Group scheduling with independent setup times, ready times, and deteriorating job processing times

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  • Int J Adv Manuf Technol (2012) 60:643649DOI 10.1007/s00170-011-3639-1

    ORIGINAL ARTICLE

    Group scheduling with independent setup times, readytimes, and deteriorating job processing times

    Ji-Bo Wang Xue Huang Yu-Bin Wu Ping Ji

    Received: 24 February 2009 / Accepted: 13 September 2011 / Published online: 5 October 2011 Springer-Verlag London Limited 2011

    Abstract In this paper we investigate a single ma-chine scheduling problem with deteriorating jobs andgroup technology assumption. By deteriorating jobsand group technology assumption, we mean that jobprocessing times are simple linear functions of itsstarting times. The group setup times are assumedto be known and fixed. We attempt to minimize themakespan with ready times of the jobs. For a specialcase, we show that the problem can be solved in poly-nomial time when deterioration and group technologyare considered simultaneously.

    Keywords Scheduling Single machine Deteriorating jobs Group technology Ready time

    1 Introduction

    Traditional scheduling problems usually involve jobswith constant, independent processing times. In prac-tice, however, we often encounter settings in which thejob processing times vary with time. Hence, there is agrowing interest in the literature to study scheduling

    J.-B. Wang (B) X. Huang Y.-B. WuSchool of Science, Shenyang Aerospace University,Shenyang 110136, Chinae-mail: wangjibo75@yahoo.com.cn

    X. Huange-mail: huangxuebj@126.com

    P. JiDepartment of Industrial and Systems Engineering,The Hong Kong Polytechnic University,Hung Hom, Kowloon, Hong Kong, Chinae-mail: mfpji@inet.polyu.edu.hk

    problems involving deteriorating jobs, i.e., jobs whoseprocessing times are increasing functions of their start-ing times. Job deterioration appears, e.g., in schedulingmaintenance jobs, steel production, national defense,emergency medicine, or cleaning assignments, whereany delay in processing a job is penalized by incurringadditional time for accomplishing the job. Extensivesurveys of different scheduling models and problemsinvolving jobs with start time-dependent processingtimes can be found in Alidaee and Womer [1], Chenget al. [2], and Gawiejnowicz [3]. More recent paperswhich have considered scheduling jobs with deteriora-tion effects include Cheng et al. [4, 5], Wang and Xia [68], Janiak and Kovalyov [9], Gawiejnowicz et al. [10],Gawiejnowicz [11], Kang and Ng [12], Wang [13, 29],Wang et al. [14, 18, 21, 23, 24, 26], Wu et al. [15, 16], Wuand Lee [17], Toksar and Guner [19, 20], Lee et al. [22],Yin et al. [25], Wang and Wang [27], Huang et al. [28],Wang and Sun [30], Bahalke et al. [31], Xu et al. [32],Shen et al. [33], Ozturkoglu and Bulfin [34], and Zhanget al. [35].

    On the other hand, an important class of schedul-ing problem is characterized by the group technologyassumption, i.e., the jobs are classified into groups bythe similar production requirements, no machine setupsare needed between two consecutively scheduled jobsfrom the same group, although an independent setupis required between jobs of different groups. In grouptechnology, it is conventional to schedule continuouslyall jobs from the same group. Group technology thatgroups similar products into families helps increase theefficiency of operations and decrease the requirementof facilities. Hence, the scheduling in group technol-ogy environment results in a new stream of research[36, 37]. To the best of our knowledge, only a few results

  • 644 Int J Adv Manuf Technol (2012) 60:643649

    concerning scheduling problems with deteriorating jobsand group technology simultaneously are known. Wanget al. [14] considered the makespan minimization prob-lem and total completion time minimization problemunder group technology assumes that the setup timesare constant, and the actual processing time of a jobis a general linear decreasing function of its startingtime. They showed that these problems can be solvedin polynomial time. Since longer setup or preparationmight be necessary as food quality deteriorates or apatients condition worsens, Wu et al. [16] and Wu andLee [17] considered a situation where the setup timegrows and jobs deteriorate as they wait for process-ing, i.e., group setup times and job processing timesare both described by a linear deterioration function.For single-machine group scheduling, they proved thatthe makespan minimization problem can be solved inpolynomial time. Wang et al. [18] considered single-machine group scheduling problems with deteriorationjobs, i.e., the group setup times and job processingtimes are both described by a function which is pro-portional to a linear function of time. They showedthat the makespan minimization problem and the totalweighted completion time minimization problem re-main polynomially solvable.

    Dessouky [38] pointed out the importance of readytimes (release dates) that in semiconductor manufac-turing where it is common to find newer, more modernmachines running side by side with older, less efficientmachines which are kept in operation because of highreplacement cost to identify a schedule if one existsin which each job cannot be started before its readytime and must not be completed after its due date.Moreover, all of the above works considered schedul-ing problems without ready times. Thus, in this paper,we consider the single machine scheduling with readytimes of the jobs under the group technology assump-tion and deteriorating job processing times. The groupsetup times are assumed to be known and fixed. Theremaining part of the paper is organized as follows. Inthe next section, a precise formulation of the problemis given. The problem of minimization of the makespanis given in Section 3. The last section contains someconclusions.

    2 Problem formulation

    In this section, we first define the notation that is usedthroughout this paper, followed by the description ofthe problem.

    m The number of groups (m 2)

    Gi Group i, i = 1, 2, . . . , mni The number of jobs belonging to group

    Gi, i = 1, 2, . . . , mn The total number of jobs, i.e., n1 + n2 +

    . . . + nm = nJij Job j in group Gi, i = 1, 2, . . . , m, j = 1,

    2, . . . , niij The deterioration rate of job Jij, i = 1,

    2, . . . , m, j = 1, 2, . . . , nirij The ready (arrival) time of job Jij, i =

    1, 2, . . . , m, j = 1, 2, . . . , nisi The setup time of group Gi, i = 1, 2,

    . . . , mpij The actual processing time of job Jij, i =

    1, 2, . . . , m, j = 1, 2, . . . , ni A job sequence of n jobsCij = Cij() The completion time for job Jij in Cmax The makespan of a given permutation,

    i.e., Cmax = max{Cij|i = 1, 2, . . . , m, j =1, 2, . . . , ni}

    There are n jobs grouped into m groups, and thesen jobs are to be processed on a single machine. Asetup time is required if the machine switches fromone group to another and all setup times of groupsfor processing at time t0 > 0. We also assume that theprocessing of a job may not be interrupted. Let ni be thenumber of jobs belonging to group Gi, thus, n1 + n2 +. . . + nm = n. Let Jij denote the jth job in group Gi,i = 1, 2, . . . , m; j = 1, 2, . . . , ni, rij > 0 denote the ready(arrival) time of job Jij. Let pij be the actual processingtime of the jth job in the group Gi, si be the setuptime of group Gi. As in Mosheiov [39], we consider thefollowing model

    pij = ijt,where ij is the deterioration rate of the jth job in thegroup Gi, t is its start time. The objective is to minimizethe makespan, i.e., the maximum completion time of alljobs.

    For a given schedule , Cij() represents the com-pletion time of job Jij in group Gi under schedule . Cmax = max{Cij|i = 1, 2, . . . , m; j = 1, 2, . . . , ni} rep-resent makespan of a given schedule. In the remainingpart of the paper, all the problems considered will bedenoted using the three-field notation schema ||introduced by Graham et al. [40].

    3 Makespan minimization scheduling problem

    In this section, we consider a single machine schedulingproblem with deteriorating jobs and ready times of

  • Int J Adv Manuf Technol (2012) 60:643649 645

    the jobs. The objective function is to minimize themakespan of all jobs.

    Theorem 1 For the problem 1|rij, pij = ijt, si, grouptechnology (GT)|Cmax, if the schedule of groups is given,then the optimal schedule that jobs within each groupcan be obtained by scheduling the jobs in nondecreasingorder of rij, i = 1, 2, . . . , m, j = 1, 2, . . . , ni.

    Proof Suppose that and are two job schedules.The difference between and is a pairwise in-terchange of two adjacent jobs Jiu and Jiv in groupGi, i.e., = (S1, Jiu, Jiv, S2) and = (S1, Jiv, Jiu, S2),where S1 and S2 denote a partial sequence. In addition,let A denote the completion time of the last job inS1. Under , the completion times of jobs Jiu andJiv are

    Ciu() = max {A, riu} + iu max {A, riu}

    = max {A, riu}(1 + iu) (1)

    and

    Civ() = max{Ciu, riv}(1 + iv)

    = max{A(1 + iu)(1 + iv),

    riu(1 + iu)(1 + iv), riv(1 + iv)}. (2)

    Similarly, the completion times of jobs Jiv and Jiu in are respectively

    Civ(

    ) = max {A, riv}(1 + iv) (3)

    and

    Ciu(

    ) = max{A(1 + iu)(1 + iv),

    riv(1 + iu)(1 + iv), riu(1 + iu)} (4)

    Suppose riu riv , it is easily verified that:First term in (4) = first term in (2);Second term in (4) second term in (2);Second term in (4) third term in (2).

    Hence, we have

    Ciu( ) Civ()

    = max{A(1 + iu)(1 + iv), riv(1 + iu)(1 + iv),

    riu(1 + iu)} max{A(1 + iu)(1 + iv),

    riu(1 + iu)(1 + iv), riv(1 + iv)} 0

    We conclude that an optimal schedule can be ob-tained by sequencing the jobs in nondecreasing orderof rij. unionsq

    Next, we assumed that A denotes the completiontime of the (i 1)th group and ri(1) ri(2) ri(ni)is satisfied in the ith group Gi. Then, the completiontime of the ith group Gi are:

    Ci(ni)(Gi)

    = max{

    (A + si)ni

    l=1

    (1 + i(l)

    ), ri(1)

    ni

    l=1

    (1 + i(l)

    ),

    ri(2)ni

    l=2

    (1 + i(l)

    ), , ri(ni)

    (1 + i(ni)

    )}

    = max{

    A + si, rB(i)B(i)1l=1

    (1 + i(l)

    )

    }ni

    l=1

    (1 + i(l)

    )

    (5)

    where rB(i)ni

    l=B(i)(1 + i(l)) = max{ri(1)ni

    l=1(1 + i(l)),ri(2)

    nil=2(1+i(l)), , ri(ni)(1+i(ni))}, B(i) {1, 2, ,

    ni} and 0l=1(1 + i(l)) = 1.In addition, we assumed that the setup time

    rB(i)ni

    l=B(i)(1 + i(l)), si and sini

    l=1(1 + il) are agree-able. That is, if rB(i)

    nil=B(i)(1 + i(l)) rB( j)

    n jl=B( j)(1 +

    j(l)) it implies that si s j and si nil=1(1 + il) s j

    n jl=1(1 + jl).

    Theorem 2 For the problem 1|rij, pij =ijt, si, GT|Cmax,if rB(i)

    nil=B(i)(1 + i(l)), si and si

    nil=1(1 + il) are agree-

    able, i.e., if rB(i)ni

    l=B(i)(1 + i(l)) rB( j)n j

    l=B( j)(1 + j(l)) it implies that si s j and si nil=1(1 + il)

  • 646 Int J Adv Manuf Technol (2012) 60:643649

    s jn j

    l=1(1 + jl), then the groups are arranged in nonde-creasing order of

    rB(i)B(i)1

    l=1 (1 + i(l)),

    where rB(i)ni

    l=B(i)(1 + i(l)) = max{ri(1)ni

    l=1(1 + i(l)),ri (2)

    nil = 2(1 + i (l)), , ri (ni)(1 + i (ni) )}, B (i) {1, 2,

    , ni}.

    Proof Let and be two schedules where thedifference between and is a pairwise inter-change of two adjacent groups Gi and G j, that is, =[S1, Gi, G j, S2], = [S1, G j, Gi, S2], where S1 and S2are partial sequences. Furthermore, we assume that Adenote the completion time of the last job in S1. Toshow dominates , it suffices to show that C j(n j)() Ci(ni)(

    ). Under , using Eq. 5, we obtain that thecompletion time of the group Gi is

    Ci(ni)()

    = max{

    A + si, rB(i)B(i)1l=1

    (1 + i(l)

    )

    }ni

    l=1

    (1 + i(l)

    )

    and the completion time of the group G j is

    C j(n j)()

    = max{

    Ci(ni)()+s j,rB( j)

    B( j)1l=1

    (1+ j(l)

    )

    } n j

    l=1

    (1+i(l)

    )

    = max{

    (A + si)ni

    l=1

    (1 + i(l)

    ) + s j,

    rB(i)B(i)1

    l=1(1 + i(l)

    )ni

    l=1

    (1 + i(l)

    ) + s j,

    rB( j)B( j)1

    l=1(1 + j(l)

    )

    } n j

    l=1

    (1 + j(l)

    )(6)

    Under , the completion times of the groups G j andGi are

    C j(n j)()

    = max{

    A + s j, rB( j)B( j)1l=1

    (1 + j(l)

    )

    } n j

    l=1

    (1 + j(l)

    )

    and the completion time of the group Gi is

    Ci(ni)()

    = max{

    C j(n j)()+si, rB(i)B(i)1

    l=1 (1+i(l))

    }ni

    l=1(1+ i(l))

    = max{

    (A + s j)n j

    l=1

    (1 + j(l)

    ) + si,

    rB( j)B( j)1

    l=1(1 + j(l)

    )n j

    l=1

    (1 + j(l)

    ) + si,

    rB(i)B(i)1

    l=1(1 + i(l)

    )

    }ni

    l=1

    (1 + i(l)

    )(7)

    Suppose

    rB(i)B(i)1

    l=1(1 + i(l)

    ) rB( j)B( j)1l=1

    (1 + j(l)

    ) ,

    it implies that si s j and si nil=1(1 + il) s jn j

    l=1(1 + jl).

    Based on Eqs. 6 and 7, we have

    C j(n j)() Ci(ni)( )

    = max{

    (A + si)(

    ni

    l=1

    (1 + i(l)

    ) n j

    l=1

    (1 + j(l)

    )

    + s jn j

    l=1

    (1 + j(l)

    ),

    rB(i)B(i)1

    l=1(1 + i(l)

    )ni

    l=1

    (1 + i(l)) n j

    l=1

    (1 + j(l)

    ) + s jn j

    l=1

    (1 + j(l)

    ),

    rB( j)B( j)1

    l=1(1 + j(l)

    )n j

    l=1

    (1 + j(l)

    )}

    max{

    (A + s j)n j

    l=1

    (1 + j(l)

    ) ni

    l=1

    (1 + i(l)

    )

    + sini

    l=1

    (1 + i(l)

    ),

    rB( j)B( j)1

    l=1(1 + j(l)

    )n j

    l=1

    (1 + j(l)) ni

    l=1

    (1 + i(l)

    ) + sini

    l=1

    (1 + i(l)

    ),

    rB(i)B(i)1

    l=1(1 + i(l)

    )ni

    l=1

    (1 + i(l)

    )}

  • Int J Adv Manuf Technol (2012) 60:643649 647

    max{

    (A + s j)ni

    l=1

    (1 + i(l)

    ) n j

    l=1

    (1 + j(l)

    )

    + sini

    l=1

    (1 + i(l)

    ),

    rB( j)B( j)1

    l=1(1 + j(l)

    )n j

    l=1

    (1 + j(l)) ni

    l=1

    (1 + i(l)

    ) + sini

    l=1

    (1 + i(l)

    ),

    rB( j)B( j)1

    l=1(1 + j(l)

    )n j

    l=1

    (1 + j(l)

    )}

    max{

    (A + s j)n j

    l=1

    (1 + j(l)

    ) ni

    l=1

    (1 + i(l)

    )

    + sini

    l=1

    (1 + i(l)

    ),

    rB( j)B( j)1

    l=1(1 + j(l)

    )n j

    l=1

    (1 + j(l)) ni

    l=1

    (1 + i(l)

    ) + sini

    l=1

    (1 + i(l)

    ),

    rB(i)B(i)1

    l=1(1 + i(l)

    )ni

    l=1

    (1 + i(l)

    )}

    = 0.

    Therefore,

    C j(n j)() Ci(ni)(

    ).

    This completes the proof. unionsq

    By the proof of Theorems 1 and 2, we can obtain thefollowing results,

    Corollary 1 For the problem 1|rij, pij = ijt, si = s,GT|Cmax, the optimal schedule satisf ies the following:1. The job sequence in each group is in nondecreasing

    order of rij, i.e.,

    ri(1) ri(2) . . . ri(ni), i = 1, 2, . . . , m.2. If rB(i)

    nil=B(i)(1 + i(l)) and

    nil=1(1 + il) are agree-

    able, i.e., if rB(i)ni

    l=B(i)(1 + i(l)) rB( j)n j

    l=B( j)(1 + j(l)) it implies that

    nil=1(1 + il)

    n jl=1(1 + jl),

    then the groups are arranged in nondecreasingorder of

    rB(i)B(i)1

    l=1 (1 + i(l)),

    where rB(i)ni

    l=B(i)(1 + i(l)) = max{ri(1)ni

    l=1(1 + i(l)),ri(2)

    nil=2(1+ i(l)), , ri(ni)(1+ i(ni))}, B(i) {1, 2,

    , ni}.

    Corollary 2 For the problem 1|rij, pij = ijt, GT|Cmax,the optimal schedule satisf ies the following:

    1. The job sequence in each group is in nondecreasingorder of rij, i.e.,

    ri(1) ri(2) . . . ri(ni), i = 1, 2, . . . , m.2. The groups are arranged in nondecreasing order of

    rB(i)B(i)1

    l=1 (1 + i(l)),

    where rB(i)ni

    l=B(i)(1 + i(l)) = max{ri(1)ni

    l=1(1 + i(l)),ri (2)

    nil = 2(1+ i (l)), , ri ( ni) (1+ i ( ni))}, B (i) {1, 2,

    , ni}.

    From Theorems 1 and 2, if rB(i)ni

    l=B(i)(1 + i(l)),si and si

    nil=1(1 + il) are agreeable, i.e., if

    rB(i)ni

    l=B(i)(1 + i(l)) rB( j)n j

    l=B( j)(1 + j(l)) it impliesthat si s j and si nil=1(1 + il) s j

    n jl=1(1 + jl), then

    the problem 1|rij, pij = ijt, si, GT|Cmax can be solvedby the algorithm presented in the following:

    Algorithm 1

    Step 1. Jobs in each group scheduled in nondecreasingorder of rij, i.e.,

    ri(1) ri(2) . . . ri(ni), i = 1, 2, . . . , m.Step 2. Let rB(i)

    nil=B(i)(1 + i(l)) = max{ri(1)

    nil=1(1 +

    i(l)), ri(2)ni

    l=2 (1 + i(l)), , ri(ni)(1 + i(ni))},B(i) {1, 2, , ni}, calculate B(i) and

    rB(i)B(i)1l=1 (1+i(l))

    , i = 1, 2, , m.Step 3. Groups scheduled in nondecreasing order of

    (Gi) = rB(i)B(i)1l=1 (1 + i(l))

    .

    Obviously, the...

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