The complexity of machine scheduling for stability with a ... - CiteSeerX

Operations Research Letters 33 (2005) 151 – 156

Operations Research Letters www.elsevier.com/locate/dsw

The complexity of machine scheduling for stability with a single disrupted job Roel Leus 1 , Willy Herroelen∗ Operations Management Group, Department of Applied Economics, Katholieke Universiteit Leuven, Naamsestraat 69, B-3000 Leuven, Belgium Received 30 October 2002; accepted 22 April 2004

Abstract A stable schedule is a robust schedule that will change little when uncertain events occur. The purpose of this paper is to investigate the complexity status of a number of machine scheduling problems with stability objective, when the duration of a single job is anticipated to be disrupted. c 2004 Elsevier B.V. All rights reserved.  Keywords: Scheduling; Uncertainty; Robustness; Stability; Complexity

1. Introduction Manufacturing schedules are rarely executed in a ‘vacuum’ environment, and regularly su6er disruptions from a variety of sources like resource unavailability, tardy deliveries of material or sub-assemblies, altered work content of some jobs, etc. Incorporation of uncertainty in the planning stage can be done in multiple ways. A :rst option is to eliminate the use of schedules altogether and construct scheduling policies that will determine dynamically which jobs to dispatch at what time instances. We refer to Part 2 of [20] for a survey in machine scheduling and to [23] for a broader project scheduling setting. Alternatively, ∗ Corresponding author. Tel.: +32-16-326970; fax: +32-16326732. E-mail addresses: [email protected] (R. Leus), [email protected] (W. Herroelen). 1 Postdoctoral fellow of the Fund for Scienti:c Research, Flanders (Belgium) (F.W.O.).

a schedule can be constructed despite the uncertainty inherent to the scheduling environment. We refer to [3,18] for a general discussion of the usefulness of such a pre-schedule. When disruptions occur during schedule execution, this pre-schedule needs to be rescheduled. A pre-schedule is robust if it is built with an eye on protection against the manifestation of variability during its execution. Logically, robust scheduling generally incorporates assumptions about the rescheduling strategy that will be followed. The term stability refers to the situation where there is little deviation between the pre-schedule and the executed schedule. Stability can be strived for during rescheduling, and is then alternatively referred to as minimally disruptive, minimal perturbation and minimum deviation scheduling; for examples we refer to [2,4,21,22,24]. The option that we explore in this paper is to introduce stability already into the pre-schedule, thus constituting a particular form of schedule robustness that has also been

c 2004 Elsevier B.V. All rights reserved. 0167-6377/$ - see front matter
doi:10.1016/j.orl.2004.04.008

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named solution robust or predictable scheduling. Examples from literature are sparse, we mention [18,19]. 2. Notation, problem denition and general outline Uncertainty during schedule execution is modeled by variability in job durations. In light of the diFculty of scheduling with continuous duration distributions, a number of studies have resorted to modeling duration variability by means of discrete scenarios, e.g. [6,7,12]. This will also be our choice, and we additionally evade inherent complexity due to the possible combinations of job duration realizations by optimizing for the situation in which a single job deviates from its pre-schedule duration. The resulting restricted model is useful when disturbances are sparse and spread throughout time, such that the number of interactions is limited. This is especially applicable when resources are not machines but human beings, such that job durations are not mere realizations of nature but rather manageable to a certain extent. In a project scheduling context, [11] presents computational evidence that schedules which are optimized for a single disruption, perform very well also when the actual number of disruptions increases. Studies in which comparable suggestions have been made are [1] (a single deterministic or stochastic breakdown), [17] (one disruption on a single fallible machine in a job shop) and [18] (minimize the distance from a schedule with all jobs disrupted). In a reactive rather than proactive (robust) setting, we :nd similar considerations in [10] (analysis of schedule disruptions caused by the arrival of a single set of new jobs). Finally, a reasoning quite akin to ours in a graph coloring context can be found in [25], where the robustness of a coloring is measured as the probability of the coloring remaining valid after one random complementary edge is added to the edge set. We assume that a set of jobs N , |N | = n, with deterministic durations pi , i ∈ N , is to be scheduled on a set of m identical parallel machines (with special attention also to case m = 1), a solution being a pre-schedule S that speci:es starting times si (S) and machine allocations for all jobs. We impose a (common) deadline d on the pre-schedule: si (S) + pi 6 d, ∀i ∈ N . A prob-

ability
of disruption qi is associated with every job i ∈ N , N qi = 1. Random variable Li denotes the increase in pre-schedule duration pi for job i if i is disturbed. Li is assumed discrete with probability mass function gi (·), which associates non-zero probability with positive values lik ∈ i , where i denotes the set
of disturbance scenarios for the duration of job i; k∈ i gi (lik ) = 1. For encoding reasons, we require that all values qi be rational numbers represented by two integers, and that gi map into the set of rational numbers. Disruption lengths lik are assumed to be integers. Non-negative rational number wi represents the cost incurred per unit time overrun on the start time of job i, to penalize the resulting system nervousness and shop co-ordination diFculties. Logically, the actual starting time of job i is a random variable Si (S), which is dependent on the pre-schedule. Jobs are not started earlier than planned, i.e. si (S) 6 Si (S), ∀i ∈ N , which guarantees that the pre-schedule is realized if all goes as planned (no disruptions). Disruption information becomes available only when the pre-schedule is executed. The expected weighted deviation in start times in the realized schedule from those in the pre-schedule is used as stability measure for
a schedule S: the objective function we minimize is N wj (ESj (S) − sj (S)), with E the expectation operator. Some deterministic scheduling problems with minimum makespan objective are NP-complete, which means that in the same scheduling environment, veri:cation of the existence of a schedule respecting d and a bound on any performance measure is also NP-complete, provided a succinct encoding for trivial (because unimportant) variability characteristics is possible. Remark that this does not preclude the possibility of a polynomial-time algorithm for subproblems requiring longer encoding for duration variability (when maxi | i | is not polynomially bounded in n). In view of the foregoing, we pay attention to the special case where the imposed scheduling deadline d is not ‘restrictive’, meaning that d is at least as large as the deterministic minimum makespan. In this paper, we examine scheduling with the stability objective introduced above and also scheduling for makespan protection under the form of minimizing expected excess over d. Note that, whereas the second performance measure is regular, the :rst one is not.

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3. Single machine results We start by studying the single machine scheduling problem. We reschedule simply by right-shifting the remaining jobs on the machine with the disrupted job without re-sequencing. If we de:ne [k; i] to be the job that is scheduled in the ith position on machine k, then S[1; 1] (S) = s[1; 1] (S) and S[1; i] (S) = max{s[1; i] (S); S[1; i−1] (S) + X[1; i−1] }, i = 2; : : : ; n, with Xi a stochastic variable representing the actual duration of i according to the disruption scheme described above. We show that the following theorem holds. Theorem 1. The single-disruption stability problem on a single machine is ordinarily NP-hard. Proof. We denote the associated decision problem by .  is clearly in NP. We describe a polynomial trans formation
 from  , the decision problem version of P2 wj Cj , which is ordinarily NP-complete ([5], by reduction from KNAPSACK). For an instance of  , we have input parameters pi , the durations of the jobs to be scheduled, and wi , the weights of the jobs. Cj are the job completion times. For set N  of jobs to be scheduled, the expression to be minimized by  is  i∈N


wi Ci =

nk 2   k=1 i=1

 w[k; i]

i−1  j=1

 p[k; j] +



wi pi ;

(1)

i∈N


with nk the total number of jobs assigned to machine k, so solution of  boils down to answering the question whether there exists a schedule such that the :rst expression in the second term of (1) does not exceed a given number K  . The corresponding parameters of  are as follows: N = N  , but we choose durations pi = 1 for each job. Each job has a single
disruption scenario li1 = 1 and probability qi = pi = N
pj , and we choose d = n + 1, so a feasible schedule always exists. The cost coeFcients are selected as wi = wi , i ∈ N. De:ne pairwise Noat
Fnij = sj − si − pi , i; j ∈ N , and [1; i] = [i]. Constraint i=0 F[i][i+1] = 1 is imposed by the choice for d: for a feasible schedule, we have to divide the Noat of one time unit across the time bu6ers F[i][i+1] (i = 0; : : : ; n), with F[0][1] and F[n][n+1] the Noat before the :rst and after the last job, respectively. We :rst show that for a given order of the jobs, scheduling

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this available Noat contiguously is always a dominant decision for . We also have (EL is the expectation operator with respect to L): ES[j] = s[j] + 

j−1 

q[z] EL[z]

z=1



max 0; L[z] −

= s[j] +

j−1  z=1

 q[z]

j−1 

 F[x][x+1]

x=z

1−

j−1 

 F[x][x+1]

;

x=z


so the objective function can be seen to be N wj
j−1
n
n (ESj − sj ) = j=2 w[j] z=1 q[z] − x=2 F[x−1][x]
x−1
n ( j=x w[j] z=1 q[z] ), and all q[i] and w[i] are constant for a :xed job order. We see that it is never a dominated solution to assign all Noat to the bu6er with highest coeFcient in the second term of the objective expression. In conclusion, we restrict the search for an optimal schedule to solutions with a bu6er of size 1 somewhere in the schedule, and given the disruption lengths of 1, this divides the jobs into 2 blocks that do not inNuence each other. We denote the job at the kth position in the ith block by [i; k] and the number of jobs in block b by nb . Solution of the thus constructed -instance veri:es whether a schedule exists for which, for a rational number K:  n nb i−1 2       w[b; i] p[b; j] pi 6 K : b=1 i=1

j=1

i=1


We see that choice of threshold K = K  = j pj allows us to solve  by means of . In conclusion,  is ordinarily NP-complete and the theorem follows. It is interesting to see that it is not the sequencing part itself but rather the interplay between sequencing and bu6er allocation that induces the complexity status of this problem: an adjacent interchange argument
shows that the stability problem for the case d = j pj is solvable in polynomial time, even for multiple disruption scenarios per job, by ordering the jobs with shortest weighted expected disruption length (qi ELi Li =wi ) :rst. Also, makespan protection is trivially solvable: all semi-active or ‘left-justi:ed’ schedules are optimal.

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We next turn our attention to unequal ready times. The makespan protection problem is still easily solved in polynomial time; an optimal schedule is obtained by starting the jobs in non-decreasing order of ready time, and as early as possible. This is straightforwardly shown by an adjacent interchange argument (it does not trivially follow from 1|ri |Cmax because we have to account for disruption lengths). We also have the following result: Theorem 2. The single-disruption stability problem with unequal ready times on a single machine is strongly NP-hard. Proof. We represent the decision problem version by , which is easily seen to be in NP. We describe a pseudo-polynomial transformation to  from the strongly NP-complete decision problem  corre
sponding with 1|ri | Cj (see [16]). We start by the choice of a deadline d, which is selected such that at least one optimal solution to  exists with makespan no larger than d, a safe value being the largest ri -value plus the sum of all job durations. N consists of the jobs from  with corresponding
duration, collected in set P, augmented with (d − i∈P pi ) enforcer jobs of duration 1, which are gathered in set Q. Disruption lengths for all jobs in N equal their duration; all disruption probabilities are equal to (1=|P ∪ Q|) = k. The cost coeFcients wi are zero for i ∈ Q and 1 for i ∈ P; the ready times ri are zero for i ∈ Q and equal to the original ready times for i ∈ P. For an arbitrary schedule to a resulting -instance, the expected increase in starting time for any job j is proportional to its scheduled starting time with proportionality constant k, so if K  is a numerical bound  on the objective function
of  ,  solves the instance  with bound k(K − i∈P pi ). The transformation can be seen to be pseudo-polynomial, since it is polynomial in the magnitude of the largest number and the size of the instance of  , and the largest number in  is polynomially bounded in the largest number and the size of  . In this way, a polynomial bound on the largest number in  translates to a polynomial bound on the largest number in , which is required for strong NP-completeness. The presence of precedence constraints is an extra complication for the scheduling problems at hand;

such constraints can be represented by a precedence graph G(N; A). Even without resource constraints, we are still faced with a scheduling question, which can be solved in polynomial time (studied in [11]). An important complication when precedence constraints are imposed is that starting time disruptions can be the result of the disruption of the duration of either a direct or transitive machine predecessor (as before) or a direct or transitive predecessor in G, but we will not be faced with this complication explicitly in the following. The makespan protection problem is polynomially solvable, similarly as the case without precedence constraints. Remark that 1|ri ; prec|Cmax is also polynomially solvable ([13]) and the same holds for makespan protection with ready times and precedence constraints. Theorem 3. The single-disruption stability problem with precedence constraints on a single machine is strongly NP-hard. Proof. We provide a simple pseudo-polynomial transformation from  , the decision version of
1|prec| Cj , which is strongly NP-complete ([14,15]). For any instance of  , we construct an instance of  with the same jobs with durations equal to 1, d equal to the number of jobs, all disruption probabilities equal, and single disruption lengths equal to the original durations. All wi = 1, and the precedence constraints are maintained. The reduction is then immediate by noting that no Noat can be inserted because of the choice of the deadline d and the fact that the expected disruption of a job is the sum of the original durations of the preceding jobs (disruption lengths), times a constant (for normalizing the probabilities). The deadline is non-restrictive: any linear extension of the partial ordering implied by the precedence constraints de:nes a feasible schedule. Note that the proofs of Theorems 1–3 have been formulated for a non-restrictive deadline, and generalize to the restrictive case. 4. Parallel machine results Let us now turn to the parallel machine case. We reschedule without re-sequencing nor

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changing machines, so S[k; 1] (S) = s[k; 1] (S), k = 1; : : : ; m, and S[k; i] (S) = max{s[k; i] (S); S[k; i−1] (S) + X[k; i−1] }, k = 1; : : : ; m and i = 2; : : : ; nk . Obviously, this restriction facilitates analysis; practical motivation can be the fact that tooling or other peripheral equipment has already been put into place, or other preparatory setup restrictions hold, or that there are non-negligible distances between machines or other communication delays. P2Cmax is ordinarily NP-hard, so the complexity status of makespan protection for two machines with possibly restrictive deadline is immediate. The case with non-restrictive deadline is still ordinarily NP-hard via reduction from P2Cmax by giving all jobs a disruption length equal to d minus the makespan threshold of the latter problem and decision problem bound 0. In [9], the following problem is shown to be NP-complete in the strong sense: 3-PARTITION. Input: a :nite set Q of 3h elements (h ∈ IN), a bound B ∈ IN, and an integer size za for each a ∈ Q, such
that each za satis:es B=4 ¡ za ¡ B=2 and such that a∈Q za = hB. Question: can Q be partitioned into h disjoint sets Q1 , Q2 ; : : : ; Qh such
that, for 1 6 i 6 h; a∈Qi za = B? PCmax (with free number of parallel machines) is strongly NP-hard, since 3-PARTITION can be considered a special case, with immediate result for both scheduling for stability as well as for makespan protection in the same environment, when the deadline can be restrictive. For non-restrictive d, makespan protection remains strongly NP-hard by a similar reasoning as for the two-machine case—the necessity of a reduction is an artifact of the non-restrictiveness constraint, but rather straightforward. For the stability objective, we present the following result. Theorem 4. The single-disruption stability problem with free number of parallel machines is strongly NP-hard. Proof. We prove that it is strongly NP-complete to decide if the optimal objective function value of the problem under study does not exceed a threshold K; we refer to this decision problem as .  is clearly in NP. We describe a pseudo-polynomial transformation from 3-PARTITION to . We start from an arbitrary partition of Q into 3-element subsets and determine the size ( of the largest subset, and we then construct

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a corresponding h-machine scheduling problem. N = H ∪ Q◦ , with |Q◦ | = |Q| and H containing h enforcer jobs indexed k = 3h + 1; : : : ; 4h; all 4h jobs have equal disruption probability 1=(4h): ∀i ∈ Q◦ , we have wi =1, | i | = 1, li1 = ( − B + 1 and pi = zi . For the jobs in H , wi = 3hl11 + 1, | i | = 1, li1 = 3hl11 + d and pi = 1. We set d = ( + 2 and K = 3hl11 =(4h). Given our choice of d, a feasible schedule for the corresponding instance of  always exists, so d is not restrictive. If K is to be respected, each machine should carry exactly one of the extra jobs, and this job should be scheduled last on the machine. More speci:cally, it can always be scheduled from time d − 1 to d. K can be respected if each H -job can be completely protected from disruptions in its machine predecessors. We can always schedule all Q◦ -jobs such that each processor is contiguously occupied from time 0, and then time window [B; ( + 1] should not be used by any job, which leaves us with h separate time blocks on the h processors, each of length exactly B. Since this is just enough time in total to accommodate all the jobs in Q◦ , each block must be completely :lled; these blocks therefore play the same role as the sets Q1 ; Q2 ; : : : ; Qh in the desired partition of Q. We remind the reader that, since B=4 ¡ za ¡ B=2 for each item a, :lling length B by means of 2 or 4 jobs (or less or more, respectively), is impossible. Thus, the answer to a 3-PARTITION instance can be seen to be ‘yes’ if and only if the answer to the corresponding -instance is ‘yes’. The ‘if’ part of this statement is easy to see, and the ‘only if’ part follows essentially because the objective function for every ‘valid’ 3-PARTITION is already 3hl11 =(4h). The jobs have integer durations and disruption lengths and as a consequence, if the extra job on one machine cannot be protected, the increment in the objective function is at least (3hl11 + 1)=(4h). In [8], problem P2|chains|Cmax is proved to be strongly NP-hard, which allows for an easy reduction to makespan protection for two machines with precedence constraints and possibly restrictive deadline. The non-restrictive case is also easily proven strongly NP-hard. Finally, we brieNy discuss the problem of allocating m parallel machines to a set of jobs when a feasible starting time schedule has been produced already and a machine assignment needs to be developed in a way

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that is compatible with the input starting times. Case m=1 is trivial, while m=2 for makespan protection can be shown to be ordinarily NP-hard by reduction from PARTITION by insertion of two ‘enforcer’ jobs in parallel at the start of a schedule with all jobs (items) in series, both with disruption length equal to the sum of the durations (sizes) divided by 2, and asking whether a schedule exists with objective zero when d is the sum of the durations. Makespan protection reduces to stability by insertion of two enforcer jobs in parallel at the end of the schedule. 5. Concluding remarks Extensions to a speci:c number of disruptions higher than one seem interesting but will probably all prove intractable but for the most simple cases. We point out that all complexity proofs have been formulated for the case with only one disruption scenario per job. References [1] I. Adiri, J. Bruno, E. Frostig, A.H.G. Rinnooy Kan, Single machine Now-time scheduling with a single breakdown, Acta Inform. 36 (1989) 679–696. [2] M.S. Akturk, E. Gorgulu, Match-up scheduling under a machine breakdown, European J. Oper. Res. 112 (1999) 81–97. [3] H. Aytug, M.A. Lawley, K. McKay, S. Mohan, R. Uzsoy, Executing production schedules in the face of uncertainties: a review and some future directions, European J. Oper. Res. (2004), to appear. [4] J.C. Bean, J.R. Birge, J. Mittenthal, C.E. Noon, Match-up scheduling with multiple resources, release dates and disruptions, Oper. Res. 39 (1991) 470–483. [5] J. Bruno, E.G. Co6mann Jr., R. Sethi, Scheduling independent tasks to reduce mean :nishing time, Comm. ACM 17 (1974) 382–387. [6] R.L. Daniels, J.E. Carrillo, *-robust scheduling for single-machine systems with uncertain processing times, IIE Trans. 29 (1997) 977–985. [7] R.L. Daniels, P. Kouvelis, Robust scheduling to hedge against processing time uncertainty in single-stage production, Management Sci. 41 (1995) 363–376.

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