gp phi khi tnh ton.

Xt bi ton vi cc ph mi nh trn nh sau :

(**)

th ta c :

nh l :

Gi s x = [xij]m.nl phng n vn chuyn ti u ca (**) th khi :

1. Nu xij = 0 (i,j) E th xl phng n vn chuyn ti u ca bi ton vn ti cng cm (*)

2. Nu tn ti xkl E m xkl > 0 th bi ton vn ti c ng cm (**) khng c nhimchp nhn c.

d- Bi ton vn ti km ch bin trung gian

Gi s rng trong m hnh vn ti c mt s im ngun, tc l im sn xut, cho ramt s sn phm cn phi ch bin trc khi n im cu. Gi s c =1k im chbin vi kh nng ch bin l a n v sn phm tng ng. Gi cc ph vn chuynmt n v bn sn phm t i n l ci

' v chuyn mt n v sn phm t n j l ci''

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. Bi ton t ra l lp k hoch vn chuyn tt c cc sn phm qua ch bin n tt ccc im cu sao cho cc ph nh nht.

Gi xij l lng sn phm t i qua ri qua j, ta cn tm x=[ xij]mkn sao cho :

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Bi ton dng trn mngM u

Nhiu bi ton quy hoch tuyn tnh c th quy v bi ton lm cc tiu ph tn vnchuyn hng trong mt mng (gm cc nt v cc cung ng) sao cho m boc cc nhu cu mt s nt sau khi bit ngun cung cp ti mt s nt khc. Ccbi ton nh vy c gi l cc bi ton dng trn mng hay bi ton chuyn vn(TransShipment Problem). y l lp bi ton quan trng nht v hay gp nht trongquy hoch tuyn tnh. Lp ny bao gm cc bi ton quen thuc trong thc t nh :

- Bi ton vn ti

- Bi ton mng in

- Bi ton mng giao thng

- Bi ton qun l

- Bi ton phn b vt t

- Bi ton b nhim

- Bi ton k hoch ti chnh

- Bi ton ng ngn nht

- Bi ton dng ln nht

- .................

V l mt bi ton quy hoch tuyn tnh nn cc bi ton dng trn mng c th giic bng bt k thut ton no gii c bi ton quy hoch tuyn tnh, chng hnbng thut ton n hnh nh bit . Tuy nhin, nu tn dng nhng cu trc c bitca cc bi ton dng trn mng s lm cho phng php n hnh n gin hn vc thc hin nhanh hn.

Pht biu bi ton dng trn mng

Mng l mt th c hng k hiu G=(N,A), N l tp cc nt, A l tp cc cung, cngmt s thng tin v s lng b sung nh sau :

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. bi (iN) biu th ngun t ngoi vo nt i, gi tt l ngun

. uij biu th ti nng ca cung (i,j)A

. cij biu th cc ph cho mt n v ca dng trn cung (i,j)A

. xij biu th lng vn chuyn ca dng trn cung (i,j)A

Gi tr tuyt i |bi| c gi l nhu cu ca nt i. Nu bi>0 th nt i c gi l imngun, nu bi

MATHEDUCARE.COM

trong cc tiu ly trn mi dng chp nhn c. Nh vy ta nhn cmt bi ton quy hoch tuyn tnh nh sau :

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Quy hoch tuyn tnhM u

Quy hoch nguyn (Integer Programming) , vit tt l IP, l bi ton quy hoch mtrong tt c hoc mt phn cc bin b rng buc ch ly gi tr nguyn. Trng hpth nht c gi l quy hoch nguyn hon ton (Pure Integer Programming PIP),trng hp th hai c gi l quy hoch nguyn b phn (Mixed Integer Programming MIP). Tuy vy thut ng quy hoch nguyn c dng chung cho c hai trnghp.

Mng cc bi ton c v n gin nht m cng l quan trng nht trong lp cc bi tonquy hoch nguyn l cc bi ton chn cc quyt nh (chn/khng chn). Chng hnnh bi ton b nhim, bin quyt nh vic b nhim nhn gi tr nh sau :

V cc bin quyt nh thng ch nhn mt trong hai gi tr nn bi ton ny cn cgi l bi ton quy hoch nguyn nh phn (Binary Integer Programming) .

Mt tng t nhin gii bi ton quy hoch nguyn l c gii nh mt bi tonquy hoch tuyn tnh tng qut tm b qua rng buc bin phi nguyn. Khi tm cphng n ti u th s lm trn n c phng n ti u nguyn gn ng. Phngphp ny c th p dng trong thc t nhng phi ch n hai nguy c sau y :

- Mt l phng n ti u c lm trn khng chp nhn c i vi bi ton quyhoch nguyn.

- Hai l phng n ti u c lm trn chp nhn c nhng c th gi tr mctiu tng ng l rt xa vi mc tiu ti u ca bi ton quy hoch tuyn tnh nguyn.

Bi ton quy hoch nguyn trong thc t

a- Bi ton bal

Mt nh thm him mang theo mt bal ch cha c mt trng lng khng qu b.C n loi vt dng phi mang theo. Mi vt loi vt i c trng lng l ai v gi tr sdng l ci. Hi ng ta phi chn la cc vt mang theo nh th no c gi tr s dngl ln nht ?

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Gi xi (i=1n) l s lng vt loi i m ng ta mang theo th m hnh ton ca bi tonbal ny l quy hoch nguyn nh sau :

V mt ton hc th nu hm mc tiu l min z hoc rng buc l ng thc th bi toncng gi l bi ton bal. Bi ton bal c dng c bit v n gin v ch c mt rngbuc ngoi rng buc du v tnh nguyn. Ngi ta nghin cu c nhiu cch giiring cho bi ton v a bi ton quy hoch nguyn v bi ton bal gii.

b- Bi ton sn xut c l ph c nh

Gi s mt nh my c k hoch s sn xut n sn phm. Chi ph sn xut sn phmj=1n gm l ph c nh kj , khng ph thuc vo s lng sn phm j, v cc ph cji vi mi n v sn phm j.

Gi xj 0 l lng sn phm j=1n s sn xut th chi ph sn xut sn phm j s l :

mc tiu sn xut vi chi ph cc tiu s l :

Trong trng hp ny hm mc tiu z l hm phi tuyn vi cc i s l xj (j=1n) mcd cc rng buc thc t nh nguyn liu, th trung,.... u l tuyn tnh nn bi tonrt kh gii. Ngi ta c th a bi ton ny v bi ton quy hoch tuyn tnh nguynb phn bng cch a vo cc bin ph nh phn nh sau :

(1)

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biu th yj (j=1n) l bin nh phn c lp, khng ph thuc vo xj nh trong (1)ngi ta a vo mt rng buc tuyn tnh nh sau :

xj Myj (j=1n)

y M>0 v rt ln rng buc xj l tha. Khi hm mc tiu v rng buctrn tr thnh :

(2)

Tht vy :

- Nu xj > 0 th yj khng th bng 0 nn yj =1

- Nu xj = 0 th yj = 0 hoc yj=1

Nhng v kj>0 ( nu kj= 0 th khng cn a vo bin ph yj) v hm mc tiu l minz nn thut ton tm phng n ti u lun ly yj=0 v phng n vi xj=0 v yj=1khng th l ti u. Khi vit cc rng buc tuyn tnh khc vo ta c bi ton quyhoch tuyn tnh nguyn b phn.

CU HI CHNG 4

1- Trnh by chin lc b tri hn.

2- Trnh by chin lc MaxiMin v MiniMax.

3- Xy dng quy hoch tuyn tnh trong trng hp khng c nghim n nh.

4- Trnh by cc giai on gii bi ton vn ti.

BI TP CHNG 4

1- Tm phng n ti u cho bi ton l thuyt tr chi c ma trn im c cho nhsau :

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2 3 -2 -1

-1 5 4 -2

-2 -5 0 3

2- Gii bi ton vn ti c ma trn cc ph

60 70 40 30

100 2 1 4 3

80 5 3 2 6

20 6 2 1 5

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cng CNG MN HC

MN : QUY HOCH TUYN TNH

M MN HC : TH 431

S N V HC TRNH : 2

HC K : 5

MC CH YU CU

Sau khi hc xong mn quy hoch tuyn tnh sinh vin phi bit cch xy dng m hnhton cho bi ton thc t n gin, p dng thnh tho gii thut n hnh gii lpbi ton quy hoch tuyn tnh v lp trnh c trn my tnh.

KIN THC NN CN THIT

KIN THC TON CN THIT

TM TT NI DUNG MN HC

Mn hc c m u bng vic gii thiu vi vn thc t dn n m hnh quyhoch tuyn tnh. Trng tm ca mn hc l phn trnh by gii thut n hnh ccmc s dng khc nhau. L thuyt i ngu c trnh by mt cch n gin. Phn

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ng ca quy hoch tuyn tnh c trnh by sau cng thy s ng dng rng ri caquy hoch tuyn tnh

CNG CHI TIT CC CHNG

CHNG I : L THUYT C BN V QUY HOCH TUYN TNH

I- GII THIU BI TON QUY HOCH TUYN TNH

1- Bi ton vn u t

2- Bi ton lp k hoch sn xut

3- Bi ton vn ti

II- NH NGHA V NHNG KT QU C BN

1- Quy hoch tuyn tnh tng qut

2- Quy hoch tuyn tnh dng chnh tc

3- Phng n

4- a din li cc phng n kh thi - Phng php hnh hc

III- MT V D M U

IV- DU HIU TI U

1- Ma trn c s - Phng n c s - Suy bin

2- Du hiu ti u

CHNG II : GII THUT N HNH

I- GII THUT N HNH C BN

1- C s l thuyt

2- nh l v s hi t

3- Gii thut n hnh c bn

4- Ch trong trng hp suy bin

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II- GII THUT N HNH CI TIN

1- Mt cch tnh ma trn nghch o

2- Quy hoch tuyn tnh dng chun

3- Gii thut n hnh ci tin

4- Php tnh trn dng - Bng n hnh

III- PHNG PHP BIN GI CI BIN

1- Bi ton ci bin

2- Phng php hai pha

3- Phng php M v cng ln

CHNG III : BI TON I NGU

I- KHI NIM V I NGU

1- i ngu ca quy hoch tuyn tnh dng chnh tc

2- nh ngha i ngu trong trng hp quy hoch tng qut

3- Cc nh l v s i ngu

II- GII THUT I NGU

CHNG IV : NG DNG QUY HOCH TUYN TNH

I- M U

II- BI TON TR CHI

1- Tr chi c nghim n nh

2- Tr chi khng c nghim n nh

III- BI TON VN TI

1- M u

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2- Cc khi nim c bn

3- Bi ton vn ti cn bng thu pht

4- Cc bi ton c a v bi ton vn ti

IV- BI TON DNG TRN MNG

1- M u

2- Pht biu bi ton dng trn mng

V- QUY HOCH NGUYN

1- M u

2- Bi ton quy hoch nguyn trong thc t

TI LIU THAM KHO

[ Ban - 1998]

Ph Mnh Ban Quy Hoch Tuyn Tnh

Nh xut bn Gio Dc ( ti bn ln 2)

[ Hn - xxxx]

ng Hn Quy Hoch Tuyn Tnh

i hc Kinh t TP H Ch Minh ( lu hnh ni b )

[ Khnh-Nng - 2000]

Phan Quc Khnh Trn Hu Nng Quy Hoch Tuyn Tnh

Nh xut bn Gio Dc

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Bi tp tng hpI- X nghip sn xut giy c 3 phn xng. Do trang b k thut khc nhau nn mchao ph tre g, axit sn xut mt tn giy thnh phm cng khc nhau. Mc hao phc cho trong bng di y :

S lng tre g c trong nm l 1.500.000 tn, Axit l 100.000 tn.

Yu cu

1. Xy dng m hnh sao cho tng s giy sn xut trong nm ca x nghip l nhiunht.

2. Xy dng m hnh bi ton i ngu vi m hnh ton ca cu 1.

3. Tm phng n ti u ng vi m hnh ton cu 1. T suy ra s tn giy ca miphn xng cn sn xut trong nm.

4. p dng kt qu bi ton i ngu t bng n hnh ti u cu 3 suy ra phng nti u cho bi ton i ngu cu 2.

II- Mt x nghip c th sn xut bn loi mt hng xut khu H1, H2, H3, H4. snxut 4 loi mt hng ny, x nghip s dng 2 loi nguyn liu N1, N2. S nguyn liuti a m x nghip huy ng c tng ng l 600kg v 800kg. Mc tiu hao mi loinguyn liu sn xut mt mt hng v li nhun thu c c cho trong bng sau :

Yu cu

1- Lp m hnh x nghip sn xut t li nhun cao nht.

2- Xy dng bi ton i ngu ng vi m hnh ton cu 1.

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3- p dng thut ton n hnh ci tin v kt qu i ngu tm cc phng n ti ucho c 2 m hnh.

III- X nghip c kh Hng Vng c 32 cng nhn nam v 20 cng nhn n. X nghipc 2 loi my : ct v tin. Nng sut trung bnh ca cc cng nhn i vi mi loi myc cho trong bng bn di y :

Bit rng trong ngy ct c bao nhiu chi tit th tin ht by nhiu chi tit

Yu cu

1- Lp m hnh x nghip sn xut c nhiu sn phm nht.

2- Lp m hnh i ngu ng vi m hnh cu 1.

3- p dng thut ton n hnh ci tin v kt qu i ngu tm phng n ti u choc 2 m hnh ton trn.

IV- Mt cng ty chuyn sn xut 3 loi sn phm A, B, C. Trong nguyn liu snxut ra 3 loi sn phm trn c nhp v t 2 ngun N1, N2. Chi ph cho mi n vnguyn liu nhp t ngun N1 l 100000 USD v ngun N2 l 90000 USD.

Cc loi sn phm sn xut cn cc n v nguyn liu ca tng ngun c cho trongbng sau :

S lng ti thiu sn phm loi A cn sn xut trong thi gian ti l 20000 , sn phmloi B l 18000, sn phm loi C l 15000.

Yu cu

1- Lp m hnh tng chi ph sn xut m cng ty b ra l nh nht m vn m boyu cu v sn phm.

2- Lp m hnh cng ty sn xut t doanh thu cao nht

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3. Tm phng n ti u cho c 2 m hnh.

V- Mt c s d nh sn xut ti a trong mt ngy 500 bnh m di v 500 bnhm trn, mun t li nhun nhiu nht, vi nhng iu kin nh sau :

- Gi bn mt bnh m di lm t 400 gam bt l 325 ng, mt bnh m trn lm t250 gam bt l 220 ng.

- S lng bt c cung cp ti a trong ngy l 225 kg vi gi mi kg l 300 ng.

- L nng bnh cho php nng 75 bnh m di hay 100 bnh m trn trong mt ginhng khng th nng hai loi cng mt lc. L nng hot ng ti a 8 gi trongmt ngy.

Yu cu

1- Lp m hnh cho bi ton nu trn.

2- Xy dng bi ton i ngu cho bi ton trn.

3- Tm phng n ti u cho c hai bi ton.

VI- Ba x nghip A, B, C cng c th sn xut o vt v qun. Kh nng sn xut cami x nghip nh sau : Khi u t 1000USD vo x nghip A th thu c 35 o vt v45 qun ; vo x nghip B th thu c 40 o vt v 42 qun ; vo x nghip C th thuc 43 o vt v 30 qun. Lng vi v v gi cng sn xut c cho trong bngsau :

Tng s vi huy ng c l 10000m.

Tng s gi cng huy ng c l 52000 gi.

Theo hp ng th ti thiu phi c 1500 b qun o, nu l b th qun l d bn.

Hy lp k hoch u t vo mi x nghip bao nhiu vn :

- Hon thnh hp ng

- Khng kh khn v tiu th

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- Khng b ng v vi v gi cng

- Tng s vn u t l nh nht

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Tham gia ng gp

Ti liu: Gio trnh quy hoch tuyn tnh

Bin tp bi: thang leduc

URL: http://voer.edu.vn/c/78021439

Giy php: http://creativecommons.org/licenses/by/3.0/

Module: Thng tin v tc gi

Cc tc gi: thang leduc

URL: http://www.voer.edu.vn/m/e5488080


Module: Gii thiu bi ton quy hoch tuyn tnh


URL: http://www.voer.edu.vn/m/8dfb947a


Module: Quy hoch tuyn tnh tng qut v chnh tc


URL: http://www.voer.edu.vn/m/1ee7339a


Module: c im ca cc tp hp cc phng n


URL: http://www.voer.edu.vn/m/989b7bfb


Module: L thuyt c bn v quy hoch tuyn tnh-Mt s v d m u


URL: http://www.voer.edu.vn/m/0c13b50d


Module: Du hiu ti u


URL: http://www.voer.edu.vn/m/51e4793a

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Module: Gii thut n hnh c bn


URL: http://www.voer.edu.vn/m/7d2b1e8a


Module: Phng php bin gi ci bin


URL: http://www.voer.edu.vn/m/8f0a87bb


Module: Quy hoch tuyn tnh suy bin


URL: http://www.voer.edu.vn/m/8d60d974


Module: Khi nim v i ngu


URL: http://www.voer.edu.vn/m/60cbc806


Module: Gii thut i ngu


URL: http://www.voer.edu.vn/m/d484dbe1


Module: ng dng quy hoch tuyn tnh-M u


URL: http://www.voer.edu.vn/m/5b1a74ae


Module: Bi ton vn ti


URL: http://www.voer.edu.vn/m/a36107d4


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Module: Bi ton dng trn mng


URL: http://www.voer.edu.vn/m/9a304ac1


Module: Quy hoch tuyn tnh


URL: http://www.voer.edu.vn/m/503d9533


Module: cng


URL: http://www.voer.edu.vn/m/f71b7e1a


Module: Bi tp tng hp


URL: http://www.voer.edu.vn/m/c73c33dd


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Chng trnh Th vin Hc liu M Vit Nam

Chng trnh Th vin Hc liu M Vit Nam (Vietnam Open Educational Resources VOER) c h tr bi Qu Vit Nam. Mc tiu ca chng trnh l xy dng khoTi nguyn gio dc M min ph ca ngi Vit v cho ngi Vit, c ni dung phongph. Cc ni dung u tun th Giy php Creative Commons Attribution (CC-by) 4.0do cc ni dung u c th c s dng, ti s dng v truy nhp min ph trcht trong trong mi trng ging dy, hc tp v nghin cu sau cho ton x hi.

Vi s h tr ca Qu Vit Nam, Th vin Hc liu M Vit Nam (VOER) tr thnhmt cng thng tin chnh cho cc sinh vin v ging vin trong v ngoi Vit Nam. Mingy c hng chc nghn lt truy cp VOER (www.voer.edu.vn) nghin cu, hctp v ti ti liu ging dy v. Vi hng chc nghn module kin thc t hng nghntc gi khc nhau ng gp, Th Vin Hc liu M Vit Nam l mt kho tng ti liukhng l, ni dung phong ph phc v cho tt c cc nhu cu hc tp, nghin cu cac gi.

Ngun ti liu m phong ph c trn VOER c c l do s chia s t nguyn ca cctc gi trong v ngoi nc. Qu trnh chia s ti liu trn VOER tr ln d dng nhm 1, 2, 3 nh vo sc mnh ca nn tng Hanoi Spring.

Hanoi Spring l mt nn tng cng ngh tin tin c thit k cho php cng chng ddng chia s ti liu ging dy, hc tp cng nh ch ng pht trin chng trnh gingdy da trn khi nim v hc liu m (OCW) v ti nguyn gio dc m (OER) . Khinim chia s tri thc c tnh cch mng c khi xng v pht trin tin phongbi i hc MIT v i hc Rice Hoa K trong vng mt thp k qua. K t , phongtro Ti nguyn Gio dc M pht trin nhanh chng, c UNESCO h tr v cchp nhn nh mt chng trnh chnh thc nhiu nc trn th gii.

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Thng tin v tc giGii thiu bi ton quy hoch tuyn tnhQuy hoch tuyn tnh tng qut v chnh tcc im ca cc tp hp cc phng nL thuyt c bn v quy hoch tuyn tnh-Mt s v d m uDu hiu ti uGii thut n hnh c bnPhng php bin gi ci binQuy hoch tuyn tnh suy binKhi nim v i nguGii thut i ngung dng quy hoch tuyn tnh-M uBi ton vn tiGiai on 1 : Quy 0 cc ph cc chnGiai on 2 : Kim tra tnh ti uGiai on 3 : Xy dng phng n mi tt hnGiai on 1 : Quy 0 cc ph chnGiai on 2 : Kim tra tnh ti u

Bi ton dng trn mngQuy hoch tuyn tnh cngBi tp tng hpTham gia ng gp

[math-educare]_giáo trình quy hoạch tuyến tính_le duc thang

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