Inapproximability Results for Scheduling with Interval and Resource Restrictions
aa r X i v : . [ c s . CC ] J u l Inapproximability Results for Scheduling withInterval and Resource Restrictions
Marten Maack
Department of Computer Science, Kiel University, Kiel, [email protected]
Klaus Jansen
Department of Computer Science, Kiel University, Kiel, [email protected]
Abstract
In the restricted assignment problem, the input consists of a set of machines and a set of jobs eachwith a processing time and a subset of eligible machines. The goal is to find an assignment ofthe jobs to the machines minimizing the makespan, that is, the maximum summed up processingtime any machine receives. Herein, jobs should only be assigned to those machines on which theyare eligible. It is well-known that there is no polynomial time approximation algorithm with anapproximation guarantee of less than 1.5 for the restricted assignment problem unless P=NP. Inthis work, we show hardness results for variants of the restricted assignment problem with particulartypes of restrictions.For the case of interval restrictions—where the machines can be totally ordered such that jobsare eligible on consecutive machines—we show that there is no polynomial time approximationscheme (PTAS) unless P=NP. The question of whether a PTAS for this variant exists was statedas an open problem before, and PTAS results for special cases of this variant are known.Furthermore, we consider a variant with resource restriction where the sets of eligible machinesare of the following form: There is a fixed number of (renewable) resources, each machine has acapacity, and each job a demand for each resource. A job is eligible on a machine if its demand isat most as big as the capacity of the machine for each resource. For one resource, this problem hasbeen intensively studied under several different names and is known to admit a PTAS, and for tworesources the variant with interval restrictions is contained as a special case. Moreover, the versionwith multiple resources is closely related to makespan minimization on parallel machines with a lowrank processing time matrix. We show that there is no polynomial time approximation algorithmwith a rate smaller than 48 / ≈ .
02 or 1 . Theory of computation → Scheduling algorithms; Theory ofcomputation → Problems, reductions and completeness
Keywords and phrases
Scheduling, Restricted Assignment, Approximation, Inapproximability, PTAS
Funding
This work was partially supported by the German Academic Exchange Service (DAAD)and by the German Research Foundation (DFG) project JA 612/15-2.
Acknowledgements
We thank Malin Rau and Lars Rohwedder for helpfull discussions on the prob-lem.
Consider the restricted assignment problem: Given a set of machines M and a set of jobs J each with a processing time or size p j and a subset of eligible machines M ( j ) ⊆ M , thegoal is to find a schedule σ : J → M with σ ( j ) ∈ M ( j ) for each job j and minimizing themakespan C max ( σ ) = max i ∈M P j ∈ σ − ( i ) p j . Inapproximability Results for Scheduling with Interval and Resource Restrictions
In a seminal work, Lenstra, Shmoys and Tardos [25] presented a 2-approximation forrestricted assignment and also showed that there is no polynomial time approximation al-gorithm with rate smaller than 1 . M ( j ) = M for each job j , we have the classical problem of makespanminimization on identical parallel machines (machine scheduling) which is already stronglyNP-hard. On the other hand, machine scheduling is well-known to admit a polynomial timeapproximation scheme (PTAS) due to a classical result by Hochbaum and Shmoys [13]. Inrecent years, the approximability of special cases of restricted assignment has been inten-sively studied (see, e.g., [5, 10, 16, 20]) with one line of research focusing on the existenceof approximation schemes (see, e.g., [11, 17, 28, 29]). The present work seeks to contributein this research direction. Interval Restrictions.
Arguably one of the most natural variants of the restricted assign-ment problem is the case of scheduling with interval restrictions (RAI). In this variant, themachines are totally ordered and each job is eligible on consecutive machines. More precisely,we have M = { M , . . . , M m } , and for each job j we have M ( j ) = { M ℓ , . . . , M r } for someindices ℓ, r ∈ [ m ]. Several special cases of RAI are known to admit a PTAS: the hierarchicalcase [29], where for each job the interval of eligible machines starts with the first machine;the nested case [28, 11], where M ( j ) ⊆ M ( j ′ ), M ( j ′ ) ⊆ M ( j ) or M ( j ) ∩ M ( j ′ ) = ∅ foreach pair of jobs ( j, j ′ ); and the inclusion-free case [33, 23], where M ( j ) ⊆ M ( j ′ ) impliesthat j and j ′ share either their first or last eligible machine. Furthermore, for general RAI,a 2 − / (max j ∈J p j )-approximation due to Schwarz [33] is known (assuming integral pro-cessing times); and the special case with two distinct processing times is even polynomialtime solvable [38]. Note that the problem has also been studied in the context of onlinealgorithms (see [24, 27]).The question of whether there is a PTAS for RAI has been posed by several authors[21, 33, 38]. As the main result of the present work, we resolve this question in the negative: ◮ Theorem 1.
There is no PTAS for scheduling with interval restrictions unless P = NP. Resource Restrictions.
The second variant considered in this work, is the problem ofscheduling with resource restrictions with R resources (RAR( R )). Herein, a set R or R (renewable) resources is given, each machine i has a resource capacity c r ( i ) and each job j has a resource demand d r ( j ) for each r ∈ R . Job j is eligible on machine i if d r ( j ) ≤ c r ( i )for each resource r . For R = 1, the problem is equivalent to the mentioned hierarchicalcase and has been studied intensively [26, 27]. Furthermore, it is not hard to see that RAIis properly placed between RAR(1) and RAR(2) (see Section 3) and hence there is a closerelationship between the two problems. For arbitrary R , the problem was mentioned in awork by Bhaskara et el. [2] under the name of geometrically restricted scheduling but tothe best of our knowledge it has not been further studied up to now. There is, however, aclose relationship to the low rank version of makespan minimization on unrelated parallelmachines (unrelated scheduling) introduced in [2]. In the problem of unrelated scheduling,the processing time of each job is dependent on the machine it is scheduled on, that is, There is a paper [22] claiming to have found a PTAS for RAI. However, according to [35], the result isnot correct and the authors published a revised version of the paper [23] claiming a less general result,namely, a PTAS for the inclusion-free case. The demands d ( j ) and capacities c ( i ) may be interpreted as points in R -dimensional space. arten Maack and Klaus Jansen 3 a processing time matrix ( p ij ) j ∈J ,i ∈M is given in the input. Unrelated scheduling is aclassical problem, and the 2-approximation by Lensta et al. [25] was actually formulated forthis problem. Restricted assignment can be seen as a special case of unrelated schedulingby setting p ij = p j for i ∈ M ( j ) and p ij = ∞ otherwise. In the rank D version of unrelatedscheduling (LRS( D )), the processing time matrix has a rank of at most D , or, equivalently [6],we may assume that there are D -dimensional size vectors s ( j ) for each job j and speed vectors v ( i ) for each machine i such that p ij = P Dk =1 s k ( j ) · v k ( i ). Considering the latter definition,scheduling with resource restrictions may intuitively be seen as the restricted assignmentequivalent of low rank unrelated scheduling. It is not hard to see that formally alreadyfor RAR(1) instances the processing time matrix can have rank |M| . However, LRS( D )includes approximations of any RAR( D −
1) instance with arbitrary precision (see Section 3for details). The case with D = 1 of LRS( D ) is equivalent with the classical problem ofmakespan minimization on uniformly related machines and well known to admit a PTAS[14]. Bhaskara et el. [2] gave a quasi-polynomial time approximation scheme (QPTAS) for D = 2, and showed that there is no PTAS or approximation algorithm with rate smallerthan 1 . D ≥ D ≥ D = 4 to D = 3 by Chen et al. [6] and from D = 7 to D = 4 by Chen, Ye and Zhang [7].We present similar inapproximability results for scheduling with resource restrictions: ◮ Theorem 2.
There is no approximation algorithm with rate less than / ≈ . or . for scheduling with resource restrictions with or resources, respectively, unless P = NP.
Santa Claus.
The problems of restricted assignment and unrelated scheduling are also stud-ied with the reverse objective of maximizing the minimal machine load min i ∈M P j ∈ σ − ( i ) p ij .Usually these variants are described in a more game theoretical context with players insteadof machines, goods instead of jobs, and values instead of processing times, and sometimesunrelated scheduling with the reverse objective is called the Santa Claus problem. In this pa-per, we will mostly stick to the scheduling notation but denote the variants of the consideredproblems with reverse objective as the Santa Claus version of the respective problem.For the Santa Claus version of the restricted assignment problem a 13-approximationdue to Annamalai, Kalaitzis and Svensson [1] is known, which has been improved to a rateof 6 + ε by both Cheng and Mao [8] and Davies, Rothvoss and Zhang [9]. PTAS results areknown for the case without restrictions [40] and the inclusion-free interval case [23].Our results can be directly transferred to the Santa Claus versions of the respectiveproblems: ◮ Theorem 3.
Unless P = NP, there is no PTAS for the Santa Claus version of schedulingwith interval restrictions and no approximation algorithm with rate less than / or for the Santa Claus version of scheduling with resource restrictions with or resources,respectively. Paper structure.
In the remainder of this section, we discuss some further related literatureand present preliminary considerations needed throughout the paper. In Section 2, wepresent our results for RAI; in Section 3, we discuss the problem of RAR( R ); and lastly, inSection 4, we present some open problems and possible future research directions. Further Related Work.
First note that if the number of machines is constant, there is afully polynomial time approximation scheme (FPTAS) already for unrelated scheduling [15].Furthermore, for some broad overview concerning parallel machine scheduling with different
Inapproximability Results for Scheduling with Interval and Resource Restrictions hierarchical =RAR(1) = chaintree-hierarchical nested bipartite permutation= inclusion-freebi-cograph RAI = convexconstant cliquewidth RAR(2)
Figure 1
An overview of the inclusion structure of several of the discussed variants of restrictedassignment. If two problems are connected, the upper includes the lower one. The dashed problemsdo not admit a PTAS, unless P=NP (see Theorem 1), the remaining ones do. kinds of restrictions in the context of online and approximation algorithms, we refer to thesurveys by Lee et al. [24] and Leung and Li [26, 27].We already discussed many variants of restricted assignment that admit a PTAS. Inparticular, Ou, Leung and Li [29] presented a PTAS for the hierarchical case; Epstein andLevin [11] and Muratore, Schwarz and Woeginger [28] for the nested case; and Schwarz [33]and Khodamoradi et al. [23] for the inclusion-free case. Another case that has been studiedin the literature is the tree-hierarchical case, where the machines can be arranged in a rootedtree such that for each job the set of eligible machines corresponds to a path starting at theroot. It was shown to admit a PTAS by Epstein and Levin [11] and Schwarz [34]. It isnot hard to see that all of the above cases contain the hierarchical case as a subcase, andthat the tree-hierarchical, nested and inclusion-free case are distinct. There is, however, avariant admitting a PTAS that covers both the nested and the tree-hierarchical case: Foreach instance of the restricted assignment problem the corresponding incidence graph is abipartite graph whose nodes are given by the jobs and machines and a job j is adjacent to amachine i if j is eligible on i . Jansen, Maack and Solis-Oba [17] showed that there is PTASfor restricted assignment for the case that the clique- or rank-width of the incidence graph isconstant. Furthermore, if the incidence graph is a bi-cograph the clique-width is well-knownto be small and this case covers the nested and tree-hierarchical case. The inclusion-freecase, on the other hand, is equivalent to the case that the incidence graph is a bipartitepermutation graph [23] which does not have a bounded clique-width [3]. Note that RAR(1)or RAI are equivalent to the cases that the incidence graph is a chain [12] or convex graph[22], respectively. For an overview of the discussed cases, we refer to Figure 1.Lastly, there has been a series of promising results in recent years concerning restrictedassignment and variants thereof, and we highlight a few of them. In a breakthrough result,Svensson [36] showed that a certain integer linear program modeling the problem has anintegrality gap of at most 33 /
17, which implies an algorithm approximating the optimalobjective value with rate 33 / ε for any ε > /
6, and in [19] thesame authors provide a quasi-polynomial approximation algorithm with rate 11 / ε thatalso outputs a corresponding schedule. For the special case of restricted assignment withonly two distinct processing times (not counting ∞ ) an approximation algorithm due toChakrabarty, Khanna and Li [4] with a rate slightly below 2 is known. Furthermore, thecase in which the set of eligible machines for each job has cardinality at most 2 has beenstudied under the name of graph balancing. Ebenlendr, Krcál and Sgall [10] presented a arten Maack and Klaus Jansen 5 . . . . Preliminaries.
In the following, we deal with satisfiability problems like the classical 3-SAT problem where a logical formula over variables x , . . . , x n is given. The formula is aconjunction of clauses, each clause is a disjunction of three literals, and a literal is either avariable or its negation. The goal is to decide whether there is a fulfilling truth assignment,that is, an assignment of the variables to the truth values “true” and “false”, denoted by ⊤ and ⊥ , respectively, such that the formula evaluates to “true”.We consider polynomial time approximation algorithms: Given an instance I of an op-timization problem, an α -approximation A for this problem produces a solution in timepoly( | I | ), where | I | denotes the input length. For the objective function value A ( I ) of thissolution it is guaranteed that A ( I ) ≤ α opt ( I ), in the case of an minimization problem, or A ( I ) ≥ (1 /α ) opt ( I ), in the case of an maximization problem, where opt ( I ) is the value ofan optimal solution. We call α the approximation guarantee or rate of the algorithm. Insome cases a polynomial time approximation scheme (PTAS) can be achieved, that is, an(1 + ε )-approximation for each ε >
0. If for such a family of algorithms the running time ispolynomial in both 1 /ε and | I | it is called fully polynomial (FPTAS).Nearly all the reductions in this work follow the same pattern: Given an instance I ofthe starting problem, we construct an instance I ′ of the variant of the restricted assignmentproblem considered in the respective case. For I ′ , all job sizes are integral and upperbounded by some constant T such that the overall size of the jobs equals |M| T . Obviously,if for such an instance a machine receives jobs with overall size more or less than T , themakespan of the schedule is greater than T . Then we show that that there exists a schedulewith makespan T for I ′ , if and only if I is a yes-instance. This rules out the existence ofan approximation algorithm with rate smaller than ( T + 1) /T and a PTAS in particular.Furthermore, for the Santa Claus version, approximation algorithms with rate smaller than T / ( T −
1) are ruled out.
The sole goal of this section is to prove Theorem 1, that is, the non-existence of a PTASfor RAI (given P =NP). Our starting point for the reduction is a satisfiability problem 3-SAT ∗ that we tailor to our needs. We show that 3-SAT ∗ is NP-hard via a straight forwardreduction from the 1-in-3-SAT problem, which is well-know to be NP-complete [31] anddiscussed in more detail below. Next, we provide a reduction from 3-SAT ∗ to the classicalrestricted assignment problem (with arbitrary sets of eligible machines). This reductionintroduces some of the needed gadgets and ideas for the main result. Lastly, we show howthe reduction can be refined for RAI, and this is the most elaborate step. Starting Point.
An instance of 1-in-3-SAT is a conjunction of clauses with 3 literals each.Each clause is a formula depending on 3 literals that is satisfied if and only if exactly oneof its literals takes the value ⊤ . We call such formulas 1-in-3-clauses in the following anddefine 2-in-3-clauses correspondingly. Inapproximability Results for Scheduling with Interval and Resource Restrictions
An instance of the problem 3-SAT ∗ also is a conjunction of clauses with exactly 3 literalseach. However, each of the clauses is either a 1-in-3-clause or a 2-in-3-clause and there areas many clauses of the first as of the second type. Furthermore, we require that each literaloccurs exactly twice . In the following, we denote a 1-in-3-clause or 2-in-3-clause with literals z , z and z by ( z , z , z ) or ( z , z , z ) , respectively.To see that 3-SAT ∗ is NP-hard, consider an instance of 1-in-3-SAT with n variables x , . . . , x n and m clauses. We now construct an equivalent 3-SAT ∗ instance. Let d i bethe number of times the variable x i occurs in the given 1-in-3-SAT formula. For eachvariable x i , we introduce new variables x i, , . . . , x i,d i and y i, , . . . , y i,d i along with clauses( x i, , ¬ x i, , y i, ) , . . . , ( x i,d i − , ¬ x i,d i , y i,d i − ) , ( x i,d i , ¬ x i, , y i,d i ) and clauses ( y i,j , ¬ y i,j , ¬ y i,j ) for each j ∈ [ d i ]. Note that each variable y i,j has to take the value ⊤ in a fulfilling assign-ment, due to the clause ( y i,j , ¬ y i,j , ¬ y i,j ) . The remaining clauses ensure, that for each i thevariables x i, , . . . x i,d i have the same value in a fulfilling assignment. Furthermore, for eachof the clauses of the original problem, we introduce one 1-in-3-clause and one 2-in-3-clause.The 1-in-3-clauses are obtained by exchanging the j -th occurrence of each variable x i with x i,j . Moreover, the 2-in-3-clauses are obtained by copying the new 1-in-3-clauses, negatingall the literals and turning them into a 2-in-3-clause. Hence, each 2-in-3-clauses evaluates to ⊤ , if and only if its corresponding 1-in-3-clause does. It is not hard to verify the correctnessof the reduction. Similar constructions are widely used, see, e.g., [37] or [6]. The remarkableaspect of the present construction lies in its symmetrical structure which helps to avoidadditional dummy gadgets in the following reductions. Simple Reduction.
In the following, we assume that an instance of 3-SAT ∗ with m C , . . . , C m , m C m +1 , . . . , C m and n variables x , . . . , x n is given.Note that we have 2 m clauses with 3 literals each, and 4 n occurring literals in total, hence3 m = 2 n . In addition to the ordering of the variables and clauses, we fix an ordering ofthe literals belonging to each clause, and an ordering of the occurrences of each variableby assigning an index t ∈ [4] to each of them. In particular, for each variable x j , t = 1 , t = 3 , x j . Furthermore, let κ : [ n ] × [4] → [2 m ] × [3] be the bijection defined asfollows: κ ( j, t ) = ( i, s ) implies that the t -th occurrence of x j is positioned in clause C i onposition s .We now define the restricted assignment instance. For some of the machines, we introduce private loads which is a synonym for jobs of the corresponding size that have to be scheduledon the respective machine because its the only eligible one. The sizes and sets of eligiblemachines of the introduced jobs are presented in Table 1 and the target makespan is givenby T = 322.For each clause C i , there are three clause machines CMach i,s with s ∈ [3] correspondingto its three literals, as well as three clause jobs CJob ◦ s ′ i,s ′ with s ′ ∈ [3] and ◦ s ′ ∈ {⊤ , ⊥} .We have ◦ = ⊤ and ◦ = ⊥ , as well as ◦ = ⊥ if C i is a 1-in-3 clause, and ◦ = ⊤ otherwise. Furthermore, each clause machine has a private load of 111.For each variable x j , there are two truth assignment machines TMach j,q with q ∈ [2]corresponding to the positive ( q = 1) and negative ( q = 2) literal of x j , as well as 2 truthassignment jobs TJob ◦ j with ◦ ∈ {⊤ , ⊥} .For each variable x j , there are eight variable jobs VJob ◦ j,t with t ∈ [4] and ◦ ∈ {⊤ , ⊥} corresponding to the two occurrences of the positive ( t ∈ { , } ) and negative ( t ∈ { , } )literal of x j .First note: arten Maack and Klaus Jansen 7 Table 1
The sizes and sets of eligible machines of the jobs in the simple reduction. The entryfor
CMach i,s marks the private load of the machine. The target makespan is given by T = 322.Job Size Eligible Machines CMach i,s
CMach i,s
CJob ⊤ i,s ′ CMach i, , CMach i, , CMach i, CJob ⊥ i,s ′ CMach i, , CMach i, , CMach i, TJob ⊤ j TMach j, , TMach j, TJob ⊥ j TMach j, , TMach j, VJob ⊤ j,t TMach j, ⌈ t/ ⌉ , CMach κ ( j,t ) VJob ⊥ j,t TMach j, ⌈ t/ ⌉ , CMach κ ( j,t ) ⊲ Claim 4.
The overall size of all the jobs is exactly |M| T . Proof.
We have 6 m + 2 n = 6 n machines (since 3 m = 2 n ), and—taking into account that wehave as many 1-in-3 as 2-in-3 clauses—the overall job size equals:6 m ·
111 + m (3 ·
100 + 3 · n (100 + 102 + 4 ·
111 + 4 · n = 322 · n = |M| T ⊳ We will show that there is a fulfilling truth assignment for the 3-SAT ∗ instance if and onlyif there is a schedule in which each machine receives jobs with load exactly T .For any job Job ◦ with ◦ ∈ {⊤ , ⊥} , we refer to ◦ as its truth configuration and saythat Job ◦ has ◦ -configuration. The rationale of the reduction is as follows: Each clausemachine CMach i,s should receive exactly one variable job corresponding to the literal placedin position s in the clause. The truth configuration of this variable job should correspondto the truth value the variable contributes to the clause. To ensure that the jobs VJob ⊤ j,t belonging to variable x j contribute consistent truth values, the truth assignment jobs andmachines are introduced. In the following, we sometimes talk about the truth assignmentgadget and thus refer to these jobs and machines. Similarly, the clause machines and jobsare sometimes called the clause gadget. In the appendix, we provide an example 3-SAT ∗ instance and the corresponding restricted assignment instance produced in the reduction.Next, we present a sequence of easy claims concerning the properties of a schedule forthe above instance with makespan T . ⊲ Claim 5.
Each machine receives exactly 3 jobs (including private loads).
Proof.
Since the overall size of the jobs is |M| T , we know that each machine has to receivejobs with overall size T = 322. Each job or private load has a size of at least 100 and atmost 111. ⊳ Since each digit of each occurring size is upper bounded by 2, the above claim implies thatthere can be no carryover when adding up job sizes of jobs scheduled on each machine.Hence the digits of the numbers involved can be considered independently, e.g., there can beat most two jobs with a 1 in the third (or second) digit of its size scheduled on any machine.This together with the given job restrictions already implies: ⊲ Claim 6.
Each truth assignment machine receives exactly one truth assignment and twovariable jobs; and each clause machine receives exactly one clause and one variable job.
Inapproximability Results for Scheduling with Interval and Resource Restrictions
Table 2
Each set indicates one of the possible job assignments for each machine in a schedulewith makespan T .Machine Possible Schedules TMach j, { TJob ⊤ j , VJob ⊤ j, , VJob ⊤ j, } , { TJob ⊥ j , VJob ⊥ j, , VJob ⊥ j, } TMach j, { TJob ⊤ j , VJob ⊤ j, , VJob ⊤ j, } , { TJob ⊥ j , VJob ⊥ j, , VJob ⊥ j, } CMach i,s (1-in-3-clause) { VJob ⊤ κ − ( i,s ) , CJob ⊤ i, } , { VJob ⊥ κ − ( i,s ) , CJob ⊥ i, } , { VJob ⊥ κ − ( i,s ) , CJob ⊥ i, } CMach i,s (2-in-3-clause) { VJob ⊤ κ − ( i,s ) , CJob ⊤ i, } , { VJob ⊤ κ − ( i,s ) , CJob ⊤ i, } , { VJob ⊥ κ − ( i,s ) , CJob ⊥ i, } TMach j, TMach j, TJob ⊤ j VJob ⊤ j, VJob ⊤ j, TJob ⊥ j VJob ⊥ j, VJob ⊥ j, VJob ⊥ j, VJob ⊥ j, VJob ⊤ j, VJob ⊤ j, TMach j, TMach j, TJob ⊥ j VJob ⊥ j, VJob ⊥ j, TJob ⊤ j VJob ⊤ j, VJob ⊤ j, VJob ⊤ j, VJob ⊤ j, VJob ⊥ j, VJob ⊥ j, Figure 2
The truth assignment gadget: There are two possible schedules of the truth assignmentmachines
TMach j, and TMach j, that already determine the schedule of the variable jobs. ⊲ Claim 7.
The jobs scheduled on a truth assignment or clause machine all have the sametruth configuration (excluding private loads). ⊲ Claim 8.
Let j ∈ [ n ]. The truth configuration of any job scheduled on TMach j, is distinctfrom the truth configuration of any job scheduled on TMach j, .The resulting possible schedules for each machine are summed up in Table 2, and Figure 2depicts the resulting two possible schedules for each pair of truth assignment machines.Lastly, we have: ⊲ Claim 9.
For each i ∈ [2 m ], the three clause machines corresponding to i receive exactlyone variable job with ⊤ -configuration if C i is a 1-in-3-clause and exactly two such jobs if C i is a 2-in-3-clause. Proof.
The overall load on each triplet of clause machines has to be 3 T = 966 and the privateloads and clause jobs that have to be scheduled on the triplet have summed up load 635,in case of a 1-in-3-clause, and 634, in case of a 2-in-3-clause. The only other jobs eligibleon the clause machines are variable jobs with size 111 in ⊤ -configuration and 110 otherwise.This implies the claim. ⊳ Using the above claims, we can easily show: ◮ Proposition 10.
There is a fulfilling truth assignment for the given ∗ instance ifand only if there is a schedule with makespan T for the constructed restricted assignmentinstance. arten Maack and Klaus Jansen 9 Proof.
Let there be a schedule with makespan T for the constructed instance. For eachvariable x j and occurrence t ∈ [4], let VJob ◦ j,t j, be the variable job scheduled on CMach κ ( j,t ) (see Table 2). We choose the truth value of x j to be ◦ j, . The variable x j occurs exactly fourtimes in the formulas, namely as a positive literal on the positions κ ( j,
1) and κ ( j,
2) andas a negative literal at position κ ( j,
3) and κ ( j, ◦ j, = ◦ j, and ◦ j, = ◦ j, = ◦ j, . Hence, for each variable x j andoccurrence t ∈ [4], the truth configuration VJob ◦ j,t j, corresponds exactly to the truth value x j contributes to the clause given by κ ( j, t ). Lastly, for each clause C i , there are exactlythree variable jobs scheduled on the corresponding clause machines, and exactly one or twoof these has ⊤ -configuration, if C i is a 1-in-3-clause or 2-in-3-clause respectively (Claim 9).Hence, C i is fulfilled.Next, we consider the case that a fulfilling truth assignment is given. For each variable x j , let ⊳ j be the corresponding truth value and ⊲ j its negation. We set ◦ j,t = ⊳ j for t ∈ { , } and ◦ j,t = ⊲ j for t ∈ { , } and assign VJob ◦ j,t j, to CMach κ ( j,t ) . All the other jobs are assignedas indicated by Table 2 and Figure 2. It is easy to verify, that all jobs are assigned and eachmachine has a load of T . ◭ The basic approach of using some kind of truth assignment and clause gadget for reductionsin the context of restricted assignment and unrelated scheduling has been used before, see,e.g., [6, 10].
Refined Reduction.
When trying to adapt the above reduction to the more restrictedproblem of RAI, we obviously have less latitude when defining the restrictions. To dealwith this, we introduce additional gadgets and encode much more information into thejob sizes. The idea of the reduction can be described as follows. We arrange the truthassignment gadgets on the left and the clause gadgets on the right. Consider the case thata truth assignment decision is made in the left most truth assignment gadget. Informationabout this decision—called signal in the following—has to be passed on to the proper clausegadgets passing multiple other truth assignment and clause gadgets on the way. This signalin the simple reduction simply corresponds to a variable job that is to be scheduled onits corresponding clause machine, and in order to prevent interaction with other gadgets,we could encode information about the corresponding variable into the size of the variablejob. However, this would lead to a super constant number of job sizes. To avoid this, weintroduce a new gadget called the bridge and highway gadget. Very roughly speaking, thesignal is passed on to the highway via gateways ; the highway passes each following truthassignment gadget using bridges and carries the signal to the proper clauses. Next, we givea detailed description and analysis of the refined reduction.We adopt all the machines and jobs introduced in the simple reduction, but change thesizes and sets of eligible machines and introduce additional jobs and machines as well asprivate loads for every machine. We introduce the following jobs and machines:For each j ∈ [ n ] and t ∈ [4], we introduce one gateway machine GMach j,t .For each j ∈ [ n ], t ∈ [4] and j ′ ∈ { j + 1 , . . . n } , we introduce two bridge machines BMachIn j,t,j ′ and BMachOut j,t,j ′ . Furthermore, we introduce two bridge jobs BJob ⊤ j,t,j ′ and BJob ⊥ j,t,j ′ .For each j ∈ [ n ], t ∈ [4] and j ′ ∈ { j, . . . n } , we introduce two highway jobs HJob ⊤ j,t,j ′ and HJob ⊥ j,t,j ′ .In order to define the intervals of eligible machines, we first need a total order of the machines.We partition the machines into blocks, define an internal order for each block, and then definean order of the blocks. Remember that κ : [ n ] × [4] → [2 m ] × [3] is a bijection indicating the positions of the occurrences of variables in the clauses. In particular, κ ( j,
1) = ( i, s )indicates that the first positive occurrence of variable x j is in clause C i on position s , and κ ( j, κ ( j, κ ( j,
4) indicate analogue information for the second positve, first negative,and second negative occurrence of x j .For each j ∈ [ n ], we have a truth assignment block T j containing the truth assignmentmachines TMach j, and TMach j, in this order.For each i ∈ [2 m ], we have a clause block C i containing the clause machines CMach i,s foreach s ∈ [3] and ordered increasingly by s .For each j ∈ [ n ], we have a successor block S j containing the gateway machines GMach j,t for each t ∈ [4] and the bridge machines BMachOut j ′ ,t,j for each t ∈ [4] and j ′ 1) and κ ( j, t ) = (2 , GMach j,t precedes BMachOut j ,t ,j which in turn precedes BMachOut j ,t ,j .For each j ∈ [ n ] with j > 1, we have a predecessor block P j containing the bridgemachines BMachIn j ′ ,t,j for each t ∈ [4] and j ′ < j . Machine BMachIn j ′ ,t,j has index κ ( j ′ , t ) and the machines are ordered by the increasing lexicographical ordering of theirindices.The blocks are ordered as follows:( T , S , P , T , S , . . . , P n , T n , S n , C , . . . , C m )The sets of eligible machines are specified in Table 3 and the job sizes in Table 4 . In theappendix, we provide an example instance together with a figure (Figure 6) visualizing theordering of the machines and the eligibility constraints. Furthermore, Figure 3 gives someintuition on the overall structure. We have: ⊲ Claim 11. The overall size of the jobs is exactly |M| T . Proof. This can be verified by basic arithmetic using Table 4. For simplicity, this can alsobe done digit by digit. We look at the last digit as an example. Note that we have 4 n + 6 n machines and the last digit of the makespan is 2. Summing up the last digits of the jobsizes, on the other hand, yields 2 n + 2 n + n + n + n + n + 2 n ( n − 1) + 2 n ( n + 1) + 4 n + n + n + n + n + 2 n ( n − 1) + 2 n ( n − 1) = 8 n + 12 n . ⊳ Like for the simple reduction, we proof a sequence of easy claims concerning the propertiesof a schedule for the constructed instance with makespan T . ⊲ Claim 12. Each machine receives exactly 4 jobs if it is a truth assignment machine andexactly 3 jobs otherwise (including private loads). Proof. This follows directly from Claim 11 and the job sizes defined in Table 4. ⊳ Since each machine receives at most 4 jobs and each digit in the job sizes is bounded by 2,we may consider each digit of the involved numbers independently, e.g., if two jobs and themakespan have a 1 at the ℓ -th digit, we already know that these jobs cannot be scheduledon the same machine. This already implies a series of claims: Note that we have prioritized comprehensibility over small sizes. For instance, it is not hard to seethat the columns in Table 4 corresponding to the highway and clause jobs could be deleted and thereduction would still work. arten Maack and Klaus Jansen 11 Table 3 The sets of eligible machines for each job or job type, defined by the first and lasteligible machine in the ordering. Note that in case of the highway jobs all four combinations of firstand last machine are possible.Job First machine Last machineClause job CJob ◦ s i,s CMach i, CMach i, Truth assignment job TJob ◦ j TMach j, TMach j, Variable job VJob ◦ j,t TMach j, ⌈ t/ ⌉ GMach j,t Bridge job BJob ◦ j,t,j ′ BMachIn j,t,j ′ BMachOut j,t,j ′ Highway job HJob ◦ j,t,j ′ BMachOut j,t,j ′ , if j ′ > j , GMach j,t , if j ′ = j BMachIn j,t,j ′ +1 if j ′ < n , CMach κ ( j,t ) , if j ′ = n Table 4 Table of job and machine types with job sizes and private loads and the makespan. Thesecond column states the number of jobs and machines of the respective types. Each horizontalsequence of numbers following the second column indicates the size of the respective job or privateload. Each of the corresponding columns serves a function in the reduction: the first bounds thenumber of jobs on each machines; the following eight implement restrictions for the bridge, highway,clause, truth assignment and variable jobs; and the last encodes truth values. B H C T V V V VCJob ⊤ i,s m = 2 n CJob ⊥ i,s m = 2 n TJob ⊤ j n TJob ⊥ j n VJob ⊤ j, n VJob ⊤ j, n VJob ⊤ j, n VJob ⊤ j, n VJob ⊥ j, n VJob ⊥ j, n VJob ⊥ j, n VJob ⊥ j, n BJob ⊤ j,t,j ′ n ( n − 1) 1 1 0 0 0 0 0 0 0 0 BJob ⊥ j,t,j ′ n ( n − 1) 1 1 0 0 0 0 0 0 0 1 HJob ⊤ j,t,j ′ n ( n + 1) 1 0 1 0 0 0 0 0 0 1 HJob ⊥ j,t,j ′ n ( n + 1) 1 0 1 0 0 0 0 0 0 0 CMach i,s m = 4 n TMach j, n TMach j, n GMach j, n GMach j, n GMach j, n GMach j, n BMachIn j,t,j ′ n ( n − 1) 1 0 0 1 1 1 1 1 1 1 BMachOut j,t,j ′ n ( n − 1) 1 0 0 1 1 1 1 1 1 1Makespan T S n − P n T n S n C C m VJob ◦ n, VJob ◦ n, Figure 3 The bridge and highway gadget. The intervals of eligible machines of highway, bridgeand variable jobs are depicted in blue, red and orange, respectively. In this example, variable x n occurs for the second time in its positive form in the last clause at the first position, and for thefirst time in its negative form in the first clause at the first position. ⊲ Claim 13. The jobs TJob ⊤ j and TJob ⊥ j can exclusively be scheduled on TMach j, and TMach j, , for each j ∈ [ n ], and each of the two machines receives exactly one of the two jobs. ⊲ Claim 14. The jobs VJob ⊤ j,t and VJob ⊥ j,t can exclusively be scheduled on TMach j, ⌈ t/ ⌉ and GMach j,t , for each j ∈ [ n ] and t ∈ [4], and each of the two machines receives exactly one ofthe two jobs. ⊲ Claim 15. Bridge jobs can exclusively be scheduled on bridge machines and each bridgemachine receives exactly one bridge job. ⊲ Claim 16. Highway jobs can exclusively be scheduled on bridge, gateway and clausemachines and each such machine receives exactly one highway job. ⊲ Claim 17. Each clause machine CMach i,s receives exactly one of the corresponding clausejobs CJob ◦ s ′ i,s ′ with s ′ ∈ [3].At this point, we already know that variable (and truth assignment) jobs can exclusively bescheduled on the first or last machine of their respective interval of eligible machines. Thenext step is to show that the same holds for highway and bridge jobs. To do so, the orderingof the bridge and highway machines is of critical importance. ⊲ Claim 18. The jobs BJob ⊤ j,t,j ′ and BJob ⊥ j,t,j ′ can exclusively be scheduled on BMachIn j,t,j ′ and BMachOut j,t,j ′ , for each j ∈ [ n ], j ′ ∈ { j + 1 , . . . , n } and t ∈ [4], and each of the twomachines receives exactly one of the two jobs. Proof. The claim can be proved with a simple inductive argument: Let j ′ ∈ { , . . . , n } and,furthermore, ( j ℓ , t ℓ ) ∈ [ j ′ − × [4] denote the ℓ -th element from [ j ′ − × [4] when ordering thepairs ( j, t ) ∈ [ j ′ − × [4] by the increasing lexicographical ordering of the pairs κ ( j, t ). Con-sidering the ordering of the machines and the job restrictions, BJob ⊤ j ,t ,j ′ and BJob ⊥ j ,t ,j ′ arethe only bridge jobs that can be scheduled on BMachIn j ,t ,j ′ and BMachOut j ,t ,j ′ (see Figure3). Hence, the claim has to hold for ( j , t ). But then again BJob ⊤ j ,t ,j ′ and BJob ⊥ j ,t ,j ′ arethe only remaining bridge jobs that can be scheduled on BMachIn j ,t ,j ′ and BMachOut j ,t ,j ′ ,and so on. ⊳⊲ Claim 19. The jobs HJob ⊤ j,t,j ′ and HJob ⊥ j,t,j ′ can exclusively be scheduled on machine X and Y , for each j ∈ [ n ], j ′ ≥ j , and t ∈ [4]; where X = BMachOut j,t,j ′ if j ′ > j , and X = GMach j,t otherwise, and Y = BMachIn j,t,j ′ +1 if j ′ < n , and Y = CMach κ ( j,t ) otherwise.Furthermore, each of the two machines receives exactly one of the two jobs. arten Maack and Klaus Jansen 13 Table 5 For each machine there are only few possible jobs that may be assigned to it in aschedule with makespan T . Each set corresponds to one of the possible schedules.Machine Possible Schedule TMach j, { TJob ⊤ j , VJob ⊤ j, , VJob ⊤ j, } , { TJob ⊥ j , VJob ⊥ j, , VJob ⊥ j, } TMach j, { TJob ⊤ j , VJob ⊤ j, , VJob ⊤ j, } , { TJob ⊥ j , VJob ⊥ j, , VJob ⊥ j, } GMach j,t { VJob ⊤ j,t , HJob ⊤ j,t,j } , { VJob ⊥ j,t , HJob ⊥ j,t,j } BMachIn j,t,j ′ { BJob ⊤ j,t,j ′ , HJob ⊤ j,t,j ′ − } , { BJob ⊥ j,t,j ′ , HJob ⊥ j,t,j ′ − } BMachOut j,t,j ′ { BJob ⊤ j,t,j ′ , HJob ⊤ j,t,j ′ } , { BJob ⊥ j,t,j ′ , HJob ⊥ j,t,j ′ } CMach i,s (1-in-3) { CJob ⊤ i, , HJob ⊤ κ − ( i,s ) ,n } , { CJob ⊥ i, , HJob ⊥ κ − ( i,s ) ,n } , { CJob ⊥ i, , HJob ⊥ κ − ( i,s ) ,n } CMach i,s (2-in-3) { CJob ⊤ i, , HJob ⊤ κ − ( i,s ) ,n } , { CJob ⊤ i, , HJob ⊤ κ − ( i,s ) ,n } , { CJob ⊥ i, , HJob ⊥ κ − ( i,s ) ,n } Proof. We can use the same argument (with reversed orderings) as we did in the last claim.It is only slightly more complicated, because more machine types are involved. ⊳ Summing up, each job except for clause jobs may only be scheduled on the first or lastmachine of their interval of eligible machines, and each of these machines receives either therespective job in ⊤ - or ⊥ -configuration. Considering this distribution of the jobs and thelast digit of the size vectors, we get the following two claims: ⊲ Claim 20. For any machine, the jobs assigned to this machine all have the same truthconfiguration (excluding private loads). ⊲ Claim 21. For each i ∈ [2 m ], the three clause machines corresponding to i receive exactlyone highway job with ⊤ -configuration, if C i is a 1-in-3-clause, and exactly two such jobs, if C i is a 2-in-3-clause.The former property together with the possible job distribution determined so far impliesthat there are only few possible schedules for each machine. We summarize these schedulesin Table 5. Furthermore, we can infer that the truth assignment gadget works essentiallythe same as before (see Figure 2): ⊲ Claim 22. Let j ∈ [ n ]. The truth configuration of any job scheduled on TMach j, isdistinct from the truth configuration of any job scheduled on TMach j, .Lastly, we can show that the bridge and highway gadget works as well: ⊲ Claim 23. Let j ∈ [ n ] and t ∈ [4]. The variable job scheduled on TMach j, ⌈ t/ ⌉ and thehighway job scheduled on CMach κ ( j,t ) have the same truth configuration. Proof. Note that the truth configuration of the variable job scheduled on GMach j,t comparedwith the one of the variable job scheduled on TMach j, ⌈ t/ ⌉ is reversed. Hence, the highwayjob scheduled on GMach j,t also has the reversed truth-configuration while the highway jobthat is passed on again has the original truth-configuration. This argument can be repeatedwith the bridge and highway jobs in the following, yielding the asserted claim. ⊳ Using the above claims, we can conclude the proof of Theorem 1 via the following Lemma: ◮ Lemma 24. There is a fulfilling truth assignment for the given ∗ instance, if andonly if there is a schedule with makespan T for the constructed RAI instance. Proof. First, we consider the case that a schedule with makespan T for the constructedRAI instance is given. For each variable x j and occurrence t ∈ [4], let HJob ◦ j,t j,t,n be thehighway job scheduled on CMach κ ( j,t ) (see Table 5). We choose the truth value of x j to be ◦ j, . Considering the distribution of jobs on the truth assignment machines (see Table 5), aswell as Claim 22 and 23, we know that for each variable x j and occurrence t ∈ [4], the truthconfiguration ◦ j,t corresponds exactly to the truth value x j contributes to the clause givenby κ ( j, t ). Furthermore, we know that for each clause C i , there are exactly three variablejobs scheduled on the corresponding clause machines, and exactly one or two of these has ⊤ -configuration, if C i is a 1-in-3-clause or 2-in-3-clause, respectively (Claim 21). Hence, C i is fulfilled.Now, let there be a fulfilling truth assignment, and ⊳ j be the corresponding truth valueof variable x j and ⊲ j its negation. We set △ j,t = ⊳ j for t ∈ { , } and △ j,t = ⊲ j for t ∈ { , } and assign HJob △ j,t j,t,n to CMach κ ( j,t ) . Let ▽ jt be the negation of △ jt . All the other jobs areassigned as indicated by the claims and Table 5 in particular: Each machine receives itsprivate load; CMach κ ( j,t ) additionally receives one of the eligible remaining clause jobs with △ j,t -configuration (this can be done because the truth assignment is fulfilling); BMachOut j,t,j ′ receives HJob ▽ j,t j,t,j ′ and BJob ▽ j,t j,t,j ′ ; BMachIn j,t,j ′ receives HJob △ j,t j,t,j ′ − and BJob △ j,t j,t,j ′ ; GMach j,t receives HJob ▽ j,t j,t,j and VJob ▽ j,t j,t ; TMach j, receives VJob △ j, j, , VJob △ j, j, and TJob △ j, j ; and TMach j, receives VJob △ j, j, , VJob △ j, j, as well as TJob △ j, j . It is easy to verify, that all jobsare assigned and each machine has a load of T . ◭ In this section, we first present some preliminary observations concerning RAR( R ) anddiscuss the relationship of the problem with RAI and LRS( D ). Next, we revisit establishedreductions for the restricted assignment problem and show that they can be modeled withonly few resources. This already gives the result for 4 resources in Theorem 2. Lastly,we study the cases with 2 and 3 resources. We first give a reduction for R = 3 and thenrefine the result to work for R = 2 as well thereby concluding the proof of Theorem 2 andTheorem 3. Preliminaries. Recall that in the problem of scheduling with resource restrictions with R resources (RAR( R )), a set R of R (renewable) resources is given, each machine i has aresource capacity c r ( i ) and each job j has a resource demand d r ( j ) for each r ∈ R . Job j is eligible on machine i , if d r ( j ) ≤ c r ( i ) for each resource r . We allow arbitrary real valuesfor the capacities and demands but it is not hard to see that relatively small integer valuessuffice. Indeed, given an instance of RAR( R ), we may perform the following two steps:First, we increase for each job j and each resource r the demand d r ( j ) to the smallest valueincluded in { c r ( i ) | i ∈ M , c r ( i ) ≥ d r ( j ) } (if this set is empty the job cannot be processedanywhere). Afterwards, there are at most m = |M| distinct demand or capacity values foreach resource, and we can change the smallest value to 1 the second to 2 and so on. Thisyields an instance with the same restrictions and the property that all the capacities anddemands are included in [ m ].Technically, there are two versions of the problem RAR( R ) depending on whether theresources, demands and capacities are explicitly given or not. In the second variant recogni-tion is an issue that we do not address in this work since our results work for both versionsof the problem. However, note that the proof of the following lemma gives some intuitionconcerning this: arten Maack and Klaus Jansen 15 ◮ Lemma 25. Each restricted assignment instance with m machines is also a RAR( m ) instance; and for each m ∈ N there is a RAR( m ) instance with m machines (and m jobs)that is not a RAR( R ) instance for any R < m . Proof. Given a restricted assignment instance with m machines, we define resources, de-mands and capacities that model the given restrictions. First, we identify each machinewith a resource, that is, we set R = M . Furthermore, we set the capacities of machine i ∈ M concerning resource r ∈ R to be c r ( i ) = 1 if r = i and c i ( i ) = 0; and the demand ofjob j ∈ J concerning r to be d r ( i ) = 0 if r ∈ M ( j ), and d r ( i ) = 1 otherwise. It is easy tocheck that for each job j and machine i , we have d r ( j ) ≤ c r ( i ) for each resource r , if andonly if i ∈ M ( j ).The next goal is to construct a simple RAR( m ) instance for each m ∈ N . We first collectsome simple observations. Given some instance of RAR( R ) with j ∈ J and i ′ ∈ M \ M ( j ),we know that there is a resource r ( j, i ′ ) ∈ R such that c r ( j,i ′ ) ( i ′ ) < d r ( j,i ′ ) ( j ) (otherwise i ′ ∈ M ( j )). One could say that r ( j, i ′ ) separates i ′ from j or M ( j ). On the other hand,we know that d r ( j ) ≤ c r ( i ) for each i ∈ M ( j ) and r ∈ R . Let i ∈ M ( j ). Now, considerthe case that we have another job j ′ with i ′ ∈ M ( j ′ ) and i ∈ M \ M ( j ′ ). Then we have r ( j, i ′ ) = r ( j ′ , i ), because otherwise: d r ( j ′ ,i ) ( j ) ≤ c r ( j ′ ,i ) ( i ) < d r ( j ′ ,i ) ( j ′ ) ≤ c r ( j ′ ,i ) ( i ′ ) = c r ( j,i ′ ) ( i ′ ) < d r ( j,i ′ ) ( j ) = d r ( j ′ ,i ) ( j )Using this insight, we construct the instance as follows: We set M = [ m ] and J = (cid:8) j ⊆M (cid:12)(cid:12) | j | = m − (cid:9) with M ( j ) = j . We may assume unit processing times. This instancehas exactly m jobs and machines and we know that it is a RAR( m ) instance because ofthe first part of the proof. Now, given any resources R along with capacities and demandsthat model the restrictions for the above instance, we show that |R| ≥ m . For each j ∈ J let i j ∈ M be the single machine that is restricted to process j , i.e., i j ∈ M \ M ( j ). Thisimplies M = { i j | j ∈ J } and due to the above observation, we have r ( j, i j ) = r ( j ′ , i j ′ )for each pair of distinct jobs j, j ′ ∈ J . Hence, { r ( j, i j ) | j ∈ J } = R ′ ⊆ R and |R ′ | = m concluding the proof. ◭ The relationship between scheduling with resource and interval restrictions is discussedin the following lemma: ◮ Lemma 26. Each RAR(1) instance is also a RAI instance and there is a RAI instancethat is not a RAR(1) instance. Moreover, each RAI instance is also a RAR(2) instance andthere is a RAR(2) instance that is not a RAI instance. With a slight abuse of notation, wemay write: RAR(1) ⊂ RAI ⊂ RAR(2) . Proof. Given an instance of RAR(1), we may sort the machines based on their capacityvalues decreasingly. Than each job j that can be processed on any machine, can also beprocessed on any predecessor of this machine. Hence, M ( j ) corresponds to an interval ofmachines starting with the first machine. On the other hand, consider an instance with twomachines and two jobs. The first job is (exclusively) eligible on the first machine and thesecond one on the second. This instance is a RAI but not a RAR(2) instance.Given an instance of RAI, we may assume wlog. M = [ m ] and that the ordering of themachines is the natural ordering. We set R = [2]. Furthermore, for each machine i , we set c ( i ) = ( i, ( m +1) − i ); and for each job j with M ( j ) = { ℓ, . . . , r } , we set d ( j ) = ( ℓ, ( m +1) − r )(see Figure 4). If i ∈ M ( j ), we have c ( i ) = i ≥ ℓ = d ( j ) and c ( i ) = ( m + 1) − i ≥ ( m + 1) − r = d ( j ); and if i ∈ M \ M ( j ), we have either i < ℓ or i > r which implies c ( i ) < d ( j ) or c ( i ) < d ( j ) respectively. 11 22 33 44 Figure 4 The left picture visualizes that each RAI instance can be seen as a RAR(2) instanceand the right one depicts an RAR(2) instance that is not a RAI instance. In both pictures, eachdimension corresponds to a resource, the squares mark the capacities of machines and the circlesthe demands of jobs. If the capacity of a machine is at least as big as the demand of a job in bothdimension, the job is eligible on the machine. Table 6 A simple example of a RAR(2) instance that is not a RAI instance. Note that alldemands could be rounded up to the next integer value without changing the construction.Machine Capacity Job Demand1 (3 , { , , , } (0 . , . , { , } (2 . , . , { , } (1 . , . , { , } (0 . , . Lastly, we construct an instance of RAR(2) that is not a RAI instance. Let M = [4], R = [2] and J = (cid:8) { , , , } , { , } , { , } , { , } (cid:9) . We may assume unit job sizes. Theresource capacities and demands are given in Table 6 and the construction is illustrated inFigure 4. It is easy to see that M ( j ) = j for each j ∈ J . In any total ordering of themachines in which each job is eligible on consecutive machine, the machine 1 has to be adirect neighbor of 2, 3 and 4. This is not possible. ◭ We already mentioned in the introduction that there is a close relationship betweenscheduling with resource restrictions and low rank unrelated scheduling. Remember that inthe rank D unrelated scheduling problem (LRS( D )) the processing time matrix ( p ij ) i ∈M ,j ∈J has a rank of at most D , or, equivalently, there is a D dimensional size or speed vector s ( j )or v ( i ) for each job j or machine i respectively, and the processing time p ij is given by P d ∈ [ D ] s d ( j ) v d ( i ). It is easy to construct for any m ∈ N a RAR(1) instance that is aLRS( m ) instance as well: The instance with J = M = [ m ], R = { } , d ( k ) = c ( k ) = k for each k ∈ [ m ] and unit processing times suffices (assuming that the number ∞ isinterpreted as some sufficiently big number). On the other hand, any RAR( R ) instance canbe approximated with arbitrary precision by LRS( R + 1) instances in the following sense: ◮ Lemma 27. Let I be a RAR( R ) instance. For any ε, K > , there is a LRS( R + 1) instance I ′ with the same jobs and machines and the following property: Let p ′ ij be theprocessing time of job j on machine i in instance I ′ . We have p j ≤ p ′ ij ≤ p j + ε if i ∈ M ( j ) and p ′ ij ≥ p j + K otherwise. Proof. Wlog. we assume R = [ R ]. Let δ = ε/R and N = max { K/δ, } . We define the size arten Maack and Klaus Jansen 17 and speed vectors of I ′ as follows: For each job j we set s ′ r ( j ) = δN d r ( j ) for each r ∈ [ R ], aswell as s ′ R +1 ( j ) = p j . Moreover, for each machine i we set v ′ r ( j ) = N − c r ( j ) for each r ∈ [ R ],as well as v ′ R +1 ( j ) = 1. Then p ′ ij = P r ∈ [ R +1] s r ( j ) v r ( i ) = p j + P r ∈ [ R ] δN d r ( j ) − c r ( j ) andtherefore p ′ ij ≥ p j in any case. Furthermore, if i ∈ M ( j ), we have d r ( j ) ≤ c r ( j ) for each r ∈ [ R ], and hence p ′ ij ≤ p j + Rδ = p j + ε . If, on the other hand, i 6∈ M ( j ), then there is an r ∈ [ R ] such that d r ( j ) > c r ( j ) yielding p ′ ij ≥ p j + δN ≥ p j + K . ◭ The above lemma implies that from the perspective of approximation algorithms RAR( R ) isessentially included in LRS( R + 1). We could use this lemma and Theorem 1 or Theorem 2to show that there is no PTAS for LRS(3) unless P=NP. While this is already known [6],the resulting construction may be more accessible. Established Reductions Revisited. In the following, we first present the classical reductionby Lenstra et al. [25] showing 1 . . . A , B and C with | A | = | B | = | C | = n ∈ N , as well as a set of triplets E ⊆ (cid:8) { a, b, c } (cid:12)(cid:12) a ∈ A, b ∈ B, c ∈ C (cid:9) . Thegoal is to decide whether there is a subset F ⊆ E that perfectly covers A , B and C , thatis, for each x ∈ A ∪ B ∪ C there is exactly one triplet e ∈ F with x ∈ e . The set F iscalled a 3D-matching. We assume that the elements of A , B and C are indexed, that is, A = { a , a , . . . , a n } , B = { b , b , . . . , b n } and C = { c , c , . . . , c n } . Furthermore, we assumethat for each x ∈ A ∪ B ∪ C there is at least one e ∈ E with x ∈ E (otherwise the problemis trivial). Via a reduction from 3-DM to the restricted assignment problem, Lenstra et al.[25] showed: ◮ Theorem 28 ([25]) . There is no polynomial time approximation algorithm for restrictedassignment with rate smaller than . unless P = NP. Proof. Given an instance of 3-DM, we set M = E and E ( x ) = { e ∈ E | x ∈ e } for each x ∈ A ∪ B ∪ C . For each a ∈ A , we introduce | E ( a ) | − ≥ e ∈ E ( a ). Moreover, for each x ∈ B ∪ C we introduce an elementjob with size 1 eligible on machines e ∈ E ( x ). Note that the overall size of the jobs is givenby 2 n + 2 P a ∈ A ( | E ( a ) | − 1) = 2 n + 2( | E | − n ) = 2 |M| .If there is a schedule with makespan 2 for this instance, then each machine either pro-cesses two element or one dummy job. For each x ∈ B ∪ C , we have x ∈ e for the machine e processing the corresponding element job, and, furthermore, for each a ∈ A , there is exactlyone machine e with a ∈ e that does not process a dummy job (and therefore processeselement jobs). Hence, we get a 3D-matching by selecting the machines that process elementjobs.If, on the other hand, there is a 3D-matching F for the 3-DM instance, than we canschedule the element job corresponding to x ∈ B ∪ C on the machine e ∈ F with x ∈ E and the dummy jobs corresponding to a ∈ A on the | E ( a ) | − e ∈ E ( a ) \ F . Thisyields a schedule with makespan 2. ◭ We reproduce the restrictions in the above reduction using six resources and get: ◮ Corollary 29. There is no polynomial time approximation algorithm for RAR(6) with ratesmaller than . unless P = NP. Proof. We set R = (cid:8) ( X, k ) (cid:12)(cid:12) X ∈ { A, B, C } , k ∈ [2] (cid:9) . Let e ∈ E and X ∈ { A, B, C } . We setthe resource capacities c ( X, ( e ) = i and c ( X, ( e ) = ( n + 1) − i . Let x j be the element withindex j in X . We set the resource demand of a (element or dummy) job J correspondingto x j as follows: d ( X, ( J ) = j , d ( X, ( J ) = ( n + 1) − j , as well as d ( Y,k ) ( J ) = 0 for each Y ∈ { A, B, C } \ { X } and k ∈ [2]. It is easy to see that J can exclusively be scheduled onmachines e with x j ∈ e . ◭ In the classical 3-SAT problem, a conjunction of m clauses is given and each clause is adisjunction of at most three literals of variables x , . . . , x n . In the result due to Ebenlendret al. [10], the modified 3-SAT problem, where each variable occurs exactly three and eachliteral at most two times in the formula, is reduced to the graph balancing problem, that is,restricted assignment with the additional property that each job is eligible on at most twomachines. To show that the modified 3-SAT problem is NP-hard, we can use techniquesalready applied in Section 2: We may replace the d j occurrences of variable x j with newvariables z j , . . . , z jd j and add new clauses ( z j ∧ ¬ z j ), . . . ( z jd j − ∧ ¬ z jd j ), ( z jd j ∧ ¬ z j ). ◮ Theorem 30 ([10]) . There is no polynomial time approximation algorithm with ratesmaller than . for the graph balancing problem unless P = NP. Proof. Given an instance of modified 3-SAT, we introduce clause machines v i correspondingto the clauses C i , and literal machines u j, and u j, corresponding to the literals x j and ¬ x j .Furthermore, we introduce truth assignment jobs e j for each variable x j with size 2 andeligible on u j, and u j, ; and clause jobs f i,j,α for each clause C i and literal y j occurring in C i with α = 1 if y j = x j and α = 0 if y j = ¬ x j . The job f i,j,α has size 1 and is eligible on v i and u j,α . Lastly, we introduce a dummy job d i for each clause C i with less then threeliterals. Its size is 1 if C i contains two literals, and 2 if C i contains only one literal.In a schedule with makespan 2, there is at least one clause job f i,j,α for each v i that isscheduled on u j,α and not on v i . Hence, the job e j has to be scheduled on u j, | α − | . Now,it is easy to see that there is a schedule with makespan 2, if and only if there is a fulfillingassignment. The construction works as follows: Given a schedule with makespan 2, we setvariable x j to ⊤ if e j is scheduled on u j, , and to ⊥ otherwise. Moreover, given a fulfillingtruth assignment we assign the truth assignment jobs correspondingly, and the machines u j,α that did not receive a truth assignment job receive all eligible clause jobs (at mosttwo). ◭ We reproduce the restrictions in the above reduction using four resources and get: ◮ Corollary 31. There is no polynomial time approximation algorithm for RAR(4) with ratesmaller than . unless P = NP. Proof. We set R = [4]. The clause machine v i has a resource capacity vector of (2 n +1 , n + 1 , i, ( m + 1) − i ), and the literal machine u j,α has capacity vectors (2 j − α, (2 n + 1) − (2 j − α ) , m + 1 , m + 1). Furthermore, the truth assignment job e j has a resource demandvector of (2 j − , (2 n + 1) − j, m + 1 , m + 1); the clause job f i,j,α has a demand vector of(2 j − α, (2 n + 1) − (2 j − α ) , i, ( m + 1) − i ); and the dummy job d i has a demand vector of(2 n + 1 , n + 1 , i, ( m + 1) − i ). It is easy to verify that the resulting sets of eligible machinesare the same as described in Theorem 30. ◭ arten Maack and Klaus Jansen 19 Three Resources. We present a reduction from 3-DM to RAR(3). The reduction is basedon the classical result by Lenstra et al. [25] and very similar to a reduction by Bhaskara etal. [2] for LRS(4). However, there is a problem with the choice of processing times in thelatter reduction (see Appendix B), and the present result can be used to fix it.Given an instance ( A, B, C, E ) of 3-DM, let n = | A | and E ( x ) = { e ∈ E | x ∈ e } foreach x ∈ A ∪ B ∪ C . Furthermore, we set α A = 12, α B = 13, α C = 22, β A = 14, β B = 15and β C = 18. Let R = { A, B, C } and M = E . For each machine e , we define the resourcecapacities as follows. Let X ∈ { A, B, C } and x i ∈ X ∩ e be the element of x with index i .We set c X ( e ) = i . Furthermore, for each element x i ∈ X with index i in X ∈ { A, B, C } ,we introduce one element job with size α X and | E ( x ) | − β X . Theresource demand for each of these jobs is given by d ( i ) with d X ( i ) = i and d Y ( i ) = 0 for Y ∈ { A, B, C } \ { X } . ⊲ Claim 32. We have α A + α B + α C = 47 = β A + β B + β C ; any four numbers takenfrom Γ = { α A , α B , α C , β A , β B , β C } = { , , , , , } sum up to a value bigger than47; any selection of less than 3 numbers sums up to a value smaller than 47; and for anythree numbers γ , γ , γ ∈ Γ with γ ≤ γ ≤ γ and γ + γ + γ = 47, we have either( γ , γ , γ ) = ( α A , α B , α C ) or ( γ , γ , γ ) = ( β A , β B , β C ). Proof. The first three assertions are obvious, and the fourth holds due to a simple caseanalysis:If γ > 15, we have γ ≥ 18, and hence 47 = γ + γ + γ ≥ · γ = 54: a contradiction.Note that γ ≥ ( γ + γ ) / − γ ) / 2. Hence, γ ≤ 15 implies γ ≥ 16 and therefore γ ∈ { , } .If we have γ = 22 = α C , then γ ≤ ( γ + γ ) / − γ ) / . 5. Hence, γ = 12 = α A and γ = 13 = α B .If we have γ = 18 = β C , than γ ≥ ( γ + γ ) / − γ ) / . 5. Hence, γ ∈ { , } . If γ = 15 = β B , then γ = 14 = β A , and if γ = 18, then γ = 11 / ∈ Γ.This concludes the proof of the claim. ⊳ By brute force, it can be verified that 47 is the smallest value such that suitable numbers α A , α B , α C , β A , β B and β C exist and the above claim holds. ⊲ Claim 33. The summed up size of all the element and dummy jobs is 47 |M| . Proof. We have exactly n element jobs with size α A , α B and α C , respectively, yielding anoverall load of 47 n . The dummy jobs have an overall load of: β A X a ∈ A ( | E ( a ) | − 1) + β B X b ∈ B ( | E ( b ) | − 1) + β C X c ∈ C ( | E ( b ) | − β A + β B + β C )( | E | − n ) = 47( |M| − n )In this equation, we used the simple fact that { E ( x ) | x ∈ X } is a partition of E for each X ∈ { A, B, C } , and hence | E | = P x ∈ X | E ( x ) | . ⊳ These two claims imply: ⊲ Claim 34. In any schedule for the constructed instance with makespan 47, each ma-chine receives exactly three jobs with sizes γ , γ , γ such that ( γ , γ , γ ) = ( α A , α B , α C ) or( γ , γ , γ ) = ( β A , β B , β C ).Using these claims, we can show: Table 7 The resource demands and capacities for the different job (types) and machines.Jobs Resources Machines Resources a i (2 i, { a i , b j , c j } (2 i, n + j ) a ′ i (2 i − , { a ′ i , b j , c j } (2 i − , n + j ) b j (0 , n + j ) { a i , b ′ i , c ′ i } (2 i, i ) c j (0 , n + j ) { a ′ i , b ′ i , c ′ ζ ( i ) } (2 i − , ζ ( i )) b ′ i (2 i − , c ′ i (0 , i ) ◮ Proposition 35. There is a perfect matching for the given instance, if and only ifthere is a schedule with makespan for the constructed RAR(3) instance. Proof. Let F be a perfect matching for the 3-DM instance. For each x ∈ A ∪ B ∪ C we assignthe corresponding element job to the machine e with x ∈ e and e ∈ F . Furthermore, thedummy jobs corresponding to x ∈ X with X ∈ { A, B, C } , are distributed to the machines e with x ∈ e and e / ∈ F such that each machine receives exactly one job. Hence, each machine e ∈ E receives exactly three eligible jobs either with sizes α A , α B and α C (if e ∈ F ) or β A , β B and β C (otherwise).Next, we assume that there is a schedule with makespan 47 for the scheduling instance.For each X ∈ { A, B, C } , there are exactly |M| many jobs with size α X or β X , and due tothe above claims, we know that each machine receives exactly one of these jobs. For each j ∈ [ n ], let x j ∈ X be the element with index j in X ∈ { A, B, C } . The machines S nj = i E ( x j )are the only machines that may process jobs corresponding to x i , . . . , x n for each i ∈ [ n ]and we have exactly P nj = i | E ( x j ) | many such jobs. Hence, the machines from E ( x i ) receiveexactly the jobs corresponding to x i . Now, considering this and Claim 34, we get a perfectmatching by selecting the machines that process three element jobs. ◭ Two Resources. We are able to refine the result for three resources to work for two resourcesas well by using another variant of 3-DM as the starting point of the reduction. Theproblem 3-DM ∗ was introduced by Chen et al. [7] to get an improved lower bound forthe approximation ratio of rank four unrelated scheduling. In this problem, a set of sixdisjoint sets E = { A, A ′ , B, B ′ , C, C ′ } is given. For each X ∈ E , we have | X | = 3 n forsome n ∈ N and the sets are indexed by [3 n ], e.g., A = { a , a , . . . , a n } . Furthermore,there are two sets of triplets E ⊆ (cid:8) { a i , b j , c j } , { a ′ i , b j , c j } (cid:12)(cid:12) i ∈ [3 n ] , j ∈ [3 n ] (cid:9) and E = (cid:8) { a i , b ′ i , c ′ i } , { a ′ i , b ′ i , c ′ ζ ( i ) } (cid:12)(cid:12) i ∈ [3 n ] (cid:9) with ζ (3 k + 1) = 3 k + 2, ζ (3 k + 2) = 3 k + 3 and ζ (3 k + 3) = 3 k + 1 for each k ∈ { , . . . , n − } . Note that the second set of triplets is alreadydetermined by the element sets in the input. Similar to the classical 3-DM problem, thegoal is to decide whether there is a subset F ⊆ E ∪ E that perfectly covers the element set,that is, for each x ∈ S X ∈E X there is exactly one triplet e ∈ F with x ∈ e . Furthermore,we assume that for each x ∈ S X ∈E X there is at least one e ∈ E with x ∈ E (otherwise theproblem is trivial).Let α A = α A ′ = 12, α B = α B ′ = 13, α C = α C ′ = 22, β A = β A ′ = 14, β B = β B ′ = 15and β C = β C ′ = 18. We set M = E ∪ E and R = [2]. The corresponding resourcecapacity vectors are presented in Table 7. Furthermore, for each element x ∈ X in X ∈ E ,we introduce one element job with size α X and | E ( x ) | − β X . Thevector of resource demands for each such job is given in Table 7. Note that Claim 32, 33and 34 hold for this reduction as well and with the same reasoning. Furthermore, a simple arten Maack and Klaus Jansen 21 case analysis yields: ⊲ Claim 36. For each x ∈ S X ∈E X , a (dummy or element) job corresponding to x is eligibleon each machine e with x ∈ e .Using these claims, we can conclude the proof of Theorem 2: ◮ Lemma 37. There is a perfect matching for the given ∗ instance, if and only if thereis a schedule with makespan for the constructed RAR(2) instance. Proof. Let F be a perfect matching for the 3-DM ∗ instance. For each x ∈ S X ∈E X , weassign the corresponding element job to the machine e with x ∈ e and e ∈ F . Furthermore,the dummy jobs corresponding to x ∈ X with X ∈ E , are distributed to the machines e with x ∈ e and e / ∈ F such that each such machine receives exactly one job. Hence, each machine e ∈ E receives exactly three eligible jobs either with sizes α A , α B and α C or β A , β B and β C .Next, we assume that there is a schedule with makespan 47 for the scheduling instance.There are exactly |M| many jobs with size α A = α A ′ or β A = β A ′ corresponding to elementsof A ∪ A ′ , and due to Claim 34 we know that each machine receives exactly one of these jobs.The machines corresponding to triplets from E ( a n ) are the only ones that can process the | E ( a n ) | jobs corresponding to a n , and hence each of these machines receives exactly one ofthese jobs. Now, the machines corresponding to triplets from E ( a ′ n ) are the only remaining ones that can process the | E ( a ′ n ) | jobs corresponding to a ′ n . Iterating this argument, weget that each machine e receives exactly one job corresponding to some x ∈ A ∪ A ′ with x ∈ e . Note that the above argument was based on the first resource value. Considering thesecond resource value yields the same result for each x ∈ C ∪ C ′ . For the elements x ∈ B ∪ B ′ both resource values have to be considered, namely the second for b ∈ B and the first for b ′ ∈ B ′ , but the argument stays the same. Summing up, each machine e = { x, y, z } receivesexactly three jobs corresponding to x , y and z . Now, considering this and Claim 34, we geta perfect matching by selecting the triplets e that processes three element jobs. ◭ In this paper we provided hardness of approximation results for scheduling with interval andresource restrictions. We list some possible future research directions:From the perspective of complexity, tighter hardness results seem plausible. In particu-lar, we have the same inapproximability results for RAR(2) and RAR(3) and it would beinteresting to find a better result for RAR(3).From the algorithmic perspective, it remains open whether any of the studied problemsand RAI in particular admits an approximation algorithm with a rate smaller than 2. Therehave been some results [38, 33] for RAI using promising linear programming relaxationsthat may be useful in this context. Another possibility is the application of the local searchtechniques originally used by Svensson [36] for the restricted assignment problem. Thisapproach recently yielded a breakthrough for the graph balancing problem [20].Finally, while a PTAS for RAR(1) is known [29], it is unclear whether the problemadmits a so called efficient PTAS with a running time of the form f (1 /ε )poly( | I | ) for somecomputable function f . References Chidambaram Annamalai, Christos Kalaitzis, and Ola Svensson. Combinatorial algorithmfor restricted max-min fair allocation. 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Lett. , 20(4):149–154, 1997. doi:10.1016/S0167-6377(96)00055-7 . arten Maack and Klaus Jansen 25 A Examples for Section 2 Example Simple Reduction. The following formula is an instance of 3-SAT ∗ with minimalsize:( ¬ x , ¬ x , x ) ∧ ( ¬ x , x , ¬ x ) ∧ ( x , ¬ x , x ) ∧ ( ¬ x , x , x ) Note that the formula is fulfilled if all the variables take the value ⊤ . The correspondingrestricted assignment instance is depicted in Figure 5. CM , CM , CM , CJ ⊤ , CJ ⊥ , CM , CM , CM , CJ ⊤ , CJ ⊥ , CM , CM , CM , CJ ⊤ , CJ ⊥ , CM , CM , CM , CJ ⊤ , CJ ⊥ , CJ ⊥ , CJ ⊥ , CJ ⊤ , CJ ⊤ , TM , TM , TJ ⊤ TJ ⊥ VJ ⊤ , VJ ⊥ , VJ ⊤ , VJ ⊥ , VJ ⊤ , VJ ⊥ , VJ ⊤ , VJ ⊥ , TM , TM , TJ ⊤ TJ ⊥ VJ ⊤ , VJ ⊥ , VJ ⊤ , VJ ⊥ , VJ ⊤ , VJ ⊥ , VJ ⊤ , VJ ⊥ , TM , TM , TJ ⊤ TJ ⊥ VJ ⊤ , VJ ⊥ , VJ ⊤ , VJ ⊥ , VJ ⊤ , VJ ⊥ , VJ ⊤ , VJ ⊥ , Figure 5 The restricted assignment instance constructed for a minimal example instance. Thehatched rectangles represent private loads, and the connecting lines indicate eligibility. If these linesend at a dashed rectangle, the eligibility information concerns everything within the rectangle. Wechose a short notation for the jobs and machines writing, e.g., VJ ◦ j,t instead of VJob ◦ j,t . Example Refined Reduction. We consider the same formula as above for the refined reduc-tion. The values of κ for the occurrences of the first two variables together with the resultingincreasing lexicographical ordering is depicted in Table 8. Furthermore, in Figure 6 the truthassignment as well as the bridge and highway gadget for the first two variables are depicted. Table 8 The occurrences of the first two values in the clauses and the resulting increasinglexicographical ordering of the occurrences.( j, t ) (1,1) (1,2) (1,3) (1,4) (2,1) (2,2) (2,3) (2,4) κ ( j, t ) (1,3) (3,3) (2,1) (4,1) (2,2) (4,3) (1,1) (3,2)Ordering ( j, t ) (2,3) (1,1) (1,3) (2,1) (2,4) (1,2) (1,4) (2,2) I n a pp r o x i m a b ili t y R e s u l t s f o r S c h e du li n g w i t h I n t e r v a l a nd R e s o u r c e R e s t r i c t i o n s T S P T S TM , TM , GM , GM , GM , GM , BMI , , BMI , , BMI , , BMI , , TM , TM , GM , BMO , , BMO , , GM , GM , BMO , , BMO , , GM , TJob ◦ TJob ◦ VJob ◦ , VJob ◦ , VJob ◦ , VJob ◦ , HJob ◦ , , HJob ◦ , , HJob ◦ , , HJob ◦ , , VJob ◦ , BJob ◦ , , BJob ◦ , , VJob ◦ , VJob ◦ , BJob ◦ , , BJob ◦ , , VJob ◦ , HJob ◦ , , HJob ◦ , , HJob ◦ , , HJob ◦ , , HJob ◦ , , HJob ◦ , , HJob ◦ , , HJob ◦ , , Figure 6 The truth assignment as well as the bridge and highway gadget for the first two variables of an example instance. The colored lines mark theintervals of eligible machines for the respective jobs. In the picture, we use a more compact notation for the machines, and write, e.g., TM , instead of TMach , . arten Maack and Klaus Jansen 27 B Reduction by Bhaskara et al. [2] We state the reduction by Bhaskara et al. [2] from 3-DM to LRS(4) and show that it is notsound. We remark that the construction can be repaired with not too much effort usingprocessing times similar to the ones presented in the reduction for RAR(3) in Section 3.Given an instance ( A, B, C, E ) of 3-DM and some ε > 0, let n = | A | , N = n/ε and E ( x ) = { e ∈ E | x ∈ e } for each x ∈ A ∪ B ∪ C . We identify the set of machines M with the set of triplets E , i.e., M = E . The speed vector of e = { a i , b j , c k } ∈ M is givenby ( N i , N j , N k , x ∈ A ∪ B ∪ C , we introduce one element and | E ( x ) | − Element Dummy a i ∈ A ( εN − i , , , 1) ( εN − i , , , . b j ∈ B (0 , εN − j , , 1) (0 , εN − j , , . c k ∈ C (0 , , εN − k , 1) (0 , , εN − k , . In [2] the authors show that there is a schedule with makespan at most 3 + 3 ε if there isa 3D-matching. Furthermore, two Lemmata (Lemma 2.1 and 2.2 in [2]) are used to showthat the existence of a schedule with makespan at most 3 . 09 + 3 ε implies the existence of a3D-matching. We present a counter example. Let n = 3 and: E = (cid:8) { a , b , c } , { a , b , c } , { a , b , c } , { a , b , c } , { a , b , c } (cid:9) Hence, we have: x a a a b b b c c c | E ( x ) | In order to match a and a , a 3D-matching would have to contain the first two tripletsmatching c twice. Hence, there is no 3D-matching for this instance.On the other hand, we define a schedule with makespan 3 + 3 ε ≤ . 09 + 3 ε . Machine { a , b , c } { a , b , c } { a , b , c } { a , b , c } { a , b , c } Jobs a , a , b (element) a , b , c (dummy) a , b , c (dummy) a , b , c (element) b , c , c (element)Load 3 + (2 +1 /N ) ε /N ) ε ε ε /N ))