A New Algorithm Linear Programming

Transcription

1 A New Algorthm ner Progrmmng Dhnnjy P. ehendle Sr Prshurmhu College, Tlk Rod, Pune-400, Ind Astrct In ths pper we propose two types of new lgorthms for lner progrmmng. The frst type of these new lgorthms uses lgerc methods whle the second type of these new lgorthms uses geometrc methods. The frst type of lgorthms s sed on tretng the ojectve functon s prmeter. In ths method, we form mtrx usng coeffcents n the system of equtons consstng ojectve equton nd equtons otned from nequltes defnng constrnt y ntroducng slck/surplus vrles. We otn reduced row echelon form for ths mtrx contnng only one vrle, nmely, the ojectve functon tself s n unknown prmeter. We nlyze ths mtrx n the reduced row echelon form nd develop cler cut method to fnd the optml soluton for the prolem t hnd, f nd when t exsts. We see tht the entre optmzton process cn e developed through the proper nlyss of the sd mtrx n the reduced row echelon form. The second type of lgorthms tht we propose for lner progrmmng re nspred y geometrcl consdertons nd use smple mthemtcs relted to fndng ntersectons of lnes nd plnes. All these lgorthms hve common m: they ll try to pproch closer nd closer to centrod or some centrlly locted nteror pont for speedng up the process of rechng n optml soluton! Imgne the lne prllel to vector C, where C T x denotes the ojectve functon to e optmzed, nd further suppose tht ths lne s lso pssng through the pont representng optml soluton. The new lgorthms tht we propose n ths pper essentlly try to rech t some fesle nteror pont whch s n the close vcnty of ths lne, n successve steps. When one wll e le to rrve fnlly t pont elongng to smll neghorhood of some pont on ths lne then y movng from ths pont prllel to vector C one cn rech to the pont elongng to the suffcently smll neghorhood of the pont representng optml soluton. We then proceed to show tht the lgerc method developed ove for lner progrmmng nturlly extends to nonlner nd nteger progrmmng prolems. For nonlner nd nteger progrmmng prolems we use the technque of Groner ses (snce Groner ss s n equvlent of reduced row echelon form for system of nonlner equtons) nd the methods of solvng lner Dophntne equtons (snce the nteger progrmmng prolem demnds for optml nteger soluton) respectvely.. Introducton: There re two types of lner progrms (lner progrmmng prolems):. xmze: C T x Suject to: Or. nmze: C T x Ax x 0

2 Suject to: Ax x 0 where x s column vector of sze n of unknown vrles. We cll these vrles the prolem vrles where C s column vector of sze n of proft (for mxmzton T prolem) or cost (for mnmzton prolem) coeffcents, nd C s row vector of sze n otned y mtrx trnsposton of C. where A s mtrx of constrnts coeffcents of sze m n. where s column vector of constnts of sze m representng the oundres of constrnts. By ntroducng the pproprte slck vrles (for mxmzton prolem) nd surplus vrles (for mnmzton prolem), the ove mentoned lner progrms gets converted nto stndrd form s: xmze: C T x Suject to: Ax s = (.) x 0, s 0 where s s slck vrle vector of sze m. Ths s mxmzton prolem. Or nmze: C T x Suject to: Ax s = (.) x 0, s 0 where s s surplus vrle vector of sze m. Ths s mnmzton prolem. In geometrcl lnguge, the constrnts defned y the nequltes form regon n the form of convex polyhedron, regon ounded y the constrnt plnes, Ax =, nd the coordnte plnes. Ths regon s clled fesle regon nd t s strghtforwrd to estlsh tht there exsts t lest one vertex of ths polyhedron t whch the optml soluton for the prolem s stuted when the prolem t hnd s well defned,.e. nether nconsstent, 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 constrnt plnes s prllel to the ojectve plne we my hve multtude of optml solutons. The ponts on n entre plne or n entre edge cn consttute the optml soluton set. These prolems re hndled most populrly y usng the well known smplex lgorthm or some of ts vrnt. Despte ts theoretcl exponentl complexty the smplex method works qute effcently for most of the prctcl prolems. However, there re few computtonl dffcultes ssocted wth smplex lgorthm. In order to vew them n nutshell we egn wth sttng some common notons nd defntons tht re prevlent n the lterture. A vrle x s clled sc vrle n gven equton f t ppers wth unt coeffcent n tht equton nd wth zero coeffcents n ll other equtons. A vrle whch s not sc s clled nonsc vrle. A sequence of elementry row opertons tht chnges gven system of lner equtons nto n equvlent system (hvng the sme soluton set) nd n whch gven nonsc vrle cn

3 e mde sc vrle s clled pvot operton. An equvlent system contnng sc nd nonsc vrles otned y pplcton of sutle elementry row opertons s clled cnoncl system. At tmes, the ntroducton of slck vrles for otnng stndrd form utomtclly produces cnoncl system, contnng t lest one sc vrle n ech equton. Sometmes sequence of pvot opertons s needed to e performed to get cnoncl system. The soluton otned from cnoncl system y settng the nonsc vrles to zero nd solvng for the sc vrles s clled sc soluton nd n ddton when ll the vrles hve nonnegtve vlues the soluton stsfyng ll the mposed constrnts s clled sc fesle soluton. Smplex method cnnot strt wthout n ntl sc fesle soluton. The process of fndng such soluton, whch s necessty n mny of prctcl prolems, s clled Phse I of the smplex lgorthm. Smplex method strts ts Phse II wth n ntl sc fesle soluton n cnoncl form t hnd. Then smplex tests whether ths soluton s optml y checkng whether ll the vlues of reltve profts (profts tht result due to unt chnge n the vlues of nonsc vrles) of ll the nonsc vrles re nonpostve. When not optml, the smplex method otns n djcent sc fesle soluton y selectng nonsc vrle hvng lrgest reltve proft to ecome sc. Smplex then determnes nd crres out the extng of sc vrle, y the so clled mnmum rto rule, to chnge t nto nonsc vrle ledng to formton of new cnoncl system. On ths new cnoncl system the whole procedure s repeted tll one rrves t n optml soluton. The mn computtonl dffcultes of the smplex method whch my cuse the reducton n ts computtonl effcency re s follows: ] There cn e more thn one nonsc vrle wth lrgest vlue for reltve proft nd so te cn tke plce whle selectng nonsc vrle to ecome sc. The choce t ths stuton s done rtrrly nd so the choce mde t ths stge cusng lrgest possle per unt mprovement s not necessrly the one tht gves lrgest totl mprovement n the vlue of the ojectve functon nd so not necessrly mnmzes the numer of smplex tertons. ] Whle pplyng mnmum rto rule t s possle for more thn one constrnt to gve the sme lest rto cusng te n the selecton of sc vrle to leve for ecomng nonsc. Ths degenercy cn cuse further complcton, nmely, the smplex method cn go on wthout ny mprovement n the ojectve functon nd the method my trp nto n nfnte loop nd fl to produce the desred optml soluton. Ths phenomenon s clled cyclng whch enforces modfcton n the lgorthm y ntroducng some ddtonl tme consumng rules tht reduce the effcency of the smplex lgorthm. ] Smplex s not effcent on theoretcl grounds sclly ecuse t serches djcent sc fesle solutons only nd ll other smplex vrnts whch exmne nondjcent solutons s well hve not shown ny pprecle chnge n the overll effcency of these modfed smplex lgorthms over the orgnl lgorthm. Becuse of the fr gret prctcl mportnce of the lner progrms nd other smlr prolems n the opertons reserch t s most desred thng to hve n lgorthm whch works n sngle step, f not, n s few steps s possle. No method hs een found whch wll yeld n optml soluton to lner progrm n sngle step ([], Pge 9). We m to propose n lgorthm for

4 lner progrmmng whch ms t fulfllng ths requrement n est possle nd novel wy.. Two Types of New Algorthms for ner Progrmmng: In ths secton we propose nd dscuss two types of new lgorthms for solvng lner progrmmng prolems. The frst type of lgorthms use lgerc methods whle the second type of lgorthms use geometrc methods. We egn wth frst type of lgorthms sed on tretng ojectve functon s prmeter nd ms t fndng ts optml vlue. We strt wth the followng equton: C T x = d (.) where d s n unknown prmeter, nd cll t the ojectve equton. The (prmetrc) plne defned y ths equton wll e clled ojectve plne. Plese note tht we re dscussng frst the mxmzton prolems. A smlr pproch for mnmzton prolems wll e dscussed next. Gven mxmzton prolem, we frst construct the comned system of equtons contnng the ojectve equton nd the equtons defned y the constrnts mposed y the prolem under consderton, comned 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 column vector F to mtrx E s lst column. We then fnd R, the reduced row echelon 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 reduced row echelon form R contns only one prmeter, nmely, d nd ll other entres re constnts. From R we cn determne the soluton set S for every fxed x d, S = { /( fxed) d rels} s. The suset of ths soluton set of vectors x whch lso stsfes the nonnegtvty constrnts s the set of ll fesle s solutons for tht d. It s cler tht ths suset cn e empty for prtculr choce of d tht s mde. The mxmzton prolem of lner progrmmng s to determne the unque d whch provdes fesle soluton nd hs mxmum 4

5 vlue for d,.e., to determne the unque d,.e. the unque optml vlue for the ojectve functon, whch cn e used to otn n optml soluton. In the cse of n unounded lner progrm there s no upper (lower, n the cse of mnmzton prolem) lmt for the vlue of d, whle n the cse of n nfesle lner progrm the set of fesle solutons s empty. The steps tht wll e executed to determne the optml soluton should lso tell y mplcton when such optml soluton does not exst n the cse of n unounded or nfesle prolem. The generl form of the mtrx R representng the reduced row echelon form s R = ( m m ) m ( m ) n n n mn ( m ) n m ( m ) m ( m ) m m m mm ( m ) n c ( m ) c d e c m c d e d e d e c d e m ( m ) The frst n columns of the ove mtrx represent the coeffcents of the prolem vrles (.e. vrles defned n the lner progrm) x, x,, xn. The next m columns represent the coeffcents of the slck vrles s, s,, s m used to convert nequltes nto equltes to otn the stndrd form of the lner progrm. The lst column represents the trnsformed rght hnd sde of the equton (.) durng the process ( sutle sequence of trnsformtons) tht s crred out to otn the reduced row echelon form. Note tht the lst column of R contns the lner form d s prmeter whose optml vlue s to e determned such tht the nonnegtvty constrnts remn vld,.e. x 0, n nd s j 0, j m. Among frst ( n m) columns of R certn frst columns correspond to sc vrles (columns tht re unt vectors) nd the remnng ones to nonsc vrles (columns tht re not unt vectors). For solvng lner progrm y our wy we proceed wth nlyss of the R. We m to fnd tht vlue of prmeter d whch s optml. To cheve ths tsk we my sometmes need to trnsform ths system further y ether rerrngng the constrnt equtons y sutly permutng these equtons such tht (s fr s possle) the vlues n the columns of our strtng mtrx A ( m n ) get rerrnged n the followng wy: Ether they re rsng nd then ecomng sttonry, or fllng nd then ecomng sttonry, or fllng ntlly up to certn length of the column vector nd then rsng gn. After ths rerrngement to form trnsformed A ( m n ) we gn proceed to form ts correspondng [E, F] nd gn fnd ts R = rref ([E, F]) whch wll most lkely hve the desred representton n whch columns for nonsc vrles contn 5

6 nonnegtve vlues. The de of rrngement of the columns of A ( m n ) s mentoned ove s purely heurstc nd s sed on the fvorle outcome oserved durng pplyng ths de to lner progrmmng prolems whle tcklng them. We hve not tred to dscover theoretcl resonng ehnd chevng fvorle form for R n most of the cses nd hve left ths prolem for the reder to solve nd to fnd out the theoretcl resonng ehnd gettng ths fvorle form. Now, even fter rerrngement of the columns of A ( m n ) s mentoned ove f stll some negtve entres remn present n the columns of R correspondng to some nonsc vrles then we crry out sutle elementry row trnsformtons on the otned R = rref ([E, F]) so tht the columns of coeffcents ssocted wth these nonsc vrles ecome nonnegtve. We re dong ths ecuse s wll e seen elow we cn then put zero vlue for these nonsc vrles nd cn determne the vlues of ll the sc vrles nd the lner progrm wll then e solved completely. It s esy to check tht for lner progrm f ll the coeffcents of prmeter d n the lst column of R re postve then the lner progrm t hnd s unounded snce n such cse the prmeter d cn e ncresed rtrrly wthout voltng the nonnegtvty constrnts on vrles x, s j. Also, for lner progrm f ll the coeffcents of some nonsc vrle represented y column of R re nonpostve nd re strctly negtve n those rows hvng negtve coeffcent to prmeterd tht ppers n the lst column of these rows then gn the prolem elongs to the ctegory of unounded prolems snce we cn ncrese the vlue of d to ny hgh vlue wthout voltng the nonnegtvty constrnts for the vrles y ssgnng suffcently hgh vlue to ths nonsc slck vrle. Note tht the rows of R ctully offer expressons for sc vrles n terms of nonsc vrles nd terms of type ckd ek, k =,, ( m ) contnng the prmeter d on the rght sde. The rows wth postve coeffcent for the prmeter d represent those equtons n whch the prmeter d cn e ncresed rtrrly wthout voltng the nonnegtvty constrnts on vrles x,. So, these equtons s j wth postve coeffcent for the prmeter d re not mplyng ny upper ound on the mxmum possle vlue of prmeter d. However, these rows re useful n certn stutons s they re useful to fnd lower ound on the vlue of prmeter d. The rows wth negtve coeffcent for the prmeter d represent those equtons n whch the prmeter d cnnot e ncresed rtrrly wthout voltng the nonnegtvty constrnts on vrles x,. So, these equtons s j wth negtve coeffcent 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 ctully every row of R s offerng us vlue for prmeter d whch cn e otned y equtng to zero ech term of the type ckd ek, k =,, ( m ). Those vlues of d tht we otn n ths wy wll e denoted s d or d when then vlue of c k s negtve or postve 6

7 7 respectvely. We denote y mn{ d }, the mnmum vlue mong d, nd we denote y mx{ d } the mxmum vlue mong the d. We now proceed to fnd out the sumtrx of R, sy N R, mde up of ll columns of R nd contnng those rows j of R for whch the coeffcents j c of the prmeter d re negtve. et k c c c,,, coeffcent of d n the rows of R whch re negtve. We collect these rows wth negtve coeffcent for d to form the mentoned sumtrx, N R, of R gven elow. Wth ths t s cler tht coeffcents of d n ll other rows of R re greter thn or equl to zero. = k k k k k k k k m n m n m n m n N e d c e d c e d c e d c R It should e cler to see tht f N R s empty (.e. not contnng sngle row) then the prolem t hnd s unounded. Among the frst ) ( m n columns of N R frst n columns represent the coeffcents of prolem vrles nd next m columns represent the coeffcents of slck vrles. There re certn columns strtng from frst column nd pper n successons whch re unt vectors. These columns whch re unt vectors correspond to sc vrles. The columns pperng n successons fter these columns nd not unt vectors correspond to nonsc vrles. As mentoned, mong the columns for nonsc vrles those hvng ll entres nonnegtve cn only led to decrement 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 cn sfely set the vlue of such vrles equl to zero. When ll columns correspondng to nonsc vrles n N R re hvng ll entres nonnegtve nd further f mn{ d } mx{ d } then we cn set ll nonsc vrles to zero, set d = mn{ d } n every row of R nd fnd the sc fesle soluton whch wll e optml, wth mn{ d } s optml vlue for the ojectve functon t hnd. Stll further, When ll columns correspondng to nonsc vrles n N R re hvng ll entres nonnegtve ut mn{ d } < mx{ d } then f k e > 0 then we cn stll set ll nonsc vrles to zero, set d = mn{ d } n every row of R nd fnd the sc

8 fesle soluton whch wll e optml, wth mn{ d } s optml vlue for the ojectve functon t hnd,.e. f vlue of e k > 0 n the expressons c kd ek n the rows of R other those n R N tht re hvng vlue of c k > 0 then we cn proceed on smlr lnes to fnd optml vlue for d. In R N we now proceed to consder those nonsc vrles for whch the columns of R N contn some (t lest one) postve vlues nd some negtve (t lest one) vlues. In such cse when we ssgn some postve vlue to such nonsc vrle t leds to decrese n the vlue of d n those rows n whch c k > 0 nd ncrese n the vlue of d n those rows n whch c k < 0. We now need to consder the wys of delng wth ths stuton. We del wth ths stuton s follows: In ths cse, we choose nd crry out pproprte nd legl elementry row trnsformtons on the mtrx R n the reduced row echelon form to cheve nonnegtve vlue for ll the entres n the columns correspondng to nonsc vrles n the sumtrx R N of R. The elementry row trnsformtons re chosen to produce new mtrx whch 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 equvlence we cn now set ll the nonsc vrles n ths trnsformed mtrx to zero nd otn wth justfcton d mn = mn{ d } s optml vlue for the ojectve functon nd otn sc fesle soluton s optml soluton y susttuton. et us now dscuss our new lgorthm n steps: Algorthm. (xmzton):. Express the gven prolem n stndrd form: xmze: C T x Suject to: Ax s = x 0, s 0. Construct the ugmented mtrx [E F], where T C ( n) 0( m) E = A I d, nd F = ( m n) ( m m) nd otn the reduced row echelon form: R = rref ([E, F]). If there s row (or rows) of zeroes t the ottom of R n the frst n columns nd contnng nonzero constnt n the lst column then declre tht the prolem s nconsstent nd stop. Else f the coeffcents of d n the lst column re ll postve or f there exsts column of R correspondng to some nonsc vrle wth ll entres negtve then declre tht the prolem t hnd s unounded nd stop. 8

9 4. Else f for ny vlue of d one oserves tht nonnegtvty constrnt for some vrle gets volted y t lest one of the vrles then declre tht the prolem t hnd s nfesle nd stop. 5. Else fnd the sumtrx of R, sy R N, mde up of those rows of R for whch the coeffcent of d n the lst column s negtve. 6. Check whether the columns of R N correspondng to nonsc vrles re nonnegtve. Else, rerrnge the constrnt equtons y sutly permutng these equtons such tht (s fr s possle) the vlues n the columns of our strtng mtrx A ( m n ) get rerrnged n the followng wy: Ether they re rsng nd then ecomng sttonry, or fllng nd then ecomng sttonry, or fllng ntlly up to certn length of the column vector nd then rsng gn. After ths rerrngement to form trnsformed A ( m n ) gn proceed s s done n step ove to form ts correspondng ugmented mtrx [E, F] nd gn fnd ts R = rref ([E, F]) whch wll most lkely hve the desred representton,.e. n the new R N tht one wll construct from the new R wll hve columns for nonsc vrles whch wll e contnnng nonnegtve entres. 7. Solve c r d e r = 0 for ech such term n the lst column of R N nd fnd the vlue of d = d r for r =,,, k nd fnd d mn = mn{ d } r. Smlrly, solve c r d e r = 0 for ech such term n the lst column for rows of R other thn those n R N nd fnd the vlues d = d r for r =,,, k nd fnd d mx = mx{ d } r. Check the columns of R N correspondng to nonsc vrles. If ll these columns contn only nonnegtve entres nd f mn{ d } mx{ d } then set ll nonsc vrles to zero. Susttute d = dmn n the lst column of R. Determne the sc fesle soluton whch wll e optml nd stop. Further, f columns correspondng to nonsc vrles contn only nonnegtve entres nd f mn{ d } < mx{ d } then check whether vlue of e k > 0 n the expressons c kd ek n these rows of R other those n R N tht re hvng vlue of c k > 0. If yes, then set ll nonsc vrles to zero. Susttute d = dmn n the lst column of R. Determne the sc fesle soluton whch wll e gn optml. 8. Even fter proceedng wth rerrngng columns of our strtng mtrx A ( m n ) y sutly permutng rows representng constrnt equtons s s 9

10 done n step 6, nd then proceedng s per step 7 nd chevng the vldty of mn{ d } mx{ d }, f stll there remn columns n R N correspondng to some nonsc vrles contnng postve entres n some rows nd negtve entres n some other rows then devse nd pply sutle elementry row trnsformtons on R such tht the columns representng coeffcents of nonsc vrles of new trnsformed mtrx R or t lest ts sumtrx R correspondng to new trnsformed R contn only nonnegtve entres. N And step 7 ecomes pplcle. We now proceed wth some exmples: Exmple.: xmze: Suject 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, clerly, R N = [, 0, 0, 0, /, -d6 ] [ 0, 0,, 0, /, 0-*d ] [ 0, 0, 0,,, -*d ] y For ths exmple the column formng coeffcents for nonsc vrle s contns nonegtve numers. So, we set s = 0. Clerly, d mn =. = Optml vlue for the ojectve functon. Usng ths vlue of optmum we hve x =.66, y = 0.66, s =, s =, s 0. 0 = Exmple.: We frst consder the duel of the exmple suggested y E... Bele [], whch rngs nto exstence the prolem of cyclng for the smplex method, nd provde soluton s per the ove new method whch offers t drectly wthout ny cyclng phenomenon. xmze: 0.75 x 0 x 0.5 x 6 x 4 Suject to: 0.5 x 8 x x 9 x x x 0.5 x 4 x 0 0

11 x 0 x 0 x, x, x, 4 Soluton: For ths prolem we hve the followng 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, clerly, 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 successve rows of R y R(), R(), R(), R(4). We chnge () () () 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 columns for ll nonsc vrles, whch re now those correspondng to x 4, s,s. So, y settng x 4 = s = s = 0 nd settng ll expressons of type c d e = 0 n the lst column we fnd r r d mn = mn{ d } =.5. Usng ths vlue n the lst column 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 consder n unounded prolem. The new method drectly mples the unounded nture of the prolem through the postvty of the coeffcents of d n mtrx R for the prolem. xmze: x y Suject to: x y x y 0 x y

12 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 ] [ 0, 0,, 0, -4/5, (/5)d-4/5] [ 0, 0, 0,, /5, /5(/5)d] Here, ll the coeffcents of d re postve. So, y settng vrle s = 0 we cn see tht we cn ssgn ny rtrrly lrge vlue to vrle d wthout volton of nonnegtvty constrnts for vrles. Thus, the prolem hs n unounded soluton. Exmple.4: We now consder prolem hvng n nfesle strtng ss. We see tht new lgorthm hs no dffculty to del wth t. xmze: x y Suject 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 successve rows of R y R(), R(), R(), R(4). We chnge () 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 ]

13 R, RN correspond to sc vrles x, y. Snce, The frst two columns of mn{ d } mx{ d } nd columns correspondng to nonsc vrles s,s contn nonnegtve entres n we hve s = 0 nd mn x =.857, y = R N, so we set these vrles to zero. From lst row d = 9. Also from frst nd second rows,.074 Exmple.5: We now consder n nfesle prolem. xmze: x y Suject 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 coeffcent 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 successve rows of R y R(), R(), R(), R(4). We chnge () 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 consstency of lst row d = (8/) nd usng ths vlue for d we hve y = (/). Thus, ths prolem s nfesle. Remrk.: Klee nd nty [4], hve constructed n exmple of set of lner n progrms wth n vrles for whch smplex method requres tertons to rech n optml soluton. Theoretc work of Borgwrdt [5] nd Smle [6] ndctes tht fortuntely the occurrence of prolems elongng to the clss of Klee nd nty,

14 whch don t shre the verge ehvor, s so rre s to e neglgle. We now proceed to show tht there s no prolem of effcency for new lgorthm n delng wth the prolems elongng to ths clss. Exmple.6: We now consder prolem for whch the smplex tertons re exponentl functon of the sze of the prolem. A prolem elongng to the clss n descred y Klee nd nty contnng n vrles requres smplex steps. We see tht the new method doesn t requre ny specl effort xmze: 00 x 0 x x Suject to: x 0 x x x 0 x x 0000 x, x, x 0 Soluton: The followng re the mtrces 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] We perform followng elementry row trnsformtons on R: et us denote the successve rows of R y R(), R(), R(), R(4). We chnge () () () 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, ] Snce, mn{ d } mx{ d } nd columns correspondng to nonsc vrles x,s contn nonnegtve entres n R N, so we set these vrles to zero. We get esy complete soluton s follows: 4

15 x = 0, x = 0, x = 0000, s =, s = 00, s = 0, nd the mxmum vlue of d = Exmple.7: We now consder n exmple for whch the reduced row echelon form contns y tself the sc 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 whch only nonpostve entres occur n the ottom row of the tleu representng reltve profts. Ths s understood y the nonnegtvty of entres n the columns of R N correspondng to nonsc vrles. xmze: x y Suject to: 4 x y 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, clerly, R N = [ 0, 0,, 0, 8/5, 44/5-(/5)d] [ 0, 0, 0,, 0, 4-d ] Snce, mn{ d } mx{ d } nd columns correspondng to nonsc vrle R contn nonnegtve entres n N, so we set these vrles to zero. Here, the nonsc vrle columns drectly contn nonnegtve entres ledng to decresng n proft when some postve vlue s ssgned to ths vrle nd so we set ths vrle to zero whch leds to the mxml sc fesle soluton: d = 4, x = 4, y =, s = 6, s = 0, s = 0. We now consder few exmples n whch rerrngement of constrnt equtons utomtclly produce sutle form for reduced row echelon form: 6 x y Exmple.8:. xmze: Suject to: x y 5 4x y x y 4 x, y 0 For ths prolem we get followng R, R N s 5

16 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] 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 column correspondng to nonsc vrle re mxed type,.e. entres n the frst nd thrd rows of R N re negtve whle entry n the second row of R N s postve. We now rerrnge the constrnts so tht ether entres n the new columns re rsng nd then ecomng sttonry, or fllng nd then ecomng sttonry, or fllng ntlly up to certn length of the column vector nd then rsng gn, s mentoned ove. Thus, we just rewrte the prolem s: 6 x y xmze: Suject to: x y 5 x y 4 4x y x, y 0 then form new A ( m n ) mtrx whch produce nd gn proceed to fnd reduced row echelon 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] Snce, mn{ d } mx{ d } nd columns correspondng to nonsc vrle R contn nonnegtve entres n N, so we set these vrles to zero. Here, the nonsc vrle columns drectly contn nonnegtve entres ledng to decresng n proft when some postve vlue s ssgned to ths vrle nd so we set ths vrle to zero whch leds to the mxml sc fesle soluton: d =, x = /, y = 8 /, s = 0, s = /, s 0. 7 = x y Exmple.9: xmze: Suject to: x y 4 s 6

17 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 ] [ 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 column correspondng to nonsc vrle re mxed type,.e. entres n the frst nd thrd rows of R N re negtve whle entry n the second row of R N s postve. We now rerrnge the constrnts so tht ether entres n the new columns re rsng nd then ecomng sttonry, or fllng nd then ecomng sttonry, or fllng ntlly up to certn length of the column vector nd then rsng gn, s mentoned ove. Thus, we just rewrte the prolem s: x y xmze: Suject to: x y 5 x 4y x y 4 x, y 0 then we form new A ( m n ) new mtrx whch produce nd gn proceed to fnd reduced row echelon 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] Snce, mn{ d } mx{ d } nd column correspondng to nonsc vrle R contn nonnegtve entres n N, so we set ths vrle to zero. Here, the nonsc vrle columns drectly contn nonnegtve entres ledng to decresng n proft when some postve vlue s ssgned to ths vrle nd so we set ths vrle to zero whch leds to the mxml sc fesle soluton: d = 9, s 7

18 x =, y =, s = 0, s = 9, s 0. = 4 x y Exmple.0: xmze: Suject to: x.5 y 9 x y 8 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 second lst column correspondng to nonsc vrle contns nonnegtve entres the nequlty mn{ d } mx{ d } s nvld! So, s frst ttempt, efore strtng to crry out elementry row trnsformton on R to cheve nonnegtvty of entres n the columns of R correspondng nonsc vrles, let us frst try rerrngng the nequltes so tht ether entres n the new columns re rsng nd then ecomng sttonry, or fllng nd then ecomng sttonry, or fllng ntlly up to certn length of the column vector nd then rsng gn, s mentoned ove. Thus, we just rewrte the prolem s: xmze: 4 x y Suject to: x.5 y 9 x y 6 x y 8 x, y 0 then we form new A ( m n ) new mtrx whch produce nd gn proceed to fnd reduced row echelon 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 second lst column correspondng to nonsc vrle contns nonnegtve entres nd lso the nequlty mn{ d } mx{ d } s now 8

19 vld! So, we set nonsc vrle s equl to zero whch leds to mxml sc fesle soluton: d =7. 667, x =.665, y =.667, s = 0, s =.665, s Exmple.: xmze: Suject to: 4 = x y x y x y 5 x 4y 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 constrnts s: (I) constrnt constrnt, (II) constrnt constrnt, nd (III) constrnt constrnt 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 ] Snce, mn{ d } mx{ d } nd column correspondng to nonsc vrle R contn nonnegtve entres n N, so we set ths vrle to zero nd we drectly get the optml sc fesle soluton from R = : d = d mn = 9, x =, y =, s = 0, s = 9, s = 0. We now see tht we cn proceed wth n exctly smlr wy nd del successfully wth mnmzton lner progrmmng prolems. A prolem of mnmzton goes lke: nmze: C T x 0 s 9

20 Suject to: Ax x 0 we frst construct the comned system of equtons contnng the sme ojectve equton used n mxmzton prolem (ut ths tme we wnt to fnd mnmum vlue of prmeter d defned n the ojectve equton) nd the equtons defned y the constrnts mposed y the prolem under consderton, comned 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 column vector F to mtrx E s lst column. We then fnd R, the reduced row echelon form of the ove ugmented mtrx [E, F]. Thus, R = rref ([E, F]) (.) Note tht the ugmented mtrx [E, F] s well s ts reduced row echelon form R contns only one prmeter, nmely, d nd ll other entres re constnts. From R we cn determne the soluton set S for every fxed x d, S = { /( fxed) d rels} s. The suset of ths soluton set of vectors x whch lso stsfes the nonnegtvty constrnts s the set of ll fesle s solutons for tht d. It s cler tht ths suset cn e empty for prtculr choce of d tht s mde. The mnmzton prolem of lner progrmmng s to determne the unque d whch provdes fesle soluton nd hs mnmum vlue,.e., to determne the unque d whch provdes n optml soluton. In the cse of n unounded mnmzton lner progrm there s no lower ound for the vlue of d, whle n the cse of n nfesle lner progrm the set of fesle solutons s empty. The steps tht wll e executed to determne the optml soluton should lso tell y mplcton when such optml soluton does not exst n the cse of n unounded or nfesle prolem. The generl form of the mtrx R representng the reduced row echelon form s smlr s prevously dscussed mxmzton cse: 0

21 R = ( m m ) m ( m ) n n n mn ( m ) n m ( m ) m ( m ) m m m mm ( m ) n c c d e c m ( m ) c d e d e d e c d e m ( m ) The frst n columns of the ove mtrx represent the coeffcents of the prolem vrles (.e. vrles defned n the lner progrm) x, x,, xn. The next m columns represent the coeffcents of the surplus vrles s, s,, s m used to convert nequltes nto equltes to otn the stndrd form of the lner progrm. The lst column represents the trnsformed rght hnd sde of the equton (.) durng the process ( sutle sequence of trnsformtons) tht s crred out to otn the reduced row echelon form. Note tht the lst column of R contns the lner form d s prmeter whose optml vlue s to e determned such tht the nonnegtvty constrnts remn vld,.e. x 0, n nd s j 0, j m. Among frst ( n m) columns of R certn frst columns correspond to sc vrles (columns tht re unt vectors) nd the remnng ones to nonsc vrles (columns tht re not unt vectors). For solvng the lner progrm we need to determne the vlues of ll nonsc vrles nd the optml vlue of d, from whch we cn determne the vlues of ll the sc vrles y susttuton nd the lner progrm s thus solved completely. For lner progrm f ll the coeffcents of prmeter d n the lst column of R re negtve then the lner progrm t hnd s unounded (snce, the prmeter d cn e decresed rtrrly wthout voltng the nonnegtvty constrnts on vrles x, s j ). For lner progrm f ll the coeffcents of some nonsc slck vrle represented y column of R re nonpostve nd re strctly negtve n those rows hvng postve coeffcent to prmeter d tht ppers n the lst column of these rows then we cn decrese the vlue of d to ny low vlue wthout voltng the nonnegtvty constrnts for the vrles y ssgnng suffcently hgh vlue to ths nonsc slck vrle nd the prolem s gn elongs to the ctegory of unounded prolems. Note tht the rows of R ctully represent equtons wth vrles x, =,, n nd vrles, j =,, m on left sde nd expressons of type s j ckd ek, k =,, ( m ) contnng the vrle d on the rght sde. The rows wth negtve coeffcent for the prmeter d represent those equtons n whch the prmeter d cn e decresed rtrrly wthout voltng the nonnegtvty constrnts on vrles x,. So, these equtons wth negtve s j coeffcent 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 know out upper ound on prmeter d. The rows wth postve coeffcent for the prmeter d represent those equtons n whch the prmeter d cnnot e decresed rtrrly wthout voltng the nonnegtvty constrnts on vrles j s x,. So, these equtons wth postve coeffcent for the prmeter d re mplyng lower ound on the mnmum possle vlue of prmeter d nd so mportnt ones n ths respect. So, we now proceed to fnd out the sumtrx of R, sy P R, mde up of ll columns of R nd contnng those rows j of R for whch the coeffcents j c of the prmeter d re postve. et k c c c,,, re ll nd re only postve rel numers n the rows collected n P R gven elow nd ll other coeffcents of d n other rows of R re greter thn or equl to zero. = k k k k k k k k n n m n m n m n P e d c e d c e d c e d c R If P R s empty (.e. contnng not sngle row) then the prolem t hnd s unounded. There re certn columns strtng from frst column nd pper n successons whch re unt vectors. These columns whch re unt vectors correspond to sc vrles. The columns pperng n successons fter these columns nd not unt vectors correspond to nonsc vrles. As mentoned, mong the columns of P R for nonsc vrles those hvng ll entres nonnegtve cn only led to ncrese 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 such vrles equl to zero. When ll columns correspondng to nonsc vrles n P R re hvng ll entres nonnegtve nd further f the nequlty mn{ d } mx{ d } holds then we cn set ll nonsc vrles to zero nd set d = mx{ d } n every row of R nd fnd the sc fesle soluton whch wll e optml, wth mx{ d } s optml vlue for the ojectve functon t hnd. Stll further, When ll columns correspondng to nonsc vrles n P R re hvng ll entres nonnegtve ut mn{ d } < mx{ d } then f k e > 0 then we cn stll set ll nonsc vrles to zero, set d = mx{ d } n every row of R nd fnd the sc

23 fesle soluton whch wll e optml, wth mx{ d } s optml vlue for the ojectve functon t hnd,.e. f vlue of e k > 0 n the expressons c kd ek n the rows of R rther those n R P tht re hvng vlue of c k > 0 then we cn proceed on smlr lnes to fnd optml vlue for d. In R P we now proceed to consder those nonsc vrles for whch the columns of R P contn some (t lest one) postve vlues nd some negtve (t lest one) vlues. In such cse when we ssgn some postve vlue to such nonsc vrle t leds to decrese n the vlue of d n those rows n whch c k > 0 nd ncrese n the vlue of d n those rows n whch c k < 0. We now need to consder the wys of delng wth ths stuton. We del wth ths stuton s follows: In ths cse, we choose nd crry out pproprte nd legl elementry row trnsformtons on the mtrx R n the reduced row echelon form to cheve nonnegtve vlue for ll the entres n the columns correspondng to nonsc vrles n the sumtrx R P of R. The elementry row trnsformtons re chosen to produce new mtrx whch 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 equvlence we cn now set ll the nonsc vrles n ths trnsformed mtrx to zero nd otn wth justfcton d mx = mx{ d } s optml vlue for the ojectve functon nd otn sc fesle soluton s optml soluton y susttuton. Algorthm. (nmzton):. Express the gven prolem n stndrd form: xmze: C T x Suject to: Ax s = x 0, s 0. Construct the ugmented mtrx [E F], where T C ( n) 0( m) d E =, nd F = A( m n) I ( m m) nd otn the reduced row echelon form: R = rref ([E, F]). If there s row (or rows) of zeroes t the ottom of R n the frst n columns nd contnng nonzero constnt n the lst column then declre tht the prolem s nconsstent nd stop. Else f the coeffcents of d n the lst column re ll postve or f there exsts column of R correspondng to some nonsc vrle wth ll entres negtve then declre tht the prolem t hnd s unounded nd stop.

24 4. Else f for ny vlue of d one oserves tht nonnegtvty constrnt for some vrle gets volted y t lest one of the vrles then declre 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 whch the coeffcent of d n the lst column s postve. 6. Check whether the columns of R P correspondng to nonsc vrles re nonnegtve. Else, rerrnge the constrnt equtons y sutly permutng these equtons such tht (s fr s possle) the vlues n the columns of our strtng mtrx A ( m n ) get rerrnged n the followng wy: Ether they re rsng nd then ecomng sttonry, or fllng nd then ecomng sttonry, or fllng ntlly up to certn length of the column vector nd then rsng gn. After ths rerrngement to form trnsformed A ( m n ) gn proceed s s done n step ove to form ts correspondng ugmented mtrx [E, F] nd gn fnd ts R = rref ([E, F]) whch wll most lkely hve the desred representton,.e. n the new R P tht one wll construct from the new R wll hve columns for nonsc vrles whch wll e contnng nonnegtve entres. 7. Solve c r d e r = 0 for ech such term n the lst column of R P nd fnd the vlue of d = d r for r =,,, k nd fnd d mn = mn{ d } r. Smlrly, solve c r d e r = 0 for ech such term n the lst column for rows of R other thn those n R N nd fnd the vlues d = d r for r =,,, k nd fnd d mx = mx{ d } r. Check the columns of R P correspondng to nonsc vrles. If ll these columns contn only nonnegtve entres nd f mn{ d } mx{ d } then set ll nonsc vrles to zero. Susttute d = d mx n the lst column of R. Determne the sc fesle soluton whch wll e optml nd stop. Further, f columns correspondng to nonsc vrles contn only nonnegtve entres nd f mn{ d } < mx{ d } then check whether vlue of e k > 0 n the expressons c kd ek n these rows of R other those n R P tht re hvng vlue of c k > 0. If yes, then set ll nonsc vrles to zero. d = d n the lst column of R. Determne the sc fesle Susttute mx soluton whch wll e gn optml. 8. Even fter proceedng wth rerrngng columns of our strtng mtrx A ( m n ) y sutly permutng rows representng constrnt equtons s s 4

25 done n step 6, nd then proceedng s per step 7 nd chevng the vldty of mn{ d } mx{ d }, f stll there remn columns n R P correspondng to some nonsc vrles contnng postve entres n some rows nd negtve entres n some other rows then devse nd pply sutle elementry row trnsformtons on R such tht the columns representng coeffcents of nonsc vrles of new trnsformed mtrx R or t lest ts sumtrx R P correspondng to new trnsformed R contn only nonnegtve entres so tht step 7 ecomes pplcle. We now consder few exmples for mnmzton prolems: Exmple.: Ths exmple for mnmzton s lke Exmple.6 for mxmzton n whch the reduced row echelon form contns y tself the sc 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 whch only nonnegtve entres occur n the ottom row of the tleu representng reltve costs. Ths s understood y the nonnegtvty of entres n the columns of R P correspondng to nonsc vrles. nmze: x x x Suject 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 cler from seprtely equtng ech entry n the lst column of R tht the expected nequlty, mn{ d } mx{ d }, holds good. Also, R = [ 0, 0, 0,,, 0,, 6d] P Snce ll the entres n the columns correspondng to nonsc 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 consder mnmzton prml lner progrm for whch nether the prml nor the dul hs fesle soluton. nmze: x y Suject 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 cler tht (for ny nonnegtve vlue of nonsc vrle, s ) the vlue of s s negtve nd so the prolem s thus nfesle. Smlrly, f we consder the followng dul, vz, x y x y x y x, y 0 xmze: Suject to: then we hve R = [, 0, 0,, -d] [ 0,, 0,, -4d] [ 0, 0,,, - ] R = [, 0, 0,, -d] N [ 0,, 0,, -4d] whch mples zero vlue for cler tht (even for ny nonnegtve vlue of nonsc 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 consder mnmzton lner progrm for whch the prml s unounded nd the dul s nfesle. nmze: x y Suject 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 nonsc vrle s we cn vod negtvty of s, so the prolem s unounded. Also, let us pply followng elementry trnsformtons on R: et us denote the successve rows of R y R(), R(), R(), R(4). We chnge () () 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 cn check tht ths prolem n unounded elow. Now, f we consder the followng dul, vz, xmze: 5x 5y Suject to: We hve x y 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 cler tht even for ny nonnegtve vlue of nonsc vrle, s the vlue of s s negtve nd so the prolem s nfesle. Exmple.5: nmze: 4 Suject 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 Snce t s cler from seprtely equtng ech entry n the lst column of R tht the expected nequlty, mn{ d } mx{ d }, holds good. Also, snce ll the entres n the columns correspondng to nonsc 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 Remrk.: It s cler from the dscusson mde so fr tht our m should e to cheve the stuton n whch ll the columns correspondng to nonsc vrles n R N, R P contn nonnegtve entres. Ths stuton corresponds to drectly hvng the possesson of mxml/mnml sc fesle soluton. By chevng ths one gets drectly the optml sc fesle soluton y smply settng ll the nonsc vrles to zero nd fndng the sc soluton. Ths s the stuton n sense of hvng drectly the optml smplex tleu for whch one sets ll the nonsc vrles to zero s they n fct led to decrement/ncrement n the otherwse mxml/mnml vlue of d for mxmzton/mnmzton lner progrmmng prolems under consderton. Thus, we cn fnd the soluton of ny mxmzton/mnmzton lner progrm y properly nlyzng the R, R N, R P mtrces, nd tkng the relevnt ctons s per the outcomes of the nlyss. Remrk.: Insted of shufflng rows of mtrx A, to form new E, F wthout chngng the content of the prolem, we cn lso cheve the sme effect of rerrngement of constrnt y smply shufflng rows of dentty mtrx I m m n E nd proceed wth formton of new E, F wthout chngng the content of the prolem nd chevng sme effect n the sense tht the correspondng reduced row echelon form wll utomtclly produce sme effect. Remrk.: The nonnegtvty of the entres tht re present n the columns of nonsc vrles of the concerned reduced row echelon form R N or R P s n effect smlr to otn the optml smplex tleu,.e. the tleu t whch the smplex lgorthm termntes nd where the sc fesle soluton represents the optml soluton. As mentoned n the egnnng of ths secton we now proceed wth the dscusson of second type of lgorthms tht use geometrc methods. We now dscuss some novel lgorthms for lner progrmmng nspred y geometrcl consdertons. All these lgorthms hve common m: they ll try to pproch closer nd closer to centrod or some centrlly locted nteror pont for speedng up the process of rechng n optml soluton! Imgne the lne prllel to vector C, where C T x denotes the ojectve functon to e optmzed, nd further suppose tht ths lne s lso pssng through the pont representng optml soluton. The new lgorthms tht we propose n ths pper essentlly try to rech t some fesle nteror pont whch s n the close vcnty of ths lne, n successve steps. When one wll e le to rrve fnlly t pont elongng to smll neghorhood of some pont on ths lne then y movng from ths pont prllel to vector C one cn rech to the pont elongng to the suffcently smll neghorhood of the pont representng optml soluton. 8

29 The method tht we propose now ttempts to utlze the geometrcl structure of the lner progrmmng prolem. As mentoned ove, n geometrcl lnguge every lner progrmmng prolem defnes convex polyhedron formed y ntersectng constrnt plnes nd coordnte plnes. The regon nsde of ths convex polyhedron s clled the fesle regon. It s mde up of fesle nteror ponts lyng nsde ths convex polyhedron. These so clled fesle nteror ponts stsfy ll oundry constrnts nd re nonnegtve. Solvng the mxmzton (mnmzton) lner progrmmng prolem n geometrcl terms essentlly conssts of pushng ths ojectve plne outwrds (nwrds) n the drecton of C such tht ths ojectve plne wll rech to the extreme end of the polyhedron nd wll contn the extreme vertex. Ths extreme vertex s ctully the optml soluton of the prolem t hnd. In geometrcl terms solvng lner progrmmng prolem mens to rrve t the pont representng extreme vertex, to determne coordntes of ths pont, nd further to fnd vlue of the ojectve functon (the so clled optml ojectve vlue) t ths pont. Becuse of the fr gret prctcl mportnce of the lner progrms nd other smlr prolems n the opertons reserch t s most desred thng to hve n lgorthm whch works n sngle step, f not, n s few steps s possle. No method hs so fr een found whch wll yeld n optml soluton to lner progrm n sngle step ([], Pge 9). We wsh to emphsze n ths pper tht when one s lucky enough to rrve t some nteror fesle pont elongng to the lne prllel to vector C whch s lso pssng through the pont representng optml soluton, where C T x denotes the ojectve functon to e optmzed, then one cn fnd the pont representng optml soluton to the lner progrm under consderton n sngle step y just movng long ths lne tll one reches the desred extreme vertex on ths lne t the oundry of the convex polyhedron!! We wsh to develop some NEW lgorthms for solvng lner progrmmng prolems. We now proceed wth ref descrpton of these lgorthms. We wll try to cpture the essence of the methods wth the help two geometrcl fgures, FIG. nd FIG. gven elow. We hope tht these fgures wll mke explct the geometrcl motvton tht les ehnd these lgorthms. Coordnte Plnes Strtng nteror pont C lne prllel to vector C through Optml Soluton Constrnt Plnes Interor pont ner oundry C per New strtng nteror pont Ojectve Plne Optml soluton Interor pont ner oundry FIG. 9