research article rescheduling problems with agreeable job
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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
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
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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
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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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
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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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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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
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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