Chuong8_QuyHoachTuyenTinh2

7/22/2019 Chuong8_QuyHoachTuyenTinh2

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Tnh ton khoa hc

Chng 8

QUY HOCH TUYN TNH

Linear Programming


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Tnh ton khoa hc

Bi ton quy hoch ton hc

Rtnhiubi ton thctc thphtbiudidngbi ton cctrsau:

Bi ton (1)-(4) cgil bi ton quy hochton hc

f(x) l hm mctiu, gi,hj l cc hm rngbuc. Tp

Gil tprngbuc,hay minchpnhnc. Mivectx thucD cgil ligiichpnhnchay lphngn chpnhnc


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Tnh ton khoa hc

Bi ton quy hoch ton hc

Phngn chpnhncx*thamn

cgil pa tiuhay ligiicabi ton, khi gi tr

cgil gi trtiucabi ton


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Tnh ton khoa hc

Mt s m hnh thc t

Bi ton lp k hoch sn xut cho mt nh my

Bi ton khu phn n

Bi ton vn ti


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Bi ton lp k hoch sn xut

Mt nh my c kh nng sn xut n loi sn phm. snxut cc sn phm ny cn phi s dng m loi nguyn liu.

Bit :

Hy xy dng mt chin lc sn xut mang li nhiu li nhun nht. Gi l s sn phm loi j, k hoch sn xut l

l tng chi ph nguyn liu i do

Tng li nhun thu c l Khi m hnh ton hc ca bi ton k hoch sn xut c pht biu

di dng nh sau.Tnh ton khoa hc

ij l- ng nguyn liu loi i cn thit sn xut ra mt n vsn phm loi j;

d tr nguyn liu loi i

tin l i t vic bn mt n vsn phm lo i n vsn phm lo i j

(i=1, ; j=1, )

i

j

a

b

c

m n

0jx 1 2( , ,..., )nx x x x

1

*

n

ij j

j

a x

1

*n

ij j i

j

a x b

1

n

j j

j

c x


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Tnh ton khoa hc

Bi ton lp k hoch sn xut

Bi ton lp k hoch sn xut cho mt nh myTm cc i can

1 2 n j jj=1

n

ij i

j=1

j

f(x ,x ,...,x )= c x

v i iu kin

a b , i=1,m

x 0, j =1,n


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Tnh ton khoa hc

Ni dung

I. Thut ton n hnh3.1.1. Bi ton QHTT dng chnh tc v dng chun

3.1.2. Phng n c s chp nhn c

3.1.3. Cng thc s gia hm mc tiu. Tiu chun ti u

3.1.4. Thut ton n hnh dng ma trn nghch o

3.1.5. Thut ton n hnh dng bng3.1.6. Tnh hu hn ca thut ton n hnh

3.1.7. Thut ton n hnh hai pha

II. L thuyt i ngu

3.2.1. Xy dng bi ton i ngu3.2.2. Cc nh l i ngu

3.2.3. Mt s ng dng ca l thuyt i ngu


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Tnh ton khoa hc

I. THUT TON N HNH


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Tnh ton khoa hc

Ni dung

1. Bi ton QHTT dng chnh tc v dng chun2. Phng n c s chp nhn c

3. Cng thc s gia hm mc tiu. Tiu chun ti u

4. Thut ton n hnh dng ma trn nghch o

5. Thut ton n hnh dng bng

6. Tnh hu hn ca thut ton n hnh

7. Thut ton n hnh hai pha


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Tnh ton khoa hc

1. Bi ton QHTT dng chnh tc v dng

chun


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Tnh ton khoa hc

Bi ton QHTT tng qut

Bi ton QHTT tng qut l bi ton ti u ho m trong chng ta phi

tm cc i (cc tiu) ca hm mc tiu tuyn tnh vi iu kin bin sphi tho mn h thng phng trnh v bt phng trnh tuyn tnh. Mhnh ton hc ca bi ton c th pht biu nh sau

K hiu xj 0 ch ra rng binxjkhng c i hi trc tip v du.

1 2

1

1

1

( , ,..., ) min(max), (1)

, 1,2,..., ( ) (2)

, 1, 2,..., (3)

0, 1,2,..., ( )

n

n j j

j

n

ij j i

j

n

ij j i

j

j

f x x x c x

a x b i p p m

a x b i p p m

x j q q n (4)

0, 1, 2,..., (5) jx j q q n

vi iu kin


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Tnh ton khoa hc

Bi ton QHTT tng qut

Rng buc:

1

, 1,...,n

ij j i

j

a x b i p

c gi l rng buc c bn dng ng thc.

Rng buc:

1

, 1,...,n

ij j i

j

a x b i p m

c gi l rng buc c bn dng bt ng thc.

Rng buc:0, 1,...,jx j q

c gi l rng buc v du ca bin s.


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Tnh ton khoa hc

Bi ton QHTT dng chnh tc

Ta gi bi ton QHTT dng chnh tc l bi ton sau:

1 2

1

1

( , ,..., ) min,

, 1,2,...,

0, 1,2,...,

n

n j j

j

n

ij j i

j

j

f x x x c x

a x b i m

x j n


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Tnh ton khoa hc

Bi ton QHTT dng chun

Ta gi bi ton QHTT dng chun l bi ton sau:

1 2

1

1

( , ,..., ) min,

, 1,2,...,

0, 1,2,...,

n

n j j

j

n

ij j i

j

j

f x x x c x

a x b i m

x j n


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Tnh ton khoa hc

a BT QHTT tng qut v dng chnh tc

R rng Bi ton QHTT dng chnh tc l trng hp ring ca

QHTT tng qut.

Mt khc, mt bi ton QHTT bt k lun c th a v dngchnh tc nh cc php bin i sau:

1 2

1

1

1

( , ,..., ) min(max), (1)

, 1,2,..., ( ) (2)

, 1, 2,..., (3)

0, 1,2,..., ( )

n

n j j

j

n

ij j ij

n

ij j i

j

j

f x x x c x

a x b i p p m

a x b i p p m

x j q q n (4)

0, 1, 2,..., (5) jx j q q n

1 2

1

1

( , ,..., ) min,

, 1, 2,...,

0, 1,2,...,

n

n j j

j

n

ij j i

j

j

f x x x c x

a x b i m

x j n


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Tnh ton khoa hc


a) arng bucbtngthcdng vdng.

Btphngtrnh tuyntnh

l tngngvibtphngtrnh tuyntnh sau

1

n

ij j i

j

a x b

1

n

ij j i

j

a x b


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Tnh ton khoa hc


b)a rng buc dng= v dng

.Phng trnh tuyn tnh

l tng ng vi h gm 2 bt phng trnh tuyntnh sau

1

n

ij j i

j

a x b

1

1

n

ij j i

j

n

ij j i

j

a x b

a x b


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Tnh ton khoa hc


c)a rng buc dng v dng =.

Bt phng trnh tuyn tnh

l tng ng vi h gm 1 phng trnh tuyn tnh v mtiu kin khng m i vi bin s sau y

Tng ng hiu theo ngha: Nu (x1,x2, ...,xn,yi) l nghimca h th (x1,x2, ...,xn) l nghim ca bt phng trnh.

Binyic gi lbin b(hay bin ph).

1

n

ij j i

j

a x b

1

0

n

ij j i i

j

i

a x y b

y


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Tnh ton khoa hc


d) Thay mibinkhng c iukinduxjbihiuhai binc iukinv

du:

e)abi ton tm ccivbi ton tm cctiu.Bi ton tiuho

max {f(x):x D}

l tngngvibi ton tiuho

min {-f(x):x D}

theo ngha: Li giicabi ton ny cng l ligiica bi ton kia vngcli,ngthita c ngthc:

max {f(x):x D} = - min {-f(x):x D}

,

0, 0.

j j j

j j

x x x

x x


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f)a rng buc dng v dng =.Bt phng trnh tuyntnh

l tng ng vi h gm 1 phng trnh tuyn tnh v mtiu kin khng m i vi bin s sau y

Tng ng hiu theo ngha: Nu (x1,x2, ...,xn,yi) l nghimca h th (x1,x2, ...,xn) l nghim ca bt phng trnh.

Binyic gi lbin b(hay bin ph).

0

1

i

ii

n

j

jij

y

byxa

i

n

j

jij bxa 1

Tnh ton khoa hc


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Tnh ton khoa hc

V d

Bi ton QHTTx1 + 2x2 - 3x3 + 4x4max,

x1 + 5x2 + 4x3+ 6x415,

x1 + 2x2 - 3x3+ 3x4= 9,x1 ,x2 ,x40, x3< >0,

l tng ng vi bi ton QHTT dng chnhtc sau:


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Tnh ton khoa hc

V d

1 2 3 3 4

1 2 3 3 4 5

1 2 3 3 4

1 2 3 3 4 5

2 3( ) 4 min,

5 4( ) 6 15,

2 3( ) 3 = 9, , , , , , 0,

x x x x x

x x x x x x

x x x x xx x x x x x


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Tnh ton khoa hc

Gii bi ton QHTT trong mt phng


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Tnh ton khoa hc


Bi ton QHTT dng chun 2 bin sf(x1,x2) = c1x1 + c2x2min,

vi iu kin

ai1x1 + ai2x2 bi, i= 1, 2, ..., m

K hiuD = {(x1,x2): ai1x1 + ai2x2 bi, i= 1, 2, ..., m}

l min rng buc.


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Tnh ton khoa hc


T ngha hnh hc, mibtphng trnhtuyntnh

ai1x1 + ai2x2 bi, i= 1, 2, ..., m

xc nhmtnamtphng.

Min rng bucD xc nhnh giao ca m

namtphng s l mta gic li trn mtphng.


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Tnh ton khoa hc


Phngtrnhc1x1+ c2x2=

c vectphp tuynl (c1,c2)

khi thay isxc nhcc ngthngsong songvi nhau m ta sgi l cc ngmc (vigi trmc).

Mi im u=(u1,u2)D s nm trn ng mc vi

mcu= c1u1+ c2u2=f(u1,u2)


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Tnh ton khoa hc


Bi ton t ra c th pht biu nh sau: trong s cc ngmc ct tpD, tm ng mc c gi tr mc nh nht.

Ta nhn thy, nu dch chuyn song song cc ng mc theohng vector php tuyn c = (c1,c2) th gi r mc s tng, nudch chuyn theo hng ngc li th gi tr mc s gim.

Do , bi ton t ra c th tin hnh nh sau: Bt u t mtng mc ctD, ta dch chuyn song song cc ng mctheo hng ngc hng vi vector c = (c1,c2) cho n khino vic dch chuyn tip theo lm cho ng mc khng cn

ctDna th dng. Cc im caDnm trn ng mc cui cng ny s l cc

li gii cn tm, cn gi tr ca n chnh l gi tr ti u ca biton.


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Tnh ton khoa hc

V d 1

Gii bi ton QHTT sau:x1x2min

2x1+x22,

x1x27,x1+x22,

x10, x

20.


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Tnh ton khoa hc

Min rng buc:D = M1M2M3M4M5

x1x2min2x1+x22,x1x27,x

1+x

22,

x10, x20.

x1x2=7

x1x2= 0

Phng n ti u


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Tnh ton khoa hc

V d 1

Gii theo phng php hnh hc va m t ta thucligiitiucabi ton tngngviim

M2(0,7):x* = (0,7), vigi trtiulf*= -7.

Nuthay hm mctiu cabi tonbix1+x2min,

th gi trtiusl -2 v ttccc imnm trnonM3M4 u l phng n ti u ca bi ton.

Chnghn,c thlyphngn tiucabi tonlx*=(2,0) (tngngviimM4).


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Tnh ton khoa hc

Phng n ti u: M3M4

x1+x2min2x1+x22,x1x27,x

1

+x2

2,x10, x20.

c

x1 + x2= 0

x1 + x2= 7


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Tnh ton khoa hc

Nhn xt

Trong c hai trng hp ta lun tm cphngn tiul mtnhno caminrngbuc.

Bi ton QHTT trong mtphng lun cphngn tiul nhcaminrngbuc.

Nhn xt hnh hc quan trng ny dn tivic xut thut ton n hnh gii biton QHTT.


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Tnh ton khoa hc

V d 2

1 2

1 2

1 2

1 2

1 2

13 23 max

5 15 480

4 4 160

35 20 1190

, 0

x x

x x

x x

x x

x x


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Tnh ton khoa hc

Min rng buc

(34, 0)

(0, 32)

5x1+ 15x2 480

(12, 28)

(26, 14)

(0, 0)

4x1+ 4x2 160 35x1+ 20x2 11901 2

1 2

1 2

1 2

1 2

13 23 max

5 15 480

4 4 160

35 20 1190

, 0

x x

x x

x x

x x

x x


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Tnh ton khoa hc

Hm mc tiu

(34, 0)

(0, 32)

(12, 28)

13x1+ 23x2= 1600

13x1+ 23x2= 442

(26, 14)

(0, 0)

13x1+ 23x2= 800


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Tnh ton khoa hc(34, 0)

(0, 32)

(12, 28)

(26, 14)

(0, 0)

nh ca agic rng buc

ngha hnh hc

Tn ti li gii ti u l nh ca a gic rng buc


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V d 3

Gii bi ton QHTT sau:x1x2min

2x1+x2


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Tnh ton khoa hc

ngha hnh hc ca QHTT

Min rng buc l tp li a din. Li: nuyvzl pacn, th y+(1- )zcng l pacn vi mi 01. nh:pacnxm khng th biu din di dng y+(1- )z, 0


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Tnh ton khoa hc

ngha hnh hc

Ktlun: Nubi tonc pa tiuth n lunc pa tiul nhcamin rng buc vn

ngcho nhiuchiu. Ch cn tm pa ti u

trong s hu hnphngn.


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Tnh ton khoa hc

Thut ton n hnh

Simplex Algorithm.(Dantzig 1947) Thut ton thc hin

dch chuyn t mt

nh sang mt nh ktt hn cho n khin c nh ti u.

Thut ton l hu hn

nhng c phc tphm m.


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Tnh ton khoa hc

Mt s k hiu v nh ngha

Trong cc phn tip theo ta s ch lm vicvi bi ton QHTT dng chnh tc:

Tm cc tiu:

f (x1,x2,...,xn)= nj=1 cjxjmin,vi iu kin

ni=1 aijxj = bi, i = 1,2,...,m

xj 0, j = 1,2,...,n.


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Tnh ton khoa hc

K hiu v nh ngha

a vo cc k hiu:x=(x1,x2, ...,xn)

Tvect bin s

c=(c1, c2, ..., cn)

T

vect h s hm mc tiuA= (aij)mnma trn rng buc

b=(b1,...,bm)T

- vect rng buc (v phi)


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Tnh ton khoa hc

K hiu v nh ngha Ta c th vit li bi ton di dng ma trn:

f(x) = cTxmin,Ax = b,x 0

hay

min{f(x) = cT

x:Ax = b,x

0}

Bt ng thc vect:y= (y1,y2, ...,yk) 0

c hiu theo ngha tng thnh phn:yi0 , i= 1, 2, ..., k.

hi h h


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Tnh ton khoa hc

K hiu v nh ngha K hiu cc tp ch s:

J = {1,2,...,n} tp ch s ca cc bin sI = {1,2,...,m} tp ch s ca cc rng buc

Khi ta s dng cc k hiu sau

x=x(J) = {xj:jJ} - vect bin s;c= c(J) = {cj:jJ}vect h s hm mc tiu;

A =A(I, J) = {aij: iI,jJ}ma trn rng buc

Aj = (aij: iI)vect ct thjca ma trnA.

H phng trnh rng buc c bn ca bi tonQHTT dng chnh tc cn c th vit di dng:

A1x1+A2x2+...+Anxn= b

hi h h


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Tnh ton khoa hc

Tp

D= {x:Ax= b,x0}c gi l min rng buc (min chp nhn c)

x c gi l phng n chp nhn c.

Phng n chp nhn c x*

em li gi tr nhnht cho hm mc tiu, tc l

cTx*cTx vi mi xDc gi l phng n ti u ca bi ton v khi

gi trf*= cTx*

c gi l gi tr ti u ca bi ton

K hiu v nh ngha


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Tnh ton khoa hc

2. PHNG N C S CHP NHN C

Khi nim phng n c s chp nhn c (pacscn)lkhi nim trung tm trong thut ton n hnh

Ph h h


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Tnh ton khoa hc

Trchtta githitrng

rank (A) = m (*)nghal hphngtrnh rng buccbngm

mphngtrnh clptuyntnh.

Ch : Trn thctgithit(*) l tngngvigi thithphng trnh tuyn tnh Ax = b

c nghim.

Vsau ta sgbcc githitny.

Phng n c s chp nhn c

Ph h h


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Tnh ton khoa hc

nhngha1. Tagicscama trnA l mtbgmm vector ctclptuyntnhcan.

Gis

Khi vectorx= (x1,x2,,xn ) thamn:

scgilphngn cstngngvicsB.Ccbin cgil bincscn l binphics.

1 2, ,...,

nj j jB A A A

1( , ), trong J ,..., l mt c s ca ma trn .B B mB A I J J J A

1

0, \ ;

l thnh phn th ca vector ( 1,..., ).

j N B

jk

x j J J J

x k B b k m

,j Bx j J ,j Nx j J




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Tnh ton khoa hc

Nh vy, nu k hiu

xB =x(JB),xN =x(JN)th phng n c sxtng ng vi c sBc th

xc nh nh th tc sau:

1. t xN=0.

2. Xc nh xBt h phng trnh BxB= b.

T gi thit (*) suy ra bi ton lun c phng n

c s.




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Tnh ton khoa hc

Gi sx = (xB ,xN ) l phng n c s tng ng

vi c sB.Khi bi ton QHTT dng chnh tc

c th vit li nh sau:

f(xB,xN)=cBxB + cNxNminBxB+ NxN= b,

xB, x

N0,

trong N= (Aj:jJN) c gi l ma trn phi c

s.




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Tnh ton khoa hc

Xt bi ton QHTT6x1+ 2x25x3+x4+ 4x53x6+ 12x7min

x1+ x2+ x3+x4 = 4

x1 + x5 = 2x3 +x6 = 3

3x2+ x3 +x7 = 6

x1, x2, x3, x4, x5, x6, x7 0




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Tnh ton khoa hc

Di dng ma trn:c= (6, 2,5, 1, 4,3, 12)T;

b = (4, 2, 3, 6)T;

1 1 1 1 0 0 0

1 0 0 0 1 0 0A= 0 0 1 0 0 1 0

0 3 1 0 0 0 1

A1A2 A3 A4A5A6A7




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Tnh ton khoa hc

Xt c s

B= {A4,A5,A6,A7} =E4 Phng n c sx= (x1,x2, ...,x7)tng ng vi n

thu c bng cch t:

x1 = 0,x2 = 0,x3 = 0

v cc gi tr caxB= (x4,x5,x6,x7)thu c bngvic gii h phng trnh

BxB= bhayE4xB=b

T ta tm c: xB= (4, 2, 3, 6). Vy pacs tng ng vi c s Bl

x= (0, 0, 0, 4, 2, 3, 6)




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Tnh ton khoa hc

Xt c s

B1= {A2,A5,A6,A7}

Phng n c sy= (y1,y2, ...,y7) tng ng vi n thu c bng ccht:

y1 = 0;y3 = 0,y4 = 0

v cc gi tr cayB= (y2,y5,y6,y7) thu c bng vic gii h

B1yB= b

hayy2 = 4

y5 = 2

y6 = 3

3y2 +y7 = 6T ta tm c: yB= (4, 2, 3,6).

Vy pacs tng ng vi c sB1l

y= (0, 4, 0, 0, 2, 3,6)




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Tnh ton khoa hc


D thy:

pacs tng ng vi c sBx= (0, 0, 0, 4, 2, 3, 6)

l phng n chp nhn c cnpacs tng ng vi c sB1

y= (0, 4, 0, 0, 2, 3, -6)khng l phng n chp nhn c

nh ngha.Phng n c s c gi lphngn c s chp nhn c (li gii c s chp nhnc)nu nh n l phng n chp nhn c.



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Tnh ton khoa hc

Bi ton QHTT c bao nhiu pacscnd?

S pacscnd S c s C(n,m)

Vy mt bi ton QHTT ch c th c mt s hu

hn pacscnd




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Tnh ton khoa hc

Bi ton QHTT lun c pacscnd?

Khng ng!

V d: Nu bi ton QHTT khng c phng n chp nhn

c th r rng n cng khng c pacscnd!

Tuy nhin ta c th chng minh kt qu sau:

nh l 1.Nu bi ton QHTT c pacn th n cngc pacscnd



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Tnh ton khoa hc

3. Cng thc s gia hm mc tiu


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Tnh ton khoa hc

Cng thc s gia hm mc tiu

Gi sxl pacscnd vi c s tng ng lB=(Aj:jJB). K

hiu:

JB={j1,j2, ...,jm}tp ch s bin c s;

JN=J \ JBtp ch s bin phi c s;

B=(Aj:jJB)ma trn c s;N=(Aj:jJN)ma trn phi c s;

xB=x(JB) = {xj:jJB},xN=x(JN) = {xj:jJN} vect binc s v phi c s;

cB= c(JB) = {cj:jJB}, cN= c(JN) = {cj:jJN}vect h shm mc tiu ca bin c s v phi c s;


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Tnh ton khoa hc


Xt pacndz=x+x, trong x = (x1

,x2

,...,xn) vect gia s cabin s. Ta tm cngthctnh sgia cahm mctiu:

f=cTzcTx = cTx.

Dox, zul pacnd nnAx=b vAz=b.V vygia sx phi tho mn iu kin Ax = 0,hay l:

BxB+ NxN= 0,trong xB= (xj:jJB), xN= (xj: jJN).

C i i


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Tnh ton khoa hc


Suy ra

xB=B-1NxN. (1.10) T ta c

cTx = cTBxB+ cTNxN=(cTBB-1NcTN) xN .

K hiu:

u= cTBB-1vect th vN= (j: jJN) = uN- cTNvect c lng.

ta thu c cng thc:f= cTzcTx = NxN = jxj

jJN

Cng thc thu c gi l cng thc s gia hm mc tiu


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Tnh ton khoa hc

Tiu chun ti u

Ti h ti


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Tnh ton khoa hc

Tiu chun ti u

nhngha.Pacscnd xcgi l khng thoi honunhttccc thnhphncscan l khc 0.Bi ton QHTT cgil khng thoi ho nunhttccc pacscnd can l khng thoi ho.

nhl 2.(Tiu chuntiu)Btngthc

N0 (j0,j JN) (1.13)

l iukinv trong trnghpkhng thoi hocngl iukincnpacscndxl tiu.

Ti h ti


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Tnh ton khoa hc

Tiu chun ti u

Ti h ti


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Tnh ton khoa hc

Tiu chun ti u

Tiu chun ti u


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Tnh ton khoa hc

Tiu chun ti u

i ki h ti kh b h d i


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iu kin hm mc tiu khng b chn di

Tnh ton khoa hc


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Tnh ton khoa hc

4. THUT TON N HNHdng ma trn nghch o

B l h h


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Tnh ton khoa hc

Bc lp n hnh

Ta tiptcphn tchpacscnx vicstngngB. Xt trnghpkhi

tiu chuntiuv ciukinhm mctiu khng bchndi

khng cthchin. Khi phitm cchsj0sao chojo>0.

Tphnchngminh iukincncatiu chuntiuta thyrngc th

xy dngcphngn chpnhncx vigi trhm mctiu nh

hn. Ta nhclicng thcxy dngx :

x = x + x,

trong vect gia s x c xc nh nh sauxjo = , xj= 0,jj0, jJN,

xB= - B-1Ajo

B c lp n hnh


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Tnh ton khoa hc

Bc lp n hnh

Khi ta thu c

xN= xN, xB= xB- B-1Ajo,v gi tr hm mc tiu tix s l

R rng:cng ln th gi tr hm mc tiu gim c cng nhiu. Ta quan

tm n gi tr ln nht c th c. Do cn chn sao chox l phngn chp nhn c nn n phi tho mn h bt phng trnh tuyn tnh sau:

xB= xB- B-1Ajo0, 0.

K hiuB-1Ajo = {xi,jo: iJB}, ta vit li h bt phng trnh trn di dng

xixi,jo 0, iJB,

0.

0jcxxc

Bc lp n hnh


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Tnh ton khoa hc

Bc lp n hnh

K hiu

Ta c nghim ca h bt phng trnh l

0 0.

T suy ra gi tr ln nht c th chn l 0. Khi nu thayxbi

phng nx (0) = x +x, trong vect gia s c xy dng vi =0,

gi tr hm mc tiu gim i c mt lng l iojo> 0.

0 0

0

0 0 0 00

/ , khi 0,

, khi 0

/ min { : }

i ij ij

i B

ij

i i i j i B

x x xi J

x

x x i J

Thut ton n hnh dng ma trn nghch o


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Tnh ton khoa hc


Trong cc tnh ton ca mtbc lp nhnh, ma trnB-1 gi mt vai tr quan trng.

Thutton diyscm ttrong ngn

ngcama trnnghchoB-1. Bckhito.

Tm mtpacscndxvicstngnglB.

TnhB-1.



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Tnh ton khoa hc




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Tnh ton khoa hc



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Tnh ton khoa hc

5. Thut ton n hnh dng bng

Thut ton n hnh dng bng


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Tnh ton khoa hc


thuntincho nhngtnh ton bngtay tas m t mt dng khc ca thut ton n

hnh l thutton nhnh dngbng.

Gista c pacscndxvicstngnglB. Cc k hiuvnginguyn nhtrc.

Ta gibngnhnh tngngvipacscnd

xlbngsau y

Bng n hnh


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Tnh ton khoa hc

Bng n hnh

cj

c s

C

s

Phng

n

c1 ... cj ... cn

A1 ... Aj ... An

cj1 Aj1 xj1 xj1j j1

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

ci Ai xi xij i

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

cjm Ajm xjm xjmj jm

1 ... j ... n

Bng n hnh


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Tnh ton khoa hc

Ct u tin ghi h s hm mc tiu ca cc bin c s.

Ct th hai dnh ghi tn ca cc ct c s.

Ct th ba ghi cc gi tr ca cc bin c s (cc thnh phn ca vectxB={xj:jJB} =B1b).

Cc phn txij, iJBtrong cc ct tip theo c tnh theo cng thc:

{xij, iJB} =B1Aj, j=1,2,...,n.

Ct cui cng ghi cc t s i..

Dng u tin ca bng ghi h s hm mc tiu ca cc bin (cj).

Dng tip theo ghi tn ca cc ctA1,...,An.

Dng cui cng gi l dng c lng:

j

= ci

xij

cj

,j=1,2,...,n.iJB

C th thy rng j= 0,jJB .



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Tnh ton khoa hc


Tm ct xoay


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Tnh ton khoa hc

Tm ct xoay

cj

cs

C

s

Phng

n

c1 ... cj0 ... cn

A1 ... Aj0 ... An

cj1 Aj1 xj1 xj1j0 j1

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

ci Ai xi xij0 i

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

cjm Ajm xjm xjmj0 jm

1 ... j0 ... n



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Tnh ton khoa hc


Tm dng xoay


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Tnh ton khoa hc

Tm dng xoay

cj

c s

C

s

Phng

n

c1 ... cj0 ... cn

A1 ... Aj0 ... An

cj1 Aj1 xj1 xj1j0 j1

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

ci0 Ai0 xi0 xi0j0 i0

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

cjm Ajm xjm xjmj0 jm

1 ... j0 ... n

Php bin i n hnh


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Tnh ton khoa hc

Php bin i n hnh



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Tnh ton khoa hc


Qui tc hnh ch nht


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Tnh ton khoa hc

Qui tc hnh ch nht

cj

cs

C

s

Phng

n

c1 ... cj0 ... cn

A1 ... Aj0 ... An

cj1 Aj1 xj1 xj11 xj1j0 xj1n j1

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

ci0 Ai0 xi0 xi01 xi0j0 xi0n i0

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

cjm Ajm xjm xjm1 xjmj0 xjm n jm

1 ... j0 ... n

V d


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Tnh ton khoa hc

V d

Bng n hnh


87/172

Tnh ton khoa hc

Bng n hnh

cj

cs

C

s

Phng

n

1 -6 32 1 1 10 100

A1 A2 A3 A4 A5 A6 A7

1 A4 9 1 0 0 1 0 6 0 9

100 A7 2 3 1 -4 0 0 2 1 2/3

1 A5 6 1 2 0 0 1 2 0 6

301 108 -432 0 0 198 0

Tm ct xoay: Ct xoay l Ct c c lng ln nht

Bng n hnh


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Tnh ton khoa hc

Bng n hnh

cj

cs

C

s

Phng

n

1 -6 32 1 1 10 100

A1 A2 A3 A4 A5 A6 A7

1 A4 9 1 0 0 1 0 6 0 9

100 A7 2 3 1 -4 0 0 2 1 2/3

1 A5 6 1 2 0 0 1 2 0 6

301 108 -432 0 0 198 0

Tm dng xoay: Tnh cc t s i. Dng xoay l dng c t s nh nht

Bin i bng n hnh


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Tnh ton khoa hc

Bin i bng n hnh

cj

cs

C

s

Phng

n

1 -6 32 1 1 10 100

A1 A2 A3 A4 A5 A6 A7

1 A4

1 A1 2/3 1 1/3 -4/3 0 0 2/3 1/3 2

1 A5

Bin i bng: Cc phn t trn dng xoay bng mi = Cc phn ttng ng trong bng c chia cho phn t xoay.

Bin i bng n hnh


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Tnh ton khoa hc

Bin i bng n hnh

cj

cs

C

s

Phng

n

1 -6 32 1 1 10 100

A1 A2 A3 A4 A5 A6 A7

1 A4

1 A1 2/3 1 1/3 -4/3 0 0 2/3 1/3 2

1 A5

Bin i bng: Cc phn t trn cc ct c s l cc vect n v.

Bin i bng n hnh


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Tnh ton khoa hc

Bin i bng n hnh

cj

cs

C

s

Phng

n

1 -6 32 1 1 10 100

A1 A2 A3 A4 A5 A6 A7

1 A4 0 1 0

1 A1 2/3 1 1/3 -4/3 0 0 2/3 1/3

1 A5 0 0 1

0 0 0

Bin i bng: Cc phn t trn cc ct c s l cc vect n v.

Bng n hnh bc 2


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Tnh ton khoa hc

Bng n hnh bc 2

cj

cs

C

s

Phng

n

1 -6 32 1 1 10 100

A1 A2 A3 A4 A5 A6 A7

1 A4 25/3 0 -1/3 4/3 1 0 16/3 -1/3 -

1 A1 2/3 1 1/3 -4/3 0 0 2/3 1/3 2

1 A5 16/3 0 5/3 4/3 0 1 4/3 -1/3 16/5

0 23/3 -92/3 0 0 -8/3 -301/3

Bin i bng: Cc phn t cn li tnh theo qui tc hnh ch nht.

Bng n hnh bc 3


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Tnh ton khoa hc

Bng n hnh bc 3

cj

cs

C

s

Phngn

1 -6 32 1 1 10 100

A1 A2 A3 A4 A5 A6 A7

1 A4 9 1 0 0 1 0 6 0

-6 A2 2 3 1 -4 0 0 2 1

1 A5 2 -5 0 8 0 1 -2 -2

-23 0 0 0 0 -18 -108

Tiu chun ti u c tho mn. Thut ton kt thc.

Phng n ti u: x*= (0, 2, 0, 9, 2, 0, 0). Gi tr ti u: f*= -1


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Tnh ton khoa hc

6. TNH HU HN CA THUT TON N HNH

Tnh hu hn ca thut ton n hnh


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Tnh ton khoa hc

Tnh hu hn ca thut ton n hnh


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Tnh ton khoa hc

7. THUT TON N HNH HAI PHA

Thut ton n hnh hai pha


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Tnh ton khoa hc

Bi ton ph


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Tnh ton khoa hc

p

Bi ton xut pht

Khng gim tng qut ta gi thit l

bi0, i = 1, 2, ..., m, (1.26)

bi v: nu c bi< 0 th ch cn nhn hai v phng trnh tng ng vi -1.

Theo cc thng s ca bi ton cho ta xy dng bi ton ph sau y

Vectxu= (xn+1,xn+2, ...,xn+m) c gi l vect bin gi.

1 1

min{ : , 1,2,..., , 0, 1,2,..., } (1.25)n n

j j ij j i j

j j

c x a x b i m x j n

1

1

min,

, 1,2,..., , (1.27)

0, 1,2,..., , 1,..., .

m

n i

i

n

ij j n i i

j

j

x

a x x b i m

x j n n n m

Bi ton ph


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Tnh ton khoa hc

Pha th 1: Gii bi ton ph


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Tnh ton khoa hc

i vi bi ton ph ta c ngay mt pacscn l

(x=0,xu=b)

vi c s tng ng l

B= {An+1,An+2,... ,An+m} =Em

trong An+il vect ct tng ng vi bin gixn+i, i=1, 2, ..., m.

V vyta c thp dngthutton nhnh giibi tonph. Vicgii

bi tonphbngthutton nhnh cgil pha thnhtcathut

ton n hnh hai pha gii bi ton qui hoch tuyn tnh dng chnh tc(1.25), v bi tonphcn cgil bi ton pha thnht



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Tnh ton khoa hc



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Tnh ton khoa hc

Bng n hnh


104/172

Tnh ton khoa hc

g

cj

cs

C

s

Phng

n

c*1 ... c*j ... c*n+m

A1 ... Aj ... An+m

c*j(1) Aj(1) xj(1) xj(1)j

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

c*i* Ai* xi* xi*j

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

c*j(m) Aj(m) xj(m) xj(m)j

1 ... j ... n+m



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Tnh ton khoa hc



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Tnh ton khoa hc

Ln lt im din tt c cc thnh phnbin gi

trong cs theo th tcva lm ivixi* ta si

nbngnhnh mim trong khng cn thnh

phnbingitrong cs,tcl nctrnghp

ii), ngthitrong qu trnh ny ta cngloibctt c cc rng bucph thuc tuyn tnh trong h

Ax=b.

T bng thu c ta c th bt u thc hin pha thhai ca thut ton n hnh hai pha.



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Tnh ton khoa hc

Nh vy thut ton n hnh hai pha p dng i vi

mt QHTT bt k ch c th kt thc mt trong batnh hung sau y:

1) Bi ton khng c pacnd.

2) Bi ton c hm mc tiu khng b chn.3) Tm c phng n c s ti u cho bi ton.

ng thi trong qu trnh thc hin thut ton ta cng

pht hin v loi b c tt c cc rng buc ph

thuc tuyn tnh trong h rng buc c bnAx=b.

Mt s kt qu l thuyt


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Tnh ton khoa hc

nh l 1.H bi ton QHTT c pacn th n

cng c pacscn.

Chng minh.

p dng thut ton n hnh hai pha i vi

bi ton t ra, kt thc pha th nht ta thu

cphngn cschpnhnccho biton.

Mt s kt qu l thuyt


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Tnh ton khoa hc

nh l 1.6.Nu bi ton QHTT c phng n ti u

th n cng c phng n c s ti u.

Chng minh.

Gi s bi ton c phng n ti u. Khi , thutton nhnh hai pha p dnggiibi ton tra

ch c th kt thc tnh hung 3), tc l thu c

phngn cstiucho n.

Mt s kt qu l thuyt


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Tnh ton khoa hc

nh l 1.7. iu kin cn v bi ton QHTT c

phng n ti u l hm mc tiu ca n bchn di trnminrng buckhc rng.

Chngminh.

iukincn.Gisx* lphngn tiucabi ton. Khi,f(x) f(x*), xD, tcl hm mctiubchndi.

iukin.Nubi ton c hm mctiubchnditrn

minrngbuckhc rng,th p dngthutton nhnh hai

pha giin ta chc thktthc tnh hung3), tcl tm

cphngn cstiucho n.

V d


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Tnh ton khoa hc

V d


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Tnh ton khoa hc

V d: Pha th nht


113/172

Tnh ton khoa hc

0


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Tnh ton khoa hc


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Tnh ton khoa hc


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Tnh ton khoa hc

V d: Pha th hai


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Tnh ton khoa hc

Kt qu


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Tnh ton khoa hc

Phng n ti u:

x*= (0, 2, 0, 3, 0);

Gi tr ti u:

f*

= 2.

Phng php nh thu (M - phng php)


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Tnh ton khoa hc

y l phng php n hnh m rng

Trong thut ton n hnh 2 pha ta phi p dng thutton n hnh 2 ln: Ln th nht gii bi ton pha 1 tm pacscn xut pht.

Ln th hai pha 2 tm phng n ti u cho bi tont ra.

C th kt hp 2 pha ny li v ch cn p dng thutton n hnh mt ln.

C nhiu cch kt hp khc nhau v M phngphp l mt trong cc cch kt hp .

Phng php nh thu


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Tnh ton khoa hc

Xt bi ton QHTT dng chnh tc (1.25), gi thit

rng (1.26) ng. Xy dng bi ton b tr - bi tonM sau:

vi M l mt s dng v cng ln. Cc bin

gi l bin gi, k hiuxu l vector bin gi.

,...,,2,1,0

)28.1(,...,,2,1,

min

1

11

mnjx

mibxxa

xMxc

j

iin

n

j

jij

m

i

in

n

j

jj

mix in ...,,2,1,

Phng php nh thu


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Tnh ton khoa hc

p dng thut ton n hnh cho (1.28), bt u t

pacscnvi c s tng ng l ma trn n v cp m.

Ta vit jdi dng

Trong bng n hnh ta dng c lng chia lm 2dng: mt dng cha v mt dng cha

xt du ca jv so snh chng vi nhau ta sdng quy tc sau:

),1,;,1,0( mibxnjx iinj

Mjjj

j j

Phng php nh thu


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Tnh ton khoa hc

Kt thc thut ton gii bi ton (1.28) ta i n 1trong cc kh nng sau:

(i) Thu c mt phng n ti u (x*

,xu*) vixu* = 0. Khi r rngx* l pa ti u ca bi ton xut pht 1.25.

qpqp

qpqpqp

jjjjj

v

hoc,tycnlhocnu

vlhoc,tycnlhocnu

,

0000

Phng php nh thu


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Tnh ton khoa hc

(ii) Thu c mt phng n ti u (x*,xu*) vixu* 0. Khi

d dng thy bi ton xut pht khng c pacn.(iii) Pht hin iu kin hm mc tiu khng b chn di.

Khi nu bi ton xut pht c pacnd th n cng c hmmc tiu khng b chn di.

xc nh xem bi ton xut pht c pacnd hay khng cnphi thc hin pha th nht ca thut ton n hnh hai pha,tc l coi h s hm mc tiu ca cc bin thtxjvij= 1, ...,n

bng 0. V sau ch lm vic vi cc dng c lng th hai(dng cha cc h s ).

j

Hiu qu ca thut ton n hnh


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Tnh ton khoa hc

Mt im yu ca TTH l v l thuyt, n c thi gian tnh

hm m. iu ny c Klee-Minty ch ra bng v d sau:

1

1 1

1

10 max,

2 10 100 , 1,2,..., ,

0, 1,2,...,

nn j

j

j

i i j i

j i

j

j

x

x x i n

x j n

gii bi ton ny, Thut ton n hnh i hi 2n-1bc lp.

Hiu qu ca thut ton n hnh


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Tnh ton khoa hc

V d Klee-Minty vi n=3:

1 2 3

1

1 2

1 2 3

1 2 3

100 10 max

1

20 100

200 20 10000 , , 0

x x x

x

x x

x x xx x x

Trn thc t: Thut ton n hnh c thi gian tnhO(m3)

Thut ton thi gian tnh a thc gii QHTTPolynomial Algorithms


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Tnh ton khoa hc

Ellipsoid. (Khachian 1979, 1980)

i hi thi gian: O(n4L).

n= s bin s

L= s bt cn thit biu din d liu vo

y l kt qu mang tnh t ph v mt l thuyt.

Cha c hiu qu thc t.

Karmarkar's algorithm. (Karmarkar 1984) O(n3.5L).

C thi gian a thc v c th ci t hiu qu.

Cc thut ton im trong (Interior point algorithms).

O(n3L).

C th snh vi thut ton n hnh! Vt tri so vi tt n hnh khi gii cc bi ton kch thc ln.

Phng php ny c m rng gii cc bi ton tng qut hn.

Ellipsoid Method


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Tnh ton khoa hc

Thut ton hm ro(Barrier Function Algorithms)


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Tnh ton khoa hc

Simplex solution path

Barrier central path

oPredictor

oCorrector

Ti u

Interior Point Methods


129/172

Tnh ton khoa hc

II. L THUYT I NGU

L thuyt i ngu ca QHTT l lnh vc nghin cu trong biton QHTT c kho st thng qua mt bi ton QHTT b tr lin

h cht ch vi n gi l bi ton i ngu

Ni dung


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Tnh ton khoa hc

1. Bi ton i ngu ca QHTT tng qut.

2. nh l i ngu.

3. Gii qui hoch tuyn tnh trn MATLAB


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Tnh ton khoa hc

1. Bi ton i ngu

Bi ton QHTT tng qut


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Tnh ton khoa hc

Xt bi ton QHTT tng qut

1 2

1

1

1

1 1

1 1

( , ,..., ) min,

, 1,2,..., ( )

, 1, 2,...,

0, 1,2,..., ( )

0, 1, 2,...,

n

n j j

j

n

ij j ij

n

ij j i

j

j

j

f x x x c x

a x b i p p m

a x b i p p m

x j n n n

x j n n n

Bi ton QHTT tng qut


133/172

Tnh ton khoa hc

a vo cc k hiu:

11 12 1

21 22 2

1 2

n

n

m m mn

a a a

a a a

A

a a a

- ma trn rng buc,

Bi ton QHTT tng qut


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Tnh ton khoa hc

Xy dng bi ton i ngu


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Tnh ton khoa hc

Khi bi ton QHTT tng qut c th vit li di

dng sau y

( ) ' min,

, ,, ,

0, ,

0, .

i i

i i

j

j

f x c x

a x b i M a x b i M

x j N

x j N

Bi ton QHTT tng qut


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Tnh ton khoa hc

Binibi ton QHTT tngqut vdngchnh tc

bngcch:

a vo cc binphxischuyn rng buc dng

btngthcvdngngthc,

thay mibinkhng c iukindubihiuhai

binc iukindu:xj=xj+xj, khi mict

Ajsthaybihai ctAjvAj,

ta thu cbi ton dngchnh tcsau y:

Xy dng bi ton i ngu


137/172

Tnh ton khoa hc

Bi ton (2.3):

trong

:six i M

,0

(2.3),

min,

x

bxA

xc

Xy dng bi ton i ngu


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Tnh ton khoa hc

Xy dng bi ton i ngu


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Tnh ton khoa hc

Xy dng bi ton i ngu


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Tnh ton khoa hc

Xy dng bi ton i ngu


141/172

Tnh ton khoa hc

Khi vect0 1By c B

lphngn chpnhnccabi ton ingu.Nuby

gita nhnghi hm mctiu cabi ton ingul:

ybmax,

th y0khng nhng l pacnm cn l phngn tiucho

bi ton ingu.

Cc kt qu va nu s c pht biu chnh xc trong ccnhnghav nhl diy.

Bi ton i ngu


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Tnh ton khoa hc

nhngha.Gischo bi ton QHTT tngqut

(cgil bi tongc). Khi bi ton ingucan l bi ton QHTT sau y

Bi ton gc (Primal Prob) Bi ton i ngu (Dual Prob)

( ) ' min, ( ) ' max,, , 0,

, , 0,

0, , ,

0,

i i i

i i i

j j j

j j

f x c x g y y ba x b i M y

a x b i M y

x j N yA c

x j N yA

,jc


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Tnh ton khoa hc

2. nh l i ngu

nh l i ngu


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Tnh ton khoa hc

Trc ht ta chng minh b sau:

B 2.1(nh l i ngu yu). Gi s{x, y} l cp pacn ca bi ton gc-ingu. Khi

f(x) = cxyb=g(y). (2.8)

( ) ' min, ( ) ' max,

, , 0,

, , 0,

0, , ,

0, ,

i i i

i i i

j j j

j j j

f x c x g y y b

a x b i M y

a x b i M y

x j N yA c

x j N yA c

nh l i ngu


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Tnh ton khoa hc

nh l i ngu


146/172

Tnh ton khoa hc

nh l i ngu


147/172

Tnh ton khoa hc

( ) ' min, ( ) ' max,

, , 0,, , 0,

0, , ,

0, ,

i i i

i i i

j j j

j j j

f x c x g y y b

a x b i M ya x b i M y

x j N yA c

x j N yA c

nh l i ngu


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Tnh ton khoa hc

i vi mt bi ton QHTT bt k c th xy ra

ba kh nng:

1) Bi ton c phng n ti u;

2) Bi ton c hm mc tiu khng b chn;

3) Bi ton khng c phng n chp nhn c.

V vy, i vi cp bi ton QHTT gc-i ngu

c th xy ra 9 tnh hung m t trong bng 2.1sau y:

nh l i ngu


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Tnh ton khoa hc

nh l i ngu


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Tnh ton khoa hc

nh l i ngu


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Tnh ton khoa hc

nh l i ngu


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Tnh ton khoa hc

nh l i ngu


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Tnh ton khoa hc

nh l v lch b


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Tnh ton khoa hc

( ) ' min, ( ) ' max,

, , 0,

, , 0,0, , ,

0, ,

i i i

i i i

j j j

j j j

f x c x g y y b

a x b i M y

a x b i M yx j N yA c

x j N yA c

nh l v lch b


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Tnh ton khoa hc

nh l 2.4.Cp phng n chp nhn c ca bi ton gc-

i ngu {x, y} l ti u khi v ch khi thc hin cc iu kinsau y:

(aixbi)yi= 0, i= 1,2,...,m;

xj(cjyAj) = 0,j= 1,2,...,n.

Chng minh.tui= (aixbi)yi, i= 1,2,...,m;

vj=xj(cjyAj) ,j= 1,2,...,n.

Do {x, y} l cp pacn, nn

ui0, i= 1,2,...,m; vj 0,j= 1,2,...,n.

( ) ' min, ( ) ' max,

, , 0,

, , 0,

0, , ,

0, ,

i i i

i i i

j j j

j j j

f x c x g y y b

a x b i M y

a x b i M y

x j N yA c

x j N yA c

nh l v lch b


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Tnh ton khoa hc

t

V d


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Tnh ton khoa hc

Xt bi ton quy hochtuyntnh

C pa tiux* =(0, , 0, 5/2, 3/2), gi trtiuf* = 9/2.

Tm pa tiucabi ton ingu.

5...,,2,1,0

45235

123

min

5321

4321

321

54321

jx

xxxxxxxx

xxx

xxxxx

j

V d

Bi ton i ngu


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Tnh ton khoa hc

Bi ton ingu

Giy* =(y*1,y*2,y*3) l pa tiucabi ton ingu,khiy*phil pa chpnhncv x*, y* thamn (2.10),(2.11)

3,2,1,0

1

1

1

152

1253max43

3

2

321

321

321

321

iy

y

y

yyy

yyy

yyyyyy

i

H qu

*


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Tnh ton khoa hc

Hqu. Phngn chpnhncx*lphngn

tiucabi ton QHTT khi v chkhi hphngtrnh v btphng trnh tuyn tnh sau y l c

nghim:

*

*

( ) 0, ,

( ) 0, ,

0, ,

, ,

, .

i i i

j j j

i

j j

j j

a x b y i M

c yA x j N

y i M

yA c j N

yA c j N

( ) ' min, ( ) ' max,

, , 0,

, , 0,

0, , ,

0, ,

i i i

i i i

j j j

j j j

f x c x g y y b

a x b i M y

a x b i M y

x j N yA c

x j N yA c

V d

Xt bi t QHTT


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Tnh ton khoa hc

Xt bi ton QHTT

Kim tra tnh ti u ca vect

x

*

= (0, 0, 16, 31, 14)i vi bi ton QHTT cho

V d

D d ki * l h h


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Tnh ton khoa hc

D dng kim tra c rngx*l phng n chp

nhn c ca bi ton cho:A=[1 -4 2 -5 9; 0 1 -3 4 -5; 0 1 -1 1 -1];

x=[0;0;16;31;14]; A*x

ans =

3

6

1

Theo hqu,x*

lphngn tiukhi v chkhi hphng trnh v btphng trnh sau y l cnghim

V d

(y +2) x* = 0 Bi ton i ngu


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Tnh ton khoa hc

(y1+2)x 1 = 0

(4y1+y2+y3+ 6)x*2 = 0

(2y13y2y3 5)x*3 = 0

(5y1+ 4y2+y3+ 1)x*4 = 0

(9y15y2y3 + 4)x*5 = 0

y1 24y1+ y2+y36

2y13y2y35

5y1+ 4y2+y31

9y15y2y3 4

Bi ton i ngu

3y1+ 6y2+y3min

y1 2

4y1+ y2+y36

2y13y2y35

5y1+ 4y2+y31

9y15y2y3 4

x*1= 0

x*

2

= 0

x*3= 16

x*4= 31

x*5= 14

V d

H i l t i h


163/172

Tnh ton khoa hc

H cui cng l tng ng vi h sau:

y1 24y1+ y2+ y36

2y13y2y3 = 5

5y1+ 4y2+y3 =19y15y2y3 =4

H ba phng trnh cui cng c nghim duy nhty*= (-1, 1, -10).

(A=[2 -3 -1;-5 4 1; 9 -5 -1]; b=[5;-1;-4]; y=A\b)

Ddng kimtra crngy*thomn haibtphngtrnh u. Do y*

l nghim cahphng trnh v btphng trnh trn. Theo hqu,

iuchngtx*lphngn tiucabi ton QHTT tra.


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Tnh ton khoa hc

3.Gii qui hoch tuyn tnh trn MATLAB

Hm LINPROG

MATLAB h li ii bi t QHTT


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Tnh ton khoa hc

MATLAB cung cp hmlinprog gii bi ton QHTT.

Di y l mt s cch s dng hm ny X=LINPROG(f,A,b)

X=LINPROG(f,A,b,Aeq,beq)

X=LINPROG(f,A,b,Aeq,beq,LB,UB)

X=LINPROG(f,A,b,Aeq,beq,LB,UB,X0)

X=LINPROG(f,A,b,Aeq,beq,LB,UB,X0,OPTIONS)

[X,FVAL]=LINPROG(...)

[X,FVAL,EXITFLAG] = LINPROG(...)

[X,FVAL,EXITFLAG,OUTPUT] = LINPROG(...)

[X,FVAL,EXITFLAG,OUTPUT,LAMBDA]=LINPROG(...)

Hm LINPROG

Lnh X=LINPROG(f A b) gii bi ton QHTT:


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Tnh ton khoa hc

LnhX=LINPROG(f,A,b)gii bi ton QHTT:

min { f 'x : A x


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Tnh ton khoa hc

Lnh X=LINPROG(f,A,b,Aeq,beq,LB,UB,X0) xc nh thm

phng n xut phtX0.

Ch :La chn ny ch c chp nhn nu s dng thut ton tp

tch cc. Phng php ngm nh gii l thut ton im trongs

khng chp nhn im xut pht.

LnhX=LINPROG(f,A,b,Aeq,beq,LB,UB,X0,OPTIONS)thc

hin gii vi cc thng s ti u c xc nh bi bin c cu trc

OPTIONS,c to bi hm OPTIMSET.

Gnoption=optimset('LargeScale','off',

'Simplex','on') chn thut ton n hnh gii bi ton.

Hy g help OPTIMSET bit chi tit.

Hm LINPROG

Lnh [ X FVAL]=LINPROG( ) tr li thm gi tr hm mc


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Tnh ton khoa hc

Lnh [ X,FVAL]=LINPROG(...) tr li thm gi tr hm mc

tiu ti phng nX: FVAL = f'*X. Lnh[X,FVAL,EXITFLAG]=LINPROG(...)tr liEXITFLAG m t iu kin kt thc caLINPROG. Cc gi trcaEXITFLAG c ngha sau

1 LINPROG hi t n li gii X. 0 t n gii hn s bc lp.

-2 Khng tm c pacnd.

-3 Bi ton c hm mc tiu khng b chn. -4 gi trNaNxut hin trong qu trnh thc hin thut ton.

-5 C hai bi ton gc v i ngu u khng tng thch. -7 Hng tm kim qu nh, khng th ci thin c na.

Hm LINPROG

Lnh [X FVAL EXITFLAG OUTPUT] = LINPROG( ) tr li bin


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Tnh ton khoa hc

Lnh [X,FVAL,EXITFLAG,OUTPUT] = LINPROG(...)tr li bin

cu trc OUTPUTvi

OUTPUT.iterations- s bc lp phi thc hin

OUTPUT.algorithm- thut ton c s dng

OUTPUT.message thng bo

Lnh [X,FVAL,EXITFLAG,OUTPUT,LAMBDA]=LINPROG(...)

tr li nhn t Lagrangian LAMBDA , tng ng vi li gii ti u:

LAMBDA.ineqlintng ng vi rng buc bt ng thcA,

LAMBDA.eqlintng ng vi rng buc ng thc Aeq,

LAMBDA.lower tng ng vi LB,

LAMBDA.upper tng ng vi UB.

V d

Gii bi ton qui hoch tuyn tnh:


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Tnh ton khoa hc

Gii bi ton qui hoch tuyn tnh:

2x1

+x2

+ 3x3

min

x1+x2+ x3 + x4 + x5 = 5

x1+x2+ 2x3 + 2x4 + 2x5 = 8

x1+x2 = 2

x3 + x4 + x5 = 3

x1, x2, x3 , x4 , x5 0 f=[2 1 3 0 0]; beq=[5; 8; 2; 3];

Aeq=[1 1 1 1 1; 1 1 2 2 2;1 1 0 0 0;0 0 1 1 1];

A=[]; b=[]; LB=[0 0 0 0 0]; UB=[];X0=[];

[X,FVAL,EXITFLAG,OUTPUT,LAMBDA]=linprog(f,A,b,Aeq,beq,LB,UB,X0)

Kt qu

X = OUTPUT =


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Tnh ton khoa hc

X =

0.00002.0000

0.0000

1.5000

1.5000

FVAL =

2.0000

EXITFLAG =

1

OUTPUT

iterations: 5algorithm: 'large-scale: interior point'

cgiterations: 0

message: 'Optimization terminated.'

LAMBDA =

ineqlin: [0x1 double]

eqlin: [4x1 double]

upper: [5x1 double]

lower: [5x1 double]

V d

S dng thut ton n hnh:


172/172

S dng thut ton n hnh:

opt=optimset('LargeScale','off','Simplex','on')[X,FVAL,EXITFLAG,OUTPUT]=LINPROG(f,A,b,Aeq,beq,LB,UB,X0,opt)

ta thu c kt qu: X = [0 2 0 3 0]

FVAL = 2

EXITFLAG = 1

OUTPUT =

iterations: 1

algorithm: 'medium scale: simplex'

cgiterations: []

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Chuong8_QuyHoachTuyenTinh2