Lý Thuyết Điều Khiển Nâng Cao_Nguyễn Doãn Phước

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Li ni u c c ln xut bn th ba ny vi ni dung v cht lng tt hn hai ln xut

bn trc (ln u l vo nm 2005), tc gi xin c gi li cm n v nhng nhn xt, gp ca bn c, cc bn sinh vin, hc vin cao hc, nghin cu sinh gi ti cho tc gi.

Quyn sch ny c vit ra t cc bi ging trong nhiu nm ca tc gi ti Trng i hc Bch khoa H Ni v L thuyt iu khin, gm bn phn chnh:

iu khin ti u,

Nhn dng i tng iu khin,

iu khin bn vng v

iu khin thch nghi.

Mc ch ca tc gi khi vit quyn sch ny ch n gin l mong mun cung cp cho cc bn sinh vin ang theo hc cc ngnh iu khin t ng, o lng v Tin hc cng nghip, T ng ha, thm mt ti liu b tr cho vic hiu k, hiu su bi ging cng nh h tr vic t hc ca sinh vin, hc vin cao hc, nghin cu sinh thuc cc ngnh lin quan.

Quyn sch c vit vi s cm thng, chu ng rt to ln ca gia nh tc gi. N cng c c hon thnh nh s c v, khuyn khch v to iu kin thun li ca cc ng nghip trong B mn iu khin T ng, Trng i hc Bch khoa, ni tc gi ang cng tc. Tc gi xin c gi ti gia nh v cc bn li cm n chn thnh.

Mc d rt n lc, song chc khng th khng c thiu st. Do tc gi rt mong nhn c nhng gp sa i, b sung thm ca bn c hon thin. Th gp xin gi v:

Trng i hc Bch khoa H Ni Khoa in, B mn iu khin T ng

[email protected]

H Ni, ngy 28 thng 5 nm 2007

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Mc lc

1 iu khin ti u tnh 11 1.1 Nhp mn 11

1.1.1 Th no l bi ton iu khin ti u tnh? ..................................................................... 11

1.1.2 Phn loi bi ton ti u.................................................................................................. 15 Bi ton ti u tuyn tnh/phi tuyn .............................................................................. 15 Bi ton cn ti u (suboptimal) ................................................................................... 16 Bi ton ti u c rng buc/khng rng buc............................................................. 18 Nghim ti u a phng/ton cc .............................................................................. 18

1.1.3 Cng c ton hc: Tp li v hm li.............................................................................. 19

1.2 Nhng bi ton ti u in hnh 23 1.2.1 Bi ton ti u li............................................................................................................. 23

1.2.2 Bi ton ti u ton phng ............................................................................................ 25

1.2.3 Bi ton ti u hyperbol .................................................................................................. 27

1.3 Tm nghim bng phng php l thuyt 29 1.3.1 Mi quan h gia bi ton ti u v bi ton im yn nga ......................................... 29

1.3.2 Phng php KuhnTucker ............................................................................................ 31 1.3.3 Phng php Lagrange................................................................................................... 34

1.4 Tm nghim bng phng php s 36 1.4.1 Bi ton ti u tuyn tnh v phng php n hnh (simplex)....................................... 36

1.4.2 Phng php tuyn tnh ha tng on .......................................................................... 40

1.4.3 Phng php NewtonRaphson ..................................................................................... 41

1.5 Tm nghim bng phng php hng n cc tr 44 1.5.1 Nguyn l chung.............................................................................................................. 44

1.5.2 Xc nh bc tm ti u.................................................................................................. 46 Xc nh bng phng php gii tch ........................................................................... 46 Xc nh bng phng php s.................................................................................... 47 Thut ton nht ct vng .............................................................................................. 47

1.5.3 Phng php GaussSeidel ........................................................................................... 49 1.5.4 Phng php gradient..................................................................................................... 51

1.5.5 K thut hm pht v hm chn ..................................................................................... 53 K thut hm pht......................................................................................................... 53 K thut hm chn........................................................................................................ 56

1.6 Mt s v d ng dng 57 1.6.1 Xc nh tham s ti u cho b iu khin PID .............................................................. 57

1.6.2 Nhn dng tham s m hnh i tng tin nh ............................................................ 60 Nhn dng tham s m hnh khng lin tc ................................................................. 60 Nhn dng tham s m hnh lin tc ............................................................................ 62

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1.6.3 ng dng vo iu khin bn vng trong khng gian trng thi .................................... 63 Pht biu bi ton......................................................................................................... 63 Phng php Roppenecker.......................................................................................... 65 Phng php Konigorski............................................................................................... 68

1.6.4 ng dng vo iu khin thch nghi................................................................................ 73 Mc ch ca iu khin thch nghi............................................................................... 73 Vai tr ca iu khin ti u tnh trong iu khin thch nghi ....................................... 77

Cu hi n tp v bi tp 78

2 iu khin ti u ng 81 2.1 Nhp mn 81

2.1.1 Th no l bi ton iu khin ti u ng?................................................................... 81 Bi ton ti u ng lin tc ......................................................................................... 81 Bi ton iu khin ti u khng lin tc...................................................................... 83

2.1.2 Phn loi bi ton ti u ng ........................................................................................ 84

2.2 Phng php bin phn 86

2.2.1 Hm Hamilton, phng trnh EulerLagrange v iu kin cn ..................................... 86 2.2.2 Bn thm v hm Hamilton............................................................................................. 92

2.2.3 Phng trnh vi phn Riccati v b iu khin ti u khng dng cho i tng tuyn tnh (trng hp thi gian hu hn) ............................................................. 94

Pht biu bi ton v tm nghim nh phng php bin phn................................... 94 Tm nghim ti u t phng trnh vi phn Riccati ....................................................... 96 Thit k b iu khin ti u, phn hi trng thi, khng dng.................................... 99

2.2.4 Phng trnh i s Riccati v b iu khin ti u tnh, phn hi trng thi cho i tng tuyn tnh (trng hp thi gian v hn) B iu khin LQR.............. 100

Pht biu bi ton....................................................................................................... 100 Li gii ca bi ton B iu khin ti u LQR phn hi dng ............................. 101 B iu khin ti u LQR phn hi m....................................................................... 103

2.3 Nguyn l cc i 104 2.3.1 iu khin i tng na tuyn tnh, bit trc im trng thi u v

khong thi gian xy ra qu trnh ti u........................................................................ 105

2.3.2 iu khin ti u tc ng nhanh i tng tuyn tnh................................................. 108 Nguyn l cc i ....................................................................................................... 108 Xy dng qu o trng thi ti u............................................................................. 112 nh l Feldbaum v s ln chuyn i gi tr v ngha ng dng .......................... 118

2.3.3 Nguyn l cc i dng tng qut: iu kin cn, iu kin honh............................. 123 iu kin cn .............................................................................................................. 123 iu kin honh (iu kin trc giao) ......................................................................... 126 Bi ton ti u c khong thi gian c nh v cho trc........................................... 131 Bi ton ti u c i tng khng autonom.............................................................. 132

2.3.4 V ngha vector bin ng trng thi ......................................................................... 133

2.4 Phng php quy hoch ng (Bellman) 137 2.4.1 Ni dung phng php.................................................................................................. 139

Nguyn l ti u ca Bellman ..................................................................................... 139 Hai vng tnh ca phng php: Vng ngc (k thut nhng) v vng xui............ 140

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2.4.2 M rng cho trng hp hm mc tiu khng dng tng .......................................... 144

2.4.3 M rng cho trng hp im cui khng c nh........................................................ 147

2.4.4 M rng cho h lin tc v phng trnh HamiltonJacobiBellman............................ 148


3 iu khin ti u ngu nhin 159 3.1 Mt s khi nim nhp mn 159

3.1.1 Qu trnh ngu nhin..................................................................................................... 159 nh ngha v m t chung ......................................................................................... 159 Qu trnh ngu nhin dng ......................................................................................... 161 Qu trnh ngu nhin egodic....................................................................................... 161 Hm mt ph v nh Laplace ca qu trnh ngu nhin egodic ........................... 162

3.1.2 H ngu nhin v m hnh ton hc trong min phc .................................................. 163 Php bin i Fourier ................................................................................................. 163 Xc nh m hnh hm truyn t ............................................................................... 164

3.1.3 Bi ton iu khin ti u ngu nhin........................................................................... 165

3.2 iu khin ti u ngu nhin tnh 167 3.2.1 Nhn dng trc tuyn tham s m hnh khng lin tc................................................. 167

3.2.2 Nhn dng trc tuyn (on-line) m hnh tuyn tnh lin tc .......................................... 169 Nhn dng trc tuyn m hnh khng tham s........................................................... 169 Nhn dng trc tuyn tham s m hnh i tng khng c thnh phn vi phn ............................................................................................................................ 172 Nhn dng trc tuyn tham s m hnh i tng khng c thnh phn tch phn ............................................................................................................................ 173

3.2.3 Nhn dng ch ng (off-line) m hnh tuyn tnh khng lin tc ................................ 174 Nhn dng ch ng tham s m hnh AR (Autoregressive)..................................... 174 S dng thut ton Levinson tm nghim phng trnh Yule-Walker .................... 176 Nhn dng off-line tham s m hnh MA v ARMA.................................................... 183

3.3 iu khin ti u ngu nhin ng 186 3.3.1 B lc Wiener ................................................................................................................ 186

Mc ch ca b lc .................................................................................................... 186 Cc bc thit k ........................................................................................................ 188

3.3.2 B quan st trng thi Kalman (lc Kalman) ................................................................ 191 Mc ch ca b quan st........................................................................................... 191 Thit k b quan st trng thi cho i tng tuyn tnh............................................ 192

3.3.3 B iu khin LQG (Linear Quadratic Gaussian) ......................................................... 195 Ni dung b iu khin LQG ...................................................................................... 195 Nguyn l tch (separation principle) ......................................................................... 199


4 iu khin ti u RH (iu khin bn vng) 203 4.1 Khng gian chun Hardy 203

4.1.1 Khng gian chun L2 v H2 (RH2)................................................................................. 203

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Khng gian L2 ............................................................................................................. 203 Khng gian H2 v RH2 ................................................................................................ 204 M rng cho ma trn hm phc (h MIMO) ............................................................... 206 Cch tnh chun bc hai ............................................................................................. 206

4.1.2 Khng gian chun H v RH....................................................................................... 208 Khi nim khng gian H v RH .............................................................................. 208 Tnh chun v cng .................................................................................................... 209

4.2 Tham s ha b iu khin 212 4.2.1 H c cc khu SISO.................................................................................................... 212

Trng hp i tng l n nh................................................................................. 212 Trng hp i tng khng n nh.......................................................................... 214 Thut ton tm nghim phng trnh Bezout .............................................................. 216 Tng kt: Thut ton xc nh tp cc b iu khin n nh.................................... 222

4.2.2 H c cc khu MIMO................................................................................................... 225 Khi nim hai ma trn nguyn t cng nhau.............................................................. 225 Phn tch ma trn truyn t thnh cp cc ma trn nguyn t cng nhau............... 227 Xc nh tp cc b iu khin lm n nh h thng................................................ 230 Thut ton tm nghim h phng trnh Bezout ......................................................... 232 Tng kt: Thut ton tham s ha b iu khin n nh .......................................... 236

4.2.3 ng dng trong iu khin n nh ni ......................................................................... 238 Khi nim n nh ni ................................................................................................. 238 Tnh n nh ni c (internal stabilizable) ............................................................... 240 B iu khin n nh ni ........................................................................................... 243

4.3 iu khin ti u RH 244

4.3.1 Nhng bi ton iu khin RH in hnh .................................................................... 244 Bi ton cn bng m hnh......................................................................................... 244 Bi ton cc tiu nhy vi sai lch m hnh .......................................................... 245

Bi ton ti u RH mu (standard) ........................................................................... 246 Bi ton n nh bn vng vi sai lch m hnh ......................................................... 249

4.3.2 Trnh t thc hin bi ton ti u RH .......................................................................... 251 Bc 1: Chuyn thnh bi ton cn bng m hnh .................................................... 251 Bc 2: Tm nghim bi ton cn bng m hnh ........................................................ 252

4.3.3 Kh nng tn ti nghim ca bi ton cn bng m hnh ............................................ 252

4.3.4 Phng php 1: Tm nghim bi ton cn bng m hnh nh ton t Hankel v nh l Nehari ........................................................................................................... 255

Phn tch hm trong v hm ngoi............................................................................. 255 Ton t Hankel ........................................................................................................... 257 nh l Nehari v nghim ca bi ton (4.73) ............................................................ 259 Thut ton xc nh nghim bi ton cn bng m hnh ........................................... 260

4.3.5 Phng php 2: Tm nghim bi ton cn bng m hnh nh php ni suy NevannlinnaPick ......................................................................................................... 262

Ni suy NevannlinnaPick.......................................................................................... 263 Tm gi tr chn di ln nht ..................................................................................... 266 Tng kt: Thut ton tm nghim bi ton cn bng m hnh .................................... 268

4.3.6 Nghim cn ti u (suboptimal) .................................................................................... 270

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5 iu khin thch nghi v bn vng 275 5.1 L thuyt Lyapunov 275

5.1.1 Tiu chun n nh Lyapunov v nh l LaSalle ......................................................... 275 T tng chung........................................................................................................... 278 Tiu chun Lyapunov v hm Lyapunov.................................................................... 279 nh l LaSalle ............................................................................................................ 282 p dng cho h tuyn tnh v phng trnh Lyapunov ............................................... 286

5.1.2 Thit k b iu khin GAS nh hm iu khin Lyapunov (CLF)............................... 290 Khi nim hm iu khin Lyapunov.......................................................................... 290 Thit k hm iu khin Lyapunov cho h affine ....................................................... 292 Thit k cun chiu (backstepping) hm CLF cho h truyn ngc .......................... 295 Thit k cun chiu (backstepping) hm CLF cho h tam gic ................................. 297 Thit k hm CLF cho h truyn ngc cht nh php i bin vi phi .................... 301 Thit k hm CLF cho h affinetruyn ngc nh php i bin vi phi ................. 306 iu khin tuyn tnh ha chnh xc gn im cc cho h tam gic.......................... 309

5.2 iu khin thch nghi t chnh (STR) 312 5.2.1 Tng qut v c cu nhn dng tham s m hnh, phng php bnh phng

nh nht v m hnh hi quy ......................................................................................... 313 Phng php bnh phng nh nht .......................................................................... 313 Nhn dng tham s m hnh khng lin tc ............................................................... 315 Nhn dng tham s m hnh lin tc .......................................................................... 316

5.2.2 C cu xc nh tham s b iu khin t m hnh i tng ...................................... 316 Xc nh tham s b iu khin PI theo phng php ti u ln .......................... 317 Xc nh tham s b iu khin PID theo phng php ti u i xng.................... 317 Xc nh tham s b iu khin ti u theo nhiu...................................................... 318 Thit k b iu khin phn hi, tnh, theo nguyn tc cho trc im cc .............. 319 Thit k b iu khin ng, phn hi tn hiu ra c im cc cho trc .................. 321 Thit k b iu khin vi m hnh mu (model following)......................................... 323 Xc nh tham s b iu khin khng lin tc .......................................................... 330

5.2.3 S dng m hnh mu nh mt thit b theo di: iu khin thch nghi t chnh trc tip ................................................................................................................ 332

Xc nh trc tip tham s b iu khin khng lin tc ............................................ 332 Xc nh trc tip tham s b iu khin lin tc ....................................................... 335

5.3 iu khin thch nghi c m hnh theo di (MRAC) 336 5.3.1 Hiu chnh tham s b iu khin theo lut MIT........................................................... 338

Ni dung phng php ............................................................................................... 338 nh gi cht lng c cu chnh nh....................................................................... 343

5.3.2 Hiu chnh tham s b iu khin nh cc tiu ha hm mc tiu hp thc (xc nh dng)............................................................................................................ 345

5.4 iu khin n nh ISS v iu khin bt nh, thch nghi khng nhiu 356 5.4.1 t vn ..................................................................................................................... 356

5.4.2 iu khin thch nghi i tng phi tuyn c tham s hng bt nh ........................... 358 Phng php gi nh r (certainty equivalence) ....................................................... 358

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Thit k cun chiu (backstepping) b iu khin thch nghi gi nh r cho i tng truyn ngc ............................................................................................... 364 Thit k cun chiu (backstepping) b iu khin bm thch nghi cho i tng tam gic c tham s hng bt nh .................................................................. 369

5.4.3 iu khin thch nghi i tng phi tuyn c tham s bt nh ph thuc thi gian................................................................................................................................ 372

Phng php nn min hp dn (damping) ............................................................... 373 Khi nim n nh ISS, hm ISSLyapunov v hm ISSCLF.................................. 376 Phng php iu khin ISS thch nghi khng nhiu................................................. 383 Thit k cun chiu (backstepping) hm ISSCLF.................................................... 389

5.5 S dng phng php iu khin trt 393 5.5.1 Xut pht im ca phng php iu khin trt ....................................................... 393

5.5.2 Thit k b iu khin trt n nh bn vng.............................................................. 396

5.5.3 Thit k b iu khin trt bm bn vng................................................................... 401

5.6 iu khin thch nghi b bt nh 402 5.6.1 iu khin thch nghi b bt nh i tng tuyn tnh ................................................. 402

B bt nh bng phn hi tn hiu ra......................................................................... 402 B bt nh bng phn hi trng thi.......................................................................... 406

5.6.2 iu khin thch nghi b bt nh i tng phi tuyn .................................................. 407 iu khin tuyn tnh ha chnh xc h c mt u vo............................................ 407 iu khin tuyn tnh ha chnh xc h c nhiu u vo ......................................... 415 iu khin thch nghi b bt nh i tng phi tuyn affine...................................... 430


6 Mt s khi nim c bn ca iu khin v nhng vn b sung 439 6.1 Nhng khi nim c bn 439

6.1.1 Cu trc i s .............................................................................................................. 439 Nhm .......................................................................................................................... 439 Vnh............................................................................................................................ 440 Trng......................................................................................................................... 441 Khng gian vector....................................................................................................... 441 a tp tuyn tnh......................................................................................................... 443 i s.......................................................................................................................... 444 Ideale .......................................................................................................................... 445

6.1.2 i s ma trn v m hnh h tuyn tnh ...................................................................... 445 Cc php tnh vi ma trn........................................................................................... 446 Hng ca ma trn ....................................................................................................... 447 nh thc ca ma trn................................................................................................. 448 Ma trn nghch o ..................................................................................................... 449 Vt ca ma trn .......................................................................................................... 450 Ma trn l mt nh x tuyn tnh ................................................................................ 450 Php bin i tng ng......................................................................................... 452 Gi tr ring v vector ring ........................................................................................ 453 M hnh trng thi h tuyn tnh ................................................................................. 455

6.1.3 Khng gian hm s v m hnh h phi tuyn................................................................ 458 Khng gian metric....................................................................................................... 458

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Khng gian ............................................................................................................. 459 Khng gian compact ................................................................................................... 460 Khng gian chun....................................................................................................... 460 Khng gian Banach..................................................................................................... 462 Khng gian Hilbert ...................................................................................................... 462 Khng gian cc nh x lin tc................................................................................... 463 M hnh trng thi h phi tuyn .................................................................................. 465

6.2 L thuyt hm bin phc 468 6.2.1 nh ngha, khi nim hm lin tc, hm gii tch......................................................... 468

6.2.2 Hm bo gic (conform)................................................................................................ 469

6.2.3 Tch phn phc v nguyn l cc i modulus ............................................................. 472

6.3 L thuyt n nh Kharitonov 474 6.3.1 Ni dung nh l Kharitonov .......................................................................................... 474

6.3.2 Thit k b iu khin n nh bn vng cho i tng tuyn tnh c tham s bt nh.......................................................................................................................... 479

6.4 ng dng hnh hc vi phn vo iu khin 483 6.4.1 H affine ........................................................................................................................ 483

6.4.2 Cc php tnh c bn .................................................................................................... 485 o hm ca hm v hng (o hm Lie) ............................................................... 485 Php nhn Lie, hay o hm ca vector hm............................................................ 486 Hm m rng (distribution) ......................................................................................... 487

6.4.3 Php i bin vi phi a h affine v dng chun ...................................................... 490 Nguyn l chung ......................................................................................................... 490 ng dng trong thit k b iu khin hai cp .......................................................... 496

Ti liu tham kho 498

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1 iu khin ti u tnh

1.1 Nhp mn

1.1.1 Th no l bi ton iu khin ti u tnh?

Trong qu trnh iu khin h thng, ta thng hay gp phi loi bi ton chn cc tham s iu khin trong s nhng tham s thch hp sao cho h thng t c cht lng mt cch tt nht. Nu s dng k hiu:

tp cc tham s iu khin thch hp l P,

cc tham s iu khin cn chn l vector p =(p1 , p2 , ,pn )T , tc l bi ton

c n tham s, v

cht lng h thng do b tham s p mang li l Q( p )

th cc bi ton nu trn c vit chung li thnh:

Q( p )Pp

min *p = argPp

min Q( p ) (1.1)

hoc Q( p )Pp

max *p = argPp

max Q( p ) (1.2)

nh ngha 1.1: Vector tham s iu khin h thng c gi l ti u nu:

a) N tha mn cc yu cu ca bi ton iu khin (phng n thch hp).

b) N mang li cho h thng mt cht lng iu khin tt nht.

V d 1.1: (u t c li nhun cao nht)

Trong mt nh my c n phn xng A1 , , An . Nh my hin c nguyn vt liu

vi s lng a trong kho v c d nh chia cho cc phn xng sn xut. Gi pk ,

k=1, ,n l s vt liu m phn xng Ak nhn c cng nh hk (pk ) l li sut m

n mang li cho phn xng. Bi ton t ra cho nh my l phi phn chia nguyn vt liu nh th no cho cc phn xng cui cng tng li nhun ca nh my l ln nht.

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R rng, tham s iu khin y l (c gp chung li thnh vector):

p =

np

p1

v tp cc tham s iu khin thch hp c dng:

P = { p Rn =

n

kkp

1 a v pk0 , k=1,2, ,n } .

Li nhun ca nh my do phng n p mang li c tnh bng:

Q( p ) = =

n

kkk ph

1)(

Vy, nhim v ca nh my l phi tm c cch chia ti u *p l nghim ca:

*p = argPp

max Q( p )

V d 1.2: (Chn tham s ti u cho b iu khin PI)

Hnh 1.1a) m t h thng gm i tng iu khin c hm truyn t:

S (s ) =)3)(1(

2ss ++

v b iu khin PI vi hm truyn t:

R (s ) = kp (1+ sTI

1) =

sT

sTk

I

Ip )1( +

Nhim v bi ton iu khin t ra l phi xc nh tham s kp , TI cho b iu

khin PI trong khong:

0 < pk kp +pk cng nh 0 I

p

T

k

2

v nh vy, tp P gm b tham s iu khin thch hp c dng nh sau:

P = { p =

I

p

T

kR2 pk kp

+pk ,

IT TI

+IT v 3+2kp>

I

p

T

k

2}

Tip tc, nh cng thc Parseval, hm o cht lng Q( p ) cho trong (1.3) c vit

li thnh:

)( pQ =

djE

2)(21

=

j

jdssEsE

j)()(

21

trong (xem thm bng 1.1 ca mc 1.6.1):

E (s ) =)()(1

1sRsS+

s1

=32

2

4)23(2

)43(

sTsTskTk

ssT

IIpIp

I

++++++

l nh Laplace ca sai lch e ( t )=w ( t )y ( t ) . T y suy ra:

)( pQ =p

I

kT

4

ppI

Ippp

kkT

Tkkk

2)23(4

3620)23(2

++++

Vy dng chun (1.1) ca bi ton xc nh tham s ti u PI s l:

y ( t )w ( t ) e ( t )PI S (s )

y(t)

t

1

Hnh 1.1: Minh ha v d 1.2.

a) b)

14

p * =

*

*

I

p

T

k= arg

Ppmin

p

I

kT

4

ppI

Ippp

kkT

Tkkk

2)23(4

3620)23(2

++++

.

V d 1.3: (B iu khin c gi thnh thp nht)

Cho i tng tuyn tnh (gi s khng n nh) c m t bi m hnh trng thi dng chun iu khin nh sau:

dt

xd=

210

100010

aaa

x +

100

u

vi x =

3

2

1

x

x

x

l k hiu ch vector cc bin trng thi x1 , x2 , x3. i tng c iu

khin phn hi trng thi bng b iu khin (tnh)

kT = (k1 k2 k3)

Vi b iu khin ny, h kn (hnh 1.2) s c phng trnh c tnh

A(s) = (a0+k1 )+(a1+k2 )s+(a2+k3 )s2

Do n n nh khi v ch khi

k1 > a0 k2 > a1 v k3 > a2

Ni cch khc, tp P l tp con trong khng gian ba chiu R3 gm cc phng n iu

khin thch hp (cc phng n chn tham s k1 , k2 , k3 ) c dng

P ={ p =

3

2

1

k

k

k

R3 k1 > a0 , k2 > a1 v k3 > a2 }

Gi thnh chi ph cho b iu khin kT c tnh theo tng knh. Knh mt k1 s

c gi thnh l 12 ng trn mt n v, knh hai k2 l 18 ng/n v v knh ba k3 l

15 ng/n v. Suy ra

)( pQ = 12 k1 +18 k2 +15 k3

l gi thnh ca b iu khin kT .

)( pQ

i tng iu khin

x uw

( k1 , k2 , k3 )


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Ngi thit k b iu khin c nhim v l phi chn mt phng n p *P c

Q( p ) nh nht, tc l

p * =

*3

*2

*1

k

k

k

= argPp

min (12 k1 +18 k2 +15 k3 ) .

Nhn li c ba v d nu trn ta thy chng c mt im chung. l tp P cc

tham s iu khin v hm o cht lng Q( p ) u c dng i s (khng cha ton t

tch phn, vi phn). Nhng dng bi ton iu khin ti u nh vy c gi l iu khin ti u tnh.

nh ngha 1.2: Bi ton iu khin ti u tnh l bi ton (1.1) hoc (1.2), trong tp

P ca cc tham s iu khin thch hp p v mc tiu cht lng Q( p ) u c m

t bng cc hm i s.

1.1.2 Phn loi bi ton ti u

Trc ht ta thy, vic cho rng bi ton ti u ch c dng (1.1) s khng lm mt tnh tng qut ca bi ton iu khin ti u. V nu ngc li, tham s iu khin ti

u *p thc s phi lm cho *)( pQ l ln nht, tc l bi ton (1.2), th ta ch cn thay

)( pQ trong (1.2) bi )(~

pQ = )( pQ l s li tr v bi ton (1.1) nu trn.

Bi ton ti u tuyn tnh/phi tuyn

nh ngha 1.3: Bi ton iu khin ti u (1.1) c gi l tuyn tnh, nu tp P ca cc

tham s iu khin thch hp p v hm mc tiu Q( p ) u c m t bng cc bt

phng trnh hoc phng trnh tuyn tnh. Ngc li th n c gi l bi ton iu khin ti u phi tuyn.

V d 1.4: (Bi ton ti u tuyn tnh)

Bi ton ti u (1.1) vi

P = { a1 1p1 + a1 2p2 + + a1 npn b1

a2 1p1 + a2 2p2 + + a2 npn b2

am 1p1 + am 2p2 + + am npn bm}

)( pQ

16

v )( pQ = c1p1 + c2p2 + + cnpn

l bi ton ti u tuyn tnh.

Nh vy, quyt nh mt bi ton iu khin ti u l tuyn tnh hay phi tuyn, khng th ch cn c vo m hnh h thng m phi da vo dng chun (1.1) ca n. in hnh nht l bi ton nu trong v d 1.2. R rng h thng l tuyn tnh v c m t bi m hnh hm truyn t G (s ) , song bi ton xc nh tham s ti u ng vi n li l bi ton iu khin ti u tnh v phi tuyn.

Trong quyn sch ny, chng ta s tp trung ch yu vo cc phng php gii bi ton ti u phi tuyn v xem bi ton ti u tuyn tnh nh l mt trng hp ring.

Bi ton cn ti u (suboptimal)

Tip theo, ta bn n tn gi ti u ca bi ton (1.1). Gi s p * l tham s iu

khin ti u. Vy th gi tr Q( p *) phi l nh nht. iu ny ch rng s khng c cc

khi nim "ti u hn" hay "ti u nht". Tht vy nu nh bn cnh p * cn c mt

phng n p~ "ti u hn", tc l vi n Q( p~ ) cn c gi tr nh hn *)( pQ th iu ny

li mu thun vi gi thit rng p * l ti u.

Nh vy bn thn tn gi ti u mang ngha nht. Tuy nhin cng phi ch

rng ngha nht ca ti u l ni rng hm Q( p ) ch c mt gi tr nh nht *)( pQ ch

khng phi tham s iu khin ti u p * l duy nht. C th mt bi ton iu khin

ti u c nhiu vector tham s ti u *1

p , *2

p . Nhng vector tham s ny u mang

n cho h thng mt cht lng ging nhau l hm mc tiu Q( p ) nhn gi tr nh

nht ti (hnh 1.3):

)( *1

pQ = )( *2

pQ = = Qmin =Pp

min )( pQ .

)( pQ

pQmin

*1

p *2

p *3

pHnh 1.3: Nghim ca bi ton iu khin ti u.

17

Khi cp n vn mt bi ton iu khin ti u c th c nhiu nghim th cng cn bn n nhng bi ton iu khin ti u khng c nghim. minh ha cho trng hp ny ta xt v d 1.5 di y:

V d 1.5: (Bi ton ti u v nghim)

v d 1.3 ta xt bi ton tm b iu khin phn hi trng thi c gi thnh thp nht n nh i tng v cng i n dng chun ca n nh sau

p * =

*3

*2

*1

kkk

= argPp

min )( pQ (1.4)

trong

P ={ p =

3

2

1

k

k

k

R 3 k1>a0 , k2>a1 v k3>a2}

v )( pQ = 12 k1 +18 k2 +15 k3 .

Hin nhin rng bi ton (1.4) vi cc hng s dng a0 ,a1 ,a2 v nghim. Tht

vy, ta gi s rng tn ti p * ti u, hay:

*)( pQ = 12 *1k +18*2k +15

*3k

l gi tr nh nht ca )( pQ . Gi p~ l mt phng n khc vi ba thnh phn

1~k , 2

~k , 3

~k c xc nh nh sau:

1~k =

20

*1 ak + , 2

~k =

21

*2 ak + , 3

~k =

22

*3 ak +

Khi th do:

1~k < *1k , 2

~k < *2k , 3

~k < *3k

nn:

Q( p~ ) = 12 1~k +18 2

~k +15 3

~k < Q( p *) = 12 *1k +18

*2k +15

*3k .

v l iu phi l.

Bng v d 1.5 ta ch ra s tn ti cc bi ton iu khin ti u v nghim, tc

l khng c mt phng n iu khin p *P no lm cho )( pQ c gi tr nh nht.

Trng hp ny thng hay gp phi loi bi ton c P l tp h nhng o hm ca

)( pQ li khc 0 trong P (chng hn c nghim nm trn bin ca P).

Tuy nhin, nu nh )( pQ vi p P li c gi tr chn di ln nht (infimum):

18

q = )(inf pQp

v khi cho trc mt hng s dng nh ty ta lun tm c p~ P sao cho

|Q( p~ ) q | <

th phng n p~ c gi l cn ti u.

nh ngha 1.4: Mt phng n iu khin p~ thuc tp cc phng n iu khin thch

hp P s c gi l li gii cn ti u ca bi ton iu khin ti u (1.1) nu:

|Q( p~ ) )(inf pQp

| <

trong l hng s dng nh ty nhng cho trc.

Mc d y nghim p~ P khng phi l ti u thc s theo ng ngha ca n,

nhng li l mt phng n iu khin chp nhn c trong min sai s cho php, tha mn cc yu cu ca bi ton, nn n vn c xem nh l li gii gn ng ca bi ton ti u (1.1).

Bi ton ti u c rng buc/khng rng buc

Cc bi ton ti u dng (1.10) khng bt buc phi c iu kin rng buc p P.

Ni cch khc khng bt buc phi c tp cc tham s iu khin thch hp P. Nhng

bi ton khng c iu kin rng buc cho p , hay P=Rn , c gi l bi ton khng b

rng buc (unconstrained).

nh ngha 1.5: Bi ton (1.10) c P=Rn c gi l bi ton iu khin ti u tnh

khng b rng buc (unconstrained). Ngc li nu PRn (tp con thc trong Rn ) th n c gi l bi ton ti u b rng buc (constrained).

Nghim ti u a phng/ton cc

Nghim ca bi ton (1.1) c gi l nghim ton cc (global). Tn gi ny hm

ch rng ti *p hm Q( p ) c gi tr nh nht v iu ng trong ton b tp P. Tuy

nhin trong kh nhiu phng php gii bi ton ti u (1.1) ngi ta thng phi i vng qua nghim ti u a phng (local). l nhng nghim ch tha mn:

*p = argUp

min )( pQ

vi UP l mt ln cn nh ca *p (hnh 1.4).

19

nh ngha 1.6: Xt bi ton ti u (1.1). Nu c mt im 0

p P tha mn

)(0

pQ )( pQ

vi mi p thuc mt ln cn UP no ca 0

p th 0

p c gi l nghim a

phng (local) ca bi ton.

1.1.3 Cng c ton hc: Tp li v hm li

nh ngha 1.7: Tp hp L c gi l li, nu on thng ni hai im

x1 , x2 bt k ca L, c k hiu l

[x1 ,x2 ] , cng s nm hon ton

trong L . Ni cch khc nu c

x1 ,x2L th cng c:

ax1+bx2L

vi a , b l hai s thc dng tha mn a+b=1.

Theo nh ngha trn th tp rng cng l mt tp li. Hnh 1.5 minh ha cho nh ngha 1.7. nghin cu v tp li, bn cnh nh ngha 1.6, ngi ta cn thng s dng cc khi nim sau:

1) Tp t hp tuyn tnh li: Cho n phn t x1 , x2 , , xn ca mt khng gian vector

X. Tp t hp tuyn tnh li M ca n c hiu l:

M = { y = =

n

iii xa

1 aiR+ , i=1,2, ,n v

=

n

iia

1=1 }

Nh vy, khc vi khi nim bao tuyn tnh, y cn c thm iu kin =

n

iia

1=1

trong ai ch l nhng s thc dng (hnh 1.6a).

2) Bao li: Cho tp hp M (cha cn phi l li). Bao li L ca M l tp li nh nht cha M (hnh 1.6b).

Tp khng liTp li

x1x2

x1

x2

Hnh 1.5: Minh ha nh ngha 1.7.

Q( p )

pHnh 1.4: Nghim ti u a phng/ton cc.

Nghim a phngNghim ton cc

20

3) im bin xb ca tp li M: L im m khng biu din c di dng:

xb = ax1+bx2 , trong x1 , x2L , x1 x2 v a ,b>0 vi a+b=1.

4) Ri tuyn tnh: Cho hai tp hp con M1 v M2 (cha cn phi l li) ca khng gian

vector X. Chng c gi l ri tuyn tnh vi nhau, nu nh tn ti mt phng (a tp tuyn tnh):

D = { cTx = k cRn , kR v xiX , i=1,2, ,n }

tha mn (hnh 1.6c v 1.6d):

xM1 cTxk v xM2 c

Txk .

V tp li ta c nhng pht biu sau:

tp con M ca khng gian Rn l tp li, th cn v l mi tp li tuyn tnh,

to bi cc phn t ca M li thuc M. Ni cch khc nu c x1 , x2 , , xmM

th cng c =

m

iii xa

1M vi mi aiR+ v

=

m

iia

1=1.

Tng M=M1+M2={ x1+x2 x1M1 , x2M2 } ca hai tp li l mt tp li.

Tch M=M1M2={ x=

2

1x

x x1M1 , x2M2 } ca hai tp li l mt tp li.

Tp giao M=k

kM ca (v s) cc tp li Mk l mt tp li.

Tp L gm tt c cc t hp li tuyn tnh ca cc phn t x1 , x2 , , xn ca tp

M (cha cn li) s l bao li ca M.

Phn t xbM l im bin ca tp li M khi v ch khi M\{xb } cng l tp li.

y

x1

x3 x2

a) b)

M M1 M2 M1 M2

Dc) d)

Hnh 1.6: Mt s khi nim c s dng nghin cu tp li.

Ri tuyn tnh Khng ri tuyn tnh Bao liTp t hp tuyn tnh li

21

Nu MRn l tp li ng, khng rng, vi 0M th s tn ti t nht mt mt

phng cTx=k (k>0) t xM suy ra c cTx>k.

Nu MRn l tp li ng, khng rng, vi 0M th s tn ti t nht mt mt

phng cTx=0 t xM suy ra c cTx0 .

Nu M1 v M2 l hai tp li tha mn M1M2= , th chng s ri tuyn tnh vi

nhau, tc l s tn ti t nht mt mt phng cTx=k nu xM1 th cTxk v

nu xM2 th cTxk .

Nu M1 v M2 l hai tp li ng tha mn M1M2= , th chng s ri tuyn

tnh vi nhau theo ngha cht, tc l s tn ti t nht mt mt phng cTx=k

nu xM1 th cTxk .

nh ngha 1.8: Xt hm y= f (x ) xc nh trn tp li M (xM). Khi n s c gi l:

a) hm li nu:

f (ax1+bx2 ) af (x1 )+bf (x2 )

vi mi x1 ,x2M , a+b=1 v a ,b>0.

c) hm li cht nu:

f (ax1+bx2 ) < af (x1 )+bf (x2 ) , x1 ,x2M , x1x2 , a+b=1 v a ,b>0

d) hm lm (hay lm cht) nu f (x ) l hm li (hay li cht).

Hnh 1.7 minh ha nh ngha 1.8 thng qua th hm s. hm li, on ni hai im bt k ca th hm s khng c nm di ng th hm s. Vi hm li cht th ng ni phi lun nm pha trn ng th hm s.

Nu nh ngha tp epi ca hm s y= f (x ) l tp tt c cc im

y

x~ c y

~ f (x ) ,

tc l (hnh 1.8):

y y

Hm li Hm khng li

x

M

x

M

a) b)

Hnh 1.7: a) Hm li b) Hm khng li

yepi( f )

x

Hnh 1.8: Minh ha tp epi ca hm

22

epi( f )={

y

x~ y

~ f (x ) }

th r rng hm y= f (x ) li khi v ch khi tp epi( f ) ca n l tp li.

V hm li ta c nhng pht biu sau:

1) Nu f i (x ) , i=1,2, ,n l cc hm li th hm f (x )= =

n

iii xfa

1)( vi ai0, i=1,2, ,n

cng l hm li.

2) Nu f i (x ) , i=1,2, ,n l cc hm li v ng vi mi im x chng u b chn

trn, th f (x )=ni1

max f i (x ) cng l hm li.

3) Cho hm y= f (x ) xc nh v kh vi trn tp li (m) M. f (x ) l hm li th cn v l

( x 2 x 1 )T gradf ( x 1 ) f (x2 ) f (x1 ) , vi mi x1 , x2M ,

trong

gradf ( x ) = T

nxxf

xxf

)(, ,

)(

1

4) Cho hm y= f (x ) xc nh trn tp li M v kh vi hai ln ti . Ma trn Hesse c nh ngha l:

H (x )=(hi j )=(ji xx

xf

)(2) phn t hng i ct j l

ji xxxf

)(2

.

Khi th:

a) f (x ) s l hm li khi v ch khi H (x ) xc nh bn dng (positiv semidefinit) vi mi xM , tc l khi v ch khi H (x ) l ma trn i xng v tt c cc gi tr ring ca n c phn thc khng m vi mi xM .

b) f (x ) s l hm li cht khi v ch khi H (x ) xc nh dng (positiv definit) vi mi xM , tc l khi v ch khi H (x ) l ma trn i xng v tt c cc gi tr ring ca n c phn thc dng vi mi xM .

5) Cho m hm li f i (x ) , i=1,2, ,m cng xc nh trn tp li M . Vy th:

a) Hoc h bt phng trnh f i (x )

23

1.2 Nhng bi ton ti u in hnh

1.2.1 Bi ton ti u li

nh ngha 1.9: Nu bi ton ti u (1.1) c )( pQ l hm li v tp gii hn:

P={ p Rn )( pgi 0 , i=1,2, ,m }

c )( pgi , i=1, ,m cng l cc hm li, th n c gi l bi ton ti u li.

Nhiu bi ton ti u tnh, ch cn mt vi bin i nh, l ta c th chuyn c n v dng bi ton ti u li. Ngoi ra, do hm tuyn tnh cng l hm li, nn bi ton ti u li nh ngha nh trn bao gm lun c lp bi ton (1.1) c tp tham s iu khin P dng:

P={ p Rn )( pgi 0 , )( ph j =bj , i=1,2, ,m ; j=1,2, ,q }

vi )( pgi , i=1,2, ,m l nhng hm li, v )( ph j , j=1,2, ,q l nhng hm

tuyn tnh, tc l )( ph j = paTj , trong

Tja =(aj , 1 , , a j , n ).

nh l 1.1: Bi ton ti u li (1.1) c nhng tnh cht c bn sau:

a) Tp P cc tham s iu khin thch hp l mt tp li.

b) Mi nghim ti u a phng ca bi ton ti u li cng s l nghim ton cc, hay bi ton khng c nghim ti u a phng.

c) Nu bi ton c nhiu nghim th tp ca tt c cc nghim ti u *p l mt

tp li.

d) Nu P khng rng v gii ni th bi ton lun c nghim. Nu Q( p ) cn l

hm li cht th bi ton s c nghim duy nht.

Chng minh:

a) Gi Mk={ p Rn )( pgk 0 } . Khi P=

n

kkM

1= nn s nu ta ch ra c Mk

l tp li. Gi s 1

p ,2

p Mk . Vy th )( 1pgk 0 v )( 2pgk 0. T y, cng vi tnh

li ca )( pgk ta suy ra c )( 21 pbpagk + )( 1pagk + )( 2pbgk 0, trong a+b=1,

v a ,b>0. Vy )(21

pbpa + cng thuc Mk , hay Mk l tp li (.p.c.m).

b) Gi *p l nghim ton cc. Gi s 0

p l nghim a phng vi *p 0

p .Vy th

*)( pQ )(0

pQ . Suy ra vi mi cp gi tr a ,b>0 tha mn a+b=1 c:

24

)*(0

pbpaQ + a *)( pQ +b )(0

pQ 0 tha mn a+b=1 ta s c t tnh li ca )( pQ :

)(21

pbpaQ + a )(1

pQ +b )(2

pQ = )(1

pQ =Qmin (1.5)

Ngoi ra, do 1

p l nghim ti u nn cn phi c:

)(21

pbpaQ + )(1

pQ =Qmin (1.6)

Suy ra phi c )(21

pbpaQ + =Qmin. iu ny ch rng 1pa + 2pb cng l nghim ti

u, hay tp ca tt c cc nghim ti u *p l mt tp li (.p.c.m).

d) Do P l tp ng, nn hin nhin khi khng rng v gii ni, bi ton s c nghim.

Gi s rng bi ton c hai nghim 1

p 2

p . Nu )( pQ l hm li cht th bt ng thc

(1.5) s c thay bng

)(21

pbpaQ + k1 . Nu )( pQ kh vi, tc l tn ti

vector gradient, k hiu bi grad )( pQ , th do tnh cht ca vector gradient l lun

vung gc vi ng ng mc v ch chiu tng gi tr ca hm, nn ti im 1

p trn

ng ng mc )( pQ =k1 , hai vector grad )( 1pQ v ( 2p 1p ) phi to vi nhau mt

gc khng ln hn 90 (hnh 1.9). iu ny ch rng:

Nu c )(1

pQ < )(2

pQ th cng phi c (2

p 1

p )Tgrad )(1

pQ 0.

25

T y, ta suy ra c:

nh l 1.2: Nu bi ton ti u (1.1) c )( pQ kh vi trong P th:

a) *p l nghim ti u ca bi ton khi v ch khi:

( p *p )Tgrad *)( pQ 0 vi mi p P (1.7)

b) Nu *p l im trong ca P th iu kin (1.7) s c thay bng:

grad *)( pQ =0

V d 1.6: (Minh ha nh l 1.2)

Xt bi ton ti u argPp

min )( pQ c:

)( pQ = 122

41 4 ppp ++

v P={ p =

2

1

pp

p10, p20, p1+p26, p2 ( p11)2 }

y l bi ton ti u li. N c nghim ton cc *p =

10

. khng nh iu ta s

dng nh l 1.2. V ti *p hm )( pQ c grad *)( pQ =

24

nn:

( p *p )Tgrad *)( pQ = 4p1+2(p21) 221p 0 .

1.2.2 Bi ton ti u ton phng

nh ngha 1.10: Bi ton ti u *p = argPp

min )( pQ c gi l ton phng, nu:

)( pQ = pApT + paT v P={ p =

np

p1 pB b , pi 0 , i=1, ,n }

trong aRn , bRm , BRm n v ARn n l ma trn i xng.

1p

2p

)( pQ =k1

)( pQ =k2>k1

grad )(1

pQ

Hnh 1.9: ng ng mc v vector gradient.

26

Bi ton ti u ton phng nu trn bao gm lun c lp cc bi ton c tp tham s iu khin thch hp dng:

P={ p Rn pB b } (1.8)

hoc P={ p Rn pB =b , pi 0 , i=1, ,n } (1.9)

Tht vy, nu P c dng (1.8) th ta c th thay:

pi=qir i vi qi0, r i0, i=1, ,n

s c c bi ton ton phng cho 2n bin qi , r i , i=1, ,n . Hoc nu P c dng

(1.9) th ta thm n bin mi qi0, i=1, ,n v vit li:

b pB q =

B

IB~

) (

qp

, trong q =

nq

q1.

Hm tuyn tnh l hm li. Do hm Q( p )= paT + pApT c ma trn Hesse:

)( pH =

2

2

1

2

1

2

21

2

nn

n

p

Qpp

Q

ppQ

p

Q

=AT+A=2A

nn n s l hm li khi v ch khi A xc nh bn dng (mc 1.1.3). Suy ra:

nh l 1.3: bi ton ti u ton phng trong nh ngha 1.10 l bi ton li th cn v l ma trn A xc nh bn dng. Nu A cn l ma trn xc nh dng th

hm mc tiu )( pQ s l hm li cht.

T y, v vi nh l 1.1 cng nh 1.2 ca bi ton ti u li, ta c c:

nh l 1.4: Bi ton ti u cho trong nh ngha 1.10, vi A l ma trn xc nh bn dng, tc l bi ton ti u li, ton phng, c nhng tnh cht c bn sau:

a) Mi nghim a phng ca n cng l nghim ton cc.

b) *p l nghim ti u ca bi ton khi v ch khi:

( p *p )T ( *2 pA +a )0 vi mi p P . (1.10)

c) Nu *p l im trong ca P th iu kin (1.10) s c thay bng:

*2 pA +a =0. (1.11)

27

V d 1.7: (Nghim ca bi ton ti u ton phng khng b rng buc)

Cho bi ton ti u khng b rng buc (unconstrained) vi:

)( pQ = = =

n

i

n

jjiij pp

1 1 +

=

n

iii pa

1

y l bi ton ti u ton phng c min tham s iu khin thch hp P l ton b

khng gian Rn . chuyn v dng chnh tc nh trong nh ngha 1.10, ta t:

A~

=(ai j ) vi ai j = 2jiij +

v a=(a1 , a2 , , an )T

s c

)( pQ = p

A

AAp TT )~~

( + + paT = pApT + paT vi p =(p1 , p2 , ,pn )T

Ma trn A l i xng. Gi thit A xc nh dng. Vy th A s khng suy bin.

Theo nh l 1.3, y l bi ton ti u ton phng, li, c nghim *p bn trong P. Do

nghim ny c tnh theo (1.11) nh sau:

*2 pA +a =0. *p =21 A 1a .

1.2.3 Bi ton ti u hyperbol

nh ngha 1.11: Bi ton ti u (1.1) c gi l hyperbol, nu )( pQ c dng:

)( pQ =)(

)(

pv

pu, )( pv >0 vi mi p P (1.12)

v

P={ p Rn )( pgi 0 , i=1,2, ,m }

Dng bi ton ti u hyperbol l kh ph bin trong thc t, nht l trong cc bi ton chn tham s iu khin ti u cho b iu khin (xem li v d 1.2 v chn tham s b iu khin PI). Hn na, kt hp hai mc tiu ti u vi nhau, chng hn kt

hp mc tiu chi ph nh nht )( pu min, vi li nhun cao nht )( pv max, ngi ta

vn thng lp hm mc tiu chung )( pQ dng (1.12).

Do trong nh ngha 1.11 cha c s rng buc rng hm mc tiu )( pQ phi l

hm li, nn ni chung bi ton ti u hyperbol cha phi l bi ton ti u li. c

28

th hn na, ngi ta chia bi ton ti u hyperbol thnh cc lp bi ton con nh sau:

1) Bi ton ti u hyperbol lilm, nu )( pu >0 l hm li, )( pv >0 l hm lm vi

mi p P .

2) Bi ton ti u hyperbolton phng, nu:

)( pu = pApT + paT +a0 v )( pv = pBpT + pbT +b0

trong )( pu >0, )( pv >0 vi mi p P , a ,bRn , A ,BRn n v A l ma trn xc

nh dng.

3) Bi ton hyperboltuyn tnh, nu:

)( pu = paT +a0 , )( pv = pbT +b0

P={ p Rn pC d }

trong a ,bRn , dRm , A ,BRn n v CRm n .

nh l 1.5: Mi bi ton ti u hyperboltuyn tnh u chuyn c v dng bi ton ti u tuyn tnh (x , t )*=arg

Xxmin f (x , t ) vi

x= pt , f (x , t )= aTx + a0 t

X={

t

xRn + 1 Cxdt 0 , t > 0 v bTx+b0 t =1 }

Chng minh:

t tx

= p vi t>0 th bi ton hyperpoltuyn tnh c vit li thnh:

min ),(~

txQ =min{tbxb

taxaT

T

0

0

+

+

t

x P~ }

P~

={

t

xRn + 1 Cxdt 0 , t > 0 }

Nu

**

t

x l nghim bi ton hyperpoltuyn tnh th

**

t

x

vi >0 cng l nghim

ca n, nn cng vi )( pv >0 ta suy ra c iu phi chng minh.

29

1.3 Tm nghim bng phng php l thuyt

S m bo cho bi ton ti u (1.1) c nghim l nh l sau:

nh l 1.6: (Weierstrass) Nu min P l ng, gii ni v hm mc tiu )( pQ l lin tc

trn P, tc l thuc khng gian C [P ] , th )( pQ lun c gi tr cc i v gi tr cc

tiu trong P.

1.3.1 Mi quan h gia bi ton ti u v bi ton im yn nga

Xt bi ton ti u:

{ }

* arg min ( )

( ) 0 vi 1,2, ,

p P

ni

p Q p

P p g p i m

=

= =

R (1.13)

Ta lp hm:

),( qpf = )( pQ + )( pgqT vi )( pg =

)(

)(1

pg

pg

m

, q =

mq

q1 v qi0 (1.14)

nh ngha 1.12: im *)*,( qp c

gi l im yn nga (hnh 1.10) ca hm (1.14) nu *p l im

cc tiu ca *),( qpf v *q l

im cc i ca )*,( qpf , tc l:

)*,( qpf *)*,( qpf *),( qpf

nh l 1.7: Xt bi ton ti u (1.13) v hm (1.14).

a) (iu kin ): Nu *)*,( qp l

im yn nga ca hm ),( qpf tnh theo (1.14) th *p s l nghim ti u

ca bi ton cho.

b) (iu kin cn v ): Gi s )( pQ , )( pgi , i=1,2, ,m l cc hm li, tc l

bi ton cho l bi ton ti u li, v tp P c t nht mt im trong, tc

l tn ti t nht mt im p P c )( pgi

30

*)*,( qp l im yn nga ca hm ),( qpf khi v ch khi *p l nghim ti

u ca bi ton ti u li (1.13).

Chng minh:

a) Trc ht, do *)*,( qp l im yn nga ca ),( qpf nn theo nh ngha 1.12 c:

*)( pgqT *)(*)( pgq T vi mi q 0

Thay q = *q +e i , trong e i l vector n v th i ca Rm , tc l vector c phn t th

i bng 1, cc phn t khc bng 0, s c *)( pgi 0 . Vy *p l phn t thuc P.

Tip tc, cng do *)*,( qp l im yn nga ca ),( qpf v v vector *q c cc phn

t khng m, k hiu l *q 0 cng nh )( pg 0 nn cng theo nh ngha 1.12 li c:

)( pQ )( pQ + )(*)( pgq T = *),( qpf *)*,( qpf )*,( qpf = *)( pQ + *)( pgqT

ng vi mi q 0. Bi vy, vi q =0 ta c c )( pQ *)( pQ (.p.c.m).

b) iu kin c chng minh cu a) vi iu kin yu hn. Ta ch cn chng

minh iu kin cn. Do *p l nghim ca bi ton ti u li nn )( pQ *)( pQ cng l

mt hm li. Xt h gm m+1 cc bt phng trnh li:

)( pgi

31

Ngoi ra, hin nhin vi mi q 0 (c hiu l cc phn t ca vector q l khng

m) cn c:

*)( pQ *)( pQ + *)( pgqT = )*,( qpf

Suy ra:

)*,( apf *)( pQ ),( qpf

Thay q =a v p = *p ta c *)( pQ = )*,( apf v iu ny dn n:

)*,( apf )*,( apf ),( apf

Vy )*,( ap chnh l im yn nga ca (1.14) (.p.c.m).

1.3.2 Phng php KuhnTucker

nh l 1.7 cho thy vic xc nh nghim *p ca bi ton ti u (1.13) c th c

thay bng vic xc nh im yn nga *)*,( qp ca hm (1.14). u im chnh ca cch

gii ny l ta chuyn hon ton bi ton ti u thnh bi ton gii h cc bt phng trnh v phng trnh.

Cng c h tr vic xc nh im yn nga *)*,( qp ca hm (1.14) l nh l

KuhnTucker pht biu di y. Thc cht, ni dung nh l KuhnTucker mang nhiu nt ca mt thut ton hn l mt nh l ton hc. Chnh v l m ta gi n l phng php KuhnTucker.

nh l 1.8: Cho bi ton ti u (1.13) c )( pQ , )( pgi , i=1,2, ,m l cc hm kh vi.

Lp hm ),( qpf theo (1.14). Khi :

a) (iu kin cn): Nu *)*,( qp l im yn nga ca ),( qpf th n s tha mn h

cc bt phng trnh v phng trnh:

*)*,(grad qpfp =0 (1.16)

*)*,(grad qpfq 0 (1.17)

( *q )T *)*,(grad qpfq =0 (1.18)

trong ),(grad qpfp =T

npf

pf

, ,1

32

v ),(grad qpfq =T

mqf

qf

, ,1

b) (iu kin cn v ): Nu )( pQ , )( pgi , i=1,2, ,m cn l cc hm li (bi

ton ti u li) v tp P c t nht mt im trong, th pht biu a) s l iu kin cn v .

Chng minh:

a) V *)*,( qp l im yn nga ca ),( qpf nn theo nh ngha 1.12, *p l im cc

tiu ca *),( qpf = )(~

pf . Do bi ton min )(~

pf khng b rng buc (unconstained) nn

ti *p c grad *)(~

pf =0 v iu ny tng ng vi (1.16). Mt khc, cng v *)*,( qp l

im yn nga ca ),( qpf nn *q l im cc i ca )*,( qpf = )(qf , tc l im cc

tiu ca )(qf . Do )(qf tuyn tnh theo q nn bi ton *q = argUq

min ))(( qf l bi

ton ti u li b rng buc U={ q Rm q 0 } . Theo nh l 1.2 th c hai kh nng

xy ra: Nu *q l im trong ca min U th grad *)(qf =0 v iu ny tng ng

vi (1.17) v (1.18). Ngc li, nu *q l im bin ca min U, tc l *q =0, th hin

nhin c (1.18), cn khng nh (1.17) c suy ra t nh l 1.2 m c th l bt ng thc (1.7):

( q *q )Tgrad *)(qf 0, q *q =0 grad *)(qf 0

b) iu kin cn c chng minh cu a), ta ch cn phi chng minh iu kin .

Do )( pQ , )( pgi , i=1,2, ,m l cc hm li nn ),( qpf cng l hm li theo p . Theo

tnh cht hm li (mc 1.1.3) ta c:

*),( qpf *)*,( qpf ( p *p )T *)*,(grad qpfp =0

*),( qpf *)*,( qpf vi mi p

Mt khc, v ),( qpf tuyn tnh theo q , nn n cng l hm li theo q . Vy:

)*,( qpf *)*,( qpf ( q *q )T *)*,(grad qpfq

= Tq *)*,(grad qpfq Tq*)( *)*,(grad qpfq 0

)*,( qpf *)*,( qpf vi mi *q 0

iu ny chng t *)*,( qp chnh l im yn nga ca ),( qpf .

33

V d 1.8: (Minh ha ng dng phng php KuhnTucker)

Xc nh khong cch t gc ta ti tp li G (hnh 1.11):

G={ p =

2

1

p

pR2 p1+p24 v 2p1+p25 }.

Bi ton trn c vit li thnh:

*p = argPp

min

)pQ(

pp )( 2221 +

P={ p R2 4p1p20, 52p1p20}

R rng y l bi ton ti li v tp P c cha t nht

mt im trong, chng hn l im p =

24

. Lp

hm:

),( qpf = )( pQ + )( pgqT = )25()4( 21221122

21 ppqppqpp +++

trong q10, q20. Khi , theo nh l 1.8, *p s c tm thng qua vic gii h

phng trnh v bt phng trnh:

fpgrad =0

==02

022

212

211

qqp

qqp (1.19)

fqgrad 0

025

04

21

21

pp

pp (1.20)

Tq fqgrad =0 q1 (4p1p2 )+q2 (52p1p2 )=0 (1.21)

gii h cc phng trnh v bt phng trnh trn cho 4 n s p1 , p2 , q1 , q2 ,

trc ht ta thy do c (1.20) v q10, q20 nn (1.21) thay c bng:

==

0)25(

0)4(

212

211

ppq

ppq (1.22)

Trng hp 1: Nu q1=q2=0 th p1=p1=0 v iu ny mu thun vi (1.20) nn b loi.

Trng hp 2: Nu q1=0, q2>0 th t (1.19) v (1.21) ta suy ra c p1=q2=2,

q2=1. Song kt qu ny li mu thun vi (1.20) nn cng b loi.

Trng hp 3: Nu q1>0, q2=0 th t (1.22) c p2=4p1 v t (1.19) c

2p1q1=0 cng nh 2p1q1=8. Vy p1=p2=2, q1=4, q2=0. Kt qu ny

p2

p1

2

2 4

G


34

tha mn (1.20) nn l nghim ca h tt c cc phng trnh v bt phng trnh

(1.19)(1.22). Ni cch khc, bi ton ti u c mt nghim l *p =

22

.

Ta khng cn phi xt tip trng hp cui q1>0, q2>0 v bi ton ti u li ny

c )( pQ l hm li cht nn theo nh l 1.1, n ch c th c nghim duy nht.

1.3.3 Phng php Lagrange

Cho bi ton ti u:

{ }

* arg min ( )

( ) 0 vi 1,2, ,

p P

ni

p Q p

P p h p i m

=

= = =

R (1.23)

trong cc hm )( pQ , )( phi =0, i=1,2, ,m c gi thit l kh vi (trong mt

min h), tc l thuc C1 [P ] .

Nu P khng rng v gii ni th theo nh l 1.6, bi ton (1.23) chc chn c nghim.

Tng t nh phng php KuhnTucker, ta lp hm:

),( apf = )( pQ + =

m

iii pha

1)( = )( pQ + )( phaT (1.24)

nhng vi a1 , a2 , , am l m s thc ty (khng bt buc phi khng m). K hiu

a v )( ph trong (1.24) c hiu l:

a=

ma

a1, )( ph =

)(

)(1

ph

ph

m

Gia hm ),( apf v bi ton (1.23) c mi quan h nh sau:

nh l 1.9: Gi s bi ton (1.23) c nghim ti u *p . Khi

a) Hoc nghim *p tha mn:

)*,(grad apfp =0 ppQ

*)(

+aTp

ph

*)(

=0T (1.25)

nu phng trnh c nghim aT=(a1 , a2 , , am ) .

35

b) Hoc *p tha mn:

aTp

ph

*)(

=0T (1.26)

vi v s cc gi tr aT .

K hiu p

s dng trong hai cng thc (1.25) v (1.26) l ch php tnh xc

nh ma trn Jacobi ca hm hoc vector hm nhiu bin:

pQ

=(1p

Q

, ,np

Q

) v ph

=

n

mm

n

ph

ph

ph

ph

1

1

1

1

D thy, h phng trnh tuyn tnh (1.25) c nghim aT th phi c (xem thm

phn ph lc v nh x v nh x tuyn tnh, mc 6.1.1):

Rank

p

ph *)(= Rank

p

pQp

ph

*)(

*)(

cng nh (1.26) c v s nghim aT th ma trn

p

ph *)( phi c hng nh hn m.

nh l 1.9 ch l iu kin cn. Tuy nhin, n vn thng c s dng nh mt

thut ton xc nh nghim *p qua cc bc nh sau:

Tm tt c cc im 1

p , ,l

p lm cho ma trn p

ph

)(

b st hng.

Tm tt c cc im 1+l

p , ,k

p l nghim ca (1.25) v )( ph =0.

Xc nh *p = argki1

min )(i

pQ .

V d 1.9: (Minh ha ng dng phng php Lagrange)

Xc nh khong cch ngn nht t im p =

52

ti ng parabol p2=214

1p .

Chuyn v dng bi ton ti u chun c:

36

*p = argPp

min

)(

])5()2[( 222

1

pQ

pp +

trong

P={ p R2)(

4 221

ph

pp =0 }

Trc ht ta lp hm:

),( apf =(p12)2+(p25)

2+a ( 221 4 pp )

sau thc hin ln lt cc bc:

Xc nh ma trnp

ph

)(

= (2p1 , 4). Ma trn ny lun c hng bng m=1 vi mi

p , tc l khng b st hng vi mi p .

Tnh

=

=

0)(

0),(grad

ph

apfp

=

==+

04

04)5(2

02)2(2

221

2

11

pp

ap

app

1

p =

44

v 2

p =

12

.

So snh )(1

pQ =5< )(2

pQ =32 ta rt ra c *p =1

p =

44

.

1.4 Tm nghim bng phng php s

1.4.1 Bi ton ti u tuyn tnh v phng php n hnh (simplex)

Bi ton ti u (1.1) c gi l tuyn tnh chun (NOPnormal linear optimal problem) nu:

{ }1 1 2 2( )

v 0

Tn n

n

Q p q p q p q p q p

P p Ap b p

= = + + +

= =

R (1.27)

C th mt s bi ton tuyn tnh ban u cha c dng chun (1.27). Song ta lun chuyn chng v c dng (1.27). Chng hn nh t:

P={ p Rn pA b v p 0 }

hay

ak 1p1 + ak 2p2 + + ak npn bk

th bng cch thm vo cc bin mi pn + k 0 ta s c:

p2

p1

54

4 2


37

ak 1p1 + ak 2p2 + + ak npn + pn + k = bk

Cng nh vy nu nh cha c iu kin p 0 th ta s th bin p = pp ~ vi p~ 0 v

p 0 li c dng (1.27).

Tuy rng vic trnh by cc phng php gii bi ton ti u tuyn tnh khng phi l nhim v chnh ca quyn sch ny, song p dng cho bi ton ti u phi tuyn, ta cng nn bit n phng php n hnh (simplex) ca Danzig. N bao gm cc bc:

1) Gi s Rank(A )=r . Chn tt c r vector ct c lp tuyn tnh ca A. Khng mt

tnh tng qut, nu ta cho rng l r ct u tin. Biu din cc bin pr+1 , pr+2 , , pn qua cc bin p1 , p2 , , pr sau thay vo Q v P s c

)( pQ = d1p1+d2p2 + + drpr

P={ p Rn

+

n

r

p

p 1=

rrnrn

r

ee

ee

,1,

,11,1

rp

p1+

rnf

f1 v p 0 }

Nu biu din di dng bng s c:

p1 p2 pr T

pr+1 e1,1 e1,2 e1,r f1

pr+2 e2,1 e2,2 e2,r f2

pn en-r,1 en-r,2 en-r,r fn-r

Q d1 d2 dr 0

2) i ch mt hng v mt ct cho nhau hng cui cng gm ton s khng m, khng i hng Q v ct T. V d mun i ch ct k v hng l ca:

pl = + apk + + bpm +

ps = + cpk + + dpm +

cho nhau th vi bin ph =a1

(a 0) s c:

pk = + pl bpm +

ps = + cpl + + (dcb )pm +

Biu din kt qu trn di dng bng th c:

38

Trc khi i hng/ct

pk pm

pl a b = 1/a

c d

Sau khi i hng/ct

pl pm

pk b

c dcb

3) Vi n phn t pi v Rank(A )=r th nhiu nht ch c rnC php i cthng. Gi s

sau mt ln i hngct, bng ma trn c dng:

y1 y2 yr T

yr+1 h1,1 h1,2 h1,r Tr+1

y r+2 h2,1 h2,2 h2,r Tr+2

yn hn-r,1 hn-r,2 hn-r,r Tn

Q m1 m2 mr M

trong T1 , T2 , , Tn - r l nhng gi tr khng m.

Cui cng, da theo nh l () v nguyn tc n hnh ta c cc kt lun sau:

a) Nu mk 0 , k = 1 , 2 , , r im ti u s l im c cc phn t yk=0 vi

k=1,2, ,r v ym = Tm , m=r+1,r+2, ,n . Gi tr hm mc tiu ti

l M.

b) Nu c mt phn t mk

39

V d 1.10: (Chuyn bi ton tuyn tnh thng thng v dng chun)

Cho bi ton ti u tuyn tnh *p = argPp

min )( pQ vi:

)( pQ = 6p19p3

P={ p R4 p1+p28, p3+p48, 2p1p2+3p3p4=0

v p10, p20, p30, p40 }

a c v dng chun (1.27) ta nh ngha thm cc bin ph mi p50, p60 v bin i P thnh:

p1+p2+p5=8 v p3+p4+p6=8

Khi bi ton s tr thnh:

)( pQ = 6p19p3

P={ p R6 p1+p2+p5=8, p3+p4+p6=8, 2p1p2+3p3p4=0 v p 0}

V d 1.11: (Minh ha phng php n hnh)

Cho bi ton NOP (1.27) vi:

)( pQ = 6p19p3 , P={ p R6

A

001312101100010011

p =

b

088

v p 0 }

Ma trn A c 3 ct u c lp tuyn tnh. Biu din p4 , p5 , p6 v )( pQ theo p1 ,

p2 , p3 . Sau lp bng ma trn v sau 2 ln i hngct s c:

p1 p2 p3 T p1 p2 p6 T

p5 1 1 0 8 p5 1 1 0 8 1

p6 2 1 4 8 1/4 p3 1/2 1/4 1/4 2

p4 2 1 3 0 p4 1/2 1/4 3/4 6

Q 6 0 9 0 M 3/2 9/4 9/4 18

p1 p5 p6 T

p2 1 1 0 8

p3 3/4 1/4 1/4 4

p4 3/4 1/4 3/4 4

M 3/4 9/4 9/4 36

40

Cui cng ta c kt lun sau v nghim ti u:

p1=p5=p6=0; p2=8 ; p3=p4=4 v Qm i n=36.

Ch : Bi ton v d 1.11 chnh l dng chun ca bi ton ti u tuyn tnh cho trong v d 1.10. T nghim ca bi ton dng chun ca n v d 1.11 ta cng s c c nghim ca bi ton trong v d 1.10 bng cch b i tt c nhng bin ph thm vo. C th, nghim ca bi ton ti u v d 1.10 s l:

p1=0; p2=8 ; p3=4 v p4=4.

1.4.2 Phng php tuyn tnh ha tng on

p dng c nhng phng php gii bi ton ti u tuyn tnh, chng hn nh phng php n hnh, cho cc bi ton ti u phi tuyn, th trc ht ta cn phi xp x bi ton phi tuyn thnh tuyn tnh. Cng vic tuyn tnh ha bi ton phi tuyn

*p = argPp

min )( pQ c thc hin hai cng on:

tuyn tnh ha iu kin rng buc P,

v tuyn tnh ha hm mc tiu )( pQ .

Hnh 1.13a) l v d minh ha. , bi ton ti u phi tuyn *p = argPp

min )( pQ

ban u c tuyn tnh ha (tng on) thnh ba bi ton ti u tuyn tnh con l

*p = argkPp

min )( pQk , k=1,2,3, trong Pk phi l cc min nh thuc P sao cho

trong hm phi tuyn )( pQ c th xp x c bng cc hm tuyn tnh:

)( pQk = )( kpQ +( kpp )Tgrad )(

kpQ (1.28)

v pk l mt im ty thuc Pk . Cc min Pk khng c php giao nhau i mt v

phi ph kn P. Ngoi ra, chng cng nh th vic xp x )( pQ tng on theo (1.28) s

cng chnh xc.

Thng thng, ngi ta hay tuyn tnh ha iu kin rng buc P thnh cc iu

kin rng buc tuyn tnh Pk di dng (hnh 1.13b):

Pk ={ p Rnai k pk b i k vi i=1,2, ,n }

tc l chia nh min P thnh cc min (siu din) Pk bng cc "tm li" c cnh song

song vi cc trc ta .

41

Sau khi tuyn tnh ha bi ton ti u *p = argPp

min )( pQ ban u thnh m bi

ton ti u tuyn tnh con *p = argkPp

min )( pQk , k=1,2, ,m , th vic gii bi ton ti

u phi tuyn s c thay bng vic gii m bi ton ti u tuyn tnh vi cc bc:

1) Tm nghim bi ton ti u tuyn tnh kPp

min )( pQk . Gi nghim l kp~ .

2) Xc nh *p =kmk

p~min1

.

1.4.3 Phng php NewtonRaphson

Phng php NewtonRaphson tin hnh vic tm nghim *p = argPp

min )( pQ theo

nguyn l lp (iterative) qua nhiu bc tnh. Bt u l im khi pht 0

p P , n tm

mt im1

p P , sao cho c c )(1

pQ < )(0

pQ . Nu sai s | )(1

pQ )(0

pQ | vn cn

qu ln th n thc hin li bc tnh trn nhng t im xut pht mi l 1

p P c

2p P sao cho )(

2pQ < )(

1pQ . Nu sai s | )(

2pQ )(

1pQ | vn ln th li tm

3p P t

2p P . C nh vy, qua nhiu bc tnh, phng php s a ra c mt dy cc

gi tr {k

p } v chc chn dy gi tr s tim cn ti nghim ti u *p cn tm, tc l

chc chn c k

limk

p = *p nu nh mi nghim a phng ca bi ton cng l

nghim ton cc, chng hn nh bi ton ti u li (nh l 1.1 v 1.4).

Trong nhiu ti liu, phng php ny cn c tn l gi Newton (Quasi Newton).

p1

p2

P

P1

Q (p )

p

Hnh 1.13: Minh ha phng php tuyn tnh ha.

P2 P3

P

Q1

Q2

Q3

a) b)

42

Xt bi ton ti u khng b rng buc (uncontrained) *p =arg min )( pQ c )( pQ

kh vi v li. Theo nh l 1.2, ti *p c grad *)( pQ =0, tc l c:

*)( pfi =ip

pQ

*)(

=0, i=1,2, ,n

Gi s )( pfi l hm gii tch, vy th ti mt im kp thuc ln cn *p , n phn

tch c thnh chui Taylor:

0= *)( pfi = )( ki pf +( *p kp )Tgrad )(

kipf + (1.29)

Nu b qua tt c thnh phn bc cao v phi, th tt nhin (1.29) ch cn l cng thc xp x:

0 )(ki

pf +( *p k

p )Tgrad )(ki

pf

song vn c c quan h ng thc, ta c th thay *p bng im xp x 1+k

p . Khi

s c:

0 = )(ki

pf +(1+k

p k

p )Tgrad )(ki

pf vi i=1,2, ,n

0= grad )(k

pQ +Hk ( 1+kp kp )

trong Hk l k hiu ch ma trn Hesse ca hm mc tiu )( pQ ti im kp , tc l:

Hk =

2

2

1

2

1

2

21

2

)()(

)()(

n

k

n

k

n

kk

p

pQ

pp

pQ

pp

pQ

p

pQ

(1.30)

T y suy ra:

1+k

p =k

p 1kH grad )( kpQ (1.31)

v chnh l cng thc xc nh xp x 1+k

p *p t k

p . Da vo (1.31) ta c c cc

bc tm *p nh sau:

1) Chn mt im xut pht 0

p v mt s dng e nh ty .

2) Thc hin ln lt cc bc sau vi k=0,1,

43

a) Tnh 1+k

p t k

p theo (1.31).

b) Nu | )(1+k

pQ )(k

pQ | e th gn k:=k+1 ri quay li bc a). Ngc li th

chuyn sang bc 3).

3) Dng vi p s: *p 1+k

p .

iu kin p dng c phng php NewtonRaphson l hm )( pQ phi kh

vi hai ln, v khi ta mi c ma trn Hesse ca n.

Ngoi ra, phng php cn c nhng tnh cht sau:

S cho ra nghim ton cc, nu )( pQ l hm li.

Phng php s c tc hi t tt nu ma trn Hk l xc nh dng, v:

gradT )(k

pQ (1+k

p k

p ) = gradT )(k

pQ 1kH grad )( kpQ

44

Ly im khi pht 0

p =

b

a bt k, tnh vector grad )( pQ v ma trn Hesse H0 ti :

grad )(0

pQ =

)2(8)1(6

b

a, H0=

8006

,

ri thay vo (1.30) s c:

1

p =0

p 10H grad )(

0pQ =

b

a

6008

481

)2(8)1(6

b

a=

21

= *p .

Nh vy, r rng l ch sau mt bc tnh ta n c im ti u *p v iu ny

hon ton khng ph thuc vo im xut pht.

1.5 Tm nghim bng phng php hng n cc tr

1.5.1 Nguyn l chung

V nguyn tc, ging nh phng php NewtonRaphson, cc phng php hng n cc tr l phng php tm *p = arg

Ppmin )( pQ theo nguyn l lp (iterative),

tc l tm ln lt 1+k

p P t k

p P vi k=0,1, sao cho c c )(1+k

pQ < )(k

pQ ,

cho ti khi t c sai s cho php | )(1+k

pQ )(k

pQ | < e . im khc ca chng so vi

NewtonRaphson l 1+k

p c tm t k

p vi mt hng tm hk c chn trc sao

cho i dc trn n bng khong cch bc tm sk ta lun tm c im:

1+k

p =k

p + sk kh (1.32)

tha mn )(1+k

pQ < )(k

pQ , tc l ti 1+k

p ng ng mc ca hm )( pQ c gi tr

nh hn l ti k

p (hnh 1.14a).

tm 1+k

p t im k

p dc theo hng tm hk nh cng thc (1.32), ta cn phi

cn n khong cch bc tm sk . N c th l mt hng s cho trc, song cng c th

c chn ti u theo ngha:

sk=args

min )( kk hspQ + = arg smin f (s ) (1.33)

45

vi iu kin rng buc kk hsp + P (hnh 1.14b). Ch rng khi cho trc sk ngi ta

s khng cn phi gii bi ton ti u con (1.33), song vic chn trc khong cch bc

tm ny nh hng kh nhiu n tc hi t ca thut ton. Nu im k

p cn cch

kh xa im ti u *p th tc hi t s cng tt khi sk c chn cng ln, nhng

nu k

p n gn *p th sk cng nh, nghim tm c s cng chnh xc.

Cc phng php hng n cc tr c dng chung nh sau:

1) Chn im khi pht0

p v mt s dng e nh ty .

2) Thc hin ln lt cc bc sau vi k=0,1,

a) Chn hng tm hk .

b) Chn khong cch bc tm sk . C hai cch chn sk : (i) hoc l hng s cho

trc, (ii) hoc l nghim ca bi ton ti u hm mt bin (1.33).

c) Tnh 1+k

p =k

p + skhk .

d) Nu | )(1+k

pQ )(k

pQ | e th gn k:=k+1 ri quay li bc a). Ngc li th

chuyn sang bc 3).

3) Dng vi p s: *p 1+k

p .

im khc nhau c th ca tng phng php hng n cc tr ch nm ch xc

nh hng tm hk (ph thuc kp ). Chng hn nh:

Phng php GaussSeidel c hng tm hk song song vi trc ta ca khng

gian Rn cha iu kin rng buc P.

Phng php gradient c hng tm hk ngc vi hng ca vector gradient ca

hm mc tiu ti k

p l grad )(k

pQ .

2p

1p

*p

0p

s0h0

s1h1

1+kp k

p

)( pQ =k1

)( pQ =k2>k1

0

hk

sk

Hnh 1.14: Minh ha nguyn l chung ca phng php hng n cc tr.

Cc ng ng mc

a) b)

46

Ngoi ra, ta c th thy thm rng phng php rt d cho ra nghim a phng,

v iu ny ph thuc vo vic chn im xut pht 0

p (hnh 1.15). Tuy nhin, nu bi

ton ti u cho l bi ton ti u li th khng ph thuc im xut pht, phng php lun cho ra nghim ton cc (nh l 1.1 v 1.4).

1.5.2 Xc nh bc tm ti u

Thc cht ca vic chn khong cch bc tm ti u l gii bi ton ti u hm

mt bin (1.33) vi iu kin rng buc k

p +shkP . Khng mt tnh tng qut nu ta

cho rng bi ton c dng:

s* = arg10

mins

f (s ) , tc l ch vi 0s1 (1.34)

C th d dng thy c f (s )= )( kk hspQ + l hm li v nghim ti u s* chnh l

im tip xc ca vector hng tm hk vi mt ng ng mc ca )( pQ . C th l:

hm f (s ) gim t s=0 n im cc tiu s* v sau tng vi s*

47

Xc nh bng phng php s

Tip theo, ta s lm quen vi mt phng php s tm kim s* d ci t m khng cn phi c gi thit v tnh kh vi ca f (s ) , tc l khng cn phi tnh o hm ca f (s ) . l phng php thu nh khong nghim.

Ban u, nu ta k hiu s0=0 v s1=1 th khong cha nghim s* s l [s0 ,s1 ] .

thu nh khong cha nghim ta ly hai im s2 v s3 vi s2

48

im s2 c chn sao cho n chia khong nghim [s0 ,s1 ] theo nguyn tc: t l ca on ngn trn on di bng t l on di trn ton khong. Ni cch khc

im s2 phi tha mn (hnh 1.17): 21

02

ssss

=01

21

ssss

im s3 c ly i xng vi

s2 qua tm ca khong nghim

[s0 ,s1 ] . Khi s3 cng s chia

khong nghim theo ng nguyn tc t l ca on ngn trn on di bng t l on di trn ton khong, tc l

cng c: 03

31

ssss

=01

03

ssss

Trong bc u tin c s0=0, s1=1 nn 2

2

1 ss

= 1s2 hay s2= 253 0,382.

iu c bit ca phng thc nht ct vng l hai im s2 ,s3 lun nm i xng

qua tm ca khong [s0 ,s1 ] . Do , k t nhng bc sau ta khng cn phi tnh li hai

im s2 ,s3 m ch cn ly im i xng qua tm ca khong [s0 ,s1 ] vi mt trong hai

im s2 ,s3 l s c im cn li.

Kt hp chung vi thut ton nu, ta c thut ton nht ct vng gm cc bc sau (hnh 1.18):

1) t s0=0, s1=1, s2=0,382 v chn mt hng s dng e nh.

2) Ly im s3 i xng vi s2 qua tm ca khong [s0 ,s1 ] .

s0 s2 s1 s3

s0 s1 s3

Khong nghim mi

s2

s0 s1

Khong nghim mi

s2 s3

Khong nghim mi

Hnh 1.18: Minh ha cc bc thc hin ca thut ton nht ct vng.

Bc k1

Bc k

Bc k+1

Khong ngn

s0=0 s2 s1=1 s3

Hnh 1.17: Chn im chia khong nghim theo nguyn l nht ct vng.

Khong di

49

3) Thc hin ln lt cc bc sau:

a) Nu f (s2 )< f (s3 ) th gn s1 :=s3 ri ly im s3 khc i xng vi s2 qua tm

ca khong nghim mi [s0 ,s1 ] . Ngc li th gn s0 :=s2 ri ly im s2

khc i xng vi s3 qua tm ca khong nghim mi [s0 ,s1 ] .

b) Nu khong nghim |s1s0 | vn cn ln, tc l |s1s0 | e th quay li bc a). Ngc li th chuyn sang bc 4).

4) p s s*sk vi sk l mt im ty thuc khong nghim [s0 ,s1 ] .

1.5.3 Phng php GaussSeidel

phng php GaussSeidel, hng tm c chn ln lt song song vi cc trc ta ca khng gian Rn , trong n l s chiu ca vector p . Nu k hiu e1 , e2 , ,

en l cc vector c s ca Rn tc l cc vector to thnh h trc ta ca Rn

e i =(0 , ,0 ,1 ,0, ,0)T

th hng tm hk ti bc th k=0,1, s c xc nh nh sau:

hk =

Pp

Pp

nu 0

nu 0 (1.37)

Khi , nu ta lp hm mc tiu mi xc nh vi mi p Rn :

),( pH = )( pQ + )( pS

trong mt s dng thch hp, th gia nghim

*p = arg min ),( pH = arg min [ )( pQ + )( pS ] (1.38)

ca bi ton ti u khng b rng buc (1.38) v nghim *p ca bi ton ti u b rng

buc (1.36) c quan h sau:

nh l 1.10: Gi *p l nghim ca bi ton (1.36) v *p l nghim ca bi ton (1.38).

Khi :

a) Nu *p P th *p = *p .

b) Lun tn ti s dng ln c *p P, tc l lun c lim *p = *p .

Chng minh:

Khng nh a) l hin nhin v vi )*( pS =0, lun c:

)*( pQ = ),*( pH ),( pH = )( pQ vi mi p P

chng minh khng nh b) ta s dng hnh minh ha 1.21. R rng t tnh cht

(1.37) ca hm )( pS ta lun tm c s dng ln gi tr hm ),( pH s rt

ln khi p P . Nh vy nghim *p ca (1.38) phi thuc P.

Hn na, v hm )( pS tha mn (1.37) gip cho ),( pH nhn gi tr rt ln khi

p P nn ngi ta gi n l hm pht, tc l n s pht khi vector p vt ra ngoi

min P.

55

Nu c hai hm mt bin:

s1 (z )=

=>>

0 nu 0

0 nu 0

z

z v s2 (z )=

==>

0 nu 0

0 nu 0

z

z

th mt s cc hm pht )( pS thng c s dng cho bi ton ti u (1.38) l:

a) )( pS = ))((max 1 pgs ii

+ ))((max 2 phs jj

b) )( pS = =

m

ii pgs

11 ))(( +

=

q

ji phs

12 ))((

V d 1.15: (Minh ha k thut hm pht)

Cho bi ton ti u mt bin, b rng buc p*=arg Pp

min Q(p) vi:

Q (p ) = p2 v P={ pR (1p )0 v (p24)0 }

C th thy bi ton c nghim p*=2.

nh ngha hm lin tc:

S (p )= max2{0, (1p ) } + max2{0, (p24)}

ta thy S (p ) tha mn tnh cht (1.37). Lp hm:

H (p , )= p2+[max2{0, (1p ) } + max2{0, (p24)}]

trong l s dng ln, s c:

H(p,)=

[ ]

>+

56

V *p = arg minH (p , ) l bi ton khng b rng buc nn nghim *p ca n s

c tm theo dp

pdH ),( =0. Nhng v

dppdH ),(

57

Mt hm no lm cho ),( pH xc nh trn P vi gi tr cng ln khi p cng

tin ti gn bin ca P c gi l hm chn. S dng hm mt bin:

b1 (z )=z r (r>0) hoc b2 (z )= lnz

th cc hm sau y s l hm chn:

a) )( pB = ))((max 1 pgb ii

b) )( pB = =

m

ii pgb

1))((

tc l vi chng, hm:

),( pH = )( pQ + )( pB

s c gi tr cng ln khi p cng tin ti gn bin ca P. Gia nghim *p (nm bn

trong min P) ca bi ton ti u *p = arg Ppmin ),( pH v nghim *p (c th nm

trn bin ca P) ca bi ton ti u b rng buc (1.1) c quan h sau:

0

lim

*p = *p

1.6 Mt s v d ng dng

1.6.1 Xc nh tham s ti u cho b iu khin PID

Mt trong nhng b iu khin c s dng rng ri nht trong thc t l b PID vi hm truyn t:

R (s )=

++ sT

sTk D

Ip

11

trong : kp l hng s khuch i,

TI l hng s thi gian tch phn,

TD l hng s thi gian vi phn.

Nguyn l iu khin bng PID l phn hi tn hiu ra (hnh 1.23), trong , ph

thuc vo i tng, cc tham s kp , TI , TD cn phi c chn sao cho h kn c c

cht lng nh mong mun, chng hn nh n nh, qu iu chnh nh, thi gian

qu ngn, khng c sai lch tnh .

e( t ) u( t ) y( t ) w( t )

Hnh 1.23: iu khin vi PID

S(s) PID

58

Mt mc tiu cht lng kt hp hi ha tt c cc ch tiu cht lng ni trn l

chn kp , TI , TD sao cho:

Q =

0

2 )( dtte min (1.39)

trong e ( t ) l tn hiu sai lch.B tham s kp , TI , TD tha mn (1.39) c gi l b

tham s ti u. Nh vy, tm c b tham s ti u kp , TI , TD ta cn phi tin

hnh cc bc:

Xc nh s ph thuc ca Q vo vector tham s Tp =(kp ,TI ,TD ) , tc l xc

nh Q= )( pQ .

Gii bi ton ti u *p = arg min )( pQ .

minh ha vic xc nh Q= )( pQ ta xt trng hp c th vi:

E (s )=1

0 1 1

0 1

nn

nn

c c s c s

d d s d s

+ + +

+ + +, dn0

trong E (s ) l nh Laplace ca e ( t ) . Khi , nh cng thc Parseval:

Q=

0

2 )( dtte =

djE

2(21

=

j

j

dssEsEj

)()(21

hm mc tiu Q s c tnh qua hai bc [35]:

Xc nh tt c cc im cc sk ca E (s ) .

Tnh Q=ks

Res [E (s )E (s ) ]

Vi hai bc tnh trn ta i n mt s kt qu cho n=1,2,3,4 nh sau [20]:

Bng 1.1: Cng thc tnh hm mc tiu Q cho mt s trng hp n=1,2,3,4:

n )( pQ

1 10

20

2 ddc

2 210

2200

21

2 ddddcdc +

3 )(2

)2(

302130

32203020

2110

22

ddddddddcddcccddc

++

4 )(2

)()2()2()(

421

23032140

241432

2043020

2141031

223

20210

23

ddddddddd

dddddcdddcccdddcccdddddc

+++

59

Ngoi ra, t cng thc:

E (s )=s1

)()(11

sSsR+

th r rng cc tham s c0 , ,cn - 1 , d0 ,d1 , ,dn ca E (s ) l nhng hm s ca

Tp =(kp ,TI ,TD ) . Thay cc hm s )(0 pc , , )(1 pcn , )(0 pd , , )( pdn vo cng

thc bng trn ta c Q= )( pQ .

V d 1.16: (Xc nh tham s ti u cho b iu khin tch phn)

Cho h kn m t hnh 1.24, trong

R (s )=sTI

1 v S (s )=

2)21(

5,0

s+.

H c nhiu n ( t ) v nhim v iu khin t ra y l xc nh tham s TI cho b

iu khin h c kh nng khng nhiu tc thi n ( t )=1( t ) tt nht theo ngha:

Q =

0

2 )( dtte min.

Trc ht, c nh Laplace E (s ) ca sai lch e ( t ) , ta xc nh hm truyn t tn hiu nhiu n ( t ) ti u ra y ( t ) :

Gn y (s )= )()(1)(

sSsRsS

+=

5,044

5,023 +++ sTsTsT

sT

III

I

T y, vi w ( t )=0 v n ( t )=1( t ) , tc l N (s )=s1

ta c nh Laplace ca sai lch e ( t ) :

E (s )= Y (s )= Gn y (s )N (s )=5,044

5,023 +++

sTsTsT

T

III

I

Tra bng cho trng hp n=3 vi:

c0=0,5TI , c1=c2=0,

d0=0,5, d1=TI , d2=d3=4TI ,

ta i n:

Q=24

2

II

TT

.

Suy ra

Q=24

2

II

TT

min TI =1.

e( t )n( t )

y( t )w( t )


S(s) sTI

1

60

1.6.2 Nhn dng tham s m hnh i tng tin nh

Trong iu khin ta rt hay gp phi bi ton xy dng m hnh ton hc m t i tng trn c s quan st (o) cc tn hiu vo ra ca n. Bi ton c tn gi l nhn dng i tng iu khin. N c Zadeh nh ngha nh sau [36]:

nh ngha 1.13: Nhn dng (identification) l phng php xc nh m hnh ton hc c th trong lp cc m hnh thch hp cho trn c s quan st cc tn hiu vo ra ca i tng sao cho sai lch gia m hnh tm c vi i tng thc l nh nht.

nh ngha cho thy bi ton nhn dng c nt ca mt bi ton ti u. Cng vic nhn dng lun c bt u vi nhng thng tin hiu bit mang tnh gi , tuy cn c th kh s ng, v i tng. Cc thng tin c gi l thng tin Apriori. Chng hn thng tin Apriori cho bit rng i tng l tuyn tnh hay phi tuyn, lin tc hay ri rc .

Xt i tng SISO vi tn hiu vo u v tn hiu ra y . Nu thng tin Apriori cho bit i tng l tuyn tnh c cu trc, th m hnh ton hc ca n s thuc lp cc hm phc, thchu t , hp thc vi cu trc bit trc:

G (z ) cho i tng khng lin tc.

G (s ) cho i tng lin tc.

trong t s v mu s ca hm truyn t l hai a thc nguyn t cng nhau (khng c chung nghim), ng thi c bc xc nh l n ,m . Nhim v nhn dng t ra y ch cn l xc nh tham s cho hai a thc sai lch gia G (z ) hoc G (s ) vi i tng c nh nht. Bi ton nhn dng c tn gi l nhn dng tham s m hnh i tng.

Nhn dng tham s m hnh khng lin tc

Xt i tng SISO khng lin tc c m hnh thuc lp:

G (z ) =)(

)(1

1

zA

zB=

nn

mm

zazaa

zbzb

+++

+++

11

10

11 (1.40)

Nu gi uk , yk , k=0,1, l dy gi tr tn hiu vo/ra quan st c vi chu k

ly mu T, tc l:

uk=u (kT ) , yk=y (kT ) , k=0,1,

th khi m hnh (1.40) l tuyt i chnh xc, ta s c:

uk+ =

m

iikiub

1=

=

n

iiki ya

0

61

Song do m hnh (1.40) cn cn phi c xc nh nn ng thc trn khng cn ng. Gia hai v ca n c mt sai lch. Sai lch ny c tn gi l sai lch d bo tuyn tnh ti thi im t=kT gia m hnh v i tng (hnh 1.25):

ek = uk+ =

m

iikiub

1

=

n

iiki ya

0

ek = uk pfTk

trong p =

n

m

a

a

b

b

0

1

v k

f =

nk

k

mk

k

y

y

u

u 1

.

T y ta lp hm mc tiu Q l hm m t sai s chung gia m hnh v i tng cho ton b khong thi gian quan st, tc l tng bnh phng ca tng sai lch ti cc

thi im t=kT , k=0,1, ,N nh sau:

)( pQ = =

N

kke

0

2 = eTe = (h pF )T (h pF )

= p

A

FFp TT )( p

b

FhT

T )2( +c

hhT = pApT pbT +c (1.41)

vi e=

Ne

e0, h=

Nu

u0 v F=

TN

T

f

f0

R rng A l ma trn xc nh bn dng, do hm mc tiu )( pQ cho trong cng

thc (1.41) l hm ton phng, li. Bi vy bi ton nhn dng vector tham s *p cho

m hnh (1.40) chnh l bi ton ti u li, ton phng khng b rng buc:

*p = arg min )( pQ 2A *p b=0 (xem nh l 1.4) (1.42)

Ch : Trong m hnh (1.40) ta gi thit b0=10. Nu i tng li c m hnh

vi b0=0 th phi c a00, v a thc t s v a thc mu s l nguyn t cng nhau.

Do a00, nn ta c th cho rng i tng c m hnh thuc lp:

G (z )=n

n

mm

zaza

zbzbb

+++

+++

1

1

1

110 , (n m) (1.43)

i tng

ek

yk uk

Hnh 1.25: Sai lch d bo tuyn tnh.

B(z 1 ) A(z 1 )

62

Khi , lm tng t nh vi m hnh (1.40) ta cng s i n bi ton ti u li, ton phng, khng b rng buc (1.42) vi:

)( pQ = (h pF )T(h pF )= pA

FFp TT )( p

b

FhT

T )2( +c

hhT

v p =

m

n

b

b

a

a

0

1

, k

f =

mk

k

nk

k

u

u

y

y 1

, F=

TN

T

f

f0

, h=

Ny

y0.

Nhn dng tham s m hnh lin tc

Xt i tng SISO tuyn tnh vi tn hiu vo u ( t ) v tn hiu ra y ( t ) c hm truyn t thuc lp cc m hnh khng c thnh phn vi phn:

G(s) =)()(

sUsY

=n

n

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