Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2013 Article ID 138240 7 pageshttpdxdoiorg1011552013138240

Research ArticleRescheduling Problems with AgreeableJob Parameters to Minimize the Tardiness Costs underDeterioration and Disruption

Zhang Xingong12 and Wang Yong1

1 College of Economics and Business Administration Chongqing University Chongqing 400030 China2 College of Mathematical Sciences Chongqing Normal University Chongqing 400047 China

Correspondence should be addressed to Wang Yong wangyongcq126com

Received 21 May 2013 Accepted 18 July 2013

Academic Editor Yunqiang Yin

Copyright copy 2013 Z Xingong and W Yong This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

This paper considers single-machine rescheduling problems with agreeable job parameters under deterioration and disruptionDeteriorating jobs mean that the processing time of a job is defined by an increasing function of its starting time Reschedulingmeans that after a set of original jobs has already been scheduled a new set of jobs arrives and creates a disruptionWe consider fourcases of minimization of the total tardiness costs with agreeable job parameters under a limit of the disruptions from the originaljob sequence We propose polynomial-time algorithms or some dynamic programming algorithms under sequence disruption andtime disruption

1 Introduction

Scheduling problems are very important in manufacturingsystems Hence numerous scheduling problems have beenstudied for many years In the classical scheduling theoryjob processing times are assumed to be known and fixedfrom the first job to be processed until the last job to becompleted However there aremany situations in which a jobthat is processed later consumes more time than the same jobwhen processed earlier Scheduling in this setting is known asscheduling deteriorating jobs

Significant contributions towards addressing or solvingdeteriorating job scheduling problems on a single machineinclude among others the following Browne and Yechiali[1] cited applications concerning the control of queues andcommunication systems where jobs deteriorate as they awaitprocessing Kunnathur and Gupta [2] and Mosheiov [3] gaveseveral other real-life situations where deteriorating jobsoccurThese include the search for an object underworseningweather or performance of medical treatments under dete-riorating health conditions Comprehensive discussion of

scheduling problems with time-dependent processing timesof jobs can be found in Cheng et al [4] and Gawiejnowicz[5] Recently Biskup and Herrmann [6] observed that thesum of the processing times of the jobs processed beforea job contributes to the actual processing time of the joband they cite equipment wearout (eg a drill) as a real-lifeexample of their observation Wang and Guo [7] considereda single-machine scheduling problem with the effects oflearning and deterioration The goal is to determine anoptimal combination of the due date and schedule so as tominimize the sum of earliness tardiness and due-date costsNg et al [8] considered a two-machine flow shop schedulingproblemwith linearly deteriorating jobs tominimize the totalcompletion time Cheng et al [9] considered scheduling withdeteriorating jobs in which the actual processing time of a jobis a function of the logarithm of the total processing time ofthe jobs processed before it (to avoid the unrealistic situationwhere the jobs scheduled lately will incur excessively longprocessing times) and the setup times are proportional to theactual processing times of the already scheduled jobs

2 Mathematical Problems in Engineering

Rescheduling involves adjusting a previously plannedpossibly optimal scheduling to account for a disruptionExamples of common disruptions include the arrival ofnew orders order cancelations changes in order priorityprocessing delays changes in release dates machine break-downs and the unavailability of raw materials personnel ortools There are several papers on rescheduling approachesin manufacturing systems Raman et al [10] developed abranch-and-bound procedure to reschedule a flexible man-ufacturing system in the presence of dynamic job arrivalsChurch andUzsoy [11] addressed a similar problem for whichthey describe periodic rescheduling policies and analyzetheir error bounds Jain and Elmaraghy [12] used geneticalgorithms to develop heuristic approaches for reschedulinga flexible manufacturing system Vieira et al [13] providedan extensive review of rescheduling problems Yang [14]studied the single-machine rescheduling with new jobsarrivals and processing time compression Hall and Potts[15] considered the problem of rescheduling of a singlemachine with newly arrived jobs to minimize the maximumlateness and the total completion time under a limit of thedisruption from the original scheduling Yuan and Mu [16]considered the rescheduling problem for jobs on a singlemachine with release dates to minimize makespan under alimit on the maximum sequence disruption Zhao and Tang[17] presented two single-machine rescheduling problemswith linear deteriorating jobs under disruption deterioratingjobs mean that the actual processing time of the job is anincreasing function of its starting time They considered therescheduling problem to minimize the total completion timeunder a limit of the disruption from the original schedulingHoogeveen et al [18] tackled several simple setup timeconfigurations yielding different scheduling problems forwhich they propose optimal polynomial time algorithms orprovide NP-hardness proofs They also present the problemof enumerating the set of strict Pareto optima for the sum ofsetup times and disruption cost criteria

Based on the motivation of Hall and Potts [15] and Zhaoand Tang [17] we consider some rescheduling problems withthe criterion minimizing the total tardiness costs under alimit of the disruption from the original schedule in thispaper The rest of the paper is organized as follows In thenext section we give the problem description In Section 3we consider single-machine scheduling problems The lastsection is the conclusion

2 Problem Definition and Notation

By the terminology of Hall and Potts [15] our researchfulproblem can be stated as follows Let 119869

0= 1198691 119869

1198990 denote

a set of original jobs to be processed non preemptively ona single machine In the presented model we assume thatthese jobs have been scheduled optimally to minimize someclassical objective and that120587lowast is an optimal job sequence withno idle time between the jobs Let 119869

119873= 1198691198990+1

1198691198990+119899119873

denote a set of new jobs that arrive together We assume thatthese jobs arrive at time zero after a schedule for the jobs of1198690has been determined but before processing begins There

is no loss of generality in this assumption if the jobs arriveafter time zero then the fully processed jobs of 119869

0are removed

from the problem any partly processed jobs are processedto completion and 119869

0and 119899

0are updated accordingly Let

119869 = 1198690⋃119869119873and 119899 = 119899

0+ 119899119873 Each job 119869

119895isin 119869 has an integral

normal processing time 119901119895and a deteriorating rate 119887 gt 0 the

actual processing time of job 119869119895is 119901119895(119886 + 119887119904

119895) where 119904

119895(ge0)

is the starting time of job 119869119895and 119886 (gt0) is constant For any

schedule 120590 of the jobs in 119869 we define the following variables

119904119895(120590) is the time at which job 119869

119895isin 119869 starts its

processing in schedule 120590119889119895(120590) is the due date of job 119869

119895isin 119869 in schedule 120590

119862119895(120590) is the time at which job 119869

119895isin 119869 is completed in

schedule 120590119879119895(120590) = max119862

119895(120590) minus 119889

119895 0 is the tardiness value of

job 119869119895isin 119869

119863119895(120587lowast 120590) is the sequence disruptions of job 119869

119895isin 1198690

that is if 119869119895is the 119909th job in 120587

lowast and the 119910th job in 120590respectively then119863

119895(120587lowast 120590) = |119910 minus 119909|

Δ119895(120587lowast 120590) = |119862

119895(120590)minus119862

119895(120587lowast)| is the time disruption of

job 119869119895isin 1198690

When there is no ambiguity we simplify the above sym-bols and write 119904

119895 119889119895119862119895 119879119895119863119895(120587lowast) and Δ

119895(120587lowast) respectively

We consider only the single-scheduling problems withthe following constraints on the amount of disruption where119896 ge 0 is a known integer

119863max(120587lowast) le 119896 max119863

119895(120587lowast)119869119895

isin 1198690 le 119896 the max-

imum sequence disruption of the jobs cannot exceed119896sum119863119895(120587lowast) le 119896 sum

119869119895isin1198690119863119895(120587lowast) le 119896 the total sequence

disruption of the jobs cannot exceed 119896Δmax(120587

lowast) le 119896 maxΔ

119895(120587lowast)119869119895

isin 1198690 le 119896 the max-

imum time disruption of the jobs cannot exceed 119896sumΔ119895(120587lowast) le 119896 sum

119869119895isin1198690Δ119895(120587lowast) le 119896 the total time

disruption of the jobs cannot exceed 119896

Since the 1 | sum119879119895problem is NP-hard in Du and Leung

[19] the 1 | Γ le 119896 | sum119879119895problem is also NP-hard where Γ isin

119863max(120587lowast) sum119863

119895(120587lowast) Δmax(120587

lowast) sumΔ

119895(120587lowast) In this paper we

consider a special case the processing time and due date ofjobs are agreeable that is 119901

119894le 119901119895rArr 119889119894le 119889119895for all jobs 119869

119894

and 119869119895

Using the three-field notation [20] the considered prob-lems can be denoted as

1 | 119901119894le 119901119895rArr 119889119894le 119889119895 119863max(120587

lowast) le 119896 119901

119895(119886 + 119887119904

119895) |

sum119879119895

1 | 119901119894le 119901119895rArr 119889119894le 119889119895 sum119863119895(120587lowast) le 119896 119901

119895(119886 + 119887119904

119895) |

sum119879119895

1 | 119901119894le 119901119895rArr 119889119894le 119889119895 Δmax(120587

lowast) le 119896 119901

119895(119886 + 119887119904

119895) |

sum119879119895

1 | 119901119894le 119901119895rArr 119889119894le 119889119895 sumΔ

119895(120587lowast) le 119896 119901

119895(119886 + 119887119904

119895) |

sum119879119895

Mathematical Problems in Engineering 3

3 Minimum Tardiness Problem withAgreeable Job Parameters

We start with the following result by Kononov and Gaw-iejnowicz [21]

Lemma 1 For the 1 | 119901119895(119886 + 119887119904

119895) | 119862max problem if

120587 = 1198691 119869

119899 and the starting time of the job 119869

1is 119905 then

makespan is sequence independent and

119862max (120587) = 119905

119899

prod

119894=1

(1 + 119887119901119894) +

119886

119887(

119899

prod

119894=1

(1 + 119887119901119894) minus 1) (1)

For notational convenience we assume that the jobs areindexed by agreeable order that is 119901

1le sdot sdot sdot le 119901

1198990and 119889

1le

sdot sdot sdot le 1198891198990 Thus 120587lowast = 119869

1 119869

1198990 with no idle time between

jobsWe now show that the EDD or SPT rule applies to servalof the rescheduling problems we consider

Lemma 2 Problems 1 | 119901119894le 119901119895rArr 119889119894le 119889119895 119863max(120587

lowast) le

119896 119901119895(119886 + 119887119904

119895) | sum119879

119895 and 1 | 119901

119894le 119901119895rArr 119889119894le 119889119895 Δmax(120587

lowast) le

119896 119901119895(119886+119887119904

119895) | sum119879

119895have an optimal schedule with no idle time

between jobs and

(a) a schedule for problem 1 | 119901119894

le 119901119895

rArr 119889119894

le 119889119895

119863max(120587lowast) le 119896 119901

119895(119886 + 119887119904

119895) | sum119879

119895is feasible if the

number of jobs of 119869119873scheduled before the last job of

1198690is less than or equal to 119896

(b) a schedule for problem 1 | 119901119894

le 119901119895

rArr 119889119894

le 119889119895

Δmax(120587lowast) le 119896 119901

119895(119886 + 119887119904

119895) | sum119879

119895is feasible if the total

actual processing time of jobs of 119869119873scheduled before the

last job of 1198690is less than or equal to 119896

Proof The proof is similar to that of Lemma 1 in Hall andPotts [15]

Lemma 3 For problems 1 | 119901119894le 119901119895

rArr 119889119894le 119889119895 Γ le 119896

and 119901119895(119886 + 119887119904

119895) | sum119879

119895 where Γ isin 119863max(120587

lowast) sum119863

119895(120587lowast)

Δmax(120587lowast) sumΔ

119895(120587lowast) there exists an optimal schedule in

which the jobs of 1198690are sequenced in the EDD or SPT rule as in

120587lowast the jobs of 119869

119873are sequenced in the EDD or SPT rule and

there is no idle time between jobs

Proof We first analyze the jobs of 1198690 Consider an optimal

schedule 120590lowast in which the jobs of 119869

0are not sequenced in

the EDD or SPT rule as in 120587lowast Let 119869

119894be the job with the

smallest index that appears later relative to the other jobs of1198690in 120590lowast than in 120587

lowast and let 119869119895(119895 gt 119894) be the last job of 119869

0

that precedes job 119869119894in 120590lowast Because 120587lowast is an optimal sequence

119901119895and 119889

119895are agreeable Assume that the starting time of job

119869119895in 120590lowast is 1199050 then 119862

119895(120590lowast) = 119905

0+ 119901119895(119886 + 119887119905

0) Perform an

interchange on jobs 119869119895and 119869119894 and get a new schedule 1205901015840 In 120590

1015840the starting time of job 119869

119894is 1199050 then119862

119894(1205901015840) = 1199050+119901119894(119886+119887119905

0) lt

1199050+ 119901119895(119886 + 119887119905

0) = 119862

119895(120590lowast) From Lemma 1 119862

119895(1205901015840) = 119862

119894(120590lowast)

Thus the jobs between job 119869119894and 119869119895are completed earlier in

1205901015840 than in 120590

lowast Next we consider the total tardiness of jobs 119869119894

and 119869119895in 1205901015840 and in 120590

lowast

The total tardiness of jobs 119869119894and 119869119895in 120590lowast is as follows

119879119894(120590lowast) + 119879119895(120590lowast) = max 119862

119894(120590lowast) minus 119889119894 0

+max 119862119895(120590lowast) minus 119889119895 0

(2)

The total tardiness of jobs 119869119894and 119869119895in 1205901015840 is as follows

119879119894(1205901015840) + 119879119895(1205901015840) = max 119862

119894(1205901015840) minus 119889119894 0

+max 119862119895(1205901015840) minus 119889119895 0

(3)

To compare the total tardiness of jobs 119869119894and 119869

119895in 120590lowast

and in 1205901015840 we divide it into two cases In the first case when

119862119895(120590lowast) le 119889119895 we have 119879

119894(120590lowast) + 119879119895(120590lowast) = max119862

119894(120590lowast) minus 119889119894 0

Suppose that neither119879119894(1205901015840) nor119879

119895(1205901015840) is zero Note that this is

themost restrictive case since it comprises the case that eitherone or both 119879

119894(1205901015840) and 119879

119895(1205901015840) are zero From Lemma 1 and

119901119894le 119901119895 119889119894le 119889119895 we have 119879

119894(120590lowast)+119879119895(120590lowast)minus119879119894(1205901015840)+119879119895(1205901015840) =

119862119894(120590lowast) minus119862119894(1205901015840) minus119862119895(1205901015840) +119889119895= 119889119895minus119862119894(1205901015840) ge 119889119895minus119862119895(120590lowast) ge 0

In the second case when 119862119895(120590lowast) gt 119889119895 we have 119879

119894(120590lowast) +

119879119895(120590lowast) = 119862119894(120590lowast)+119862119895(120590lowast)minus119889119894minus119889119894 Suppose that neither119879

119894(1205901015840)

nor 119879119895(1205901015840) is zero From Lemma 1 and 119901

119894le 119901119895 119889119894le 119889119895 we

have

119879119894(120590lowast) + 119879119895(120590lowast) minus 119879

119894(1205901015840) + 119879119895(1205901015840)

= 119862119894(120590lowast) + 119862119895(120590lowast) minus 119862119895(1205901015840) minus 119862119894(1205901015840) ge 0

(4)

Now we have proved that the total tardiness of 1205901015840 is less thanor equal to that of 120590lowast

Let the position of job 119869119894in 120587lowast be 119896

1and let the position

of job 119869119895in 120587lowast be 119896

2and in 120590

1015840 be 1198963 If 1198963

ge 1198962 then

119863119895(120587lowast 1205901015840) = 119896

3minus 1198962 119863119894(120587lowast 120590lowast) = 119896

3minus 1198961 Since 119894 lt 119895

implies 1198961

lt 1198962 we have 119863

119895(120587lowast 1205901015840) lt 119863

119894(120587lowast 120590lowast) If 119896

3lt

1198962 then 119863

119895(120587lowast 1205901015840) = 119896

2minus 1198963and 119863

119895(120587lowast 120590lowast) = 119896

2minus

(1198963minus ℎ) where ℎ is the difference between the position of

job 119869119894and 119869119895in 120590lowast So 119863

119895(120587lowast 1205901015840) lt 119863

119895(120587lowast 120590lowast) Hence we

have 119863max(120587lowast 1205901015840) lt 119863max(120587

lowast 120590lowast) In either case because

119863119894(120587lowast 1205901015840) = 119863

119894(120587lowast 120590lowast) minus ℎ and119863

119895(120587lowast 1205901015840) le 119863

119895(120587lowast 120590lowast) + ℎ

Hence sum119863119895(120587lowast 1205901015840) le sum119863

119895(120587lowast 120590lowast)

Moreover if 119862119895(1205901015840) ge 119862119895(120587lowast) then Δ

119895(120587lowast 1205901015840) = 119862119895(1205901015840) minus

119862119895(120587lowast) SinceΔ

119894(120587lowast 120590lowast) = 119862119894(120590lowast)minus119862119894(120587lowast) = 119862119895(1205901015840)minus119862119894(120587lowast)

and 119862119894(120587lowast) lt 119862

119895(120587lowast) therefore Δ

119895(120587lowast 1205901015840) lt Δ

119894(120587lowast 120590lowast) If

119862119895(1205901015840) lt 119862119895(120587lowast) then Δ

119895(120587lowast 1205901015840) = 119862119895(120587lowast) minus 119862119895(1205901015840) because

Δ119895(120587lowast 120590lowast) = 119862

119895(120587lowast) minus 119862

119895(120590lowast) Δ119895(120587lowast 1205901015840) lt Δ

119895(120587lowast 120590lowast)

Thus we have Δmax(120587lowast 1205901015840) lt Δmax(120587

lowast 120590lowast) In either case

because Δ119894(120587lowast 1205901015840) = Δ

119894(120587lowast 120590lowast) minus ℎ

1015840 and Δ119895(120587lowast 1205901015840) le

Δ119895(120587lowast 120590lowast) + ℎ1015840 where ℎ1015840 = 119862

119894(120590lowast) minus 119862119894(1205901015840) Then we deduce

that sumΔ119895(120587lowast 1205901015840) le sumΔ

119895(120587lowast 120590lowast) Thus for either problem

1205901015840 is feasible and optimal We can show that there exists an

optimal schedule in which the jobs of 1198690are sequenced in the

EDD or SPT order as in 120587lowast by finite numbers of repetitions

of the argument A similar interchange argument establishes

4 Mathematical Problems in Engineering

that the jobs of 119869119873can also be obtained by sequencing in the

EDD or SPT order The same EDD or SPT ordering of thejobs of 119869

0in 120587lowast and an optimal schedule show that there is

no idle time in this optimal schedule Otherwise removingthis idle time maintains feasibility and decreases the totaltardiness

We refer to the (EDD EDD) property when a schedule isconstructed using Lemmas 2 and 3We first consider problem1 | 119901119894le 119901119895rArr 119889119894le 119889119895 119863max(120587

lowast) le 119896 119901

119895(119886 + 119887119904

119895) | sum119879

119895

From Lemmas 2 and 3 there are at most 119896 jobs of 119869119873that can

be sequenced before the last job of 1198690 and these jobs have the

smallest due datesThus we propose the following algorithmunder the maximum sequence disruption constraint (seeBox 1)

Algorithm 4 Consider the following steps

Step 1 Index the job of 119869119873in the EDD order

Step 2 Schedule jobs 1 1198990+ 119896 in the EDD rule in the first

1198990+119896 position and schedule jobs 119899

0+119896+1 119899

0+119899119873in the

EDD order in the final 119899119873

minus 119896 positions

Theorem 5 For the 1 | 119901119894le 119901119895

rArr 119889119894le 119889119895 119863max(120587

lowast) le

119896 119901119895(119886 + 119887119904

119895) | sum119879

119895problem Algorithm 4 finds an optimal

schedule in 119874(119899 + 119899119873log 119899119873) time

Proof From Lemmas 2 and 3 the constraint 119863max(120587lowast) le 119896

allows at most 119896 jobs of 119869119873to be sequenced before the final

of 1198690 and these are the jobs of 119869

119873with the smallest due

dates Classical schedule theory shows that the jobs of thisfirst group are sequenced in the EDD order while Lemma 3establishes that the remaining 119899

119873minus 119896 jobs of 119869

119873are also

sequenced in the EDD orderNext we note that the Step 1 for the jobs of 119869

119873requires

119874(119899119873log 119899119873) time Step 2 is executed in119874(119899) time bymerging

the first 119896 jobs of the EDD ordered jobs of 119869119873with the jobs of

1198690as sequenced in 120587

lowast and then placing the last 119899119873minus 119896 jobs of

the EDD order ordered jobs of 119869119873at the end of the schedule

Next we consider problem 1 | 119901119894

le 119901119895

rArr 119889119894

le

119889119895 sum119863

119895(120587lowast) le 119896 119901

119895(119886 + 119887119904

119895) | sum119879

119895 From Lemmas 1 2 and

3 there is the total sequence disruption of the jobs of 119869119873which

is less than or equal to 119896 and can be sequenced before the lastjob of 119869

0 and these jobs have the smallest due dates Thus

we propose the following algorithm under the total sequencedisruption constraint (see Box 2)

Let119891(119894 119895 120575) beminimum total tardiness value of a partialschedule for jobs 119869

1 119869

119894and 1198691198990+1

1198691198990+119895

where the totalsequence disruption is equal to 120575The dynamic programmingprocedure can now be stated as follows

Algorithm 6 Consider the following steps

Step 1 (Initialization) Set 119891(119894 119895 120575) = 0 for (119894 119895 120575) = (0 0 0)

and 119891(119894 119895 120575) = infin for (119894 119895 120575) = (0 0 0) 119894 = 1 1198990 119895 =

1 119899119873 119894 = 0 119896

Step 2 (Recurrence Relation)

119891 (119894 119895 120575) = min119891 (119894 minus 1 119895 120575 minus 119895) +max 119862

119894minus 119889119894 0

119891 (119894 119895 minus 1 120575) +max 1198621198990+119895

minus 1198891198990+119895

0

(5)

where 119862119895denotes the completion time of job 119869

119895

Step 3 (Optimal Solution) Calculate the optimal solutionvalue min

0le120575le119896119891(1198990 119899119873 120575)

In the recurrence relation the first term in the minimiza-tion corresponds to the case where the partial schedule endswith job 119869

119894isin 1198690 Because 119869

119895jobs of 119869

119873appear before job

119869119894in such a partial schedule the increase in total sequence

disruption is equal to 119895 The second term corresponds to thecase where the partial schedule ends with job 119869

1198990+119895isin 119869119873

In addition we demonstrate the result of Algorithm 6 inthe following example

Example 7 1198990

= 3 119899119873

= 3 1199050

= 0 1198690

= 1198691 1198692 1198693 119869119873

=

1198694 1198695 1198696 1199011= 32 119889

1= 4 119901

2= 1 119889

2= 24 119901

3= 2 119889

3=

3 1199014= 15 119889

4= 3 119901

5= 4 119889

5= 5 119901

6= 3 119889

6= 47 119886 = 02

119887 = 04 and 119896 = 5Solution According to Algorithm 4 and Lemmas 2 and

3 Because the total sequence disruption of the jobs 1198691 1198692 1198693

can not exceed 119896 = 5 By dynamic programming algorithmwe obtain job sequence and the total tardiness cost as follows

if 119896 = 0 the optimal sequence is [1198692

rarr 1198693

rarr

1198691

rarr 1198694

rarr 1198696

rarr 1198695] and the total tardiness cost is

268007

if 119896 = 0 the job sequence is [1198692

rarr 1198693

rarr 1198691

rarr

1198694

rarr 1198696

rarr 1198695] and the total tardiness cost is

268007

if 119896 = 1 the job sequence is [1198692

rarr 1198693

rarr 1198694

rarr

1198691

rarr 1198696

rarr 1198695] and the total tardiness cost is

258007

if 119896 = 2 the job sequence is [1198692

rarr 1198694

rarr 1198693

rarr

1198691

rarr 1198696

rarr 1198695] and the total tardiness cost is

258007

if 119896 = 3 the job sequence is [1198694

rarr 1198692

rarr 1198693

rarr

1198691

rarr 1198696

rarr 1198695] and the total tardiness cost is

258007

if 119896 = 4 the job sequence is [1198694

rarr 1198692

rarr 1198693

rarr

1198696

rarr 1198691

rarr 1198695] and the total tardiness cost is

268223

if 119896 = 5 the job sequence is [1198694

rarr 1198692

rarr 1198696

rarr

1198693

rarr 1198691

rarr 1198695] and the total tardiness cost is

285223

Furthermore if total sequence disruption 119896 is equal to1 2 3 then minimizing total tardiness cost is 258007 andoptimal job sequence is [119869

2rarr 1198693rarr 1198694rarr 1198691rarr 1198696rarr 1198695]

[1198692

rarr 1198694

rarr 1198693

rarr 1198691

rarr 1198696

rarr 1198695] and [119869

4rarr 1198692

rarr

1198693rarr 1198691rarr 1198696rarr 1198695]

Mathematical Problems in Engineering 5

Input Given 119901119895 119889119895for 119895 = 1 119899 119896 and 120587

lowast where 119896 le 119899119873

Indexing Index the jobs of 119869119873in the EDD order

Schedule Construction Schedule jobs 1 1198990+ 119896 in the EDD order in the first 119899

0+ 119896 positions

Schedule jobs 1198990+ 119896 + 1 119899

0+ 119899119873in EDD order in the final 119899

119873minus 119896 positions

Box 1

Input Given 119901119895 119889119895for 119895 = 1 119899 119896 and 120587

lowast where 119896 le 1198990119899119873

Indexing Index the jobs of 119869119873in the EDD order

Box 2

Theorem 8 For the 1 | 119901119894le 119901119895rArr 119889119894le 119889119895 sum119863119895(120587lowast) le

119896 119901119895(119886 + 119887119904

119895) | sum119879

119895problem Algorithm 6 finds an optimal

schedule in 119874(1198992

0n2119873) time

Proof FromLemmas 1 2 and 3 it only remains to enumerateall possible ways of merging the EDD ordered lists of jobsof 1198690and 119869119873 Algorithm 6 does so by comparing the cost of

all possible state transitions and therefore finds an optimalschedule

Because 119894 le 1198990 119895le119899119873

and 120575 le 119896 le 1198990119899119873 there are119874(119899

2

01198992

119873)

values of the state variables Step 1 requires119874(119899119873log 119899119873) Step

2 requires constant time for each set of values of the statevariables Thus the overall time complexity of Algorithm 6is 119874(119899

2

01198992

119873)

Now we consider problem 1 | 119901119894

le 119901119895

rArr 119889119894

le

119889119895 Δmax(120587

lowast) le 119896 119901

119895(119886 + 119887119904

119895) | sum119879

119895 From Lemmas 1 2

and 3 the maximum time disruption of jobs of 1198690is at most

119896 and jobs of 119869119873before the last job of 119869

0have the smallest

due dates Thus we propose the following algorithm underthe maximum time disruption constraint (see Box 3)

Let 119891(119894 119895 120575) be minimum total tardiness value of apartial schedule for jobs 119869

1 119869

119894and 1198691198990+1

1198691198990+119895

wherethe maximum time disruption is equal to 120575 The dynamicprogramming procedure can now be stated as follows

Algorithm 9 Consider the following steps

Step 1 (Initialization) Set 119891(119894 119895 120575) = 0 for (119894 119895 120575) = (0 0 0)

and 119891(119894 119895 120575) = infin for (119894 119895 120575) = (0 0 0) 119894 = 1 1198990 119895 =

1 119899119873 and 119894 = 0 119896

Step 2 (Recurrence Relation)

119891 (119894 119895 120575) = min119891 (119894 minus 1 119895 120575 minus 119875

ℎ) +max 119862

119894minus 119889119894 0

119891 (119894 119895 minus 1 120575) +max 1198621198990+119895

minus 1198891198990+119895

0

(6)

where 119875ℎis the sum of actual processing time of the new jobs

of 119869119873between 119869

119894minus1and 119869119894and119862

119895denotes the completion time

of job 119869119895

Step 3 (Optimal Solution) Calculate the optimal solutionvalue min

0le120575le119896119891(1198990 119899119873 120575)

In the recurrence relation the first term in the minimiza-tion corresponds to the case where the partial schedule endswith job 119869

119894isin 1198690 Because 119869

119895jobs of 119869

119873appear before job 119869

119894

in such a partial schedule the increase in the maximum timedisruption is equal to 119875

ℎThe second term corresponds to the

case where the partial schedule ends with job 1198691198990+119895

isin 119869119873

Similar to Example 7 by Algorithm 9 we have(i) the maximum time disruption 119896 is in [17237

257919] the job sequence and the total tardiness costare the same to Example

(ii) the maximum time disruption 119896 is in [0 17237) thejob sequence is [119869

2rarr 1198693rarr 1198691rarr 1198694rarr 1198696rarr 1198695]

and the total tardiness cost is 268007

Theorem 10 For the 1 | 119901119894le 119901119895rArr 119889119894le 119889119895 Δmax(120587

lowast) le

119896 119901119895(119886 + 119887119904

119895) | sum119879

119895problem Algorithm 9 finds an optimal

schedule in 119874(1198990119899119873119862max + 119899

119873log 119899119873) time

Proof From Lemmas 1 2 and 3 Δ le 119896 means that the totalactual processing time of the new jobs of 119869

119873before the last job

of 1198690is at most 119896 and these are the jobs of 119869

119873with the smallest

due dates Hence Algorithm 9 schedules the jobs accordingto the (EDD EDD) property

Because 119894 le 1198990 119895 le 119899

119873and 120575 le 119896 lt 119862max there

are 119874(1198990119899119873119862max) values of the state variables Step 1 requires

119874(119899119873log 119899119873) Step 2 requires constant time for each set of

values of the state variablesThus the overall time complexityof Algorithm 9 is 119874(119899

0119899119873119862max + 119899

119873log 119899119873)

Now we consider problem 1 | 119901119894

le 119901119895

rArr 119889119894

le

119889119895 sumΔ

119895(120587lowast) le 119896 119901

119895(119886 + 119887119904

119895) | sum119879

119895 From Lemmas 1 2

and 3 there is the total time disruption of jobs of 119869119873which

is less than or equal to 119896 and can be sequenced before thelast job of 119869

0 and these jobs have the smallest due dates

The following dynamic programming algorithm performs anoptimal merging of jobs of 119869

0and 119869

119873in a way similar to

Algorithm 6 (see Box 4)Let119891(119894 119895 120575) beminimum total tardiness value of a partial

schedule for jobs 1198691 119869

119894and 1198691198990+1

1198691198990+119895

where the total

6 Mathematical Problems in Engineering

Input Given 119901119895 119889119895for 119895 = 1 119899 119896 and 120587

lowast where 119896 lt 119862max (see Lemma 1)Indexing Index the jobs of 119869

119873in the EDD order

Box 3

Input Given 119901119895 119889119895for 119895 = 1 119899 119896 and 120587

lowast where 119896 le 1198990119862max

Indexing Index the jobs of 119869119873in the EDD order

Box 4

time disruption is equal to 120575 The dynamic programmingprocedure can now be stated as follows

Algorithm 11

Step 1 (Initialization) Set 119891(119894 119895 120575) = 0 for (119894 119895 120575) = (0 0 0)

and 119891(119894 119895 120575) = infin for (119894 119895 120575) = (0 0 0) 119894 = 1 1198990 119895 =

1 119899119873 and 119894 = 0 119896

Step 2 (Recurrence Relation)

119891 (119894 119895 120575)= min

119891(119894 minus 1 119895 120575 minus

1198990+119895

sum

ℎ=1198990+1

119901[ℎ]

) +max 119862119894minus 119889119894 0

119891 (119894 119895 minus 1 120575) +max 1198621198990+119895

minus 1198891198990+119895

0

(7)

where 119901[ℎ]

is the actual processing time of job 119869ℎ 119862119895denotes

the completion time of job 119869119895

Step 3 (Optimal Solution) Calculate the optimal solutionvalue min

0le120575le119896119891(1198990 119899119873 120575)

In the recurrence relation the first term in the minimiza-tion corresponds to the case where the partial schedule endswith job 119869

119894isin 1198690 Because 119869

119895jobs of 119869

119873appear before job 119869

119894in

such a partial schedule the increase in total time disruptionis equal to sum

1198990+119895

ℎ=1198990+1119901[ℎ] The second term corresponds to the

case where the partial schedule ends with job 1198691198990+119895

isin 119869119873

Theorem 12 For the 1 | 119901119894le 119901119895rArr 119889119894le 119889119895 sumΔ

119895(120587lowast) le

119896 119901119895(119886 + 119887119904

119895) | sum119879

119895problem Algorithm 11 finds an optimal

schedule in 119874(1198992

0119899119873119862max) time

Proof The proof of optimality of Algorithm 11 is similar toTheorem 8 regarding the time complexity Because 119894 le 119899

0

119895 le 119899119873 and 120575 le 119896 le 119899

0119862max there are 119874(119899

2

0119899119873119862max) values

of the state variables Step 1 requires 119874(119899119873log 119899119873) Similar

arguments to those in the proof of Theorem 8 show that theoverall time complexity of Algorithm 11 is119874(119899

2

0119899119873119862max)

4 Conclusions

In this paper we studied the issue of rescheduling to allowthe unexpected arrival of new jobs and took into account

the effect of the disruption on a previously planned opti-mal schedule The main contribution of this paper is thatwe develop the machine rescheduling scheduling problemsagreeable job parameters under deterioration and disruptionRescheduling means to schedule the jobs again togetherwith a set of new jobs Deteriorating job means that theactual processing time of a job is an increasing functionof its starting time When the processing time and duedate of jobs are agreeable we considered some problems tominimize total tardiness under a limit on two disruptionconstraints sequence disruption and time disruption Weproposed polynomial time algorithms or some dynamicprogramming algorithms for each problem Future researchmay stimulate rescheduling models to mitigate the effectsof the disruptions that occur frequently in manufacturingpractice

Acknowledgments

The authors would like to thank the authors of the referencesfor enlightening themThis paper was supported by NationalNatural Science Foundation of China (71272085) TianyuanFund for Mathematics (11226237) the Humanities and SocialSciences Program of the Ministry of Education of China(12YJA630135) the Foundation of Chongqing EducationCommission (KJ120624) and the Key Project Fundation ofChongqing Normal University (2011XLZ05)

References

[1] S Browne and U Yechiali ldquoScheduling deteriorating jobs on asingle processorrdquo Operations Research vol 38 no 3 pp 495ndash498 1990

[2] A S Kunnathur and S K Gupta ldquoMinimizing the makespanwith late start penalties added to processing times in a singlefacility scheduling problemrdquo European Journal of OperationalResearch vol 47 no 1 pp 56ndash64 1990

[3] G Mosheiov ldquoScheduling jobs under simple linear deteriora-tionrdquoComputers andOperations Research vol 21 no 6 pp 653ndash659 1994

[4] T C E Cheng Q Ding and B M T Lin ldquoA concise surveyof schedulingwith time-dependent processing timesrdquoEuropeanJournal of Operational Research vol 152 no 1 pp 1ndash13 2004

[5] S Gawiejnowicz Time-Dependent Scheduling Springer BerlinGermany 2008

Mathematical Problems in Engineering 7

[6] D Biskup and JHerrmann ldquoSingle-machine scheduling againstdue dates with past-sequence-dependent setup timesrdquo Euro-pean Journal of Operational Research vol 191 no 2 pp 587ndash5922008

[7] J-B Wang and Q Guo ldquoA due-date assignment problem withlearning effect and deteriorating jobsrdquo Applied MathematicalModelling vol 34 no 2 pp 309ndash313 2010

[8] C T Ng J-B Wang T C E Cheng and L L Liu ldquoAbranch-and-bound algorithm for solving a two-machine flowshop problemwith deteriorating jobsrdquoComputers ampOperationsResearch vol 37 no 1 pp 83ndash90 2010

[9] T C E Cheng W-C Lee and C-C Wu ldquoSingle-machinescheduling with deteriorating jobs and past-sequence-dependent setup timesrdquo Applied Mathematical Modelling vol35 no 4 pp 1861ndash1867 2011

[10] N Raman F B Talbot and R V Rachamadugu ldquoDue datebased scheduling in a general flexible manufacturing systemrdquoJournal of Operations Management vol 8 no 2 pp 115ndash1321989

[11] L K Church and R Uzsoy ldquoAnalysis of periodic and event-driven rescheduling policies in dynamic shopsrdquo InternationalJournal of Computer Integrated Manufacturing vol 5 pp 153ndash163 1992

[12] A K Jain and H A Elmaraghy ldquoProduction schedul-ingrescheduling in flexible manufacturingrdquo International Jour-nal of Production Research vol 35 no 1 pp 281ndash309 1997

[13] G E Vieira J W Herrmann and E Lin ldquoReschedulingmanufacturing systems a framework of strategies policies andmethodsrdquo Journal of Scheduling vol 6 no 1 pp 39ndash62 2003

[14] B B Yang ldquoSingle machine rescheduling with new jobs arrivalsand processing time compressionrdquo International Journal ofAdvanced Manufacturing Technology vol 34 no 3-4 pp 378ndash384 2007

[15] N G Hall and C N Potts ldquoRescheduling for new ordersrdquoOperations Research vol 52 no 3 pp 440ndash453 2004

[16] J J Yuan and Y D Mu ldquoRescheduling with release dates tominimize makespan under a limit on the maximum sequencedisruptionrdquo European Journal of Operational Research vol 182no 2 pp 936ndash944 2007

[17] C L Zhao and H Y Tang ldquoRescheduling problems withdeteriorating jobs under disruptionsrdquo Applied MathematicalModelling vol 34 no 1 pp 238ndash243 2010

[18] H Hoogeveen C Lente and V Trsquokindt ldquoRescheduling for neworders on a single machine with setup timesrdquo European Journalof Operational Research vol 223 no 1 pp 40ndash46 2012

[19] J Du and J Y-T Leung ldquoMinimizing total tardiness on onemachine is NP-hardrdquo Mathematics of Operations Research vol15 no 3 pp 483ndash495 1990

[20] R L Graham E L Lawler J K Lenstra and A H GRinnooy Kan ldquoOptimization and approximation in determin-istic sequencing and scheduling a surveyrdquo Annals of DiscreteMathematics vol 5 pp 287ndash326 1979

[21] A Kononov and S Gawiejnowicz ldquoNP-hard cases in schedulingdeteriorating jobs on dedicated machinesrdquo Journal of the Oper-ational Research Society vol 52 no 6 pp 708ndash717 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Mathematical Problems in Engineering

Rescheduling involves adjusting a previously plannedpossibly optimal scheduling to account for a disruptionExamples of common disruptions include the arrival ofnew orders order cancelations changes in order priorityprocessing delays changes in release dates machine break-downs and the unavailability of raw materials personnel ortools There are several papers on rescheduling approachesin manufacturing systems Raman et al [10] developed abranch-and-bound procedure to reschedule a flexible man-ufacturing system in the presence of dynamic job arrivalsChurch andUzsoy [11] addressed a similar problem for whichthey describe periodic rescheduling policies and analyzetheir error bounds Jain and Elmaraghy [12] used geneticalgorithms to develop heuristic approaches for reschedulinga flexible manufacturing system Vieira et al [13] providedan extensive review of rescheduling problems Yang [14]studied the single-machine rescheduling with new jobsarrivals and processing time compression Hall and Potts[15] considered the problem of rescheduling of a singlemachine with newly arrived jobs to minimize the maximumlateness and the total completion time under a limit of thedisruption from the original scheduling Yuan and Mu [16]considered the rescheduling problem for jobs on a singlemachine with release dates to minimize makespan under alimit on the maximum sequence disruption Zhao and Tang[17] presented two single-machine rescheduling problemswith linear deteriorating jobs under disruption deterioratingjobs mean that the actual processing time of the job is anincreasing function of its starting time They considered therescheduling problem to minimize the total completion timeunder a limit of the disruption from the original schedulingHoogeveen et al [18] tackled several simple setup timeconfigurations yielding different scheduling problems forwhich they propose optimal polynomial time algorithms orprovide NP-hardness proofs They also present the problemof enumerating the set of strict Pareto optima for the sum ofsetup times and disruption cost criteria

Based on the motivation of Hall and Potts [15] and Zhaoand Tang [17] we consider some rescheduling problems withthe criterion minimizing the total tardiness costs under alimit of the disruption from the original schedule in thispaper The rest of the paper is organized as follows In thenext section we give the problem description In Section 3we consider single-machine scheduling problems The lastsection is the conclusion

2 Problem Definition and Notation

By the terminology of Hall and Potts [15] our researchfulproblem can be stated as follows Let 119869

0= 1198691 119869

1198990 denote

a set of original jobs to be processed non preemptively ona single machine In the presented model we assume thatthese jobs have been scheduled optimally to minimize someclassical objective and that120587lowast is an optimal job sequence withno idle time between the jobs Let 119869

119873= 1198691198990+1

1198691198990+119899119873

denote a set of new jobs that arrive together We assume thatthese jobs arrive at time zero after a schedule for the jobs of1198690has been determined but before processing begins There

is no loss of generality in this assumption if the jobs arriveafter time zero then the fully processed jobs of 119869

0are removed

from the problem any partly processed jobs are processedto completion and 119869

0and 119899

0are updated accordingly Let

119869 = 1198690⋃119869119873and 119899 = 119899

0+ 119899119873 Each job 119869

119895isin 119869 has an integral

normal processing time 119901119895and a deteriorating rate 119887 gt 0 the

actual processing time of job 119869119895is 119901119895(119886 + 119887119904

119895) where 119904

119895(ge0)

is the starting time of job 119869119895and 119886 (gt0) is constant For any

schedule 120590 of the jobs in 119869 we define the following variables

119904119895(120590) is the time at which job 119869

119895isin 119869 starts its

processing in schedule 120590119889119895(120590) is the due date of job 119869

119895isin 119869 in schedule 120590

119862119895(120590) is the time at which job 119869

119895isin 119869 is completed in

schedule 120590119879119895(120590) = max119862

119895(120590) minus 119889

119895 0 is the tardiness value of

job 119869119895isin 119869

119863119895(120587lowast 120590) is the sequence disruptions of job 119869

119895isin 1198690

that is if 119869119895is the 119909th job in 120587

lowast and the 119910th job in 120590respectively then119863

119895(120587lowast 120590) = |119910 minus 119909|

Δ119895(120587lowast 120590) = |119862

119895(120590)minus119862

119895(120587lowast)| is the time disruption of

job 119869119895isin 1198690

When there is no ambiguity we simplify the above sym-bols and write 119904

119895 119889119895119862119895 119879119895119863119895(120587lowast) and Δ

119895(120587lowast) respectively

We consider only the single-scheduling problems withthe following constraints on the amount of disruption where119896 ge 0 is a known integer

119863max(120587lowast) le 119896 max119863

119895(120587lowast)119869119895

isin 1198690 le 119896 the max-

imum sequence disruption of the jobs cannot exceed119896sum119863119895(120587lowast) le 119896 sum

119869119895isin1198690119863119895(120587lowast) le 119896 the total sequence

disruption of the jobs cannot exceed 119896Δmax(120587

lowast) le 119896 maxΔ

119895(120587lowast)119869119895

isin 1198690 le 119896 the max-

imum time disruption of the jobs cannot exceed 119896sumΔ119895(120587lowast) le 119896 sum

119869119895isin1198690Δ119895(120587lowast) le 119896 the total time

disruption of the jobs cannot exceed 119896

Since the 1 | sum119879119895problem is NP-hard in Du and Leung

[19] the 1 | Γ le 119896 | sum119879119895problem is also NP-hard where Γ isin

119863max(120587lowast) sum119863

119895(120587lowast) Δmax(120587

lowast) sumΔ

119895(120587lowast) In this paper we

consider a special case the processing time and due date ofjobs are agreeable that is 119901

119894le 119901119895rArr 119889119894le 119889119895for all jobs 119869

119894

and 119869119895

Using the three-field notation [20] the considered prob-lems can be denoted as

1 | 119901119894le 119901119895rArr 119889119894le 119889119895 119863max(120587

lowast) le 119896 119901

119895(119886 + 119887119904

119895) |

sum119879119895

1 | 119901119894le 119901119895rArr 119889119894le 119889119895 sum119863119895(120587lowast) le 119896 119901

119895(119886 + 119887119904

119895) |

sum119879119895

1 | 119901119894le 119901119895rArr 119889119894le 119889119895 Δmax(120587

lowast) le 119896 119901

119895(119886 + 119887119904

119895) |

sum119879119895

1 | 119901119894le 119901119895rArr 119889119894le 119889119895 sumΔ

119895(120587lowast) le 119896 119901

119895(119886 + 119887119904

119895) |

sum119879119895

Mathematical Problems in Engineering 3

3 Minimum Tardiness Problem withAgreeable Job Parameters

We start with the following result by Kononov and Gaw-iejnowicz [21]

Lemma 1 For the 1 | 119901119895(119886 + 119887119904

119895) | 119862max problem if

120587 = 1198691 119869

119899 and the starting time of the job 119869

1is 119905 then

makespan is sequence independent and

119862max (120587) = 119905

119899

prod

119894=1

(1 + 119887119901119894) +

119886

119887(

119899

prod

119894=1

(1 + 119887119901119894) minus 1) (1)

For notational convenience we assume that the jobs areindexed by agreeable order that is 119901

1le sdot sdot sdot le 119901

1198990and 119889

1le

sdot sdot sdot le 1198891198990 Thus 120587lowast = 119869

1 119869

1198990 with no idle time between

jobsWe now show that the EDD or SPT rule applies to servalof the rescheduling problems we consider

Lemma 2 Problems 1 | 119901119894le 119901119895rArr 119889119894le 119889119895 119863max(120587

lowast) le

119896 119901119895(119886 + 119887119904

119895) | sum119879

119895 and 1 | 119901

119894le 119901119895rArr 119889119894le 119889119895 Δmax(120587

lowast) le

119896 119901119895(119886+119887119904

119895) | sum119879

119895have an optimal schedule with no idle time

between jobs and

(a) a schedule for problem 1 | 119901119894

le 119901119895

rArr 119889119894

le 119889119895

119863max(120587lowast) le 119896 119901

119895(119886 + 119887119904

119895) | sum119879

119895is feasible if the

number of jobs of 119869119873scheduled before the last job of

1198690is less than or equal to 119896

(b) a schedule for problem 1 | 119901119894

le 119901119895

rArr 119889119894

le 119889119895

Δmax(120587lowast) le 119896 119901

119895(119886 + 119887119904

119895) | sum119879

119895is feasible if the total

actual processing time of jobs of 119869119873scheduled before the

last job of 1198690is less than or equal to 119896

Proof The proof is similar to that of Lemma 1 in Hall andPotts [15]

Lemma 3 For problems 1 | 119901119894le 119901119895

rArr 119889119894le 119889119895 Γ le 119896

and 119901119895(119886 + 119887119904

119895) | sum119879

119895 where Γ isin 119863max(120587

lowast) sum119863

119895(120587lowast)

Δmax(120587lowast) sumΔ

119895(120587lowast) there exists an optimal schedule in

which the jobs of 1198690are sequenced in the EDD or SPT rule as in

120587lowast the jobs of 119869

119873are sequenced in the EDD or SPT rule and

there is no idle time between jobs

Proof We first analyze the jobs of 1198690 Consider an optimal

schedule 120590lowast in which the jobs of 119869

0are not sequenced in

the EDD or SPT rule as in 120587lowast Let 119869

119894be the job with the

smallest index that appears later relative to the other jobs of1198690in 120590lowast than in 120587

lowast and let 119869119895(119895 gt 119894) be the last job of 119869

0

that precedes job 119869119894in 120590lowast Because 120587lowast is an optimal sequence

119901119895and 119889

119895are agreeable Assume that the starting time of job

119869119895in 120590lowast is 1199050 then 119862

119895(120590lowast) = 119905

0+ 119901119895(119886 + 119887119905

0) Perform an

interchange on jobs 119869119895and 119869119894 and get a new schedule 1205901015840 In 120590

1015840the starting time of job 119869

119894is 1199050 then119862

119894(1205901015840) = 1199050+119901119894(119886+119887119905

0) lt

1199050+ 119901119895(119886 + 119887119905

0) = 119862

119895(120590lowast) From Lemma 1 119862

119895(1205901015840) = 119862

119894(120590lowast)

Thus the jobs between job 119869119894and 119869119895are completed earlier in

1205901015840 than in 120590

lowast Next we consider the total tardiness of jobs 119869119894

and 119869119895in 1205901015840 and in 120590

lowast

The total tardiness of jobs 119869119894and 119869119895in 120590lowast is as follows

119879119894(120590lowast) + 119879119895(120590lowast) = max 119862

119894(120590lowast) minus 119889119894 0

+max 119862119895(120590lowast) minus 119889119895 0

(2)

The total tardiness of jobs 119869119894and 119869119895in 1205901015840 is as follows

119879119894(1205901015840) + 119879119895(1205901015840) = max 119862

119894(1205901015840) minus 119889119894 0

+max 119862119895(1205901015840) minus 119889119895 0

(3)

To compare the total tardiness of jobs 119869119894and 119869

119895in 120590lowast

and in 1205901015840 we divide it into two cases In the first case when

119862119895(120590lowast) le 119889119895 we have 119879

119894(120590lowast) + 119879119895(120590lowast) = max119862

119894(120590lowast) minus 119889119894 0

Suppose that neither119879119894(1205901015840) nor119879

119895(1205901015840) is zero Note that this is

themost restrictive case since it comprises the case that eitherone or both 119879

119894(1205901015840) and 119879

119895(1205901015840) are zero From Lemma 1 and

119901119894le 119901119895 119889119894le 119889119895 we have 119879

119894(120590lowast)+119879119895(120590lowast)minus119879119894(1205901015840)+119879119895(1205901015840) =

119862119894(120590lowast) minus119862119894(1205901015840) minus119862119895(1205901015840) +119889119895= 119889119895minus119862119894(1205901015840) ge 119889119895minus119862119895(120590lowast) ge 0

In the second case when 119862119895(120590lowast) gt 119889119895 we have 119879

119894(120590lowast) +

119879119895(120590lowast) = 119862119894(120590lowast)+119862119895(120590lowast)minus119889119894minus119889119894 Suppose that neither119879

119894(1205901015840)

nor 119879119895(1205901015840) is zero From Lemma 1 and 119901

119894le 119901119895 119889119894le 119889119895 we

have

119879119894(120590lowast) + 119879119895(120590lowast) minus 119879

119894(1205901015840) + 119879119895(1205901015840)

= 119862119894(120590lowast) + 119862119895(120590lowast) minus 119862119895(1205901015840) minus 119862119894(1205901015840) ge 0

(4)

Now we have proved that the total tardiness of 1205901015840 is less thanor equal to that of 120590lowast

Let the position of job 119869119894in 120587lowast be 119896

1and let the position

of job 119869119895in 120587lowast be 119896

2and in 120590

1015840 be 1198963 If 1198963

ge 1198962 then

119863119895(120587lowast 1205901015840) = 119896

3minus 1198962 119863119894(120587lowast 120590lowast) = 119896

3minus 1198961 Since 119894 lt 119895

implies 1198961

lt 1198962 we have 119863

119895(120587lowast 1205901015840) lt 119863

119894(120587lowast 120590lowast) If 119896

3lt

1198962 then 119863

119895(120587lowast 1205901015840) = 119896

2minus 1198963and 119863

119895(120587lowast 120590lowast) = 119896

2minus

(1198963minus ℎ) where ℎ is the difference between the position of

job 119869119894and 119869119895in 120590lowast So 119863

119895(120587lowast 1205901015840) lt 119863

119895(120587lowast 120590lowast) Hence we

have 119863max(120587lowast 1205901015840) lt 119863max(120587

lowast 120590lowast) In either case because

119863119894(120587lowast 1205901015840) = 119863

119894(120587lowast 120590lowast) minus ℎ and119863

119895(120587lowast 1205901015840) le 119863

119895(120587lowast 120590lowast) + ℎ

Hence sum119863119895(120587lowast 1205901015840) le sum119863

119895(120587lowast 120590lowast)

Moreover if 119862119895(1205901015840) ge 119862119895(120587lowast) then Δ

119895(120587lowast 1205901015840) = 119862119895(1205901015840) minus

119862119895(120587lowast) SinceΔ

119894(120587lowast 120590lowast) = 119862119894(120590lowast)minus119862119894(120587lowast) = 119862119895(1205901015840)minus119862119894(120587lowast)

and 119862119894(120587lowast) lt 119862

119895(120587lowast) therefore Δ

119895(120587lowast 1205901015840) lt Δ

119894(120587lowast 120590lowast) If

119862119895(1205901015840) lt 119862119895(120587lowast) then Δ

119895(120587lowast 1205901015840) = 119862119895(120587lowast) minus 119862119895(1205901015840) because

Δ119895(120587lowast 120590lowast) = 119862

119895(120587lowast) minus 119862

119895(120590lowast) Δ119895(120587lowast 1205901015840) lt Δ

119895(120587lowast 120590lowast)

Thus we have Δmax(120587lowast 1205901015840) lt Δmax(120587

lowast 120590lowast) In either case

because Δ119894(120587lowast 1205901015840) = Δ

119894(120587lowast 120590lowast) minus ℎ

1015840 and Δ119895(120587lowast 1205901015840) le

Δ119895(120587lowast 120590lowast) + ℎ1015840 where ℎ1015840 = 119862

119894(120590lowast) minus 119862119894(1205901015840) Then we deduce

that sumΔ119895(120587lowast 1205901015840) le sumΔ

119895(120587lowast 120590lowast) Thus for either problem

1205901015840 is feasible and optimal We can show that there exists an

optimal schedule in which the jobs of 1198690are sequenced in the

EDD or SPT order as in 120587lowast by finite numbers of repetitions

of the argument A similar interchange argument establishes

4 Mathematical Problems in Engineering

that the jobs of 119869119873can also be obtained by sequencing in the

EDD or SPT order The same EDD or SPT ordering of thejobs of 119869

0in 120587lowast and an optimal schedule show that there is

no idle time in this optimal schedule Otherwise removingthis idle time maintains feasibility and decreases the totaltardiness

We refer to the (EDD EDD) property when a schedule isconstructed using Lemmas 2 and 3We first consider problem1 | 119901119894le 119901119895rArr 119889119894le 119889119895 119863max(120587

lowast) le 119896 119901

119895(119886 + 119887119904

119895) | sum119879

119895

From Lemmas 2 and 3 there are at most 119896 jobs of 119869119873that can

be sequenced before the last job of 1198690 and these jobs have the

smallest due datesThus we propose the following algorithmunder the maximum sequence disruption constraint (seeBox 1)

Algorithm 4 Consider the following steps

Step 1 Index the job of 119869119873in the EDD order

Step 2 Schedule jobs 1 1198990+ 119896 in the EDD rule in the first

1198990+119896 position and schedule jobs 119899

0+119896+1 119899

0+119899119873in the

EDD order in the final 119899119873

minus 119896 positions

Theorem 5 For the 1 | 119901119894le 119901119895

rArr 119889119894le 119889119895 119863max(120587

lowast) le

119896 119901119895(119886 + 119887119904

119895) | sum119879

119895problem Algorithm 4 finds an optimal

schedule in 119874(119899 + 119899119873log 119899119873) time

Proof From Lemmas 2 and 3 the constraint 119863max(120587lowast) le 119896

allows at most 119896 jobs of 119869119873to be sequenced before the final

of 1198690 and these are the jobs of 119869

119873with the smallest due

dates Classical schedule theory shows that the jobs of thisfirst group are sequenced in the EDD order while Lemma 3establishes that the remaining 119899

119873minus 119896 jobs of 119869

119873are also

sequenced in the EDD orderNext we note that the Step 1 for the jobs of 119869

119873requires

119874(119899119873log 119899119873) time Step 2 is executed in119874(119899) time bymerging

the first 119896 jobs of the EDD ordered jobs of 119869119873with the jobs of

1198690as sequenced in 120587

lowast and then placing the last 119899119873minus 119896 jobs of

the EDD order ordered jobs of 119869119873at the end of the schedule

Next we consider problem 1 | 119901119894

le 119901119895

rArr 119889119894

le

119889119895 sum119863

119895(120587lowast) le 119896 119901

119895(119886 + 119887119904

119895) | sum119879

119895 From Lemmas 1 2 and

3 there is the total sequence disruption of the jobs of 119869119873which

is less than or equal to 119896 and can be sequenced before the lastjob of 119869

0 and these jobs have the smallest due dates Thus

we propose the following algorithm under the total sequencedisruption constraint (see Box 2)

Let119891(119894 119895 120575) beminimum total tardiness value of a partialschedule for jobs 119869

1 119869

119894and 1198691198990+1

1198691198990+119895

where the totalsequence disruption is equal to 120575The dynamic programmingprocedure can now be stated as follows

Algorithm 6 Consider the following steps

Step 1 (Initialization) Set 119891(119894 119895 120575) = 0 for (119894 119895 120575) = (0 0 0)

and 119891(119894 119895 120575) = infin for (119894 119895 120575) = (0 0 0) 119894 = 1 1198990 119895 =

1 119899119873 119894 = 0 119896

Step 2 (Recurrence Relation)

119891 (119894 119895 120575) = min119891 (119894 minus 1 119895 120575 minus 119895) +max 119862

119894minus 119889119894 0

119891 (119894 119895 minus 1 120575) +max 1198621198990+119895

minus 1198891198990+119895

0

(5)

where 119862119895denotes the completion time of job 119869

119895

Step 3 (Optimal Solution) Calculate the optimal solutionvalue min

0le120575le119896119891(1198990 119899119873 120575)

In the recurrence relation the first term in the minimiza-tion corresponds to the case where the partial schedule endswith job 119869

119894isin 1198690 Because 119869

119895jobs of 119869

119873appear before job

119869119894in such a partial schedule the increase in total sequence

disruption is equal to 119895 The second term corresponds to thecase where the partial schedule ends with job 119869

1198990+119895isin 119869119873

In addition we demonstrate the result of Algorithm 6 inthe following example

Example 7 1198990

= 3 119899119873

= 3 1199050

= 0 1198690

= 1198691 1198692 1198693 119869119873

=

1198694 1198695 1198696 1199011= 32 119889

1= 4 119901

2= 1 119889

2= 24 119901

3= 2 119889

3=

3 1199014= 15 119889

4= 3 119901

5= 4 119889

5= 5 119901

6= 3 119889

6= 47 119886 = 02

119887 = 04 and 119896 = 5Solution According to Algorithm 4 and Lemmas 2 and

3 Because the total sequence disruption of the jobs 1198691 1198692 1198693

can not exceed 119896 = 5 By dynamic programming algorithmwe obtain job sequence and the total tardiness cost as follows

if 119896 = 0 the optimal sequence is [1198692

rarr 1198693

rarr

1198691

rarr 1198694

rarr 1198696

rarr 1198695] and the total tardiness cost is

268007

if 119896 = 0 the job sequence is [1198692

rarr 1198693

rarr 1198691

rarr

1198694

rarr 1198696

rarr 1198695] and the total tardiness cost is

268007

if 119896 = 1 the job sequence is [1198692

rarr 1198693

rarr 1198694

rarr

1198691

rarr 1198696

rarr 1198695] and the total tardiness cost is

258007

if 119896 = 2 the job sequence is [1198692

rarr 1198694

rarr 1198693

rarr

1198691

rarr 1198696

rarr 1198695] and the total tardiness cost is

258007

if 119896 = 3 the job sequence is [1198694

rarr 1198692

rarr 1198693

rarr

1198691

rarr 1198696

rarr 1198695] and the total tardiness cost is

258007

if 119896 = 4 the job sequence is [1198694

rarr 1198692

rarr 1198693

rarr

1198696

rarr 1198691

rarr 1198695] and the total tardiness cost is

268223

if 119896 = 5 the job sequence is [1198694

rarr 1198692

rarr 1198696

rarr

1198693

rarr 1198691

rarr 1198695] and the total tardiness cost is

285223

Furthermore if total sequence disruption 119896 is equal to1 2 3 then minimizing total tardiness cost is 258007 andoptimal job sequence is [119869

2rarr 1198693rarr 1198694rarr 1198691rarr 1198696rarr 1198695]

[1198692

rarr 1198694

rarr 1198693

rarr 1198691

rarr 1198696

rarr 1198695] and [119869

4rarr 1198692

rarr

1198693rarr 1198691rarr 1198696rarr 1198695]

Mathematical Problems in Engineering 5

Input Given 119901119895 119889119895for 119895 = 1 119899 119896 and 120587

lowast where 119896 le 119899119873

Indexing Index the jobs of 119869119873in the EDD order

Schedule Construction Schedule jobs 1 1198990+ 119896 in the EDD order in the first 119899

0+ 119896 positions

Schedule jobs 1198990+ 119896 + 1 119899

0+ 119899119873in EDD order in the final 119899

119873minus 119896 positions

Box 1

Input Given 119901119895 119889119895for 119895 = 1 119899 119896 and 120587

lowast where 119896 le 1198990119899119873

Indexing Index the jobs of 119869119873in the EDD order

Box 2

Theorem 8 For the 1 | 119901119894le 119901119895rArr 119889119894le 119889119895 sum119863119895(120587lowast) le

119896 119901119895(119886 + 119887119904

119895) | sum119879

119895problem Algorithm 6 finds an optimal

schedule in 119874(1198992

0n2119873) time

Proof FromLemmas 1 2 and 3 it only remains to enumerateall possible ways of merging the EDD ordered lists of jobsof 1198690and 119869119873 Algorithm 6 does so by comparing the cost of

all possible state transitions and therefore finds an optimalschedule

Because 119894 le 1198990 119895le119899119873

and 120575 le 119896 le 1198990119899119873 there are119874(119899

2

01198992

119873)

values of the state variables Step 1 requires119874(119899119873log 119899119873) Step

2 requires constant time for each set of values of the statevariables Thus the overall time complexity of Algorithm 6is 119874(119899

2

01198992

119873)

Now we consider problem 1 | 119901119894

le 119901119895

rArr 119889119894

le

119889119895 Δmax(120587

lowast) le 119896 119901

119895(119886 + 119887119904

119895) | sum119879

119895 From Lemmas 1 2

and 3 the maximum time disruption of jobs of 1198690is at most

119896 and jobs of 119869119873before the last job of 119869

0have the smallest

due dates Thus we propose the following algorithm underthe maximum time disruption constraint (see Box 3)

Let 119891(119894 119895 120575) be minimum total tardiness value of apartial schedule for jobs 119869

1 119869

119894and 1198691198990+1

1198691198990+119895

wherethe maximum time disruption is equal to 120575 The dynamicprogramming procedure can now be stated as follows

Algorithm 9 Consider the following steps

Step 1 (Initialization) Set 119891(119894 119895 120575) = 0 for (119894 119895 120575) = (0 0 0)

and 119891(119894 119895 120575) = infin for (119894 119895 120575) = (0 0 0) 119894 = 1 1198990 119895 =

1 119899119873 and 119894 = 0 119896

Step 2 (Recurrence Relation)

119891 (119894 119895 120575) = min119891 (119894 minus 1 119895 120575 minus 119875

ℎ) +max 119862

119894minus 119889119894 0

119891 (119894 119895 minus 1 120575) +max 1198621198990+119895

minus 1198891198990+119895

0

(6)

where 119875ℎis the sum of actual processing time of the new jobs

of 119869119873between 119869

119894minus1and 119869119894and119862

119895denotes the completion time

of job 119869119895

Step 3 (Optimal Solution) Calculate the optimal solutionvalue min

0le120575le119896119891(1198990 119899119873 120575)

In the recurrence relation the first term in the minimiza-tion corresponds to the case where the partial schedule endswith job 119869

119894isin 1198690 Because 119869

119895jobs of 119869

119873appear before job 119869

119894

in such a partial schedule the increase in the maximum timedisruption is equal to 119875

ℎThe second term corresponds to the

case where the partial schedule ends with job 1198691198990+119895

isin 119869119873

Similar to Example 7 by Algorithm 9 we have(i) the maximum time disruption 119896 is in [17237

257919] the job sequence and the total tardiness costare the same to Example

(ii) the maximum time disruption 119896 is in [0 17237) thejob sequence is [119869

2rarr 1198693rarr 1198691rarr 1198694rarr 1198696rarr 1198695]

and the total tardiness cost is 268007

Theorem 10 For the 1 | 119901119894le 119901119895rArr 119889119894le 119889119895 Δmax(120587

lowast) le

119896 119901119895(119886 + 119887119904

119895) | sum119879

119895problem Algorithm 9 finds an optimal

schedule in 119874(1198990119899119873119862max + 119899

119873log 119899119873) time

Proof From Lemmas 1 2 and 3 Δ le 119896 means that the totalactual processing time of the new jobs of 119869

119873before the last job

of 1198690is at most 119896 and these are the jobs of 119869

119873with the smallest

due dates Hence Algorithm 9 schedules the jobs accordingto the (EDD EDD) property

Because 119894 le 1198990 119895 le 119899

119873and 120575 le 119896 lt 119862max there

are 119874(1198990119899119873119862max) values of the state variables Step 1 requires

119874(119899119873log 119899119873) Step 2 requires constant time for each set of

values of the state variablesThus the overall time complexityof Algorithm 9 is 119874(119899

0119899119873119862max + 119899

119873log 119899119873)

Now we consider problem 1 | 119901119894

le 119901119895

rArr 119889119894

le

119889119895 sumΔ

119895(120587lowast) le 119896 119901

119895(119886 + 119887119904

119895) | sum119879

119895 From Lemmas 1 2

and 3 there is the total time disruption of jobs of 119869119873which

is less than or equal to 119896 and can be sequenced before thelast job of 119869

0 and these jobs have the smallest due dates

The following dynamic programming algorithm performs anoptimal merging of jobs of 119869

0and 119869

119873in a way similar to

Algorithm 6 (see Box 4)Let119891(119894 119895 120575) beminimum total tardiness value of a partial

schedule for jobs 1198691 119869

119894and 1198691198990+1

1198691198990+119895

where the total

6 Mathematical Problems in Engineering

Input Given 119901119895 119889119895for 119895 = 1 119899 119896 and 120587

lowast where 119896 lt 119862max (see Lemma 1)Indexing Index the jobs of 119869

119873in the EDD order

Box 3

Input Given 119901119895 119889119895for 119895 = 1 119899 119896 and 120587

lowast where 119896 le 1198990119862max

Indexing Index the jobs of 119869119873in the EDD order

Box 4

time disruption is equal to 120575 The dynamic programmingprocedure can now be stated as follows

Algorithm 11

Step 1 (Initialization) Set 119891(119894 119895 120575) = 0 for (119894 119895 120575) = (0 0 0)

and 119891(119894 119895 120575) = infin for (119894 119895 120575) = (0 0 0) 119894 = 1 1198990 119895 =

1 119899119873 and 119894 = 0 119896

Step 2 (Recurrence Relation)

119891 (119894 119895 120575)= min

119891(119894 minus 1 119895 120575 minus

1198990+119895

sum

ℎ=1198990+1

119901[ℎ]

) +max 119862119894minus 119889119894 0

119891 (119894 119895 minus 1 120575) +max 1198621198990+119895

minus 1198891198990+119895

0

(7)

where 119901[ℎ]

is the actual processing time of job 119869ℎ 119862119895denotes

the completion time of job 119869119895

Step 3 (Optimal Solution) Calculate the optimal solutionvalue min

0le120575le119896119891(1198990 119899119873 120575)

In the recurrence relation the first term in the minimiza-tion corresponds to the case where the partial schedule endswith job 119869

119894isin 1198690 Because 119869

119895jobs of 119869

119873appear before job 119869

119894in

such a partial schedule the increase in total time disruptionis equal to sum

1198990+119895

ℎ=1198990+1119901[ℎ] The second term corresponds to the

case where the partial schedule ends with job 1198691198990+119895

isin 119869119873

Theorem 12 For the 1 | 119901119894le 119901119895rArr 119889119894le 119889119895 sumΔ

119895(120587lowast) le

119896 119901119895(119886 + 119887119904

119895) | sum119879

119895problem Algorithm 11 finds an optimal

schedule in 119874(1198992

0119899119873119862max) time

Proof The proof of optimality of Algorithm 11 is similar toTheorem 8 regarding the time complexity Because 119894 le 119899

0

119895 le 119899119873 and 120575 le 119896 le 119899

0119862max there are 119874(119899

2

0119899119873119862max) values

of the state variables Step 1 requires 119874(119899119873log 119899119873) Similar

arguments to those in the proof of Theorem 8 show that theoverall time complexity of Algorithm 11 is119874(119899

2

0119899119873119862max)

4 Conclusions

In this paper we studied the issue of rescheduling to allowthe unexpected arrival of new jobs and took into account

the effect of the disruption on a previously planned opti-mal schedule The main contribution of this paper is thatwe develop the machine rescheduling scheduling problemsagreeable job parameters under deterioration and disruptionRescheduling means to schedule the jobs again togetherwith a set of new jobs Deteriorating job means that theactual processing time of a job is an increasing functionof its starting time When the processing time and duedate of jobs are agreeable we considered some problems tominimize total tardiness under a limit on two disruptionconstraints sequence disruption and time disruption Weproposed polynomial time algorithms or some dynamicprogramming algorithms for each problem Future researchmay stimulate rescheduling models to mitigate the effectsof the disruptions that occur frequently in manufacturingpractice

Acknowledgments

The authors would like to thank the authors of the referencesfor enlightening themThis paper was supported by NationalNatural Science Foundation of China (71272085) TianyuanFund for Mathematics (11226237) the Humanities and SocialSciences Program of the Ministry of Education of China(12YJA630135) the Foundation of Chongqing EducationCommission (KJ120624) and the Key Project Fundation ofChongqing Normal University (2011XLZ05)

References

[1] S Browne and U Yechiali ldquoScheduling deteriorating jobs on asingle processorrdquo Operations Research vol 38 no 3 pp 495ndash498 1990

[2] A S Kunnathur and S K Gupta ldquoMinimizing the makespanwith late start penalties added to processing times in a singlefacility scheduling problemrdquo European Journal of OperationalResearch vol 47 no 1 pp 56ndash64 1990

[3] G Mosheiov ldquoScheduling jobs under simple linear deteriora-tionrdquoComputers andOperations Research vol 21 no 6 pp 653ndash659 1994

[4] T C E Cheng Q Ding and B M T Lin ldquoA concise surveyof schedulingwith time-dependent processing timesrdquoEuropeanJournal of Operational Research vol 152 no 1 pp 1ndash13 2004

[5] S Gawiejnowicz Time-Dependent Scheduling Springer BerlinGermany 2008

Mathematical Problems in Engineering 7

[6] D Biskup and JHerrmann ldquoSingle-machine scheduling againstdue dates with past-sequence-dependent setup timesrdquo Euro-pean Journal of Operational Research vol 191 no 2 pp 587ndash5922008

[7] J-B Wang and Q Guo ldquoA due-date assignment problem withlearning effect and deteriorating jobsrdquo Applied MathematicalModelling vol 34 no 2 pp 309ndash313 2010

[8] C T Ng J-B Wang T C E Cheng and L L Liu ldquoAbranch-and-bound algorithm for solving a two-machine flowshop problemwith deteriorating jobsrdquoComputers ampOperationsResearch vol 37 no 1 pp 83ndash90 2010

[9] T C E Cheng W-C Lee and C-C Wu ldquoSingle-machinescheduling with deteriorating jobs and past-sequence-dependent setup timesrdquo Applied Mathematical Modelling vol35 no 4 pp 1861ndash1867 2011

[10] N Raman F B Talbot and R V Rachamadugu ldquoDue datebased scheduling in a general flexible manufacturing systemrdquoJournal of Operations Management vol 8 no 2 pp 115ndash1321989

[11] L K Church and R Uzsoy ldquoAnalysis of periodic and event-driven rescheduling policies in dynamic shopsrdquo InternationalJournal of Computer Integrated Manufacturing vol 5 pp 153ndash163 1992

[12] A K Jain and H A Elmaraghy ldquoProduction schedul-ingrescheduling in flexible manufacturingrdquo International Jour-nal of Production Research vol 35 no 1 pp 281ndash309 1997

[13] G E Vieira J W Herrmann and E Lin ldquoReschedulingmanufacturing systems a framework of strategies policies andmethodsrdquo Journal of Scheduling vol 6 no 1 pp 39ndash62 2003

[14] B B Yang ldquoSingle machine rescheduling with new jobs arrivalsand processing time compressionrdquo International Journal ofAdvanced Manufacturing Technology vol 34 no 3-4 pp 378ndash384 2007

[15] N G Hall and C N Potts ldquoRescheduling for new ordersrdquoOperations Research vol 52 no 3 pp 440ndash453 2004

[16] J J Yuan and Y D Mu ldquoRescheduling with release dates tominimize makespan under a limit on the maximum sequencedisruptionrdquo European Journal of Operational Research vol 182no 2 pp 936ndash944 2007

[17] C L Zhao and H Y Tang ldquoRescheduling problems withdeteriorating jobs under disruptionsrdquo Applied MathematicalModelling vol 34 no 1 pp 238ndash243 2010

[18] H Hoogeveen C Lente and V Trsquokindt ldquoRescheduling for neworders on a single machine with setup timesrdquo European Journalof Operational Research vol 223 no 1 pp 40ndash46 2012

[19] J Du and J Y-T Leung ldquoMinimizing total tardiness on onemachine is NP-hardrdquo Mathematics of Operations Research vol15 no 3 pp 483ndash495 1990

[20] R L Graham E L Lawler J K Lenstra and A H GRinnooy Kan ldquoOptimization and approximation in determin-istic sequencing and scheduling a surveyrdquo Annals of DiscreteMathematics vol 5 pp 287ndash326 1979

[21] A Kononov and S Gawiejnowicz ldquoNP-hard cases in schedulingdeteriorating jobs on dedicated machinesrdquo Journal of the Oper-ational Research Society vol 52 no 6 pp 708ndash717 2001

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 3

3 Minimum Tardiness Problem withAgreeable Job Parameters

We start with the following result by Kononov and Gaw-iejnowicz [21]

Lemma 1 For the 1 | 119901119895(119886 + 119887119904

119895) | 119862max problem if

120587 = 1198691 119869

119899 and the starting time of the job 119869

1is 119905 then

makespan is sequence independent and

119862max (120587) = 119905

119899

prod

119894=1

(1 + 119887119901119894) +

119886

119887(

119899

prod

119894=1

(1 + 119887119901119894) minus 1) (1)

For notational convenience we assume that the jobs areindexed by agreeable order that is 119901

1le sdot sdot sdot le 119901

1198990and 119889

1le

sdot sdot sdot le 1198891198990 Thus 120587lowast = 119869

1 119869

1198990 with no idle time between

jobsWe now show that the EDD or SPT rule applies to servalof the rescheduling problems we consider

Lemma 2 Problems 1 | 119901119894le 119901119895rArr 119889119894le 119889119895 119863max(120587

lowast) le

119896 119901119895(119886 + 119887119904

119895) | sum119879

119895 and 1 | 119901

119894le 119901119895rArr 119889119894le 119889119895 Δmax(120587

lowast) le

119896 119901119895(119886+119887119904

119895) | sum119879

119895have an optimal schedule with no idle time

between jobs and

(a) a schedule for problem 1 | 119901119894

le 119901119895

rArr 119889119894

le 119889119895

119863max(120587lowast) le 119896 119901

119895(119886 + 119887119904

119895) | sum119879

119895is feasible if the

number of jobs of 119869119873scheduled before the last job of

1198690is less than or equal to 119896

(b) a schedule for problem 1 | 119901119894

le 119901119895

rArr 119889119894

le 119889119895

Δmax(120587lowast) le 119896 119901

119895(119886 + 119887119904

119895) | sum119879

119895is feasible if the total

actual processing time of jobs of 119869119873scheduled before the

last job of 1198690is less than or equal to 119896

Proof The proof is similar to that of Lemma 1 in Hall andPotts [15]

Lemma 3 For problems 1 | 119901119894le 119901119895

rArr 119889119894le 119889119895 Γ le 119896

and 119901119895(119886 + 119887119904

119895) | sum119879

119895 where Γ isin 119863max(120587

lowast) sum119863

119895(120587lowast)

Δmax(120587lowast) sumΔ

119895(120587lowast) there exists an optimal schedule in

which the jobs of 1198690are sequenced in the EDD or SPT rule as in

120587lowast the jobs of 119869

119873are sequenced in the EDD or SPT rule and

there is no idle time between jobs

Proof We first analyze the jobs of 1198690 Consider an optimal

schedule 120590lowast in which the jobs of 119869

0are not sequenced in

the EDD or SPT rule as in 120587lowast Let 119869

119894be the job with the

smallest index that appears later relative to the other jobs of1198690in 120590lowast than in 120587

lowast and let 119869119895(119895 gt 119894) be the last job of 119869

0

that precedes job 119869119894in 120590lowast Because 120587lowast is an optimal sequence

119901119895and 119889

119895are agreeable Assume that the starting time of job

119869119895in 120590lowast is 1199050 then 119862

119895(120590lowast) = 119905

0+ 119901119895(119886 + 119887119905

0) Perform an

interchange on jobs 119869119895and 119869119894 and get a new schedule 1205901015840 In 120590

1015840the starting time of job 119869

119894is 1199050 then119862

119894(1205901015840) = 1199050+119901119894(119886+119887119905

0) lt

1199050+ 119901119895(119886 + 119887119905

0) = 119862

119895(120590lowast) From Lemma 1 119862

119895(1205901015840) = 119862

119894(120590lowast)

Thus the jobs between job 119869119894and 119869119895are completed earlier in

1205901015840 than in 120590

lowast Next we consider the total tardiness of jobs 119869119894

and 119869119895in 1205901015840 and in 120590

lowast

The total tardiness of jobs 119869119894and 119869119895in 120590lowast is as follows

119879119894(120590lowast) + 119879119895(120590lowast) = max 119862

119894(120590lowast) minus 119889119894 0

+max 119862119895(120590lowast) minus 119889119895 0

(2)

The total tardiness of jobs 119869119894and 119869119895in 1205901015840 is as follows

119879119894(1205901015840) + 119879119895(1205901015840) = max 119862

119894(1205901015840) minus 119889119894 0

+max 119862119895(1205901015840) minus 119889119895 0

(3)

To compare the total tardiness of jobs 119869119894and 119869

119895in 120590lowast

and in 1205901015840 we divide it into two cases In the first case when

119862119895(120590lowast) le 119889119895 we have 119879

119894(120590lowast) + 119879119895(120590lowast) = max119862

119894(120590lowast) minus 119889119894 0

Suppose that neither119879119894(1205901015840) nor119879

119895(1205901015840) is zero Note that this is

themost restrictive case since it comprises the case that eitherone or both 119879

119894(1205901015840) and 119879

119895(1205901015840) are zero From Lemma 1 and

119901119894le 119901119895 119889119894le 119889119895 we have 119879

119894(120590lowast)+119879119895(120590lowast)minus119879119894(1205901015840)+119879119895(1205901015840) =

119862119894(120590lowast) minus119862119894(1205901015840) minus119862119895(1205901015840) +119889119895= 119889119895minus119862119894(1205901015840) ge 119889119895minus119862119895(120590lowast) ge 0

In the second case when 119862119895(120590lowast) gt 119889119895 we have 119879

119894(120590lowast) +

119879119895(120590lowast) = 119862119894(120590lowast)+119862119895(120590lowast)minus119889119894minus119889119894 Suppose that neither119879

119894(1205901015840)

nor 119879119895(1205901015840) is zero From Lemma 1 and 119901

119894le 119901119895 119889119894le 119889119895 we

have

119879119894(120590lowast) + 119879119895(120590lowast) minus 119879

119894(1205901015840) + 119879119895(1205901015840)

= 119862119894(120590lowast) + 119862119895(120590lowast) minus 119862119895(1205901015840) minus 119862119894(1205901015840) ge 0

(4)

Now we have proved that the total tardiness of 1205901015840 is less thanor equal to that of 120590lowast

Let the position of job 119869119894in 120587lowast be 119896

1and let the position

of job 119869119895in 120587lowast be 119896

2and in 120590

1015840 be 1198963 If 1198963

ge 1198962 then

119863119895(120587lowast 1205901015840) = 119896

3minus 1198962 119863119894(120587lowast 120590lowast) = 119896

3minus 1198961 Since 119894 lt 119895

implies 1198961

lt 1198962 we have 119863

119895(120587lowast 1205901015840) lt 119863

119894(120587lowast 120590lowast) If 119896

3lt

1198962 then 119863

119895(120587lowast 1205901015840) = 119896

2minus 1198963and 119863

119895(120587lowast 120590lowast) = 119896

2minus

(1198963minus ℎ) where ℎ is the difference between the position of

job 119869119894and 119869119895in 120590lowast So 119863

119895(120587lowast 1205901015840) lt 119863

119895(120587lowast 120590lowast) Hence we

have 119863max(120587lowast 1205901015840) lt 119863max(120587

lowast 120590lowast) In either case because

119863119894(120587lowast 1205901015840) = 119863

119894(120587lowast 120590lowast) minus ℎ and119863

119895(120587lowast 1205901015840) le 119863

119895(120587lowast 120590lowast) + ℎ

Hence sum119863119895(120587lowast 1205901015840) le sum119863

119895(120587lowast 120590lowast)

Moreover if 119862119895(1205901015840) ge 119862119895(120587lowast) then Δ

119895(120587lowast 1205901015840) = 119862119895(1205901015840) minus

119862119895(120587lowast) SinceΔ

119894(120587lowast 120590lowast) = 119862119894(120590lowast)minus119862119894(120587lowast) = 119862119895(1205901015840)minus119862119894(120587lowast)

and 119862119894(120587lowast) lt 119862

119895(120587lowast) therefore Δ

119895(120587lowast 1205901015840) lt Δ

119894(120587lowast 120590lowast) If

119862119895(1205901015840) lt 119862119895(120587lowast) then Δ

119895(120587lowast 1205901015840) = 119862119895(120587lowast) minus 119862119895(1205901015840) because

Δ119895(120587lowast 120590lowast) = 119862

119895(120587lowast) minus 119862

119895(120590lowast) Δ119895(120587lowast 1205901015840) lt Δ

119895(120587lowast 120590lowast)

Thus we have Δmax(120587lowast 1205901015840) lt Δmax(120587

lowast 120590lowast) In either case

because Δ119894(120587lowast 1205901015840) = Δ

119894(120587lowast 120590lowast) minus ℎ

1015840 and Δ119895(120587lowast 1205901015840) le

Δ119895(120587lowast 120590lowast) + ℎ1015840 where ℎ1015840 = 119862

119894(120590lowast) minus 119862119894(1205901015840) Then we deduce

that sumΔ119895(120587lowast 1205901015840) le sumΔ

119895(120587lowast 120590lowast) Thus for either problem

1205901015840 is feasible and optimal We can show that there exists an

optimal schedule in which the jobs of 1198690are sequenced in the

EDD or SPT order as in 120587lowast by finite numbers of repetitions

of the argument A similar interchange argument establishes

4 Mathematical Problems in Engineering

that the jobs of 119869119873can also be obtained by sequencing in the

EDD or SPT order The same EDD or SPT ordering of thejobs of 119869

0in 120587lowast and an optimal schedule show that there is

no idle time in this optimal schedule Otherwise removingthis idle time maintains feasibility and decreases the totaltardiness

We refer to the (EDD EDD) property when a schedule isconstructed using Lemmas 2 and 3We first consider problem1 | 119901119894le 119901119895rArr 119889119894le 119889119895 119863max(120587

lowast) le 119896 119901

119895(119886 + 119887119904

119895) | sum119879

119895

From Lemmas 2 and 3 there are at most 119896 jobs of 119869119873that can

be sequenced before the last job of 1198690 and these jobs have the

smallest due datesThus we propose the following algorithmunder the maximum sequence disruption constraint (seeBox 1)

Algorithm 4 Consider the following steps

Step 1 Index the job of 119869119873in the EDD order

Step 2 Schedule jobs 1 1198990+ 119896 in the EDD rule in the first

1198990+119896 position and schedule jobs 119899

0+119896+1 119899

0+119899119873in the

EDD order in the final 119899119873

minus 119896 positions

Theorem 5 For the 1 | 119901119894le 119901119895

rArr 119889119894le 119889119895 119863max(120587

lowast) le

119896 119901119895(119886 + 119887119904

119895) | sum119879

119895problem Algorithm 4 finds an optimal

schedule in 119874(119899 + 119899119873log 119899119873) time

Proof From Lemmas 2 and 3 the constraint 119863max(120587lowast) le 119896

allows at most 119896 jobs of 119869119873to be sequenced before the final

of 1198690 and these are the jobs of 119869

119873with the smallest due

dates Classical schedule theory shows that the jobs of thisfirst group are sequenced in the EDD order while Lemma 3establishes that the remaining 119899

119873minus 119896 jobs of 119869

119873are also

sequenced in the EDD orderNext we note that the Step 1 for the jobs of 119869

119873requires

119874(119899119873log 119899119873) time Step 2 is executed in119874(119899) time bymerging

the first 119896 jobs of the EDD ordered jobs of 119869119873with the jobs of

1198690as sequenced in 120587

lowast and then placing the last 119899119873minus 119896 jobs of

the EDD order ordered jobs of 119869119873at the end of the schedule

Next we consider problem 1 | 119901119894

le 119901119895

rArr 119889119894

le

119889119895 sum119863

119895(120587lowast) le 119896 119901

119895(119886 + 119887119904

119895) | sum119879

119895 From Lemmas 1 2 and

3 there is the total sequence disruption of the jobs of 119869119873which

is less than or equal to 119896 and can be sequenced before the lastjob of 119869

0 and these jobs have the smallest due dates Thus

we propose the following algorithm under the total sequencedisruption constraint (see Box 2)

Let119891(119894 119895 120575) beminimum total tardiness value of a partialschedule for jobs 119869

1 119869

119894and 1198691198990+1

1198691198990+119895

where the totalsequence disruption is equal to 120575The dynamic programmingprocedure can now be stated as follows

Algorithm 6 Consider the following steps

Step 1 (Initialization) Set 119891(119894 119895 120575) = 0 for (119894 119895 120575) = (0 0 0)

and 119891(119894 119895 120575) = infin for (119894 119895 120575) = (0 0 0) 119894 = 1 1198990 119895 =

1 119899119873 119894 = 0 119896

Step 2 (Recurrence Relation)

119891 (119894 119895 120575) = min119891 (119894 minus 1 119895 120575 minus 119895) +max 119862

119894minus 119889119894 0

119891 (119894 119895 minus 1 120575) +max 1198621198990+119895

minus 1198891198990+119895

0

(5)

where 119862119895denotes the completion time of job 119869

119895

Step 3 (Optimal Solution) Calculate the optimal solutionvalue min

0le120575le119896119891(1198990 119899119873 120575)

In the recurrence relation the first term in the minimiza-tion corresponds to the case where the partial schedule endswith job 119869

119894isin 1198690 Because 119869

119895jobs of 119869

119873appear before job

119869119894in such a partial schedule the increase in total sequence

disruption is equal to 119895 The second term corresponds to thecase where the partial schedule ends with job 119869

1198990+119895isin 119869119873

In addition we demonstrate the result of Algorithm 6 inthe following example

Example 7 1198990

= 3 119899119873

= 3 1199050

= 0 1198690

= 1198691 1198692 1198693 119869119873

=

1198694 1198695 1198696 1199011= 32 119889

1= 4 119901

2= 1 119889

2= 24 119901

3= 2 119889

3=

3 1199014= 15 119889

4= 3 119901

5= 4 119889

5= 5 119901

6= 3 119889

6= 47 119886 = 02

119887 = 04 and 119896 = 5Solution According to Algorithm 4 and Lemmas 2 and

3 Because the total sequence disruption of the jobs 1198691 1198692 1198693

can not exceed 119896 = 5 By dynamic programming algorithmwe obtain job sequence and the total tardiness cost as follows

if 119896 = 0 the optimal sequence is [1198692

rarr 1198693

rarr

1198691

rarr 1198694

rarr 1198696

rarr 1198695] and the total tardiness cost is

268007

if 119896 = 0 the job sequence is [1198692

rarr 1198693

rarr 1198691

rarr

1198694

rarr 1198696

rarr 1198695] and the total tardiness cost is

268007

if 119896 = 1 the job sequence is [1198692

rarr 1198693

rarr 1198694

rarr

1198691

rarr 1198696

rarr 1198695] and the total tardiness cost is

258007

if 119896 = 2 the job sequence is [1198692

rarr 1198694

rarr 1198693

rarr

1198691

rarr 1198696

rarr 1198695] and the total tardiness cost is

258007

if 119896 = 3 the job sequence is [1198694

rarr 1198692

rarr 1198693

rarr

1198691

rarr 1198696

rarr 1198695] and the total tardiness cost is

258007

if 119896 = 4 the job sequence is [1198694

rarr 1198692

rarr 1198693

rarr

1198696

rarr 1198691

rarr 1198695] and the total tardiness cost is

268223

if 119896 = 5 the job sequence is [1198694

rarr 1198692

rarr 1198696

rarr

1198693

rarr 1198691

rarr 1198695] and the total tardiness cost is

285223

Furthermore if total sequence disruption 119896 is equal to1 2 3 then minimizing total tardiness cost is 258007 andoptimal job sequence is [119869

2rarr 1198693rarr 1198694rarr 1198691rarr 1198696rarr 1198695]

[1198692

rarr 1198694

rarr 1198693

rarr 1198691

rarr 1198696

rarr 1198695] and [119869

4rarr 1198692

rarr

1198693rarr 1198691rarr 1198696rarr 1198695]

Mathematical Problems in Engineering 5

Input Given 119901119895 119889119895for 119895 = 1 119899 119896 and 120587

lowast where 119896 le 119899119873

Indexing Index the jobs of 119869119873in the EDD order

Schedule Construction Schedule jobs 1 1198990+ 119896 in the EDD order in the first 119899

0+ 119896 positions

Schedule jobs 1198990+ 119896 + 1 119899

0+ 119899119873in EDD order in the final 119899

119873minus 119896 positions

Box 1

Input Given 119901119895 119889119895for 119895 = 1 119899 119896 and 120587

lowast where 119896 le 1198990119899119873

Indexing Index the jobs of 119869119873in the EDD order

Box 2

Theorem 8 For the 1 | 119901119894le 119901119895rArr 119889119894le 119889119895 sum119863119895(120587lowast) le

119896 119901119895(119886 + 119887119904

119895) | sum119879

119895problem Algorithm 6 finds an optimal

schedule in 119874(1198992

0n2119873) time

Proof FromLemmas 1 2 and 3 it only remains to enumerateall possible ways of merging the EDD ordered lists of jobsof 1198690and 119869119873 Algorithm 6 does so by comparing the cost of

all possible state transitions and therefore finds an optimalschedule

Because 119894 le 1198990 119895le119899119873

and 120575 le 119896 le 1198990119899119873 there are119874(119899

2

01198992

119873)

values of the state variables Step 1 requires119874(119899119873log 119899119873) Step

2 requires constant time for each set of values of the statevariables Thus the overall time complexity of Algorithm 6is 119874(119899

2

01198992

119873)

Now we consider problem 1 | 119901119894

le 119901119895

rArr 119889119894

le

119889119895 Δmax(120587

lowast) le 119896 119901

119895(119886 + 119887119904

119895) | sum119879

119895 From Lemmas 1 2

and 3 the maximum time disruption of jobs of 1198690is at most

119896 and jobs of 119869119873before the last job of 119869

0have the smallest

due dates Thus we propose the following algorithm underthe maximum time disruption constraint (see Box 3)

Let 119891(119894 119895 120575) be minimum total tardiness value of apartial schedule for jobs 119869

1 119869

119894and 1198691198990+1

1198691198990+119895

wherethe maximum time disruption is equal to 120575 The dynamicprogramming procedure can now be stated as follows

Algorithm 9 Consider the following steps

Step 1 (Initialization) Set 119891(119894 119895 120575) = 0 for (119894 119895 120575) = (0 0 0)

and 119891(119894 119895 120575) = infin for (119894 119895 120575) = (0 0 0) 119894 = 1 1198990 119895 =

1 119899119873 and 119894 = 0 119896

Step 2 (Recurrence Relation)

119891 (119894 119895 120575) = min119891 (119894 minus 1 119895 120575 minus 119875

ℎ) +max 119862

119894minus 119889119894 0

119891 (119894 119895 minus 1 120575) +max 1198621198990+119895

minus 1198891198990+119895

0

(6)

where 119875ℎis the sum of actual processing time of the new jobs

of 119869119873between 119869

119894minus1and 119869119894and119862

119895denotes the completion time

of job 119869119895

Step 3 (Optimal Solution) Calculate the optimal solutionvalue min

0le120575le119896119891(1198990 119899119873 120575)

In the recurrence relation the first term in the minimiza-tion corresponds to the case where the partial schedule endswith job 119869

119894isin 1198690 Because 119869

119895jobs of 119869

119873appear before job 119869

119894

in such a partial schedule the increase in the maximum timedisruption is equal to 119875

ℎThe second term corresponds to the

case where the partial schedule ends with job 1198691198990+119895

isin 119869119873

Similar to Example 7 by Algorithm 9 we have(i) the maximum time disruption 119896 is in [17237

257919] the job sequence and the total tardiness costare the same to Example

(ii) the maximum time disruption 119896 is in [0 17237) thejob sequence is [119869

2rarr 1198693rarr 1198691rarr 1198694rarr 1198696rarr 1198695]

and the total tardiness cost is 268007

Theorem 10 For the 1 | 119901119894le 119901119895rArr 119889119894le 119889119895 Δmax(120587

lowast) le

119896 119901119895(119886 + 119887119904

119895) | sum119879

119895problem Algorithm 9 finds an optimal

schedule in 119874(1198990119899119873119862max + 119899

119873log 119899119873) time

Proof From Lemmas 1 2 and 3 Δ le 119896 means that the totalactual processing time of the new jobs of 119869

119873before the last job

of 1198690is at most 119896 and these are the jobs of 119869

119873with the smallest

due dates Hence Algorithm 9 schedules the jobs accordingto the (EDD EDD) property

Because 119894 le 1198990 119895 le 119899

119873and 120575 le 119896 lt 119862max there

are 119874(1198990119899119873119862max) values of the state variables Step 1 requires

119874(119899119873log 119899119873) Step 2 requires constant time for each set of

values of the state variablesThus the overall time complexityof Algorithm 9 is 119874(119899

0119899119873119862max + 119899

119873log 119899119873)

Now we consider problem 1 | 119901119894

le 119901119895

rArr 119889119894

le

119889119895 sumΔ

119895(120587lowast) le 119896 119901

119895(119886 + 119887119904

119895) | sum119879

119895 From Lemmas 1 2

and 3 there is the total time disruption of jobs of 119869119873which

is less than or equal to 119896 and can be sequenced before thelast job of 119869

0 and these jobs have the smallest due dates

The following dynamic programming algorithm performs anoptimal merging of jobs of 119869

0and 119869

119873in a way similar to

Algorithm 6 (see Box 4)Let119891(119894 119895 120575) beminimum total tardiness value of a partial

schedule for jobs 1198691 119869

119894and 1198691198990+1

1198691198990+119895

where the total

6 Mathematical Problems in Engineering

Input Given 119901119895 119889119895for 119895 = 1 119899 119896 and 120587

lowast where 119896 lt 119862max (see Lemma 1)Indexing Index the jobs of 119869

119873in the EDD order

Box 3

Input Given 119901119895 119889119895for 119895 = 1 119899 119896 and 120587

lowast where 119896 le 1198990119862max

Indexing Index the jobs of 119869119873in the EDD order

Box 4

time disruption is equal to 120575 The dynamic programmingprocedure can now be stated as follows

Algorithm 11

Step 1 (Initialization) Set 119891(119894 119895 120575) = 0 for (119894 119895 120575) = (0 0 0)

and 119891(119894 119895 120575) = infin for (119894 119895 120575) = (0 0 0) 119894 = 1 1198990 119895 =

1 119899119873 and 119894 = 0 119896

Step 2 (Recurrence Relation)

119891 (119894 119895 120575)= min

119891(119894 minus 1 119895 120575 minus

1198990+119895

sum

ℎ=1198990+1

119901[ℎ]

) +max 119862119894minus 119889119894 0

119891 (119894 119895 minus 1 120575) +max 1198621198990+119895

minus 1198891198990+119895

0

(7)

where 119901[ℎ]

is the actual processing time of job 119869ℎ 119862119895denotes

the completion time of job 119869119895

Step 3 (Optimal Solution) Calculate the optimal solutionvalue min

0le120575le119896119891(1198990 119899119873 120575)

In the recurrence relation the first term in the minimiza-tion corresponds to the case where the partial schedule endswith job 119869

119894isin 1198690 Because 119869

119895jobs of 119869

119873appear before job 119869

119894in

such a partial schedule the increase in total time disruptionis equal to sum

1198990+119895

ℎ=1198990+1119901[ℎ] The second term corresponds to the

case where the partial schedule ends with job 1198691198990+119895

isin 119869119873

Theorem 12 For the 1 | 119901119894le 119901119895rArr 119889119894le 119889119895 sumΔ

119895(120587lowast) le

119896 119901119895(119886 + 119887119904

119895) | sum119879

119895problem Algorithm 11 finds an optimal

schedule in 119874(1198992

0119899119873119862max) time

Proof The proof of optimality of Algorithm 11 is similar toTheorem 8 regarding the time complexity Because 119894 le 119899

0

119895 le 119899119873 and 120575 le 119896 le 119899

0119862max there are 119874(119899

2

0119899119873119862max) values

of the state variables Step 1 requires 119874(119899119873log 119899119873) Similar

arguments to those in the proof of Theorem 8 show that theoverall time complexity of Algorithm 11 is119874(119899

2

0119899119873119862max)

4 Conclusions

In this paper we studied the issue of rescheduling to allowthe unexpected arrival of new jobs and took into account

the effect of the disruption on a previously planned opti-mal schedule The main contribution of this paper is thatwe develop the machine rescheduling scheduling problemsagreeable job parameters under deterioration and disruptionRescheduling means to schedule the jobs again togetherwith a set of new jobs Deteriorating job means that theactual processing time of a job is an increasing functionof its starting time When the processing time and duedate of jobs are agreeable we considered some problems tominimize total tardiness under a limit on two disruptionconstraints sequence disruption and time disruption Weproposed polynomial time algorithms or some dynamicprogramming algorithms for each problem Future researchmay stimulate rescheduling models to mitigate the effectsof the disruptions that occur frequently in manufacturingpractice

Acknowledgments

The authors would like to thank the authors of the referencesfor enlightening themThis paper was supported by NationalNatural Science Foundation of China (71272085) TianyuanFund for Mathematics (11226237) the Humanities and SocialSciences Program of the Ministry of Education of China(12YJA630135) the Foundation of Chongqing EducationCommission (KJ120624) and the Key Project Fundation ofChongqing Normal University (2011XLZ05)

References

[1] S Browne and U Yechiali ldquoScheduling deteriorating jobs on asingle processorrdquo Operations Research vol 38 no 3 pp 495ndash498 1990

[2] A S Kunnathur and S K Gupta ldquoMinimizing the makespanwith late start penalties added to processing times in a singlefacility scheduling problemrdquo European Journal of OperationalResearch vol 47 no 1 pp 56ndash64 1990

[3] G Mosheiov ldquoScheduling jobs under simple linear deteriora-tionrdquoComputers andOperations Research vol 21 no 6 pp 653ndash659 1994

[4] T C E Cheng Q Ding and B M T Lin ldquoA concise surveyof schedulingwith time-dependent processing timesrdquoEuropeanJournal of Operational Research vol 152 no 1 pp 1ndash13 2004

[5] S Gawiejnowicz Time-Dependent Scheduling Springer BerlinGermany 2008

Mathematical Problems in Engineering 7

[6] D Biskup and JHerrmann ldquoSingle-machine scheduling againstdue dates with past-sequence-dependent setup timesrdquo Euro-pean Journal of Operational Research vol 191 no 2 pp 587ndash5922008

[7] J-B Wang and Q Guo ldquoA due-date assignment problem withlearning effect and deteriorating jobsrdquo Applied MathematicalModelling vol 34 no 2 pp 309ndash313 2010

[8] C T Ng J-B Wang T C E Cheng and L L Liu ldquoAbranch-and-bound algorithm for solving a two-machine flowshop problemwith deteriorating jobsrdquoComputers ampOperationsResearch vol 37 no 1 pp 83ndash90 2010

[9] T C E Cheng W-C Lee and C-C Wu ldquoSingle-machinescheduling with deteriorating jobs and past-sequence-dependent setup timesrdquo Applied Mathematical Modelling vol35 no 4 pp 1861ndash1867 2011

[10] N Raman F B Talbot and R V Rachamadugu ldquoDue datebased scheduling in a general flexible manufacturing systemrdquoJournal of Operations Management vol 8 no 2 pp 115ndash1321989

[11] L K Church and R Uzsoy ldquoAnalysis of periodic and event-driven rescheduling policies in dynamic shopsrdquo InternationalJournal of Computer Integrated Manufacturing vol 5 pp 153ndash163 1992

[12] A K Jain and H A Elmaraghy ldquoProduction schedul-ingrescheduling in flexible manufacturingrdquo International Jour-nal of Production Research vol 35 no 1 pp 281ndash309 1997

[13] G E Vieira J W Herrmann and E Lin ldquoReschedulingmanufacturing systems a framework of strategies policies andmethodsrdquo Journal of Scheduling vol 6 no 1 pp 39ndash62 2003

[14] B B Yang ldquoSingle machine rescheduling with new jobs arrivalsand processing time compressionrdquo International Journal ofAdvanced Manufacturing Technology vol 34 no 3-4 pp 378ndash384 2007

[15] N G Hall and C N Potts ldquoRescheduling for new ordersrdquoOperations Research vol 52 no 3 pp 440ndash453 2004

[16] J J Yuan and Y D Mu ldquoRescheduling with release dates tominimize makespan under a limit on the maximum sequencedisruptionrdquo European Journal of Operational Research vol 182no 2 pp 936ndash944 2007

[17] C L Zhao and H Y Tang ldquoRescheduling problems withdeteriorating jobs under disruptionsrdquo Applied MathematicalModelling vol 34 no 1 pp 238ndash243 2010

[18] H Hoogeveen C Lente and V Trsquokindt ldquoRescheduling for neworders on a single machine with setup timesrdquo European Journalof Operational Research vol 223 no 1 pp 40ndash46 2012

[19] J Du and J Y-T Leung ldquoMinimizing total tardiness on onemachine is NP-hardrdquo Mathematics of Operations Research vol15 no 3 pp 483ndash495 1990

[20] R L Graham E L Lawler J K Lenstra and A H GRinnooy Kan ldquoOptimization and approximation in determin-istic sequencing and scheduling a surveyrdquo Annals of DiscreteMathematics vol 5 pp 287ndash326 1979

[21] A Kononov and S Gawiejnowicz ldquoNP-hard cases in schedulingdeteriorating jobs on dedicated machinesrdquo Journal of the Oper-ational Research Society vol 52 no 6 pp 708ndash717 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Mathematical Problems in Engineering

that the jobs of 119869119873can also be obtained by sequencing in the

EDD or SPT order The same EDD or SPT ordering of thejobs of 119869

0in 120587lowast and an optimal schedule show that there is

no idle time in this optimal schedule Otherwise removingthis idle time maintains feasibility and decreases the totaltardiness

We refer to the (EDD EDD) property when a schedule isconstructed using Lemmas 2 and 3We first consider problem1 | 119901119894le 119901119895rArr 119889119894le 119889119895 119863max(120587

lowast) le 119896 119901

119895(119886 + 119887119904

119895) | sum119879

119895

From Lemmas 2 and 3 there are at most 119896 jobs of 119869119873that can

be sequenced before the last job of 1198690 and these jobs have the

smallest due datesThus we propose the following algorithmunder the maximum sequence disruption constraint (seeBox 1)

Algorithm 4 Consider the following steps

Step 1 Index the job of 119869119873in the EDD order

Step 2 Schedule jobs 1 1198990+ 119896 in the EDD rule in the first

1198990+119896 position and schedule jobs 119899

0+119896+1 119899

0+119899119873in the

EDD order in the final 119899119873

minus 119896 positions

Theorem 5 For the 1 | 119901119894le 119901119895

rArr 119889119894le 119889119895 119863max(120587

lowast) le

119896 119901119895(119886 + 119887119904

119895) | sum119879

119895problem Algorithm 4 finds an optimal

schedule in 119874(119899 + 119899119873log 119899119873) time

Proof From Lemmas 2 and 3 the constraint 119863max(120587lowast) le 119896

allows at most 119896 jobs of 119869119873to be sequenced before the final

of 1198690 and these are the jobs of 119869

119873with the smallest due

dates Classical schedule theory shows that the jobs of thisfirst group are sequenced in the EDD order while Lemma 3establishes that the remaining 119899

119873minus 119896 jobs of 119869

119873are also

sequenced in the EDD orderNext we note that the Step 1 for the jobs of 119869

119873requires

119874(119899119873log 119899119873) time Step 2 is executed in119874(119899) time bymerging

the first 119896 jobs of the EDD ordered jobs of 119869119873with the jobs of

1198690as sequenced in 120587

lowast and then placing the last 119899119873minus 119896 jobs of

the EDD order ordered jobs of 119869119873at the end of the schedule

Next we consider problem 1 | 119901119894

le 119901119895

rArr 119889119894

le

119889119895 sum119863

119895(120587lowast) le 119896 119901

119895(119886 + 119887119904

119895) | sum119879

119895 From Lemmas 1 2 and

3 there is the total sequence disruption of the jobs of 119869119873which

is less than or equal to 119896 and can be sequenced before the lastjob of 119869

0 and these jobs have the smallest due dates Thus

we propose the following algorithm under the total sequencedisruption constraint (see Box 2)

Let119891(119894 119895 120575) beminimum total tardiness value of a partialschedule for jobs 119869

1 119869

119894and 1198691198990+1

1198691198990+119895

where the totalsequence disruption is equal to 120575The dynamic programmingprocedure can now be stated as follows

Algorithm 6 Consider the following steps

Step 1 (Initialization) Set 119891(119894 119895 120575) = 0 for (119894 119895 120575) = (0 0 0)

and 119891(119894 119895 120575) = infin for (119894 119895 120575) = (0 0 0) 119894 = 1 1198990 119895 =

1 119899119873 119894 = 0 119896

Step 2 (Recurrence Relation)

119891 (119894 119895 120575) = min119891 (119894 minus 1 119895 120575 minus 119895) +max 119862

119894minus 119889119894 0

119891 (119894 119895 minus 1 120575) +max 1198621198990+119895

minus 1198891198990+119895

0

(5)

where 119862119895denotes the completion time of job 119869

119895

Step 3 (Optimal Solution) Calculate the optimal solutionvalue min

0le120575le119896119891(1198990 119899119873 120575)

In the recurrence relation the first term in the minimiza-tion corresponds to the case where the partial schedule endswith job 119869

119894isin 1198690 Because 119869

119895jobs of 119869

119873appear before job

119869119894in such a partial schedule the increase in total sequence

disruption is equal to 119895 The second term corresponds to thecase where the partial schedule ends with job 119869

1198990+119895isin 119869119873

In addition we demonstrate the result of Algorithm 6 inthe following example

Example 7 1198990

= 3 119899119873

= 3 1199050

= 0 1198690

= 1198691 1198692 1198693 119869119873

=

1198694 1198695 1198696 1199011= 32 119889

1= 4 119901

2= 1 119889

2= 24 119901

3= 2 119889

3=

3 1199014= 15 119889

4= 3 119901

5= 4 119889

5= 5 119901

6= 3 119889

6= 47 119886 = 02

119887 = 04 and 119896 = 5Solution According to Algorithm 4 and Lemmas 2 and

3 Because the total sequence disruption of the jobs 1198691 1198692 1198693

can not exceed 119896 = 5 By dynamic programming algorithmwe obtain job sequence and the total tardiness cost as follows

if 119896 = 0 the optimal sequence is [1198692

rarr 1198693

rarr

1198691

rarr 1198694

rarr 1198696

rarr 1198695] and the total tardiness cost is

268007

if 119896 = 0 the job sequence is [1198692

rarr 1198693

rarr 1198691

rarr

1198694

rarr 1198696

rarr 1198695] and the total tardiness cost is

268007

if 119896 = 1 the job sequence is [1198692

rarr 1198693

rarr 1198694

rarr

1198691

rarr 1198696

rarr 1198695] and the total tardiness cost is

258007

if 119896 = 2 the job sequence is [1198692

rarr 1198694

rarr 1198693

rarr

1198691

rarr 1198696

rarr 1198695] and the total tardiness cost is

258007

if 119896 = 3 the job sequence is [1198694

rarr 1198692

rarr 1198693

rarr

1198691

rarr 1198696

rarr 1198695] and the total tardiness cost is

258007

if 119896 = 4 the job sequence is [1198694

rarr 1198692

rarr 1198693

rarr

1198696

rarr 1198691

rarr 1198695] and the total tardiness cost is

268223

if 119896 = 5 the job sequence is [1198694

rarr 1198692

rarr 1198696

rarr

1198693

rarr 1198691

rarr 1198695] and the total tardiness cost is

285223

Furthermore if total sequence disruption 119896 is equal to1 2 3 then minimizing total tardiness cost is 258007 andoptimal job sequence is [119869

2rarr 1198693rarr 1198694rarr 1198691rarr 1198696rarr 1198695]

[1198692

rarr 1198694

rarr 1198693

rarr 1198691

rarr 1198696

rarr 1198695] and [119869

4rarr 1198692

rarr

1198693rarr 1198691rarr 1198696rarr 1198695]

Mathematical Problems in Engineering 5

Input Given 119901119895 119889119895for 119895 = 1 119899 119896 and 120587

lowast where 119896 le 119899119873

Indexing Index the jobs of 119869119873in the EDD order

Schedule Construction Schedule jobs 1 1198990+ 119896 in the EDD order in the first 119899

0+ 119896 positions

Schedule jobs 1198990+ 119896 + 1 119899

0+ 119899119873in EDD order in the final 119899

119873minus 119896 positions

Box 1

Input Given 119901119895 119889119895for 119895 = 1 119899 119896 and 120587

lowast where 119896 le 1198990119899119873

Indexing Index the jobs of 119869119873in the EDD order

Box 2

Theorem 8 For the 1 | 119901119894le 119901119895rArr 119889119894le 119889119895 sum119863119895(120587lowast) le

119896 119901119895(119886 + 119887119904

119895) | sum119879

119895problem Algorithm 6 finds an optimal

schedule in 119874(1198992

0n2119873) time

Proof FromLemmas 1 2 and 3 it only remains to enumerateall possible ways of merging the EDD ordered lists of jobsof 1198690and 119869119873 Algorithm 6 does so by comparing the cost of

all possible state transitions and therefore finds an optimalschedule

Because 119894 le 1198990 119895le119899119873

and 120575 le 119896 le 1198990119899119873 there are119874(119899

2

01198992

119873)

values of the state variables Step 1 requires119874(119899119873log 119899119873) Step

2 requires constant time for each set of values of the statevariables Thus the overall time complexity of Algorithm 6is 119874(119899

2

01198992

119873)

Now we consider problem 1 | 119901119894

le 119901119895

rArr 119889119894

le

119889119895 Δmax(120587

lowast) le 119896 119901

119895(119886 + 119887119904

119895) | sum119879

119895 From Lemmas 1 2

and 3 the maximum time disruption of jobs of 1198690is at most

119896 and jobs of 119869119873before the last job of 119869

0have the smallest

due dates Thus we propose the following algorithm underthe maximum time disruption constraint (see Box 3)

Let 119891(119894 119895 120575) be minimum total tardiness value of apartial schedule for jobs 119869

1 119869

119894and 1198691198990+1

1198691198990+119895

wherethe maximum time disruption is equal to 120575 The dynamicprogramming procedure can now be stated as follows

Algorithm 9 Consider the following steps

Step 1 (Initialization) Set 119891(119894 119895 120575) = 0 for (119894 119895 120575) = (0 0 0)

and 119891(119894 119895 120575) = infin for (119894 119895 120575) = (0 0 0) 119894 = 1 1198990 119895 =

1 119899119873 and 119894 = 0 119896

Step 2 (Recurrence Relation)

119891 (119894 119895 120575) = min119891 (119894 minus 1 119895 120575 minus 119875

ℎ) +max 119862

119894minus 119889119894 0

119891 (119894 119895 minus 1 120575) +max 1198621198990+119895

minus 1198891198990+119895

0

(6)

where 119875ℎis the sum of actual processing time of the new jobs

of 119869119873between 119869

119894minus1and 119869119894and119862

119895denotes the completion time

of job 119869119895

Step 3 (Optimal Solution) Calculate the optimal solutionvalue min

0le120575le119896119891(1198990 119899119873 120575)

In the recurrence relation the first term in the minimiza-tion corresponds to the case where the partial schedule endswith job 119869

119894isin 1198690 Because 119869

119895jobs of 119869

119873appear before job 119869

119894

in such a partial schedule the increase in the maximum timedisruption is equal to 119875

ℎThe second term corresponds to the

case where the partial schedule ends with job 1198691198990+119895

isin 119869119873

Similar to Example 7 by Algorithm 9 we have(i) the maximum time disruption 119896 is in [17237

257919] the job sequence and the total tardiness costare the same to Example

(ii) the maximum time disruption 119896 is in [0 17237) thejob sequence is [119869

2rarr 1198693rarr 1198691rarr 1198694rarr 1198696rarr 1198695]

and the total tardiness cost is 268007

Theorem 10 For the 1 | 119901119894le 119901119895rArr 119889119894le 119889119895 Δmax(120587

lowast) le

119896 119901119895(119886 + 119887119904

119895) | sum119879

119895problem Algorithm 9 finds an optimal

schedule in 119874(1198990119899119873119862max + 119899

119873log 119899119873) time

Proof From Lemmas 1 2 and 3 Δ le 119896 means that the totalactual processing time of the new jobs of 119869

119873before the last job

of 1198690is at most 119896 and these are the jobs of 119869

119873with the smallest

due dates Hence Algorithm 9 schedules the jobs accordingto the (EDD EDD) property

Because 119894 le 1198990 119895 le 119899

119873and 120575 le 119896 lt 119862max there

are 119874(1198990119899119873119862max) values of the state variables Step 1 requires

119874(119899119873log 119899119873) Step 2 requires constant time for each set of

values of the state variablesThus the overall time complexityof Algorithm 9 is 119874(119899

0119899119873119862max + 119899

119873log 119899119873)

Now we consider problem 1 | 119901119894

le 119901119895

rArr 119889119894

le

119889119895 sumΔ

119895(120587lowast) le 119896 119901

119895(119886 + 119887119904

119895) | sum119879

119895 From Lemmas 1 2

and 3 there is the total time disruption of jobs of 119869119873which

is less than or equal to 119896 and can be sequenced before thelast job of 119869

0 and these jobs have the smallest due dates

The following dynamic programming algorithm performs anoptimal merging of jobs of 119869

0and 119869

119873in a way similar to

Algorithm 6 (see Box 4)Let119891(119894 119895 120575) beminimum total tardiness value of a partial

schedule for jobs 1198691 119869

119894and 1198691198990+1

1198691198990+119895

where the total

6 Mathematical Problems in Engineering

Input Given 119901119895 119889119895for 119895 = 1 119899 119896 and 120587

lowast where 119896 lt 119862max (see Lemma 1)Indexing Index the jobs of 119869

119873in the EDD order

Box 3

Input Given 119901119895 119889119895for 119895 = 1 119899 119896 and 120587

lowast where 119896 le 1198990119862max

Indexing Index the jobs of 119869119873in the EDD order

Box 4

time disruption is equal to 120575 The dynamic programmingprocedure can now be stated as follows

Algorithm 11

Step 1 (Initialization) Set 119891(119894 119895 120575) = 0 for (119894 119895 120575) = (0 0 0)

and 119891(119894 119895 120575) = infin for (119894 119895 120575) = (0 0 0) 119894 = 1 1198990 119895 =

1 119899119873 and 119894 = 0 119896

Step 2 (Recurrence Relation)

119891 (119894 119895 120575)= min

119891(119894 minus 1 119895 120575 minus

1198990+119895

sum

ℎ=1198990+1

119901[ℎ]

) +max 119862119894minus 119889119894 0

119891 (119894 119895 minus 1 120575) +max 1198621198990+119895

minus 1198891198990+119895

0

(7)

where 119901[ℎ]

is the actual processing time of job 119869ℎ 119862119895denotes

the completion time of job 119869119895

Step 3 (Optimal Solution) Calculate the optimal solutionvalue min

0le120575le119896119891(1198990 119899119873 120575)

In the recurrence relation the first term in the minimiza-tion corresponds to the case where the partial schedule endswith job 119869

119894isin 1198690 Because 119869

119895jobs of 119869

119873appear before job 119869

119894in

such a partial schedule the increase in total time disruptionis equal to sum

1198990+119895

ℎ=1198990+1119901[ℎ] The second term corresponds to the

case where the partial schedule ends with job 1198691198990+119895

isin 119869119873

Theorem 12 For the 1 | 119901119894le 119901119895rArr 119889119894le 119889119895 sumΔ

119895(120587lowast) le

119896 119901119895(119886 + 119887119904

119895) | sum119879

119895problem Algorithm 11 finds an optimal

schedule in 119874(1198992

0119899119873119862max) time

Proof The proof of optimality of Algorithm 11 is similar toTheorem 8 regarding the time complexity Because 119894 le 119899

0

119895 le 119899119873 and 120575 le 119896 le 119899

0119862max there are 119874(119899

2

0119899119873119862max) values

of the state variables Step 1 requires 119874(119899119873log 119899119873) Similar

arguments to those in the proof of Theorem 8 show that theoverall time complexity of Algorithm 11 is119874(119899

2

0119899119873119862max)

4 Conclusions

In this paper we studied the issue of rescheduling to allowthe unexpected arrival of new jobs and took into account

the effect of the disruption on a previously planned opti-mal schedule The main contribution of this paper is thatwe develop the machine rescheduling scheduling problemsagreeable job parameters under deterioration and disruptionRescheduling means to schedule the jobs again togetherwith a set of new jobs Deteriorating job means that theactual processing time of a job is an increasing functionof its starting time When the processing time and duedate of jobs are agreeable we considered some problems tominimize total tardiness under a limit on two disruptionconstraints sequence disruption and time disruption Weproposed polynomial time algorithms or some dynamicprogramming algorithms for each problem Future researchmay stimulate rescheduling models to mitigate the effectsof the disruptions that occur frequently in manufacturingpractice

Acknowledgments

The authors would like to thank the authors of the referencesfor enlightening themThis paper was supported by NationalNatural Science Foundation of China (71272085) TianyuanFund for Mathematics (11226237) the Humanities and SocialSciences Program of the Ministry of Education of China(12YJA630135) the Foundation of Chongqing EducationCommission (KJ120624) and the Key Project Fundation ofChongqing Normal University (2011XLZ05)

References

[1] S Browne and U Yechiali ldquoScheduling deteriorating jobs on asingle processorrdquo Operations Research vol 38 no 3 pp 495ndash498 1990

[2] A S Kunnathur and S K Gupta ldquoMinimizing the makespanwith late start penalties added to processing times in a singlefacility scheduling problemrdquo European Journal of OperationalResearch vol 47 no 1 pp 56ndash64 1990

[3] G Mosheiov ldquoScheduling jobs under simple linear deteriora-tionrdquoComputers andOperations Research vol 21 no 6 pp 653ndash659 1994

[4] T C E Cheng Q Ding and B M T Lin ldquoA concise surveyof schedulingwith time-dependent processing timesrdquoEuropeanJournal of Operational Research vol 152 no 1 pp 1ndash13 2004

[5] S Gawiejnowicz Time-Dependent Scheduling Springer BerlinGermany 2008

Mathematical Problems in Engineering 7

[6] D Biskup and JHerrmann ldquoSingle-machine scheduling againstdue dates with past-sequence-dependent setup timesrdquo Euro-pean Journal of Operational Research vol 191 no 2 pp 587ndash5922008

[7] J-B Wang and Q Guo ldquoA due-date assignment problem withlearning effect and deteriorating jobsrdquo Applied MathematicalModelling vol 34 no 2 pp 309ndash313 2010

[8] C T Ng J-B Wang T C E Cheng and L L Liu ldquoAbranch-and-bound algorithm for solving a two-machine flowshop problemwith deteriorating jobsrdquoComputers ampOperationsResearch vol 37 no 1 pp 83ndash90 2010

[9] T C E Cheng W-C Lee and C-C Wu ldquoSingle-machinescheduling with deteriorating jobs and past-sequence-dependent setup timesrdquo Applied Mathematical Modelling vol35 no 4 pp 1861ndash1867 2011

[10] N Raman F B Talbot and R V Rachamadugu ldquoDue datebased scheduling in a general flexible manufacturing systemrdquoJournal of Operations Management vol 8 no 2 pp 115ndash1321989

[11] L K Church and R Uzsoy ldquoAnalysis of periodic and event-driven rescheduling policies in dynamic shopsrdquo InternationalJournal of Computer Integrated Manufacturing vol 5 pp 153ndash163 1992

[12] A K Jain and H A Elmaraghy ldquoProduction schedul-ingrescheduling in flexible manufacturingrdquo International Jour-nal of Production Research vol 35 no 1 pp 281ndash309 1997

[13] G E Vieira J W Herrmann and E Lin ldquoReschedulingmanufacturing systems a framework of strategies policies andmethodsrdquo Journal of Scheduling vol 6 no 1 pp 39ndash62 2003

[14] B B Yang ldquoSingle machine rescheduling with new jobs arrivalsand processing time compressionrdquo International Journal ofAdvanced Manufacturing Technology vol 34 no 3-4 pp 378ndash384 2007

[15] N G Hall and C N Potts ldquoRescheduling for new ordersrdquoOperations Research vol 52 no 3 pp 440ndash453 2004

[16] J J Yuan and Y D Mu ldquoRescheduling with release dates tominimize makespan under a limit on the maximum sequencedisruptionrdquo European Journal of Operational Research vol 182no 2 pp 936ndash944 2007

[17] C L Zhao and H Y Tang ldquoRescheduling problems withdeteriorating jobs under disruptionsrdquo Applied MathematicalModelling vol 34 no 1 pp 238ndash243 2010

[18] H Hoogeveen C Lente and V Trsquokindt ldquoRescheduling for neworders on a single machine with setup timesrdquo European Journalof Operational Research vol 223 no 1 pp 40ndash46 2012

[19] J Du and J Y-T Leung ldquoMinimizing total tardiness on onemachine is NP-hardrdquo Mathematics of Operations Research vol15 no 3 pp 483ndash495 1990

[20] R L Graham E L Lawler J K Lenstra and A H GRinnooy Kan ldquoOptimization and approximation in determin-istic sequencing and scheduling a surveyrdquo Annals of DiscreteMathematics vol 5 pp 287ndash326 1979

[21] A Kononov and S Gawiejnowicz ldquoNP-hard cases in schedulingdeteriorating jobs on dedicated machinesrdquo Journal of the Oper-ational Research Society vol 52 no 6 pp 708ndash717 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 5

Input Given 119901119895 119889119895for 119895 = 1 119899 119896 and 120587

lowast where 119896 le 119899119873

Indexing Index the jobs of 119869119873in the EDD order

Schedule Construction Schedule jobs 1 1198990+ 119896 in the EDD order in the first 119899

0+ 119896 positions

Schedule jobs 1198990+ 119896 + 1 119899

0+ 119899119873in EDD order in the final 119899

119873minus 119896 positions

Box 1

Input Given 119901119895 119889119895for 119895 = 1 119899 119896 and 120587

lowast where 119896 le 1198990119899119873

Indexing Index the jobs of 119869119873in the EDD order

Box 2

Theorem 8 For the 1 | 119901119894le 119901119895rArr 119889119894le 119889119895 sum119863119895(120587lowast) le

119896 119901119895(119886 + 119887119904

119895) | sum119879

119895problem Algorithm 6 finds an optimal

schedule in 119874(1198992

0n2119873) time

Proof FromLemmas 1 2 and 3 it only remains to enumerateall possible ways of merging the EDD ordered lists of jobsof 1198690and 119869119873 Algorithm 6 does so by comparing the cost of

all possible state transitions and therefore finds an optimalschedule

Because 119894 le 1198990 119895le119899119873

and 120575 le 119896 le 1198990119899119873 there are119874(119899

2

01198992

119873)

values of the state variables Step 1 requires119874(119899119873log 119899119873) Step

2 requires constant time for each set of values of the statevariables Thus the overall time complexity of Algorithm 6is 119874(119899

2

01198992

119873)

Now we consider problem 1 | 119901119894

le 119901119895

rArr 119889119894

le

119889119895 Δmax(120587

lowast) le 119896 119901

119895(119886 + 119887119904

119895) | sum119879

119895 From Lemmas 1 2

and 3 the maximum time disruption of jobs of 1198690is at most

119896 and jobs of 119869119873before the last job of 119869

0have the smallest

due dates Thus we propose the following algorithm underthe maximum time disruption constraint (see Box 3)

Let 119891(119894 119895 120575) be minimum total tardiness value of apartial schedule for jobs 119869

1 119869

119894and 1198691198990+1

1198691198990+119895

wherethe maximum time disruption is equal to 120575 The dynamicprogramming procedure can now be stated as follows

Algorithm 9 Consider the following steps

Step 1 (Initialization) Set 119891(119894 119895 120575) = 0 for (119894 119895 120575) = (0 0 0)

and 119891(119894 119895 120575) = infin for (119894 119895 120575) = (0 0 0) 119894 = 1 1198990 119895 =

1 119899119873 and 119894 = 0 119896

Step 2 (Recurrence Relation)

119891 (119894 119895 120575) = min119891 (119894 minus 1 119895 120575 minus 119875

ℎ) +max 119862

119894minus 119889119894 0

119891 (119894 119895 minus 1 120575) +max 1198621198990+119895

minus 1198891198990+119895

0

(6)

where 119875ℎis the sum of actual processing time of the new jobs

of 119869119873between 119869

119894minus1and 119869119894and119862

119895denotes the completion time

of job 119869119895

Step 3 (Optimal Solution) Calculate the optimal solutionvalue min

0le120575le119896119891(1198990 119899119873 120575)

In the recurrence relation the first term in the minimiza-tion corresponds to the case where the partial schedule endswith job 119869

119894isin 1198690 Because 119869

119895jobs of 119869

119873appear before job 119869

119894

in such a partial schedule the increase in the maximum timedisruption is equal to 119875

ℎThe second term corresponds to the

case where the partial schedule ends with job 1198691198990+119895

isin 119869119873

Similar to Example 7 by Algorithm 9 we have(i) the maximum time disruption 119896 is in [17237

257919] the job sequence and the total tardiness costare the same to Example

(ii) the maximum time disruption 119896 is in [0 17237) thejob sequence is [119869

2rarr 1198693rarr 1198691rarr 1198694rarr 1198696rarr 1198695]

and the total tardiness cost is 268007

Theorem 10 For the 1 | 119901119894le 119901119895rArr 119889119894le 119889119895 Δmax(120587

lowast) le

119896 119901119895(119886 + 119887119904

119895) | sum119879

119895problem Algorithm 9 finds an optimal

schedule in 119874(1198990119899119873119862max + 119899

119873log 119899119873) time

Proof From Lemmas 1 2 and 3 Δ le 119896 means that the totalactual processing time of the new jobs of 119869

119873before the last job

of 1198690is at most 119896 and these are the jobs of 119869

119873with the smallest

due dates Hence Algorithm 9 schedules the jobs accordingto the (EDD EDD) property

Because 119894 le 1198990 119895 le 119899

119873and 120575 le 119896 lt 119862max there

are 119874(1198990119899119873119862max) values of the state variables Step 1 requires

119874(119899119873log 119899119873) Step 2 requires constant time for each set of

values of the state variablesThus the overall time complexityof Algorithm 9 is 119874(119899

0119899119873119862max + 119899

119873log 119899119873)

Now we consider problem 1 | 119901119894

le 119901119895

rArr 119889119894

le

119889119895 sumΔ

119895(120587lowast) le 119896 119901

119895(119886 + 119887119904

119895) | sum119879

119895 From Lemmas 1 2

and 3 there is the total time disruption of jobs of 119869119873which

is less than or equal to 119896 and can be sequenced before thelast job of 119869

0 and these jobs have the smallest due dates

The following dynamic programming algorithm performs anoptimal merging of jobs of 119869

0and 119869

119873in a way similar to

Algorithm 6 (see Box 4)Let119891(119894 119895 120575) beminimum total tardiness value of a partial

schedule for jobs 1198691 119869

119894and 1198691198990+1

1198691198990+119895

where the total

6 Mathematical Problems in Engineering

Input Given 119901119895 119889119895for 119895 = 1 119899 119896 and 120587

lowast where 119896 lt 119862max (see Lemma 1)Indexing Index the jobs of 119869

119873in the EDD order

Box 3

Input Given 119901119895 119889119895for 119895 = 1 119899 119896 and 120587

lowast where 119896 le 1198990119862max

Indexing Index the jobs of 119869119873in the EDD order

Box 4

time disruption is equal to 120575 The dynamic programmingprocedure can now be stated as follows

Algorithm 11

Step 1 (Initialization) Set 119891(119894 119895 120575) = 0 for (119894 119895 120575) = (0 0 0)

and 119891(119894 119895 120575) = infin for (119894 119895 120575) = (0 0 0) 119894 = 1 1198990 119895 =

1 119899119873 and 119894 = 0 119896

Step 2 (Recurrence Relation)

119891 (119894 119895 120575)= min

119891(119894 minus 1 119895 120575 minus

1198990+119895

sum

ℎ=1198990+1

119901[ℎ]

) +max 119862119894minus 119889119894 0

119891 (119894 119895 minus 1 120575) +max 1198621198990+119895

minus 1198891198990+119895

0

(7)

where 119901[ℎ]

is the actual processing time of job 119869ℎ 119862119895denotes

the completion time of job 119869119895

Step 3 (Optimal Solution) Calculate the optimal solutionvalue min

0le120575le119896119891(1198990 119899119873 120575)

In the recurrence relation the first term in the minimiza-tion corresponds to the case where the partial schedule endswith job 119869

119894isin 1198690 Because 119869

119895jobs of 119869

119873appear before job 119869

119894in

such a partial schedule the increase in total time disruptionis equal to sum

1198990+119895

ℎ=1198990+1119901[ℎ] The second term corresponds to the

case where the partial schedule ends with job 1198691198990+119895

isin 119869119873

Theorem 12 For the 1 | 119901119894le 119901119895rArr 119889119894le 119889119895 sumΔ

119895(120587lowast) le

119896 119901119895(119886 + 119887119904

119895) | sum119879

119895problem Algorithm 11 finds an optimal

schedule in 119874(1198992

0119899119873119862max) time

Proof The proof of optimality of Algorithm 11 is similar toTheorem 8 regarding the time complexity Because 119894 le 119899

0

119895 le 119899119873 and 120575 le 119896 le 119899

0119862max there are 119874(119899

2

0119899119873119862max) values

of the state variables Step 1 requires 119874(119899119873log 119899119873) Similar

arguments to those in the proof of Theorem 8 show that theoverall time complexity of Algorithm 11 is119874(119899

2

0119899119873119862max)

4 Conclusions

In this paper we studied the issue of rescheduling to allowthe unexpected arrival of new jobs and took into account

the effect of the disruption on a previously planned opti-mal schedule The main contribution of this paper is thatwe develop the machine rescheduling scheduling problemsagreeable job parameters under deterioration and disruptionRescheduling means to schedule the jobs again togetherwith a set of new jobs Deteriorating job means that theactual processing time of a job is an increasing functionof its starting time When the processing time and duedate of jobs are agreeable we considered some problems tominimize total tardiness under a limit on two disruptionconstraints sequence disruption and time disruption Weproposed polynomial time algorithms or some dynamicprogramming algorithms for each problem Future researchmay stimulate rescheduling models to mitigate the effectsof the disruptions that occur frequently in manufacturingpractice

Acknowledgments

The authors would like to thank the authors of the referencesfor enlightening themThis paper was supported by NationalNatural Science Foundation of China (71272085) TianyuanFund for Mathematics (11226237) the Humanities and SocialSciences Program of the Ministry of Education of China(12YJA630135) the Foundation of Chongqing EducationCommission (KJ120624) and the Key Project Fundation ofChongqing Normal University (2011XLZ05)

References

[1] S Browne and U Yechiali ldquoScheduling deteriorating jobs on asingle processorrdquo Operations Research vol 38 no 3 pp 495ndash498 1990

[2] A S Kunnathur and S K Gupta ldquoMinimizing the makespanwith late start penalties added to processing times in a singlefacility scheduling problemrdquo European Journal of OperationalResearch vol 47 no 1 pp 56ndash64 1990

[3] G Mosheiov ldquoScheduling jobs under simple linear deteriora-tionrdquoComputers andOperations Research vol 21 no 6 pp 653ndash659 1994

[4] T C E Cheng Q Ding and B M T Lin ldquoA concise surveyof schedulingwith time-dependent processing timesrdquoEuropeanJournal of Operational Research vol 152 no 1 pp 1ndash13 2004

[5] S Gawiejnowicz Time-Dependent Scheduling Springer BerlinGermany 2008

Mathematical Problems in Engineering 7

[6] D Biskup and JHerrmann ldquoSingle-machine scheduling againstdue dates with past-sequence-dependent setup timesrdquo Euro-pean Journal of Operational Research vol 191 no 2 pp 587ndash5922008

[7] J-B Wang and Q Guo ldquoA due-date assignment problem withlearning effect and deteriorating jobsrdquo Applied MathematicalModelling vol 34 no 2 pp 309ndash313 2010

[8] C T Ng J-B Wang T C E Cheng and L L Liu ldquoAbranch-and-bound algorithm for solving a two-machine flowshop problemwith deteriorating jobsrdquoComputers ampOperationsResearch vol 37 no 1 pp 83ndash90 2010

[9] T C E Cheng W-C Lee and C-C Wu ldquoSingle-machinescheduling with deteriorating jobs and past-sequence-dependent setup timesrdquo Applied Mathematical Modelling vol35 no 4 pp 1861ndash1867 2011

[10] N Raman F B Talbot and R V Rachamadugu ldquoDue datebased scheduling in a general flexible manufacturing systemrdquoJournal of Operations Management vol 8 no 2 pp 115ndash1321989

[11] L K Church and R Uzsoy ldquoAnalysis of periodic and event-driven rescheduling policies in dynamic shopsrdquo InternationalJournal of Computer Integrated Manufacturing vol 5 pp 153ndash163 1992

[12] A K Jain and H A Elmaraghy ldquoProduction schedul-ingrescheduling in flexible manufacturingrdquo International Jour-nal of Production Research vol 35 no 1 pp 281ndash309 1997

[13] G E Vieira J W Herrmann and E Lin ldquoReschedulingmanufacturing systems a framework of strategies policies andmethodsrdquo Journal of Scheduling vol 6 no 1 pp 39ndash62 2003

[14] B B Yang ldquoSingle machine rescheduling with new jobs arrivalsand processing time compressionrdquo International Journal ofAdvanced Manufacturing Technology vol 34 no 3-4 pp 378ndash384 2007

[15] N G Hall and C N Potts ldquoRescheduling for new ordersrdquoOperations Research vol 52 no 3 pp 440ndash453 2004

[16] J J Yuan and Y D Mu ldquoRescheduling with release dates tominimize makespan under a limit on the maximum sequencedisruptionrdquo European Journal of Operational Research vol 182no 2 pp 936ndash944 2007

[17] C L Zhao and H Y Tang ldquoRescheduling problems withdeteriorating jobs under disruptionsrdquo Applied MathematicalModelling vol 34 no 1 pp 238ndash243 2010

[18] H Hoogeveen C Lente and V Trsquokindt ldquoRescheduling for neworders on a single machine with setup timesrdquo European Journalof Operational Research vol 223 no 1 pp 40ndash46 2012

[19] J Du and J Y-T Leung ldquoMinimizing total tardiness on onemachine is NP-hardrdquo Mathematics of Operations Research vol15 no 3 pp 483ndash495 1990

[20] R L Graham E L Lawler J K Lenstra and A H GRinnooy Kan ldquoOptimization and approximation in determin-istic sequencing and scheduling a surveyrdquo Annals of DiscreteMathematics vol 5 pp 287ndash326 1979

[21] A Kononov and S Gawiejnowicz ldquoNP-hard cases in schedulingdeteriorating jobs on dedicated machinesrdquo Journal of the Oper-ational Research Society vol 52 no 6 pp 708ndash717 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Mathematical Problems in Engineering

Input Given 119901119895 119889119895for 119895 = 1 119899 119896 and 120587

lowast where 119896 lt 119862max (see Lemma 1)Indexing Index the jobs of 119869

119873in the EDD order

Box 3

Input Given 119901119895 119889119895for 119895 = 1 119899 119896 and 120587

lowast where 119896 le 1198990119862max

Indexing Index the jobs of 119869119873in the EDD order

Box 4

time disruption is equal to 120575 The dynamic programmingprocedure can now be stated as follows

Algorithm 11

Step 1 (Initialization) Set 119891(119894 119895 120575) = 0 for (119894 119895 120575) = (0 0 0)

and 119891(119894 119895 120575) = infin for (119894 119895 120575) = (0 0 0) 119894 = 1 1198990 119895 =

1 119899119873 and 119894 = 0 119896

Step 2 (Recurrence Relation)

119891 (119894 119895 120575)= min

119891(119894 minus 1 119895 120575 minus

1198990+119895

sum

ℎ=1198990+1

119901[ℎ]

) +max 119862119894minus 119889119894 0

119891 (119894 119895 minus 1 120575) +max 1198621198990+119895

minus 1198891198990+119895

0

(7)

where 119901[ℎ]

is the actual processing time of job 119869ℎ 119862119895denotes

the completion time of job 119869119895

Step 3 (Optimal Solution) Calculate the optimal solutionvalue min

0le120575le119896119891(1198990 119899119873 120575)

In the recurrence relation the first term in the minimiza-tion corresponds to the case where the partial schedule endswith job 119869

119894isin 1198690 Because 119869

119895jobs of 119869

119873appear before job 119869

119894in

such a partial schedule the increase in total time disruptionis equal to sum

1198990+119895

ℎ=1198990+1119901[ℎ] The second term corresponds to the

case where the partial schedule ends with job 1198691198990+119895

isin 119869119873

Theorem 12 For the 1 | 119901119894le 119901119895rArr 119889119894le 119889119895 sumΔ

119895(120587lowast) le

119896 119901119895(119886 + 119887119904

119895) | sum119879

119895problem Algorithm 11 finds an optimal

schedule in 119874(1198992

0119899119873119862max) time

Proof The proof of optimality of Algorithm 11 is similar toTheorem 8 regarding the time complexity Because 119894 le 119899

0

119895 le 119899119873 and 120575 le 119896 le 119899

0119862max there are 119874(119899

2

0119899119873119862max) values

of the state variables Step 1 requires 119874(119899119873log 119899119873) Similar

arguments to those in the proof of Theorem 8 show that theoverall time complexity of Algorithm 11 is119874(119899

2

0119899119873119862max)

4 Conclusions

In this paper we studied the issue of rescheduling to allowthe unexpected arrival of new jobs and took into account

the effect of the disruption on a previously planned opti-mal schedule The main contribution of this paper is thatwe develop the machine rescheduling scheduling problemsagreeable job parameters under deterioration and disruptionRescheduling means to schedule the jobs again togetherwith a set of new jobs Deteriorating job means that theactual processing time of a job is an increasing functionof its starting time When the processing time and duedate of jobs are agreeable we considered some problems tominimize total tardiness under a limit on two disruptionconstraints sequence disruption and time disruption Weproposed polynomial time algorithms or some dynamicprogramming algorithms for each problem Future researchmay stimulate rescheduling models to mitigate the effectsof the disruptions that occur frequently in manufacturingpractice

Acknowledgments

The authors would like to thank the authors of the referencesfor enlightening themThis paper was supported by NationalNatural Science Foundation of China (71272085) TianyuanFund for Mathematics (11226237) the Humanities and SocialSciences Program of the Ministry of Education of China(12YJA630135) the Foundation of Chongqing EducationCommission (KJ120624) and the Key Project Fundation ofChongqing Normal University (2011XLZ05)

References

[1] S Browne and U Yechiali ldquoScheduling deteriorating jobs on asingle processorrdquo Operations Research vol 38 no 3 pp 495ndash498 1990

[2] A S Kunnathur and S K Gupta ldquoMinimizing the makespanwith late start penalties added to processing times in a singlefacility scheduling problemrdquo European Journal of OperationalResearch vol 47 no 1 pp 56ndash64 1990

[3] G Mosheiov ldquoScheduling jobs under simple linear deteriora-tionrdquoComputers andOperations Research vol 21 no 6 pp 653ndash659 1994

[4] T C E Cheng Q Ding and B M T Lin ldquoA concise surveyof schedulingwith time-dependent processing timesrdquoEuropeanJournal of Operational Research vol 152 no 1 pp 1ndash13 2004

[5] S Gawiejnowicz Time-Dependent Scheduling Springer BerlinGermany 2008

Mathematical Problems in Engineering 7

[6] D Biskup and JHerrmann ldquoSingle-machine scheduling againstdue dates with past-sequence-dependent setup timesrdquo Euro-pean Journal of Operational Research vol 191 no 2 pp 587ndash5922008

[7] J-B Wang and Q Guo ldquoA due-date assignment problem withlearning effect and deteriorating jobsrdquo Applied MathematicalModelling vol 34 no 2 pp 309ndash313 2010

[8] C T Ng J-B Wang T C E Cheng and L L Liu ldquoAbranch-and-bound algorithm for solving a two-machine flowshop problemwith deteriorating jobsrdquoComputers ampOperationsResearch vol 37 no 1 pp 83ndash90 2010

[9] T C E Cheng W-C Lee and C-C Wu ldquoSingle-machinescheduling with deteriorating jobs and past-sequence-dependent setup timesrdquo Applied Mathematical Modelling vol35 no 4 pp 1861ndash1867 2011

[10] N Raman F B Talbot and R V Rachamadugu ldquoDue datebased scheduling in a general flexible manufacturing systemrdquoJournal of Operations Management vol 8 no 2 pp 115ndash1321989

[11] L K Church and R Uzsoy ldquoAnalysis of periodic and event-driven rescheduling policies in dynamic shopsrdquo InternationalJournal of Computer Integrated Manufacturing vol 5 pp 153ndash163 1992

[12] A K Jain and H A Elmaraghy ldquoProduction schedul-ingrescheduling in flexible manufacturingrdquo International Jour-nal of Production Research vol 35 no 1 pp 281ndash309 1997

[13] G E Vieira J W Herrmann and E Lin ldquoReschedulingmanufacturing systems a framework of strategies policies andmethodsrdquo Journal of Scheduling vol 6 no 1 pp 39ndash62 2003

[14] B B Yang ldquoSingle machine rescheduling with new jobs arrivalsand processing time compressionrdquo International Journal ofAdvanced Manufacturing Technology vol 34 no 3-4 pp 378ndash384 2007

[15] N G Hall and C N Potts ldquoRescheduling for new ordersrdquoOperations Research vol 52 no 3 pp 440ndash453 2004

[16] J J Yuan and Y D Mu ldquoRescheduling with release dates tominimize makespan under a limit on the maximum sequencedisruptionrdquo European Journal of Operational Research vol 182no 2 pp 936ndash944 2007

[17] C L Zhao and H Y Tang ldquoRescheduling problems withdeteriorating jobs under disruptionsrdquo Applied MathematicalModelling vol 34 no 1 pp 238ndash243 2010

[18] H Hoogeveen C Lente and V Trsquokindt ldquoRescheduling for neworders on a single machine with setup timesrdquo European Journalof Operational Research vol 223 no 1 pp 40ndash46 2012

[19] J Du and J Y-T Leung ldquoMinimizing total tardiness on onemachine is NP-hardrdquo Mathematics of Operations Research vol15 no 3 pp 483ndash495 1990

[20] R L Graham E L Lawler J K Lenstra and A H GRinnooy Kan ldquoOptimization and approximation in determin-istic sequencing and scheduling a surveyrdquo Annals of DiscreteMathematics vol 5 pp 287ndash326 1979

[21] A Kononov and S Gawiejnowicz ldquoNP-hard cases in schedulingdeteriorating jobs on dedicated machinesrdquo Journal of the Oper-ational Research Society vol 52 no 6 pp 708ndash717 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 7

[6] D Biskup and JHerrmann ldquoSingle-machine scheduling againstdue dates with past-sequence-dependent setup timesrdquo Euro-pean Journal of Operational Research vol 191 no 2 pp 587ndash5922008

[7] J-B Wang and Q Guo ldquoA due-date assignment problem withlearning effect and deteriorating jobsrdquo Applied MathematicalModelling vol 34 no 2 pp 309ndash313 2010

[8] C T Ng J-B Wang T C E Cheng and L L Liu ldquoAbranch-and-bound algorithm for solving a two-machine flowshop problemwith deteriorating jobsrdquoComputers ampOperationsResearch vol 37 no 1 pp 83ndash90 2010

[9] T C E Cheng W-C Lee and C-C Wu ldquoSingle-machinescheduling with deteriorating jobs and past-sequence-dependent setup timesrdquo Applied Mathematical Modelling vol35 no 4 pp 1861ndash1867 2011

[10] N Raman F B Talbot and R V Rachamadugu ldquoDue datebased scheduling in a general flexible manufacturing systemrdquoJournal of Operations Management vol 8 no 2 pp 115ndash1321989

[11] L K Church and R Uzsoy ldquoAnalysis of periodic and event-driven rescheduling policies in dynamic shopsrdquo InternationalJournal of Computer Integrated Manufacturing vol 5 pp 153ndash163 1992

[12] A K Jain and H A Elmaraghy ldquoProduction schedul-ingrescheduling in flexible manufacturingrdquo International Jour-nal of Production Research vol 35 no 1 pp 281ndash309 1997

[13] G E Vieira J W Herrmann and E Lin ldquoReschedulingmanufacturing systems a framework of strategies policies andmethodsrdquo Journal of Scheduling vol 6 no 1 pp 39ndash62 2003

[14] B B Yang ldquoSingle machine rescheduling with new jobs arrivalsand processing time compressionrdquo International Journal ofAdvanced Manufacturing Technology vol 34 no 3-4 pp 378ndash384 2007

[15] N G Hall and C N Potts ldquoRescheduling for new ordersrdquoOperations Research vol 52 no 3 pp 440ndash453 2004

[16] J J Yuan and Y D Mu ldquoRescheduling with release dates tominimize makespan under a limit on the maximum sequencedisruptionrdquo European Journal of Operational Research vol 182no 2 pp 936ndash944 2007

[17] C L Zhao and H Y Tang ldquoRescheduling problems withdeteriorating jobs under disruptionsrdquo Applied MathematicalModelling vol 34 no 1 pp 238ndash243 2010

[18] H Hoogeveen C Lente and V Trsquokindt ldquoRescheduling for neworders on a single machine with setup timesrdquo European Journalof Operational Research vol 223 no 1 pp 40ndash46 2012

[19] J Du and J Y-T Leung ldquoMinimizing total tardiness on onemachine is NP-hardrdquo Mathematics of Operations Research vol15 no 3 pp 483ndash495 1990

[20] R L Graham E L Lawler J K Lenstra and A H GRinnooy Kan ldquoOptimization and approximation in determin-istic sequencing and scheduling a surveyrdquo Annals of DiscreteMathematics vol 5 pp 287ndash326 1979

[21] A Kononov and S Gawiejnowicz ldquoNP-hard cases in schedulingdeteriorating jobs on dedicated machinesrdquo Journal of the Oper-ational Research Society vol 52 no 6 pp 708ndash717 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of