An efficient pseudo-polynomial algorithm for finding a lower bound on the makespan for the Resource Constrained Project Scheduling Problem

an author's

https://oatao.univ-toulouse.fr/21875

https://doi.org/10.1016/j.ejor.2018.11.005

Arkhipov, Dmitry I. and Battaïa, Olga and Lazarev, Alexander A. An efficient pseudo-polynomial algorithm for

finding a lower bound on the makespan for the Resource Constrained Project Scheduling Problem. (2019) European

Journal of Operational Research, 275 (1). 35-44. ISSN 0377-2217

Discrete

Optimization

An

efficient

pseudo-polynomial

algorithm

for

finding

a

lower

bound

on

the

makespan

for

the

Resource

Constrained

Project

Scheduling

Problem

R

Dmitry

Arkhipov

a,b,∗

_,

_Olga

_Battaïa

a

_,

_Alexander

_Lazarev

b,c,d,e a ISAE-SUPAERO, Université de Toulouse, Toulouse, France

b V.A. Trapeznikov Institute of Control Sciences of Russian Academy of Sciences, Moscow, Russia c Lomonosov State University, Moscow, Russia

d National Research University Higher School of Economics, Moscow, Russia e Moscow Physical and Technical Institute (State University), Dolgoprudnyi, Russia

Keywords: Project scheduling Scheduling RCPSP Lower bounds Pseudo-polynomial algorithms

a

b

s

t

r

a

c

t

Severalalgorithms for finding alower bound onthe makespan for the ResourceConstrainedProject SchedulingProblem(RCPSP)wereproposed inthe literature.However, fastcomputable lowerbounds usuallydonotprovidethebestestimationsandthemethodsthatobtainbetterboundsaremainlybased onthecooperationbetweenlinearandconstraintprogrammingandthereforearetime-consuming.Inthis paper,anewpseudo-polynomialalgorithmisproposedtofindamakespanlowerboundforRCPSPwith time-dependentresourcecapacities.Itsideaisbasedonaconsecutiveevaluationofpairsofresourcesand theircumulatedworkload.Usingtheproposedalgorithm,severalboundsforthePSPLIBbenchmarkwere improved.Theresultsforindustrialapplicationsarealsopresented wherethealgorithmcouldprovide goodboundsevenforverylargeprobleminstances.

1. Introduction

The Resource Constrained Project SchedulingProblem (RCPSP) is a well-known problem in scheduling theory. This problem is knowntobeNP-hardinthestrongsense(Garey&Johnson,1975). Inthisresearch,we considerageneralizedstatementofthe prob-lem witha time-dependentresourcecapacityfunction definedas follows.ThereisasetoftasksNandasetofrenewableresources R.TheamountofresourceX_∈Rwhichcanbeusedbytasksofset N duringtime slot [t,t+1

)

is defined by capacity function cX(t).

The statementofa constantresourcecapacityisa particularcase ofthisformulation.Foranytaskj_∈N,thefollowingparametersare given:

• pj– processingtime;

• ajX – requiredamountofresourceX∈R.

R _{This research is supported by the Russian Foundation for Basic Research (grant}

18-37-00295 mol_a ) and the Foundation of ISAE-SUPAERO.

∗ _{Corresponding author at: ISAE-SUPAERO, Universitéde Toulouse, Toulouse,}

Precedence relations between tasks are given by directed acyclicgraphG(N,E).Ifanedge(i→j)∈Eexists,itmeansthattask imustbefinishedbeforethestartingtimeoftaskj.

Further, the following parameters can be calculated for each tasktakingintoaccounttheprecedenceandresourceconstraints:

• r_j – release time, the earliest time from which task j can be started;

• Dj – deadline, the latest time for finishing task j if a global

deadlineTfortheprojectisknown.

Itshouldbenotedthatthesetaskparameterscanbealsogiven asinitialdataandincorporatedinthealgorithmdescribedbelow.

Theobjectiveistofindaschedulewiththelowestmakespani.e. withtheshortestprojectduration.

Aschedule

π

isfeasibleforthesetsofresourcesRandtasksN, ifforanyj∈NstartingtimeSj(

π

)≥ rjisdefinedandallprecedence

andresourcecapacityconstraintsaresatisfied.Thesetofall feasi-bleschedulesisnotedby



(N,R).Theobjectiveistofindafeasible schedulewiththeminimalmakespanvaluei.e.

min

π∈(N,R)maxj∈N Cj

(

π

)

,

where C_j

(

π

)

₌S_j

(

π

)

₊p_{j≤ Dj} – the completion time of task j, whereDj– deadlineoftaskj.

France.

E-mail addresses: miptrafter@gmail.com (D. Arkhipov), olga.battaia@isae.fr (O. Battaïa), jobmath@mail.ru (A. Lazarev).

In the literature, a number of algorithms to calculate lower boundson the makespan were proposed. Their overviewis pre-sented in Section 2, more details can be found in the follow-ingcomprehensivesurveys(Neronetal.,2006)and(Knust,2015). Inthis paper, a novel pseudo-polynomial algorithm isdeveloped whichextends the relaxation ofRCPSP to a Cumulative Schedul-ingProblemby consideringpairs ofresources.Ourapproachuses ”time-tabling” techniques to adjust the capacity function of re-sourcesfirstandthenitcalculatesalowerboundonthemakespan byevaluatinghighestpossibleresourceloadsforeachtimeslot.

The rest of thepaper is organizedasfollows. The main algo-rithmforlowerboundcalculationispresentedinSection3.The re-sultsofthenumericalexperimentsarediscussedinSection4. Con-cludingremarksandresearchperspectivesaregiveninSection5. 2. Previousresearch

There isalargenumberofpublicationsdevotedtothe discus-sion on lower bounds for RCPSP. Recent comprehensive reviews canbe found in Neronet al.(2006) andKnust (2015). The anal-ysisofthe computationalcomplexityofsome algorithms andthe qualityof the obtainedboundsare discussed inGafarov, Lazarev, andWerner(2010).Themajorityofefficientalgorithmscanbe re-ferredto as”destructive” methods.Such analgorithm startswith adefined projectdeadline T andtriesto finda feasible schedule forit.Ifafeasiblesolutiondoesnotexist,thedeadlineisincreased (usuallybyincrementingthedeadlinewith1unitoftime)andthe calculationprocedure restarts.The calculation continuesuntilthe algorithmcannot revealanycontradictionwiththedefined dead-lineor until the endof the allocated calculation time. Similarly, constraintprogrammingmethodscanbeappliedasforexamplein

Schutt,Feydy,Stuckey,andWallace(2011)andLaborie(2003). Here below are shortly discussed the mosteffective methods forfinding a lower bound on the makespanfor RCPSP.For more detailsonthesealgorithms,thereadercanconsulttherecent com-prehensivereviews(Neronetal.,2006)andKnust(2015).

1. Disjunctivebounds

In the study ofBaptiste and Pape (2000),each renewable re-source is considered as a system of several identical processors with the number of processors equal to the capacity of the re-source.The problemis formulated usingmixed integer program-mingandheuristicsareusedtosolveit.

Several algorithms (Applegate & Cook, 1991;Baptiste, Pape, & Nuijten,1999;Carlier&Pinson,1989;Carlier&Pinson,1990; Car-lier & Pinson, 1994) aim to facilitate the lower bound calcula-tionbyconstructingcomplementaryprecedencerelationswiththe verification of the assumptions that certain requirements of set A must / cannot be satisfied before / after some considered set ofrequirementsB.Thesealgorithms havepolynomial runtimefor oneiteration,butthenumberofiterationsincreasesexponentially whenthenumberofrequirementsAandBincreases.

2. Cumulativebounds

The algorithm of Carlier and Latapie (1991) is based on the identificationofsuchsetsoftasksforwhichtheavailable amount ofresourceisinsufficientfortheirparallelexecution.Eachresource isconsideredasasystemofseveralidenticalprocessorswherethe numberofprocessorsperresourceislessthanitscapacity.Inthe furtherstudiesofCarlierandPinson(1998)andCarlierandPinson (2004),it isassumedthat each processor canexecute morethan onetaskperunit oftime andthesametaskcanbecompletedby severalprocessors.

Polynomial time algorithms were proposed to solve a relaxed problem(Brucker,2002; Haouari &Gharbi, 2003; Tercinet, Lente, &Neron,2004)wheretheplanninghorizon(betweenthestarting pointandthedeadline)wassplitintointervalsandeachresource wasconsidered as a multiprocessor system with the number of

processors equal to the capacityof the resource.Further, the in-terruptionoftasksattheboundariesofintervalswasallowed.

TherelaxationtosocalledCumulativeSchedulingProblemwas alsoexplored. It isobtainedfrom theinitial problemby ignoring allresourcesexceptoneandreplacingtheprecedencerelationsby releasetimesanddeadlines.Theoptimalmakespanforthis prob-lemprovidesa lowerbound forthe initialproblem.However, the obtained Cumulative Scheduling Problem is also NP-hard in the strongsense.Nevertheless,methods developedforthe calculation ofalowerboundonthemakespanforsuchaformulationprovide a lower bound for the makespan of the initial problem as well (Brucker & Knust, 2000; Carlier & Neron,2000; Carlier& Neron, 2003; Mingozzi, Maniezzo, Ricciardelli, & Bianco, 1998). Satisfia-bility tests SAT can also be performed by dividing the planning horizonintointervalsandcheckingtheamountoftheavailable re-source in each of the considered intervals (Baptiste et al., 1999; Erschler,Lopez,&Thuriot,1991;Lopez,Erschler,&Esquirol, 1992; Schwindt,2005).

3. MethodsbasedonConstraintProgramming

AmongthestudiesusingdifferenttechniquesofConstraint Pro-gramming,forexampleSchuttetal.(2011)andLaborie(2003),the techniquesdeveloped inNuijten (1994) andCaseau andLaburthe (1996)arebasedonreducingthetimeintervalscalculatedforeach taskduetotheanalysisoftheavailable amountofeach resource forachosensetoftasks.Suchalgorithmsaretimeconsuming be-causeofthelargenumberofpossiblesetsunderconsideration.

4. Algorithms based on the exploration of multi-resource con-straints

Such algorithms (Baptiste & Demassey,2004; Garaix,Artigues, & Demassey, 2005; Laborie, 2005) are based on the research of ”criticalsets” (MCS-minimumcriticalset,FS-forbiddenset)i.e. setsoftasks thatcannot beperformedsimultaneouslybecauseof resource capacity constraints, while anysubset ofsuch a critical setdoesnotviolateresourceconstraintsandcanbeperformed si-multaneously.

5. Linearprogrammingrelaxations

A lower bound can also be obtained by a relaxation of the initial problem to a linear programming problem (Christofides, Alvarez-Valdes,&Tamarit,1987;Pritsker,Watters,&Wolfe.,1969). Adetailedanalysisofthesealgorithms canbe foundinrecent surveys (Neron et al., 2006 and Knust, 2015). The conclusion of this analysis is that fast algorithms usually do not provide the bestlowerboundsandthemethodsthatobtainbetterboundsare mainly based on the cooperation between linear and constraint programmingand therefore are time-consuming. Inthis paper, a newpseudo-polynomialalgorithmisproposedwhichaimsto pro-videgoodlowerboundsinacceptablecomputationaltime. 3. Anovelalgorithmforfindingalowerboundonthe makespan

Theconsidered decisionversionofRCPSPisformulated as fol-lows:

Problem 1. Givensetof tasksN, setofresources R anddeadline (timehorizon)T,doesanyfeasibleschedule

π

∈



(N,R)existwith amakespaninferiororequaltoT,i.e.

max

j∈N Cj

(

π

)

≤ T. (1)

Without loss of generality, it is assumed that the project can bestartedattimet₌0.Weintroducetwodummytasks0_,n₊1_∈ N which represent the start andthe endof the project,i.e. r0= rn+1=0, p0=pn+1=0,D0=Dn+1=T andforany j∈N

\

{

0,n+ 1

}

precedences0→jand j→n+1exist.

Thegeneralschemeoffindingalowerboundonthemakespan isbasedonthefourfollowingprocedures:

Fig. 1. Compulsory part of a time interval of task j .

Procedure1. Pre-processing. This procedure updates the release times and deadlines for tasks under condition that themakespanisinferiororequaltoT.Ifduringthese calculationsoneoftheexistingconstraintscannotbe satisfied, there is no feasible solution with such a boundonthemakespan.

Procedure2. Thisprocedurecalculatesanupperboundonthe re-source consumption by set of tasks N during time interval [0,t+1

)

considering all pairs of resources X, Y. Precedence constraints are replaced by release times and deadlines. In this way, also precedence constraintswithtimelagscanbetakenintoaccount. Procedure3. Procedure2isapplied fororiginalprecedencegraph G(N,E)andthegraphwithreversedprecedence rela-tionsG

(

N,E

)

.Theobjectiveistocompare,forany re-sourceX_∈R,thesumofupperboundsonitsamount consumedinintervals[0, t) and(t,T]withthetotal amountofresourcerequiredforalltasks_j∈NajXpj.

Ifthelatterislower thantheformer,theconsidered problemisconsideredinfeasible.

Procedure4. Finally,thebinarysearchpartchangestimehorizonT andthenthecalculationisrestarted.

Inthefollowing,each partofthealgorithmisdiscussedin de-tails.

3.1. Procedure1:pre-processing

Wedenotethelength ofalongestpathfromi∈Ntoj∈N byPij

ifthere is a pathfrom i toj in graphG(N, E). The calculation of P_{ij≥ 0}forall pairs oftasks i, j_∈N havinga path fromi toj in G canbedoneusingDijkstra’salgorithm(Dijkstra,1959).

Let us consider all pairs of tasks i, j∈N such that Pij≥ 0 and

updatereleasetimesanddeadlinesusingformulae

rj:=max

{

rj,ri+Pi j

}

,

Di:=min

{

Di,Dj− Pi j

}

.

If forany j∈N, holds Dj− rj<pj,then inequality (1)is violated

andthe algorithm terminates. Otherwise,the compulsory partof the time interval (betweenthe releasetime anddeadline)is cal-culatedforeachtaskj∈N[CPs

j,CPej

)

usingformulaeCPsj=Dj− pj,

CPe

j=rj+pj(Fig.1).ThisideawasformulatedinLahrichi(1982).

IfCPs

j<CPej,then underanyschedule

π

,whichsatisfies given

releasedatesanddeadlines,taskjconsumesexactlyajXofresource

X∈Rateachmomentoftimet∈[CPs

j,CPej

)

.Therefore,theamount

ofresourceX∈Rthatcanbeusedbyother tasksateachmoment oftimet∈[CPs

j,CPje

)

isnot morethan cX

(

t

)

− ajX.Thisidealeads

ustoreplacecapacityfunctioncX(t)byfunctioncX

(

t

)

representing

theamount ofresourceX∈Rwhichcan beused toperform non-compulsorypartsoftasks(Fig.2)

c_X

(

t

)

=cX

(

t

)

−



j∈N|t∈[CPs j,CPej)

ajX.

If foranyX∈R andt=0,...,T− 1 inequality c_X

(

t

)

<0 holds, thereisnofeasibleschedulewhichsatisfiesdeadline T.The num-ber of breakpoints of function c_X

(

t

)

is not superior to 2n+m,

Fig. 2. Amount of resource X ∈ R which can be used to perform non-compulsory parts of tasks.

Fig. 3. Update of deadlines.

wherem is thenumber ofbreakpoints ofcX(t) inhorizonT. The

complexityof c_X

(

t

)

calculation for all X∈R can be estimated by O

((

n+m

)

r

)

operations, wheren=

|

N

|

and r=

|

R

|

. Note that the calculation of c_X

(

t

)

is similar to the Resource profile calculation presentedinFox(1990);Pape(1988).

The idea of the following algorithm is close to sweep algo-rithmspresented inBeldiceanuandCarlsson(2001,2002);Letort, Beldiceanu, and Carlsson (2012), the difference lies in the uti-lization offunction c_X

(

t

)

which isactively used anddynamically changedinouralgorithm.

Foreachtaskj∈NandresourceX∈R,itsdemandinresourceX iscomparedwiththeavailabilityofresourceXi.e.thevalueof ca-pacityfunctionc_X

(

t

)

forallmbreakpointsoffunctionc_X

(

t

)

which doesnotbelongtocompulsoryinterval[CPs

j,CPej

)

.Ifforanysetof

breakpointst0,...,tm,cX

(

t

)

<ajX i.e.theamountofresourceXis

notsufficienttoperformtaskj,thefollowingupdatesarerealized:

• if for any l∈

{

0,...,m− 1

}

such that tl<max

{

CPsj,CPje

}

and

rj≤ tl+1holdscX

(

tl

)

<ajX,updaterj:=tl+1; • ifforanyl∈

{

1,...,m

}

suchthatmax

{

CPs

j,CPej

}

<tl andtl−1< DjholdscX

(

tl−1

)

<ajX,updateDj:=tl−1 (Fig.3).

Ifforanytaskj∈N,itsreleasedateordeadlineisupdated,the preprocessingpart isrestarted withthe newvalues ofrj andDj.

Otherwise,thepreprocessingalgorithmterminatessuccessfully. Lemma 1. The complexity of the preprocessing part is O

(

n2

₍

_n+ m

)

Tr

)

operations, where n is the number of tasks,m is the highest numberofbreakpointsoftheresourcecapacityfunction,Tisthetime horizonandristhenumberofresources.

Proof. Thecalculation ofPij foralli,j∈N takesO

(

n

|

E

|

+n2logn

)

operations, where |E| is the number of edges in graph G. Each iteration of the first round for release and deadline calculation takesO(n2_{)operationsforcheckingallpathsandO}

₍

_n

₍

_n+m

₎

_r

₎

_for resource inequalities verification. The number of iterationsis no morethannTsinceeachtaskcannothavemorethanTreleasetime ordeadlineupdates.Therefore,thetotalcomplexityofthe prepro-cessingpartcanbeestimatedbyO

(

n2

₍

_n+m

₎

_Tr

₎

_{operations. } 3.2.Innercycle:relativeresourceloadcalculation

Letus consider two resources: X and Y. The earliest possible momentoftimewhenj∈NcanstarttouseresourcesX,Y∈Risrj.

Foranyt≤ pj intimeinterval [rj,rj+t

)

,the amountofresources

XandYconsumedby taskj cannot bemorethan t_{· ajX} andt_{· ajY} respectively.If[CPs

j,CPej

)

=∅taskjusesexactlyajXandajYineach

oftimeslots[CPs

j,CPsj+1

)

,...,[CPej− 1,CPje

)

.

Let

Fig. 4. A jX ( t ) – the highest possible amount of resource X used by the non-

compulsory part of task j in interval [0 , t + 1) .

– bethehighestpossibleamountofresourceX usedbythe non-compulsorypartoftaskjininterval[0_,t₊1

)

(Fig.4).

Theinnercycleprocedureprocessestimeslotbytimeslot start-ing atthe moment t=0. In each time slot, the highest possible consumption of resources X and Y by all tasks of set N is esti-matedby takingintoaccountonlynon-compulsorypartsoftasks. The amount of resource X used by non-compulsory part of task j∈N ininterval [0,t+1

)

isdenoted by ujX(t), andthe total

con-sumptionofresourceXbyalltasksininterval[0,t+1

)

isdenoted byUX

(

t

)

=j∈NujX

(

t

)

.

The main idea behindthe developed algorithmis to calculate anupperboundonthepossibleconsumptionofresourcesXandY takingintoaccountthenon-compulsorypartsofthetasksthatcan beassignedtoeachtimeinterval.Inageneralcase,thedemandin resourcesofallsuchtaskswillbe superiortotheresource capac-ity,butnot necessarilyin the sameproportion forresource X as forresourceY.Theoriginalityoftheproposed approachistotake intoaccount the fixed proportion ofusage ofdifferent resources byeachtask.Tocalculatetheupperbound,alinearcombinationof fractionalpartsofsuchtasksthatusethehighestavailableamount ofbothresourcesisresearched.Amongdifferentcombinations us-ingthetotalityoftheavailableresources,ageometricalgorithmis usedtochoosethecombinationforwhichthevalidityofthelower boundonthemakespanisprovenbyTheorems1and2.For exam-ple,inageneralcase,thisalgorithmwillpreferthecombinationof thetasksusingbothresourcestothecombinationusingthetasks requiringonlyoneresource.Thisisdoneinordertoprovidemore flexibilityintheresourceusageforremainingtimeintervals.

For resources X and Y, the consumption scheme

ϕ

is defined whennon-compulsory usedamountsofresources ujX(t) andujY(t)

are known for any task j_∈N and time slot t₌0_,...,T_{− 1}. The consumption scheme is valid if for any task j∈N and time slot t=0,...,T− 1thefollowingconditionshold:

ujX

(

t

)

≤ AjX

(

t

)

, ujY

(

t

)

≤ AjY

(

t

)

, ujX

(

t

)

ujY

(

t

)

= ajX ajY.

Thefirstandthe second inequalitiesare associatedwiththe def-initionsof A_jX(t) andA_jY(t), respectively. The last equalityis very important,sinceitrequiresthattheproportionofresourcesXand Yusedbytaskj∈Nremainsthesameintheconsidered consump-tion scheme asin any feasible schedule.Note that each feasible schedulewithdeadlineTpossessesvalidconsumptionschemesfor allresources.

All time slotst=0,...,T− 1are consideredone by oneinan iterative way and for each of them, the following optimization problemissolved:

Problem 2. For each j∈N values ujX

(

t− 1

)

and ujY

(

t− 1

)

are

given and functions AjX(t), AjX(t) are defined. The objective is to

determineujX

(

t

)

≥ ujX

(

t− 1

)

andujY

(

t

)

≥ ujY

(

t− 1

)

forall tasks

j_∈N such that UX(t) and UY(t) reach the highest possible value

(sinceweareinterestedinanupperboundonresource consump-tion).Thefollowingconstraintsshouldbetakenintoaccount:

ujX

(

t

)

− ujX

(

t− 1

)

ujY

(

t

)

− ujY

(

t− 1

)

= ajX ajY, ujX

(

t

)

≤ AjX

(

t

)

, ujY

(

t

)

≤ AjY

(

t

)

,  j∈N

(

ujX

(

t

)

− ujX

(

t− 1

))

≤ cX

(

t

)

,  j∈N

(

ujY

(

t

)

− ujY

(

t− 1

))

≤ cY

(

t

)

.

Ifforanytime slotthereismore thanonesolutionwhich satisfy theseconditions, thesolution will be chosen usingthe following criterion:

min

j∈N



(

ujX

(

t

)

− ujX

(

t− 1

))

2+

(

ujY

(

t

)

− ujY

(

t− 1

))

2. (2)

Thenecessityofthiscriterionisexplained byTheorem1.This problemcanbereformulatedintermsofvectors.

Problem 3. For time slot t we have a set of two-dimensional vectors

v

1=

(

A1X

(

t

)

− u1X

(

t− 1

)

,A1Y

(

t

)

− u1Y

(

t− 1

))

,...,

v

n=

(

AnX

(

t

)

− unX

(

t− 1

)

,AnY

(

t

)

− unY

(

t− 1

))

associated with all

tasks of set N. The objective is to find a set of coefficients

{

α1

,...,

α

n

}

∈[0,1]suchthatthelinearcombination

L=

α

1

v

1+· · · +

α

n

v

n

has the highest possible projections on the axes (it corresponds to the highestusage of the resources) andsatisfies the inequali-tiesLX≤ cX

(

t

)

andLY ≤ cY

(

t

)

.Ifthere ismorethan onesolution,

choosetheonewiththelowestsumofthevectorslengthsi.e.(it correspondstocriterion2):

min

j∈N

α

j

|

v

j

|

. (3)

Lemma2. Problems2and3areequivalent.

Proof. Inproblem2wehavetofindujX(t),ujY(t)suchthatAjX

(

t

)

≥

ujX

(

t

)

≥ ujX

(

t− 1

)

,AjY

(

t

)

≥ ujY

(

t

)

≥ ujY

(

t− 1

)

and

ujX

(

t

)

− ujX

(

t− 1

)

ujY

(

t

)

− ujY

(

t− 1

)

=

ajX

ajY.

SincevaluesujX

(

t− 1

)

,ujY

(

t− 1

)

,AjX(t)andAjY(t)aregiven,each

pair of values ujX(t), ujY(t) can be associated with a vector

v

j=

(

ujX

(

t

)

− ujX

(

t− 1

)

, ujY

(

t

)

− ujY

(

t− 1

))

where ujX

(

t

)

=ujX

(

t−

1

)

+

α

j

(

AjX

(

t

)

− ujX

(

t− 1

))

and ujY

(

t

)

=ujY

(

t− 1

)

+

α

j

(

AjY

(

t

)

−

ujY

(

t− 1

))

,

α

j∈[0, 1]. Therefore linear combination L=

α1

v

1+ · · · +

α

n

v

nhasprojections LX=  j∈N

α

j

(

ujX

(

t

)

− ujX

(

t− 1

))

=UX

(

t

)

− UX

(

t− 1

)

, LY=  j∈N

α

j

(

ujY

(

t

)

− ujY

(

t− 1

))

=UY

(

t

)

− UY

(

t− 1

)

onaxesOXandOYrespectively. SinceUX

(

t− 1

)

andUY

(

t− 1

)

are

fixed,thehighestpossiblevaluesofUX(t)andUY(t)correspondto

thehighestvaluesofLXandLY(thehighestusageoftheresources).

Fig. 5. Polygon construction.

Fig. 6. Geometric algorithm: subcase 2a.

aboveconditions,thesecondobjective(3)isappliedtochoosethe solution.Note,that

 j∈N

α

j

|

v

j

|

=  j∈N



(

ujX

(

t

)

− ujX

(

t− 1

))

2+

(

ujY

(

t

)

− ujY

(

t− 1

))

2, hence(3)isequivalentto(2). 

The following geometric algorithm is designed to solve opti-mallyProblem3.

1. Construct the convexcentrally symmetricpolygon of possible linear combinations of vectors

v

1,. . .,

v

n with coefficients in

[0,1] as follows. Let OV=

v

1+· · · +

v

n. The upper and lower

borders of this polygon are associated with the sequences of vectorsplacedindescendingandascendingordersoftangents ofthe angleformed withthe abscissaaxis(Fig.5). Further,it isassumed that thesevectorsare alreadysorted inascending orderoftangents.

2. ConsiderpointC

(

c_X

(

t

)

,c_Y

(

t

))

.IfCisoutsidethepolygon,three followingsubcasesarepossible.

(a) CbelongstozoneZ1,i.e.CX≥ VX,CY≥ VY.Theprocedure

re-turns

α

j=1foreachj∈N(Fig.6).

(b) C belongs to zone Z2, i.e. CY<VY andthe projection of C

on the axis of ordinates intersects the polygon. The pro-cedure returns a set of coefficients

α

j, such as j∈N

α

j

v

j

corresponds to the rightmost intersection of polygon and Y₌c_Y

(

t

)

(Fig.7).

(c) CbelongstozoneZ3,i.e.CX<VX andtheprojectionofCon

theaxeofabscissaintersectsthepolygon.Theprocedure re-turnsasetofcoefficients

α

_j,suchas_j∈N

α

_j

v

jcorresponds

tothehighestintersectionofpolygonandX=c_X

(

t

)

(Fig.8). 3. If point C is inside the polygon (zoneZ4), we make a trans-lationof thelower border onvector OC

(

c_X

(

t

)

,c_Y

(

t

))

and find thesetofcoefficients

{

β1

,...,

β

n

}

whichdefinesthepathfrom

pointCtoV(dashedlineinFig.9),thetranslatedlowerborder

Fig. 7. Geometric algorithm: subcase 2b.

Fig. 8. Geometric algorithm: subcase 2c.

Fig. 9. Geometric algorithm: step 3.

and theborders of the initial polygon,i.e.

β1

v

1+· · · +

β

n

v

n=

OV− OC.Thentheprocedurereturnsthesetofcoefficients

{

1−

β1

,. . .,1₋

β

n

}

that corresponds toa polyline which isshown

onFig.9.

The following lemma is requiredto prove the correctness of the geometricalgorithm.

Lemma 3. Let two sets of vectors A=

{

v

A

1,...,

v

Al

}

and B=

{

v

B 1,...,

v

Bk

}

such that l j=1

v

Aj= k

j=1

v

Bj andthepolygonassociated

withAbetotallyincludedintothepolygonassociatedwithB(Fig.10). Thenthetotal lengthof vectors ofset Ais not superior to the total

Fig. 10. Lemma 3 . Polygon associated with A be totally included into the polygon associated with B .

Fig. 11. Proof of Lemma 3 .

lengthofvectorsofsetB,i.e.

l  j=1

|

v

A j

|

≤ k  j=1

|

v

B j

|

.

IfA_=B,thentheinequalityisstrict.

Proof.Let us compare polygons A and B vector by vector. If we find a difference on vector

v

i, let us do an additional

construc-tion as on Fig. 11, extending vector

v

i to the intersection with

thepolygonB.Then, letusmakea centrallysymmetric construc-tion forupper border vectorsto obtain newpolygon B, such as A⊂ B⊂ B.Then, letusreplacepolygonBbypolygonB.Obviously, sucha changereducestheperimeter.Repeatingitno moretimes thanthe numberof edges of polygonA, we obtain two identical

polygons. 

Lemma4. Theproposedgeometricalgorithmfinds anoptimal solu-tionforProblem3inO(n2_)operations.

Proof.Let us show that in each case algorithm finds the set of vectors which sum has the greatest possible projections on the axes (which correspond to the highest possible consumptions of theavailableresources).IfpointCisoutsidethepolygon,thenthe coefficientsassociatedeitherwiththeinitialsetofvectors(2a)or withtheintersectionoflineY₌CY withalowerborderlineofthe

polygon(2b)or withthe intersectionof lineX=CX withan

up-perborderlineofthepolygon(2c)isreturned.Allthispointshave thehighestpossiblecoordinates amongallpointsofthe polygon, whichcoordinatesare not higherthanCX andCY.If Claysinside

thepolygon,then algorithmreturnsthesetofvectorswhichsum hasthecoordinates(CX,CY).

Now let us show that the obtained set of vectors

{

α1

v

1,. . .,

α

n

v

n

}

hastheshortesttotallengthamongallthosewith

themaximalsumofthecoordinates.InthecaseswherepointCis locatedin zonesZ1, Z2andZ3, there is onlyone solution which satisfiesthiscondition.WhenCliesinzoneZ4,thealgorithmfinds thesetofvectors

{

β1

v

1,. . .,

β

n

v

n

}

thatcorresponds topolylineCV

withthelongestpossiblelength.Finally,thealgorithmreturnsthe setofcoefficients

{

α1

,...,

α

n

}

=

{

(

1−

β1

)

,...,

(

1−

β

n

)

}

.Since

 j∈N

|

α

j

v

j

|

=  j∈N

|

v

j

|

−  j∈N

|

β

j

v

j

|

,

set of vectors

{

α1

v

1,. . .,

α

n

v

n

}

has the shortest possible sum of

vectorlengths.Thereforeobtainedsolutionisoptimalwithrespect tocriterion3.ThusLemma4isverified.

The greatest number of operations is required for the case whenCliesinzoneZ4.In thiscase, alowerbordertranslationis required,the intersectionpoint withthe polygonborderline can befoundinO(n2_{)operations. }

ThefollowinglemmasshouldbeprovedaheadTheorem1. Lemma 5. Let us have a set of two-dimensional vectors A=

{

v

1,...,

v

m

}

placed in tangents ascending order and a point V

which belongs to the polygonassociated with A. Suppose thatA=

{

α1

v

1,...,

α

m

v

m

}

is a setof vectors, suchthat

∀

j=1,...,m:

α

j∈

[0,1],m

j=1

α

m

v

m=OV andthesumofvectorlengthsmj=1

α

m

|

v

m

|

is minimal.Then, for any setof vectors B=

{

β1

v

1,...,

β

m

v

m

}

, such

that

β

∈[0,1]and  βjvj∈B

β

j

v

j=  αjvj∈A

α

j

v

j=OV,

the polygon associated with A belongs to the polygon associated withB.

Proof. Letusassume thecontrary.Suppose that thereisa setof vectorsBwhichsatisfiesthe Lemma’sconditionsbutthepolygon associatedwithAdoesnotbelongtothepolygonassociatedwith B.Lemma3impliesthatthepolygonassociatedwithBcannotfully belong to the polygon associated with A. Therefore, we have to dealonlywiththesituationwheretheconsideredpolygonsare in-tersected.Hence,polygons’lowerborderlineshaveatleastfour in-tersectionpointsincludingOandV.Letustakealookattwo con-secutive intersectionpointsK andL, such that lower border seg-mentKLA _liesunderKLB_{.SincebothpolylinesOV}A _andOVB_consist

of vectorsplaced in theascending order of tangents, thevectors whichconstitute polyline KLB _{cannot belong tothe setofvectors}

whichconstituteOKA _andLVA_{.Hence,wecanreplaceKL}A _byLBB

andthusdecreasetheperimeterofthepolygonassociatedwithA. ThisviolatestheassumptionthatthesumofvectorslengthsofA isminimal.Lemma5isproved. 

Lemma 6. Suppose that there are two sets of two-dimensional vectors A=

{

v

A

1,...,

v

Am

}

and B=

{

v

B1,...,

v

Bk

}

such that



j∈A

v

Aj=



j∈B

v

Bj andthepolygonassociated withsetA istotallyincluded in

thepolygonassociated withset B.Therefore,thereis a setof coeffi-cients

α

1

1,...,

α

k1,...,

α

m

1,...,

α

km∈[0,1],whichsatisfiesthe

follow-ing: m  j=1

α

j 1=1, ... m  j=1

α

j k=1, k  i=1

α

1 i

v

B k= v A 1, ... k  i=1

α

m i

v

Bk=

v

Am.

Proof. Let us find coefficients

α

1

1,...,

α

k1 explicitly, using the

graphic approach described in Figs. 12–14. The polygon asso-ciated with A is totally included in the polygon associated with B, which is included in the polygon associated with B.

Fig. 12. Lemma 6 .

Fig. 13. Finding vA

1 =  ki=1 αi1 vBk .

Fig. 14. New polygons A  _{and B}  _.

Therefore the polygons associated with sets A=A

\

{

v

A

1

}

and B₌

{

v

B 1−

α

11

v

B1,...,

v

Bk−

α

1 k

v

B

k

}

satisfy the initial conditions of

Lemma 6. We can iteratethisprocedure to find all requiredsets ofcoefficientswhichcorrespondtoallvectorsofsetA. 

Thepresentedgeometricalgorithmconsidersthetimeslotsone by one in an iterativeway. At each step,an optimalsolution for

Problem 2isfoundforeach pairofresourcesX andY.LetUX|Y(t)

andU_Y|X(t) berespectivelytheamountsofresources XandYused bysetoftasksNintimeinterval[0,t+1

)

.Thefollowingtheorem proves that UX|Y(t) andUY|X(t) provideupper boundsonthe

con-sumptionofresourcesXandYduringtimeinterval[0_,t₊1

)

. Theorem1. Underanyvalidconsumptionscheme,theamountof re-sourcesXandYconsumedininterval[0_,t₊1

)

isnotmorethan

UX|Y

(

t

)

+ t  t₌₀

(

cX

(

t

)

− cX

(

t

))

and UY|X

(

t

)

+ t  t=0

(

cY

(

t

)

− cY

(

t

))

respectively.

Proof. Assume the contrary. Suppose that there is a consump-tion scheme

ϕ

∗ which violates the initial assumption and uses morethanU_X|Y

(

t

)

+t

t=0

(

cX

(

t

)

− cX

(

t

))

resourceXintime

inter-val[0,t+1

)

.Ifthereismorethan oneofsuchschemes,consider

Fig. 15. The highest possible consumption subject to C ∈ Z 2 .

theone which uses the highesttotal amount ofresources X and Yintimeinterval [0,t+1

)

.Letu∗_jX

(

t

)

be theamountofresource X used by taskj underconsumption scheme

ϕ

∗ in time interval [0,t+1

)

.

Let us consequently consider the resource consumption at u∗_jX

(

t

)

for t₌0_,...,t. Suppose t is the first moment of time whichsatisfiesu∗_jX

(

t

)

=ujX

(

t

)

oru∗jY

(

t

)

=ujY

(

t

)

forsomej∈N.

We consider polygon OV associated with the set of vectors

v

j=

(

AjX

(

t

)

− ujX

(

t− 1

)

,AjY

(

t

)

− ujY

(

t− 1

))

corresponding to the

consumptionsofresourcesbyeachtaskj∈N.ThevertexofOVwith thehighestcoordinatesXandYisdenotedbyV.Wealsoconsider pointC

(

c_X

(

t

)

,c_Y

(

t

))

.

Thereare four possiblecases ofpositioning Cin zonesZ1, Z2, Z3,Z4inrelationtopolygonOV.

1. C∈Z1. Each resource cannot be totally used, i.e.

(

j∈N

v

j

)

X <

c_X

(

t

)

and

(

_j∈N

v

j

)

Y <cY

(

t

)

.Inthiscase,foranyj∈N,the

fol-lowing conditions hold ujX

(

t

)

=AjX

(

t

)

and ujY

(

t

)

=AjY

(

t

)

.

Therefore, we can change the consumption during time slot [t,t+1

)

for

ϕ

∗ using the full amounts of resources aswell asunder

ϕ

withoutviolationofanyassumption.

2. C_∈Z2. Resource Ycannot be totallyused. Ifunder

ϕ

∗ not full amount of resource X is used, we can use it by one of task j∈N which ajX>0 andujX(t)<AjX(t). Such a change will not

decrease the values of functions U∗_jY

(

t

)

and U∗_jX

(

t

)

for any t. Therefore, we can consider only the case where U∗_jX

(

t

)

− U∗_jX

(

t− 1

)

=c_X

(

t

)

. Note that the highest resource consump-tion can be achieved only by using a linear combination of vectors withthe highestpossible ratio a_ajY

jX (Fig. 15). Hence, if

there is a difference inresource consumption between

ϕ

and

ϕ

∗_{, then there is a task j∈N such that ujY}

₍

_t

₎

_>u∗

jY

(

t

)

and

thereisataski_∈N whichholdsu_iY

(

t

)

_<u∗_iY

(

t

)

and a_ajY

jX <

a_iY a_iX .

Thismeansthatwecanreplacethepartoftaskiusedintime slot [t,t+1

)

by j under

ϕ

∗ without violation of any con-straint.Suchachangewillnotdecreasethevaluesoffunctions U∗_jY

(

t

)

andU∗_jX

(

t

)

foranyt.Letusapplythesamechangesuntil ujY

(

t

)

=u∗jY

(

t

)

doesnotholdforanyj∈N.

3. C∈Z3.Thiscaseissimilartothepreviousone.Weonlyneedto swapresourcesXandYinthedescriptionofcase2.

4. C∈Z4.This meansthat under

ϕ

the fullcapacities ofboth re-sourcescanbeachieved.

Suppose that under

ϕ

∗ full capacities of resources X and Y are achievedintimeslot[t,t+1

)

.AccordingtoLemma4we obtain that the polygon related to the resource consumption in [t,t+1

)

under

ϕ

has the lowest perimeter of all poly-gons related to the highest consumption of resources X and Y. Lemma 5implies that it is totally included in the polygon

Fig. 16. Master algorithm.

relatedto

ϕ

∗.Therefore,wecan changeresourceconsumption under

ϕ

∗ intime slot[t,t+1

)

to theconsumption used un-der

ϕ

by takingrequiredpartsof

α

j

v

j fromthe future

times-lots ofinterval [t₊1_,T

)

.Lemma 6implies that itis possible to replacecorrectlyallpartsofvectors

v

1,. . .,

v

n usedforthis

procedure by linear combinations of vectors

v

∗

1,. . .,

v

∗n, which

wereusedin

ϕ

∗previously.Thus,wecanmakeachangein

ϕ

∗ without violating the conditions of the Theorem and we ob-tain equal consumptionsof

ϕ

∗ and

ϕ

for time slot [t,t+1

)

without increasing or decreasing any amount of resources X andY beingused inanytime slot [t₊1_,t₊2

)

_,...,[t_,t₊1

)

. ForthecasewherefullcapacitiesofresourcesXandYarenot achieved together in time slot [t,t+1

)

under

ϕ

∗, the con-sumptionscheme

ϕ

∗ismodifiedsimilarly.

Depending onthe casewe faceintime slot [t,t+1

)

we ap-ply the procedure which does not decrease the values of func-tions UX(t) or UY(t). After having been proceeded with all time

slots, we obtain ujX

(

t

)

=u∗jX

(

t

)

and ujY

(

t

)

=u∗jY

(

t

)

for any j∈N

andt=0,...,T− 1. 

3.3.Maincycle:masteralgorithm

The masterpartofouralgorithm usesProcedure2forG(N,E) andthegraphwithreversed precedencerelationsG

(

N_,E

)

to com-pare,foranyresource X∈R, asumofupperboundson its possi-bleconsumed amountin intervals[0, t) and(t, T] withthe total amountofresourcerequiredforall tasks_j∈Na_jXp_j.Ifthelatter islowerthattheformer,theconsideredproblemisconsidered in-feasiblefortimehorizonT.

Then, this verification is made for all moments of times t= 0,1,2,...,T− 1 forall pairsofresourcesX,Y∈R,forwhich func-tionsUX|Y(t) andUY|X(t) are calculated.Eachfeasibleschedule

de-fines a valid consumption scheme. Theorem 1 implies that for eachresourceX_∈Randanyt,anupperboundoftheconsumption ofresource X by tasks in non-compulsory parts of time interval [0,t+1

)

underany valid consumption scheme can be estimated byfunction

UBX

(

t

)

=min Y∈RUX|Y

(

t

)

.

Further,thesameprocedures,includingpreprocessing,areapplied tosetoftasksNbutforthegraphwithreversedprecedence rela-tionsG

(

N_,E

)

_,the valuesof functionsU_X_|Y

(

t

)

are calculated. Asa result,foreachresourceX∈Rweobtainafunction

UBX

(

t

)

=min_Y

∈RU



X|Y

(

t

)

whichisanupperboundontheconsumptionofresourceXin non-compulsorypartsoftimeinterval

(

T_{− t− 1,}T]forthetasksofset N.

Afterthat,thealgorithmrepeatsthesamecycleonallmoments oftimet=1,...,T− 1tocheckifforanyresourceX∈Rasumof upperbounds on its available capacityin intervals [0,t+1

)

and

(

t+1,T] (Fig. 16) is not lower than the sumof the demands in

thisresourcebyalltasks,i.e.  j∈N ajXpj≥ UBX

(

t

)

+UBX

(

T− t− 2

)

+ T−1  t=0

(

cX

(

t

)

− cX

(

t

))

.

Ifthisconditionisviolated,thentheproblemisinfeasiblefortime horizonT.

Theorem2. Suppose thatthemaster algorithmwas used forsetof tasksN,set ofresources R andtime horizon T. Iffor any X∈Rand t=0,...,T− 1,thefollowinginequalitydoesnothold:

 j∈N ajXpj≤ UBX

(

t

)

+UBX

(

T− t− 2

)

+ T−1  t=0

(

cX

(

t

)

− cX

(

t

))

, (4)

thenthereis nofeasibleschedulewithmakespaninferiororequalto T.

Proof: According to Theorem 1 we obtain that for any feasi-bleschedule

π

∈



(N,R), theamountofresourceX usedbytasks intime interval[0,t+1

)

doesnotexceedUBX

(

t

)

+tt=0

(

cX

(

t

)

−

c_X

(

t

))

.The amount of resourceX usedin non-compulsory parts of interval

(

t+1,T] for tasks does not exceed UB_X

(

T− t− 2

)

+ T−1

t=t+1

(

cX

(

t

)

− cX

(

t

))

.Therefore,takingintoaccountcompulsory

partsforanyfeasibleschedule

π

, theamountofresourceX used inhorizon[0,T]doesnotexceed

UBX

(

t

)

+UBX

(

T− t− 2

)

+ T−1



t=0

(

cX

(

t

)

− cX

(

t

))

.

If inequality (4) is violated, then for each feasible schedule

π

∈



(N,R) the amount ofresource X requiredforprocessing all tasks of set N cannot be used during time interval [0, T]. This provesthestatementoftheTheorem.

3.4. Binarysearch

Inthispart,a simplebinarysearch isusedtofindthe highest possiblevalueofthetimehorizonTwhichsatisfiestheconditions inTheorem2.

Theorem 3. The developed algorithm finds a lower bound on the makespan in O

(

n2_r

₍

_n+m+r

₎

_TlogT

₎

_{operations, where n is the} numberoftasks,Tisthetimehorizon,risthenumberofresources,m is thehighest numberofbreakpointsof thecapacity functionofone resource.

Proof. Thenumberofbi-sectionsearchiterationscanbeestimated by O(logT) operations. At each iteration, the preprocessing takes O

(

n2

₍

_n+m

₎

_rT

₎

_{operations. The master part takes O(n}2_T) opera-tions for each pair of resources. Number of pairs of resources is O(r2_{).Therefore,thetotalcomplexityofthealgorithmcanbe} esti-matedbyO

(

n2_r

₍

_n+m+r

₎

_TlogT

₎

_{operations. }

4. Numericalexperiments

ThealgorithmwasimplementedinC++.Twoseriesof numeri-calexperimentswere carriedout usingIntel Corei72.8gigahertz CPU with16 gigabytes RAM. In the first one, the algorithm was testedon thewell-known PSPLIBbenchmark(Kolisch&Sprecher, 1997).Inthesecondone,thealgorithmwasappliedtolarge-scaled RCPSPinstancesbasedonrealdataprovided byKuznetsovDesign Bureau. The results of thetests are presentedin Tables 1 and2, respectively.

The first series of tests was performed for the problem in-stances from PSPLIB benchmark. The objective was to compare theresults providedby ourapproach withthe bestknown lower bounds (BKLB), presented at PSPLIB website (consulted in July 2017).TheresultsaregiveninTable1).Theyshowthatfor66%of

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