最速下降法例题

例题用最速下降法求解2212min{}(a0,b0,ab)xxab+?0axb轾=轾轾r解：，，2020aQb轾轾=轾轾轾轾122()2xafxxb=022g轾=轾轾r1000()xxtg=+-rrr选取..，.[][]00000222220222220TTggabtgQgabab轾轾轾===+轾轾轾轾轾轾轾轾轾rrrr1()22()aabaababxbbbaabab-轾轾-轾轾+=+=轾轾轾--+轾轾轾轾+轾r.第一次迭代12()02()ababgbaab-轾轾+=轾-轾轾+轾rr2111(),xxtg=+-rrr111112()2()2()2()22()02()2()22()0TTababbaabbaababggabtabgQgabbaaabbaababbab-轾轾--轾+轾轾-++轾轾轾+轾==-轾轾轾轾--轾+轾轾轾-++轾轾轾轾轾+轾轾rrrrabab=+11()2()()(2)()()2()()(2)aabababatabababxxtgtbbabababtababab----轾轾轾轾轾轾+++=+-=-=轾轾轾----轾轾轾轾轾轾+++轾轾轾rrrabtab=+1t可通过求极小获得228(){()}0()abababtabab-=--+=+22()(2)()(2)()[]/[]/abatbabttabababj----=+++2222()(2)()(2)()4[]/4[]/()()abatbabttabababj----?=--++()tj同理可得此例题有三点结论22222()()2()()()2()()()aabaababababababxbbabaabbbaababab轾---轾轾轾轾轾+++轾=-=轾轾轾--+-轾轾轾轾轾++轾轾+轾轾r()()()()kkkkkaababxbbaab轾-轾+轾=轾-轾+轾轾r2()()2()()kkkkkababgbaab轾-轾+轾=轾-轾+轾轾r例2用最速下降法求解取0(1,1)Tx=-解：21,12Q轾=轾轾121224()()21xxfxgxxx++轾?=轾+-轾rrr[]030Tg=r0(1,1)Tx=-1000()xxtg=+-rrr[][][]03300912136230301203t轾轾轾===轾轾轾轾轾轾轾轾轾10,03/2g轾=轾-轾rr2212121212min(,)41fxxxxxxxx=+++-+,迭代两次.1135/211012x---轾轾轾=+=轾轾轾轾轾轾r或[]131Txt=--r2111(),xxtg=+-rrr[][]1003/23/29/412109/2203/2123/2t轾-轾-轾===轾轾-轾轾-轾轾25/205/21,13/27/42x--轾轾轾=+=轾轾轾轾轾轾r23/400g轾=?轾轾rr15/21()0,,12ttxj-===2()(13)1(13)4(13)11ttttj=--++--+---+()6(13)312189tttj?=-----=-例3用最速下降法求解33121212min(,)3fxxxxxx=+-010x轾=轾轾r迭代二次判别所得的点是否为极小点212121233()33xxfxxx轾-?轾-+轾r解：112263(,)36xGxxx-轾=轾-轾033g轾=轾-轾r01111,101tpxtt---轾轾轾轾==+=轾轾轾轾轾轾轾轾rr332()(1)3(1)661tttttttj=-+--=-+11/21()1260,,.1/22tttxj轾=-===轾轾r13/40,3/4g-轾=?轾-轾rr不是极小点，继续迭代1xr取111/211/2,11/211/2tpxtt+轾轾轾轾==+=轾轾轾轾+轾轾轾轾rr（舍去）是驻点，G正定，221063,,(1,1).1036xgG-轾轾轾===轾轾轾-轾轾轾rr2xr是严格局部极小。2xr211()6(1/2)6(1/2)0,,22tttttj=+-+===-32()2(1/2)3(1/2)tttj=+-+例4：已知2212121212min(,)41fxxxxxxxx=+++-+取0(1,1)Tx=-用Newton法求最优解。解：21,12Q轾=轾轾121224()()21xxfxgxxx++轾?=轾+-轾rrr0(1,1)Tx=-[]030Tg=r1121312131112011203x----轾轾轾轾轾轾=-=-轾轾轾轾轾轾-轾轾轾轾轾轾gr16311323--轾轾轾=-=轾轾轾-轾轾轾()0gx=grrr，Q正定，xgr是极小点。例5已知33121212min(,)3fxxxxxx=+-,从010x轾=轾轾r用Newton法迭代一次，判别迭代点是否最优解：212121233()33xxfxxx轾-?轾-+轾r112263(,)36xGxxx-轾=轾-轾033g轾=轾-轾r06330G-轾=轾-轾11163310331030303639x--轾轾轾轾轾轾=-=+轾轾轾轾轾轾---轾轾轾轾轾轾r1901.0919-轾轾轾=+=轾轾轾--轾轾轾13()0,3gx轾=?轾轾rrr不是极小点例6已知2212121212min(,)41fxxxxxxxx=+++-+取0(1,1)Tx=-用Fletcher－Reeves共轭梯度法迭代两次。解：21,12Q轾=轾轾121224()()21xxfxgxxx++轾?=轾+-轾rrr0033,00gp-轾轾==轾轾轾轾rr15/2,1x-轾=轾轾r（见最速下降法）10,03/2g轾=轾-轾rr1100,pgpa=-+rrr210209/4194gga===rr1033/41,3/203/24p--轾轾轾=+=轾轾轾轾轾轾r取12p-轾=轾轾r//13/43/2p-轾=轾轾r211xxtp=+rrr[][][]11103/22312110212121223TTgptpQp-轾--轾轾=-===-轾轾轾--轾轾轾轾轾轾rrrr25/2131.1222x---轾轾轾=+=轾轾轾轾轾轾r200g轾=轾轾r2xr为最优点。例7已知2212121212min(,)41fxxxxxxxx=+++-+取0(1,1)Tx=-解：21,12Q轾=轾轾121224()()21xxfxgxxx++轾?=轾+-轾rrr用D.F.P算法迭代两次。0(1,1)Tx=-[]030Tg=r010,01H轾=轾轾03,0p-轾=轾轾r15/2,1x-轾=轾轾r10,3/2g轾=轾-轾r0103/2,0sxx-轾=-=轾轾rrr0000001000000TTTTssHyyHHHsyyHy=+-rrrrrrrr0103,3/2ygg-轾=-=轾-轾rrr[]00393/20,3/22Tsy-轾=-=轾-轾rr[]00034533/23/24TyHy-轾=--=轾-轾rr[]003/29/403/20000Tss-轾轾=-=轾轾轾轾rr[]0000399/233/23/29/29/4THyyH-轾轾=--=轾轾-轾轾rr109/4099/27/102/524009/29/42/54/5945HH-轾轾轾=+-=轾轾轾-轾轾轾1117/102/503/52/54/53/26/5pHg-轾轾轾=-=-=-轾轾轾---轾轾轾rr//12p-轾=轾轾r215/212txxpt--轾=+=轾+轾rrr()2(5/2)2(5/2)(12)4(12)6=6t-3=0,t=1/2tttttf=---+---+++-22()(5/2)(5/2)(12)(12)4(5/2)(12)1tttttttf=--+--++++---++25/2131,1222x---轾轾轾=+=轾轾轾轾轾轾r220,0gx轾=轾轾rr为最优点验证12.HQ-=1201101/20,.13/2sxxygg-轾轾=-==-=轾轾轾轾rrrrrr1111112111111TTTTTssHyyHHHsyyHy=+-rrrrrrrr[]11031/21,3/22Tsy轾=-=轾轾rr[]111/21/41/21/2111/21Tss--轾轾=-=轾轾-轾轾rr1117/102/50390.2/54/53/225TyHy-轾轾轾==轾轾轾-轾轾轾rr117/102/503/52/54/53/26/5Hy--轾轾轾==轾轾轾-轾轾轾r11113/53/5129()()6/56/52425TTHyHy---轾轾轾==轾轾轾-轾轾轾rr2/31/32111/32/3123--轾轾==轾轾--轾轾112121112123Q---轾轾==轾轾-轾轾27/102/51/41/2122592/54/51/21243925H---轾轾轾=+-轾轾轾---轾轾轾