最速下降法例題ppt課件

1、1例題例題 用最速下降法求解2212min (a0,b0,ab)xxab0axb r解： ， ，2020aQb 122( )2xaf xxb 022g r1000()xxtgrrr選取 . ， . 200000222220222220TTggabtg Qgabab rrrr1()22()a abaababxbb baabab r.第一次迭代312()02()ababgbaabrr2111(),xxtgrrr111112()2()2()2()22()02()2()22()0TTababbaabbaababg gabtabg Qgabbaaabbaababbabrrrrabab411()2()()

2、(2 )()()2()()(2 )a ababab atabababxxtgtb bababa btabababrrrabtab1t可通過(guò)求極小獲得228()() 0()ababab tab ab 22()(2 )()(2 )( ) / /ab atba bttababab2222() (2 )() (2 )( )4/4/()()abatbabttababab ( ) t5同理可得此例題有三點(diǎn)結(jié)論22222()()2()()()2()()()a aba ababababababxb babaabb baabababr()()()()kkkkka ababxb baabr2()()2()()kk

3、kkkababgbaabr6例2 用最速下降法求解取0( 1,1)Tx 解：21,12Q121224( )( )21xxf xg xxxrr r030Tg r0( 1,1)Tx 1000()xxtgrrr03300912136230301203t 10, 03/2grr2212121212min (,)41f x xxxx xxx,迭代兩次.1135/211012xr7或131Txt r2111(),xxtgrrr1003/23/29/412109/2203/2123/2t25/205/21,13/27/42xr23/400grr15/21( )0, , 12ttxr2( )( 13 )1(

4、13 )4( 13 )1 1tttt ( )6( 13 )312189ttt 8例3 用最速下降法求解33121212min (,)3f x xxxx x010 x r迭代二次判別所得的點(diǎn)是否為極小點(diǎn)212121233( )33xxf xxxr解：112263( ,)36xG x xx033gr01111,101tpxtt rr332( )(1)3(1)661tttt ttt11/21( )1260,.1/22tttxr913/4 0,3/4grr 不是極小點(diǎn)，繼續(xù)迭代1xr取111/211/2,11/211/2tpxtt rr（舍去）是駐點(diǎn)，G正定， 221063,(1,1).1036xgG

5、 rr2xr是嚴(yán)格局部極小。2xr211( )6(1/2)6(1/2)0,22ttttt 32( )2(1/2)3(1/2)ttt10例4：已知2212121212min (,)41f x xxxx xxx取0( 1,1)Tx 用Newton 法求最優(yōu)解。解：21,12Q121224( )( )21xxf xg xxxrr r0( 1,1)Tx 030Tg r1121312131112011203x gr16311323()0g xgrr r， Q 正定，xgr是極小點(diǎn)。11例5 已知33121212min (,)3f x xxxx x, 從010 x r 用Newton 法迭代一次，判別迭代

6、點(diǎn)是否最優(yōu)解：212121233( )33xxf xxxr112263( ,)36xG x xx033gr06330G11163310331030303639x r1901.0919 13( )0,3g x rr r不是極小點(diǎn)12例6 已知2212121212min (,)41f x xxxx xxx取0( 1,1)Tx 用FletcherReeves共軛梯度法迭代兩次。解：21,12Q121224( )( )21xxf xg xxxrr r0033,00gp rr15/2, 1xr（見(jiàn)最速下降法）10, 03/2grr1100, pgp rrr210209/4194ggrr1033/41,3

7、/203/24pr13取12pr/13/43/2pr211xxt prrr11103/22312110212121223TTgptp Qp rrrr25/2131.1222xr200g r2xr為最優(yōu)點(diǎn)。14例7 已知2212121212min (,)41f x xxxx xxx取0( 1,1)Tx 解：21,12Q121224( )( )21xxf xg xxxrr r用D.F.P 算法迭代兩次。0( 1,1)Tx 030Tg r010,01H03,0pr15/2,1xr10 ,3/2gr0103/2,0sxxrrr0 000001000000TTTTs sH y y HHHsyyH yr

8、rr rrrrr0103,3/2yggrrr1500393/20,3/22Tsy r r00034533/23/24Ty H y rr003/29/403/20000Ts sr r0000399/233/23/29/29/4TH y y Hr r109/4099/27/102/524009/29/42/54/5945HH1117/102/503/52/54/53/26/5pH g rr/12pr16215/212txxptrrr( )2( 5/2)2( 5/2)(12 )4(12 )6 =6t-3=0 , t=1/2 ttttt 22( )( 5/2)( 5/2)(12 )(12 ) 4( 5/2)(12 )1ttttttt 25/2131,1222xr220, 0gx rr為最優(yōu)點(diǎn)驗(yàn)證12.HQ1201101/20,.13/2sxxyggrrrrrr171 111112111111TTTTTs sH y y HHHs yy H yr rr rr rrr11031/21,3/22Ts y r r1 11/21/41/21/2111/21Ts sr r11 17/102/50390.2/54/53/225Ty H yrr