iMc lc

Mc lc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . i

Li cm n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii

M u . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Chng 1. Cc kin thc c bn v tp li v hm li . . . . . . . . . 2

1.1.Tp li . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2.Hm li . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

Chng 2. iu kin cc tiu hm li . . . . . . . . . . . . . . . . . . . . . . . . 12

2.1.Bi ton quy hoch li . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.1.1. Cc khi nim . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.1.2. S tn ti nghim ti u . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.1.3. iu kin ti u . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.2.Ti u c rng buc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.2.1. i ngu Lagrange. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.2.2. iu kin ti u . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

Chng 3.Mt s phng php gii bi ton quy hoch li . . 27

3.1.Cc thut ton s dng o hm bc nht . . . . . . . . . . 273.1.1. Thut ton gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.1.2. Phng php chiu Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.1.3. Thut ton chiu di gradient xp x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323.1.4. Thut ton Frank-Wolfe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.2.Phng php Newton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3.3.Phng php hm pht . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463.3.1. Phng php hm pht im ngoi. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463.3.2. Phng php hm pht im trong . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

Kt lun. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

Ti liu tham kho . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

ii

LI CM N

Trc khi trnh by ni dung chnh ca kha lun, em xin by t lng

bit n su sc ti GS.TSKH. L Dng Mu ngi tn tnh hng dn

v gip em trong sut qu trnh hc tp v nghin cu em c th

hon thnh kha lun ny.

Em cng xin by t lng bit n chn thnh ti qu thy, c gio Vin

Ton hc - Vin Khoa hc v Cng ngh Vit Nam ging dy v gip

em hon thnh kha hc.

Nhn dp ny em cng xin chn thnh cm n Ban Gim hiu, cc bn

ng nghip Trng i hc Cng ngh Thng tin v Truyn thng - i

hc Thi Nguyn, gia nh v bn b lun ng vin, gip v to

iu kin cho em v mi mt trong sut qu trnh hc tp v thc hin

kha lun tt nghip.

Mc d c nhiu c gng nhng Lun vn kh trnh khi nhng

thiu st. Tc gi rt mong nhn c kin ng gp ca qu thy, c

v bn c lun vn c hon thin hn.

Xin trn trng cm n!

H Ni, ngy 30 thng 08 nm 2011

Tc gi

Quch Th Mai Lin

1M U

Quy hoch li l mt lp bi ton c bn ca ti u ha. Mt c

im c bn nht ca lp bi ton ny l mi im cc tiu a phng u

l cc tiu tuyt i. Tnh cht quan trng ny cho php cc l thuyt c

tnh a phng, nh gii hn, vi phn,... c th p dng trc tip vo quy

hoch li. L thuyt v bi ton quy hoch li c quan tm nghin

cu nhiu v thu c nhiu kt qu quan trng da trn l thuyt

ca gii tch li v ti u ha; v phng din tnh ton, c kh nhiu

phng php hu hiu cho lp bi ton ny. Cc phng php c

gii thiu trong cun sch Ti u li (Convex Optimization) ca cc tc

gi Stephen Boyd and Lieven Vandenberghe do nh xut bn Cambridge

University Press in nm 2004.

Mc ch ca bn lun vn ny l trnh by mt s phng php

c bn nht cho bi ton quy hoch li. C th lun vn trnh by cc

phng php sau: cc phng php s dng o hm bc nht, phng

php Newton v cc phng php hm pht.

Lun vn gm c 3 chng:

Chng 1: Gii thiu cc kin thc c bn nht v gii tch li, cbit ch trng vo php chiu vung gc ln mt tp li ng v tnh di

vi phn ca hm li; chng c s dng trong cc chng tip theo.

Chng 2: Trnh by i ngu Lagrange v p dng nh l Krush -Kuhn - Tucker, nh l Kuhn - Tucker gii bi ton quy hoch li v

cc nh l v s tn ti nghim ti u ca bi ton quy hoch li.

Chng 3: Trnh by cc phng php gii bi ton quy hoch li nh:phng php dng o hm bc nht gradient, chiu gradient v trng

hp tng qut ca n l chiu di gradient xp x, thut ton Frank -

Wolf, phng php Newton dng o hm bc hai v cc phng php

hm pht.

2Chng 1

Cc kin thc c bn v tp li vhm li

Trong lun vn ny, ta ch xt khng gian hu hn chiu IRn vi tch

v hng c k hiu l ., . v chun tng ng c k hiu l ..Chng ny trnh by mt s kin thc c bn nht ca gii tch li

s c s dng chng sau. Ni dung ca chng c trch dn ch

yu t ti liu tham kho [1] v [3].

1.1. Tp li

nh ngha 1.1. Cho hai im a, b IRn.(i) Mt ng thng i qua hai im a, b l tp hp c dng

{x IRn : x = a+ b, , IR, + = 1}.

(ii) on thng ni hai im a, b trong IRn c dng

{x IRn : x = a+ b, 0, 0, + = 1}.

nh ngha 1.2. Mt tp D c gi l tp affine nu D cha mi ng

thng i qua hai im bt k x, y D, tc lx, y D, IR x+ (1 )y D.

Mnh 1.1. Tp D 6= l tp affine khi v ch khi n c dng D =M + a vi M l mt khng gian con ca IRn v a IRn. Khng gian Mc xc nh duy nht v c gi l khng gian con song song ca D.

3nh ngha 1.3. Th nguyn (hay chiu) ca mt tp affine D l th

nguyn ca khng gian con song song vi D v c k hiu l dimD.

nh ngha 1.4. Siu phng trong khng gian IRn l mt tp hp cc

im c dng

{x IRn : aTx = },trong a IRn l mt vect khc 0 v IR.nh ngha 1.5. Cho a IRn l mt vect khc 0 v IR. Tp

{x : aTx },c gi l na khng gian ng v tp

{x : aTx > }gi l na khng gian m.

nh ngha 1.6. Mt tp D c gi l mt tp li nu

a, b D v 0 1 a+ (1 )b D.nh l 1.1. Tp li l ng vi php giao, php cng, php nhn vi mt

s thc. Tc l, nu C v D l hai tp li trong IRn th C D, C + Dcng l cc tp li.

nh ngha 1.7. Ta ni x l t hp li ca cc im (vect) x1, ..., xk nu

x =kj=1

jxj, j 0 (j = 1, ..., k),kj=1

j = 1.

Mnh 1.2. Tp hp D l li khi v ch khi n cha mi t hp li ca

cc im ca n. Tc l, D li khi v ch khi

k IN, 1, ..., k 0 :kj=1

j = 1,x1, ..., xk D kj=1

jxj D.

nh ngha 1.8. Mt tp c gi l tp li a din nu n l giao ca

mt s hu hn cc na khng gian ng.

4nh ngha 1.9. Bao li ca mt tp D l giao ca tt c cc tp li

cha D. Bao li ca tp D c k hiu l coD.

Bao li ca mt tp D l tp li nh nht cha D.

nh ngha 1.10. Th nguyn ca mt tp li D c cho bi th nguyn

ca a tp affine nh nht cha D. a tp affine ny c gi l bao affine

ca D v c k hiu l aff D. Th nguyn ca tp li D s c k hiu

l dimD.

nh ngha 1.11. Mt im a ca mt tp li D gi l im trong tng

i nu vi mi x D u c mt s > 0 cho a+(xa) D. Tpcc im trong tng i ca D c k hiu riD.

nh ngha 1.12. Mt tp D c gi l nn nu

> 0,x D x D.Mt nn c gi l nn nhn nu n khng cha ng thng. Mt nn

c gi l nn li nu n ng thi l tp li. Nu nn li ny li l mt

tp li a din th ta ni n l nn li a din.

nh ngha 1.13. Cho D IRn l mt tp li v x0 D.(i) Tp

ND(x0) := { IRn : , x x0 0, x D}.

gi l nn php tuyn ngoi ca D ti x0 v tp ND(x0) c gil nn php tuyn trong ca D ti x0.

(ii) Tp

N D(x0) := { IRn : , x x0 , x D}

c gi l nn php tuyn ca D ti x0.

Hin nhin 0 ND(x0) v dng nh ngha ta c ND(x0) l mt nn ling.

Trong chng 2 v chng 3, ta s s dng cc nh l tch tp li, y

cng l nhng nh l c bn nht ca gii tch li.

5nh ngha 1.14. Cho hai tp C v D, ta ni rng siu phng

H := {x : v, x = }(i) tch hai tp C v D nu

v, a v, b, a C, b D.

(ii) tch cht C v D nu:

v, a < < v, b, a C, b D.

(iii) tch mnh C v D nu:

supxAv, x < < inf

yBv, y.

nh l 1.2 (nh l tch 1). Cho C v D l hai tp li khc rng trong

IRn sao cho C D = . Khi c mt siu phng tch C v D.nh l 1.3 (nh l tch 2). Cho C v D l hai tp li khc rng trong

IRn sao cho C D = . Gi s c t nht mt tp l compc. Khi haitp C v D c th tch mnh c bi mt siu phng.

H qu 1.1 (B Farkas). Cho a IRn v A l ma trn cp m n.Khi a, x 0 vi mi x tha mn Ax 0, khi v ch khi tn ti y 0thuc IRm sao cho a = ATy.

ngha hnh hc ca b Farkas: siu phng i qua gc ta

a, x = 0 nn Ax 0 v mt pha ca n khi v ch khi vect phptuyn a ca siu phng nm trong nn sinh bi cc hng ca ma trn A.

nh ngha 1.15. Cho D 6= (khng nht thit li) v y l mt vect btk, t:

dD(y) := infxDx y.

Ta ni dD(y) l khong cch t y n D. Nu tn ti pi D sao chodD(y) = y pi, th ta ni pi l hnh chiu (vung gc) ca y trn D vk hiu l pi = PD(y).

6Theo nh ngha, ta thy rng hnh chiu PD(y) ca y trn D l

nghim ca bi ton ti u

minxD

{1

2x y2 : x D

}.

Ni cch khc vic tm hnh chiu ca y trn D c th a v vic tm cc

tiu ca hm ton phng ||x y||2 trn D. Nu D 6= th dD(y) huhn, v

0 dD(y) x y, x D.Mnh 1.3. Cho D l mt tp li ng khc rng. Khi

(i) Vi mi y IRn, pi D hai tnh cht sau l tng ng:a) pi = PD(y),

b) y pi ND(pi).(ii) Vi mi y IRn, hnh chiu PD(y) ca y trn D lun tn ti v duy

nht.

(iii) PD(x) PD(y) x y, x, y IRn (tnh khng gin),(iv) PD(x)PD(y)2 PD(x)PD(y), xy, x, y IRn (tnh ng bc).

1.2. Hm li

Trong phn ny ta ch xt nhng hm f khng nhn gi tr .

nh ngha 1.16. Mt hm s f xc nh trn tp li D c gi l

(i) li trn D nu

f(x+ (1 )y) f(x) + (1 )f(y),x, y D, 0 < < 1.

(ii) li cht nu

f(x+ (1 )y) < f(x) + (1 )f(y),x, y D, 0 < < 1.

(iii) lm (lm cht) nu f l li (li cht).

7nh l 1.4. Cho f v g l cc hm li trn tp li C v D tng ng.

Khi cc hm s f+g, (, 0) v max{f, g} cng li trn CD.

Mt hm li c th khng lin tc ti mt im trn bin min xc

nh ca n. Tuy nhin, n lin tc ti mi im trong ca tp theo

nh l sau:

nh l 1.5. Mt hm li xc nh trn tp li D th lin tc ti mi im

trong ca D.

Tnh cht sau y c trng cho mt hm li kh vi, v thun li

kim tra tnh li ca mt hm s. K hiu f (a) hoc f(a) l o hmca f ti a.

nh l 1.6. Cho f : D IR l mt hm kh vi trn tp li m D. iukin cn v f li trn D l

f(x) + f(x), y x f(y), x, y D.

Nu f kh vi hai ln th iu kin cn v f li trn D l vi mi

x A ma trn Hessian H(x) ca f ti x xc nh khng m, tc l

yTH(x)y 0,x D, y IRn.

Nh vy, mt dng ton phng xTQx l mt hm li khi v ch khi Q

xc nh khng m. Mt dng ton phng l mt hm li cht khi v ch

khi ma trn ca n xc nh dng.

Tnh kh vi ca mt hm li gi vai tr quan trng trong cc phng

php ti u ha. Lp cc hm li c nhng tnh cht kh vi rt p m

cc lp hm khc khng c. Gi s f : IRn IR {+} l hm li. Tac cc khi nim sau

nh ngha 1.17. Cho > 0. Mt vc t w IRn c gi l mt di gradient ca f ti x0 IRn nu:

w, x x0 f(x) f(x0) + , x IRn.

8Tp hp tt c cc di gradient gi l di vi phn ca hm f tix0, k hiu l

f(x0) := {w IRn : w, x x0 f(x) f(x0) + , x IRn}.

nh ngha 1.18. Vct w IRn c gi l di gradient ca f tix0 IRn nu:

w, x x0 f(x) f(x0), x IRn.

Tp hp tt c cc di gradient ca hm f ti x0 c gi l di vi phn

ca f ti x0, k hiu l:

f(x0) := {w IRn : w, x x0 f(x) f(x0), x IRn}.

Hm f c gi l kh di vi phn ti x0 nu f(x0) 6= .V d 1.1. Cho D l mt tp li, khc rng ca khng gian IRn. Xt hm

ch trn tp D

D(x) :=

{0 nu x D,+ nu x / D.

Vi mi x0 D ta c:

w D(x0) D(x) D(x0) w, x x0,x D

0 w, x x0,x D w ND(x0).Chng t

D(x0) = ND(x

0),x0 D.Cng c trng hp tn ti nhng im x ti f khng c di vi

phn, ngha l tp f(x) c th l mt tp rng. Tuy nhin, i vi hmli, ta c nh l sau:

nh l 1.7. Cho f l mt hm li (hu hn) trn tp li D. Lc f c

di vi phn ti mi im thuc riD.

T nh l ny suy ra rng nu f l mt hm li trn ton khng gian

IRn th n c di vi phn ti mi im, v riIRn = IRn.

9nh ngha 1.19. Ta gi o hm theo hng d ca mt hm s f (khng

nht thit l li) ti im x l i lng

f (x, d) := lim0+

f(x+ d) f(x)

nu gii hn ny tn ti.

nh l 1.8. Nu f l mt hm li trn tp li D th vi mi x D vmi d sao cho x+ d D, o hm theo hng d ca f ti x lun tn tiv nghim ng

f (x, d) f(x+ d) f(x).Ngoi ra vi mi im x c nh, f (x, .) l mt hm li trn tp li{d : x+ d D}.

T nh l ny d dng suy ra rng nu f kh vi th

f (x, d) = f(x), d,d. (1.1)

Ni chung mt hm li khng nht thit kh vi ti mi im. Di vi

phn l mt khi nim m rng ca o hm trong trng hp hm khng

kh vi. Trong trng hp f(x) ch gm duy nht mt im th f kh viti x. Trong phn tip theo, ta s nh ngha v - nghim v - chiuca mt hm li.

nh ngha 1.20. Cho D IRn l tp li, f : D IR l hm li v 0. Xt bi ton:

minxD

f(x). (P )

Mt im x D c gi l - nghim ca bi ton (P ) nu:

f(x) minxD

f(x) + .

Mnh 1.4. Vect x D l - nghim ca bi ton (P ) khi v ch khi0 f(x).

10

Chng minh. Gi s x D l - nghim ca bi ton (P ). Khi

f(x) f(x) + , x D.

Suy ra

0, x x f(x) f(x) + , x D 0 (f(x)) .

Ngc li, nu 0 (f(x)) th ta c:

0, x x f(x) f(x) + , x D.

Chng t x l nghim ca bi ton (P ).

nh ngha 1.21. Cho D l mt tp li ng khc rng trong IRn, x IRnv 0 . Mt im px D c gi l - chiu ca x trn D nu px lmt - nghim ca bi ton

minyD

{1

2x y2

}(Q)

ngha l:1

2x px2 1

2x PD(x)2 + ,

trong PD(x) l hnh chiu ca x trn D.

Mnh 1.5. Cho D l tp li ng khc rng. Khi px l - chiu

ca x trn D khi v ch khi

x px, px y , y D. (1.2)

Chng minh. Gi s px l chiu ca x trn D. Ta c

minyD

1

2x y2 min

{1

2x y2 + D(y)

}(1.3)

trong D(y) l hm ch ca y trn tp D. t

f(y) :=1

2x y2, x IRn.

11

Theo nh ngha 1.21, px l nghim ca bi ton (1.3). T Mnh 1.4 ta c

0 [f(px) + D(px)] = f(px) + D(px). (1.4)

Theo V d 1.1, D(px) = ND(px) nn t (1.4) ta c:

0 {x+ px}+N D(px).

Suy ra

(x px) N D(px) x px, px , D.

Ngc li, gi s c (1.2). Ta c

x PD(x)2 = x px2 + 2x px, px PD(x)+ px PD(x)2 x px2 + 2x px, px PD(x).

Suy ra

x PD(x)2 x px2 2.Chng t px l chiu ca x trn D.

12

Chng 2

iu kin cc tiu hm li

Chng ny trnh by mt s kin thc quan trng phc v cho

chng 3. l i ngu Lagrange v p dng vo gii bi ton ti u li;

cc nh l c bn nh nh l Karush - Kuhn - Tucker, nh l Kuhn -

Tucker. Ni dung ca chng ch yu c trch dn t ti liu tham kho

[1].

2.1. Bi ton quy hoch li

2.1.1. Cc khi nim

Cho D IRn v f : IRn IR. Xt bi ton quy hoch ton hc

min{f(x) : x D}. (P )

Bi ton ny c hiu l hy tm mt im x D sao cho f(x) f(x)vi mi x D. Mi im x D c gi l mt phng n chp nhnc ca bi ton (P ). Tp D c gi l min (tp) chp nhn c, f

c gi l hm mc tiu ca bi ton (P ). Thng thng, tp D c

cho nh l tp nghim ca mt h bt ng thc hoc ng thc c dng

D := {x X : gj(x) 0, hi(x) = 0, j = 1, ...,m, i = 1, ..., p}, (2.1)

trong 6= X IRn v gj, hi : IRn IR (j = 1, ...m, i = 1, ...p). Biton (P ) vi D cho bi (2.1) gi l trn nu c hm mc tiu v cc rng

buc u trn (kh vi).

13

Bi ton (P ) c rt nhiu ng dng trong cc lnh vc khc nhau. V

d, trong kinh t n l bi ton xc nh phng n sn xut sao cho chi

ph thp nht. Trong v d ny, x l phng n sn xut m mi ta

xj ca n l s lng sn phm loi j cn sn xut, cn f(x) l chi ph

ng vi phng n x. Bi ton (P ) trong m hnh ny c ngha l tm mt

phng n sn xut trong tp hp cc phng n chp nhn c D sao

cho chi ph sn xut ng vi phng n ny l thp nht.

nh ngha 2.1. im x D c gi l li gii ti u a phng cabi ton (P) nu tn ti mt ln cn U ca x sao cho

f(x) f(x), x U D.v x gi l li gii ti u ton cc ca (P) nu

f(x) f(x), x D.

2.1.2. S tn ti nghim ti u

Xt bi ton ti u ton cc (P ). C 4 trng hp tn ti nghim ti

u ca bi ton ny

D = (Khng c nghim). f khng b chn di trn D( inf

xDf(x) = ).

infxD

f(x)

14

Chng minh. Nu x l nghim ti u th F+(D) = [f(x),+] ng (lphn b ca mt tp m) v b chn di.

Ngc li, gi s F+(D) b chn di. t t = inf F+(D) th t > .Do F+(D) ng, t F+D nn tn ti x D sao cho f(x) = t. Chngt x l mt im cc tiu ca f trn D.

nh l 2.2 (Weierstrass). Nu D l tp compact v f na lin tc di

trn D, th bi ton (P ) c nghim ti u.

Chng minh. t := infxD f(x). Theo nh ngha c mt dy {xk} D sao cho limk+ f(xk) = . Do D compact nn c mt dy con hi tv x0 D, khng gim tnh tng qut c th coi xk x0. V f na lintc di nn > . Nhng x0 D nn theo nh ngha ca , ta phic f(x0) . Vy f(x0) = .

nh l 2.3. Nu f na lin tc di trn D v tha mn iu kin bc

sau

f(x) + khi x D, x +th f c im cc tiu trn D.

Chng minh. t D(a) := {x D : f(x) f(a)} vi a D. R rng,D(a) ng v b chn nn f c im cc tiu trn D(a) v im cng

chnh l im cc tiu ca f trn D.

2.1.3. iu kin ti u

nh l 2.4. Gi s D l tp li v f l hm li, kh di vi phn trn

D. Khi x l nghim ti u ca bi ton (P) nu v ch nu

0 f(x) +ND(x), (2.2)

trong ND(x) k hiu nn php tuyn ca D ti x.

Chng minh. : Gi s c (2.2). Khi tn ti p sao cho

p f(x) (ND(x)).

15

Do p f(x) nn

p, x x f(x) f(x), x.

v v p ND(x) nn

p, x x 0, x D.

Vy

f(x) f(x) 0, x D.Chng t x l nghim ti u ca bi ton (P ). : Gi s x l nghim ti u. Bng cch ly khng gian affine caD, ta c th gi s D l mt tp c s chiu y . Do D l tp li,

intD 6= . Xt hai tp sau

E := {(t, x) IR IRn : t > f(x) f(x), x D}; G := {0} D.

C E v G u l tp li (do D v f li). Hn na, G E = . p dngnh l siu phng tch tn ti (u0, u) 6= 0 IR IRn sao cho

u0t+ uTx u00 + uTy, (t, x) E, y D (2.3)

T (2.3), cho t +, ta thy u0 0. Cng t (2.3) nu u0 = 0 th

u, x y 0, x, y D. (2.4)

Hin nhin, 0 D v bng php tnh tin ta c th gi s 0 intD. Theo(2.4) ta c u = 0 (khng xy ra v u0 = 0). Do u0 < 0. Chia c 2 v

ca (2.3) cho u0 > 0, ta c

t+ uTx uTy, x, y D.

Cho t f(x) f(x), ta c

[f(x) f(x)] + uTx uTy, x, y D. (2.5)

Thay y = x vo (2.5), ta c

[f(x) f(x)] + uTx uTx, x D.

16

Do

f(x) f(x) + uT (x x) 0, x D. (2.6)Nu x 6 C, bng cch ly f(x) = nn t (2.6) ta cng nhn c

f(x) f(x) + uT (x x) 0, x X.ngha l u f(x). Mt khc, thay x = x vo (2.5), ta c

uT (y x) 0, y D.Suy ra u ND(x). Kt hp vi u f(x), ta c 0 f(x) +ND(x

).

H qu 2.1. Vi cc gi thit nh nh l 2.4, nu x intD l nghimti u ca bi ton (P ) th 0 f(x). Hn na, nu f kh vi v D = IRnth 0 = f(x).

2.2. Ti u c rng buc

2.2.1. i ngu Lagrange

i ngu l mt phn quan trng ca bi ton ti u. C rt nhiu

kiu i ngu, nhng i ngu Lagrange c s dng rng ri hn c.

i ngu Lagrange da trn c s hm Lagrange. Ta xt bi ton

min{f(x) : x X, gj(x) 0, j = 1, ...,m} (P )trong X IRn l tp li khc rng. T bi ton trn, ta nh nghabi ton ti u khc c dng

max{d(y) : y Y } (D)trong Y IRm.nh ngha 2.2. Bi ton (D) c gi l i ngu ca bi ton (P ) nu

vi mi im chp nhn c x ca (P ) v mi y chp nhn c ca

(D), ta c

f(x) d(y).

17

Bi ton (D) c gi l i ngu chnh xc ca bi ton (P ) nu (D) l

bi ton i ngu ca (P ) v tn ti x (P ), y (D) sao cho

f(x) d(y).

T nh ngha 2.2 ta thy rng, nu bi ton (D) l i ngu chnh

xc ca bi ton (P ) th f(x) = d(y) v hin nhin (P ) cng l i nguchnh xc ca (D).

Xt bi ton (P ), ta nh ngha hm Lagrange

L(x, y) := f(x) +mj=1

yjgj(x).

Ly hm mc tiu ca bi ton i ngu l

d(y) := infxX

L(x, y). (LD)

v min rng buc ca (LD) bng IRm+ . Khi bi ton i ngu tr thnh

supy0

d(y) := supy0

infxX

L(x, y).

nh l 2.5. Bi ton (LD) l i ngu ca bi ton (P ).

Chng minh. Ta c

d(y) = infxX

L(x, y) f(x) +mj=1

yjgj(x) f(x), x X, y Y.

Chng t (LD) l i ngu ca bi ton (P ).

Nhn xt 2.1. Nhn chung, mt cp i ngu cha chc l i ngu

chnh xc nh v d sau y s ch ra.

V d 2.1. Xt bi ton

min{f(x) = x2, x X = [0, 2], x 1 0},

18

Ta thy

f(1) = min f = 1.Hm Lagrange ca bi ton l

L(x, y) = x2 + y(x 1), y 0, x X = [0, 2].

Ta thy

maxy0

d(y) = 0.

Vy cp i ngu l khng chnh xc.

Vy cn thm iu kin g hai bi ton (P ) v (LD) l cp i ngu

chnh xc? Ta c nh l sau:

nh l 2.6 (i ngu chnh xc). Gi s

(i) (P ) c mt li gii ti u,

(ii) Cc hm f v gj, (j = 1, ...,m) li v lin tc trn X.

(iii) iu kin Slater tha mn, ngha l c x0 sao cho gj(x0) < 0 vi mi

j.

Khi (P ) v (LD) l cp i ngu chnh xc.

Chng minh. Gi s

g(x) := (g1(x), . . . , gm(x))T .

Ly

A := {(t, z) IR IRn : t > f(x), z g(x), x X}.Do f v gj (j = 1, ...,m) li nn tp A l li. Gi s x

l nghim ti uca bi ton (P ) th (f(x), 0) 6 A. Theo nh l tch tn ti (, y) 6= 0thuc IR IRn sao cho:

t+ yTz f(x), (t, z) A. (2.7)

19

Do cc hm f v gj lin tc, bt ng thc trn ng vi mi (t, z) A(A l bao ng ca tp A). M (f(x), g(x)) A,

f(x) + yTg(x) f(x), x X. (2.8)

Ta ch ra rng (, y) 0. Thc vy, gi s c ch s j tha mn yj < 0.Ly (t0, z0) A. im (t0, z) (t0, z0 + ej) A vi mi 0 (ej lvect n v th j). Thay (t, z) = (t0, z) vo (2.7) v cho +. Ta cv tri bng m v phi hu hn. Mu thun ny chng t y 0.Chng minh tng t ta ch ra c 0. Hn na, > 0 v nu = 0th y 6= 0. Trong trng hp ny t (2.8) ta c:

yTg(x) 0, x X.

Mu thun vi iu kin Slater. Chia c hai v ca (2.7) cho v theo

nh ngha ca d ta c

d(y

) f(x).

Do (LD) l i ngu ca bi ton (P ) v theo nh l 2.5, (P ) v (LD)

l cp i ngu chnh xc.

2.2.2. iu kin ti u

Xt bi ton (P ) nh ngha bi

min f(x)

vi iu kin

x D := {x X : gj(x) 0, hi(x) = 0, j = 1, ...,m, i = 1, ..., k}.

trong 6= X IRn v f, gj, hi : IRn IR (j, i). Ta gi bi ton (P )l bi ton li nu X l tp li ng v cc hm f , gj l li, hi l hm

affine.

nh l 2.7 (Karush-Kuhn-Tucker). Gi s (P ) l bi ton li. Nu x

l mt nghim ti u ca bi ton (P ) th tn ti i 0 (i = 0, 1, ...,m)

20

v j (j = 1, ..., k) khng ng thi bng 0 sao cho

L(x, , ) = minxX

L(x, , ) (iu kin o hm trit tiu),

i gi(x) = 0 (i = 1, ..,m) (iu kin b).

Hn na, nu intX 6= v iu kin Slater

x0 D : gi(x0) < 0 (i = 1, ...,m)

c tha mn v cc hm affine hi (i = 1, .., k) c lp tuyn tnh trn

X th 0 > 0 v cc iu kin o hm trit tiu v iu kin b l iukin im chp nhn c x l nghim ti u ca bi ton (P ).

Chng minh. Gi s x l mt nghim ti u ca bi ton (P ). t

C := {(0, 1, , m, 1, , k) : (x X) :

f(x)f(x) < 0, gi(x) i, hj(x) = j, (i = 1, ...,m; j = 1, ..., k)}.Do X 6= li, f , gi l cc hm li v hj l hm affine trn X, nn C ltp li ng, khc rng trong IRm+k+1.

Hn na 0 6 C v nu 0 C th tn ti mt im chp nhn c x sao chof(x) < f(x). iu ny mu thun vi gi thit x l nghim ti u cabi ton (P ). Theo nh l siu phng tch, c cc s i (i = 0, 1, ...,m),j (j = 1, ..., k) khng ng thi bng 0 sao cho

mi=0

ii +kj=1

jj 0, (0, , m, 1, , k) C. (2.9)

Nu 0, , m > 0, th thay x = x, ta c:

(0, , m, 0, ..., 0) C

V th c th xem 0, 1, , m 0. Hn na, vi > 0 v x X, ta

ly:

0 = f(x) f(x) + , i = gi(x)(i = 1, ...,m), j = hj(x)(i = 1, ..., k)

21

v thay vo (2.9), cho 0 ta c:

0f(x)+mi=1

i gi(x)+ki=1

ihi(x) 0f(x)+mi=1

i gi(x)+

ki=1

ihi(x), x X

(2.10)

hay

L(x, , ) L(x, , ), x X.iu kin o hm trit tiu c chng minh. chng minh iu kin

b, ta thy, do x l im chp nhn c, gj(x) 0 vi mi j. Nu cch s i sao cho gi(x

) = < 0, th vi mi > 0, ta c

(, ..., , , ...., , 0, ..., 0) C ( v tr i+ 1).Thay vo (2.9) v cho 0, ta c i 0. Nhng v < 0, nn i 0.Do , i = 0.

By gi ta chng minh iu kin . Trc ht, ta c 0 > 0. Tht vy,nu 0 = 0, do iu kin o hm trit tiu v iu kin b, ta c

0 =mi=1

i gi(x) +

kj=1

jhj(x)

mi=1

i gi(x) +kj=1

jhj(x), x X.

(2.11)

Do 0 = 0, nn xy ra 2 trng hp: Trng hp 1: Tn ti ch s i sao cho i > 0. Khi thay th x = x0

vo bt ng thc (2.11) ta c:

0 =mi=1

i gi(x) +

kj=1

jhj(x)

mi=1

i gi(x0) +

kj=1

jhj(x0) < 0 (v l).

Trng hp 2: i = 0 vi mi i v tn ti j sao cho j > 0. Ta c

0 =kj=0

jhj(x)

kj=0

jhj(x), x X.

Do intX 6= v hj l hm affine vi mi j nn ta ckj=0

jhj(x) = 0, x X.

22

Theo gi thit, cc hm hj c lp tuyn tnh trn X, nn j = 0 vi mi

j. iu ny mu thun vi gi thit i v j khng ng thi bng 0. Do

0 > 0 v chia c hai v ca (2.10) cho 0 > 0, ta c th gi s hm

Lagrange ca bi ton (P ) c dng:

L(x, , ) = f(x) +mi=1

igi(x) +kj=1

jhj(x).

S dng iu kin o hm trit tiu v iu kin b, vi mi nghim chp

nhn c x, ta c

f(x) = f0(x) +mi=1

i gi(x) +

kj=1

jhj(x)

f0(x) +mi=1

i gi(x) +kj=1

jhj(x) f(x).

iu ny chng t rng x l mt nghim ti u ca bi ton (P ).

Ch 2.1. Nu X l tp m (hn na, X l ton b khng gian), th

theo Moreau-Rockafellar, iu kin o hm trit tiu ko theo

0 f(x) +mj=1

jgj(x) +

ki=1

ihi(x).

Khi f v mi hm gj (j = 1, 2, ...m) u kh vi th iu kin trn tr thnh

0 = 0f0(x) +mj=1

jgj(x) +ki=1

ihi(x).

nh ngha 2.3. Mt vect d 6= 0 c gi l mt hng chp nhn cca tp D ti x D nu:

x + d D, > 0 nh.

K hiu D(x) l tp tt c hng chp nhn c ca D ti x v D(x)l bao ng ca n.

23

nh l 2.8. Gi s hm f kh vi trn mt tp m cha D v x l mtim cc tiu a phng ca f trn D. Khi :

dTf(x) 0, d D(x). (2.12)Chng minh. Khai trin Taylor ca f ti x l:

f(x + d) = f(x) + f(x), d+ o(||d||). (2.13)Do x l cc tiu a phng ca bi ton (P) nn

f(x + d) f(x) 0, > 0 nh.T (2.13) ta c:

dTf(x) + o(d)

0, > 0 nh.Suy ra (2.12).

Mt im x D m tha mn iu kin (2.12) c gi l im dngca f trn D. Mt im dng cha chc l im cc tiu a phng.

V d 2.2. Xt bi ton

min f(x) = x3 trn C = [1, 2].R rng x = 0 l im dng ca f(x) nhng im cc tiu ca f(x) trnC t ti x = 1.

Xt bi ton (P ) :

min f(x)

vi iu kin

x D := {x X, gj(x) 0, hi(x) = 0, j = 1, ...,m; i = 1, ..., k}.Cho x0 D v tp

A(x0) := {j : gj(x0) = 0}c gi l tp ch s tch cc. t S(x0) l tp nghim ca h phng

trnh tuyn tnh sau{ hi(x0), d = 0, i = 1, ..., kgj(x0), d = 0, j A(x0)

24

Mnh 2.1. Vi mi x0 D ta c D(x0) S(x0).

Chng minh. Cho d D(x0). Nu dTgj(x0) > 0 (j A(x0)) th theonh l 1.8 ta c

gj(x0 + d) gj(x0) > 0 hay gj(x0 + d) > gj(x0) = 0 do j A(x0).

iu ny mu thun vi gi thit d l mt hng chp nhn c. Vy

gj(x0), d = 0, j A(x0).

Chng minh tng t, ta c

hi(x0), d = 0, i = 1, ..., k.

Chng t d S(x0). V vy D(x0) S(x0). Hn na S(x0) l tp ngnn D(x0) S(x0).

nh ngha 2.4. Ta ni rng iu kin chnh quy c tha mn ti x0

nu S(x0) = D(x0).

nh l 2.9 (Kuhn-Tucker). Gi s cc hm f , gj (j = 1, ...,m), hi (i =

1, ..., k) l cc hm kh vi lin tc v x l nghim ti u a phng cabi ton (P ) tha mn iu kin chnh quy. Khi , tn ti cc nhn t

Lagrange

= (1, ..., m) 0, = (1, ..., k) khng ng thi bng 0 (2.14)

sao cho

f(x) +mj=1

jgj(x) +ki=1

ihi(x) = 0, (2.15)

jg(x) = 0 j = 1, ...,m (iu kin b). (2.16)

Ngc li, nu f , gi l cc hm li vi mi i v hi l cc hm affine vi

mi j v nu x D tha mn cc iu kin (2.14), (2.15), (2.16), th xl nghim ti u ca bi ton (P).

25

Chng minh. S dng khai trn Taylor

f(x + d) = f(x) + f(x), d+ o(d),

ta c

f(x), d 0,d D(x).Do D(x) = S(x), nn f(x), d 0 vi mi d S(x). p dng B Farkas vi ma trn A c cc dng

gj(x), j A(x),hi(x),hi(x), i = 1, ..., k.

Ta c cc s j 0 , j A(x) v i 0, i 0, i = 1, ..., k sao cho:

f(x) +

jA(x)jgj(x) +

ki=1

(i i )hi(x) = 0.

Ly j = 0 vi mi j 6 A(x) v i = i i vi mi i th ta c (2.15)v (2.16).

Gi s gi l cc hm li v hj l cc hm affine vi mi i, j. Ta s chng

minh ba iu kin (2.14), (2.15) v (2.16) l iu kin x D lnghim ti u ca bi ton (P ). Tht vy, nu x khng phi l nghimti u, th s tn ti x D sao cho f(x) < f(x). t d := x x 6= 0.Khi

f(x), d = limt0

f(x + td) f(x)t

< 0. (2.17)

Mt khc, jgj(x) = 0 vi mi j, nn j = 0 nu j 6 A(x). Vi x D,

lp lun tng t, ta c

gj(x), x x gj(x) gj(x) 0, j A(x).

Do

jgj(x), d 0, j A(x). (2.18)Theo tnh cht affine ca cc hm hi vi mi i = 1, ..., k, ta c:

hi(x), d = 0. (2.19)

26

Suy ra

i hi(x), d = 0, i = 1, ..., k.Kt hp (2.17), (2.18) v (2.19), ta c

f(x), d+mj=1

jgj(x), d+ki=1

i hi(x), d < 0.

iu ny mu thun vi (2.15). Vy x l nghim ti u ca bi ton(P ).

27

Chng 3

Mt s phng php gii bi tonquy hoch li

Chng ny trnh by mt s phng php c bn nht dng gii

bi ton quy hoch li. l cc phng php gradient, chiu gradient

v chiu di gradient xp x, phng php Frank-Wolfe, phng php

Newton, cc phng php hm pht. y l chng chnh ca bn lun

vn. Cc kin thc trnh by trong chng c tham kho ch yu t ti

liu tham kho [1], [2], [3], [4] v [5].

Xt bi ton ti u li khng rng buc

min{f(x) : x IRn}. (P )

trong

(i) Mi hng u l hng chp nhn c ca bi ton (P ).

(ii) C mt vi phng php xc nh hng gim, trong hng

gim nhanh nht l hng o hm.

3.1. Cc thut ton s dng o hm bc nht

3.1.1. Thut ton gradient

Nu hm f l kh vi th hng gradient l hng tng nhanh nht,

do hng ngc li l hng gim nhanh nht (hay hng dc nht)

gi tr ca hm f . Cho nn phng php gradient cn c gi l phng

28

php hng dc nht.

Thut ton. Chn x0 Rn v di bc k > 0. Ti mi bc lpk = 0, 1, ... c xk.

Bc 1. Tnh f(xk). Nu f(xk) = 0 th thut ton kt thc v xk lim dng. Tri li chuyn sang Bc 2.

Bc 2. Tnh

xk+1 := xk kf(xk),trong k > 0 ( di bc) sao cho f(x

k+1) f(xk). Thay k := k + 1v quay v bc lp k.

Mt s cch xc nh di bc l:

a) Quy tc chnh xc: k = argmin{f(xk + dk) : 0}.b) Quy tc Armijo: Ly s t nhin nh nht m sao cho

f(xk /2mf(xk)) f(xk) /2mf(xk)2, (A)

trong 0 < < 1, > 0 cho trc. Khi , ly k =2m .

nh l 3.1. Gi s f b chn di v gradient f l Lipschitz, ngha l

L > 0 : f(x)f(y) Lx y, x, y.

Khi thut ton gradient vi quy tc Armijo l hi t theo nghaf(xk)0 khi k +.Chng minh. Gi s dk = f(xk). Theo nh l gi tr trung bnh cx (xk, xk+1) sao cho

f(xk+1) f(xk) = f(x), xk+1 xk.

Do , t xk+1 = xk kf(xk), ta c

f(xk+1)f(xk) = kdk,f(x) = kf(xk),f(xk)f(xk)+f(x)

= kf(xk)2 + kf(xk),f(x)f(xk) kf(xk)2 + kf(xk).f(xk)f(x). (3.1)

29

S dng tnh Lipschitz ca f v x (xk, xk+1), ta c

f(xk)f(x) Lxkx Lxkxk+1 = kLf(xk). (3.2)

T (3.1) v (3.2), do k =

2t, ta c:

f(xk+1) f(xk) kf(xk)2 + (k)2Lf(xk)2

= 2tf(xk)2

(1 L

2t

), t > 0. (3.3)

T iu kin 0 < < 1, chng ta lun chn c s t nhin nh nht

t = m sao cho:

1 L2t . (3.4)

Khi

f(xk+1) f(xk) 2mf(xk)2, vi xk+1 = xk

2mf(xk). (3.5)

Do tn ti s t nhin m tha mn (3.4) nn t (3.3) ko theo (3.5). Do

m l s t nhin nh nht tha mn (3.4), ta c:

1 L2m1

< .

Suy ra

2m>(1 )

2L.

Thay vo (3.3), ta c

f(xk+1) f(xk) < (1 )2L

f(xk)2. (3.6)

Suy ra {f(xk)} n iu gim. Theo gi thit, f b chn di nn lim f(xk) >. Do f(xk+1)f(xk) 0. Ly gii hn (3.6) khi k + ta cf(xk) 0.Ch . Thc t, ta c th ly di bc k khng ph thuc vo k bng

cch ly mt s t nhin m0 sao cho

1 kL , k := 12m0

. (3.7)

30

Thc vy, do 0 < < 1 nn m0 lun tn ti. Theo chng minh trn, ta c

f(xk+1) f(xk) 2tf(xk)2(1 L

2t), t > 0.

Thay t = m0 vo bt ng thc ny, ta c

f(xk+1) f(xk) 2m0f(xk)2(1 L

2m0

)(3.8)

vi xk+1 = xk 2m0f(xk). T (3.7) v (3.8) suy ra

f(xk+1) f(xk) 2m0f(xk)2 0. (3.9)

Khi , dy {f(xk)} n iu gim v do lim[f(xk+1) f(xk)] = 0 (vf b chn di). T (3.9) ta c f(xk) 0.

3.1.2. Phng php chiu Gradient

Xt bi ton (P ). Hm f kh vi lin tc trn tp li ng D. Vi > 0,

t

z(x) := PD(x f(x)), x D.

B 3.1. Ta c cc khng nh sau

(i) Hm z(.) lin tc trn D.

(ii) im x D l im dng ca bi ton (P ) nu v ch nu x =z(x).

(iii) Nu x D th (z(x) x)Tf(x) 1z(x) x2.

Chng minh. (i) Hin nhin v f v php chiu l 2 hm lin tc.(ii) Theo tnh cht ca php chiu, ta c

x = z(x) = PD(x f(x))

khi v ch khi

x f(x) x, y x 0, y C

31

hay

f(x), y x 0, y C.iu ny chng t x l im dng ca bi ton (P ).(iii) Theo nh ngha ca php chiu, ta c

x f(x) z(x), y z(x) 0, y D.Thay y = x vo bt ng thc trn, ta c:

x f(x) z(x), x z(x) 0.Suy ra

x z(x)2 f(x), x z(x) 0 f(x), z(x) x z(x) x2.

B c chng minh.

Thut ton. Chn x0 D v , (0, 1), 0 > 0. Ti mi bc lpk = 0, 1, ... c xk.

Bc 1. Tnh f(xk). Nu f(xk) = 0 th thut ton kt thc v xk lim dng. Tri li chuyn sang Bc 2.

Bc 2. Tnh

zk := PD(xk kf(xk)).

Nu zk = xk th thut ton kt thc, xk l im dng. Tri li, ly dk =

zkxk v s dng quy tc Armijo tm s t nhin nh nht mk tha mnf(xk + mkdk) f(xk) + mk(dk)Tf(xk).

t k+1 := mk , xk+1 := xk + k+1d

k. Thay k bi k := k + 1 v quay v

bc lp k.

nh l 3.2. Gi s f kh vi lin tc trn D v x0 D sao cho tp mcLf(x

0) := {x D : f(x) f(x0)}b chn. Khi dy {xk} thu c bng phng php chiu gradient cxc nh tt. Nu thut ton khng kt thc th dy {xk} b chn v bt kim t no ca dy ny cng l im dng ca bi ton (P ).

32

Chng minh. Do f lin tc, tp Lf(x0) ng nn tp ny l compact. V

vy bi ton (P ) c nghim ti u trn tp Lf(x0) v im cng chnh

l nghim ti u ca f trn D. Gi s thut ton khng kt thc ti bc

lp k. Do f kh vi, p dng cng thc Taylor ta c:

f(xk + dk) f(xk) = (dk)Tf(xk) + 0()

trong 0 v theo B 3.1 (iii), ta c (dk)Tf(xk) < 0. Do , tnti > 0 sao cho

f(xk + dk) f(xk) (dk)Tf(xk) (3.10)

vi mi (0, ) v thut ton nu c xc nh tt. Ly = mvi m l s nguyn dng. Do s t nhin m trong thut ton lun tn

ti. Chng t dy {f(xk)} gim nn xk Lf(x0). Vy dy {xk} b chnv do c im t. S mk l s t nhin nh nht tha mn (3.10) vi

= mk nn vi mk 1 ta c

f(xk + mk1dk

) f(xk) > mk1(dk)Tf(xk),hay (

mk1)1 [

f(xk + mk1dk

) f(xk)] > (dk)Tf(xk). (3.11)Thay k = ks trong (3.11) v cho ks +, v mk1 0 khi k +ta c:

(d)Tf(x) (d)Tf(x),trong x, d tng ng l cc gii hn ca dy {xk}, {dk}.Do (0, 1) nn ta c: (d)Tf(x) 0. (3.12)Theo B 3.1 (iii) th (d)Tf(x) 0. Vy (d)Tf(x) = 0. Theo B 3.1 (i) v (iii) ta c z(x) = x. Theo B 3.1 (ii) ta c x l im dng.

3.1.3. Thut ton chiu di gradient xp x

Xt bi ton:

minxD

f(x). (P )

33

K hiu tp nghim ca bi ton (P ) l S(f,D). Nu x D l nghimca bi ton (P ) ta lun c

f(x) f(x), x D hay f(x) f(x) 0, x D.Cho , l cc tham s dng v cc dy s thc {k}, {k}, {k}, {k}

tha mn cc iu kin sau

k > , k > 0, k 0, k 0 k IN, (3.13) kk

= +,

2k < +, (3.14) kkk

< +,

k < +. (3.15)Chng hn c th ly

k =1

k, k = 1, k =

1

k, k =

1

k(k + 1).

Thut ton. Chn x0 D. Ti mi bc lp k = 0, 1, ... c xk.Bc 1: Gi s xk D. Ly gk kf(xk). Ta nh ngha:

k =kk

trong k = max{k, gk}.

Nu gk = 0 th xk l k nghim ca bi ton (P ) v dng thut tonnu k . Tri li chuyn sang Bc 2.Bc 2: Tnh xk+1 D

kgk + xk+1 xk, x xk+1 k, x D. (3.16)Nu xk+1 xk th dng v ta c mt nghim xp x.Ngc li quay v Bc 1.

Ch 3.1. Ta c:

Theo Mnh 1.5 im xk+1 c gi l k-chiu ca (xkkgk) voD. c bit, nu k = 0 th xk+1 l hnh chiu vung gc ca (xk kgk)hay xk+1 = PD(xk kgk). Nu k = k = 0, k IN th px l php chiu vung gc v khi

thut ton tr thnh thut ton chiu gradient, tiu chun dng l gk = 0

bc 1 v xk+1 = xk bc 2.

34

B 3.2. Vi mi k, ta c cc bt ng thc sau:

(i) kgk k;(ii) kxk+1 xk 2k + k.

Chng minh. (i) Theo nh ngha k ta c:

kgk = kgk

max{k, gk} k.

(ii) Thay x = xk vo (3.16) ta c:

xk+1 xk2 kgk, xk xk+1+ k kgk.xk+1 xk+ k (theo Bt ng thc Cauchy - Schwarz) kxk+1 xk+ k (theo chng minh (i).

Hay xk+1 xk2 kxk+1 xk+ k 0. (3.17)Xt phng trnh bc hai s() = 2 vi 0. Khi s() 0

ko theo

+2 + 4

2. (3.18)

Nhn c hai v ca bt ng thc (3.18) vi v s dng tnh cht

ab a2 + b2

2ta c:

21(2 +

2 + 4

) 21

(2 +

2 + 2 + 4

2

)= 2 + .

Thay vo (3.17) vi = xk+1 xk, = k, = k ta c:

kxk+1 xk 2k + k.

B c chng minh.

Mnh 3.1. Gi s bi ton (P ) c tp nghim S(f,D) khc rng. Khi

vi mi x S(f,D) v vi mi k, ta c cc khng nh sau:

35

(i) xk+1 x2 xk x2 + 2k(f(x) f(xk))+ k,

trong k = 2kk + 22k + 4k.

(ii) Dy {xk x2} hi t vi mi x S(f,D).(iii) Dy {xk} b chn.Chng minh. Ta c:

xk x2 = xk+1 xk2 2xk xk+1, x xk+1+ xk+1 x2.

Nn

xk+1 x2 = xk x2 xk+1 xk2 + 2xk xk+1, x xk+1

xk x2 + 2xk xk+1, x xk+1. (3.19)Thay x = x vo (3.16) ta c:

kgk + xk+1 xk, x xk+1 k 2xk xk+1, x xk+1 2kgk, x xk+1+ 2k. (3.20)

Thay (3.20) vo (3.19) ta c:

xk+1 x2 xk x2 + 2kgk, x xk+1+ 2k= xk x2 + 2kgk, x xk+ 2kgk, xk xk+1+ 2k

p dng Bt ng thc Cauchy-Schwarz v B 3.2 (i) cho ta bt ng

thc sau:

xk+1 x2 xk x2 + 2kgk, x xk+ 2kxk xk+1+ 2k xkx2+2kgk, xxk+22+4k (Theo B 3.2 (ii)). (3.21)

Mt khc, gk kf(xk) nn ta c:

gk, x xk f(x) f(xk) + k,

Do k > 0 nhn c hai v ca bt ng thc trn vi k ta c:

2kgk, x xk 2k(f(x) f(xk))+ 2kk. (3.22)

36

Thay (3.22) vo (3.21) ta c:

xk+1 x2 xk x2 + 2k(f(x) f(xk))+ k, (3.23)

trong k = 2kk + 22k + 4k.

(ii) Do x S(f,D) nn f(xk) f(x) hay f(x) f(xk) 0 thay vo(3.23) ta c:

xk+1 x2 xk x2 + k, (3.24)trong k = 2kk + 2

2k + 4k.

Theo (3.14), (3.15), (3.16) th

k > 0 v+k=0

k < +. (3.25)

T (3.24) v (3.25) ta c dy {xk x2} hi t.(iii) Do dy {xk x2} hi t ta suy ra dy {xk} b chn.nh l 3.3. Gi s tp S(f,D) khc rng v dy {gk} b chn. Khi

lim supk+

(f(xk) f(x)) = 0, x S(f,D).

Chng minh. Gi s x S(f,D). Theo Mnh 3.1 (i) ta c:

0 2k(f(xk) f(x)) xk x2 xk+1 x2 + k,

trong k = 2kk + 22k + 4k > 0 v

+k=0

k < + theo (3.14), (3.15),(3.16).Ly tng cc bt ng thc trn vi k = 0, 1...m ta c:

0 2mk=0

k(f(xk) f(x))

x0 x2 xm+1 x2 +mk=0

k

x0 x2 +mk=0

k.

37

Cho m + ta c:

0 2+k=0

k(f(xk) f(x))+ +

k=0

k.

V

+k=0

k < + nn 0 2+k=0

k(f(xk) f(x)) < +. (3.26)

Theo gi thit {gk} b chn, t (3.13) v (3.16) ta c tn ti L saocho gk L vi mi k IN . Do :

kk

= max{1, 1k gk} L

, k IN.

Khi , theo (3.16) th:

k =kk L

kk, k IN. (3.27)

T (3.26) v (3.27) ta c:

+k=0

kk

(f(xk) f(x)) < +.

Do+k=0

kk

= + nn t y suy ra:

lim supk+

(f(xk) f(x)) = 0, x S(f,D). 2

nh l 3.4. Gi s bi ton (P ) c tp nghim S(f,D) khc rng, dy

{gk} b chn v f na lin tc di trn tp D. Khi dy {xk} hi tn nghim ca bi ton (P ).

Chng minh. Gi s x S(f,D). Theo nh ngha ca lim sup tn timt dy con {xkj} ca {xk} sao cho:

lim supk+

(f(xk) f(x)) = lim

j+(f(xkj) f(x)) .

Theo Mnh (3.1) (iii), dy {xkj} b chn. V vy, khng mt tnh tngqut ta gi s l:

limj+

xkj = x. (3.28)

38

Theo gi thit hm f na lin tc di trn tp D v t nh l 3.3 ta c:

f(x) f(x) lim infj+

(f(x) f(xkj))

= {

lim supj+

(f(xkj) f(x))}

= {

lim supk

(f(xk) f(x))}

= 0.

Do x S(f,D) nn f(x) f(x) 0 cho nn ta c:f(x) f(x) = 0,

hay f(x) = f(x) nn x S(f,D).Khi theo nh l 3.3 (ii) dy {xk x2} hi t, kt hp vi (3.28) tac:

limk+

xk = x, x S(f,D).nh l c chng minh.

3.1.4. Thut ton Frank-Wolfe

Thut ton. Tm x0 D. Ti bc lp k(k = 0, 1, ...) c xkBc 1. Tnh f(xk). Nu f(xk) = 0 th dng thut ton v xk l imdng. Tri li, s dng phung php n hnh gii bi ton ti u tuyn

tnh

min{f(xk), x : x D}. (LPk)thu c nghim ti u c s uk. Ta xt hai trng hp:

(i) Nu

f(xk), uk xk 0,th kt thc: xk l im dng.

(ii) Nu

f(xk), uk xk < 0,ly dk := uk xk l mt hng gim. Tm di bc lp tk theo cngthc:

39

tk = argmin{f(xk + tdk) : 0 t 1}.Bc 2. Tnh xk+1 := xk + tkd

k. Thay k := k + 1 v quay li bc lp k .

nh l 3.5. (i) f(xk+1) < f(xk), k.(ii) Nu thut ton kt thc bc lp k, th xk l im dng. Nu thut

ton khng kt thc, th mi im t ca dy {xk} l im dng.(iii) Nu thm iu kin, f l hm li trn D, th f(xk) hi t gim v f

v ta c

0 f(xk) f f(xk), xk uk, k.Chng minh. (i) V theo thut ton, dk l hng gim.

(ii) Theo tiu chun dng, nu thut ton kt thc bc lp k, th

f(xk), uk xk 0.

Do uk l mt li gii ti u ca bi ton (LPk), nn ta c

f(xk), x xk f(xk), uk xk 0, x D.

Vy xk l mt im dng.

Tip theo, gi s thut ton khng kt thc. Ly x l mt im t btk ca dy {xk} khi c mt dy con {xkj} dn n x khi j .Do D c hu hn cc nh, ta c th gi s rng ukj = u vi mi j. Tf(xk+1) < f(xk) vi mi k, theo nh ngha xkj+1 v u, ta c

f(xkj+1) < f(xkj+1) f (xkj + t(u xkj)) , 0 t 1.Cho j +, t tnh lin tc ca hm f ta c

f(x) f (x + t(u x)) , 0 t 1.

Suy ra

0 limt0+

f (x + t(u x)) f(x)t

= f(x), u x.

40

Mt khc, t xkj l mt li gii ti u ca bi ton (LPkj), ta c

f(xkj), u xkj f(xkj), x xkj, x D.

Ly gii hn khi j +, ta c

f(x), x x 0, x D.

Vy x l mt im dng.(iii) Gi s rng f l li trn D. Do tnh li,

0 f(x), x x f(x) f(x), x D.

Hay x l li gii ti u ca bi ton (P). Hn na, do uk l nghim tiu ca bi ton (LPk) v f li nn ti mi bc lp k ta c

f(xk), uk xk f(xk), x xk f(x) f(xk),

hay l

f(xk) f(x) f(xk), xk uk, k.

3.2. Phng php Newton

Trong mc ny ta s trnh by phng php gii bi ton ti u khng

rng buc min{f(x) : x IRn} bng phng php Newton. tng chnhca phng php Newton l xp x bc hai bng cch s dng ma trn

Hessian i vi hm mc tiu. Theo khai trin Taylor ta c:

f(x) q(x) f(xk) +Tf(xk)(x xk) + 12

(x xk)TH(xk)(x xk),

trong H(xk) = 2f(xk), l ma trn Hessian ca f ti xk.iu kin cn cc tiu l q(x) = 0 hay

f(xk) +H(xk)(x xk) = 0.

Gi s nghch o ca H(xk) tn ti, khi ta c cng thc quy:

xk+1 = xk H(xk)1f(xk).

41

Thut ton. Chn x0 Rn v t H(x0) = 2f(x0). Ti bc lpk (k = 0, 1...) c xk

Bc 1. Tnh f(xk). Nu f(xk) = 0 th dng thut ton v xk l imdng. Tri li chuyn sang Bc 2.

Bc 2. Tnh:

xk+1 = xk H(xk)1f(xk). (N)Thay k := k + 1 v quay v Bc 1 .

nh l 3.6. Cho hm f : Rn R v x l im tha mn f(x) = 0.Gi s tn ti cc s R > 0, > 0, > 0 sao cho vi mi x, y nm trong

hnh cu tm x bn knh R, ta c

(i) 2f(x) tn ti.(ii) 2f(x)2f(y) x y.(iii) 2f(x)1 tn ti v 2f(x)1 .Khi tn ti mt s r sao cho vi im bt u x0 bt k tha mn

x0 x < r, th dy {xk} c nh ngha bi

xk+1 = xk 2f(xk)1f(xk) k = 0, 1, 2....

hi t n x vi tc siu tuyn tnh.

chng minh nh l ny ta cn kt qu ca b sau.

B 3.3. Cho hm f : Rn R. Gi s tn ti mt hnh cu B sao cho

2f(x)2f(y) x y, x, y B.

Khi vi x, y B, ta c:

f(y)f(x)2f(x)(y x) 2y x2.

Chng minh. S dng nh l Taylor ta c:

f(y)f(x) = 102f (x+ t(y x)) (y x)dt.

42

v

2f(x)(y x) = 102f(x)(y x)dt.

Tr hai phng trnh trn ta c:

f(y)f(x)2f(x)(y x)

= 102f (x+ t(y x)(y x))2f(x)(y x)dt

102f (x+ t(y x)(y x))2f(x)(y x)dt

10ty x2dt =

2y x2.

Vy ta c: f(y)f(x)2f(x)(y x) 2y x2.

Chng minh nh l.

Gi s xk tha mn cc iu kin (i) - (iii). Theo nh ngha ca bc lp,

ta c:

xk+1 x = xk x 2f(xk)1[f(xk)f(x)]= 2f(xk)1[f(x)f(xk)2f(xk)(x xk)].

Do :

xk+1 x 2f(xk)1.f(x)f(xk)2f(xk)(x xk).Theo B 3.3 ta c th vit li:

xk+1 x 2xk x2.

T ta thy nu xk x < 1

th

xk+1 x < 2

1

xk x = 1

2xk x.

V vy nu

r = min{R, 1},

th ta c xk+1x < 12xkx. Suy ra xk+1x 0 nu x X \D.

47

Chng hn c th ly

p(x) :=mj=1

max(0, gj(x))2, (3.32)

hoc

p(x) :=mj=1

max(0, gj(x)). (3.33)

D dng thy rng nu cc hm gj, (j = 1, ...,m) li, th p l li.

Vi mi t > 0, ta nh ngha bi ton pht nh sau:

min{Ft(x) := f(x) + tp(x), x X}. (Pt)nh l 3.7. Gi s bi ton (P ) c nghim vi mi t > 0. Ly dy s

dng {tk} n iu tng n + v xk l li gii ca (Ptk). Khi :(i) p(xk) p(xk+1),(ii) f(xk) hi t tng dn n f v mi im t ca dy {xk} l li gii

ti u ca bi ton gc (P ).

Chng minh. (i) Do xk l li gii ti u ca bi ton (Ptk) vi mi k ta c

f(xk) + tkp(xk) f(xk+1) + tkp(xk+1), (3.34)

f(xk+1) + tk+1p(xk+1) f(xk) + tk+1p(xk).

Cng hai bt ng thc trn v gin c ta c

(tk tk+1)(p(xk) p(xk+1)) 0V {tk} n iu tng nn p(xk) p(xk+1). Thay vo phng trnh (3.34)ta cf(xk) f(xk+1) vi mi k.(ii) Gi s x l mt li gii ti u ca bi ton (P). Do p(x) = 0, ta c

f(xk) + tkp(xk) f(x) + tkp(x) = f(x). (3.35)

Gi s u l mt im t ca dy {xk} v u / D. Khi p(u) > 0 v{tk} n iu tng n + nn

f(u) + tkp(u) f khi k ln. (3.36)

48

Ly gii hn (3.35), ta c

f(u) + lim tkp(u) f.

Mu thun vi (3.36) vy u D. V f(u) = f v theo tnh cht xcnh hm p v theo (3.34) ta c

f(xk) < f.

Do f lin tc nn qua gii hn ta c f(u) < f. Theo chng minh trn,u D nn f(u) = f. Suy ra f(xk) hi t tng dnf v u l nghimca bi ton (P ).

3.3.2. Phng php hm pht im trong

Phng php hm pht im ngoi ly xp x t bn ngoi cn phng

php ny ly xp x t bn trong. Phng php hm pht im trong c

s dng khi chng ta c th tm mt im x0 D0 .(vi D0 := {x : gj(x) < 0, j = 1, ...,m}

).

Hm pht p : D0 IR c xy dng nh sau:(a) p lin tc trn D0;

(b) Vi dy bt k {xk} D0 hi t n x 6 D0, ta c lim inf p(xk) =+.V d ta c th nh ngha

p(x) := mj=1

log(gj(x)), (3.37)

hoc

p(x) :=mj=1

1

gj(x). (3.38)

R rng p xc nh trn D0, nu cc hm gj li, th p li cht.

Vi mi t > 0 c nh, ta nh ngha bi ton pht:

min{Ft(x) := f(x) + tp(x) : x D0}. (Bt)

49

nh l 3.8. Gi s bi ton (P) c nghim. Cho dy s dng tk n

iu gim n 0 v xk l nghim ca bi ton (Btk). Khi :

(i) p(xk) p(xk+1),

(ii) f(xk) hi t gim n f v mi im t ca dy {xk} l nghim tiu ca bi ton gc (P).

Chng minh. Do xk l nghim ti u ca bi ton (Btk) vi mi k, ta c

f(xk) + tkp(xk) f(xk+1) + tkp(xk+1),

f(xk+1) + tk+1p(xk+1) f(xk) + tk+1p(xk). (3.39)

Cng hai bt ng thc trn v gin c, ta c

(tk tk+1)(p(xk) p(xk+1)) 0.

V tk n iu gim nn

p(xk) p(xk+1).

Thay vo phng trnh (3.39) ta cf(xk) f(xk+1) vi mi k. Do lim f(xk) tn ti.

Ta xt hai trng hp:

Trng hp 1. Bi ton (P) c mt nghim x D0. Do xk l nghim tiu ca bi ton B(tk), ta c

f(xk) + tkp(xk) f(x) + tkp(x). (3.40)

Cho u l im t ca dy {xk}. n gin ta gi s rng xk u. Nuu D0 th ly gii hn (3.40), ta c

f(u) f(x).

Vy u l nghim ti u ca bi ton (P). Do tnh n iu ca dy{f(xk)}, ta suy ra ton b dy {f(xk)} hi t n f.

50

Nu u 6 D0, do xk u, theo cch xy dng hm p, c mt ch s K1sao cho tkp(x

k) 0 vi mi k K1. Khi t (3.40) suy raf(xk) f(x) + tkp(x), k K1.

Cho k +, do tk 0, ta c lim f(xk) f(x). Nhng v xk D vimi k nn f(xk) f(x). Vy lim f(xk) = f(x).Trng hp 2. Bi ton (P) khng c nghim trn D0. t

:= limk+

f(xk).

Do f(xk) f vi mi k, ta c f . Nu f < , th do xk D0 vtnh lin tc ca hm f , tn ti u D0 sao cho

f < f(u) < . (3.41)

Khi

f(xk) + tkp(xk) f(u) + tkp(u).

Theo chng minh trn ta c:

f(xk) f(u) + tkp(u), k K1.Cho k + ta thu c = lim f(xk) f(u) iu ny mu thun vi(3.41). Vy lim f(xk) = f.

V d 3.2. Gii bi ton

min{f(x, y) = x 2y : (x, y) D},trong

D = {g1(x, y) = 1 x+ y2 0, g2(x, y) = y 0}.Gii: Ta xy dng hm pht

p(x, y) = [log(1 + x y2) + log y].Xt bi ton pht

min{Ft(x, y) = x 2y t[log(1 + x y2) + log y] : (x, y) D0}

51

xc nh trn D0 = {g1(x, y) = 1 x+ y2 < 0, g2(x, y) = y < 0}.Theo nh l 2.4 v iu kin ti u ta c (x, y) l nghim ti u cabi ton pht khi v ch khi

Ft(x, y) +ND0(x, y) = 0.

Do D0 l tp m v (x, y) D0, nn ND0(x, y) = {0}.Vy

Ft(x, y) = 0.Ta c

F

x(x, y) = 1 t

1 + x y2 = 0, (3.42)F

y(x, y) = 2 t( 2y

1 + x y2 +1

y)

= 0. (3.43)

T (3.42) ta ct

1 + x y2 = 1

Thay vo (3.43) ta c

2 + 2y ty

= 0 hay l 2y2 2y t = 0

Suy ra

y =1 +

1 + 2t

2hoc y =

11 + 2t2

,

Nu y = 1 +

1 + 2t

2th x =

3t 1 +1 + 2t2

.

Nu y = 1

1 + 2t

2th x =

3t 11 + 2t2

.

Cho t 0 ta c cc nghim l

(1, 0) v gi tr f = 1,

(0, 1) v gi tr f = 2.So snh cc gi tr ta c fmin = 2 ti (0, 1).

52

Xt bi ton (P)

min{f(x) : gj(x) 0, j = 1, ...,m}. (P )Gi s tp rng buc

D := {x : gj(x) 0, j = 1, ...,m}.b chn v tn ti x0 sao cho iu kin Slater gj(x

0) < 0 vi mi j c

tha mn. Gi s cc hm f v gj l li v kh vi lin tc hai ln. Do gj

lin tc nn intD 6= .Hm pht (Hm chn) c nh ngha di y c s dng rng

ri:

(x) :=

nj=1

log(gj(x)) nu gj(x) < 0 j,+ nu gj(x) 0.

Hm pht trn c nhng tnh cht sau:

(i) (x) kh vi v li cht trn intD;

(ii) (x) + khi x D (bin ca D).Do D b chn, t cc tnh cht trn suy ra min(x) lun t c ti mt

im duy nht trong D. im cc tiu duy nht c gi l tm Gii

tch ca cc hm rng buc gj (j = 1, ...,m).

S dng tham s t > 0 v xt bi ton ti u khng rng buc sau

min{Ft(x) := f(x) + t(x) : x intD}. (UP )R rng, nu hm f li v lin tc trn C, th Ft li cht. Gi s vi mi

t > 0 bi ton ny tn ti nghim (v do l duy nht) vi x := x(t) intD. Khi iu kin ti u l

Ft(x) f(x) + 1t

ni=1

gj(x)gj(x) = 0. (3.44)

ng cong {x(t) : t > 0} c gi l ng trung tm gc. ng congx(t) tin ti tm gii tch khi t +. S dng nh l (2.4) v iu kin

53

ti u ca bi ton quy hoch li ta c x l nghim ti u khi v ch khi

0 Ft(x) +ND0(x).

( y ND0(x) = {0} do D0 l mt tp m). Vy ta c c (3.44) l

iu kin cn v cho x l im cc tiu trn ton khng gian cahm Lagrange.

L(x, ) := f(x) +mj=1

jgj(x)

trong

j := j(t) =

1

tgj(x)> 0.

Do j > 0 l im chp nhn c ca bi ton i ngu v ng{(t) : t > 0} c gi l ng trung tm i ngu.

nh l 3.9. Gi s f := min{f(x) : gj(x) 0, j = 1, ...,m}, th

f f(x(t)) mt.

Chng minh. Do f(x) f(x) +mj=1 jgj(x) vi mi x C v j 0,ta c:

f infx{f(x) +

mj=1

j(t)gj(x)}

= {f(x(t)) +mj=1

j(t)gj(x(t))}

= f(x(t))m/t.

v j(t) = 1

tgj(x(t)).

Theo nh l 3.8, ta c f(x(t)) f v x(t) dn ti li gii ti u cabi ton (P) khi t +.Thut ton men theo ng trung tm. Chn mt sai s cho php

> 0, t0 > 0. Tm mt im trong x0 sao cho g(x0) < 0.

54

Bc 1. Bt u vi x0 v t = t0, tnh x(t) trn ng trung tm (c th

dng phng php Newton ). Ngha l

x(t) = argmin{Ft(x) f(x) 1t

mj=1

log(gj(x)) : x D0}.

(Do g(x0) < 0 v Ft(x) + khi gj(x) 0 vi mi j, th bc lptip theo phi tha mn g(x) < 0).

Bc 2. (Tiu chun dng) Nu m/t : Kt thc v x x(t) l mt-nghim. Ngc li ly t := t vi > 1 v quay li Bc 1 vi x0 c

thay th bi x1 x(t).Ch 3.2. Nu ta ly t0 =

m

v gii bi ton:

min

{f(x) 1

t0

mj=1

log(gj(x)) : x IRn},

th nghim ti u ca bi ton ny l mt -nghim.

Tuy nhin trong tnh ton ngi ta thng tng dn t tng bc lp

thut ton chy nhanh hn.

V d 3.3. S dng phng php im trong v nh l Kuhn - Tucker

gii bi ton (P ):

min{f(x, y, z) = x 2y + 3z + z2},

vi rng buc

x+ y + z = 2, x 0, y 0, z 0.Gii: Vit li bi ton di dng sau:

Minf(x, y, z) = x 2y + 3z + z2,

vi cc rng buc:

h(x, y, z) = x+ y + z = 2,

g1(x, y, z) = x 0,

55

g2(x, y, z) = y 0,g3(x, y, z) = z 0.

Ta c nhn xt ngay rng, min rng buc ca bi ton ny l mt tp li

a din v hm mc tiu li mnh trn tp li ny, do bi ton lun c

duy nht nghim.

Xy dng hm pht:

P (x, y, z) = 3j=1

log(gj(x, y, z)) = (log x+ log y + log z).

Khi bi ton pht l:

Min{Ft(x, y, z) = (x 2y + 3z + z2) t(log x+ log y + log z)}

vi rng buc phng trnh:

h(x, y, z) = x+ y + z 2 = 0.

Hm Lagragian l:

L(x, y, z, ) = (x2y+3z+z2)t(log x+log y+log z)+(x+y+z2).

Ch rng do hm mc tiu ca bi ton pht l li mnh vi mi t > 0

c nh, nn bi ton pht lun tn ti nghim v duy nht nghim. p

dng nh l Kuhn - Tucker cho bi ton ny, ta c:

x+ y + z = 2, (3.45)L

x(x, y, z) = 1 t

x+ = 0, (3.46)

L

y(x, y, z) = 2 t

y+ = 0, (3.47)

L

z(x, y, z) = 3 + 2z t

z+ = 0, (3.48)

x > 0; y > 0; z > 0. (3.49)

Nhng im tha mn h phng trnh trn l im Kuhn-Tuker ca bi

ton pht. Gi s x(t), y(t), z(t), (t) l nghim ca h ny. Khi cho

t 0, ta c nghim ca bi ton ban u. Tuy nhin h ny kh gii.

56

Do c th l bt k s thc no, nn ta xt tng trng hp ca c

nh, qua tm x, y v z theo h trn. T (3.46), (3.49), ta c:

Vi > 1 ta c x(t) = t1 +

> 0, t > 0. (3.50)T (3.47), (3.49), ta c

Vi > 2 ta c y(t) =t

2 > 0, t > 0. (3.51)T phng trnh (3.48) ta c

2z2 + (+ 3)z t = 0

hay

z1(t) =(+ 3) +(+ 3)2 + 8t

4; (3.52)

z2(t) =(+ 3)(+ 3)2 + 8t

4. (3.53)

Chn c nh > 2 ri cho t 0 trong (3.50), (3.51), ta c

x = 0, y = 0 v

z1 = 0; z2 =(+ 3)

2

Thay x = 0, y = 0 vo phung trnh (3.45) ta c z = 2 v = 7 < 2.Vy bi ton pht khng c nhn t Lagrange vi > 2.

Vi = 2 th thay vo phng trnh (3.46), (3.47), ta c

x =t

3 0 khi t 0, t

y= 0.

Thay = 2 vo phng trnh (3.52), (3.53) v cho t 0 ta c z = 0hoc z = 5

2< 0 (loi theo iu kin hm g3 = z 0).

Thay x = 0, z = 0 vo (3.45) ta c y = 2.

Ta c nghim (0, 2, 0) ng vi = 2 v f = 4.Vy ta c nghim ca bi ton l (0, 2, 0) v fmin = 4.

57

KT LUN

Lun vn ny trnh by cc kin thc c bn nht v bi ton quy

hoch li. C th, trc ht, lun vn gii thiu cc iu kin ti u v i

ngu Lagrange cho bi ton quy hoch li. Tip theo, lun vn trnh by

cc phng php c bn gii bi ton quy hoch li. l cc phng

php:

Xp x tuyn tnh s dng o hm bc nht, nh phng phpgradient, chiu gradient, chiu di gradient xp x v phng php Frank

- Wolfe. Thut ton gradient xy dng dy xk tin dn ti nghim trn

c s hng ngc vi hng gradient. y l mt phng php thng

dng tm cc tiu, v n n gin v c th p dng cho nhiu lp hm.

Thut ton chiu di gradient xp x cho ta nghim gn ng vi nghim

ti u ca bi ton ban u. Phng php ny p dng tt cho bi ton

vi hm mc tiu khng kh vi. Thut ton Frank - Wolfe xy dng dy

{xk} vi mc ch l hm gi tr ca xk gim dn theo k v s tin tinghim ti u ca bi ton.

Phng php Newton c p dng cho nhng lp hm hai ln khvi lin tc s dng o hm bc hai. Tc hi t n nghim v n gi

tr ti u ca phng php ny nhanh hn so vi khi dng phng php

gradient.

Cc phng php hm pht im trong v im ngoi v thut tonim trong theo hng ng trung tm gii bi ton quy hoch li s

dng quy tc thm vo bi ton ban u mt lng pht a bi ton

li c rng buc v bi ton khng rng buc. Phng php hm pht im

trong ly xp x t bn trong tp rng buc, phng php ny s dng

hm chn ngn im cc tiu ca bi ton khng rng buc ra ngoi

min chp nhn c. Trong lun vn cng gii thiu thut ton men theo

ng trung tm vi i lng pht l hm chn lga. Vi phng php

hm pht im ngoi, im cc tiu ca bi ton khng rng buc ngoi

min chp nhn c ca bi ton ban u nhng cc im t ca dy

tin ti nghim ti u ca bi ton ban u.

58

Ti liu tham kho

Ti liu Ting Vit

[1] L Dng Mu v Nguyn Vn Hin (s ra), Nhp mn Gii tch li ng

dng, NXB Khoa hc t nhin v Cng ngh.

[2] Trn V Thiu v Nguyn Th Thu Thy, Nhp mn Ti u phi tuyn,

NXB i hc Quc gia H Ni, 2011.

[3] Hong Ty, L thuyt ti u, Vin Ton hc, H Ni, 2006.

Ti liu Ting Anh

[4] Stephen Boyd and Lieven Vandenberghe, Convex Optimization, Cam-

bridge University Press, 2004.

[5] Paulo Santos and Susana Scheimberg, An inexact subgradient algorithm

for equilibrium problems, Computational & Applied Mathematics, 30

(2011), 91-107.

[6] Hoang Tuy, Convex Analysis and Global Optimization, Kluwer Aca-

demic Publishers, 1997.

Mc lcLi cm nM uCc kin thc c bn v tp li v hm li Tp li Hm li

iu kin cc tiu hm li Bi ton quy hoch liCc khi nimS tn ti nghim ti uiu kin ti u

Ti u c rng buci ngu Lagrangeiu kin ti u

Mt s phng php gii bi ton quy hoch li Cc thut ton s dng o hm bc nht Thut ton gradientPhng php chiu GradientThut ton chiu di gradient xp xThut ton Frank-Wolfe

Phng php Newton Phng php hm phtPhng php hm pht im ngoi.Phng php hm pht im trong

Kt lunTi liu tham kho