New Algorithms: Linear, Nonlinear, and Integer Programming
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- Elaine Quinn
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1 New Algorthms: ner, Nonlner, nd Integer Progrmmng Dhnnjy P. ehendle Sr Prshurmhu College, Tl Rod, Pune-400, Ind Astrt In ths pper we propose new lgorthm for lner progrmmng. Ths new lgorthm s sed on tretng the ojetve funton s prmeter. We form mtrx usng oeffents n the system of equtons onsstng ojetve equton nd equtons otned from nequltes defnng onstrnt y ntrodung sl/surplus vrles. We otn redued row ehelon form for ths mtrx ontnng only one vrle, nmely, the ojetve funton tself s n unnown prmeter. We nlyze ths mtrx n the redued row ehelon form nd develop ler ut method to fnd the optml soluton for the prolem t hnd, f nd when t exsts. We see tht the entre optmzton proess n e developed through the proper nlyss of the sd mtrx n the redued row ehelon form. From the nlyss of the sd mtrx n the redued row ehelon form t wll e ler tht n order to fnd optml soluton we my need rryng out ertn proesses le rerrngng of the onstrnt equtons n prtulr wy nd/or performng pproprte elementry row trnsformtons on ths mtrx n the redued row ehelon form. These opertons re mnly med t hevng nonnegtvty of ll the entres n the olumns orrespondng to nons vrles n ths mtrx or ts sumtrx otned y olletng ertn rows of ths mtrx (.e. sumtrx wth rows hvng negtve oeffent for prmeter d, whh stnds for the ojetve funton s prmeter for mxmzton prolem nd sumtrx wth rows hvng postve oeffent prmeter d, gn representng the ojetve funton s prmeter for mnmzton prolem). The re s to e ten so tht the new mtrx rrved t y rerrngng the onstrnt equtons nd/or y rryng out sutle elementry row trnsformtons must e equvlent to orgnl mtrx. Ths equvlene s n the sense tht ll the fesle soluton sets for the prolem vrles otned for dfferent possle vlues of d wth orgnl mtrx nd trnsformed mtrx re sme. We then proeed to show tht ths de nturlly extends to del wth nonlner nd nteger progrmmng prolems. For nonlner nd nteger progrmmng prolems we use the tehnque of Groner ses (sne Groner ss s n equvlent of redued row ehelon form for system of nonlner equtons) nd the methods of solvng lner Dophntne equtons (sne the nteger progrmmng prolem demnds for optml nteger soluton) respetvely.. Introduton: There re two types of lner progrms (lner progrmmng prolems):. xmze: C T x Sujet to: Or. nmze: C T x Ax x 0
2 Sujet to: Ax x 0 where x s olumn vetor of sze n of unnown vrles. We ll these vrles the prolem vrles where C s olumn vetor of sze n of proft (for mxmzton T prolem) or ost (for mnmzton prolem) oeffents, nd C s row vetor of sze n otned y mtrx trnsposton of C. where A s mtrx of onstrnts oeffents of sze m n. where s olumn vetor of onstnts of sze m representng the oundres of onstrnts. By ntrodung the pproprte sl vrles (for mxmzton prolem) nd surplus vrles (for mnmzton prolem), the ove mentoned lner progrms gets onverted nto stndrd form s: xmze: Sujet to: C T (.) x Ax s = x 0, s 0 where s s sl vrle vetor of sze m. Ths s mxmzton prolem. Or nmze: C T x Sujet to: Ax s = (.) x 0, s 0 where s s surplus vrle vetor of sze m. Ths s mnmzton prolem. In geometrl lnguge, the onstrnts defned y the nequltes form regon n the form of onvex polyhedron, regon ounded y the onstrnt plnes, Ax =, nd the oordnte plnes. Ths regon s lled fesle regon nd t s strghtforwrd to estlsh tht there exsts t lest one vertex of ths polyhedron t whh the optml soluton for the prolem s stuted when the prolem t hnd s well defned,.e. nether nonsstent, nor unounded, nor nfesle. There my e unque optml soluton nd sometmes there my e nfntely mny optml solutons, e.g. when one of the onstrnt plnes s prllel to the ojetve plne we my hve multtude of optml solutons. The ponts on n entre plne or n entre edge n onsttute the optml soluton set. These prolems re hndled most populrly y usng the well nown smplex lgorthm or some of ts vrnt. Despte ts theoretl exponentl omplexty the smplex method wors qute effently for most of the prtl prolems. However, there re few omputtonl dffultes ssoted wth smplex lgorthm. In order to vew them n nutshell we egn wth sttng some ommon notons nd defntons tht re prevlent n the lterture. A vrle x s lled s vrle n gven equton f t ppers wth unt oeffent n tht equton nd wth zero oeffents n ll other equtons. A vrle whh s not s s lled nons vrle. A sequene of elementry row opertons tht hnges gven system of lner equtons nto n equvlent system (hvng the sme soluton set) nd n whh gven nons vrle n
3 e mde s vrle s lled pvot operton. An equvlent system ontnng s nd nons vrles otned y pplton of sutle elementry row opertons s lled nonl system. At tmes, the ntroduton of sl vrles for otnng stndrd form utomtlly produes nonl system, ontnng t lest one s vrle n eh equton. Sometmes sequene of pvot opertons s needed to e performed to get nonl system. The soluton otned from nonl system y settng the nons vrles to zero nd solvng for the s vrles s lled s soluton nd n ddton when ll the vrles hve nonnegtve vlues the soluton stsfyng ll the mposed onstrnts s lled s fesle soluton. Smplex method nnot strt wthout n ntl s fesle soluton. The proess of fndng suh soluton, whh s neessty n mny of prtl prolems, s lled Phse I of the smplex lgorthm. Smplex method strts ts Phse II wth n ntl s fesle soluton n nonl form t hnd. Then smplex tests whether ths soluton s optml y heng whether ll the vlues of reltve profts (profts tht result due to unt hnge n the vlues of nons vrles) of ll the nons vrles re nonpostve. When not optml, the smplex method otns n djent s fesle soluton y seletng nons vrle hvng lrgest reltve proft to eome s. Smplex then determnes nd rres out the extng of s vrle, y the so lled mnmum rto rule, to hnge t nto nons vrle ledng to formton of new nonl system. On ths new nonl system the whole proedure s repeted tll one rrves t n optml soluton. The mn omputtonl dffultes of the smplex method whh my use the reduton n ts omputtonl effeny re s follows: ] There n e more thn one nons vrle wth lrgest vlue for reltve proft nd so te n te ple whle seletng nons vrle to eome s. The hoe t ths stuton s done rtrrly nd so the hoe mde t ths stge usng lrgest possle per unt mprovement s not neessrly the one tht gves lrgest totl mprovement n the vlue of the ojetve funton nd so not neessrly mnmzes the numer of smplex tertons. ] Whle pplyng mnmum rto rule t s possle for more thn one onstrnt to gve the sme lest rto usng te n the seleton of s vrle to leve for eomng nons. Ths degenery n use further omplton, nmely, the smplex method n go on wthout ny mprovement n the ojetve funton nd the method my trp nto n nfnte loop nd fl to produe the desred optml soluton. Ths phenomenon s lled ylng whh enfores modfton n the lgorthm y ntrodung some ddtonl tme onsumng rules tht redue the effeny of the smplex lgorthm. ] Smplex s not effent on theoretl grounds slly euse t serhes djent s fesle solutons only nd ll other smplex vrnts whh exmne nondjent solutons s well hve not shown ny pprele hnge n the overll effeny of these modfed smplex lgorthms over the orgnl lgorthm. Beuse of the fr gret prtl mportne of the lner progrms nd other smlr prolems n the opertons reserh t s most desred thng to hve n lgorthm whh wors n sngle step, f not, n s few steps s possle. No method hs een found whh wll yeld n optml soluton to lner progrm n sngle step ([], Pge 9). We m to propose n lgorthm for
4 lner progrmmng whh ms t fulfllng ths requrement n est possle nd novel wy.. A New Algorthm for ner Progrmmng: We strt wth the followng equton: C T x = d (.) where d s n unnown prmeter, nd ll t the ojetve equton. The (prmetr) plne defned y ths equton wll e lled ojetve plne. Plese note tht we re dsussng frst the mxmzton prolems. A smlr pproh for mnmzton prolems wll e dsussed next. Gven mxmzton prolem, we frst onstrut the omned system of equtons ontnng the ojetve equton nd the equtons defned y the onstrnts mposed y the prolem under onsderton, omned nto sngle mtrx equton, vz., C A( T ( n) 0( m) m n) I ( m m) x s d = (.) et E = C A( T ( n) 0( m) m n) I ( m m) d, nd let F = et [E, F] denote the ugmented mtrx otned y ppendng the olumn vetor F to mtrx E s lst olumn. We then fnd R, the redued row ehelon form ([], pges 7-75) of the ove ugmented mtrx [E, F]. Thus, R = rref ([E, F]) (.) Note tht the ugmented mtrx [E, F] s well s ts redued row ehelon form R ontns only one prmeter, nmely, d nd ll other entres re onstnts. From R we n determne the soluton set S for every fxed x d, S = { /( fxed) d rels} s. The suset of ths soluton set of vetors x whh lso stsfes the nonnegtvty onstrnts s the set of ll fesle s solutons for tht d. It s ler tht ths suset n e empty for prtulr hoe of d tht s mde. The mxmzton prolem of lner progrmmng s to determne the unque d whh provdes fesle soluton nd hs mxmum vlue for d,.e., to determne the unque d,.e. the unque optml vlue for the ojetve funton, whh n e used to otn n optml soluton. In the se of n unounded lner progrm there s no upper (lower, n the se of mnmzton prolem) lmt for the vlue of d, whle n the se of n nfesle 4
5 lner progrm the set of fesle solutons s empty. The steps tht wll e exeuted to determne the optml soluton should lso tell y mplton when suh optml soluton does not exst n the se of n unounded or nfesle prolem. The generl form of the mtrx R representng the redued row ehelon form s R = ( m m ) m ( m ) n n n mn ( m ) n m ( m ) m ( m ) m m m mm ( m ) n ( m ) d e m d e d e d e d e m ( m ) The frst n olumns of the ove mtrx represent the oeffents of the prolem vrles (.e. vrles defned n the lner progrm) x, x,, xn. The next m olumns represent the oeffents of the sl vrles s, s,, s m used to onvert nequltes nto equltes to otn the stndrd form of the lner progrm. The lst olumn represents the trnsformed rght hnd sde of the equton (.) durng the proess ( sutle sequene of trnsformtons) tht s rred out to otn the redued row ehelon form. Note tht the lst olumn of R ontns the lner form d s prmeter whose optml vlue s to e determned suh tht the nonnegtvty onstrnts remn vld,.e. x 0, n nd s j 0, j m. Among frst ( n m) olumns of R ertn frst olumns orrespond to s vrles (olumns tht re unt vetors) nd the remnng ones to nons vrles (olumns tht re not unt vetors). For solvng lner progrm y our wy we proeed wth nlyss of the R. We m to fnd tht vlue of prmeter d whh s optml. To heve ths ts we my sometmes need to trnsform ths system further y ether rerrngng the onstrnt equtons y sutly permutng these equtons suh tht (s fr s possle) the vlues n the olumns of our strtng mtrx A ( m n ) get rerrnged n the followng wy: Ether they re rsng nd then eomng sttonry, or fllng nd then eomng sttonry, or fllng ntlly up to ertn length of the olumn vetor nd then rsng gn. After ths rerrngement to form trnsformed A ( m n ) we gn proeed to form ts orrespondng [E, F] nd gn fnd ts R = rref ([E, F]) whh wll most lely hve the desred representton n whh olumns for nons vrles ontn nonnegtve vlues. The de of rrngement of the olumns of A ( m n ) s mentoned ove s purely heurst nd s sed on the fvorle outome oserved durng pplyng ths de to lner progrmmng prolems whle tlng them. We hve not tred to dsover theoretl resonng ehnd hevng fvorle form for R n most of the ses nd hve left ths prolem for the reder 5
6 to solve nd to fnd out the theoretl resonng ehnd gettng ths fvorle form. Now, even fter rerrngement of the olumns of A ( m n ) s mentoned ove f stll some negtve entres remn present n the olumns of R orrespondng to some nons vrles then we rry out sutle elementry row trnsformtons on the otned R = rref ([E, F]) so tht the olumns of oeffents ssoted wth these nons vrles eome nonnegtve. We re dong ths euse s wll e seen elow we n then put zero vlue for these nons vrles nd n determne the vlues of ll the s vrles nd the lner progrm wll then e solved ompletely. It s esy to he tht for lner progrm f ll the oeffents of prmeter d n the lst olumn of R re postve then the lner progrm t hnd s unounded sne n suh se the prmeter d n e nresed rtrrly wthout voltng the nonnegtvty onstrnts on vrles x, s j. Also, for lner progrm f ll the oeffents of some nons vrle represented y olumn of R re nonpostve nd re strtly negtve n those rows hvng negtve oeffent to prmeterd tht ppers n the lst olumn of these rows then gn the prolem elongs to the tegory of unounded prolems sne we n nrese the vlue of d to ny hgh vlue wthout voltng the nonnegtvty onstrnts for the vrles y ssgnng suffently hgh vlue to ths nons sl vrle. Note tht the rows of R tully offer expressons for s vrles n terms of nons vrles nd terms of type d e, =,, ( m ) ontnng the prmeter d on the rght sde. The rows wth postve oeffent for the prmeter d represent those equtons n whh the prmeter d n e nresed rtrrly wthout voltng the nonnegtvty onstrnts on vrles x,. So, these equtons s j wth postve oeffent for the prmeter d re not mplyng ny upper ound on the mxmum possle vlue of prmeter d. However, these rows re useful n ertn stutons s they re useful to fnd lower ound on the vlue of prmeter d. The rows wth negtve oeffent for the prmeter d represent those equtons n whh the prmeter d nnot e nresed rtrrly wthout voltng the nonnegtvty onstrnts on vrles x,. So, these equtons s j wth negtve oeffent for the prmeter d re mplyng n upper ound on the mxmum possle vlue of prmeter d nd so mportnt ones for mxmzton prolems. Note tht tully every row of R s offerng us vlue for prmeter d whh n e otned y equtng to zero eh term of the type d e, =,, ( m ). Those vlues of d tht we otn n ths wy wll e denoted s d or d when then vlue of s negtve or postve respetvely. We denote y mn{ d }, the mnmum vlue mong d, nd we denote y mx{ d } the mxmum vlue mong the d. We now proeed to fnd out the sumtrx of R, sy R N, mde up of ll olumns of R nd 6
7 7 ontnng those rows j of R for whh the oeffents j of the prmeter d re negtve. et,,, oeffent of d n the rows of R whh re negtve. We ollet these rows wth negtve oeffent for d to form the mentoned sumtrx, N R, of R gven elow. Wth ths t s ler tht oeffents of d n ll other rows of R re greter thn or equl to zero. = m n m n m n m n N e d e d e d e d R It should e ler to see tht f N R s empty (.e. not ontnng sngle row) then the prolem t hnd s unounded. Among the frst ) ( m n olumns of N R frst n olumns represent the oeffents of prolem vrles nd next m olumns represent the oeffents of sl vrles. There re ertn olumns strtng from frst olumn nd pper n suessons whh re unt vetors. These olumns whh re unt vetors orrespond to s vrles. The olumns pperng n suessons fter these olumns nd not unt vetors orrespond to nons vrles. As mentoned, mong the olumns for nons vrles those hvng ll entres nonnegtve n only led to derement n the vlue of d when postve vlue s ssgned to them. Ths s undesrle s we m mxmzton of the vlue of d. So, we n sfely set the vlue of suh vrles equl to zero. When ll olumns orrespondng to nons vrles n N R re hvng ll entres nonnegtve nd further f mn{ d } mx{ d } then we n set ll nons vrles to zero, set d = mn{ d } n every row of R nd fnd the s fesle soluton whh wll e optml, wth mn{ d } s optml vlue for the ojetve funton t hnd. Stll further, When ll olumns orrespondng to nons vrles n N R re hvng ll entres nonnegtve ut mn{ d } < mx{ d } then f e > 0 then we n stll set ll nons vrles to zero, set d = mn{ d } n every row of R nd fnd the s fesle soluton whh wll e optml, wth mn{ d } s optml vlue for the ojetve funton t hnd,.e. f vlue of e > 0 n the expressons e d n the
8 rows of R other those n R N tht re hvng vlue of > 0 then we n proeed on smlr lnes to fnd optml vlue for d. In R N we now proeed to onsder those nons vrles for whh the olumns of R N ontn some (t lest one) postve vlues nd some negtve (t lest one) vlues. In suh se when we ssgn some postve vlue to suh nons vrle t leds to derese n the vlue of d n those rows n whh > 0 nd nrese n the vlue of d n those rows n whh < 0. We now need to onsder the wys of delng wth ths stuton. We del wth ths stuton s follows: In ths se, we hoose nd rry out pproprte nd legl elementry row trnsformtons on the mtrx R n the redued row ehelon form to heve nonnegtve vlue for ll the entres n the olumns orrespondng to nons vrles n the sumtrx R N of R. The elementry row trnsformtons re hosen to produe new mtrx whh remns equvlent to orgnl mtrx n the sense tht the soluton set of the mtrx equton wth orgnl mtrx nd mtrx equton wth trnsformed mtrx remn sme. Due to ths equvlene we n now set ll the nons vrles n ths trnsformed mtrx to zero nd otn wth justfton d mn = mn{ d } s optml vlue for the ojetve funton nd otn s fesle soluton s optml soluton y susttuton. et us now dsuss our new lgorthm n steps: Algorthm. (xmzton):. Express the gven prolem n stndrd form: xmze: C T x Sujet to: Ax s = x 0, s 0. Construt the ugmented mtrx [E F], where T C ( n) 0( m) E = d, nd F = A( m n) I ( m m) nd otn the redued row ehelon form: R = rref ([E, F]). If there s row (or rows) of zeroes t the ottom of R n the frst n olumns nd ontnng nonzero onstnt n the lst olumn then delre tht the prolem s nonsstent nd stop. Else f the oeffents of d n the lst olumn re ll postve or f there exsts olumn of R orrespondng to some nons vrle wth ll entres negtve then delre tht the prolem t hnd s unounded nd stop. 4. Else f for ny vlue of d one oserves tht nonnegtvty onstrnt for some vrle gets volted y t lest one of the vrles then delre tht the prolem t hnd s nfesle nd stop. 8
9 5. Else fnd the sumtrx of R, sy R N, mde up of those rows of R for whh the oeffent of d n the lst olumn s negtve. 6. Che whether the olumns of R N orrespondng to nons vrles re nonnegtve. Else, rerrnge the onstrnt equtons y sutly permutng these equtons suh tht (s fr s possle) the vlues n the olumns of our strtng mtrx A ( m n ) get rerrnged n the followng wy: Ether they re rsng nd then eomng sttonry, or fllng nd then eomng sttonry, or fllng ntlly up to ertn length of the olumn vetor nd then rsng gn. After ths rerrngement to form trnsformed A ( m n ) gn proeed s s done n step ove to form ts orrespondng ugmented mtrx [E, F] nd gn fnd ts R = rref ([E, F]) whh wll most lely hve the desred representton,.e. n the new R N tht one wll onstrut from the new R wll hve olumns for nons vrles whh wll e ontnnng nonnegtve entres. 7. Solve r d e r = 0 for eh suh term n the lst olumn of R N nd fnd the vlue of d = d r for r =,,, nd fnd d mn = mn{ d } r. Smlrly, solve r d e r = 0 for eh suh term n the lst olumn for rows of R other thn those n R N nd fnd the vlues d = d r for r =,,, nd fnd d mx = mx{ d } r. Che the olumns of R N orrespondng to nons vrles. If ll these olumns ontn only nonnegtve entres nd f mn{ d } mx{ d } then set ll nons vrles to zero. Susttute d = dmn n the lst olumn of R. Determne the s fesle soluton whh wll e optml nd stop. Further, f olumns orrespondng to nons vrles ontn only nonnegtve entres nd f mn{ d } < mx{ d } then he whether vlue of e > 0 n the expressons d e n these rows of R other those n R N tht re hvng vlue of > 0. If yes, then set ll nons vrles to zero. Susttute d = dmn n the lst olumn of R. Determne the s fesle soluton whh wll e gn optml. 8. Even fter proeedng wth rerrngng olumns of our strtng mtrx A ( m n ) y sutly permutng rows representng onstrnt equtons s s done n step 6, nd then proeedng s per step 7 nd hevng the vldty of mn{ d } mx{ d }, f stll there remn olumns n R N orrespondng to some nons vrles ontnng postve entres n some rows nd 9
10 negtve entres n some other rows then devse nd pply sutle elementry row trnsformtons on R suh tht the olumns representng oeffents of nons vrles of new trnsformed mtrx R or t lest ts sumtrx R orrespondng to new trnsformed R ontn only nonnegtve entres. N And step 7 eomes pplle. We now proeed wth some exmples: Exmple.: xmze: Sujet to: x y 4 x y 4x y x, y 0 x Soluton: For ths prolem we hve R = [, 0, 0, 0, /, -d6 ] [ 0,, 0, 0, -/, -6*d ] [ 0, 0,, 0, /, 0-*d ] [ 0, 0, 0,,, -*d ] So, lerly, R N = [, 0, 0, 0, /, -d6 ] [ 0, 0,, 0, /, 0-*d ] [ 0, 0, 0,,, -*d ] y For ths exmple the olumn formng oeffents for nons vrle s ontns nonegtve numers. So, we set s = 0. Clerly, d mn =. = Optml vlue for the ojetve funton. Usng ths vlue of optmum we hve x =.66, y = 0.66, s =, s =, s 0. 0 = Exmple.: We frst onsder the duel of the exmple suggested y E... Bele [], whh rngs nto exstene the prolem of ylng for the smplex method, nd provde soluton s per the ove new method whh offers t dretly wthout ny ylng phenomenon. xmze: 0.75 x 0 x 0.5 x 6 x 4 Sujet to: 0.5 x 8 x x 9 x x x 0.5 x x 4 0 x 0 x 0 x, x, x, 4 Soluton: For ths prolem we hve the followng 0
11 R = [, 0, 0, 0, /, 8/, 4/, ( 4/)d4/ ] [0,, 0, 0, 7/4, /4, /4, ( 5/4)d/ ] [0, 0,, 0, 0, 0,, ] [0, 0, 0,, /8, /8, /9, /9 (/8)d ] So, lerly, R = [, 0, 0, 0, /, 8/, 4/, ( 4/)d4/ ] N [0,, 0, 0, 7/4, /4, /4, ( 5/4)d/4 ] [0, 0, 0,, /8, /8, /9, /9 (/8)d ] We perform followng elementry row trnsformtons on R: et us denote the suessve rows of R y R(), R(), R(), R(4). We hnge () () () R() R() (6/4)*R(4), R() R() *R(4). R(4) 8*R(4) Ths leds to new trnsformed R s follows: R = [, 0, 0,, 0, 0, 6, 6-d] [0,, 0, /4, 0, /4, 5/8, 5/8-(/)d] [0, 0,, 0, 0, 0,, ] [0, 0, 0, 8,,,, d ] In the trnsformed R we hve nonnegtve olumns for ll nons vrles, whh re now those orrespondng to x 4, s,s. So, y settng x 4 = s = s = 0 nd settng ll expressons of type d e = 0 n the lst olumn we fnd r r d mn = mn{ d } =.5. Usng ths vlue n the lst olumn of the newly otned r trnsformed R we hve: x =.0000, x = 0, x =, x 4 = 0, s = , s = 0, s = 0, nd the mxmum vlue of d =.500. Exmple.: We now onsder n unounded prolem. The new method dretly mples the unounded nture of the prolem through the postvty of the oeffents of d n mtrx R for the prolem. xmze: x y Sujet to: x y x y 0 x y x, y 0 Soluton: The followng s the mtrx R : R = [, 0, 0, 0, -/5, (/5)d-/5 ] [ 0,, 0, 0, -/5, (/5)d-/5 ]
12 [ 0, 0,, 0, -4/5, (/5)d-4/5] [ 0, 0, 0,, /5, /5(/5)d] Here, ll the oeffents of d re postve. So, y settng vrle s = 0 we n see tht we n ssgn ny rtrrly lrge vlue to vrle d wthout volton of nonnegtvty onstrnts for vrles. Thus, the prolem hs n unounded soluton. Exmple.4: We now onsder prolem hvng n nfesle strtng ss. We see tht new lgorthm hs no dffulty to del wth t. xmze: x y Sujet to: x y 4 x y 5 x 4y x, y 0 Soluton: The followng s the mtrx R : R = [, 0, 0, 0, /7, (/7)d-/7 ] [ 0,, 0, 0, -/4, /7(/4)d] [ 0, 0,, 0, /4, 7/7-(5/4)d] [ 0, 0, 0,, -/4, 6/7-(9/4)d] R = [ 0, 0,, 0, /4, 7/7-(5/4)d] N [ 0, 0, 0,, -/4, 6/7-(9/4)d] Clerly, mn{ d } mx{ d }. So, we perform followng elementry row trnsformtons on R: et us denote the suessve rows of R y R(), R(), R(), R(4). We hnge () R(4) R(4) R() Ths leds to new trnsformed R N, R s follows: R = [ 0, 0,, 0, /4, (7/7)-(5/4)d] N [ 0, 0,,, 0, (6/7) d ] R = [, 0, 0, 0, /7, (/7)d-/7 ] [ 0,, 0, 0, -/4, /7(/4)d] [ 0, 0,, 0, /4, 7/7-(5/4)d] [ 0, 0,,, 0, (6/7) d ] R, RN orrespond to s vrles x, y. Sne, The frst two olumns of mn{ d } mx{ d } nd olumns orrespondng to nons vrles s ontn nonnegtve entres n R N,s, so we set these vrles to zero. From lst row
13 we hve s = 0 nd d mn = 9. Also from frst nd seond rows, x =.857, y =.074 Exmple.5: We now onsder n nfesle prolem. xmze: x y Sujet to: x y x y 0 x, y 0 Soluton: The followng s the mtrx R : R = [, 0, 0, -/4, (/4)d ] [ 0,, 0, /8, (/8)d ] [ 0, 0,, 7/8, (-/8)d-] Here, the oeffent of d s negtve only n the lst row nd so R N = [ 0, 0,, 7/8, (-/8)d-]. We perform followng elementry row trnsformtons on R: et us denote the suessve rows of R y R(), R(), R(), R(4). We hnge () R() (/7)*R(4) R() Ths leds to R = [, 0, /7, 0, /7*d-/7] [ 0,, 0, /8, (/8)d ] [ 0, 0,, 7/8, (-/8)d-] nd R = [ 0, 0,, 7/8, (-/8)d-] N Settng s,s equl to zero, we hve for onssteny of lst row d = (8/) nd usng ths vlue for d we hve y = (/). Thus, ths prolem s nfesle. Remr.: Klee nd nty [4], hve onstruted n exmple of set of lner n progrms wth n vrles for whh smplex method requres tertons to reh n optml soluton. Theoret wor of Borgwrdt [5] nd Smle [6] ndtes tht fortuntely the ourrene of prolems elongng to the lss of Klee nd nty, whh don t shre the verge ehvor, s so rre s to e neglgle. We now proeed to show tht there s no prolem of effeny for new lgorthm n delng wth the prolems elongng to ths lss. Exmple.6: We now onsder prolem for whh the smplex tertons re exponentl funton of the sze of the prolem. A prolem elongng to the lss
14 desred y Klee nd nty ontnng n vrles requres We see tht the new method doesn t requre ny spel effort xmze: 00 x 0 x x Sujet to: x 0 x x x 0 x x 0000 x, x, x 0 Soluton: The followng re the mtres R, RN : R = [, 0, 0, 0, /0, -/00, -90(/00)d] [0,, 0, 0, -, /5, 900-(/5)d] [0, 0,, 0, 0, -, d-0000 ] [0, 0, 0,, -/0, /00, 9-(/00)d ] R = [0,, 0, 0, -, /5, 900-(/5)d ] N [0, 0, 0,, -/0, /00, 9- (/00)d] n smplex steps. We perform followng elementry row trnsformtons on R: et us denote the suessve rows of R y R(), R(), R(), R(4). We hnge () () () R() 0R() R(), nd R(4) R() R(4) R() 0*R() Ths leds to new trnsformed R N, R s follows: R N = [0,, 0, 0, 0, /0, 000-(/0)*d] [, 0, 0,, 0, 0, ] R = [0, 0, 0, 0,, -/0, -900/0*d ] [0,, 0, 0, 0, /0, 000-(/0)*d] [0, 0,, 0, 0, -, *d-0000 ] [, 0, 0,, 0, 0, ] Sne, mn{ d } mx{ d } nd olumns orrespondng to nons vrles x,s ontn nonnegtve entres n R N, so we set these vrles to zero. We get esy omplete soluton s follows: x = 0, x = 0, x = 0000, s =, s = 00, s = 0, nd the mxmum vlue of d = Exmple.7: We now onsder n exmple for whh the redued row ehelon form ontns y tself the s vrles tht re requred to e present n the optml smplex tleu,.e. the tleu tht results t the end of the smplex lgorthm, for 4
15 whh only nonpostve entres our n the ottom row of the tleu representng reltve profts. Ths s understood y the nonnegtvty of entres n the olumns of R N orrespondng to nons vrles. xmze: x y Sujet to: x y 4 x y 4 x y x, y 0 Soluton: The followng s the mtrx R : R = [, 0, 0, 0, /5, (/5)d6/5 ] [ 0,, 0, 0, -/5, -9/5(/5)d] [ 0, 0,, 0, 8/5, 44/5-(/5)d] [ 0, 0, 0,, 0, 4-d ] So, lerly, R N = [ 0, 0,, 0, 8/5, 44/5-(/5)d] [ 0, 0, 0,, 0, 4-d ] Sne, mn{ d } mx{ d } nd olumns orrespondng to nons vrle R ontn nonnegtve entres n N, so we set these vrles to zero. Here, the nons vrle olumns dretly ontn nonnegtve entres ledng to deresng n proft when some postve vlue s ssgned to ths vrle nd so we set ths vrle to zero whh leds to the mxml s fesle soluton: d = 4, x = 4, y =, s = 6, s = 0, s = 0. We now onsder few exmples n whh rerrngement of onstrnt equtons utomtlly produe sutle form for redued row ehelon form: 6 x y Exmple.8:. xmze: Sujet to: x y 5 4x y x y 4 x, y 0 For ths prolem we get followng R, R N R = [, 0, 0, 0, /, /9*d-4/] [ 0,, 0, 0, -/, 8/-/9*d] [ 0, 0,, 0, /, /-/9*d] [ 0, 0, 0,, -/, 44/-7/9*d] s 5
16 R = [ 0,, 0, 0, -/, 8/-/9*d] N [ 0, 0,, 0, /, /-/9*d] [ 0, 0, 0,, -/, 44/-7/9*d] Here some entres n the lst ut one olumn orrespondng to nons vrle re mxed type,.e. entres n the frst nd thrd rows of R N re negtve whle entry n the seond row of R N s postve. We now rerrnge the onstrnts so tht ether entres n the new olumns re rsng nd then eomng sttonry, or fllng nd then eomng sttonry, or fllng ntlly up to ertn length of the olumn vetor nd then rsng gn, s mentoned ove. Thus, we just rewrte the prolem s: 6 x y xmze: Sujet to: x y 5 x y 4 4x y x, y 0 then form new A ( m n ) mtrx whh produe nd gn proeed to fnd redued row ehelon form for new R = [, 0, 0, 0, /, -/6*d6] [ 0,, 0, 0, -, -/*d] [ 0, 0,, 0, /, -/*d] [ 0, 0, 0,, -/, -7/6*d] R = [, 0, 0, 0, /, -/6*d6] N [ 0, 0,, 0, /, -/*d] Sne, mn{ d } mx{ d } nd olumns orrespondng to nons vrle R ontn nonnegtve entres n N, so we set these vrles to zero. Here, the nons vrle olumns dretly ontn nonnegtve entres ledng to deresng n proft when some postve vlue s ssgned to ths vrle nd so we set ths vrle to zero whh leds to the mxml s fesle soluton: d =, x = /, y = 8 /, s = 0, s = /, s 0. 7 = x y Exmple.9: xmze: Sujet to: x y 4 x y 5 x 4y x, y 0 For ths prolem we get followng R, R N R = [, 0, 0, 0, /7, /7*d-/7 ] s 6
17 [ 0,, 0, 0, -/4, /7/4*d] [ 0, 0,, 0, /4, 7/7-5/4*d] [ 0, 0, 0,, -/4, 6/7-9/4*d] R N = [ 0, 0,, 0, /4, 7/7-5/4*d] [ 0, 0, 0,, -/4, 6/7-9/4*d] Here some entres n the lst ut one olumn orrespondng to nons vrle re mxed type,.e. entres n the frst nd thrd rows of R N re negtve whle entry n the seond row of R N s postve. We now rerrnge the onstrnts so tht ether entres n the new olumns re rsng nd then eomng sttonry, or fllng nd then eomng sttonry, or fllng ntlly up to ertn length of the olumn vetor nd then rsng gn, s mentoned ove. Thus, we just rewrte the prolem s: x y xmze: Sujet to: x y 5 x 4y x y 4 x, y 0 then we form new A ( m n ) new mtrx whh produe nd gn proeed to fnd redued row ehelon form for R = [, 0, 0, 0, -, d-8] [ 0,, 0, 0,, -d] [ 0, 0,, 0,, 9-d] [ 0, 0, 0,, 4, 54-5*d] R N = [ 0,, 0, 0,, -d] [ 0, 0,, 0,, 9-d] [ 0, 0, 0,, 4, 54-5*d] Sne, mn{ d } mx{ d } nd olumn orrespondng to nons vrle R ontn nonnegtve entres n N, so we set ths vrle to zero. Here, the nons vrle olumns dretly ontn nonnegtve entres ledng to deresng n proft when some postve vlue s ssgned to ths vrle nd so we set ths vrle to zero whh leds to the mxml s fesle soluton: d = 9, x =, y =, s = 0, s = 9, s 0. = 4 x y Exmple.0: xmze: Sujet to: x.5 y 9 x y 8 s 7
18 x y 6 x, y 0 For ths prolem we get followng R, R N R = [, 0, 0, 0, -, d-8] [ 0,, 0, 0, 4, 4-d] [ 0, 0,, 0, -, -575/*d] [ 0, 0, 0,,, 0-d] R = [ 0,, 0, 0, 4, 4-d] N [ 0, 0, 0,,, 0-d] Clerly, though the seond lst olumn orrespondng to nons vrle ontns nonnegtve entres the nequlty mn{ d } mx{ d } s nvld! So, s frst ttempt, efore strtng to rry out elementry row trnsformton on R to heve nonnegtvty of entres n the olumns of R orrespondng nons vrles, let us frst try rerrngng the nequltes so tht ether entres n the new olumns re rsng nd then eomng sttonry, or fllng nd then eomng sttonry, or fllng ntlly up to ertn length of the olumn vetor nd then rsng gn, s mentoned ove. Thus, we just rewrte the prolem s: xmze: 4 x y Sujet to: x.5 y 9 x y 6 x y 8 x, y 0 then we form new A ( m n ) new mtrx whh produe nd gn proeed to fnd redued row ehelon form for R = [, 0, 0, 0, /, -/*d] [ 0,, 0, 0, -, -6d] [ 0, 0,, 0, /, 5-*d] [ 0, 0, 0,, /, 0-/*d] = R N [, 0, 0, 0, /, -/*d] [ 0, 0,, 0, /, 5-*d ] [ 0, 0, 0,, /, 0-/*d ] For ths rerrngement the seond lst olumn orrespondng to nons vrle ontns nonnegtve entres nd lso the nequlty mn{ d } mx{ d } s now vld! So, we set nons vrle s equl to zero whh leds to mxml s fesle soluton: d =7. 667, x =.665, y =.667, s = 0, s =.665, s = 0 8
19 x y x y x y 5 x 4y Exmple.: xmze: Sujet to: 4 Soluton: For ths prolem we hve R = [, 0, 0, 0, /7, (/7)d-/7 ] [ 0,, 0, 0, -/4, /7 (/4)d] [ 0, 0,, 0, /4, 7/7- (5/4)d] [ 0, 0, 0,, -/4, 6/7- (9/4)d] R = [ 0, 0,, 0, /4, 7/7-(5/4)d] N [ 0, 0, 0,, -/4, 6/7-(9/4)d] But, f we permute onstrnts s: (I) onstrnt onstrnt, (II) onstrnt onstrnt, nd (III) onstrnt onstrnt nd form new A ( m n ) nd further fnd out new R, R N then we get R = [, 0, 0, 0, -, d-8 ] [ 0,, 0, 0,, -d ] [ 0, 0,, 0,, 9-d ] [ 0, 0, 0,, 4, 54-5d ] R = [ 0,, 0, 0,, -d ] N [ 0, 0,, 0,, 9-d ] [ 0, 0, 0,, 4, 54-5d ] Sne, mn{ d } mx{ d } nd olumn orrespondng to nons vrle R ontn nonnegtve entres n N, so we set ths vrle to zero nd we dretly get the optml s fesle soluton from R = : d = d mn = 9, x =, y =, s = 0, s = 9, s = 0. We now see tht we n proeed wth n extly smlr wy nd del suessfully wth mnmzton lner progrmmng prolems. A prolem of mnmzton goes le: nmze: C T x Sujet to: Ax x 0 we frst onstrut the omned system of equtons ontnng the sme ojetve equton used n mxmzton prolem (ut ths tme we wnt to fnd mnmum vlue of prmeter d defned n the ojetve equton) nd the equtons defned s 9
20 y the onstrnts mposed y the prolem under onsderton, omned nto sngle mtrx equton, vz., C A( T ( n) 0( m) m n) I( m m) x = s d (.) et E = C A( T ( n) 0( m) m n) I( m m) d, nd let F = et [E, F] denote the ugmented mtrx otned y ppendng the olumn vetor F to mtrx E s lst olumn. We then fnd R, the redued row ehelon form of the ove ugmented mtrx [E, F]. Thus, R = rref ([E, F]) (.) Note tht the ugmented mtrx [E, F] s well s ts redued row ehelon form R ontns only one prmeter, nmely, d nd ll other entres re onstnts. From R we n determne the soluton set S for every fxed x d, S = { /( fxed) d rels} s. The suset of ths soluton set of vetors x whh lso stsfes the nonnegtvty onstrnts s the set of ll fesle s solutons for tht d. It s ler tht ths suset n e empty for prtulr hoe of d tht s mde. The mnmzton prolem of lner progrmmng s to determne the unque d whh provdes fesle soluton nd hs mnmum vlue,.e., to determne the unque d whh provdes n optml soluton. In the se of n unounded mnmzton lner progrm there s no lower ound for the vlue of d, whle n the se of n nfesle lner progrm the set of fesle solutons s empty. The steps tht wll e exeuted to determne the optml soluton should lso tell y mplton when suh optml soluton does not exst n the se of n unounded or nfesle prolem. The generl form of the mtrx R representng the redued row ehelon form s smlr s prevously dsussed mxmzton se: 0
21 R = ( m m ) m ( m ) n n n mn ( m ) n m ( m ) m ( m ) m m m mm ( m ) n d e m ( m ) d e d e d e d e m ( m ) The frst n olumns of the ove mtrx represent the oeffents of the prolem vrles (.e. vrles defned n the lner progrm) x, x,, xn. The next m olumns represent the oeffents of the surplus vrles s, s,, s m used to onvert nequltes nto equltes to otn the stndrd form of the lner progrm. The lst olumn represents the trnsformed rght hnd sde of the equton (.) durng the proess ( sutle sequene of trnsformtons) tht s rred out to otn the redued row ehelon form. Note tht the lst olumn of R ontns the lner form d s prmeter whose optml vlue s to e determned suh tht the nonnegtvty onstrnts remn vld,.e. x 0, n nd s j 0, j m. Among frst ) olumns of R ertn frst olumns orrespond to s vrles ( n m (olumns tht re unt vetors) nd the remnng ones to nons vrles (olumns tht re not unt vetors). For solvng the lner progrm we need to determne the vlues of ll nons vrles nd the optml vlue of d, from whh we n determne the vlues of ll the s vrles y susttuton nd the lner progrm s thus solved ompletely. For lner progrm f ll the oeffents of prmeter d n the lst olumn of R re negtve then the lner progrm t hnd s unounded (sne, the prmeter d n e deresed rtrrly wthout voltng the nonnegtvty onstrnts on vrles x, ). For lner progrm f ll the s j oeffents of some nons sl vrle represented y olumn of R re nonpostve nd re strtly negtve n those rows hvng postve oeffent to prmeter d tht ppers n the lst olumn of these rows then we n derese the vlue of d to ny low vlue wthout voltng the nonnegtvty onstrnts for the vrles y ssgnng suffently hgh vlue to ths nons sl vrle nd the prolem s gn elongs to the tegory of unounded prolems. Note tht the rows of R tully represent equtons wth vrles x, =,, n nd vrles, j =,, m on left sde nd expressons of type s j d e, =,, ( m ) ontnng the vrle d on the rght sde. The rows wth negtve oeffent for the prmeter d represent those equtons n whh the prmeter d n e deresed rtrrly wthout voltng the nonnegtvty onstrnts on vrles x,. So, these equtons wth negtve s j oeffent for the prmeter d re not mplyng ny lower ound on the mnmum
22 possle vlue of prmeter d. However, these rows re useful to now out upper ound on prmeter d. The rows wth postve oeffent for the prmeter d represent those equtons n whh the prmeter d nnot e deresed rtrrly wthout voltng the nonnegtvty onstrnts on vrles j s x,. So, these equtons wth postve oeffent for the prmeter d re mplyng lower ound on the mnmum possle vlue of prmeter d nd so mportnt ones n ths respet. So, we now proeed to fnd out the sumtrx of R, sy P R, mde up of ll olumns of R nd ontnng those rows j of R for whh the oeffents j of the prmeter d re postve. et,,, re ll nd re only postve rel numers n the rows olleted n P R gven elow nd ll other oeffents of d n other rows of R re greter thn or equl to zero. = n n m n m n m n P e d e d e d e d R If P R s empty (.e. ontnng not sngle row) then the prolem t hnd s unounded. There re ertn olumns strtng from frst olumn nd pper n suessons whh re unt vetors. These olumns whh re unt vetors orrespond to s vrles. The olumns pperng n suessons fter these olumns nd not unt vetors orrespond to nons vrles. As mentoned, mong the olumns of P R for nons vrles those hvng ll entres nonnegtve n only led to nrese n the vlue of d when postve vlue s ssgned to them. Ths s undesrle s we m mnmzton of the vlue of d. So, we desre to set the vlues of suh vrles equl to zero. When ll olumns orrespondng to nons vrles n P R re hvng ll entres nonnegtve nd further f the nequlty mn{ d } mx{ d } holds then we n set ll nons vrles to zero nd set d = mx{ d } n every row of R nd fnd the s fesle soluton whh wll e optml, wth mx{ d } s optml vlue for the ojetve funton t hnd. Stll further, When ll olumns orrespondng to nons vrles n P R re hvng ll entres nonnegtve ut mn{ d } < mx{ d } then f e > 0 then we n stll set ll nons vrles to zero, set d = mx{ d } n every row of R nd fnd the s
23 fesle soluton whh wll e optml, wth mx{ d } s optml vlue for the ojetve funton t hnd,.e. f vlue of e > 0 n the expressons d e n the rows of R rther those n R P tht re hvng vlue of > 0 then we n proeed on smlr lnes to fnd optml vlue for d. In R P we now proeed to onsder those nons vrles for whh the olumns of R P ontn some (t lest one) postve vlues nd some negtve (t lest one) vlues. In suh se when we ssgn some postve vlue to suh nons vrle t leds to derese n the vlue of d n those rows n whh > 0 nd nrese n the vlue of d n those rows n whh < 0. We now need to onsder the wys of delng wth ths stuton. We del wth ths stuton s follows: In ths se, we hoose nd rry out pproprte nd legl elementry row trnsformtons on the mtrx R n the redued row ehelon form to heve nonnegtve vlue for ll the entres n the olumns orrespondng to nons vrles n the sumtrx R P of R. The elementry row trnsformtons re hosen to produe new mtrx whh remns equvlent to orgnl mtrx n the sense tht the soluton set of the mtrx equton wth orgnl mtrx nd mtrx equton wth trnsformed mtrx remn sme. Due to ths equvlene we n now set ll the nons vrles n ths trnsformed mtrx to zero nd otn wth justfton d mx = mx{ d } s optml vlue for the ojetve funton nd otn s fesle soluton s optml soluton y susttuton. Algorthm. (nmzton):. Express the gven prolem n stndrd form: xmze: C T x Sujet to: Ax s = x 0, s 0. Construt the ugmented mtrx [E F], where T C ( n) 0( m) d E =, nd F = A( m n) I ( m m) nd otn the redued row ehelon form: R = rref ([E, F]). If there s row (or rows) of zeroes t the ottom of R n the frst n olumns nd ontnng nonzero onstnt n the lst olumn then delre tht the prolem s nonsstent nd stop. Else f the oeffents of d n the lst olumn re ll postve or f there exsts olumn of R orrespondng to some nons vrle wth ll entres negtve then delre tht the prolem t hnd s unounded nd stop.
24 4. Else f for ny vlue of d one oserves tht nonnegtvty onstrnt for some vrle gets volted y t lest one of the vrles then delre tht the prolem t hnd s nfesle nd stop. 5. Else fnd the sumtrx of R, sy R P, mde up of those rows of R for whh the oeffent of d n the lst olumn s postve. 6. Che whether the olumns of R P orrespondng to nons vrles re nonnegtve. Else, rerrnge the onstrnt equtons y sutly permutng these equtons suh tht (s fr s possle) the vlues n the olumns of our strtng mtrx A ( m n ) get rerrnged n the followng wy: Ether they re rsng nd then eomng sttonry, or fllng nd then eomng sttonry, or fllng ntlly up to ertn length of the olumn vetor nd then rsng gn. After ths rerrngement to form trnsformed A ( m n ) gn proeed s s done n step ove to form ts orrespondng ugmented mtrx [E, F] nd gn fnd ts R = rref ([E, F]) whh wll most lely hve the desred representton,.e. n the new R P tht one wll onstrut from the new R wll hve olumns for nons vrles whh wll e ontnng nonnegtve entres. 7. Solve r d e r = 0 for eh suh term n the lst olumn of R P nd fnd the vlue of d = d r for r =,,, nd fnd d mn = mn{ d } r. Smlrly, solve r d e r = 0 for eh suh term n the lst olumn for rows of R other thn those n R N nd fnd the vlues d = d r for r =,,, nd fnd d mx = mx{ d } r. Che the olumns of R P orrespondng to nons vrles. If ll these olumns ontn only nonnegtve entres nd f mn{ d } mx{ d } then set ll nons vrles to zero. Susttute d = d mx n the lst olumn of R. Determne the s fesle soluton whh wll e optml nd stop. Further, f olumns orrespondng to nons vrles ontn only nonnegtve entres nd f mn{ d } < mx{ d } then he whether vlue of e > 0 n the expressons d e n these rows of R other those n R P tht re hvng vlue of > 0. If yes, then set ll nons vrles to zero. d = d n the lst olumn of R. Determne the s fesle Susttute mx soluton whh wll e gn optml. 8. Even fter proeedng wth rerrngng olumns of our strtng mtrx A ( m n ) y sutly permutng rows representng onstrnt equtons s s 4
25 done n step 6, nd then proeedng s per step 7 nd hevng the vldty of mn{ d } mx{ d }, f stll there remn olumns n R P orrespondng to some nons vrles ontnng postve entres n some rows nd negtve entres n some other rows then devse nd pply sutle elementry row trnsformtons on R suh tht the olumns representng oeffents of nons vrles of new trnsformed mtrx R or t lest ts sumtrx R P orrespondng to new trnsformed R ontn only nonnegtve entres so tht step 7 eomes pplle. We now onsder few exmples for mnmzton prolems: Exmple.: Ths exmple for mnmzton s le Exmple.6 for mxmzton n whh the redued row ehelon form ontns y tself the s vrles tht re requred to e present n the optml smplex tleu,.e. the tleu tht results t the end of the smplex lgorthm, for whh only nonnegtve entres our n the ottom row of the tleu representng reltve osts. Ths s understood y the nonnegtvty of entres n the olumns of R P orrespondng to nons vrles. nmze: x x x Sujet to: x x x 4x x x x x x x x, x, x 0 Soluton: For ths prolem we hve R = [, 0, 0, 0, -, 0,, -d ] [ 0,, 0, 0, -, 0,, ] [ 0, 0,, 0, -, 0,, 5-d ] [ 0, 0, 0,,, 0,, 6d ] [ 0, 0, 0, 0, 0,,, 0 ] It s ler from seprtely equtng eh entry n the lst olumn of R tht the expeted nequlty, mn{ d } mx{ d }, holds good. Also, R = [ 0, 0, 0,,, 0,, 6d] P Sne ll the entres n the olumns orrespondng to nons vrles n postve so we put, s, s, s4 = 0 d = mx( d ) = further susttutons, we hve x = 4, x =, x = 9. s. Also, we put R P re. By 5
26 Exmple.: We now onsder mnmzton prml lner progrm for whh nether the prml nor the dul hs fesle soluton. nmze: x y Sujet to: x y x y x, y 0 Soluton: For ths prolem we hve R = [, 0, 0,, -d] [ 0,, 0,, -d] [ 0, 0,,, - ] R P s n empty mtrx. So, there s no lower lmt on the vlue of d. But, Here, from the lst row of R t s ler tht (for ny nonnegtve vlue of nons vrle, s ) the vlue of s s negtve nd so the prolem s thus nfesle. Smlrly, f we onsder the followng dul, vz, x y xmze: Sujet to: x y x y x, y 0 then we hve R = [, 0, 0,, -d] [ 0,, 0,, -4d] [ 0, 0,,, - ] R = [, 0, 0,, -d] N [ 0,, 0,, -4d] whh mples zero vlue for ler tht (even for ny nonnegtve vlue of nons vrle, negtve nd so the prolem s nfesle. s nd d =. But gn from the lst row of R t s s ) the vlue of y s Exmple.4: We now onsder mnmzton lner progrm for whh the prml s unounded nd the dul s nfesle. nmze: x y Sujet to: 5 x y x y 5 x, y 0 Soluton: For ths prolem we hve R = [, 0, 0, -/, -5/-(/)d ] [ 0,, 0, /, (-/)d5/] 6
27 [ 0, 0,, -, -0 ] Clerly, R p s empty so there s no ound to the vlue of d on the lower sde nd y gvng vlue 0 to nons vrle s we n vod negtvty of s, so the prolem s unounded. Also, let us pply followng elementry trnsformtons on R: et us denote the suessve rows of R y R(), R(), R(), R(4). We hnge () () R() R() R(), nd R() R() R() R = [,, 0, 0, -d ] [ 0,, 0, /, (-/)d5/] [ 0,,, 0, -d -5 ] So, y settng = 0, s d 5 we n he tht ths prolem n unounded elow. Now, f we onsder the followng dul, vz, xmze: 5x 5y Sujet to: x y We hve x y x, y 0 R = [, 0, 0, -/, /(/0)d] [ 0,, 0, -/, /-(/0)d] [ 0, 0,,, - ] R = [ 0,, 0, -/, /-(/0)d] N Agn from the lst row of R t s ler tht even for ny nonnegtve vlue of nons vrle, s the vlue of s s negtve nd so the prolem s nfesle. Exmple.5: nmze: 4 Sujet to: x x x x4 x x 4x x x, x, x 0 We hve followng R, RP for ths prolem: R = [, 0, 0, -, -6, -4, -7*d56] [ 0,, 0,, 4,, -4*d] [ 0, 0,, 0, -7, -, 5-*d ] x 4 x x R = [ 0,, 0,, 4,, -4*d] P 7
28 Sne t s ler from seprtely equtng eh entry n the lst olumn of R tht the expeted nequlty, mn{ d } mx{ d }, holds good. Also, sne ll the entres n the olumns orrespondng to nons vrles n R P re postve so we put x s, s 0. Also, we put d = mx( d ) = 7 4, = = 7, x = 0, x = hve x 4.. By further susttutons, we Remr.: It s ler from the dsusson mde so fr tht our m should e to heve the stuton n whh ll the olumns orrespondng to nons vrles n R N, R P ontn nonnegtve entres. Ths stuton orresponds to dretly hvng the possesson of mxml/mnml s fesle soluton. By hevng ths one gets dretly the optml s fesle soluton y smply settng ll the nons vrles to zero nd fndng the s soluton. Ths s the stuton n sense of hvng dretly the optml smplex tleu for whh one sets ll the nons vrles to zero s they n ft led to derement/nrement n the otherwse mxml/mnml vlue of d for mxmzton/mnmzton lner progrmmng prolems under onsderton. Thus, we n fnd the soluton of ny mxmzton/mnmzton lner progrm y properly nlyzng the R, R N, R P mtres, nd tng the relevnt tons s per the outomes of the nlyss. A, to form new E, F wthout Remr.: Insted of shufflng rows of mtrx hngng the ontent of the prolem, we n lso heve the sme effet of rerrngement of onstrnt y smply shufflng rows of dentty mtrx I n E nd proeed wth formton of new E, F wthout hngng the ontent of the prolem nd hevng sme effet n the sense tht the orrespondng redued row ehelon form wll utomtlly produe sme effet. Remr.: The nonnegtvty of the entres tht re present n the olumns of nons vrles of the onerned redued row ehelon form R N or R P s n effet smlr to otn the optml smplex tleu,.e. the tleu t whh the smplex lgorthm termntes nd where the s fesle soluton represents the optml soluton.. A New Algorthm for Nonlner Progrmmng: We now proeed show tht we n del wth nonlner progrms (nonlner onstrned optmzton prolems) usng the sme ove gven tehnque used to del wth lner progrms. The lgorthms developed y Bruno Buherger whh trnsformed the strt noton of Groner ss nto fundmentl tool n omputtonl lger wll e utlzed. The tehnque of Groner ses s essentlly verson of redued row ehelon form (used ove to hndle the lner progrms mde up of lner polynomls) for hgher degree polynomls [7]. A typl nonlner progrm n e stted s follows: xmze/nmze: f (x) Sujet to: h j ( x) = 0, j =,,, m m m 8
29 g j ( x) 0, j = m, m,, p x 0, =,,, n Gven nonlner optmzton prolem we frst onstrut the followng nonlner system of equtons: f ( x) d = 0 (.) h j x) = 0, j =,,, m (.) g ( ( x) s = 0, j = m, m, p (.) j j, where d s the unnown prmeter whose optml vlue s to e determned sujet to nonnegtvty ondtons on prolem vrles nd sl vrles. For ths to heve we frst trnsform the system of equtons nto n equvlent system of equtons erng the sme soluton set suh tht the system s eser to solve. We hve seen so fr tht the effetve wy to del wth lner progrms s to otn the redued row ehelon form for the omned system of equtons norportng ojetve equton nd onstrnt equtons. We wll see tht for the nonlner se the effetve wy to del wth s to otn the equvlent of redued row ehelon form, nmely, the Groner ss representton for ths system of equtons (.)-(.). We then set up the equtons otned y equtng the prtl dervtves of d wth respet to prolem vrles x nd sl vrles s to zero nd utlze the stndrd theory nd methods used n lulus. We demonstrte the essene of ths method y solvng ertn exmples. where d s the unnown prmeter whose optml vlue s to e determned sujet to nonnegtvty ondtons on prolem vrles nd sl vrles. For ths to heve we frst trnsform the system of equtons nto n equvlent system of equtons erng the sme soluton set suh tht the system s eser to solve. We hve seen so fr tht the effetve wy to del wth lner progrms s to otn the redued row ehelon form for the omned system of equtons norportng ojetve equton nd onstrnt equtons. We wll see tht for the nonlner se the effetve wy to del wth s to otn the equvlent of redued row ehelon form for the set of polynomls, nmely, the Groner ss representton for ths system of equtons (.)-(.). We then set up the equtons otned y equtng the prtl dervtves of d wth respet to prolem vrles x nd sl vrles s to zero nd utlze the stndrd theory nd methods used n lulus. We demonstrte the essene of ths method y solvng n exmple: These exmples re ten from [8], [9]. These exmples suffently llustrte the power of ths new method of usng powerful tehnque of Groner ss to suessfully nd effently del wth nonlner progrmmng prolems. Exmple.: xmze: x 4x x Sujet to: x x 4 x x 5 x 4x Soluton: We uld the followng system of equtons: 9
30 x 4x x d = 0 x x s 4 = 0 x x s 5 = 0 x 4x s = 0 x x, s, s, s suh tht:, 0 We now trnsform the nonlner/lner polynomls on the left hnd sde of the ove equtons y otnng Groner ss for them s follows: 486 8d 8s 6s 6s 8ss s = 0 (..) 9 9s 5s s = 0 (..) 9 s s 9x = 0 (..) 8 4s s 9x = 0 (..4) d d Settng = 0 nd = 0 we get equtons: s s s 8s = 8 8s s = 6 d rn defent system. Note tht for mxmzton of d f we set = 0 s the vlue of s tht mxmzes d, nmely, s = ( 8/ ) (8s /), negtve vlue for ny nonnegtve vlue of s. So, we set s = 0. Smlrly, for d mxmzton of d f we set = 0 we get the vlue of s tht mxmzes d, s nmely, s = 8 4s( = 8), settng s = 0. But, y settng s = 0 n the we get seond equton ove the lrgest possle vlue for s tht one n hve (s otned y settng s = 0nd t) s 9, when s = 0. Thus, settng s = 0, s = 9 n the frst equton we get d = 9. From thrd nd fourth equton we get x =, x =. Exmple.: xmze: 8x 6x 4x 56x Sujet to: x x 4 x x 5 x 4x x, x 0 Soluton: We uld the followng system of equtons: 8x 6x 4x 56x d = 0 0
31 x s 4 x s 5 x s x = x = x 4 = We now trnsform the nonlner/lner polynomls on the left hnd sde of the ove equtons y otnng Groner ss for them s follows: 504 9d 8s 6s 56s 8s = 0 (..) 9 9s 5s s = 0 (..) 9 s s 9x = 0 (..) 8 4s s 9x = 0 (..4) from frst equton (..), n order to mxmze d, we determne the vlues of s,s s follows: d If we set = 0 s Smlrly, f we set = 0 s 7 we get the vlue of s tht mxmzes d, nmely, d 4 s =. we get the vlue of s tht mxmzes d, nmely, s =. Puttng these vlues of s,s n the frst nd seond equton we get respetvely the mxmum vlue of d = 67 nd the vlue of s = 4 =.5, x. Usng further these vlues n the thrd nd fourth equton we get x =. 75. Exmple.: nmze: ( x ) ( x 4) Sujet to: x x = Soluton: We form the ojetve equton nd onstrnt equtons s s done n the ove exmples nd then fnd the Groner ss whh yelds: 5x d x 0 = 0 x x = 0 d Settng = 0 we get the vlue of x Ths yelds = 9. 8 d nd x =. 6 Exmple.4: nmze: x x Sujet to: x x = 6 x x tht mnmzes d, nmely, = 0. x.
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