An Exact Resource Allocation Model with Hard and Soft Resource Constraints

Transcription

1 An Exac Resouce Allocaon Model wh Had and Sof Resouce Consans Sxeenh Annual Confeence of POMS, Chcago, IL, Apl 29 - May 2, Feenc Kuzslcz (kuzslc@kk.pe.hu) nvesy of Pécs, Depamen of Busness Infomacs, Rákócz ú 80, H-7623 Pécs, Hungay Phone: /33 Fax: Absac. In hs pape we pesen a mxed nege lnea pogammng (MILP) esouce allocaon model wh had and sof esouce consans fo pojecs. By defnon, a had esouce consan s no esolvable whn he gven plannng hozon, bu a sof esouce conflc may be managed by a flexble "hng-fng" saegy. A "well-balanced" opmal schedule fo he sof esouces s chaacezed by a new global bcea measue, namely he "peak esouce equemen" and he "dle me" smulaneously. The applcaon of he dle me measue fo had esouces s oponal, bu f used defnes a "smooh load" schedule. In he poposed appoach he goal funcon of he MILP model s defned as a weghed combnaon of he sngle cea measues. The MILP model s ceaed by auomaed ansfomaonal seps fom a non-lnea nal model based on he mmedae pecedence elaons of he GANTT dagam. The paccal nepeaon of he poposed model s demonsaed n an analyss of a smplfed small-scale busness sofwae developmen envonmen. In he pesened poblem we assumed ha he numbe of sofwae desgnes s a had consan whn he gven plannng hozon, bu he avalably of pogammes and eses ae sof consans, whch can be esolved by "hng-fng" n sho em base. In hs example we exploed he fac ha n a

2 small-scale busness sofwae developng pocess he usual eave "waefall" sucue mgh be eplaced by a seal "desgne-pogamme-ese" chan. Numbe: Inoducon Le P = ( A, E) denoe a pojec as a se of acves, and a se of mmedae pecedence elaons whch s an ndex subse of Acves A { A A,... } A A. The numbe of acves n pojec P s denoed by N. =, 2 A N ae non-dvdable, aomc elemens wh known esouce and me equemens. If acvy A mus be fnshed, befoe acvy A j s saed, s wen as (, j) E. We also have k+m dffeen classes of esouces wh cean consans, denoed by 2 k+ m { R, R R } R =,.... Fo evey R R esouce a R() esouce lm funcon s gven ha epesens he avalable amoun of esouce R a me. Wh hese noaons evey acvy can 2 k+ m be descbed wh a A = ( D, R, R,... R ) veco, whee D denoes he amoun of me, and R he amoun of esouce uns equed by acvy fo all R R. A schedule S = S, S,... S ) of a pojec s gven wh sa mes of A acves. Fo ( 2 N convenence le us noduce wo moe A 0 and A N + dummy acves whch epesen he begnnng and he compleon of he pojec. A 0 becomes he fs and A N + he las acvy of all schedules whou any me o esouce equemen. The followng exa mmedae pecedence elaons ae added o hese dummy asks: ( 0,) E and (, N +) E fo {,2,...N}. Wh hs noaons he S 0 sa me of A 0 s he sa me of he pojec and smlaly S N + he sa me of acvy A N + s he eales fnshng me of P. We can assume ha evey schedule sas a me S 0 =0. Wh hs assumpon he duaon of he pojec can be expessed as S N +, whch s he oal mplemenaon me of all he acves n P by he schedule S. Thoughou n hs pape 2

3 only hose schedules ae consdeed ha peseve he mmedae pecedence elaons, and he makespan of he pojec s denoed by T. Defnon Schedule S s sad o be feasble f nequaly S + D S holds fo evey (, j) E mmedae j pecedence ule. The se of all feasble schedules of pojec P s denoed by. If we wan o fnd he bes feasble schedule, some measues mus be noduced whch ae able o expess he effcency of a schedule. In hs pape a adonal peak esouce (M) measue and a new dle me (IT) one wll be used. Ou fs esouce levelng model s based only on hs IT measue, heefoe we sa wh s defnon. In case of machne schedulng poblems he dle me (IT) oened measues play he cenal ole. On he analogy of hs dea a new esouce dle me measue may be noduced fo acvy schedulng as well. As a fs appoach we can say ha a cean un of a esouce s unulzed a a pon of me, f s no assgned o any acves. The acves of allocaed esouces daw vey vaable vew dung a pojec. I s a naual equemen ha he pojec manage wans o make he load level equally smooh dung he pojec makespan. Bu because of unwaned neupons hs demand can no be sasfed n geneal. As a consequence he ulzaon dagam of some esouce ems conans hlls and valleys. Wokng and dle phases follow each ohe. Wha s he deal shape of a ulzaon dagam? nl now was commonly acceped ha ecangula shapes ae he bes. In spe of hs fom paccal expeences we found ha quasconcavy s he key wod. The mos mpoan aspec of he poposed dle me measue s ha concenaes he ulzaon aound a cean peod, by pefeng a hll shape. Ths knd of concenaed allocaon blocks can be combned n a vey flexble way, and he numbe of 3

4 neupons s educed as well. Nocng ha any ecangula shape s quas-concave we can conclude ha he new global dle me measue s a genealzaon Fgue : A ypcal quas-concave esouce ulzaon dagam n case of a sngle esouce Nex we gve he fomal defnon of a quas-concave esouce ulzaon pofle. Fo evey {, 2,..., T} le denoe he amoun of esouce ems used a me, by he schedule Defnon The {,,...,,...,, } = esouce ulzaon dagam s called quas-concave f 2 T- T ( ) fo all { 2, 3,..., T - } mn, +. S Σ. Beyond hs adonal defnon we need an equvalen fomulaon oo, whch s based on he wde envonmen of consecuve me peods. Fo hs eason we noduce he lef (and gh) sde maxmal envonmens ( MAXL, MAXR ) of MAXL = max(, 2,..., ), and MAXR max(,,..., T ). Defnon The {,,...,,...,, } 2 T- T = +. In case of evey {, 2,..., T} le = esouce ulzaon dagam s quas-concave, f he followng nequales hold: 4

5 ( MAXL MAXR ) fo all { 2, 3,..., T - } mn, +. Now le us pove he equvalence. Lemma The pevous defnons of quas-concavy of esouce uly dagams ae equvalen. Poof:. Fs le us suppose, ha he dagam s quas-concave by he defnon of maxmal envonmens. Fom hs we know, ha n case of { 2, 3,..., T - } he followng s ue: ( MAXL MAXR ) mn, +. By usng he defnon of he maxmal envonmen, o pecsely s monoony, he adonal fom can be deved decly: mn (, MAXL MAXR + ) ( MAXL, MAXR + ) = mn( max(, MAXL 2 ),max( +, MAXR )) ( max(, MAXL ),max(, MAXR )) mn(, ) mn mn Convesely s consdeed as a quas-concave dagam n he sense of he adonal defnon. We know ha fo all { 2, 3,..., T - } he ( ) +, he pevous nequaly ges he fom of mn, + holds. In case of. When he adonal defnon s appled agan and agan n deceasng ode of ndces, he esul s..., and accodngly MAXL, and he saemen mmedaely follows. If + we ge he nequaly chan +... T by smla agumens, hus MAXR+. Combnng he wo pas we ge ( MAXL MAXR ) mn, + whch was equed. 5

6 Defnon Afewads he hsogam {, 2,...,,..., T-, T } he esouce usage dagam, f: = = efeed as he quas-concave hull of = ( mn( MAXL, MAXR ) ) whee { 2, 3,..., T - } max +, T = T In fgue a ypcal non quas-concave esouce usage dagam s shown, wh s quas-concave hull. The dle esouce uns ae denoed by small squaes. In hs example all he esouces (machnes o employees) have a connuous ulzaon hough he fs 3 peods. Bu n peod 4 he second and he hd esouce uns ae whou any wokload. A smla suaon appeas fo he hd esouce un n peod 7. All ogehe he sum of all dle uns s 3 n hs sample. The dea of dle mes s spung fom machne schedulng poblems, and he penaly measue of dle uns may be fomed n dffeen ways concenng he pojec manage s pon of vew. One possble soluon fo a penaly measue s he numbe of neups (n geneal o by uns). Because we wan o exend ou fs esouce levelng model o a esouce allocaon model fo sof and had esouces, s easonable o defne he penaly measue as he numbe of dle un squaes. Le us denoe IT(S) he dffeence of he aea beween he hsogam of schedule S and s quas-concave hull. Ths defnon s especally useful when human esouces have o be scheduled, snce evey employee has he same cos, ndependenly whehe he s dle o no. I s vey easy o gve suggesve descpon of hs measue: how many un squaes have o be used o fll n he valleys of he hsogam. Whls he pevous descpve defnon s easy, s numecal calculaon s a dffcul ask. The coe of he poblem oos n he non smooh popey of he convex hull. 6

7 5 Quas-concave hull ns belong o dle esouce ems Fgue 2. Quas-concave hull of a esouce usage dagam and dle mes In spe of he numecal complexy of he IT measue s woh o use, snce has mpoan advanages:. When a esouce usage hsogam s quas-concave, hen dle mes ae concenaed a he begnnng and a he end of he ulzaon peod. Ths knd of cga shape bee goes by mulaskng n a mulpojec envonmen, snce he dffeen schedules of subpojecs may ovelap each ohe on he magns, and he makespan of he whole mulpojec may be shoened hs way. 2. The dle me based appoach a fs sgh s vey smla o he adonal (peak value o flucuaon based) measues. Howeve hey ae no he same. A pojec manage s usually no neesed n exeme values, bu ahe on effecve and adapable esouce allocaon. The IT measue ensues a connuous wokload fo evey esouce ems, and geneally educes he numbe of he neupons as well. So a schedule whee he esouce usage dagam s quas-concave paes down he expenses of he pojec. 7

8 2. Basc mxed nege lnea pogammng model In he basc model we allow moe han one esouce, bu all he esouces ae consdeed o be he same knd. No dsncon s appled. The model s fomulaed as a esouce levelng poblem, whee hee ae no esouce consans a all. The basc MILP model s developed fom an nal non lnea fomulaon hough sx smple seps. Le us noduce he followng bnay vaables descbng whehe he acvy sas a me : { AS =,2,..., N; ES,..., LS } AS = =, whee ES and LS denoes he eales (and laes) possble sa of acvy by he CPM schedule. Hee we should emak, ha dung he ccal pah mehod he esouces ae gnoed. Wh hs noaon AS, = holds f and only f he acvy sas a me peod, ohewse 0. Dung schedules acvy splng (neupng fo sho peods) s no allowed, heefoe evey acvy mus sa a a well defned peod. Ths fac may be fomulaed wh he followng ype of LS equaons: AS, = and S = = ES LS AS = ES, fo evey =,2,..., N. The mmedae pecedence elaons can be descbed wh he help of so called song-x fomulaons. [] LS = ES AS fo =,2,..., N, = 8

9 LS s= + D AS, s + AS j,s whee ES,..., LS s= ES j = and (, j) E Only he second equaon has o be explaned. The sum of he addends on he lef sde equals o 2 f and only f acvy j sas befoe he acvy s fnshed. Ths s clealy an equvalen fomulaon of he ( j) E, pecedence elaon. Defnon A max A s called oally unmodula, f all s A, elemens and subdeemnans ae equal o j +, 0 o. A poof by Schjve [5] saes, ha n case of schedulng poblems he coespondng AS, max s always oally unmodula. The heoecal mpoance of hs fac may be exploed by he help of Chaudhu s heoem of unmodulay [6]. Accodng o hs heoem nsead of AS { 0,} bnay (nege) vaables we can use 0 AS, eal vaables. The obaned elaxed lnea pogammng (LP) poblem can be solved n polynomal me. We have o noe, ha hs echnque does no wok when esouce lms have o be consdeed., Heenafe we fequenly have o efe o he ( S) equed amoun of esouce a me by he schedule S. Fo hs eason we need an auxlay noaon fo he se of ongong (acve) acves a me peod : Ac ( ) = { V S < S + D } ulzaon level can be expessed easly:. Wh he help of ongong acves he esouce = R Ac( ) = N = j= D + AS, j R We can see ha values ae lnea combnaons of AS, sang me vaables. 9

10 To mpove he eadably of he model alkave vaable names wll be used. Fo evey esouce R R and me peod {, 2,..., T} le MAXL (and MAXR ) denoe he maxmal lef (and gh) sde envonmen of he esouce ulzaon: MAXL = max(, 2,..., ) and MAXR = max(, +,..., T ). Pevously we gave a vebal defnon fo he dle me of a schedule as he numbe of un squaes have o be used o fll up he valleys of he esouce usage hsogam. Ths suggesve defnon helps us o pepae an exac fomula fo he dsance beween he acual esouce hsogam and s quas-concave hull. IT T ( S) = max( mn( MAXL, MAXR ), ) = Alhough hs fomula s a nce and compac one fom he mahemacal pon of vew, bu unfounaely has a leas wo poblemac popees: () Mnmum and maxmum ae no smooh funcons. Snce n he poposed model we wan o keep he lneay, heefoe hese funcons mus be subsued by lnea consans o by noducng addonal nege vaables. (2) The subsuon mehod suggesed above s songly dependen on he decon of opmzaon. The goal funcon of he model should be mnmzed, heefoe we have o use dffeen subsuon mehods fo he maxmum and mnmum funcons. [2] Le S Σ be a schedule whch obseves he mmedae pecedence elaons. Then he nal veson of he opmzaon poblem can be fomulaed as follows: IT( S) Mn! S Σ 0

11 Though he followng seps we pove ha by noducng appopae new vaables and consans hs model can be lneazed. The dea of he followng seps ae due o Gyögy Csébfalv []. Sep Fs o educe he compung powe of paal calculaons he duplcaons ae elmnaed fom he maxmum funcon. Evey maxmum funcon wh moe han wo agumens s eplaced by a sequence of maxmum funcons wh exacly wo agumens. MAXL = ( MAXL ) MAXL max, = fo { 2, 3,..., T - 2 } ( MAXR ) MAXR max, MAXR = T + and R R = fo { 3, 4,..., T -} T and R R Sep 2 Fo evey S Σ feasble schedule le I = denoe he sum of dle me uns a me peod fo evey { 2, 3,..., T -}. Afe mmedae applcaon of he eplacemens n sep one and he MIN = mn ( MAXL, MAXR + ) abbevaon he non smooh fom of he opmzaon poblems akes he followng shape: k T = = 2 I Mn! + I MIN fo { 2, 3,..., T -} MIN mn( MAXL, MAXR + ) S Σ and {, 2,..., k} = fo { 2, 3,..., T -} and {, 2,..., k}

12 Sep 3 Snce he mnmum of he goal funcon s waned, we can decly ead wo consequences of he specal sucues of funcons IT ( S). () Equales of he fom MAXL ( MAXL, ) = { 2, 3,..., T - 2 } max he followng nequales (lnea consans): may be eplaced by MAXL MAXL MAXL. (2) In case of equales MAXR ( MAXR, ) gve he pope esul: = { 3, 4,..., T - } max + smla eplacemens MAXR MAXR MAXR + A he end of he hd sep we sll ge a non smooh fom of he nal poblem: k T = = 2 I Mn! + I MIN fo { 2, 3,..., T -} MIN mn( MAXL, MAXR + ) and {, 2,..., k} = fo { 2, 3,..., T -} and {, 2,..., k} = fo {, 2,..., k} MAXL MAXL fo { 2, 3,..., T - 2 } MAXL MAXL and {, 2,..., k}. fo { 2, 3,..., T - 2 } and {, 2,..., k} MAXR fo { 3, 4,..., T -} MAXR MAXR T + and {, 2,..., k} fo { 3, 4,..., T -} and {, 2,..., k} MAXR = fo {, 2,..., k} S Σ T 2

13 Sep 4 Evey MIN mn( MAXL, MAXR + ) = equaon fo {,..., k } and { 2, 3,..., T - } may be eplaced by he followng non-lnea consans, whee LLTR ae a bnay (0/) vaables. The LLTR vaables ae flags, ndcang whehe he nequaly MAXL > MAXR + holds () o no ( 0 ). MAXL + ( LLTR ) MAXR + MIN = LLTR MIN MAXL MIN MAXR LLTR { 0,} + Ths anscpon s based on a smple fac ha he fomula M = mn( a, b) s equvalen wh he M = a + ( ) b, M a ; b M, { 0,} consan se. Afe eplacng all he mnmum expessons wh wo vaables wh he pope consan se, only one hng beaks he lneay. Namely he poducs conanng LLTR vaables should be elmnaed and ansfomed nex. k T = = 2 I Mn! + I MIN whee { 2, 3,..., T -} MIN MAXL and {, 2,..., k} whee { 2, 3,..., T -} and {, 2,..., k} MIN MAXR whee { 2, 3,..., T -} and {, 2,..., k} + MAXL + ( LLTR ) MAXR + MIN = LLTR 2, 3,..., T - whee { } and {, 2,..., k} LLTR { 0,} whee { 2, 3,..., T -} and {, 2,..., k} = whee {, 2,..., k} MAXL MAXL whee { 2, 3,..., T - 2 } MAXL MAXL and {, 2,..., k}. whee { 2, 3,..., T - 2 } and {, 2,..., k} MAXR whee { 3, 4,..., T -} MAXR MAXR + and {, 2,..., k} whee { 3, 4,..., T -} and {, 2,..., k} 3

14 MAXR = whee {, 2,..., k} S Σ Sep 5 T T The mehod called bg-m s especally desgned fo elmnang poduc of vaables [7]. We have only o apply hs mehod o eplace he pevously menoned non-lnea (o pecsely blnea LLTR MAXL and LLTR MAXR + consans wh lnea ones: MIN MAXL MIN MAXR + MIN = PROL + PROR + PROL MAXL LLTR PROL MAXL + MAXL LLTR MAXL PROR + MAXR + LLTR + MAXR + PROR + MAXR + MAXR + LLTR LLTR { 0,} Dung he eplacemen he unde (and ove) -lned MAXL, MAXL,. MAXR +, and MAXR + expessons denoe appopaely chosen lowe (and uppe) bounds of he coespondng MAXL and MAXR + vaables. The exac values of hese consans can be calculaed fom he paamees of he ognal poblem. Sep 6 In he las sep he S Σ em s eplaced wh s song-x fomulaon. 4

15 LS = ES LS s= AS whee =,2,..., N, = + D AS, s + AS j,s whee ES,..., LS s= ES j = and (, j) E AS, {0,} whee =,2,..., N and = ES,..., LS = N = j= D + AS j R, whee {, 2,..., T } and {, 2,..., k} Fnally he ognal esouce levelng poblem s ewen n a fom of mxed lnea pogammng poblem []: Basc IT model (MILP Mxed Inege Lnea Pogammng) k T = = 2 I Mn! + I MIN whee { 2, 3,..., T -} MIN MAXL and {, 2,..., k} whee { 2, 3,..., T -} and {, 2,..., k} MIN MAXR + whee { 2, 3,..., T -} and {, 2,..., k} MIN PROL = whee { 2, 3,..., T -} and {, 2,..., k} + PROR + PROL MAXL LLTR whee { 2, 3,..., T -} and {, 2,..., k} PROL MAXL + MAXL LLTR MAXL whee { 2, 3,..., T -} and {, 2,..., k} PROR + MAXR + LLTR + MAXR + whee { 2, 3,..., T -} and {, 2,..., k} PROR + MAXR + MAXR + LLTR MAXL whee { 2, 3,..., T -} and {, 2,..., k} = whee {, 2,..., k} MAXL whee { 2, 3,..., T - 2 } and {, 2,..., k} 5

16 . whee { 2, 3,..., T - 2 } and {, 2,..., k} MAXL MAXL MAXR whee { 3, 4,..., T -} MAXR MAXR + T and {, 2,..., k} whee { 3, 4,..., T -} and {, 2,..., k} MAXR = whee {, 2,..., k} T LLTR { 0, } whee {, 2,..., T - 2 } and {, 2,..., k} = N = j= D + AS j R, whee {, 2,..., T } and {, 2,..., k} LS = ES LS s= AS whee =,2,..., N, = + D AS, s + AS j,s whee ES,..., LS s= ES j = and (, j) E AS {0,} whee =,2,..., N and = ES,..., LS, To pove he usably of he ansfomaon descbed above a vey smple example of he basc IT model s pesened. Le us ake a vey small pojec whch conss of only 5 acves and eques only one knd of esouce. The coespondng IT model s gven wh he noaon of Lndo pogam, and s consans ae gouped by peods and ypes. Acvy ndex () Duaon (D days) Resouce equemen (R pesons) Immedae pedecesso acves (j) Table. A smple esouce levelng example pojec fo he basc IT model 6

17 R R > [2] 2[] < MIN +I2+I3 > [2] 2[] < 3[4] 4[] 5[4] SBJECT TO 3[4] 4[] 5[4] Peod Peod 2 Peod 3 Peod 4 MAXL-=0 +2AS+4A3S-=0 +I2-MIN2+2>=0 +MAXL2-2>=0 -MAXL+MAXL2>=0 +MIN2-MAXL<=0 +MIN2-MAXR3<=0 +MIN2-PROL-PROR3=0-4LLTR+PROL>=0 -MAXL-6LLTR+PROL>=-6 +4LLTR+PROR3>=4 -MAXR3+5LLTR+PROR3>=0 +2AS2+A2S2+4A3S-2=0 +I3-MIN3+3>=0 +MAXR3-3>=0 +MAXR3-MAXR4>=0 +MIN3-MAXL2<=0 +MIN3-MAXR4<=0 +MIN3-PROL2-PROR4=0-4LLTR2+PROL2>=0 -MAXL2-6LLTR2+PROL2>=-6 +4LLTR2+PROR4>=4 -MAXR4+5LLTR2+PROR4>=0 +2AS3+A2S3+A4S3-3=0 MAXR4-4=0 A2S4+4A5S4-4=0 AS+AS2+AS3= A3S= AS2+AS3+A2S2<= A2S2+A2S3+A2S4= A4S3= AS3+A2S2+A2S3<= A5S4= END INT AS AS2 AS3 A2S2 A2S3 A2S4 A3S A4S3 A5S4 LLTR LLTR2 Fgue 3. CPM and opmal schedule of he sample pojec wh s basc IT model n Lndo 3 Exended IT model fo had and sof esouces In hs secon we show how he pevous basc IT model can be exended o dffeen class of esouces. In case of sof and had esouces wo dffeen ypes of poblems should be combned ogehe. The esouce levelng poblem fo sof esouces and he esouce allocaon poblem fo had esouces can be solved by a sngle combned MILP allocaon model. 7

18 Defnon Resouce R s efeed as had consan f a any me no moe han R() amoun of hs esouce can be allocaed dung he pojec. The se of had esouces s denoed by R H. In ealsc cases hee ae cean lms fo any knd of esouces, snce he numbe of he employees o machnes s well know. In case of had esouces he ulzaon can no be hghe han he acual esouce lm, n ohe wods R() holds fo evey =,2,..., T. When an allocaon conflc s caused by a had consan, can be esolved only wh eschedulng he acves, o n he wos case exendng he eales fnshng me of he pojec. Fo sof esouces hese lms ae no aken so scly. Fo example f someme we need moe esouces han avalable, we can buy o he auxlay ems fom he make. Defnon Resouce R s efeed as sof consan f a any me moe han R() amoun of can be allocaed by applyng auxlay esouces. The se of sof esouces s denoed by R S. If sof and had esouces have o be dsngushed, we can esablsh he un cos of applyng new ems pe esouce ypes. If hs un cos fo a cean esouce s hghe han a pedefned K consan, han hs esouce s moe lkely behaves as a had esouce. Defnon When a schedule S Σ keeps all he mmedae pecedence elaons, and does no cause conflcs (ove-ulzaon) fo had esouces, s called had feasble. The esouce allocaon poblem fo he had esouces can be fomulaed as MILP poblem easly, snce only some exa consans should be added o he basc IT model of he coespondng levelng poblem. These exa consans wll ensue ha no ove-ulzaon occus. Fo smplcy le us suppose, ha had esouces pecede sof ones, so he ndces of had 8

19 esouces ae always less han he ndces of sof esouces: = { R R,... } { R R } R and, H 2 R k R S = R k+, k+ 2,... k+ m. Fo sho em pojecs we can suppose, ha esouce lms ae consans L R () dung he pojec s makespan fo all {,2,...k } had esouces. Fs of all he goal funcon of he exended model wll be dffeen. Boh he pevously noduced dle me (IT) measue and he adonal peak esouce (M) level have o be used n hs case. The new goal funcon s defned as a lnea combnaon (weghed sum) of he ognal penaly funcons. Whou he loss of genealy we can assume ha all he weghs equal o, snce he acual values ae neual fom he model pon of vew. Sep 7 The fs sep n he exenson of he basc IT model s o add new consans o ensue ha all he schedules ae equed o be no jus smply feasble, bu had feasble. R ) ( L fo {, 2,..., T} and {,2,...k} Fom he defnon of had esouces and fom he sucue of he basc IT model mmedaely follows, ha a feasble schedule of he exended model afe he 7h sep s a had feasble soluon of he basc IT model and convesely. Sep 8 Bu s no enough o pay aenon only o he pool of had esouces. sually we have cean amoun of sof esouces as well, and applyng auxlay ems n ou case s pefeed only n dsess. Theefoe s no suffcen o mnmze he dle mes of sof esouces, bu we have o lowe he peak ulzaon a he same me. (Fla and quas-concave esouce usage dagams defne bee unfom wokload.) Fo a schedule of a R R sof esouce he M (Maxmal S sage) funcon gves he pope penaly by deemnng he maxmum of he esouce usage dagam. M measue s agan a non smooh one, whch s clea fom he followng defnon: 9

20 ( S) = (,,..., ) max( MAXL MAXR ) M max 2 T =, The concuen opmzaon by he IT and M measues can be descbed as a lnea combnaon of he wo goal funcons. Fo evey { k +, k + 2,..., k + m} sof esouce a new global M uppe bound s defned, whch symbolze he appopae M ulzaon level n he goal funcon. The whole exended esouce allocaon IT+M model has he followng fom: Exended IT+M model (MILP) fo had and sof esouces Goal funcon: k+ m T k+ m I + M Mn! (C) = = 2 = k+ Consans: R () fo {, 2,..., T} M fo {, 2,..., T} and {, 2,... k} (Fa) and { k +, k + 2,... k + m} (Fb) + I MIN fo { 2, 3,..., T -} MIN MAXL and {, 2,..., k + m} (F2) fo { 2, 3,..., T -} and {, 2,..., k + m} MIN MAXR + (F3) fo { 2, 3,..., T -} and {, 2,..., k + m} MIN PROL + PROR + (F4) = fo { 2, 3,..., T -} and {, 2,..., k + m} PROL MAXL LLTR (F5) fo { 2, 3,..., T -} and {, 2,..., k + m} MAXL + MAXL LLTR MAXL (F6) PROL (F7) fo { 2, 3,..., T -} and {, 2,..., k + m} + MAXR+ LLTR + MAXR+ PROR (F8) 2, 3,..., T -, 2,..., k + m fo { } and { } + MAXR + MAXR+ LLTR PROR (F9) 2, 3,..., T -, 2,..., k + m fo { } and { } = fo {, 2,..., k + m} MAXL (F0) MAXL fo { 2, 3,..., T - 2 } MAXL MAXL and {, 2,..., k + m} (F). fo { 2, 3,..., T - 2 } és {, 2,..., k + m} (F2) 20

21 MAXR fo { 3, 4,..., T -} MAXR MAXR T + and {, 2,..., k + m} (F3) fo { 3, 4,..., T -} and {, 2,..., k + m} T (F4) MAXR = fo {, 2,..., k + m} (F5) LLTR { 0, } fo {, 2,..., T - 2 } and {, 2,..., k + m} (F6) LS = ES AS = fo =,2,..., N (F7) = N = j= D + AS, j R (F8), 2,..., T, 2,..., k + m fo { } and { } LS s= AS s + D + AS s= ESj js fo = ES,..., LS and (, j) E (F9) AS {0,} fo =,2,..., N and = ES,..., LS (F20) Fom paccal pon of vew s vey mpoan o noe, ha all he 8 seps of he ansfomaon can be algohmzed. We exploed hs popey and a compue pogam was developed o auomae hs pocess. The pogam eads he paamees of he ognal esouce levelng poblem as npu, and oupus he equvalen exended IT+M model wh he acual values of boundng consans. The fnal fom of he followng example was consuced wh he help of hs pogam as well. [4] Acvy ndex () Duaon (D days) Had esouce equemen (R pesons) Sof esouce equemen (R2 pesons) Sof esouce equemen (R3 pesons) Immedae pedecesso acves (j)

22 Table 2. A smple esouce allocaon example pojec fo he exended IT+M model Ths poblem s a smplfed veson of a sofwae developmen pojec, whee 4 desgnes, 4 pogammes and 4 eses wee wokng ogehe. In hs envonmen desgnes wee qualfed as key esouces by he advce of he pojec manage. So n he exended IT model hs knd of esouce s labeled had, and s allocaon lm s L 4 dung he whole pojec. In hs example only one knd of esouce s assgned o evey acvy. Wh hs smplfcaon he sze and he complexy of he ansfomed poblem could be educed o f n hese pages. In fgue 4 he eales schedule s shown. When had esouce lms ae gnoed he pojec s makespan s 5 days. Bu he CPM schedule allocaes moe hen 4 uns of esouce R on he fs hee days, so s no had esouce feasble Legend: Acvy ndex [Resouce equemens] R Had esouce R2, R3 Sof esouces R R > [2,0,0] 2 [0,2,0] 3 [0,0,2] 4 [3,0,0] 5 [0,3,0] 6 [0,0,3] < R [2,0,0] 8 [0,2,0] 9 [0,0,2] 0 [,0,0] [0,2,0] 2 [0,0,] 3 [2,0,0] 4 [0,3,0] 5 [0,0,2] Fgue 4. CPM schedule of he esouce allocaon and levelng poblem We ed o keep he noaon of he exended mahemacal IT model n he ansfomed Lndo 22

23 foma. The dffeen consans ae gouped by he ypes numbeed n he model descpon.! Had Levelng and Maxmal sage and Idle Tme MIN! Goal funcon (C) +RI2+RI3+RI4+RI5+RI6+RI7+RI8+RI9+RI0+RI+RI2+RI3+RI4 +R2I2+R2I3+R2I4+R2I5+R2I6+R2I7+R2I8+R2I9+R2I0+R2I+R2I2+R2I3+R2I4 +R3I2+R3I3+R3I4+R3I5+R3I6+R3I7+R3I8+R3I9+R3I0+R3I+R3I2+R3I3+R3I4 +M2 +M3 SBJECT TO! Consans of ype (Fa) and (Fb) R<=4 R2<=4 R5<=4! Consans of ype (F2) RI2-RMIN2+R2>=0 RI3-RMIN3+R3>=0 RI4-RMIN4+R4>=0! Consans of ype (F3) RMIN2-RMAXL<=0 RMIN3-RMAXL2<=0 RMIN4-RMAXL3<=0! Consans of ype (F4) RMIN2-RMAXR3<=0 RMIN3-RMAXR4<=0 RMIN4-RMAXR5<=0! Consans of ype (F5) RMIN2-RPROL-RPROR3=0 RMIN3-RPROL2-RPROR4=0 RMIN4-RPROL3-RPROR5=0! Consans of ype (F6) RPROL>=0 RPROL2>=0 RPROL3>=0-2RLLTR4+RPROL4>=0-2RLLTR5+RPROL5>=0-2RLLTR3+RPROL3>=0! Consans of ype (F7) -RMAXL2-0RLLTR2+RPROL2>=-0 -RMAXL3-0RLLTR3+RPROL3>=-0 -RMAXL3-0RLLTR3+RPROL3>=-0! Consans of ype (F8) RPROR5>=0 RPROR6>=0 RPROR5>=0 +2RLLTR+RPROR3>=2 +2RLLTR2+RPROR4>=2! Consans of ype (F9) -RMAXR3+0RLLTR+RPROR3>=0 -RMAXR4+0RLLTR2+RPROR4>=0 R2-M2<=0 R22-M2<=0 R25-M2<=0 R2I2-R2MIN2+R22>=0 R2I3-R2MIN3+R23>=0 R2I4-R2MIN4+R24>=0 R2MIN2-R2MAXL<=0 R2MIN3-R2MAXL2<=0 R2MIN4-R2MAXL3<=0 R2MIN2-R2MAXR3<=0 R2MIN3-R2MAXR4<=0 R2MIN4-R2MAXR5<=0 R2MIN2-R2PROL-R2PROR3=0 R2MIN3-R2PROL2-R2PROR4=0 R2MIN4-R2PROL3-R2PROR5=0 R2PROL>=0 R2PROL2>=0 R2PROL7>=0-3R2LLTR8+R2PROL8>=0-3R2LLTR9+R2PROL9>=0-3R2LLTR3+R2PROL3>=0 -R2MAXL+R2PROL>=0 -R2MAXL2+R2PROL2>=0 -R2MAXL3+R2PROL3>=0 -R2MAXL4-7R2LLTR4+R2PROL4>=-7 -R2MAXL5-2R2LLTR5+R2PROL5>=-2 -R2MAXL6-2R2LLTR6+R2PROL6>=-2 -R2MAXL3-2R2LLTR3+R2PROL3>=-2 R2PROR9>=0 R2PROR0>=0.. R2PROR5>=0 +3R2LLTR+R2PROR3>=3 +3R2LLTR2+R2PROR4>=3 +3R2LLTR6+R2PROR8>=3 -R2MAXR3+2R2LLTR+R2PROR3>=0 -R2MAXR4+2R2LLTR2+R2PROR4>=0 R3-M3<=0 R32-M3<=0 R35-M3<=0 R3I2-R3MIN2+R32>=0 R3I3-R3MIN3+R33>=0 R3I4-R3MIN4+R34>=0 R3MIN2-R3MAXL<=0 R3MIN3-R3MAXL2<=0 R3MIN4-R3MAXL3<=0 R3MIN2-R3MAXR3<=0 R3MIN3-R3MAXR4<=0 R3MIN4-R3MAXR5<=0 R3MIN2-R3PROL-R3PROR3=0 R3MIN3-R3PROL2-R3PROR4=0 R3MIN4-R3PROL3-R3PROR5=0 R3PROL>=0 R3PROL2>=0 R3PROL>=0-2R3LLTR2+R3PROL2>=0-2R3LLTR3+R3PROL3>=0 -R3MAXL+R3PROL>=0 -R3MAXL2+R3PROL2>=0 -R3MAXL6+R3PROL6>=0 -R3MAXL7-8R3LLTR7+R3PROL7>=-8 -R3MAXL8-8R3LLTR8+R3PROL8>=-8 -R3MAXL9-0R3LLTR9+R3PROL9>=-0 -R3MAXL0-0R3LLTR0+R3PROL0>=-0 -R3MAXL3-0R3LLTR3+R3PROL3>=-0 R3PROR3>=0 R3PROR4>=0 R3PROR5>=0 +2R3LLTR+R3PROR3>=2 +2R3LLTR2+R3PROR4>=2 +2R3LLTR0+R3PROR2>=2 -R3MAXR3+0R3LLTR+R3PROR3>=0 -R3MAXR4+0R3LLTR2+R3PROR4>=0 23

24 -RMAXR0+2RLLTR8+RPROR0>=0 -RMAXR+RPROR>=0 -RMAXR2+RPROR2>=0 -RMAXR5+RPROR5>=0 -R2MAXR2+9R2LLTR0+R2PROR2>=0 -R2MAXR3+R2PROR3>=0 -R2MAXR4+R2PROR4>=0 -R2MAXR5+R2PROR5>=0! Consans of ype (F0) RMAXL-R=0 R2MAXL-R2=0 R3MAXL-R3=0! Consans of ype (F) RMAXL2-R2>=0 RMAXL3-R3>=0 RMAXL3-R3>=0! Consans of ype (F2) -RMAXL+RMAXL2>=0 -RMAXL2+RMAXL3>=0 -RMAXL2+RMAXL3>=0! Consans of ype (F3) RMAXR3-R3>=0 RMAXR4-R4>=0 RMAXR4-R4>=0! Consans of ype (F4) RMAXR3-RMAXR4>=0 RMAXR4-RMAXR5>=0 RMAXR4-RMAXR5>=0 R2MAXL2-R22>=0 R2MAXL3-R23>=0 R2MAXL3-R23>=0 -R2MAXL+R2MAXL2>=0 -R2MAXL2+R2MAXL3>=0 -R2MAXL2+R2MAXL3>=0 R2MAXR3-R23>=0 R2MAXR4-R24>=0 R2MAXR4-R24>=0 R2MAXR3-R2MAXR4>=0 R2MAXR4-R2MAXR5>=0 R2MAXR4-R2MAXR5>=0 - R3MAXR5+0R3LLTR3+R3PROR5>=0 R3MAXL2-R32>=0 R3MAXL3-R33>=0 R3MAXL3-R33>=0 -R3MAXL+R3MAXL2>=0 -R3MAXL2+R3MAXL3>=0 -R3MAXL2+R3MAXL3>=0 R3MAXR3-R33>=0 R3MAXR4-R34>=0 R3MAXR4-R34>=0 R3MAXR3-R3MAXR4>=0 R3MAXR4-R3MAXR5>=0 R3MAXR4-R3MAXR5>=0! Consans of ype (F5) RMAXR5-R5=0 R2MAXR5-R25=0 R3MAXR5-R35=0! (F7) ípusú feléelek AS+AS2++AS7= A2S4+A2S5++A2S0= A3S7+A3S8++A3S3= A4S+A4S2++A4S7= A5S4+A5S5++A5S0=! Consans of ype (F8) 0=+2AS+3A4S+2A7S +A0S+2A3S-R 0=+2AS+2AS2+3A4S +3A4S2+2A7S+2A7S2 +A0S+A0S2+2A3S +2A3S2-R2 0=+2AS+2AS2+2AS3 +3A4S+3A4S2+3A4S3 +2A7S+ 2A7S2+2A7S3 +A0S+A0S2+A0S3 +2A3S+2A3S2+2A3S3 -R3 0=+2AS2+2AS3+2AS4 +3A4S2+3A4S3+3A4S4 +2A7S+2A7S2+2A7S3 +2A7S4+A0S2+A0S3 +A0S4+2A3S+2A3S2 +2A3S3+2A3S4-R4 0=+2AS3+2AS4+2AS5 +3A4S3+3A4S4+3A4S5 +2A7S2+2A7S3+2A7S4 +2A7S5+A0S3+A0S4 +A0S5+2A3S2+2A3S3 +2A3S4-R5 0=+2AS4+2AS5+2AS6 +3A4S4+3A4S5+3A4S6 +2A7S3+2A7S4+2A7S5 +2A7S6+A0S4+A0S5 +A0S6+2A3S3+2A3S4 -R6 0=+2AS5+2AS6+2AS7 +3A4S5+3A4S6+3A4S7 +2A7S4+2A7S5+2A7S6 A6S7+A6S8++A6S3= A7S+A7S2++A7S7= A8S5+A8S6++A8S= A9S7+A9S8++A9S3= A0S+A0S2++A0S7= 0=-R2 0=-R22 0=-R23 0=+2A2S4+3A5S4+2AS4 -R24 0=+2A2S4+2A2S5+3A5S4 +3A5S5+2A8S5+2AS4 +2AS5+3A4S5-R25 0=+2A2S4+2A2S5+2A2S6 +3A5S4+3A5S5+3A5S6 +2A8S5+2A8S6+2AS4 +2AS5+2AS6+3A4S5 +3A4S6-R26 0=+2A2S5+2A2S6+2A2S7 +3A5S5+3A5S6+3A5S7 +2A8S6+2A8S7+2AS5 +2AS6+2AS7+3A4S5 +3A4S6+3A4S7-R27 0=+2A2S6+2A2S7+2A2S8 +3A5S6+3A5S7+3A5S8 +2A8S7+2A8S8+2AS6 +2AS7+2AS8+3A4S5 +3A4S6+3A4S7+3A4S8 -R28 0=+2A2S7+2A2S8+2A2S9 +3A5S7+3A5S8+3A5S9 +2A8S8+2A8S9+2AS7 +2AS8+2AS9+3A4S6 +3A4S7+3A4S8-R29 0=+2A2S8+2A2S9+2A2S0 +3A5S8+3A5S9+3A5S0 +2A8S9+2A8S0+2AS8 +2AS9+2AS0+3A4S7 AS4+AS5++AS0= A2S7+A2S8++A2S3= A3S+A3S2++A3S4= A4S5+A4S6++A4S8= A5S9+A5S0++A5S2= 0=-R3 0=-R32 0=-R33 0=-R34 0=-R35 0=-R36 0=+2A3S7+3A6S7+2A9S7 +A2S7-R37 0=+2A3S7+2A3S8+3A6S7 +3A6S8+2A9S7+2A9S8 +A2S7+A2S8-R38 0=+2A3S7+2A3S8+2A3S9 +3A6S7+3A6S8+3A6S9 +2A9S7+2A9S8+2A9S9 +A2S7+A2S8+A2S9 +2A5S9-R39 0=+2A3S8+2A3S9+2A3S0 +3A6S8+3A6S9+3A6S0 +2A9S8+2A9S9+2A9S0 +A2S8+A2S9+A2S0 +2A5S9+2A5S0-R30 0=+2A3S9+2A3S0+2A3S +3A6S9+3A6S0+3A6S +2A9S9+2A9S0+2A9S +A2S9+A2S0+A2S +2A5S9+2A5S0+2A5S -R3 0=+2A3S0+2A3S+2A3S2 +3A6S0+3A6S+3A6S2 +2A9S0+2A9S+2A9S2 +A2S0+A2S+A2S2 +2A5S9+2A5S0+2A5S +2A5S2-R32 24

25 +2A7S7+A0S5+A0S6 +A0S7+2A3S4-R7 0=+2AS6+2AS7+3A4S6 +3A4S7+2A7S5+2A7S6 +2A7S7+A0S6+A0S7-R8 0=+2AS7+3A4S7+2A7S6 +2A7S7 +A0S7-R9 0=+2A7S7-R0 0=-R 0=-R2 0=-R3 0=-R4 0=-R5! Consans of ype (F9) (AS2+AS3++AS7)+(A2S4)<= (AS3+AS4++AS7)+(A2S4+A2S5)<= (AS4+AS5++AS7)+(A2S4+A2S5+A2S6)<= (AS5+AS6+AS7)+(A2S4+A2S5++A2S7)<= (AS6+AS7)+(A2S4+A2S5++A2S8)<= (AS7)+(A2S4+A2S5++A2S9)<= (A4S2+A4S3++A4S7)+(A5S4)<= (A4S3+A4S4++A4S7)+(A5S4+A5S5)<= (A4S4+A4S5++A4S7)+(A5S4+A5S5+A5S6)<= (A4S5+A4S6+A4S7)+(A5S4+A5S5++A5S7)<= (A4S6+A4S7)+(A5S4+A5S5++A5S8)<= (A4S7)+(A5S4+A5S5++A5S9)<= (A7S2+A7S3++A7S7)+(A8S5)<= (A7S3+A7S4++A7S7)+(A8S5+A8S6)<= (A7S4+A7S5++A7S7)+(A8S5+A8S6+A8S7)<= (A7S5+A7S6+A7S7)+(A8S5+A8S6++A8S8)<= (A7S6+A7S7)+(A8S5+A8S6++A8S9)<= (A7S7)+(A8S5+A8S6++A8S0)<= (A0S2+A0S3++A0S7)+(AS4)<= (A0S3+A0S4++A0S7)+(AS4+AS5)<= (A0S4+A0S5++A0S7)+(AS4+AS5+AS6)<= (A0S5+A0S6+A0S7)+(AS4+AS5++AS7)<= (A0S6+A0S7)+(AS4+AS5++AS8)<= (A0S7)+(AS4+AS5++AS9)<= (A3S2+A3S3+A3S4)+(A4S5)<= (A3S3+A3S4)+(A4S5+A4S6)<= (A3S4)+(A4S5+A4S6+A4S7)<= END! Consans of ype (F6) INT RLLTR INT RLLTR2 INT RLLTR3! Consans of ype (F20) INT AS INT AS2 INT AS7 INT A6S7 INT A6S8 INT A6S3 INT AS4 INT AS5 INT AS0 INT A2S4 INT A2S5 INT A2S0 INT A7S INT A7S2 INT A7S7 INT A2S7 INT A2S8 INT A2S3 +3A4S8-R20 0=+2A2S9+2A2S0+3A5S9 +3A5S0+2A8S0+2A8S +2AS9+2AS0+3A4S8 -R2 0=+2A2S0+3A5S0+2A8S +2AS0-R22 0=-R23 0=-R24 0=-R25 INT R2LLTR INT R2LLTR2 INT R2LLTR3 INT A3S7 INT A3S8 INT A3S3 INT A8S5 INT A8S6 INT A8S INT A3S INT A3S2 INT A3S4 0=+2A3S+2A3S2+2A3S3 +3A6S+3A6S2+3A6S3 +2A9S+2A9S2+2A9S3 +A2S+A2S2+A2S3 +2A5S0+2A5S+2A5S2 -R33 0=+2A3S2+2A3S3+3A6S2 +3A6S3+2A9S2+2A9S3 +A2S2+A2S3+2A5S +2A5S2-R34 0=+2A3S3+3A6S3+2A9S3 +A2S3+2A5S2 -R35 (A2S5+A2S6++A2S0)+(A3S7)<= (A2S6+A2S7++A2S0)+(A3S7+A3S8)<= (A2S7+A2S8++A2S0)+(A3S7+A3S8+A3S9)<= (A2S8+A2S9+A2S0)+(A3S7+A3S8++A3S0)<= (A2S9+A2S0)+(A3S7+A3S8++A3S)<= (A2S0)+(A3S7+A3S8++A3S2)<= (A5S5+A5S6++A5S0)+(A6S7)<= (A5S6+A5S7++A5S0)+(A6S7+A6S8)<= (A5S7+A5S8++A5S0)+(A6S7+A6S8+A6S9)<= (A5S8+A5S9+A5S0)+(A6S7+A6S8++A6S0)<= (A5S9+A5S0)+(A6S7+A6S8++A6S)<= (A5S0)+(A6S7+A6S8++A6S2)<= (A8S6+A8S7++A8S)+(A9S7)<= (A8S7+A8S8++A8S)+(A9S7+A9S8)<= (A8S8+A8S9++A8S)+(A9S7+A9S8+A9S9)<= (A8S9+A8S0+A8S)+(A9S7+A9S8++A9S0)<= (A8S0+A8S)+(A9S7+A9S8+ +A9S)<= (A8S)+(A9S7+A9S8++A9S2)<= (AS5+AS6++AS0)+(A2S7)<= (AS6+AS7++AS0)+(A2S7+A2S8)<= (AS7+AS8++AS0)+(A2S7+A2S8+A2S9)<= (AS8+AS9+AS0)+(A2S7+A2S8++A2S0)<= (AS9+AS0)+(A2S7+A2S8++A2S)<= (AS0)+(A2S7+A2S8++A2S2)<= (A4S6+A4S7+A4S8)+(A5S9)<= (A4S7+A4S8)+(A5S9+A5S0)<= (A4S8+A5S9+A5S0)+(A5S)<= INT A4S INT A4S2 INT A4S7 INT A9S7 INT A9S8 INT A9S3 INT A4S5 INT A4S6 INT A5S9 INT R3LLTR INT R3LLTR2 INT R3LLTR3 INT A5S4 INT A5S5 INT A5S0 INT A0S INT A0S2 INT A0S7 INT A5S0 INT A5S INT A5S2 The pevous model was auomacally geneaed wh ou PoMan 2. pogam ool [4], and was solved wh he Lndo sysem n 5 mnues on a P33Mhz compue. The numbe of eaons was I s woh o noe, ha despe of he lage sze of he Lndo model, s 25

26 coeffcen max s ahe spase. ROWS = 622 VARS = 46 INTEGER VARS = 35 (35 = 0/) NONZEROS = 2070 QCP = 0 DENSITY = CONSTRAINT NONZ=897 (542=+-) MIN&MAX ABSVALE= & 2 OBJ = MIN NMBER OF <: 77 NMBER OF =: 05 NMBER OF >:339 The opmum value of he IT+M goal funcon combnaon s. The coespondng schedule and s esouce usage dagams ae shown n able 3 and fgue 5. AS = 4 AS 2 = 8 AS 3 = 3 AS 4 = AS 5 = 5 AS 6 = 0 AS 7 = 7 AS 8 = AS 9 = 3 AS 0 = AS = 5 AS 2 = 0 AS 3 = 4 AS 4 = 8 AS 5 = 2 Table 3. Sang mes of acves n he opmal schedule of he sofwae developmen example R IT R =0 M R =- R IT R2 =0 M R2 =5 2 R IT R3 =0 M R3 = > 4 [3,0,0] 5 [0,3,0] 6 [0,0,3] < 0 [,0,0] [0,2,0] 2 [0,0,] 3 [2,0,0] 4 [0,3,0] 5 [0,0,2] [2,0,0] 2 [0,2,0] 3 [0,0,2] 7 [2,0,0] 8 [0,2,0] 9 [0,0,2] Fgue 5. Opmal soluon of he ansfomed sofwae developmen example 26

27 Refeences [] Gy. Csébfalv: Opmal esouce levelng models of acvy nes. Hablaon dsseaon, PTE Pécs (200) [2] Gy. Csébfalv, P. Konsannds: A new exac esouce balancng pocedue fo he mulple esouce-consaned pojec schedulng poblem. Poc. APMOD Exended Absacs, Lmasol, (998) -3 [3] Gy. Csébfalv, P. Konsannds: A new esouce levelng pocedue fo he mulple esouceconsaned pojec schedulng poblem. Decson Scences Insue 5h Inenaonal Confeence, Ahens, Geece, (999) [4] Gy. Csébfalv: PoMan v2. se's Manual. nvesy of Pécs ( ) [5] Schjve, A.: Theoy of lnea and nege pogammng. John Wley, Newyok (987) [6] S. Chaudhu, R.A. Walke, J.E. Mchell: Analyzng and explong he sucue of he consans n he ILP appoach o he schedulng poblem. IEEE Tans. Vey Lage Scale Inegaon Sys. 2 (VLSI 4 994) [7] F.A. Al-Khayyal, J.E. Falk: Jonly consaned bconvex pogammng. Mahemacs of Opeaons Reseach 8 (983)