Int J Adv Manuf Technol (2012) 60:643–649DOI 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 [6–8], 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) 60:643–649

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 apatient’s 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, . . . , ni

αij The deterioration rate of job Jij, i = 1,

2, . . . , m, j = 1, 2, . . . , ni

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

si 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

π A job sequence of n jobsCij = Cij(π) The completion time for job Jij in π

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 = αijt,

where αij 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 π , Cij(π) represents the com-pletion time of job Jij in group Gi under scheduleπ . 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 α|β|γintroduced 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) 60:643–649 645

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

Theorem 1 For the problem 1|rij, pij = αijt, 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 π and π ′ are two job schedules.The difference between π and π ′ is a pairwise in-terchange of two adjacent jobs Jiu and Jiv in groupGi, i.e., π = (S1, Jiu, Jiv, S2) and π ′ = (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 π , the completion times of jobs Jiu andJiv are

Ciu(π) = max {A, riu} + αiu max {A, riu}

= max {A, riu}(1 + αiu) (1)

and

Civ(π) = max{Ciu, riv}(1 + αiv)

= max{A(1 + αiu)(1 + αiv),

riu(1 + αiu)(1 + αiv), riv(1 + αiv)}. (2)

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

Civ(π ′) = max {A, riv}(1 + αiv) (3)

and

Ciu(π ′) = max{A(1 + αiu)(1 + αiv),

riv(1 + αiu)(1 + αiv), riu(1 + αiu)} (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(π′) − Civ(π)

= max{A(1 + αiu)(1 + αiv), riv(1 + αiu)(1 + αiv),

riu(1 + αiu)} − max{A(1 + αiu)(1 + αiv),

riu(1 + αiu)(1 + αiv), riv(1 + αiv)}≥ 0

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

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 + αi(l)

), ri(1)

ni∏

l=1

(1 + αi(l)

),

ri(2)

ni∏

l=2

(1 + αi(l)

), · · · , ri(ni)

(1 + αi(ni)

)}

= max

{

A + si,rB(i)

∏B(i)−1l=1

(1 + αi(l)

)

}ni∏

l=1

(1 + αi(l)

)

(5)

where rB(i)∏ni

l=B(i)(1 + αi(l)) = max{ri(1)

∏nil=1(1 + αi(l)),

ri(2)

∏nil=2(1+αi(l)), · · · , ri(ni)(1+αi(ni))}, B(i) ∈ {1, 2, · · · ,

ni} and∏0

l=1(1 + αi(l)) = 1.In addition, we assumed that the setup time

rB(i)∏ni

l=B(i)(1 + αi(l)), si and si∏ni

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

∏nil=B(i)(1 + αi(l)) ≤ rB( j)

∏n j

l=B( j)(1 +α j(l)) it implies that si ≤ s j and si

∏nil=1(1 + αil) ≥

s j∏n j

l=1(1 + α jl).

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

∏nil=B(i)(1 + αi(l)), si and si

∏nil=1(1 + αil) are agree-

able, i.e., if rB(i)∏ni

l=B(i)(1 + αi(l)) ≤ rB( j)∏n j

l=B( j)(1 +α j(l)) it implies that si ≤ s j and si

∏nil=1(1 + αil) ≥

646 Int J Adv Manuf Technol (2012) 60:643–649

s j∏n j

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

rB(i)∏B(i)−1

l=1 (1 + αi(l)),

where rB(i)∏ni

l=B(i)(1 + αi(l)) = max{ri(1)

∏nil=1(1 + αi(l)),

ri (2)

∏nil = 2(1 + αi (l)), · · · , ri (ni)(1 + α i (ni) )}, B (i) ∈ {1, 2,

· · · , ni}.

Proof Let π and π ′ be two schedules where thedifference between π and π ′ is a pairwise inter-change of two adjacent groups Gi and G j, that is, π =[S1, Gi, G j, S2], π ′ = [S1, G j, Gi, S2], where S1 and S2

are partial sequences. Furthermore, we assume that Adenote the completion time of the last job in S1. Toshow π dominates π ′, it suffices to show that C j(n j)(π) ≤Ci(ni)(π

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

Ci(ni)(π)

= max

{

A + si,rB(i)

∏B(i)−1l=1

(1 + αi(l)

)

}ni∏

l=1

(1 + αi(l)

)

and the completion time of the group G j is

C j(n j)(π)

= max

{

Ci(ni)(π)+s j,rB( j)

∏B( j)−1l=1

(1+α j(l)

)

} n j∏

l=1

(1+αi(l)

)

= max

{

(A + si)

ni∏

l=1

(1 + αi(l)

) + s j,

rB(i)∏B(i)−1

l=1

(1 + αi(l)

)ni∏

l=1

(1 + αi(l)

) + s j,

rB( j)∏B( j)−1

l=1

(1 + α j(l)

)

} n j∏

l=1

(1 + α j(l)

)(6)

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

C j(n j)(π′)

= max

{

A + s j,rB( j)

∏B( j)−1l=1

(1 + α j(l)

)

} n j∏

l=1

(1 + α j(l)

)

and the completion time of the group Gi is

Ci(ni)(π′)

= max

{

C j(n j)(π′)+si,

rB(i)∏B(i)−1

l=1 (1+αi(l))

}ni∏

l=1

(1+ αi(l))

= max

{

(A + s j)

n j∏

l=1

(1 + α j(l)

) + si,

rB( j)∏B( j)−1

l=1

(1 + α j(l)

)n j∏

l=1

(1 + α j(l)

) + si,

rB(i)∏B(i)−1

l=1

(1 + αi(l)

)

}ni∏

l=1

(1 + αi(l)

)(7)

Suppose

rB(i)∏B(i)−1

l=1

(1 + αi(l)

) ≤ rB( j)∏B( j)−1

l=1

(1 + α j(l)

) ,

it implies that si ≤ s j and si∏ni

l=1(1 + αil) ≥ s j∏n j

l=1(1 +α jl).

Based on Eqs. 6 and 7, we have

C j(n j)(π) − Ci(ni)(π′)

= max

{

(A + si)

(ni∏

l=1

(1 + αi(l)

) n j∏

l=1

(1 + α j(l)

)

+ s j

n j∏

l=1

(1 + α j(l)

),

rB(i)∏B(i)−1

l=1

(1 + αi(l)

)ni∏

l=1

× (1 + αi(l)

) n j∏

l=1

(1 + α j(l)

) + s j

n j∏

l=1

(1 + α j(l)

),

rB( j)∏B( j)−1

l=1

(1 + α j(l)

)n j∏

l=1

(1 + α j(l)

)}

− max

{

(A + s j)

n j∏

l=1

(1 + α j(l)

) ni∏

l=1

(1 + αi(l)

)

+ si

ni∏

l=1

(1 + αi(l)

),

rB( j)∏B( j)−1

l=1

(1 + α j(l)

)n j∏

l=1

× (1 + α j(l)

) ni∏

l=1

(1 + αi(l)

) + si

ni∏

l=1

(1 + αi(l)

),

rB(i)∏B(i)−1

l=1

(1 + αi(l)

)ni∏

l=1

(1 + αi(l)

)}

Int J Adv Manuf Technol (2012) 60:643–649 647

≤ max

{

(A + s j)

ni∏

l=1

(1 + αi(l)

) n j∏

l=1

(1 + α j(l)

)

+ si

ni∏

l=1

(1 + αi(l)

),

rB( j)∏B( j)−1

l=1

(1 + α j(l)

)n j∏

l=1

× (1 + α j(l)

) ni∏

l=1

(1 + αi(l)

) + si

ni∏

l=1

(1 + αi(l)

),

rB( j)∏B( j)−1

l=1

(1 + α j(l)

)n j∏

l=1

(1 + α j(l)

)}

− max

{

(A + s j)

n j∏

l=1

(1 + α j(l)

) ni∏

l=1

(1 + αi(l)

)

+ si

ni∏

l=1

(1 + αi(l)

),

rB( j)∏B( j)−1

l=1

(1 + α j(l)

)n j∏

l=1

× (1 + α j(l)

) ni∏

l=1

(1 + αi(l)

) + si

ni∏

l=1

(1 + αi(l)

),

rB(i)∏B(i)−1

l=1

(1 + αi(l)

)ni∏

l=1

(1 + αi(l)

)}

= 0.

Therefore,

C j(n j)(π) ≤ Ci(ni)

(π ′) .

This completes the proof. ��

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

Corollary 1 For the problem 1|rij, pij = αijt, si = s,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. If rB(i)∏ni

l=B(i)(1 + αi(l)) and∏ni

l=1(1 + αil) are agree-

able, i.e., if rB(i)∏ni

l=B(i)(1 + αi(l)) ≤ rB( j)∏n j

l=B( j)(1 +α j(l)) it implies that

∏nil=1(1 + αil) ≥ ∏n j

l=1(1 + α jl),then the groups are arranged in nondecreasingorder of

rB(i)∏B(i)−1

l=1 (1 + αi(l)),

where rB(i)∏ni

l=B(i)(1 + αi(l)) = max{ri(1)

∏nil=1(1 + αi(l)),

ri(2)

∏nil=2(1+ αi(l)), · · · , ri(ni)(1+ αi(ni))}, B(i) ∈ {1, 2,

· · · , ni}.

Corollary 2 For the problem 1|rij, pij = αijt, 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 + αi(l)),

where rB(i)∏ni

l=B(i)(1 + αi(l)) = max{ri(1)

∏nil=1(1 + αi(l)),

ri (2)

∏ nil = 2(1+ αi (l)), · · · , ri ( ni) (1+ αi ( ni))}, B (i) ∈ {1, 2,

· · · , ni}.

From Theorems 1 and 2, if rB(i)∏ni

l=B(i)(1 + αi(l)),si and si

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

rB(i)∏ni

l=B(i)(1 + αi(l)) ≤ rB( j)∏n j

l=B( j)(1 + α j(l)) it implies

that si ≤ s j and si∏ni

l=1(1 + αil) ≥ s j∏n j

l=1(1 + α jl), thenthe problem 1|rij, pij = αijt, 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)∏ni

l=B(i)(1 + αi(l)) = max{ri(1)

∏nil=1(1 +

αi(l)), ri(2)

∏nil=2 (1 + αi(l)), · · · , ri(ni)(1 + αi(ni))},

B(i) ∈ {1, 2, · · · , ni}, calculate B(i) andrB(i)∏B(i)−1

l=1 (1+αi(l)), i = 1, 2, · · · , m.

Step 3. Groups scheduled in nondecreasing order of

ρ(Gi) = rB(i)∏B(i)−1

l=1 (1 + αi(l)).

Obviously, the complexity of obtaining the optimal jobsequence within group Gi is O(nilogni) and that ofobtaining the optimal group sequence is O(mlogm).Hence, the complexity of Algorithm 1 is at mostO(nlogn), where n1 + n2 + · · · + nm = n.

Using Corollary 1, if rB(i)∏ni

l=B(i)(1 + αi(l)) and∏nil=1(1 + αil) are agreeable, i.e., if rB(i)

∏nil=B(i)

(1 + αi(l)) ≤ rB( j)∏n j

l=B( j) (1 + α j(l)) it implies that∏ni

l=1 (1 + αil) ≥ ∏n j

l=1 (1 + α jl), then the problem 1|rij,

pij = αijt, si = s, GT|Cmax can be solved byAlgorithm 1. Using Corollary 2, the problem 1|rij,

pij = αijt, GT|Cmax can also be solved by Algorithm 1.In addition, we demonstrate the algorithm in the

following example.

Example 1 Let n = 8, m = 3 and t0 = 1. Also,G1 : {J1, J2}, s1 = 3, α1 = 1

2 , α2 = 13 , r1 = 12, r2 = 3;

648 Int J Adv Manuf Technol (2012) 60:643–649

G2 : {J3, J4, J5}, s2 = 1, α3 = 1, α4 = 2, α5 = 32 , r3 = 2,

r4 = 3, r5 = 1; G3 : {J6, J7, J8}, s3 = 2, α6 = 14 , α7 = 1,

α8 = 12 , r6 = 4, r7 = 8, r8 = 2.

Solution According to Algorithm 1, we solve Example1 as follows:

Step 1: In group G1, the optimal job sequence is[J2, J1]. In group G2, the optimal job se-quence is [J5, J3, J4]. In group G3, the op-timal job sequence is [J8, J6, J7].

Step 2, 3: Next, we compute the following values foreach group:

G1 rB(1)

∏n1l=B(1) (1 + αi(l)) = max {3(1 + 1

3 )

(1 + 12 ), 12(1 + 1

2 )} = 18, B(1) = 2,

ρ(G1) = 12(1+ 1

3 )= 9;

G2 rB(2)

∏n2l=B(2) (1 + αi(l)) = max {(1 + 3

2 )

(1 + 1) (1 + 2), 2(1 + 1) (1 + 2), 3(1 +2)} = 15, B(2) = 1, ρ(G2) = 1;

G3 rB(3)

∏n3l=B(3)

(1 + αi(l)) = max {2(1 + 12 )

(1 + 14 ) (1 + 1), 4(1 + 1

4 ) (1 + 1) ,

8 (1 + 1) } = 16, B(3) = 3, ρ(G3) =8

(1 + 12 )(1 + 1

4 )= 4.2.

and

G1 :n1∏

l=1

(1 + αi(l)) = 2,

G2 :n2∏

l=1

(1 + αi(l)) = 15,

G3 :n3∏

l=1

(1 + αi(l)) = 3.75.

Since ρ(G2) = 1 < ρ(G3) = 4.2 < ρ(G1) = 4.5,

rB(2)

∏n2l=B(2) (1 + αi(l)) < rB(3)

∏n3l=B(3)

(1 + αi(l)) <

rB(1)

∏n1l=B(1)(1 + αi(l)), s2 < s3 < s1, s1

∏n1l=1(1 + αi(l)) =

6 > s3∏n3

l=1 (1 + αi(l)) = 7.5 > s2∏n2

l=1 (1 + αi(l)) = 15,

hence, from Theorems 1 and 2, the optimal groupsequence is [G2, G3, G1]. Therefore, the optimalschedule is [J5, J3, J4, J8, J6, J7, J2, J1], the optimalvalue of the makespan is 246.

4 Conclusions

In this paper, we have considered the single-machinescheduling problems with deterioration jobs and grouptechnology assumption. By deteriorating jobs andgroup technology assumption, we mean that the jobprocessing times is described by a simple linear functionof starting time. The group setup times are assumed

to be known and fixed. We showed that the makespanminimization problem with ready times can be solvedin polynomial time under an agreeable condition. Inaddition, we proposed a algorithm to solve the problem.In the future research, it is worth to consider moregeneral deterioration model and to investigate multi-machine scheduling problems.

Acknowledgements We are grateful to the editor and ananonymous referee for their helpful comments on an earlierversion of this paper. The work described in this paper waspartially supported by a grant from the Research Grants Councilof the Hong Kong Special Administrative Region, China (PolyU5170/11E). This research was also supported by the National Nat-ural Science Foundation of China (grant no. 11001181), and theScience Research Foundation of Shenyang Aerospace University(grant no. 201011Y).

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