# 6.6 The Marquardt Algorithm

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1 6.6 The Mqudt Algothm lmttons of the gdent nd Tylo expnson methods ecstng the Tylo expnson n tems of ch-sque devtves ecstng the gdent sech nto n tetve mtx fomlsm Mqudt's lgothm utomtclly combnes the gdent nd Tylo expnson methods pogm descpton nd exmple output fo the Gussn pek 6.6 : 1/9

2 ln( ) Lmttons of Pevous Methods The gdent sech woks well when the slope of ch-sque spce s steep. It "floundes" when gettng ne the mnmum becuse the gdent s ppochng zeo. A second-ode Tylo sees expnson woks best when the ntl guesses e wthn egon of ch-sque spce hvng postve cuvtue. When the ntl estmtes e n egons of negtve cuvtue the tetve pocedue dveges = nose = 0.01 nose = Gdent The Mqudt lgothm s method of utomtclly swtchng between the gdent sech nd Tylo expnson to enhnce convegence to the mnmum n ch-sque spce. 0 Tylo's Tylo's 6.6 : /9

3 Tylo Sees Devtves One of the bggest dwbcks to the second-ode Tylo expnson s the need to povde functonl fom fo the fst nd second devtves. Ths s especlly cute when wtng compute pogm. Mqudt showed tht the functons could be eplced by numec dffeentton of ch-sque. y y f, x 1 1 f, x f, x f, x f, x y c c c 1 1 y 6.6 : 3/9 Addtonlly, Mqudt ponted out the y tem n the lst equton wll sum to smll vlue s N nceses, snce the devtons hve pdf wth men of zeo , ( ) (, ) ( ) c c m m m m

4 Gdent Sech Mtx Fomlsm The second mpotnt thng ecognzed by Mqudt ws the blty to wte the gdent sech s n tetve mtx poblem, ust lke Tylo's sees expnson. The mtx soluton s, ( m) ( m, m) ( m), whee nd e the pevously defned vectos, nd s dgonl mtx contnng the step szes long ech coeffcent xs n chsque spce. Snce s untless, ech hs unts of 1/. Snce ech must hve unts of, ech step sze,,, hs to hve unts of. The most ntul choce fo the step szes s one popotonl to the coeffcent vnce, ( -1 ),. Fnlly, the step should be some smll fcton of the vnce, ( -1 ), /, whee > 1. The mtx s then obtned by multplyng the dgonl of by the constnt, whee s sevng the sme ole s the fcto f n the gdent sech. 0, ,0 1 1,1 1,1 nn, nn, 6.6 : 4/9

5 Gdent Sech Mtx Fomlsm Quck evew of the gdent sech. Gdent of s gven by 0, 1 0, 1 ˆ 0 1 ˆ Step sze s popotonl to the mgntude of the gdent, weghted by the empcl fcto of f = L f f f L But (emkbly) 1 f, x y Whee s dgonl mtx.

6 Mtx Soluton Mqudt [Jounl of the Socety fo Industl nd Appled Mthemtcs, 1963, vol. 11, pp ] demonstted tht the second-ode expnson nd the gdent sech could be combned nto one mthemtcl opeton. To do ths defne new mtx., c, c,, fo 1 fo c c When When << 1, the soluton coesponds to Tylo expnson. >> 1, the soluton coesponds to gdent sech. The clculton uses the stndd tetve pocedue, 1 ( m) (, ) ( m) mm new cu whee the subscpt, cu, denotes the cuent guesses, nd the subscpt, new, denotes the mpoved estmte of the coeffcent vlues. Iteton nvolves substtutng new guesses fo cuent guesses nd epetng the mtx lgeb. 6.6 : 5/9

7 The Algothm 1. compute ch-sque t the ntl guesses. ssemble the mtx nd vecto usng the ptl devtves 3. stt wth equl to smll numbe, sy 10-3, nd ssemble the ' mtx - the ft wll stt wth Tylo's expnson 4. solve the mtx equton fo new nd compute ch-sque t new 5. () f ( new ) ( cu ) multply by 10, essemble ' nd epet step 4 wth cu - mke the ft moe lke gdent (b) f ( new ) < ( cu ) dvde by 10, essemble ' nd epet step 4 wth cu = new - mke the ft moe lke n expnson (c) f ( new ) ( cu ) stop the tetons nd use cu s estmtes of the coeffcents - the mnmum hs been found 6. set = 0 nd compute -1 wth cu. Use the dgonl elements to obtn the vnce of the coeffcents. 6.6 : 6/9

8 Mqudt Pogm Descpton A pogm tht utomtclly computes the mnmum n chsque spce s shown n the Mthcd woksheet, "6.6 Mqudt Algothm.mcd". It uses sx functons wth vety of nputs: f(x,), nputs e the x-dt nd the ntl coeffcent guesses, ; output s the coespondng y-vlue s scl. chsq(y,x,), nputs e the x- nd y-dt, nd the - coeffcents; output s ch-sque t the locton gven by. bet(y,x,, ), nputs e the dt, coeffcents, nd the step sze, used to compute the devtves; output s the vecto whch contns lph(y,x,, ), computes the devtves; output s the mtx whch contns the second-ode devtves. set (, ), computes the ' mtx gven nd. mqut(y,x,, ), executes the Mqudt lgothm; output s vecto contnng the coeffcent vlues t the mnmum, the vlue of ch-sque t the mnmum, the fnl vlue of, nd the numbe of equed tetons. 6.6 : 7/9

9 Mqudt Pogm Output m Stt wth the ntl guesses, 0 = nd 1 = 51, nd = (the lgest vlue of gvng stble devtve). mqudty ( x 0.001) Thee tetons wee equed. = , whch mens the sech ws lwys Tylo expnson. The coeffcents hve the lowest tht we hve seen. 6.6 : 8/9 Test the functon n egon of negtve cuvtue, 0 = 6 nd 1 = mqudty ( x 0.001) The lge numbe of tetons ndctes tht the lgothm spent sgnfcnt fcton of the tme pefomng gdent sech.

10 Coeffcent Eos m The coeffcents wee set to those t the mnmum, 0 =.140 nd 1 = The lph mtx ws eclculted wth = 0. The -1 mtx yelded the dgonl elements, ( 0,0 ) -1 = nd ( 1,1 ) -1 = The estmted vnce of the ft ws s = The coeffcent vnce cn be computed fom these two tems. s s All of the methods gve sttstclly ndstngushble esults fo the lest-sques coeffcents. The Tylo nd Mqudt gve sttstclly ndstngushble stndd devtons fo the coeffcents. The gd sech s esy to mplement mnully, the Mqudt lgothm s the most tolent to bd guesses. 6.6 : 9/9

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