最速下降法例題

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 . ， . 00000222220222220ttggabtg qgabab rrrr1()22()a abaababxbb baabab r.第一次迭代12()02()ababgbaabrr2111(),xxtgrrr111112()2()2()2()22()02()2()22()0ttababbaabbaababg gabtabg qgabbaaabbaababbabrrrrabab11()2()()(2 )

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

3、abgbaabr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.1135/211012xr或131txt r2111(),xxtgrrr1003/23/29/412109/2203/2123/2t25/205/21,13/27/42xr23/400grr15/21( )0, , 12ttxr2( )( 13 )1( 13 )4(

4、13 )1 1tttt ( )6( 13 )312189ttt 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是否為極小點212121233( )33xxf xxxr解：112263( ,)36xg x xx033gr01111,101tpxtt rr332( )(1)3(1)661tttt ttt11/21( )1260,.1/22tttxr13/4 0,3/4grr 不是極小點，繼續(xù)迭代1xr取111/211/2,11/211/2tpxtt rr（舍去）是駐點，g正定， 221063,(1,1).1036xgg rr2xr是嚴格

5、局部極小。2xr211( )6(1/2)6(1/2)0,22ttttt 32( )2(1/2)3(1/2)ttt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 r16311323()0g xrr r， q 正定，xr是極小點。例5 已知33121212min (,)3f x xxxx x, 從010 x r 用newton 法迭代一次，判別迭代點是否最優(yōu)解：212121233

6、( )33xxf xxxr112263( ,)36xg x xx033gr06330g11163310331030303639x r1901.0919 13( )0,3g x rr r不是極小點例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（見最速下降法）10, 03/2grr1100, pgp rrr210209/4194ggrr1033/41,3/203/24pr取12pr/13/

7、43/2pr211xxt prrr11103/22312110212121223ttgptp qp rrrr25/2131.1222xr200g r2xr為最優(yōu)點。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 rr rrrrr0103,3/2yggrrr

8、00393/20,3/22tsy 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/12pr215/212txxptrrr( )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)點驗證12.hq1201101/20,.13/2sxxyggrrrrrr1 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 yr