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Research ArticleParallel-Batch Scheduling and Transportation Coordinationwith Waiting Time Constraint
Hua Gong Daheng Chen and Ke Xu
College of Science Shenyang Ligong University Shenyang 100159 China
Correspondence should be addressed to Daheng Chen dahengchen163com
Received 23 January 2014 Accepted 20 February 2014 Published 15 April 2014
Academic Editors J G Barbosa and D Oron
Copyright copy 2014 Hua Gong et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
This paper addresses a parallel-batch scheduling problem that incorporates transportation of rawmaterials or semifinished productsbefore processing with waiting time constraint The orders located at the different suppliers are transported by some vehicles to amanufacturing facility for further processing One vehicle can load only one order in one shipment Each order arriving at thefacility must be processed in the limited waiting time The orders are processed in batches on a parallel-batch machine where abatch contains several orders and the processing time of the batch is the largest processing time of the orders in it The goal is tofind a schedule to minimize the sum of the total flow time and the production cost We prove that the general problem is NP-hardin the strong sense We also demonstrate that the problem with equal processing times on the machine is NP-hard Furthermore adynamic programming algorithm in pseudopolynomial time is provided to prove its ordinarily NP-hardness An optimal algorithmin polynomial time is presented to solve a special case with equal processing times and equal transportation times for each order
1 Introduction
A supply chain is made of all the stages of value creationsuch as supply production and distribution In logisticsmanagement there are usually transportation of raw mate-rials or semifinished products and distribution of productsResearches on supply chainmanagement focus on developingstrategies to help companies to improve optimal chain-wide performance by proper coordination of the differentstages of supply chains For an order transportation of rawmaterials or semifinished products and production are twokey operations in the supply chain system A schedulingproblem with the limited waiting time occurs when theprocessing of orders that finished transportation has to startin a given time There are several industries where the trans-portation and production scheduling problem is influencedby the limited waiting time Examples include fresh foodand chemical and automobile industries For instance incase of fresh food production canning operationmust followcooking operation to ensure freshness after transportation ofraw materials Additional applications can be found in theadvanced automobile manufacturing environments such asjust-in-time systems Auto parts of orders are transported
from suppliers to producers for further assembling Theproducers must process the orders arrived at the factoryin a given time in order to decrease the inventory leveland increase the production level This situation is relevantto logistics and supply chain management and specificallyaddresses how to balance the transportation rate of rawmaterials and the production rate
Motivated by the aforementioned problem in logisticsmanagement this paper describes a coordinated model forscheduling transportation of raw materials or semifinishedproducts and parallel-batch production with waiting timeconsideration There is a set of orders located at the differentsuppliers The orders are transported by some vehicles to amanufacturing facility to be further processed But only oneorder can be transported at a time Each order arriving at thefacilitymust be processed in its givenwaiting timeTheordersare processed in batches on a parallel-batch machine wherea batch contains several orders and the processing time of thebatch is the largest processing time of the orders in itThe goalis to find a schedule tominimize the sumof the total flow timeand the production cost
Production scheduling problems with transportationconsiderations have been studied by a number of researchers
Hindawi Publishing Corporatione Scientific World JournalVolume 2014 Article ID 356364 8 pageshttpdxdoiorg1011552014356364
2 The Scientific World Journal
The parallel-batch machine scheduling problems have beenpreviously studied in othermanufacturing systems especiallyburn-in operations in the very large-scale integrated circuitmanufacturing (eg [1ndash3]) We briefly discuss some workrelated to integrated production scheduling and transporta-tion decisions Some research (see [4ndash7]) has been done fortwo machine flow shop problems featuring transportation ofsemifinished products and batching processing We mentionthe studies in [4ndash7] however the problems there do notconsider parallel-batch production or they consider anotherkind of batching with a constant processing timeThemodelsin [8 9] consider the coordinated scheduling problemswith two-stage transportation and machine production thatincorporate the scheduling of jobs and the pickup of the rawmaterials from the warehouse and the delivery arrangementof the finished jobs The above scheduling problems donot consider the waiting time constraints Another line ofproduction scheduling models with transportation decisionsfocuses on the delivery of finished jobs to customers witha limited number of vehicles Interested readers for thedelivery coordination are referred to the recent review in [10]The parallel-batch machine scheduling problems with batchdelivery are considered in [11 12] All of the papers reviewedin this section deal in some way with the coordination ofproduction scheduling and transportation at logistics andoperations area but none of them addresses and deals withthe problem from the limited waiting time point of viewWe not only consider the scheduling problem involvingtransportation capacity and transportation times but alsotake into account the order waiting time constraint and thebatch capacity Here the coordination of batch processingtransportation and waiting time constraints is a decisionabout whether or not to reduce production cost and improvemachine utilization
The remainder of this paper is organized as follows Inthe next section we introduce the notation to be used anddescribe the model In Section 3 we analyze the optimalproperties of the problem and prove strong NP-hardness ofthe general problem In Section 4 a dynamic programmingalgorithm is provided to solve the problem with equalprocessing times and we prove that this case is NP-hard inthe ordinary sense Section 5 deals with a case with equalprocessing times and equal transportation times and showshow to find the optimal schedule The last section contains aconclusion and some suggestions for future research
2 Description and Notation
Our problem is formally stated as follows Given a set of119899 orders (including some semifinished jobs or raw materialin one order) 119869 = 119869
1 119869
119899 which are located at the
different suppliers The orders need to be transported to amanufacturing facility and processed on a single parallel-batchmachine Only119898 vehicles are available for transportingthe orders which can load only one order in one shipmentThe vehicles are initially located at a transportation center Let119905119895denote the transportation time of order 119869
119895from the supplier
to the manufacturing facility The transportation time from
the factory back to the center is negligible In the productionpart each order 119869
119895requires a processing time of 119901
119894on the
parallel-batch machine The preemption of orders is notallowedThe parallel-batch machine can process a number oforders simultaneously as a batch as long as the total numberof orders in the batch does not exceed themachine capacity 119888The processing time of a batch is the longest processing timeof all orders in it Orders processed in the same batch havethe same completion time that is their common start time(the start time of the batch in which they are contained) plusthe processing time of the batch Suppose that loading andunloading times are included in the transportation times andthe processing times of orders Associated with each order is119908119895 which is the waiting time of an order for a period time
from arriving at the machine to starting its processing onthe machine Here 119908
119895is restricted by a constant 119882 where
max119895119901119895
le 119882 le sum 119901119895 such that the schedule has a feasible
solution at least Each batch requires a processing cost on theparallel-batch machine which reflects the production costThe goal is to find a feasible schedule to minimize the sumof the total flow time and the total processing cost
Decision VariablesConsider the following
119879119906 the total running time of vehicle u
119903119895 the arrival time of order 119869
119895on the machine
119904119895 the starting time of order 119869
119895on the machine
119908119895 the waiting time of order 119869
119895on the machine 119908
119895=
119904119895
minus 119903119895
le 119882119861119897 batch l
119887119897 the number of orders in 119861
119897
119875(119861119897) the processing time of 119861
119897on the machine
119875(119861119897) = max
119895isin119861119897119901119895
119878119897 the starting time of 119861
119897on the machine
119863119897 the completion time of 119861
119897on the machine
119862119895 the flow time of order 119869
119895on the machine
119909 the number of batches119883(119909) the total processing cost a nondecreasingfunction of x119865(120587) = sum
119899
119895=1119862119895
+ 119883(119909) the objective function of aschedule 120587
We follow the commonly used three-field notation120572|120573|120574introduced to denote the problem under study The problemis denoted by 119881119898 rarr batch|119908
119895| sum 119862119895
+ 119883(119909) where 120572 fieldindicates that the orders are first transported by 119898 vehiclesand then processed on the parallel-batch machine 120573 fielddescribes the order restrictive requirement and 120574 defines theobjective function to be minimized
Specifically the tradeoff between processing and trans-portation gives rise to the sequence of transportation andthe batch composition decisions The schedule may producemore batches in order to satisfy thewaiting time requirementsof orders This leads to the increase of the total processing
The Scientific World Journal 3
Table 1
Order 1 2 3 4 5 6 7 8119905119895
1 1 2 2 2 3 4 4119901119895
2 3 4 3 3 3 4 4
cost on the machine On the other hand if we create veryfew batches then we may not obtain a feasible scheduledue to waiting time constraint Therefore minimizing ofthe objective for the problem is achieved a feasible schedulefor sequencing and batching to tradeoff the total flow timeand the processing cost The following example gives anillustration of this observation
Example 1 Consider a set of 8 orders with their transporta-tion times and processing times as shown below for theproblem 119881119898 rarr batch|119908
119895| sum 119862119895
+ 119883(119909)
Suppose 119898 = 2 119888 = 4 119882 = 4 and 119883(119909) = 3119909 (seeTable 1)
One vehicle transports the orders in turn 1198691 1198693 1198695 and
1198697 Another vehicle transports the orders in turn 119869
2 1198694 1198696
and 1198698 The constructed schedule 120587 consists of three batches
1198611
= 1198691 1198692 1198693 1198694 1198612
= 1198695 1198696 and 119861
3= 1198697 1198698 and has
an objective value of 85 If another constructed schedule 1205871015840
contains two batches 1198611
= 1198691 1198692 1198693 1198694 1198612
= 1198695 1198696 1198697 1198698
then the waiting time of the order 1198695is 5 Although schedule
1205871015840 has a smaller processing cost it is not feasible From
these two schedules we see that a minimized objective canbe obtained from a proper combination of the sequenceof transportation and the batching of processing under thewaiting time constraints
3 The Problem 119881119898 rarr batch|119908119895| sum 119862119895
+ 119883(119909)
In this section we analyze two properties of the schedulingproblem and prove that Vmrarr batch|119908
119895| sum 119862119895
+ 119883(119909) is NP-hard in the strong sense by a reduction from the 3-PartitionProblem which is well known to be NP-hard in the strongsense [13]
3-Partition Problem (3-PP) Given 3h items 119867 = 1 2
3ℎ each item 119894 isin 119867 has a positive integer size 119886119894satisfying
1198864 lt 119886119894
lt 1198862 and sum3ℎ
119894=1119886119894
= ℎ119886 for some integer 119886The question asked is whether there are ℎ disjoint subsets1198671 1198672 119867
ℎof 119867 such that each subset contains exactly
three items and its total size sum119894isin119867119895
119886119894= 119886 119895 = 1 2 ℎ
Lemma2 For the problemVmrarr batch|119908119895| sum 119862119895+119883(119909) there
exists an optimal schedule without idle time between orders onany vehicle
The proof of Lemma 2 is straightforward and is omittedThe following result describes a candidate set of possible
processing times
Lemma3 For the problemVmrarr batch|119908119895| sum 119862119895+119883(119909) there
exists an optimal schedule in which each processing on theparallel-batch machine is made either at the arrival time of
an order on the machine or immediately when the machinebecomes available
Proof Assume that there is a process which is scheduledneither at the arrival time of an order nor at a time when themachine becomes available This process can be changed tothe latest earlier time which fits either of those conditionsSince the same orders can be processed at that earlier timeand there are no additional processes the objective value isnot increased
Theorem 4 The problem Vmrarr batch|119908119895| sum 119862119895
+ 119883(119909) isstrongly NP-hard even if 119898 = 2
Proof To show the NP-hardness of the scheduling problemwe establish the following polynomial time reduction fromthe 3-PP Given a 3-PP instance we construct an instance for119881119898 rarr batch|119908
119895| sum 119862119895
+ 119883(119909) as follows
Number of Orders Consider
119899 = 7 ℎ + 7 (1)
Ordinary Orders Consider
119905119895
= lceil
119895
6
rceil 119886lfloor(12)(119895+1)rfloor
119901119895
= 2 (lceil
119895
6
rceil + 1) 119886
119895 isin 119867 = 1 2 6ℎ
(2)
Auxiliary Orders Consider
119905119895
= 2 (119895 minus 6ℎ) 119886 119895 isin 119866 = 6ℎ + 1 7ℎ
119901119895
= 1199016ℎ+119894
= 2 (119894 + 1) (1198863119894minus2
+ 1198863119894minus1
+ 1198863119894
)
119895 isin 119866 = 6ℎ + 1 7ℎ
119905119895
= 0 119901119895
= 2119886 119895 isin 119884 = 7ℎ + 1 7ℎ + 7
(3)
Machine Capacity Consider
119888 = 7 (4)
Limited Waiting Time Consider
119882 = 2119886ℎ (5)
Processing Cost Function Consider
119883 (119909) = [
5
3
119886 (ℎ + 1) (ℎ + 2) (ℎ + 3)]
119909minusℎ
(6)
Threshold Value Consider
119910 = 4119886 (ℎ + 1) (ℎ + 2) (ℎ + 3) (7)
Clearly in a solution to this instance of 119881119898 rarr
batch|119908119895| sum 119862119895+ 119883(119909) with the objective value not exceeding
4 The Scientific World Journal
Machine Y
0 2a 6a 12a
middot middot middot
middot middot middot
middot middot middot
7h
1ndash6 7ndash12 13ndash18
ah(h + 1) a(h + 1)(h + 2)
Vehicle 1
Vehicle 2
6h + 1 6h + 2 6h + 3
1ndash6 6h + 1
(6h minus 5)ndash6h
(6h minus 11)ndash(6h minus 6) 7h minus 1 (6h minus 5)ndash6h 7h7ndash12 6h + 2
Figure 1 The schedule of Theorem 4
y we first prove the following properties (1) since 119905119895
= 0 for119895 isin 119884 all the orders of 119884 as the first batch must be processedon the machine in the time interval [0 2119886] (2) there will beexactly ℎ + 1 batches in the optimal schedule and each batchcontains 7 orders Suppose that there are q (more than ℎ + 1)batches in the scheduleThen 7ℎ + 7 orders need at least ℎ + 1
batches due to 119888 = 7 If 119902 = ℎ + 2 then the total processingcost is 119883(119909) = 119883(ℎ + 2) = [(53)119886(ℎ + 1)(ℎ + 2)(ℎ + 3)]
2gt 119910
Thus all the orders of 119867 cup 119866 are divided into ℎ batches thatis all the batches are full
We will show that the 3-PP instance has a solution if andonly if there is a schedule 120587 for the scheduling instance suchthat its objective value of 120587 is no more than y
rarrAssume that there is a solution to the 3-PP instance1198671 1198672 119867
ℎ then there is a schedule to our problem with
an objective value of nomore than119910We construct a schedule120587 for our problem as shown in Figure 1
In this schedule two vehicles begin to transport ordersat time point 0 All the orders of 119884 are processed first on themachine as the first batch In the interval [119886119897(119897 minus 1) 119886119897(119897 + 1)]vehicle 1 transports the orders of 119867
119897 119867119897
= 6119897 minus 5 6119897 minus
4 6119897 vehicle 2 transports 1198696ℎ+119897
for 119897 = 1 2 ℎ Since1199056ℎ+119897
= 2119897119886 and sum119895isin119867119897
119905119895
= 2119897119886 1198696ℎ+119897
and the last order of 119867119897
arrive at the machine simultaneously for 119897 = 1 2 ℎ then1198696ℎ+119897
and the corresponding orders of 119867119897form batch 119861
119897 The
waiting time of each order in 119861119897is smaller than119882 Hence the
starting time of 119861119897on the machine is 119886119897(119897 + 1) the completion
time of 119861119897is 119886(119897 + 1)(119897 + 2) for 119897 = 0 1 2 ℎ It is easy to
see that the above schedule is optimal and the objective valueis y
larrOn the contrary suppose that there exists a schedulefor the constructed instance with the objective value notexceeding y
Fact 1 In the schedule 119861119897contains an auxiliary order 119869
6ℎ+119897
for 119897 = 1 2 ℎAssume that there is some batch 119861
119897that contains more
than one auxiliary order Assume that 119861119897contains two
auxiliary jobs 1198696ℎ+119897
and 1198696ℎ+119896
then there must be some batch119861119896only consisting of the ordinary jobs 119896 = 119897 Consider the
following two cases
Case 1 If 119896 = 119897 minus 1 then it is shown that 119861119897minus1
=
1198696119897minus11
1198696119897minus6
1198691198951015840 | 119895
1015840isin 119867119897 The corresponding flow
time of each order in 119861119897minus1
satisfies 119862119895
= 119886119897(119897 + 1) + 2119886
for119895 isin 119861119897minus1
The flow times of orders in each batch after 119861119897minus1
will be greater than 119886(119897 + 1)(119897 + 2) + 2119886 It is easy to see thatsum 119862119895
= 4119886(ℎ + 1)(ℎ + 2)(ℎ + 3) + 14(ℎ minus 119897 + 2)119886 2 le 119897 le ℎHence 1198651015840 = sum 119862
119895+ 119883(ℎ + 1) gt 119910
Case 2 If 119896 = 119897+1 then it can be shown that two vehiclesmusttransport two auxiliary orders of batch 119861
119897 respectively Due
to the limited waiting time of orders the completion time ofbatch 119861
119897on the machine is greater than 119886(119897 + 1)(119897 + 2) + 2119886
Furthermore it is easy to see that sum 119862119895
gt 4119886(ℎ + 1)(ℎ + 2)(ℎ +
3) + 14(ℎ minus 119897 + 1)119886 1 le 119897 le ℎ minus 1 which implies 1198651015840
gt 119910
Fact 2 Each batch 119861119897must contain 6 orders of 119867
119897= 6119897 minus
5 6119897 minus 4 6119897 such that sum119895isin119867119897
119905119895
= 2119897119886 for 119897 = 1 2 ℎ
By the above argument since each batch is full we canshow that one vehicle transports the orders of 119867
119897one by
one and another vehicle transports 1198696ℎ+119897
in the time interval[119886119897(119897 minus 1) 119886119897(119897 + 1)] The orders 119869
6ℎ+119897 119867119897 as 119861119897are processed
on themachine in the time interval [119886119897(119897+1) 119886(119897+1)(119897+2)] Itmust be true thatsum
119895isin119867119897119905119895
= 2119897119886 for 119897 = 1 2 ℎ Otherwiseif sum119895isin119867119897
119905119895
lt 2119897119886 then 1199016ℎ+119897
= 2(119897 + 1) sdot (1198863119897minus2
+ 1198863119897minus1
+ 1198863119897
) lt
2(119897+1)119886We can obtain some batch119861119896whose processing time
is greater than 2(119896 + 1)119886 119896 = 119897 Hence 1198651015840
gt 119910 which is acontradiction
Therefore a partition for the set 119867 is obtained by lettingthe elements be corresponding to batch 119861
119897 Then it is easy to
see that 119902 = ℎ + 1 and ℎ batches 1198611 119861
ℎform a solution to
the 3-PP instance The proof is concluded
Theorem 4 indicates that the existence of a polynomialtime algorithm to solve the scheduling model is unlikelySince the general problem is strongly NP-hard we nextconsider two special cases of the problem
4 The Problem 119881119898 rarr batch|119901119895
= 119901
119908119895| sum 119862119895
+ 119883(119909)
We now turn our attention to a special case with identicalprocessing time condition on themachine In this section weprove that the problem119881119898 rarr batch|119901
119895= 119901 119908
119895| sum 119862119895+119883(119909)
is ordinarily NP-hard by a reduction of the Partition Problem[13] which is a known NP-hard problem and provides adynamic programming algorithm in pseudopolynomial time
The Scientific World Journal 5
Suppose a feasible schedule 120587 = 1198611 1198612 119861119909 with 119909 isin
lceil119899119888rceil 119899 batches for the problem 119881119898 rarr batch|119901119895
=
119901 119908119895| sum 119862119895
+ 119883(119909) We now present some properties for theproblem
Lemma 5 For Vmrarr batch|119901119895
= 119901 119908119895| sum 119862119895
+ 119883(119909) thereexists an optimal schedule 120587
lowastsuch that all orders assigned intothe same vehicle are scheduled in the nondecreasing sequenceof their transportation times
Proof Assume that orders 119869119894and 119869119895are assigned to the same
vehicle 119869119894is followed by 119869
119895immediately such that 119905
119894ge 119905119895in
120587lowast Let 120587
1015840 be a schedule obtained by swapping 119869119894and 119869119895 In 1205871015840
it is easy to see that 119903119894gt 1199031015840
119895and 1199031015840
119894= 119903119895 Regardless of whether
119869119894and 119869119895are processed in the same batch or not the starting
times of 119869119894and 119869119895on the machine cannot increase We have
119865(120587lowast) ge 119865(120587
1015840)
Lemma 6 For any scheduleof the problem Vmrarr batch|119901119895
=
119901 119908119895| sum 119862119895+ 119883(119909) the starting time of batch 119861
119897satisfies 119878
119897+1ge
119878119897+ 119901 for 119897 = 1 2 119909
Proof It is easily obtained based on Lemma 3
Lemma 7 For any schedule of the problem Vmrarr batch|119901119895
=
119901 119908119895| sum 119862119895
+ 119883(119909) 119861119897contains all orders which have arrived
at the machine if the number of the orders is no more than thecapacity 119888 in the time interval (119878
119897minus1 119878119897]
Proof Assume that the batches are numbered in accordancewith their start times and the number of orders in eachbatch is smaller than the machine capacity 119888 Suppose that119869119895is the first order assigned in 119861
119897+1in 120587lowast and the arrival
time of 119869119895on the machine satisfies 119903
119895le 119878119897 Let 120587
1015840 bea new schedule obtained by simply assigning 119869
119895to 119861119897
Then 1198621015840
119895= 119878119897
+ 119901 lt 119878119897+1
+ 119901 le 119862119895 The schedule
of the remaining orders in 1205871015840 are the same as in 120587
lowast Itis obvious that 119865(120587
1015840) le 119865(120587
lowast) We can see that there
exists an optimal schedule in which all batches consist ofa number of orders which finish processing contiguously
We will show that the problem with equal processingtimes is NP-hard by a reduction from the Partition Problem
Theorem 8 The problem Vmrarr batch|119901119895
= 119901 119908119895| sum 119862119895
+
119883(119909) is NP-hard even if 119898 = 2
Proof Weprove this result by reducing the Partition Problemto Vmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909) The PartitionProblem (PP) can be stated as follows given ℎ items 119867 =
1 2 ℎ each item 119894 isin 119867 has a positive integer size 119886119894
such thatsumℎ
119894=1119886119894= 2119886 for some integer 119886The question asked
is whether there are two disjoint subsets1198671and119867
2 such that
sum119894isin1198671
119886119894= sum119894isin1198672
119886119894= 119886
We construct a corresponding instance of the problemVmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909) as follows
Number of Orders Consider
119899 = 2ℎ (8)
Transportation Times Consider
119905119895
= 119886119895 119895 isin 119867 = 1 2 ℎ
119905119895
= 0 119895 isin 119884 = ℎ + 1 ℎ + 2 2ℎ
(9)
Processing Times Consider
119901119895
= 119886 119895 = 1 2 2 ℎ (10)
Limited Waiting Time Conider
119882 = 119886 (11)
Processing Cost Function Consider
119883 (119909) = 3119886ℎ119909 (12)
Machine Capacity Consider
119888 = ℎ (13)
Threshold Value Consider
119910 = 9119886ℎ (14)
First it is easy to see that in a solution to this instance ofVmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895+ 119883(119909) with the objective value
not exceeding 119910 We first prove the following property since119905ℎ+1
= sdot sdot sdot = 1199052ℎ
= 0 all the orders of 119884 as the first batch mustbe processed on the machine in the time interval [0 a] Nowwe prove that there is a solution to the constructed instanceof Vmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909) with total cost notexceeding119910 if and only if there is a solution to the PP instance
rarr If there is a solution to the PP instance we show thatthere is a schedule for the above-constructed instance withan objective value of no more than 119910 Let 119867
1and 119867
2be
subsets of119867 that solves the PP instance Vehicle 1 and Vehicle2 transport the orders of119867
1and the orders of119867
2 respectively
Let119879119906be the total running time of vehicleu for119906 = 1 2 Since
sum119895isin1198671
119905119895
= sum119895isin1198672
119905119895
= 119886 we have 1198791
= 1198792
= 119886 In fact thecompletion time of the first batch on themachine is 119886We canobtain that the waiting time of each order on the machine isnot greater than 119886 Let 119861
1= 119884 and 119861
2= 119867 It is easy to see
that 1198871
= 1198872
= ℎ and the total flow time of all the orders is3ha Hence the objective is y
larrGiven a schedule with an objective value not exceedingy then there must be a solution to the PP instance We canobtain that all the orders in 119867 are processed in the secondbatch on the machine that is all the orders are divided intotwo batches and 119887
1= 1198872
= ℎ Suppose that the number ofthe batches is greater than 2 Note that even if the waitingtime constraint is ignored the objective value is 119865 ge 119886ℎ +
sum119895isin119867
119862119895
+ 9119886ℎ gt 119910 which is a contradiction Thus 120587 =
1198611 1198612 and 119861
1= 119884 and 119861
2= 119867
6 The Scientific World Journal
If 1198791
= sum119895isin1198671
119905119895
lt 119886 then it implies 1198792
= sum119895isin1198672
119905119895
gt 119886Hence the starting time of batch 119861
2is 1198782
= 1198792 It implies that
the total flow time is sum 119862119895
= 119886ℎ + (1198782
+ 119886)ℎ gt 3119886ℎ We have119865 gt 119910 which is a contradiction Thus sum
119895isin1198671119886119895
= sum119895isin1198672
119886119895
=
119886
Let 119879 = sum119899
119895=1119905119895 In the following we derive a dynamic
programming algorithm in pseudopolynomial time to solvethe problem Vmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909)
Algorithm D 1 Index the orders in the nondecreasing trans-portation time sequence that is 119905
1le 1199052
le sdot sdot sdot le 119905119899
Define 119867119877(119895 119894 119897 119879
1 1198792 119879
119898) and 119891
119877(119895 119894 119897 119879
1 1198792
119879119898
) as the minimum objective value and the minimum totalflow time of a partial schedule of orders 119869
1 119869
119895 respec-
tively where orders 1198691 119869
119895have finished transporting and
processing by using 119897 batches on themachine and the currentbatch 119861
119897contains orders 119869
119894 119869
119895
Initial Conditions Consider
119867119909
(119895 119894 119897 1198791 1198792 119879
119898)
=
0 119895 = 0 119894 = 0 119897 = 0 1198791
= 1198792
= sdot sdot sdot 119879119898
= 0
infin otherwise
(15)
for 1198781
ge 1199051 119897 = 1 2 119909 119909 isin lceil119899119888rceil 119899 and 119879
119906=
0 1 119879
Recursive Relations Consider
119891119909
(119895 119894 119897 1198791 1198792 119879
119898)
=
min1le119906le119898
119891119909
(119895 minus 1 119894 119897 1198791 119879
119906minus 119905119895 119879
119898) + 119862
119895 if 119887
119897lt 119888
infin if 119887119897ge 119888
(16)
where 119862119895
= 119878119897+ 119901|119878119897minus1
lt 119879119906
le 119878119897
le 1198791015840
119906+ 119882 119897 = 1 2 119909
for 119878119897+ 119901 le 119878
119897+1le 1198791015840
119906+ 119882 119897 = 1 2 119909 119909 isin lceil119899119888rceil 119899
and 1198791015840
119906= min
119878119897minus1lt119879119906le119878119897119879119906
Optimal Solution Consider
119865lowast
= min 119867119909
(119899 119899 + 1 119897 1198791 1198792 119879
119898) + 119883 (119909) (17)
Theorem 9 Algorithm D1 can find an optimal schedulefor the problem Vmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909) inO(mn3Tmminus1W) time
Proof Based on Lemma 5 there exists an optimal schedulewith orders assigned to each vehicle in the nondecreasingtransportation time sequence If order 119869
119895is assigned to vehicle
u then its starting time on the machine satisfies 119903119895+ 119882 le 119879
119906
If the number of orders in the current batch 119861119897is less than c
order 119869119895can be assigned into 119861
119897The corresponding flow time
increases 119878119897
+ 119901 Based on Lemma 3 we have 119878119897minus1
lt 119879119906
le
119878119897le 1198791015840
119906+ 119882 This shows that Algorithm D1 can find optimal
solution for the problemThe time complexity of the algorithm can be established
as follows We observe that 119894 119895 119897 le 119899 119879119906
le 119879 and 119898 minus 1 of
the value 1198791 1198792 119879
119898are independent Thus the number
of different states of the recursive relations is at most n3Tmminus1For each state each use requires 119874(119898119882) Therefore theoverall time complexity of Algorithm D1 is O(mn3Tmminus1W)
The existence of such a pseudopolynomial time algorithmfor a NP-hard problemmeans that the problem is NP-hard inthe ordinary sense We have the following theorem
Theorem 10 The problem Vmrarr batch|119901119895
= 119901 119908119895| sum 119862119895
+
119883(119909) is ordinarily NP-hard
5 The Problem 119881119898 rarr batch|119901119895
= 119901 119905119895
= 119905
119908119895| sum 119862119895
+ 119883(119909)
In this section we consider a special case with identicalprocessing times and identical transportation times andderive a polynomial-time algorithm to solve it It is evidentthat the problem reduces to an optimal batching problem onthe machine for the special case Note that 119909 is a decisionvariable denoting the number of batches on the machineNote that 119899
0= lceil119899119898rceil and 119892 = 119899
0119898 minus 119899 let 119899
119906be the number
of orders transported on vehicle 119906 under a specific scheduleNote that 119899
1015840
0= lceil119899119888rceil and 119899
1015840
0le 119909 le 119899
0 if 119888 ge 119898 Hence
the number of batches 119909 on the machine can be numberedby 1198991015840
0 1198991015840
0+ 1 119899
0 Otherwise 119899
0le 119909 le 119899
1015840
0if 119888 lt 119898
Without loss of generality assume that 119888 ge 119898 in the followingdiscussion Let 120587
lowast= (1198611 1198612 119861119871) be an optimal schedule
with given 119909 = 119871 batches for the problem Vmrarr batch|119901119895
=
119901 119905119895
= 119905 119908119895| sum 119862119895
+ 119883(119909) Then we have the followinglemma
Lemma 11 For the problem Vmrarr batch|119901119895
= 119901 119905119895
=
119905 119908119895| sum 119862119895
+ 119883(119909) there exists an schedule with 119871 batches inwhich (1) 119899
0minus 1 le 119899
119906le 1198990 119906 = 1 2 119898 (2) 119887
119897= 119887119897119898 119887119897
is integral and 119897 = 1 2 119871 minus 1 and (3) 119878119897+1
= max119903119895 119878119897
+
119901 where 119903119895denotes the arrival time of some 119898 jobs on the
machine 119897 = 1 2 119871 minus 1
Proof (1) Based on Lemma 2 each vehicle has no insertedidle time during transporting Assume that there exists anoptimal schedule 120587
lowast in which the condition is not satisfiedThen there must be a pair of vehicles 119906 and V such that119899119906
ge 119899V + 2 and the last order on vehicle 119906 is processed inthe last batch If the last order on vehicle 119906 is moved to thelast position on vehicle v then the objective value will notincrease By repeating this process we can obtain a desiredoptimal schedule
(2) According to Lemma 7 in the optimal schedule 119861119897
contains all orders which finish transportation in the timeinterval (119878
119897minus1 119878119897] if the number of orders in a batch is nomore
than 119888 Since 119905119895
= 119905 for each order 119869119895 m orders arrive at the
machine together A batch contains 119898 orders or km (119896119898 le 119888)
orders if their waiting times are not greater than the limitedvalue 119882 Hence 119887
119897is integral for 119897 = 1 2 119871 minus 1
(3) It is trivial according to Lemma 3
The Scientific World Journal 7
Based on these results we can easily construct an optimalschedule for the problem
Algorithm D 2 Calculate 1198990
= lceil119899119898rceil 11989910158400
= lceil119899119888rceilFor each 119871 = 119899
1015840
0 1198991015840
0+ 1 119899
0 compute 119887
119897= 1198870
minus 1 119897 =
1 sdot sdot sdot ℎ 119887119897= 1198870 and 119897 = ℎ + 1 119871 where 119887
0= lceil1198990119871rceil and
ℎ = 1198870119871 minus 1198990
Consider the following two cases
Case 1 Consider ℎ = 0If (1198870minus1)119905 gt 119882 or119898119887
0gt 119888 then the constructed schedule
is not feasible and denote 119865(120587lowast
119871) = infin If (119887
0minus 1)119905 le 119882 and
1198981198870
le 119888 then 1198781
= 1198870119905 According to Lemma 6 there must
be 1198870119905119897 + 119905 + 119882 ge 119878
119897+ 119901 Otherwise it is not feasible Based
on Lemma 7 we can obtain
119878119897+1
= max 1198870119905 (119897 + 1) 119878
119897+ 119901 119897 = 1 2 119871 minus 1 (18)
The associated objective value can be derived as
1198651
(120587lowast
119871) = min
119871
sum
119897=1
1198981198870
(119878119897+ 119901)
minus 119892 (119878119871
+ 119901) + 119883 (119871)
(19)
Case 2 Consider ℎ gt 0If (1198870
minus 2)119905 gt 119882 or 1198981198870
gt 119888 then it is not feasible anddenote 119865(120587
lowast
119871) = infin If (119887
0minus 2)119905 le 119882 and 119898119887
0le 119888 then we
can show 1198781
= (1198870
minus 1)119905 According to Lemma 6 we have
(1198870
minus 1) 119905119897 + 119905 + 119882 ge 119878119897+ 119901 if 119897 le ℎ
(1198870
minus 1) 119905ℎ + 1198870119905 (119897 minus ℎ) + 119905 + 119882 ge 119878
119897+ 119901 if 119897 gt ℎ
(20)
Otherwise the constructed schedule is not feasible anddenote 119865(120587
lowast
119871) = infin
According to Lemma 6 we have
119878119897+1
= max (1198870
minus 1) 119905 (119897 + 1) 119878119897+ 119901 if 119897 le ℎ minus 1
119878119897+1
=max (1198870
minus 1) 119905ℎ + 1198870119905 (119897 + 1 minus ℎ) 119878
119897+ 119901 if 119897 ge ℎ
119897 = 1 2 119871 minus 1
(21)
The associated objective value can be derived as
1198652
(120587lowast
119871)
= min
ℎ
sum
119897=1
119898 (1198870
minus 1) (119878119897+ 119901) +
119871
sum
119897=ℎ+1
1198981198870
(119878119897+ 119901)
minus 119892 (119878119871
+ 119901) + 119883 (119871)
(22)
Thus the objective value is
119865 (120587lowast
119871) = min 119865
1(120587lowast
119871) 1198652
(120587lowast
119871) (23)
Theorem 12 The problem Vmrarr batch|119901119895
= 119901 119905119895
=
119905 119908119895| sum 119862119895
+ 119883(119909) can be solved by Algorithm D2 in 119874(119899119898)
time
Proof Equations (18) and (21) present that the machinecan start to process 119861
119897+1as soon as the orders assigned
into 119861119897+1
have arrived at the machine and the machine hasfinished processing119861
119897The local optimal solution in two cases
can be obtained by (19) and (22) The optimal solution isderived as (23) Hence the optimal schedule to this specialcase can be obtained by repeating the above procedure for1198991015840
0le 119871 le 119899
0and selecting the best candidate among the
solutions generated This shows that Algorithm D2 can findoptimal solution It is clear that the overall time complexityof Algorithm D2 is 119874(119899119898)
We now demonstrate the above algorithm with thefollowing numerical example
Example 13 Consider the instance with 11 orders 119898 = 2 119905 =
1 119888 = 4 119882 = 2 119901 = 2 and 119883(119909) = 5119909 Based on the abovemethod we know 119899
0= 6 119892 = 1 and 119899
1015840
0= 2 Then 119871 =
2 3 4 5 6 We have the following results
when 119871 = 2 we have 1198870
= 3 ℎ = 0 1198781
= 3 and119865(120587lowast) = 80
when 119871 = 3 we have 1198870
= 2 ℎ = 0 1198781
= 2 and119865(120587lowast) = 79
when 119871 = 4 we have 1198870
= 2 ℎ = 2 1198781
= 1 and119865(120587lowast
119871) = 101
when 119871 = 5 we have 1198870
= 2 ℎ = 4 1198870119905 sdot 3 + 119905 + 119882 lt
1198783
+ 119901 and 119865(120587lowast
119871) = infin
when 119871 = 6 we have 1198870
= 1 ℎ = 0 1198870119905 sdot 3 + 119905 + 119882 lt
1198783
+ 119901 and 119865(120587lowast
119871) = infin
Therefore we can obtain an optimal schedule120587lowast for 119871 = 3
with 1198611
= 1198691 1198692 1198693 1198694 1198612
= 1198695 1198696 1198697 1198698 and 119861
3=
1198699 11986910
11986911
6 Concluding Remarks
In this paper we address a coordinated scheduling problemwith order transportation before processing and parallel-batch production under the limited waiting time considera-tion Our goal is to optimize the sum of the total flow timeand the total processing cost First we prove that the generalproblem is NP-hard in the strong sense We also demonstratethat the problem with equal processing times on the machineisNP-hard Furthermore a dynamic programming algorithmin pseudopolynomial time is provided to prove its ordinarilyNP-hardness The problem with equal processing times andequal transportation times for each order can be solved inpolynomial time through taking into account the propertiesof the special case Our work has practical implications forthe coordination of batch-production and transportation toimprove the overall system performance in logistics andoperations management Simultaneously another importantimplication of our paper may provide a vital approach to deal
8 The Scientific World Journal
with the waiting time constraint applied to reduce the rawmaterial consumption and improve the machine utilization
There are several possible extensions to this researchFirst it is interesting to investigate the problems with otherobjective functions such as minimizing the makespan orminimizing maximum order tardinessearliness Anotherinteresting issue is to develop effective heuristics to solve thegeneral problem and investigate polynomial time algorithmsfor other special cases
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This research is partly supported by the National NaturalScience Foundation of China (Grant no 71101097) and theProgram for Liaoning Excellent Talents in University (Grantno LJQ2012017)
References
[1] P Brucker A Gladky H Hoogeveen et al ldquoScheduling abatching machinerdquo Journal of Scheduling vol 1 no 1 pp 31ndash541998
[2] C N Potts and M Y Kovalyov ldquoScheduling with batching areviewrdquo European Journal of Operational Research vol 120 no2 pp 228ndash249 2000
[3] C-Y Lee R Uzsoy and L AMartin-Vega ldquoEfficient algorithmsfor scheduling semiconductor burn-in operationsrdquo OperationsResearch vol 40 no 4 pp 764ndash775 1992
[4] L Tang J Guan and G Hu ldquoSteelmaking and refining coordi-nated scheduling problemwith waiting time and transportationconsiderationrdquo Computers and Industrial Engineering vol 58no 2 pp 239ndash248 2010
[5] C-Y Lee and V A Strusevich ldquoTwo-machine shop schedulingwith an uncapacitated interstage transporterrdquo IIE Transactionsvol 37 no 8 pp 725ndash736 2005
[6] L Tang and P Liu ldquoFlowshop scheduling problems withtransportation or deterioration between the batching and singlemachinesrdquo Computers and Industrial Engineering vol 56 no 4pp 1289ndash1295 2009
[7] H Gong and L Tang ldquoTwo-machine flowshop schedulingwith intermediate transportation under job physical spaceconsiderationrdquo Computers and Operations Research vol 38 no9 pp 1267ndash1274 2011
[8] L Tang and H Gong ldquoA hybrid two-stage transportation andbatch scheduling problemrdquo Applied Mathematical Modellingvol 32 no 12 pp 2467ndash2479 2008
[9] C-L Li and J Ou ldquoMachine scheduling with pickup anddeliveryrdquo Naval Research Logistics vol 52 no 7 pp 617ndash6302005
[10] Z-L Chen ldquoIntegrated production and outbound distributionscheduling Review and extensionsrdquo Operations Research vol58 no 1 pp 130ndash148 2010
[11] S Li J Yuan andB Fan ldquoUnbounded parallel-batch schedulingwith family jobs and delivery coordinationrdquo Information Pro-cessing Letters vol 111 no 12 pp 575ndash582 2011
[12] L Lu and J Yuan ldquoUnbounded parallel batch schedulingwith job delivery to minimize makespanrdquo Operations ResearchLetters vol 36 no 4 pp 477ndash480 2008
[13] M R Garey and D S Johnson Computers and IntractabilityA Guide To the Theory of NP-Completeness W H Freeman andCompany New York NY USA 1979
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
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Applied MathematicsJournal of
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Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
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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
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Algebra
Discrete Dynamics in Nature and Society
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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 The Scientific World Journal
The parallel-batch machine scheduling problems have beenpreviously studied in othermanufacturing systems especiallyburn-in operations in the very large-scale integrated circuitmanufacturing (eg [1ndash3]) We briefly discuss some workrelated to integrated production scheduling and transporta-tion decisions Some research (see [4ndash7]) has been done fortwo machine flow shop problems featuring transportation ofsemifinished products and batching processing We mentionthe studies in [4ndash7] however the problems there do notconsider parallel-batch production or they consider anotherkind of batching with a constant processing timeThemodelsin [8 9] consider the coordinated scheduling problemswith two-stage transportation and machine production thatincorporate the scheduling of jobs and the pickup of the rawmaterials from the warehouse and the delivery arrangementof the finished jobs The above scheduling problems donot consider the waiting time constraints Another line ofproduction scheduling models with transportation decisionsfocuses on the delivery of finished jobs to customers witha limited number of vehicles Interested readers for thedelivery coordination are referred to the recent review in [10]The parallel-batch machine scheduling problems with batchdelivery are considered in [11 12] All of the papers reviewedin this section deal in some way with the coordination ofproduction scheduling and transportation at logistics andoperations area but none of them addresses and deals withthe problem from the limited waiting time point of viewWe not only consider the scheduling problem involvingtransportation capacity and transportation times but alsotake into account the order waiting time constraint and thebatch capacity Here the coordination of batch processingtransportation and waiting time constraints is a decisionabout whether or not to reduce production cost and improvemachine utilization
The remainder of this paper is organized as follows Inthe next section we introduce the notation to be used anddescribe the model In Section 3 we analyze the optimalproperties of the problem and prove strong NP-hardness ofthe general problem In Section 4 a dynamic programmingalgorithm is provided to solve the problem with equalprocessing times and we prove that this case is NP-hard inthe ordinary sense Section 5 deals with a case with equalprocessing times and equal transportation times and showshow to find the optimal schedule The last section contains aconclusion and some suggestions for future research
2 Description and Notation
Our problem is formally stated as follows Given a set of119899 orders (including some semifinished jobs or raw materialin one order) 119869 = 119869
1 119869
119899 which are located at the
different suppliers The orders need to be transported to amanufacturing facility and processed on a single parallel-batchmachine Only119898 vehicles are available for transportingthe orders which can load only one order in one shipmentThe vehicles are initially located at a transportation center Let119905119895denote the transportation time of order 119869
119895from the supplier
to the manufacturing facility The transportation time from
the factory back to the center is negligible In the productionpart each order 119869
119895requires a processing time of 119901
119894on the
parallel-batch machine The preemption of orders is notallowedThe parallel-batch machine can process a number oforders simultaneously as a batch as long as the total numberof orders in the batch does not exceed themachine capacity 119888The processing time of a batch is the longest processing timeof all orders in it Orders processed in the same batch havethe same completion time that is their common start time(the start time of the batch in which they are contained) plusthe processing time of the batch Suppose that loading andunloading times are included in the transportation times andthe processing times of orders Associated with each order is119908119895 which is the waiting time of an order for a period time
from arriving at the machine to starting its processing onthe machine Here 119908
119895is restricted by a constant 119882 where
max119895119901119895
le 119882 le sum 119901119895 such that the schedule has a feasible
solution at least Each batch requires a processing cost on theparallel-batch machine which reflects the production costThe goal is to find a feasible schedule to minimize the sumof the total flow time and the total processing cost
Decision VariablesConsider the following
119879119906 the total running time of vehicle u
119903119895 the arrival time of order 119869
119895on the machine
119904119895 the starting time of order 119869
119895on the machine
119908119895 the waiting time of order 119869
119895on the machine 119908
119895=
119904119895
minus 119903119895
le 119882119861119897 batch l
119887119897 the number of orders in 119861
119897
119875(119861119897) the processing time of 119861
119897on the machine
119875(119861119897) = max
119895isin119861119897119901119895
119878119897 the starting time of 119861
119897on the machine
119863119897 the completion time of 119861
119897on the machine
119862119895 the flow time of order 119869
119895on the machine
119909 the number of batches119883(119909) the total processing cost a nondecreasingfunction of x119865(120587) = sum
119899
119895=1119862119895
+ 119883(119909) the objective function of aschedule 120587
We follow the commonly used three-field notation120572|120573|120574introduced to denote the problem under study The problemis denoted by 119881119898 rarr batch|119908
119895| sum 119862119895
+ 119883(119909) where 120572 fieldindicates that the orders are first transported by 119898 vehiclesand then processed on the parallel-batch machine 120573 fielddescribes the order restrictive requirement and 120574 defines theobjective function to be minimized
Specifically the tradeoff between processing and trans-portation gives rise to the sequence of transportation andthe batch composition decisions The schedule may producemore batches in order to satisfy thewaiting time requirementsof orders This leads to the increase of the total processing
The Scientific World Journal 3
Table 1
Order 1 2 3 4 5 6 7 8119905119895
1 1 2 2 2 3 4 4119901119895
2 3 4 3 3 3 4 4
cost on the machine On the other hand if we create veryfew batches then we may not obtain a feasible scheduledue to waiting time constraint Therefore minimizing ofthe objective for the problem is achieved a feasible schedulefor sequencing and batching to tradeoff the total flow timeand the processing cost The following example gives anillustration of this observation
Example 1 Consider a set of 8 orders with their transporta-tion times and processing times as shown below for theproblem 119881119898 rarr batch|119908
119895| sum 119862119895
+ 119883(119909)
Suppose 119898 = 2 119888 = 4 119882 = 4 and 119883(119909) = 3119909 (seeTable 1)
One vehicle transports the orders in turn 1198691 1198693 1198695 and
1198697 Another vehicle transports the orders in turn 119869
2 1198694 1198696
and 1198698 The constructed schedule 120587 consists of three batches
1198611
= 1198691 1198692 1198693 1198694 1198612
= 1198695 1198696 and 119861
3= 1198697 1198698 and has
an objective value of 85 If another constructed schedule 1205871015840
contains two batches 1198611
= 1198691 1198692 1198693 1198694 1198612
= 1198695 1198696 1198697 1198698
then the waiting time of the order 1198695is 5 Although schedule
1205871015840 has a smaller processing cost it is not feasible From
these two schedules we see that a minimized objective canbe obtained from a proper combination of the sequenceof transportation and the batching of processing under thewaiting time constraints
3 The Problem 119881119898 rarr batch|119908119895| sum 119862119895
+ 119883(119909)
In this section we analyze two properties of the schedulingproblem and prove that Vmrarr batch|119908
119895| sum 119862119895
+ 119883(119909) is NP-hard in the strong sense by a reduction from the 3-PartitionProblem which is well known to be NP-hard in the strongsense [13]
3-Partition Problem (3-PP) Given 3h items 119867 = 1 2
3ℎ each item 119894 isin 119867 has a positive integer size 119886119894satisfying
1198864 lt 119886119894
lt 1198862 and sum3ℎ
119894=1119886119894
= ℎ119886 for some integer 119886The question asked is whether there are ℎ disjoint subsets1198671 1198672 119867
ℎof 119867 such that each subset contains exactly
three items and its total size sum119894isin119867119895
119886119894= 119886 119895 = 1 2 ℎ
Lemma2 For the problemVmrarr batch|119908119895| sum 119862119895+119883(119909) there
exists an optimal schedule without idle time between orders onany vehicle
The proof of Lemma 2 is straightforward and is omittedThe following result describes a candidate set of possible
processing times
Lemma3 For the problemVmrarr batch|119908119895| sum 119862119895+119883(119909) there
exists an optimal schedule in which each processing on theparallel-batch machine is made either at the arrival time of
an order on the machine or immediately when the machinebecomes available
Proof Assume that there is a process which is scheduledneither at the arrival time of an order nor at a time when themachine becomes available This process can be changed tothe latest earlier time which fits either of those conditionsSince the same orders can be processed at that earlier timeand there are no additional processes the objective value isnot increased
Theorem 4 The problem Vmrarr batch|119908119895| sum 119862119895
+ 119883(119909) isstrongly NP-hard even if 119898 = 2
Proof To show the NP-hardness of the scheduling problemwe establish the following polynomial time reduction fromthe 3-PP Given a 3-PP instance we construct an instance for119881119898 rarr batch|119908
119895| sum 119862119895
+ 119883(119909) as follows
Number of Orders Consider
119899 = 7 ℎ + 7 (1)
Ordinary Orders Consider
119905119895
= lceil
119895
6
rceil 119886lfloor(12)(119895+1)rfloor
119901119895
= 2 (lceil
119895
6
rceil + 1) 119886
119895 isin 119867 = 1 2 6ℎ
(2)
Auxiliary Orders Consider
119905119895
= 2 (119895 minus 6ℎ) 119886 119895 isin 119866 = 6ℎ + 1 7ℎ
119901119895
= 1199016ℎ+119894
= 2 (119894 + 1) (1198863119894minus2
+ 1198863119894minus1
+ 1198863119894
)
119895 isin 119866 = 6ℎ + 1 7ℎ
119905119895
= 0 119901119895
= 2119886 119895 isin 119884 = 7ℎ + 1 7ℎ + 7
(3)
Machine Capacity Consider
119888 = 7 (4)
Limited Waiting Time Consider
119882 = 2119886ℎ (5)
Processing Cost Function Consider
119883 (119909) = [
5
3
119886 (ℎ + 1) (ℎ + 2) (ℎ + 3)]
119909minusℎ
(6)
Threshold Value Consider
119910 = 4119886 (ℎ + 1) (ℎ + 2) (ℎ + 3) (7)
Clearly in a solution to this instance of 119881119898 rarr
batch|119908119895| sum 119862119895+ 119883(119909) with the objective value not exceeding
4 The Scientific World Journal
Machine Y
0 2a 6a 12a
middot middot middot
middot middot middot
middot middot middot
7h
1ndash6 7ndash12 13ndash18
ah(h + 1) a(h + 1)(h + 2)
Vehicle 1
Vehicle 2
6h + 1 6h + 2 6h + 3
1ndash6 6h + 1
(6h minus 5)ndash6h
(6h minus 11)ndash(6h minus 6) 7h minus 1 (6h minus 5)ndash6h 7h7ndash12 6h + 2
Figure 1 The schedule of Theorem 4
y we first prove the following properties (1) since 119905119895
= 0 for119895 isin 119884 all the orders of 119884 as the first batch must be processedon the machine in the time interval [0 2119886] (2) there will beexactly ℎ + 1 batches in the optimal schedule and each batchcontains 7 orders Suppose that there are q (more than ℎ + 1)batches in the scheduleThen 7ℎ + 7 orders need at least ℎ + 1
batches due to 119888 = 7 If 119902 = ℎ + 2 then the total processingcost is 119883(119909) = 119883(ℎ + 2) = [(53)119886(ℎ + 1)(ℎ + 2)(ℎ + 3)]
2gt 119910
Thus all the orders of 119867 cup 119866 are divided into ℎ batches thatis all the batches are full
We will show that the 3-PP instance has a solution if andonly if there is a schedule 120587 for the scheduling instance suchthat its objective value of 120587 is no more than y
rarrAssume that there is a solution to the 3-PP instance1198671 1198672 119867
ℎ then there is a schedule to our problem with
an objective value of nomore than119910We construct a schedule120587 for our problem as shown in Figure 1
In this schedule two vehicles begin to transport ordersat time point 0 All the orders of 119884 are processed first on themachine as the first batch In the interval [119886119897(119897 minus 1) 119886119897(119897 + 1)]vehicle 1 transports the orders of 119867
119897 119867119897
= 6119897 minus 5 6119897 minus
4 6119897 vehicle 2 transports 1198696ℎ+119897
for 119897 = 1 2 ℎ Since1199056ℎ+119897
= 2119897119886 and sum119895isin119867119897
119905119895
= 2119897119886 1198696ℎ+119897
and the last order of 119867119897
arrive at the machine simultaneously for 119897 = 1 2 ℎ then1198696ℎ+119897
and the corresponding orders of 119867119897form batch 119861
119897 The
waiting time of each order in 119861119897is smaller than119882 Hence the
starting time of 119861119897on the machine is 119886119897(119897 + 1) the completion
time of 119861119897is 119886(119897 + 1)(119897 + 2) for 119897 = 0 1 2 ℎ It is easy to
see that the above schedule is optimal and the objective valueis y
larrOn the contrary suppose that there exists a schedulefor the constructed instance with the objective value notexceeding y
Fact 1 In the schedule 119861119897contains an auxiliary order 119869
6ℎ+119897
for 119897 = 1 2 ℎAssume that there is some batch 119861
119897that contains more
than one auxiliary order Assume that 119861119897contains two
auxiliary jobs 1198696ℎ+119897
and 1198696ℎ+119896
then there must be some batch119861119896only consisting of the ordinary jobs 119896 = 119897 Consider the
following two cases
Case 1 If 119896 = 119897 minus 1 then it is shown that 119861119897minus1
=
1198696119897minus11
1198696119897minus6
1198691198951015840 | 119895
1015840isin 119867119897 The corresponding flow
time of each order in 119861119897minus1
satisfies 119862119895
= 119886119897(119897 + 1) + 2119886
for119895 isin 119861119897minus1
The flow times of orders in each batch after 119861119897minus1
will be greater than 119886(119897 + 1)(119897 + 2) + 2119886 It is easy to see thatsum 119862119895
= 4119886(ℎ + 1)(ℎ + 2)(ℎ + 3) + 14(ℎ minus 119897 + 2)119886 2 le 119897 le ℎHence 1198651015840 = sum 119862
119895+ 119883(ℎ + 1) gt 119910
Case 2 If 119896 = 119897+1 then it can be shown that two vehiclesmusttransport two auxiliary orders of batch 119861
119897 respectively Due
to the limited waiting time of orders the completion time ofbatch 119861
119897on the machine is greater than 119886(119897 + 1)(119897 + 2) + 2119886
Furthermore it is easy to see that sum 119862119895
gt 4119886(ℎ + 1)(ℎ + 2)(ℎ +
3) + 14(ℎ minus 119897 + 1)119886 1 le 119897 le ℎ minus 1 which implies 1198651015840
gt 119910
Fact 2 Each batch 119861119897must contain 6 orders of 119867
119897= 6119897 minus
5 6119897 minus 4 6119897 such that sum119895isin119867119897
119905119895
= 2119897119886 for 119897 = 1 2 ℎ
By the above argument since each batch is full we canshow that one vehicle transports the orders of 119867
119897one by
one and another vehicle transports 1198696ℎ+119897
in the time interval[119886119897(119897 minus 1) 119886119897(119897 + 1)] The orders 119869
6ℎ+119897 119867119897 as 119861119897are processed
on themachine in the time interval [119886119897(119897+1) 119886(119897+1)(119897+2)] Itmust be true thatsum
119895isin119867119897119905119895
= 2119897119886 for 119897 = 1 2 ℎ Otherwiseif sum119895isin119867119897
119905119895
lt 2119897119886 then 1199016ℎ+119897
= 2(119897 + 1) sdot (1198863119897minus2
+ 1198863119897minus1
+ 1198863119897
) lt
2(119897+1)119886We can obtain some batch119861119896whose processing time
is greater than 2(119896 + 1)119886 119896 = 119897 Hence 1198651015840
gt 119910 which is acontradiction
Therefore a partition for the set 119867 is obtained by lettingthe elements be corresponding to batch 119861
119897 Then it is easy to
see that 119902 = ℎ + 1 and ℎ batches 1198611 119861
ℎform a solution to
the 3-PP instance The proof is concluded
Theorem 4 indicates that the existence of a polynomialtime algorithm to solve the scheduling model is unlikelySince the general problem is strongly NP-hard we nextconsider two special cases of the problem
4 The Problem 119881119898 rarr batch|119901119895
= 119901
119908119895| sum 119862119895
+ 119883(119909)
We now turn our attention to a special case with identicalprocessing time condition on themachine In this section weprove that the problem119881119898 rarr batch|119901
119895= 119901 119908
119895| sum 119862119895+119883(119909)
is ordinarily NP-hard by a reduction of the Partition Problem[13] which is a known NP-hard problem and provides adynamic programming algorithm in pseudopolynomial time
The Scientific World Journal 5
Suppose a feasible schedule 120587 = 1198611 1198612 119861119909 with 119909 isin
lceil119899119888rceil 119899 batches for the problem 119881119898 rarr batch|119901119895
=
119901 119908119895| sum 119862119895
+ 119883(119909) We now present some properties for theproblem
Lemma 5 For Vmrarr batch|119901119895
= 119901 119908119895| sum 119862119895
+ 119883(119909) thereexists an optimal schedule 120587
lowastsuch that all orders assigned intothe same vehicle are scheduled in the nondecreasing sequenceof their transportation times
Proof Assume that orders 119869119894and 119869119895are assigned to the same
vehicle 119869119894is followed by 119869
119895immediately such that 119905
119894ge 119905119895in
120587lowast Let 120587
1015840 be a schedule obtained by swapping 119869119894and 119869119895 In 1205871015840
it is easy to see that 119903119894gt 1199031015840
119895and 1199031015840
119894= 119903119895 Regardless of whether
119869119894and 119869119895are processed in the same batch or not the starting
times of 119869119894and 119869119895on the machine cannot increase We have
119865(120587lowast) ge 119865(120587
1015840)
Lemma 6 For any scheduleof the problem Vmrarr batch|119901119895
=
119901 119908119895| sum 119862119895+ 119883(119909) the starting time of batch 119861
119897satisfies 119878
119897+1ge
119878119897+ 119901 for 119897 = 1 2 119909
Proof It is easily obtained based on Lemma 3
Lemma 7 For any schedule of the problem Vmrarr batch|119901119895
=
119901 119908119895| sum 119862119895
+ 119883(119909) 119861119897contains all orders which have arrived
at the machine if the number of the orders is no more than thecapacity 119888 in the time interval (119878
119897minus1 119878119897]
Proof Assume that the batches are numbered in accordancewith their start times and the number of orders in eachbatch is smaller than the machine capacity 119888 Suppose that119869119895is the first order assigned in 119861
119897+1in 120587lowast and the arrival
time of 119869119895on the machine satisfies 119903
119895le 119878119897 Let 120587
1015840 bea new schedule obtained by simply assigning 119869
119895to 119861119897
Then 1198621015840
119895= 119878119897
+ 119901 lt 119878119897+1
+ 119901 le 119862119895 The schedule
of the remaining orders in 1205871015840 are the same as in 120587
lowast Itis obvious that 119865(120587
1015840) le 119865(120587
lowast) We can see that there
exists an optimal schedule in which all batches consist ofa number of orders which finish processing contiguously
We will show that the problem with equal processingtimes is NP-hard by a reduction from the Partition Problem
Theorem 8 The problem Vmrarr batch|119901119895
= 119901 119908119895| sum 119862119895
+
119883(119909) is NP-hard even if 119898 = 2
Proof Weprove this result by reducing the Partition Problemto Vmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909) The PartitionProblem (PP) can be stated as follows given ℎ items 119867 =
1 2 ℎ each item 119894 isin 119867 has a positive integer size 119886119894
such thatsumℎ
119894=1119886119894= 2119886 for some integer 119886The question asked
is whether there are two disjoint subsets1198671and119867
2 such that
sum119894isin1198671
119886119894= sum119894isin1198672
119886119894= 119886
We construct a corresponding instance of the problemVmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909) as follows
Number of Orders Consider
119899 = 2ℎ (8)
Transportation Times Consider
119905119895
= 119886119895 119895 isin 119867 = 1 2 ℎ
119905119895
= 0 119895 isin 119884 = ℎ + 1 ℎ + 2 2ℎ
(9)
Processing Times Consider
119901119895
= 119886 119895 = 1 2 2 ℎ (10)
Limited Waiting Time Conider
119882 = 119886 (11)
Processing Cost Function Consider
119883 (119909) = 3119886ℎ119909 (12)
Machine Capacity Consider
119888 = ℎ (13)
Threshold Value Consider
119910 = 9119886ℎ (14)
First it is easy to see that in a solution to this instance ofVmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895+ 119883(119909) with the objective value
not exceeding 119910 We first prove the following property since119905ℎ+1
= sdot sdot sdot = 1199052ℎ
= 0 all the orders of 119884 as the first batch mustbe processed on the machine in the time interval [0 a] Nowwe prove that there is a solution to the constructed instanceof Vmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909) with total cost notexceeding119910 if and only if there is a solution to the PP instance
rarr If there is a solution to the PP instance we show thatthere is a schedule for the above-constructed instance withan objective value of no more than 119910 Let 119867
1and 119867
2be
subsets of119867 that solves the PP instance Vehicle 1 and Vehicle2 transport the orders of119867
1and the orders of119867
2 respectively
Let119879119906be the total running time of vehicleu for119906 = 1 2 Since
sum119895isin1198671
119905119895
= sum119895isin1198672
119905119895
= 119886 we have 1198791
= 1198792
= 119886 In fact thecompletion time of the first batch on themachine is 119886We canobtain that the waiting time of each order on the machine isnot greater than 119886 Let 119861
1= 119884 and 119861
2= 119867 It is easy to see
that 1198871
= 1198872
= ℎ and the total flow time of all the orders is3ha Hence the objective is y
larrGiven a schedule with an objective value not exceedingy then there must be a solution to the PP instance We canobtain that all the orders in 119867 are processed in the secondbatch on the machine that is all the orders are divided intotwo batches and 119887
1= 1198872
= ℎ Suppose that the number ofthe batches is greater than 2 Note that even if the waitingtime constraint is ignored the objective value is 119865 ge 119886ℎ +
sum119895isin119867
119862119895
+ 9119886ℎ gt 119910 which is a contradiction Thus 120587 =
1198611 1198612 and 119861
1= 119884 and 119861
2= 119867
6 The Scientific World Journal
If 1198791
= sum119895isin1198671
119905119895
lt 119886 then it implies 1198792
= sum119895isin1198672
119905119895
gt 119886Hence the starting time of batch 119861
2is 1198782
= 1198792 It implies that
the total flow time is sum 119862119895
= 119886ℎ + (1198782
+ 119886)ℎ gt 3119886ℎ We have119865 gt 119910 which is a contradiction Thus sum
119895isin1198671119886119895
= sum119895isin1198672
119886119895
=
119886
Let 119879 = sum119899
119895=1119905119895 In the following we derive a dynamic
programming algorithm in pseudopolynomial time to solvethe problem Vmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909)
Algorithm D 1 Index the orders in the nondecreasing trans-portation time sequence that is 119905
1le 1199052
le sdot sdot sdot le 119905119899
Define 119867119877(119895 119894 119897 119879
1 1198792 119879
119898) and 119891
119877(119895 119894 119897 119879
1 1198792
119879119898
) as the minimum objective value and the minimum totalflow time of a partial schedule of orders 119869
1 119869
119895 respec-
tively where orders 1198691 119869
119895have finished transporting and
processing by using 119897 batches on themachine and the currentbatch 119861
119897contains orders 119869
119894 119869
119895
Initial Conditions Consider
119867119909
(119895 119894 119897 1198791 1198792 119879
119898)
=
0 119895 = 0 119894 = 0 119897 = 0 1198791
= 1198792
= sdot sdot sdot 119879119898
= 0
infin otherwise
(15)
for 1198781
ge 1199051 119897 = 1 2 119909 119909 isin lceil119899119888rceil 119899 and 119879
119906=
0 1 119879
Recursive Relations Consider
119891119909
(119895 119894 119897 1198791 1198792 119879
119898)
=
min1le119906le119898
119891119909
(119895 minus 1 119894 119897 1198791 119879
119906minus 119905119895 119879
119898) + 119862
119895 if 119887
119897lt 119888
infin if 119887119897ge 119888
(16)
where 119862119895
= 119878119897+ 119901|119878119897minus1
lt 119879119906
le 119878119897
le 1198791015840
119906+ 119882 119897 = 1 2 119909
for 119878119897+ 119901 le 119878
119897+1le 1198791015840
119906+ 119882 119897 = 1 2 119909 119909 isin lceil119899119888rceil 119899
and 1198791015840
119906= min
119878119897minus1lt119879119906le119878119897119879119906
Optimal Solution Consider
119865lowast
= min 119867119909
(119899 119899 + 1 119897 1198791 1198792 119879
119898) + 119883 (119909) (17)
Theorem 9 Algorithm D1 can find an optimal schedulefor the problem Vmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909) inO(mn3Tmminus1W) time
Proof Based on Lemma 5 there exists an optimal schedulewith orders assigned to each vehicle in the nondecreasingtransportation time sequence If order 119869
119895is assigned to vehicle
u then its starting time on the machine satisfies 119903119895+ 119882 le 119879
119906
If the number of orders in the current batch 119861119897is less than c
order 119869119895can be assigned into 119861
119897The corresponding flow time
increases 119878119897
+ 119901 Based on Lemma 3 we have 119878119897minus1
lt 119879119906
le
119878119897le 1198791015840
119906+ 119882 This shows that Algorithm D1 can find optimal
solution for the problemThe time complexity of the algorithm can be established
as follows We observe that 119894 119895 119897 le 119899 119879119906
le 119879 and 119898 minus 1 of
the value 1198791 1198792 119879
119898are independent Thus the number
of different states of the recursive relations is at most n3Tmminus1For each state each use requires 119874(119898119882) Therefore theoverall time complexity of Algorithm D1 is O(mn3Tmminus1W)
The existence of such a pseudopolynomial time algorithmfor a NP-hard problemmeans that the problem is NP-hard inthe ordinary sense We have the following theorem
Theorem 10 The problem Vmrarr batch|119901119895
= 119901 119908119895| sum 119862119895
+
119883(119909) is ordinarily NP-hard
5 The Problem 119881119898 rarr batch|119901119895
= 119901 119905119895
= 119905
119908119895| sum 119862119895
+ 119883(119909)
In this section we consider a special case with identicalprocessing times and identical transportation times andderive a polynomial-time algorithm to solve it It is evidentthat the problem reduces to an optimal batching problem onthe machine for the special case Note that 119909 is a decisionvariable denoting the number of batches on the machineNote that 119899
0= lceil119899119898rceil and 119892 = 119899
0119898 minus 119899 let 119899
119906be the number
of orders transported on vehicle 119906 under a specific scheduleNote that 119899
1015840
0= lceil119899119888rceil and 119899
1015840
0le 119909 le 119899
0 if 119888 ge 119898 Hence
the number of batches 119909 on the machine can be numberedby 1198991015840
0 1198991015840
0+ 1 119899
0 Otherwise 119899
0le 119909 le 119899
1015840
0if 119888 lt 119898
Without loss of generality assume that 119888 ge 119898 in the followingdiscussion Let 120587
lowast= (1198611 1198612 119861119871) be an optimal schedule
with given 119909 = 119871 batches for the problem Vmrarr batch|119901119895
=
119901 119905119895
= 119905 119908119895| sum 119862119895
+ 119883(119909) Then we have the followinglemma
Lemma 11 For the problem Vmrarr batch|119901119895
= 119901 119905119895
=
119905 119908119895| sum 119862119895
+ 119883(119909) there exists an schedule with 119871 batches inwhich (1) 119899
0minus 1 le 119899
119906le 1198990 119906 = 1 2 119898 (2) 119887
119897= 119887119897119898 119887119897
is integral and 119897 = 1 2 119871 minus 1 and (3) 119878119897+1
= max119903119895 119878119897
+
119901 where 119903119895denotes the arrival time of some 119898 jobs on the
machine 119897 = 1 2 119871 minus 1
Proof (1) Based on Lemma 2 each vehicle has no insertedidle time during transporting Assume that there exists anoptimal schedule 120587
lowast in which the condition is not satisfiedThen there must be a pair of vehicles 119906 and V such that119899119906
ge 119899V + 2 and the last order on vehicle 119906 is processed inthe last batch If the last order on vehicle 119906 is moved to thelast position on vehicle v then the objective value will notincrease By repeating this process we can obtain a desiredoptimal schedule
(2) According to Lemma 7 in the optimal schedule 119861119897
contains all orders which finish transportation in the timeinterval (119878
119897minus1 119878119897] if the number of orders in a batch is nomore
than 119888 Since 119905119895
= 119905 for each order 119869119895 m orders arrive at the
machine together A batch contains 119898 orders or km (119896119898 le 119888)
orders if their waiting times are not greater than the limitedvalue 119882 Hence 119887
119897is integral for 119897 = 1 2 119871 minus 1
(3) It is trivial according to Lemma 3
The Scientific World Journal 7
Based on these results we can easily construct an optimalschedule for the problem
Algorithm D 2 Calculate 1198990
= lceil119899119898rceil 11989910158400
= lceil119899119888rceilFor each 119871 = 119899
1015840
0 1198991015840
0+ 1 119899
0 compute 119887
119897= 1198870
minus 1 119897 =
1 sdot sdot sdot ℎ 119887119897= 1198870 and 119897 = ℎ + 1 119871 where 119887
0= lceil1198990119871rceil and
ℎ = 1198870119871 minus 1198990
Consider the following two cases
Case 1 Consider ℎ = 0If (1198870minus1)119905 gt 119882 or119898119887
0gt 119888 then the constructed schedule
is not feasible and denote 119865(120587lowast
119871) = infin If (119887
0minus 1)119905 le 119882 and
1198981198870
le 119888 then 1198781
= 1198870119905 According to Lemma 6 there must
be 1198870119905119897 + 119905 + 119882 ge 119878
119897+ 119901 Otherwise it is not feasible Based
on Lemma 7 we can obtain
119878119897+1
= max 1198870119905 (119897 + 1) 119878
119897+ 119901 119897 = 1 2 119871 minus 1 (18)
The associated objective value can be derived as
1198651
(120587lowast
119871) = min
119871
sum
119897=1
1198981198870
(119878119897+ 119901)
minus 119892 (119878119871
+ 119901) + 119883 (119871)
(19)
Case 2 Consider ℎ gt 0If (1198870
minus 2)119905 gt 119882 or 1198981198870
gt 119888 then it is not feasible anddenote 119865(120587
lowast
119871) = infin If (119887
0minus 2)119905 le 119882 and 119898119887
0le 119888 then we
can show 1198781
= (1198870
minus 1)119905 According to Lemma 6 we have
(1198870
minus 1) 119905119897 + 119905 + 119882 ge 119878119897+ 119901 if 119897 le ℎ
(1198870
minus 1) 119905ℎ + 1198870119905 (119897 minus ℎ) + 119905 + 119882 ge 119878
119897+ 119901 if 119897 gt ℎ
(20)
Otherwise the constructed schedule is not feasible anddenote 119865(120587
lowast
119871) = infin
According to Lemma 6 we have
119878119897+1
= max (1198870
minus 1) 119905 (119897 + 1) 119878119897+ 119901 if 119897 le ℎ minus 1
119878119897+1
=max (1198870
minus 1) 119905ℎ + 1198870119905 (119897 + 1 minus ℎ) 119878
119897+ 119901 if 119897 ge ℎ
119897 = 1 2 119871 minus 1
(21)
The associated objective value can be derived as
1198652
(120587lowast
119871)
= min
ℎ
sum
119897=1
119898 (1198870
minus 1) (119878119897+ 119901) +
119871
sum
119897=ℎ+1
1198981198870
(119878119897+ 119901)
minus 119892 (119878119871
+ 119901) + 119883 (119871)
(22)
Thus the objective value is
119865 (120587lowast
119871) = min 119865
1(120587lowast
119871) 1198652
(120587lowast
119871) (23)
Theorem 12 The problem Vmrarr batch|119901119895
= 119901 119905119895
=
119905 119908119895| sum 119862119895
+ 119883(119909) can be solved by Algorithm D2 in 119874(119899119898)
time
Proof Equations (18) and (21) present that the machinecan start to process 119861
119897+1as soon as the orders assigned
into 119861119897+1
have arrived at the machine and the machine hasfinished processing119861
119897The local optimal solution in two cases
can be obtained by (19) and (22) The optimal solution isderived as (23) Hence the optimal schedule to this specialcase can be obtained by repeating the above procedure for1198991015840
0le 119871 le 119899
0and selecting the best candidate among the
solutions generated This shows that Algorithm D2 can findoptimal solution It is clear that the overall time complexityof Algorithm D2 is 119874(119899119898)
We now demonstrate the above algorithm with thefollowing numerical example
Example 13 Consider the instance with 11 orders 119898 = 2 119905 =
1 119888 = 4 119882 = 2 119901 = 2 and 119883(119909) = 5119909 Based on the abovemethod we know 119899
0= 6 119892 = 1 and 119899
1015840
0= 2 Then 119871 =
2 3 4 5 6 We have the following results
when 119871 = 2 we have 1198870
= 3 ℎ = 0 1198781
= 3 and119865(120587lowast) = 80
when 119871 = 3 we have 1198870
= 2 ℎ = 0 1198781
= 2 and119865(120587lowast) = 79
when 119871 = 4 we have 1198870
= 2 ℎ = 2 1198781
= 1 and119865(120587lowast
119871) = 101
when 119871 = 5 we have 1198870
= 2 ℎ = 4 1198870119905 sdot 3 + 119905 + 119882 lt
1198783
+ 119901 and 119865(120587lowast
119871) = infin
when 119871 = 6 we have 1198870
= 1 ℎ = 0 1198870119905 sdot 3 + 119905 + 119882 lt
1198783
+ 119901 and 119865(120587lowast
119871) = infin
Therefore we can obtain an optimal schedule120587lowast for 119871 = 3
with 1198611
= 1198691 1198692 1198693 1198694 1198612
= 1198695 1198696 1198697 1198698 and 119861
3=
1198699 11986910
11986911
6 Concluding Remarks
In this paper we address a coordinated scheduling problemwith order transportation before processing and parallel-batch production under the limited waiting time considera-tion Our goal is to optimize the sum of the total flow timeand the total processing cost First we prove that the generalproblem is NP-hard in the strong sense We also demonstratethat the problem with equal processing times on the machineisNP-hard Furthermore a dynamic programming algorithmin pseudopolynomial time is provided to prove its ordinarilyNP-hardness The problem with equal processing times andequal transportation times for each order can be solved inpolynomial time through taking into account the propertiesof the special case Our work has practical implications forthe coordination of batch-production and transportation toimprove the overall system performance in logistics andoperations management Simultaneously another importantimplication of our paper may provide a vital approach to deal
8 The Scientific World Journal
with the waiting time constraint applied to reduce the rawmaterial consumption and improve the machine utilization
There are several possible extensions to this researchFirst it is interesting to investigate the problems with otherobjective functions such as minimizing the makespan orminimizing maximum order tardinessearliness Anotherinteresting issue is to develop effective heuristics to solve thegeneral problem and investigate polynomial time algorithmsfor other special cases
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This research is partly supported by the National NaturalScience Foundation of China (Grant no 71101097) and theProgram for Liaoning Excellent Talents in University (Grantno LJQ2012017)
References
[1] P Brucker A Gladky H Hoogeveen et al ldquoScheduling abatching machinerdquo Journal of Scheduling vol 1 no 1 pp 31ndash541998
[2] C N Potts and M Y Kovalyov ldquoScheduling with batching areviewrdquo European Journal of Operational Research vol 120 no2 pp 228ndash249 2000
[3] C-Y Lee R Uzsoy and L AMartin-Vega ldquoEfficient algorithmsfor scheduling semiconductor burn-in operationsrdquo OperationsResearch vol 40 no 4 pp 764ndash775 1992
[4] L Tang J Guan and G Hu ldquoSteelmaking and refining coordi-nated scheduling problemwith waiting time and transportationconsiderationrdquo Computers and Industrial Engineering vol 58no 2 pp 239ndash248 2010
[5] C-Y Lee and V A Strusevich ldquoTwo-machine shop schedulingwith an uncapacitated interstage transporterrdquo IIE Transactionsvol 37 no 8 pp 725ndash736 2005
[6] L Tang and P Liu ldquoFlowshop scheduling problems withtransportation or deterioration between the batching and singlemachinesrdquo Computers and Industrial Engineering vol 56 no 4pp 1289ndash1295 2009
[7] H Gong and L Tang ldquoTwo-machine flowshop schedulingwith intermediate transportation under job physical spaceconsiderationrdquo Computers and Operations Research vol 38 no9 pp 1267ndash1274 2011
[8] L Tang and H Gong ldquoA hybrid two-stage transportation andbatch scheduling problemrdquo Applied Mathematical Modellingvol 32 no 12 pp 2467ndash2479 2008
[9] C-L Li and J Ou ldquoMachine scheduling with pickup anddeliveryrdquo Naval Research Logistics vol 52 no 7 pp 617ndash6302005
[10] Z-L Chen ldquoIntegrated production and outbound distributionscheduling Review and extensionsrdquo Operations Research vol58 no 1 pp 130ndash148 2010
[11] S Li J Yuan andB Fan ldquoUnbounded parallel-batch schedulingwith family jobs and delivery coordinationrdquo Information Pro-cessing Letters vol 111 no 12 pp 575ndash582 2011
[12] L Lu and J Yuan ldquoUnbounded parallel batch schedulingwith job delivery to minimize makespanrdquo Operations ResearchLetters vol 36 no 4 pp 477ndash480 2008
[13] M R Garey and D S Johnson Computers and IntractabilityA Guide To the Theory of NP-Completeness W H Freeman andCompany New York NY USA 1979
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
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OptimizationJournal of
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International Journal of
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Operations ResearchAdvances in
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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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Decision SciencesAdvances in
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
The Scientific World Journal 3
Table 1
Order 1 2 3 4 5 6 7 8119905119895
1 1 2 2 2 3 4 4119901119895
2 3 4 3 3 3 4 4
cost on the machine On the other hand if we create veryfew batches then we may not obtain a feasible scheduledue to waiting time constraint Therefore minimizing ofthe objective for the problem is achieved a feasible schedulefor sequencing and batching to tradeoff the total flow timeand the processing cost The following example gives anillustration of this observation
Example 1 Consider a set of 8 orders with their transporta-tion times and processing times as shown below for theproblem 119881119898 rarr batch|119908
119895| sum 119862119895
+ 119883(119909)
Suppose 119898 = 2 119888 = 4 119882 = 4 and 119883(119909) = 3119909 (seeTable 1)
One vehicle transports the orders in turn 1198691 1198693 1198695 and
1198697 Another vehicle transports the orders in turn 119869
2 1198694 1198696
and 1198698 The constructed schedule 120587 consists of three batches
1198611
= 1198691 1198692 1198693 1198694 1198612
= 1198695 1198696 and 119861
3= 1198697 1198698 and has
an objective value of 85 If another constructed schedule 1205871015840
contains two batches 1198611
= 1198691 1198692 1198693 1198694 1198612
= 1198695 1198696 1198697 1198698
then the waiting time of the order 1198695is 5 Although schedule
1205871015840 has a smaller processing cost it is not feasible From
these two schedules we see that a minimized objective canbe obtained from a proper combination of the sequenceof transportation and the batching of processing under thewaiting time constraints
3 The Problem 119881119898 rarr batch|119908119895| sum 119862119895
+ 119883(119909)
In this section we analyze two properties of the schedulingproblem and prove that Vmrarr batch|119908
119895| sum 119862119895
+ 119883(119909) is NP-hard in the strong sense by a reduction from the 3-PartitionProblem which is well known to be NP-hard in the strongsense [13]
3-Partition Problem (3-PP) Given 3h items 119867 = 1 2
3ℎ each item 119894 isin 119867 has a positive integer size 119886119894satisfying
1198864 lt 119886119894
lt 1198862 and sum3ℎ
119894=1119886119894
= ℎ119886 for some integer 119886The question asked is whether there are ℎ disjoint subsets1198671 1198672 119867
ℎof 119867 such that each subset contains exactly
three items and its total size sum119894isin119867119895
119886119894= 119886 119895 = 1 2 ℎ
Lemma2 For the problemVmrarr batch|119908119895| sum 119862119895+119883(119909) there
exists an optimal schedule without idle time between orders onany vehicle
The proof of Lemma 2 is straightforward and is omittedThe following result describes a candidate set of possible
processing times
Lemma3 For the problemVmrarr batch|119908119895| sum 119862119895+119883(119909) there
exists an optimal schedule in which each processing on theparallel-batch machine is made either at the arrival time of
an order on the machine or immediately when the machinebecomes available
Proof Assume that there is a process which is scheduledneither at the arrival time of an order nor at a time when themachine becomes available This process can be changed tothe latest earlier time which fits either of those conditionsSince the same orders can be processed at that earlier timeand there are no additional processes the objective value isnot increased
Theorem 4 The problem Vmrarr batch|119908119895| sum 119862119895
+ 119883(119909) isstrongly NP-hard even if 119898 = 2
Proof To show the NP-hardness of the scheduling problemwe establish the following polynomial time reduction fromthe 3-PP Given a 3-PP instance we construct an instance for119881119898 rarr batch|119908
119895| sum 119862119895
+ 119883(119909) as follows
Number of Orders Consider
119899 = 7 ℎ + 7 (1)
Ordinary Orders Consider
119905119895
= lceil
119895
6
rceil 119886lfloor(12)(119895+1)rfloor
119901119895
= 2 (lceil
119895
6
rceil + 1) 119886
119895 isin 119867 = 1 2 6ℎ
(2)
Auxiliary Orders Consider
119905119895
= 2 (119895 minus 6ℎ) 119886 119895 isin 119866 = 6ℎ + 1 7ℎ
119901119895
= 1199016ℎ+119894
= 2 (119894 + 1) (1198863119894minus2
+ 1198863119894minus1
+ 1198863119894
)
119895 isin 119866 = 6ℎ + 1 7ℎ
119905119895
= 0 119901119895
= 2119886 119895 isin 119884 = 7ℎ + 1 7ℎ + 7
(3)
Machine Capacity Consider
119888 = 7 (4)
Limited Waiting Time Consider
119882 = 2119886ℎ (5)
Processing Cost Function Consider
119883 (119909) = [
5
3
119886 (ℎ + 1) (ℎ + 2) (ℎ + 3)]
119909minusℎ
(6)
Threshold Value Consider
119910 = 4119886 (ℎ + 1) (ℎ + 2) (ℎ + 3) (7)
Clearly in a solution to this instance of 119881119898 rarr
batch|119908119895| sum 119862119895+ 119883(119909) with the objective value not exceeding
4 The Scientific World Journal
Machine Y
0 2a 6a 12a
middot middot middot
middot middot middot
middot middot middot
7h
1ndash6 7ndash12 13ndash18
ah(h + 1) a(h + 1)(h + 2)
Vehicle 1
Vehicle 2
6h + 1 6h + 2 6h + 3
1ndash6 6h + 1
(6h minus 5)ndash6h
(6h minus 11)ndash(6h minus 6) 7h minus 1 (6h minus 5)ndash6h 7h7ndash12 6h + 2
Figure 1 The schedule of Theorem 4
y we first prove the following properties (1) since 119905119895
= 0 for119895 isin 119884 all the orders of 119884 as the first batch must be processedon the machine in the time interval [0 2119886] (2) there will beexactly ℎ + 1 batches in the optimal schedule and each batchcontains 7 orders Suppose that there are q (more than ℎ + 1)batches in the scheduleThen 7ℎ + 7 orders need at least ℎ + 1
batches due to 119888 = 7 If 119902 = ℎ + 2 then the total processingcost is 119883(119909) = 119883(ℎ + 2) = [(53)119886(ℎ + 1)(ℎ + 2)(ℎ + 3)]
2gt 119910
Thus all the orders of 119867 cup 119866 are divided into ℎ batches thatis all the batches are full
We will show that the 3-PP instance has a solution if andonly if there is a schedule 120587 for the scheduling instance suchthat its objective value of 120587 is no more than y
rarrAssume that there is a solution to the 3-PP instance1198671 1198672 119867
ℎ then there is a schedule to our problem with
an objective value of nomore than119910We construct a schedule120587 for our problem as shown in Figure 1
In this schedule two vehicles begin to transport ordersat time point 0 All the orders of 119884 are processed first on themachine as the first batch In the interval [119886119897(119897 minus 1) 119886119897(119897 + 1)]vehicle 1 transports the orders of 119867
119897 119867119897
= 6119897 minus 5 6119897 minus
4 6119897 vehicle 2 transports 1198696ℎ+119897
for 119897 = 1 2 ℎ Since1199056ℎ+119897
= 2119897119886 and sum119895isin119867119897
119905119895
= 2119897119886 1198696ℎ+119897
and the last order of 119867119897
arrive at the machine simultaneously for 119897 = 1 2 ℎ then1198696ℎ+119897
and the corresponding orders of 119867119897form batch 119861
119897 The
waiting time of each order in 119861119897is smaller than119882 Hence the
starting time of 119861119897on the machine is 119886119897(119897 + 1) the completion
time of 119861119897is 119886(119897 + 1)(119897 + 2) for 119897 = 0 1 2 ℎ It is easy to
see that the above schedule is optimal and the objective valueis y
larrOn the contrary suppose that there exists a schedulefor the constructed instance with the objective value notexceeding y
Fact 1 In the schedule 119861119897contains an auxiliary order 119869
6ℎ+119897
for 119897 = 1 2 ℎAssume that there is some batch 119861
119897that contains more
than one auxiliary order Assume that 119861119897contains two
auxiliary jobs 1198696ℎ+119897
and 1198696ℎ+119896
then there must be some batch119861119896only consisting of the ordinary jobs 119896 = 119897 Consider the
following two cases
Case 1 If 119896 = 119897 minus 1 then it is shown that 119861119897minus1
=
1198696119897minus11
1198696119897minus6
1198691198951015840 | 119895
1015840isin 119867119897 The corresponding flow
time of each order in 119861119897minus1
satisfies 119862119895
= 119886119897(119897 + 1) + 2119886
for119895 isin 119861119897minus1
The flow times of orders in each batch after 119861119897minus1
will be greater than 119886(119897 + 1)(119897 + 2) + 2119886 It is easy to see thatsum 119862119895
= 4119886(ℎ + 1)(ℎ + 2)(ℎ + 3) + 14(ℎ minus 119897 + 2)119886 2 le 119897 le ℎHence 1198651015840 = sum 119862
119895+ 119883(ℎ + 1) gt 119910
Case 2 If 119896 = 119897+1 then it can be shown that two vehiclesmusttransport two auxiliary orders of batch 119861
119897 respectively Due
to the limited waiting time of orders the completion time ofbatch 119861
119897on the machine is greater than 119886(119897 + 1)(119897 + 2) + 2119886
Furthermore it is easy to see that sum 119862119895
gt 4119886(ℎ + 1)(ℎ + 2)(ℎ +
3) + 14(ℎ minus 119897 + 1)119886 1 le 119897 le ℎ minus 1 which implies 1198651015840
gt 119910
Fact 2 Each batch 119861119897must contain 6 orders of 119867
119897= 6119897 minus
5 6119897 minus 4 6119897 such that sum119895isin119867119897
119905119895
= 2119897119886 for 119897 = 1 2 ℎ
By the above argument since each batch is full we canshow that one vehicle transports the orders of 119867
119897one by
one and another vehicle transports 1198696ℎ+119897
in the time interval[119886119897(119897 minus 1) 119886119897(119897 + 1)] The orders 119869
6ℎ+119897 119867119897 as 119861119897are processed
on themachine in the time interval [119886119897(119897+1) 119886(119897+1)(119897+2)] Itmust be true thatsum
119895isin119867119897119905119895
= 2119897119886 for 119897 = 1 2 ℎ Otherwiseif sum119895isin119867119897
119905119895
lt 2119897119886 then 1199016ℎ+119897
= 2(119897 + 1) sdot (1198863119897minus2
+ 1198863119897minus1
+ 1198863119897
) lt
2(119897+1)119886We can obtain some batch119861119896whose processing time
is greater than 2(119896 + 1)119886 119896 = 119897 Hence 1198651015840
gt 119910 which is acontradiction
Therefore a partition for the set 119867 is obtained by lettingthe elements be corresponding to batch 119861
119897 Then it is easy to
see that 119902 = ℎ + 1 and ℎ batches 1198611 119861
ℎform a solution to
the 3-PP instance The proof is concluded
Theorem 4 indicates that the existence of a polynomialtime algorithm to solve the scheduling model is unlikelySince the general problem is strongly NP-hard we nextconsider two special cases of the problem
4 The Problem 119881119898 rarr batch|119901119895
= 119901
119908119895| sum 119862119895
+ 119883(119909)
We now turn our attention to a special case with identicalprocessing time condition on themachine In this section weprove that the problem119881119898 rarr batch|119901
119895= 119901 119908
119895| sum 119862119895+119883(119909)
is ordinarily NP-hard by a reduction of the Partition Problem[13] which is a known NP-hard problem and provides adynamic programming algorithm in pseudopolynomial time
The Scientific World Journal 5
Suppose a feasible schedule 120587 = 1198611 1198612 119861119909 with 119909 isin
lceil119899119888rceil 119899 batches for the problem 119881119898 rarr batch|119901119895
=
119901 119908119895| sum 119862119895
+ 119883(119909) We now present some properties for theproblem
Lemma 5 For Vmrarr batch|119901119895
= 119901 119908119895| sum 119862119895
+ 119883(119909) thereexists an optimal schedule 120587
lowastsuch that all orders assigned intothe same vehicle are scheduled in the nondecreasing sequenceof their transportation times
Proof Assume that orders 119869119894and 119869119895are assigned to the same
vehicle 119869119894is followed by 119869
119895immediately such that 119905
119894ge 119905119895in
120587lowast Let 120587
1015840 be a schedule obtained by swapping 119869119894and 119869119895 In 1205871015840
it is easy to see that 119903119894gt 1199031015840
119895and 1199031015840
119894= 119903119895 Regardless of whether
119869119894and 119869119895are processed in the same batch or not the starting
times of 119869119894and 119869119895on the machine cannot increase We have
119865(120587lowast) ge 119865(120587
1015840)
Lemma 6 For any scheduleof the problem Vmrarr batch|119901119895
=
119901 119908119895| sum 119862119895+ 119883(119909) the starting time of batch 119861
119897satisfies 119878
119897+1ge
119878119897+ 119901 for 119897 = 1 2 119909
Proof It is easily obtained based on Lemma 3
Lemma 7 For any schedule of the problem Vmrarr batch|119901119895
=
119901 119908119895| sum 119862119895
+ 119883(119909) 119861119897contains all orders which have arrived
at the machine if the number of the orders is no more than thecapacity 119888 in the time interval (119878
119897minus1 119878119897]
Proof Assume that the batches are numbered in accordancewith their start times and the number of orders in eachbatch is smaller than the machine capacity 119888 Suppose that119869119895is the first order assigned in 119861
119897+1in 120587lowast and the arrival
time of 119869119895on the machine satisfies 119903
119895le 119878119897 Let 120587
1015840 bea new schedule obtained by simply assigning 119869
119895to 119861119897
Then 1198621015840
119895= 119878119897
+ 119901 lt 119878119897+1
+ 119901 le 119862119895 The schedule
of the remaining orders in 1205871015840 are the same as in 120587
lowast Itis obvious that 119865(120587
1015840) le 119865(120587
lowast) We can see that there
exists an optimal schedule in which all batches consist ofa number of orders which finish processing contiguously
We will show that the problem with equal processingtimes is NP-hard by a reduction from the Partition Problem
Theorem 8 The problem Vmrarr batch|119901119895
= 119901 119908119895| sum 119862119895
+
119883(119909) is NP-hard even if 119898 = 2
Proof Weprove this result by reducing the Partition Problemto Vmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909) The PartitionProblem (PP) can be stated as follows given ℎ items 119867 =
1 2 ℎ each item 119894 isin 119867 has a positive integer size 119886119894
such thatsumℎ
119894=1119886119894= 2119886 for some integer 119886The question asked
is whether there are two disjoint subsets1198671and119867
2 such that
sum119894isin1198671
119886119894= sum119894isin1198672
119886119894= 119886
We construct a corresponding instance of the problemVmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909) as follows
Number of Orders Consider
119899 = 2ℎ (8)
Transportation Times Consider
119905119895
= 119886119895 119895 isin 119867 = 1 2 ℎ
119905119895
= 0 119895 isin 119884 = ℎ + 1 ℎ + 2 2ℎ
(9)
Processing Times Consider
119901119895
= 119886 119895 = 1 2 2 ℎ (10)
Limited Waiting Time Conider
119882 = 119886 (11)
Processing Cost Function Consider
119883 (119909) = 3119886ℎ119909 (12)
Machine Capacity Consider
119888 = ℎ (13)
Threshold Value Consider
119910 = 9119886ℎ (14)
First it is easy to see that in a solution to this instance ofVmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895+ 119883(119909) with the objective value
not exceeding 119910 We first prove the following property since119905ℎ+1
= sdot sdot sdot = 1199052ℎ
= 0 all the orders of 119884 as the first batch mustbe processed on the machine in the time interval [0 a] Nowwe prove that there is a solution to the constructed instanceof Vmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909) with total cost notexceeding119910 if and only if there is a solution to the PP instance
rarr If there is a solution to the PP instance we show thatthere is a schedule for the above-constructed instance withan objective value of no more than 119910 Let 119867
1and 119867
2be
subsets of119867 that solves the PP instance Vehicle 1 and Vehicle2 transport the orders of119867
1and the orders of119867
2 respectively
Let119879119906be the total running time of vehicleu for119906 = 1 2 Since
sum119895isin1198671
119905119895
= sum119895isin1198672
119905119895
= 119886 we have 1198791
= 1198792
= 119886 In fact thecompletion time of the first batch on themachine is 119886We canobtain that the waiting time of each order on the machine isnot greater than 119886 Let 119861
1= 119884 and 119861
2= 119867 It is easy to see
that 1198871
= 1198872
= ℎ and the total flow time of all the orders is3ha Hence the objective is y
larrGiven a schedule with an objective value not exceedingy then there must be a solution to the PP instance We canobtain that all the orders in 119867 are processed in the secondbatch on the machine that is all the orders are divided intotwo batches and 119887
1= 1198872
= ℎ Suppose that the number ofthe batches is greater than 2 Note that even if the waitingtime constraint is ignored the objective value is 119865 ge 119886ℎ +
sum119895isin119867
119862119895
+ 9119886ℎ gt 119910 which is a contradiction Thus 120587 =
1198611 1198612 and 119861
1= 119884 and 119861
2= 119867
6 The Scientific World Journal
If 1198791
= sum119895isin1198671
119905119895
lt 119886 then it implies 1198792
= sum119895isin1198672
119905119895
gt 119886Hence the starting time of batch 119861
2is 1198782
= 1198792 It implies that
the total flow time is sum 119862119895
= 119886ℎ + (1198782
+ 119886)ℎ gt 3119886ℎ We have119865 gt 119910 which is a contradiction Thus sum
119895isin1198671119886119895
= sum119895isin1198672
119886119895
=
119886
Let 119879 = sum119899
119895=1119905119895 In the following we derive a dynamic
programming algorithm in pseudopolynomial time to solvethe problem Vmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909)
Algorithm D 1 Index the orders in the nondecreasing trans-portation time sequence that is 119905
1le 1199052
le sdot sdot sdot le 119905119899
Define 119867119877(119895 119894 119897 119879
1 1198792 119879
119898) and 119891
119877(119895 119894 119897 119879
1 1198792
119879119898
) as the minimum objective value and the minimum totalflow time of a partial schedule of orders 119869
1 119869
119895 respec-
tively where orders 1198691 119869
119895have finished transporting and
processing by using 119897 batches on themachine and the currentbatch 119861
119897contains orders 119869
119894 119869
119895
Initial Conditions Consider
119867119909
(119895 119894 119897 1198791 1198792 119879
119898)
=
0 119895 = 0 119894 = 0 119897 = 0 1198791
= 1198792
= sdot sdot sdot 119879119898
= 0
infin otherwise
(15)
for 1198781
ge 1199051 119897 = 1 2 119909 119909 isin lceil119899119888rceil 119899 and 119879
119906=
0 1 119879
Recursive Relations Consider
119891119909
(119895 119894 119897 1198791 1198792 119879
119898)
=
min1le119906le119898
119891119909
(119895 minus 1 119894 119897 1198791 119879
119906minus 119905119895 119879
119898) + 119862
119895 if 119887
119897lt 119888
infin if 119887119897ge 119888
(16)
where 119862119895
= 119878119897+ 119901|119878119897minus1
lt 119879119906
le 119878119897
le 1198791015840
119906+ 119882 119897 = 1 2 119909
for 119878119897+ 119901 le 119878
119897+1le 1198791015840
119906+ 119882 119897 = 1 2 119909 119909 isin lceil119899119888rceil 119899
and 1198791015840
119906= min
119878119897minus1lt119879119906le119878119897119879119906
Optimal Solution Consider
119865lowast
= min 119867119909
(119899 119899 + 1 119897 1198791 1198792 119879
119898) + 119883 (119909) (17)
Theorem 9 Algorithm D1 can find an optimal schedulefor the problem Vmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909) inO(mn3Tmminus1W) time
Proof Based on Lemma 5 there exists an optimal schedulewith orders assigned to each vehicle in the nondecreasingtransportation time sequence If order 119869
119895is assigned to vehicle
u then its starting time on the machine satisfies 119903119895+ 119882 le 119879
119906
If the number of orders in the current batch 119861119897is less than c
order 119869119895can be assigned into 119861
119897The corresponding flow time
increases 119878119897
+ 119901 Based on Lemma 3 we have 119878119897minus1
lt 119879119906
le
119878119897le 1198791015840
119906+ 119882 This shows that Algorithm D1 can find optimal
solution for the problemThe time complexity of the algorithm can be established
as follows We observe that 119894 119895 119897 le 119899 119879119906
le 119879 and 119898 minus 1 of
the value 1198791 1198792 119879
119898are independent Thus the number
of different states of the recursive relations is at most n3Tmminus1For each state each use requires 119874(119898119882) Therefore theoverall time complexity of Algorithm D1 is O(mn3Tmminus1W)
The existence of such a pseudopolynomial time algorithmfor a NP-hard problemmeans that the problem is NP-hard inthe ordinary sense We have the following theorem
Theorem 10 The problem Vmrarr batch|119901119895
= 119901 119908119895| sum 119862119895
+
119883(119909) is ordinarily NP-hard
5 The Problem 119881119898 rarr batch|119901119895
= 119901 119905119895
= 119905
119908119895| sum 119862119895
+ 119883(119909)
In this section we consider a special case with identicalprocessing times and identical transportation times andderive a polynomial-time algorithm to solve it It is evidentthat the problem reduces to an optimal batching problem onthe machine for the special case Note that 119909 is a decisionvariable denoting the number of batches on the machineNote that 119899
0= lceil119899119898rceil and 119892 = 119899
0119898 minus 119899 let 119899
119906be the number
of orders transported on vehicle 119906 under a specific scheduleNote that 119899
1015840
0= lceil119899119888rceil and 119899
1015840
0le 119909 le 119899
0 if 119888 ge 119898 Hence
the number of batches 119909 on the machine can be numberedby 1198991015840
0 1198991015840
0+ 1 119899
0 Otherwise 119899
0le 119909 le 119899
1015840
0if 119888 lt 119898
Without loss of generality assume that 119888 ge 119898 in the followingdiscussion Let 120587
lowast= (1198611 1198612 119861119871) be an optimal schedule
with given 119909 = 119871 batches for the problem Vmrarr batch|119901119895
=
119901 119905119895
= 119905 119908119895| sum 119862119895
+ 119883(119909) Then we have the followinglemma
Lemma 11 For the problem Vmrarr batch|119901119895
= 119901 119905119895
=
119905 119908119895| sum 119862119895
+ 119883(119909) there exists an schedule with 119871 batches inwhich (1) 119899
0minus 1 le 119899
119906le 1198990 119906 = 1 2 119898 (2) 119887
119897= 119887119897119898 119887119897
is integral and 119897 = 1 2 119871 minus 1 and (3) 119878119897+1
= max119903119895 119878119897
+
119901 where 119903119895denotes the arrival time of some 119898 jobs on the
machine 119897 = 1 2 119871 minus 1
Proof (1) Based on Lemma 2 each vehicle has no insertedidle time during transporting Assume that there exists anoptimal schedule 120587
lowast in which the condition is not satisfiedThen there must be a pair of vehicles 119906 and V such that119899119906
ge 119899V + 2 and the last order on vehicle 119906 is processed inthe last batch If the last order on vehicle 119906 is moved to thelast position on vehicle v then the objective value will notincrease By repeating this process we can obtain a desiredoptimal schedule
(2) According to Lemma 7 in the optimal schedule 119861119897
contains all orders which finish transportation in the timeinterval (119878
119897minus1 119878119897] if the number of orders in a batch is nomore
than 119888 Since 119905119895
= 119905 for each order 119869119895 m orders arrive at the
machine together A batch contains 119898 orders or km (119896119898 le 119888)
orders if their waiting times are not greater than the limitedvalue 119882 Hence 119887
119897is integral for 119897 = 1 2 119871 minus 1
(3) It is trivial according to Lemma 3
The Scientific World Journal 7
Based on these results we can easily construct an optimalschedule for the problem
Algorithm D 2 Calculate 1198990
= lceil119899119898rceil 11989910158400
= lceil119899119888rceilFor each 119871 = 119899
1015840
0 1198991015840
0+ 1 119899
0 compute 119887
119897= 1198870
minus 1 119897 =
1 sdot sdot sdot ℎ 119887119897= 1198870 and 119897 = ℎ + 1 119871 where 119887
0= lceil1198990119871rceil and
ℎ = 1198870119871 minus 1198990
Consider the following two cases
Case 1 Consider ℎ = 0If (1198870minus1)119905 gt 119882 or119898119887
0gt 119888 then the constructed schedule
is not feasible and denote 119865(120587lowast
119871) = infin If (119887
0minus 1)119905 le 119882 and
1198981198870
le 119888 then 1198781
= 1198870119905 According to Lemma 6 there must
be 1198870119905119897 + 119905 + 119882 ge 119878
119897+ 119901 Otherwise it is not feasible Based
on Lemma 7 we can obtain
119878119897+1
= max 1198870119905 (119897 + 1) 119878
119897+ 119901 119897 = 1 2 119871 minus 1 (18)
The associated objective value can be derived as
1198651
(120587lowast
119871) = min
119871
sum
119897=1
1198981198870
(119878119897+ 119901)
minus 119892 (119878119871
+ 119901) + 119883 (119871)
(19)
Case 2 Consider ℎ gt 0If (1198870
minus 2)119905 gt 119882 or 1198981198870
gt 119888 then it is not feasible anddenote 119865(120587
lowast
119871) = infin If (119887
0minus 2)119905 le 119882 and 119898119887
0le 119888 then we
can show 1198781
= (1198870
minus 1)119905 According to Lemma 6 we have
(1198870
minus 1) 119905119897 + 119905 + 119882 ge 119878119897+ 119901 if 119897 le ℎ
(1198870
minus 1) 119905ℎ + 1198870119905 (119897 minus ℎ) + 119905 + 119882 ge 119878
119897+ 119901 if 119897 gt ℎ
(20)
Otherwise the constructed schedule is not feasible anddenote 119865(120587
lowast
119871) = infin
According to Lemma 6 we have
119878119897+1
= max (1198870
minus 1) 119905 (119897 + 1) 119878119897+ 119901 if 119897 le ℎ minus 1
119878119897+1
=max (1198870
minus 1) 119905ℎ + 1198870119905 (119897 + 1 minus ℎ) 119878
119897+ 119901 if 119897 ge ℎ
119897 = 1 2 119871 minus 1
(21)
The associated objective value can be derived as
1198652
(120587lowast
119871)
= min
ℎ
sum
119897=1
119898 (1198870
minus 1) (119878119897+ 119901) +
119871
sum
119897=ℎ+1
1198981198870
(119878119897+ 119901)
minus 119892 (119878119871
+ 119901) + 119883 (119871)
(22)
Thus the objective value is
119865 (120587lowast
119871) = min 119865
1(120587lowast
119871) 1198652
(120587lowast
119871) (23)
Theorem 12 The problem Vmrarr batch|119901119895
= 119901 119905119895
=
119905 119908119895| sum 119862119895
+ 119883(119909) can be solved by Algorithm D2 in 119874(119899119898)
time
Proof Equations (18) and (21) present that the machinecan start to process 119861
119897+1as soon as the orders assigned
into 119861119897+1
have arrived at the machine and the machine hasfinished processing119861
119897The local optimal solution in two cases
can be obtained by (19) and (22) The optimal solution isderived as (23) Hence the optimal schedule to this specialcase can be obtained by repeating the above procedure for1198991015840
0le 119871 le 119899
0and selecting the best candidate among the
solutions generated This shows that Algorithm D2 can findoptimal solution It is clear that the overall time complexityof Algorithm D2 is 119874(119899119898)
We now demonstrate the above algorithm with thefollowing numerical example
Example 13 Consider the instance with 11 orders 119898 = 2 119905 =
1 119888 = 4 119882 = 2 119901 = 2 and 119883(119909) = 5119909 Based on the abovemethod we know 119899
0= 6 119892 = 1 and 119899
1015840
0= 2 Then 119871 =
2 3 4 5 6 We have the following results
when 119871 = 2 we have 1198870
= 3 ℎ = 0 1198781
= 3 and119865(120587lowast) = 80
when 119871 = 3 we have 1198870
= 2 ℎ = 0 1198781
= 2 and119865(120587lowast) = 79
when 119871 = 4 we have 1198870
= 2 ℎ = 2 1198781
= 1 and119865(120587lowast
119871) = 101
when 119871 = 5 we have 1198870
= 2 ℎ = 4 1198870119905 sdot 3 + 119905 + 119882 lt
1198783
+ 119901 and 119865(120587lowast
119871) = infin
when 119871 = 6 we have 1198870
= 1 ℎ = 0 1198870119905 sdot 3 + 119905 + 119882 lt
1198783
+ 119901 and 119865(120587lowast
119871) = infin
Therefore we can obtain an optimal schedule120587lowast for 119871 = 3
with 1198611
= 1198691 1198692 1198693 1198694 1198612
= 1198695 1198696 1198697 1198698 and 119861
3=
1198699 11986910
11986911
6 Concluding Remarks
In this paper we address a coordinated scheduling problemwith order transportation before processing and parallel-batch production under the limited waiting time considera-tion Our goal is to optimize the sum of the total flow timeand the total processing cost First we prove that the generalproblem is NP-hard in the strong sense We also demonstratethat the problem with equal processing times on the machineisNP-hard Furthermore a dynamic programming algorithmin pseudopolynomial time is provided to prove its ordinarilyNP-hardness The problem with equal processing times andequal transportation times for each order can be solved inpolynomial time through taking into account the propertiesof the special case Our work has practical implications forthe coordination of batch-production and transportation toimprove the overall system performance in logistics andoperations management Simultaneously another importantimplication of our paper may provide a vital approach to deal
8 The Scientific World Journal
with the waiting time constraint applied to reduce the rawmaterial consumption and improve the machine utilization
There are several possible extensions to this researchFirst it is interesting to investigate the problems with otherobjective functions such as minimizing the makespan orminimizing maximum order tardinessearliness Anotherinteresting issue is to develop effective heuristics to solve thegeneral problem and investigate polynomial time algorithmsfor other special cases
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This research is partly supported by the National NaturalScience Foundation of China (Grant no 71101097) and theProgram for Liaoning Excellent Talents in University (Grantno LJQ2012017)
References
[1] P Brucker A Gladky H Hoogeveen et al ldquoScheduling abatching machinerdquo Journal of Scheduling vol 1 no 1 pp 31ndash541998
[2] C N Potts and M Y Kovalyov ldquoScheduling with batching areviewrdquo European Journal of Operational Research vol 120 no2 pp 228ndash249 2000
[3] C-Y Lee R Uzsoy and L AMartin-Vega ldquoEfficient algorithmsfor scheduling semiconductor burn-in operationsrdquo OperationsResearch vol 40 no 4 pp 764ndash775 1992
[4] L Tang J Guan and G Hu ldquoSteelmaking and refining coordi-nated scheduling problemwith waiting time and transportationconsiderationrdquo Computers and Industrial Engineering vol 58no 2 pp 239ndash248 2010
[5] C-Y Lee and V A Strusevich ldquoTwo-machine shop schedulingwith an uncapacitated interstage transporterrdquo IIE Transactionsvol 37 no 8 pp 725ndash736 2005
[6] L Tang and P Liu ldquoFlowshop scheduling problems withtransportation or deterioration between the batching and singlemachinesrdquo Computers and Industrial Engineering vol 56 no 4pp 1289ndash1295 2009
[7] H Gong and L Tang ldquoTwo-machine flowshop schedulingwith intermediate transportation under job physical spaceconsiderationrdquo Computers and Operations Research vol 38 no9 pp 1267ndash1274 2011
[8] L Tang and H Gong ldquoA hybrid two-stage transportation andbatch scheduling problemrdquo Applied Mathematical Modellingvol 32 no 12 pp 2467ndash2479 2008
[9] C-L Li and J Ou ldquoMachine scheduling with pickup anddeliveryrdquo Naval Research Logistics vol 52 no 7 pp 617ndash6302005
[10] Z-L Chen ldquoIntegrated production and outbound distributionscheduling Review and extensionsrdquo Operations Research vol58 no 1 pp 130ndash148 2010
[11] S Li J Yuan andB Fan ldquoUnbounded parallel-batch schedulingwith family jobs and delivery coordinationrdquo Information Pro-cessing Letters vol 111 no 12 pp 575ndash582 2011
[12] L Lu and J Yuan ldquoUnbounded parallel batch schedulingwith job delivery to minimize makespanrdquo Operations ResearchLetters vol 36 no 4 pp 477ndash480 2008
[13] M R Garey and D S Johnson Computers and IntractabilityA Guide To the Theory of NP-Completeness W H Freeman andCompany New York NY USA 1979
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
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Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
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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
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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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
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
4 The Scientific World Journal
Machine Y
0 2a 6a 12a
middot middot middot
middot middot middot
middot middot middot
7h
1ndash6 7ndash12 13ndash18
ah(h + 1) a(h + 1)(h + 2)
Vehicle 1
Vehicle 2
6h + 1 6h + 2 6h + 3
1ndash6 6h + 1
(6h minus 5)ndash6h
(6h minus 11)ndash(6h minus 6) 7h minus 1 (6h minus 5)ndash6h 7h7ndash12 6h + 2
Figure 1 The schedule of Theorem 4
y we first prove the following properties (1) since 119905119895
= 0 for119895 isin 119884 all the orders of 119884 as the first batch must be processedon the machine in the time interval [0 2119886] (2) there will beexactly ℎ + 1 batches in the optimal schedule and each batchcontains 7 orders Suppose that there are q (more than ℎ + 1)batches in the scheduleThen 7ℎ + 7 orders need at least ℎ + 1
batches due to 119888 = 7 If 119902 = ℎ + 2 then the total processingcost is 119883(119909) = 119883(ℎ + 2) = [(53)119886(ℎ + 1)(ℎ + 2)(ℎ + 3)]
2gt 119910
Thus all the orders of 119867 cup 119866 are divided into ℎ batches thatis all the batches are full
We will show that the 3-PP instance has a solution if andonly if there is a schedule 120587 for the scheduling instance suchthat its objective value of 120587 is no more than y
rarrAssume that there is a solution to the 3-PP instance1198671 1198672 119867
ℎ then there is a schedule to our problem with
an objective value of nomore than119910We construct a schedule120587 for our problem as shown in Figure 1
In this schedule two vehicles begin to transport ordersat time point 0 All the orders of 119884 are processed first on themachine as the first batch In the interval [119886119897(119897 minus 1) 119886119897(119897 + 1)]vehicle 1 transports the orders of 119867
119897 119867119897
= 6119897 minus 5 6119897 minus
4 6119897 vehicle 2 transports 1198696ℎ+119897
for 119897 = 1 2 ℎ Since1199056ℎ+119897
= 2119897119886 and sum119895isin119867119897
119905119895
= 2119897119886 1198696ℎ+119897
and the last order of 119867119897
arrive at the machine simultaneously for 119897 = 1 2 ℎ then1198696ℎ+119897
and the corresponding orders of 119867119897form batch 119861
119897 The
waiting time of each order in 119861119897is smaller than119882 Hence the
starting time of 119861119897on the machine is 119886119897(119897 + 1) the completion
time of 119861119897is 119886(119897 + 1)(119897 + 2) for 119897 = 0 1 2 ℎ It is easy to
see that the above schedule is optimal and the objective valueis y
larrOn the contrary suppose that there exists a schedulefor the constructed instance with the objective value notexceeding y
Fact 1 In the schedule 119861119897contains an auxiliary order 119869
6ℎ+119897
for 119897 = 1 2 ℎAssume that there is some batch 119861
119897that contains more
than one auxiliary order Assume that 119861119897contains two
auxiliary jobs 1198696ℎ+119897
and 1198696ℎ+119896
then there must be some batch119861119896only consisting of the ordinary jobs 119896 = 119897 Consider the
following two cases
Case 1 If 119896 = 119897 minus 1 then it is shown that 119861119897minus1
=
1198696119897minus11
1198696119897minus6
1198691198951015840 | 119895
1015840isin 119867119897 The corresponding flow
time of each order in 119861119897minus1
satisfies 119862119895
= 119886119897(119897 + 1) + 2119886
for119895 isin 119861119897minus1
The flow times of orders in each batch after 119861119897minus1
will be greater than 119886(119897 + 1)(119897 + 2) + 2119886 It is easy to see thatsum 119862119895
= 4119886(ℎ + 1)(ℎ + 2)(ℎ + 3) + 14(ℎ minus 119897 + 2)119886 2 le 119897 le ℎHence 1198651015840 = sum 119862
119895+ 119883(ℎ + 1) gt 119910
Case 2 If 119896 = 119897+1 then it can be shown that two vehiclesmusttransport two auxiliary orders of batch 119861
119897 respectively Due
to the limited waiting time of orders the completion time ofbatch 119861
119897on the machine is greater than 119886(119897 + 1)(119897 + 2) + 2119886
Furthermore it is easy to see that sum 119862119895
gt 4119886(ℎ + 1)(ℎ + 2)(ℎ +
3) + 14(ℎ minus 119897 + 1)119886 1 le 119897 le ℎ minus 1 which implies 1198651015840
gt 119910
Fact 2 Each batch 119861119897must contain 6 orders of 119867
119897= 6119897 minus
5 6119897 minus 4 6119897 such that sum119895isin119867119897
119905119895
= 2119897119886 for 119897 = 1 2 ℎ
By the above argument since each batch is full we canshow that one vehicle transports the orders of 119867
119897one by
one and another vehicle transports 1198696ℎ+119897
in the time interval[119886119897(119897 minus 1) 119886119897(119897 + 1)] The orders 119869
6ℎ+119897 119867119897 as 119861119897are processed
on themachine in the time interval [119886119897(119897+1) 119886(119897+1)(119897+2)] Itmust be true thatsum
119895isin119867119897119905119895
= 2119897119886 for 119897 = 1 2 ℎ Otherwiseif sum119895isin119867119897
119905119895
lt 2119897119886 then 1199016ℎ+119897
= 2(119897 + 1) sdot (1198863119897minus2
+ 1198863119897minus1
+ 1198863119897
) lt
2(119897+1)119886We can obtain some batch119861119896whose processing time
is greater than 2(119896 + 1)119886 119896 = 119897 Hence 1198651015840
gt 119910 which is acontradiction
Therefore a partition for the set 119867 is obtained by lettingthe elements be corresponding to batch 119861
119897 Then it is easy to
see that 119902 = ℎ + 1 and ℎ batches 1198611 119861
ℎform a solution to
the 3-PP instance The proof is concluded
Theorem 4 indicates that the existence of a polynomialtime algorithm to solve the scheduling model is unlikelySince the general problem is strongly NP-hard we nextconsider two special cases of the problem
4 The Problem 119881119898 rarr batch|119901119895
= 119901
119908119895| sum 119862119895
+ 119883(119909)
We now turn our attention to a special case with identicalprocessing time condition on themachine In this section weprove that the problem119881119898 rarr batch|119901
119895= 119901 119908
119895| sum 119862119895+119883(119909)
is ordinarily NP-hard by a reduction of the Partition Problem[13] which is a known NP-hard problem and provides adynamic programming algorithm in pseudopolynomial time
The Scientific World Journal 5
Suppose a feasible schedule 120587 = 1198611 1198612 119861119909 with 119909 isin
lceil119899119888rceil 119899 batches for the problem 119881119898 rarr batch|119901119895
=
119901 119908119895| sum 119862119895
+ 119883(119909) We now present some properties for theproblem
Lemma 5 For Vmrarr batch|119901119895
= 119901 119908119895| sum 119862119895
+ 119883(119909) thereexists an optimal schedule 120587
lowastsuch that all orders assigned intothe same vehicle are scheduled in the nondecreasing sequenceof their transportation times
Proof Assume that orders 119869119894and 119869119895are assigned to the same
vehicle 119869119894is followed by 119869
119895immediately such that 119905
119894ge 119905119895in
120587lowast Let 120587
1015840 be a schedule obtained by swapping 119869119894and 119869119895 In 1205871015840
it is easy to see that 119903119894gt 1199031015840
119895and 1199031015840
119894= 119903119895 Regardless of whether
119869119894and 119869119895are processed in the same batch or not the starting
times of 119869119894and 119869119895on the machine cannot increase We have
119865(120587lowast) ge 119865(120587
1015840)
Lemma 6 For any scheduleof the problem Vmrarr batch|119901119895
=
119901 119908119895| sum 119862119895+ 119883(119909) the starting time of batch 119861
119897satisfies 119878
119897+1ge
119878119897+ 119901 for 119897 = 1 2 119909
Proof It is easily obtained based on Lemma 3
Lemma 7 For any schedule of the problem Vmrarr batch|119901119895
=
119901 119908119895| sum 119862119895
+ 119883(119909) 119861119897contains all orders which have arrived
at the machine if the number of the orders is no more than thecapacity 119888 in the time interval (119878
119897minus1 119878119897]
Proof Assume that the batches are numbered in accordancewith their start times and the number of orders in eachbatch is smaller than the machine capacity 119888 Suppose that119869119895is the first order assigned in 119861
119897+1in 120587lowast and the arrival
time of 119869119895on the machine satisfies 119903
119895le 119878119897 Let 120587
1015840 bea new schedule obtained by simply assigning 119869
119895to 119861119897
Then 1198621015840
119895= 119878119897
+ 119901 lt 119878119897+1
+ 119901 le 119862119895 The schedule
of the remaining orders in 1205871015840 are the same as in 120587
lowast Itis obvious that 119865(120587
1015840) le 119865(120587
lowast) We can see that there
exists an optimal schedule in which all batches consist ofa number of orders which finish processing contiguously
We will show that the problem with equal processingtimes is NP-hard by a reduction from the Partition Problem
Theorem 8 The problem Vmrarr batch|119901119895
= 119901 119908119895| sum 119862119895
+
119883(119909) is NP-hard even if 119898 = 2
Proof Weprove this result by reducing the Partition Problemto Vmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909) The PartitionProblem (PP) can be stated as follows given ℎ items 119867 =
1 2 ℎ each item 119894 isin 119867 has a positive integer size 119886119894
such thatsumℎ
119894=1119886119894= 2119886 for some integer 119886The question asked
is whether there are two disjoint subsets1198671and119867
2 such that
sum119894isin1198671
119886119894= sum119894isin1198672
119886119894= 119886
We construct a corresponding instance of the problemVmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909) as follows
Number of Orders Consider
119899 = 2ℎ (8)
Transportation Times Consider
119905119895
= 119886119895 119895 isin 119867 = 1 2 ℎ
119905119895
= 0 119895 isin 119884 = ℎ + 1 ℎ + 2 2ℎ
(9)
Processing Times Consider
119901119895
= 119886 119895 = 1 2 2 ℎ (10)
Limited Waiting Time Conider
119882 = 119886 (11)
Processing Cost Function Consider
119883 (119909) = 3119886ℎ119909 (12)
Machine Capacity Consider
119888 = ℎ (13)
Threshold Value Consider
119910 = 9119886ℎ (14)
First it is easy to see that in a solution to this instance ofVmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895+ 119883(119909) with the objective value
not exceeding 119910 We first prove the following property since119905ℎ+1
= sdot sdot sdot = 1199052ℎ
= 0 all the orders of 119884 as the first batch mustbe processed on the machine in the time interval [0 a] Nowwe prove that there is a solution to the constructed instanceof Vmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909) with total cost notexceeding119910 if and only if there is a solution to the PP instance
rarr If there is a solution to the PP instance we show thatthere is a schedule for the above-constructed instance withan objective value of no more than 119910 Let 119867
1and 119867
2be
subsets of119867 that solves the PP instance Vehicle 1 and Vehicle2 transport the orders of119867
1and the orders of119867
2 respectively
Let119879119906be the total running time of vehicleu for119906 = 1 2 Since
sum119895isin1198671
119905119895
= sum119895isin1198672
119905119895
= 119886 we have 1198791
= 1198792
= 119886 In fact thecompletion time of the first batch on themachine is 119886We canobtain that the waiting time of each order on the machine isnot greater than 119886 Let 119861
1= 119884 and 119861
2= 119867 It is easy to see
that 1198871
= 1198872
= ℎ and the total flow time of all the orders is3ha Hence the objective is y
larrGiven a schedule with an objective value not exceedingy then there must be a solution to the PP instance We canobtain that all the orders in 119867 are processed in the secondbatch on the machine that is all the orders are divided intotwo batches and 119887
1= 1198872
= ℎ Suppose that the number ofthe batches is greater than 2 Note that even if the waitingtime constraint is ignored the objective value is 119865 ge 119886ℎ +
sum119895isin119867
119862119895
+ 9119886ℎ gt 119910 which is a contradiction Thus 120587 =
1198611 1198612 and 119861
1= 119884 and 119861
2= 119867
6 The Scientific World Journal
If 1198791
= sum119895isin1198671
119905119895
lt 119886 then it implies 1198792
= sum119895isin1198672
119905119895
gt 119886Hence the starting time of batch 119861
2is 1198782
= 1198792 It implies that
the total flow time is sum 119862119895
= 119886ℎ + (1198782
+ 119886)ℎ gt 3119886ℎ We have119865 gt 119910 which is a contradiction Thus sum
119895isin1198671119886119895
= sum119895isin1198672
119886119895
=
119886
Let 119879 = sum119899
119895=1119905119895 In the following we derive a dynamic
programming algorithm in pseudopolynomial time to solvethe problem Vmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909)
Algorithm D 1 Index the orders in the nondecreasing trans-portation time sequence that is 119905
1le 1199052
le sdot sdot sdot le 119905119899
Define 119867119877(119895 119894 119897 119879
1 1198792 119879
119898) and 119891
119877(119895 119894 119897 119879
1 1198792
119879119898
) as the minimum objective value and the minimum totalflow time of a partial schedule of orders 119869
1 119869
119895 respec-
tively where orders 1198691 119869
119895have finished transporting and
processing by using 119897 batches on themachine and the currentbatch 119861
119897contains orders 119869
119894 119869
119895
Initial Conditions Consider
119867119909
(119895 119894 119897 1198791 1198792 119879
119898)
=
0 119895 = 0 119894 = 0 119897 = 0 1198791
= 1198792
= sdot sdot sdot 119879119898
= 0
infin otherwise
(15)
for 1198781
ge 1199051 119897 = 1 2 119909 119909 isin lceil119899119888rceil 119899 and 119879
119906=
0 1 119879
Recursive Relations Consider
119891119909
(119895 119894 119897 1198791 1198792 119879
119898)
=
min1le119906le119898
119891119909
(119895 minus 1 119894 119897 1198791 119879
119906minus 119905119895 119879
119898) + 119862
119895 if 119887
119897lt 119888
infin if 119887119897ge 119888
(16)
where 119862119895
= 119878119897+ 119901|119878119897minus1
lt 119879119906
le 119878119897
le 1198791015840
119906+ 119882 119897 = 1 2 119909
for 119878119897+ 119901 le 119878
119897+1le 1198791015840
119906+ 119882 119897 = 1 2 119909 119909 isin lceil119899119888rceil 119899
and 1198791015840
119906= min
119878119897minus1lt119879119906le119878119897119879119906
Optimal Solution Consider
119865lowast
= min 119867119909
(119899 119899 + 1 119897 1198791 1198792 119879
119898) + 119883 (119909) (17)
Theorem 9 Algorithm D1 can find an optimal schedulefor the problem Vmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909) inO(mn3Tmminus1W) time
Proof Based on Lemma 5 there exists an optimal schedulewith orders assigned to each vehicle in the nondecreasingtransportation time sequence If order 119869
119895is assigned to vehicle
u then its starting time on the machine satisfies 119903119895+ 119882 le 119879
119906
If the number of orders in the current batch 119861119897is less than c
order 119869119895can be assigned into 119861
119897The corresponding flow time
increases 119878119897
+ 119901 Based on Lemma 3 we have 119878119897minus1
lt 119879119906
le
119878119897le 1198791015840
119906+ 119882 This shows that Algorithm D1 can find optimal
solution for the problemThe time complexity of the algorithm can be established
as follows We observe that 119894 119895 119897 le 119899 119879119906
le 119879 and 119898 minus 1 of
the value 1198791 1198792 119879
119898are independent Thus the number
of different states of the recursive relations is at most n3Tmminus1For each state each use requires 119874(119898119882) Therefore theoverall time complexity of Algorithm D1 is O(mn3Tmminus1W)
The existence of such a pseudopolynomial time algorithmfor a NP-hard problemmeans that the problem is NP-hard inthe ordinary sense We have the following theorem
Theorem 10 The problem Vmrarr batch|119901119895
= 119901 119908119895| sum 119862119895
+
119883(119909) is ordinarily NP-hard
5 The Problem 119881119898 rarr batch|119901119895
= 119901 119905119895
= 119905
119908119895| sum 119862119895
+ 119883(119909)
In this section we consider a special case with identicalprocessing times and identical transportation times andderive a polynomial-time algorithm to solve it It is evidentthat the problem reduces to an optimal batching problem onthe machine for the special case Note that 119909 is a decisionvariable denoting the number of batches on the machineNote that 119899
0= lceil119899119898rceil and 119892 = 119899
0119898 minus 119899 let 119899
119906be the number
of orders transported on vehicle 119906 under a specific scheduleNote that 119899
1015840
0= lceil119899119888rceil and 119899
1015840
0le 119909 le 119899
0 if 119888 ge 119898 Hence
the number of batches 119909 on the machine can be numberedby 1198991015840
0 1198991015840
0+ 1 119899
0 Otherwise 119899
0le 119909 le 119899
1015840
0if 119888 lt 119898
Without loss of generality assume that 119888 ge 119898 in the followingdiscussion Let 120587
lowast= (1198611 1198612 119861119871) be an optimal schedule
with given 119909 = 119871 batches for the problem Vmrarr batch|119901119895
=
119901 119905119895
= 119905 119908119895| sum 119862119895
+ 119883(119909) Then we have the followinglemma
Lemma 11 For the problem Vmrarr batch|119901119895
= 119901 119905119895
=
119905 119908119895| sum 119862119895
+ 119883(119909) there exists an schedule with 119871 batches inwhich (1) 119899
0minus 1 le 119899
119906le 1198990 119906 = 1 2 119898 (2) 119887
119897= 119887119897119898 119887119897
is integral and 119897 = 1 2 119871 minus 1 and (3) 119878119897+1
= max119903119895 119878119897
+
119901 where 119903119895denotes the arrival time of some 119898 jobs on the
machine 119897 = 1 2 119871 minus 1
Proof (1) Based on Lemma 2 each vehicle has no insertedidle time during transporting Assume that there exists anoptimal schedule 120587
lowast in which the condition is not satisfiedThen there must be a pair of vehicles 119906 and V such that119899119906
ge 119899V + 2 and the last order on vehicle 119906 is processed inthe last batch If the last order on vehicle 119906 is moved to thelast position on vehicle v then the objective value will notincrease By repeating this process we can obtain a desiredoptimal schedule
(2) According to Lemma 7 in the optimal schedule 119861119897
contains all orders which finish transportation in the timeinterval (119878
119897minus1 119878119897] if the number of orders in a batch is nomore
than 119888 Since 119905119895
= 119905 for each order 119869119895 m orders arrive at the
machine together A batch contains 119898 orders or km (119896119898 le 119888)
orders if their waiting times are not greater than the limitedvalue 119882 Hence 119887
119897is integral for 119897 = 1 2 119871 minus 1
(3) It is trivial according to Lemma 3
The Scientific World Journal 7
Based on these results we can easily construct an optimalschedule for the problem
Algorithm D 2 Calculate 1198990
= lceil119899119898rceil 11989910158400
= lceil119899119888rceilFor each 119871 = 119899
1015840
0 1198991015840
0+ 1 119899
0 compute 119887
119897= 1198870
minus 1 119897 =
1 sdot sdot sdot ℎ 119887119897= 1198870 and 119897 = ℎ + 1 119871 where 119887
0= lceil1198990119871rceil and
ℎ = 1198870119871 minus 1198990
Consider the following two cases
Case 1 Consider ℎ = 0If (1198870minus1)119905 gt 119882 or119898119887
0gt 119888 then the constructed schedule
is not feasible and denote 119865(120587lowast
119871) = infin If (119887
0minus 1)119905 le 119882 and
1198981198870
le 119888 then 1198781
= 1198870119905 According to Lemma 6 there must
be 1198870119905119897 + 119905 + 119882 ge 119878
119897+ 119901 Otherwise it is not feasible Based
on Lemma 7 we can obtain
119878119897+1
= max 1198870119905 (119897 + 1) 119878
119897+ 119901 119897 = 1 2 119871 minus 1 (18)
The associated objective value can be derived as
1198651
(120587lowast
119871) = min
119871
sum
119897=1
1198981198870
(119878119897+ 119901)
minus 119892 (119878119871
+ 119901) + 119883 (119871)
(19)
Case 2 Consider ℎ gt 0If (1198870
minus 2)119905 gt 119882 or 1198981198870
gt 119888 then it is not feasible anddenote 119865(120587
lowast
119871) = infin If (119887
0minus 2)119905 le 119882 and 119898119887
0le 119888 then we
can show 1198781
= (1198870
minus 1)119905 According to Lemma 6 we have
(1198870
minus 1) 119905119897 + 119905 + 119882 ge 119878119897+ 119901 if 119897 le ℎ
(1198870
minus 1) 119905ℎ + 1198870119905 (119897 minus ℎ) + 119905 + 119882 ge 119878
119897+ 119901 if 119897 gt ℎ
(20)
Otherwise the constructed schedule is not feasible anddenote 119865(120587
lowast
119871) = infin
According to Lemma 6 we have
119878119897+1
= max (1198870
minus 1) 119905 (119897 + 1) 119878119897+ 119901 if 119897 le ℎ minus 1
119878119897+1
=max (1198870
minus 1) 119905ℎ + 1198870119905 (119897 + 1 minus ℎ) 119878
119897+ 119901 if 119897 ge ℎ
119897 = 1 2 119871 minus 1
(21)
The associated objective value can be derived as
1198652
(120587lowast
119871)
= min
ℎ
sum
119897=1
119898 (1198870
minus 1) (119878119897+ 119901) +
119871
sum
119897=ℎ+1
1198981198870
(119878119897+ 119901)
minus 119892 (119878119871
+ 119901) + 119883 (119871)
(22)
Thus the objective value is
119865 (120587lowast
119871) = min 119865
1(120587lowast
119871) 1198652
(120587lowast
119871) (23)
Theorem 12 The problem Vmrarr batch|119901119895
= 119901 119905119895
=
119905 119908119895| sum 119862119895
+ 119883(119909) can be solved by Algorithm D2 in 119874(119899119898)
time
Proof Equations (18) and (21) present that the machinecan start to process 119861
119897+1as soon as the orders assigned
into 119861119897+1
have arrived at the machine and the machine hasfinished processing119861
119897The local optimal solution in two cases
can be obtained by (19) and (22) The optimal solution isderived as (23) Hence the optimal schedule to this specialcase can be obtained by repeating the above procedure for1198991015840
0le 119871 le 119899
0and selecting the best candidate among the
solutions generated This shows that Algorithm D2 can findoptimal solution It is clear that the overall time complexityof Algorithm D2 is 119874(119899119898)
We now demonstrate the above algorithm with thefollowing numerical example
Example 13 Consider the instance with 11 orders 119898 = 2 119905 =
1 119888 = 4 119882 = 2 119901 = 2 and 119883(119909) = 5119909 Based on the abovemethod we know 119899
0= 6 119892 = 1 and 119899
1015840
0= 2 Then 119871 =
2 3 4 5 6 We have the following results
when 119871 = 2 we have 1198870
= 3 ℎ = 0 1198781
= 3 and119865(120587lowast) = 80
when 119871 = 3 we have 1198870
= 2 ℎ = 0 1198781
= 2 and119865(120587lowast) = 79
when 119871 = 4 we have 1198870
= 2 ℎ = 2 1198781
= 1 and119865(120587lowast
119871) = 101
when 119871 = 5 we have 1198870
= 2 ℎ = 4 1198870119905 sdot 3 + 119905 + 119882 lt
1198783
+ 119901 and 119865(120587lowast
119871) = infin
when 119871 = 6 we have 1198870
= 1 ℎ = 0 1198870119905 sdot 3 + 119905 + 119882 lt
1198783
+ 119901 and 119865(120587lowast
119871) = infin
Therefore we can obtain an optimal schedule120587lowast for 119871 = 3
with 1198611
= 1198691 1198692 1198693 1198694 1198612
= 1198695 1198696 1198697 1198698 and 119861
3=
1198699 11986910
11986911
6 Concluding Remarks
In this paper we address a coordinated scheduling problemwith order transportation before processing and parallel-batch production under the limited waiting time considera-tion Our goal is to optimize the sum of the total flow timeand the total processing cost First we prove that the generalproblem is NP-hard in the strong sense We also demonstratethat the problem with equal processing times on the machineisNP-hard Furthermore a dynamic programming algorithmin pseudopolynomial time is provided to prove its ordinarilyNP-hardness The problem with equal processing times andequal transportation times for each order can be solved inpolynomial time through taking into account the propertiesof the special case Our work has practical implications forthe coordination of batch-production and transportation toimprove the overall system performance in logistics andoperations management Simultaneously another importantimplication of our paper may provide a vital approach to deal
8 The Scientific World Journal
with the waiting time constraint applied to reduce the rawmaterial consumption and improve the machine utilization
There are several possible extensions to this researchFirst it is interesting to investigate the problems with otherobjective functions such as minimizing the makespan orminimizing maximum order tardinessearliness Anotherinteresting issue is to develop effective heuristics to solve thegeneral problem and investigate polynomial time algorithmsfor other special cases
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This research is partly supported by the National NaturalScience Foundation of China (Grant no 71101097) and theProgram for Liaoning Excellent Talents in University (Grantno LJQ2012017)
References
[1] P Brucker A Gladky H Hoogeveen et al ldquoScheduling abatching machinerdquo Journal of Scheduling vol 1 no 1 pp 31ndash541998
[2] C N Potts and M Y Kovalyov ldquoScheduling with batching areviewrdquo European Journal of Operational Research vol 120 no2 pp 228ndash249 2000
[3] C-Y Lee R Uzsoy and L AMartin-Vega ldquoEfficient algorithmsfor scheduling semiconductor burn-in operationsrdquo OperationsResearch vol 40 no 4 pp 764ndash775 1992
[4] L Tang J Guan and G Hu ldquoSteelmaking and refining coordi-nated scheduling problemwith waiting time and transportationconsiderationrdquo Computers and Industrial Engineering vol 58no 2 pp 239ndash248 2010
[5] C-Y Lee and V A Strusevich ldquoTwo-machine shop schedulingwith an uncapacitated interstage transporterrdquo IIE Transactionsvol 37 no 8 pp 725ndash736 2005
[6] L Tang and P Liu ldquoFlowshop scheduling problems withtransportation or deterioration between the batching and singlemachinesrdquo Computers and Industrial Engineering vol 56 no 4pp 1289ndash1295 2009
[7] H Gong and L Tang ldquoTwo-machine flowshop schedulingwith intermediate transportation under job physical spaceconsiderationrdquo Computers and Operations Research vol 38 no9 pp 1267ndash1274 2011
[8] L Tang and H Gong ldquoA hybrid two-stage transportation andbatch scheduling problemrdquo Applied Mathematical Modellingvol 32 no 12 pp 2467ndash2479 2008
[9] C-L Li and J Ou ldquoMachine scheduling with pickup anddeliveryrdquo Naval Research Logistics vol 52 no 7 pp 617ndash6302005
[10] Z-L Chen ldquoIntegrated production and outbound distributionscheduling Review and extensionsrdquo Operations Research vol58 no 1 pp 130ndash148 2010
[11] S Li J Yuan andB Fan ldquoUnbounded parallel-batch schedulingwith family jobs and delivery coordinationrdquo Information Pro-cessing Letters vol 111 no 12 pp 575ndash582 2011
[12] L Lu and J Yuan ldquoUnbounded parallel batch schedulingwith job delivery to minimize makespanrdquo Operations ResearchLetters vol 36 no 4 pp 477ndash480 2008
[13] M R Garey and D S Johnson Computers and IntractabilityA Guide To the Theory of NP-Completeness W H Freeman andCompany New York NY USA 1979
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
The Scientific World Journal 5
Suppose a feasible schedule 120587 = 1198611 1198612 119861119909 with 119909 isin
lceil119899119888rceil 119899 batches for the problem 119881119898 rarr batch|119901119895
=
119901 119908119895| sum 119862119895
+ 119883(119909) We now present some properties for theproblem
Lemma 5 For Vmrarr batch|119901119895
= 119901 119908119895| sum 119862119895
+ 119883(119909) thereexists an optimal schedule 120587
lowastsuch that all orders assigned intothe same vehicle are scheduled in the nondecreasing sequenceof their transportation times
Proof Assume that orders 119869119894and 119869119895are assigned to the same
vehicle 119869119894is followed by 119869
119895immediately such that 119905
119894ge 119905119895in
120587lowast Let 120587
1015840 be a schedule obtained by swapping 119869119894and 119869119895 In 1205871015840
it is easy to see that 119903119894gt 1199031015840
119895and 1199031015840
119894= 119903119895 Regardless of whether
119869119894and 119869119895are processed in the same batch or not the starting
times of 119869119894and 119869119895on the machine cannot increase We have
119865(120587lowast) ge 119865(120587
1015840)
Lemma 6 For any scheduleof the problem Vmrarr batch|119901119895
=
119901 119908119895| sum 119862119895+ 119883(119909) the starting time of batch 119861
119897satisfies 119878
119897+1ge
119878119897+ 119901 for 119897 = 1 2 119909
Proof It is easily obtained based on Lemma 3
Lemma 7 For any schedule of the problem Vmrarr batch|119901119895
=
119901 119908119895| sum 119862119895
+ 119883(119909) 119861119897contains all orders which have arrived
at the machine if the number of the orders is no more than thecapacity 119888 in the time interval (119878
119897minus1 119878119897]
Proof Assume that the batches are numbered in accordancewith their start times and the number of orders in eachbatch is smaller than the machine capacity 119888 Suppose that119869119895is the first order assigned in 119861
119897+1in 120587lowast and the arrival
time of 119869119895on the machine satisfies 119903
119895le 119878119897 Let 120587
1015840 bea new schedule obtained by simply assigning 119869
119895to 119861119897
Then 1198621015840
119895= 119878119897
+ 119901 lt 119878119897+1
+ 119901 le 119862119895 The schedule
of the remaining orders in 1205871015840 are the same as in 120587
lowast Itis obvious that 119865(120587
1015840) le 119865(120587
lowast) We can see that there
exists an optimal schedule in which all batches consist ofa number of orders which finish processing contiguously
We will show that the problem with equal processingtimes is NP-hard by a reduction from the Partition Problem
Theorem 8 The problem Vmrarr batch|119901119895
= 119901 119908119895| sum 119862119895
+
119883(119909) is NP-hard even if 119898 = 2
Proof Weprove this result by reducing the Partition Problemto Vmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909) The PartitionProblem (PP) can be stated as follows given ℎ items 119867 =
1 2 ℎ each item 119894 isin 119867 has a positive integer size 119886119894
such thatsumℎ
119894=1119886119894= 2119886 for some integer 119886The question asked
is whether there are two disjoint subsets1198671and119867
2 such that
sum119894isin1198671
119886119894= sum119894isin1198672
119886119894= 119886
We construct a corresponding instance of the problemVmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909) as follows
Number of Orders Consider
119899 = 2ℎ (8)
Transportation Times Consider
119905119895
= 119886119895 119895 isin 119867 = 1 2 ℎ
119905119895
= 0 119895 isin 119884 = ℎ + 1 ℎ + 2 2ℎ
(9)
Processing Times Consider
119901119895
= 119886 119895 = 1 2 2 ℎ (10)
Limited Waiting Time Conider
119882 = 119886 (11)
Processing Cost Function Consider
119883 (119909) = 3119886ℎ119909 (12)
Machine Capacity Consider
119888 = ℎ (13)
Threshold Value Consider
119910 = 9119886ℎ (14)
First it is easy to see that in a solution to this instance ofVmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895+ 119883(119909) with the objective value
not exceeding 119910 We first prove the following property since119905ℎ+1
= sdot sdot sdot = 1199052ℎ
= 0 all the orders of 119884 as the first batch mustbe processed on the machine in the time interval [0 a] Nowwe prove that there is a solution to the constructed instanceof Vmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909) with total cost notexceeding119910 if and only if there is a solution to the PP instance
rarr If there is a solution to the PP instance we show thatthere is a schedule for the above-constructed instance withan objective value of no more than 119910 Let 119867
1and 119867
2be
subsets of119867 that solves the PP instance Vehicle 1 and Vehicle2 transport the orders of119867
1and the orders of119867
2 respectively
Let119879119906be the total running time of vehicleu for119906 = 1 2 Since
sum119895isin1198671
119905119895
= sum119895isin1198672
119905119895
= 119886 we have 1198791
= 1198792
= 119886 In fact thecompletion time of the first batch on themachine is 119886We canobtain that the waiting time of each order on the machine isnot greater than 119886 Let 119861
1= 119884 and 119861
2= 119867 It is easy to see
that 1198871
= 1198872
= ℎ and the total flow time of all the orders is3ha Hence the objective is y
larrGiven a schedule with an objective value not exceedingy then there must be a solution to the PP instance We canobtain that all the orders in 119867 are processed in the secondbatch on the machine that is all the orders are divided intotwo batches and 119887
1= 1198872
= ℎ Suppose that the number ofthe batches is greater than 2 Note that even if the waitingtime constraint is ignored the objective value is 119865 ge 119886ℎ +
sum119895isin119867
119862119895
+ 9119886ℎ gt 119910 which is a contradiction Thus 120587 =
1198611 1198612 and 119861
1= 119884 and 119861
2= 119867
6 The Scientific World Journal
If 1198791
= sum119895isin1198671
119905119895
lt 119886 then it implies 1198792
= sum119895isin1198672
119905119895
gt 119886Hence the starting time of batch 119861
2is 1198782
= 1198792 It implies that
the total flow time is sum 119862119895
= 119886ℎ + (1198782
+ 119886)ℎ gt 3119886ℎ We have119865 gt 119910 which is a contradiction Thus sum
119895isin1198671119886119895
= sum119895isin1198672
119886119895
=
119886
Let 119879 = sum119899
119895=1119905119895 In the following we derive a dynamic
programming algorithm in pseudopolynomial time to solvethe problem Vmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909)
Algorithm D 1 Index the orders in the nondecreasing trans-portation time sequence that is 119905
1le 1199052
le sdot sdot sdot le 119905119899
Define 119867119877(119895 119894 119897 119879
1 1198792 119879
119898) and 119891
119877(119895 119894 119897 119879
1 1198792
119879119898
) as the minimum objective value and the minimum totalflow time of a partial schedule of orders 119869
1 119869
119895 respec-
tively where orders 1198691 119869
119895have finished transporting and
processing by using 119897 batches on themachine and the currentbatch 119861
119897contains orders 119869
119894 119869
119895
Initial Conditions Consider
119867119909
(119895 119894 119897 1198791 1198792 119879
119898)
=
0 119895 = 0 119894 = 0 119897 = 0 1198791
= 1198792
= sdot sdot sdot 119879119898
= 0
infin otherwise
(15)
for 1198781
ge 1199051 119897 = 1 2 119909 119909 isin lceil119899119888rceil 119899 and 119879
119906=
0 1 119879
Recursive Relations Consider
119891119909
(119895 119894 119897 1198791 1198792 119879
119898)
=
min1le119906le119898
119891119909
(119895 minus 1 119894 119897 1198791 119879
119906minus 119905119895 119879
119898) + 119862
119895 if 119887
119897lt 119888
infin if 119887119897ge 119888
(16)
where 119862119895
= 119878119897+ 119901|119878119897minus1
lt 119879119906
le 119878119897
le 1198791015840
119906+ 119882 119897 = 1 2 119909
for 119878119897+ 119901 le 119878
119897+1le 1198791015840
119906+ 119882 119897 = 1 2 119909 119909 isin lceil119899119888rceil 119899
and 1198791015840
119906= min
119878119897minus1lt119879119906le119878119897119879119906
Optimal Solution Consider
119865lowast
= min 119867119909
(119899 119899 + 1 119897 1198791 1198792 119879
119898) + 119883 (119909) (17)
Theorem 9 Algorithm D1 can find an optimal schedulefor the problem Vmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909) inO(mn3Tmminus1W) time
Proof Based on Lemma 5 there exists an optimal schedulewith orders assigned to each vehicle in the nondecreasingtransportation time sequence If order 119869
119895is assigned to vehicle
u then its starting time on the machine satisfies 119903119895+ 119882 le 119879
119906
If the number of orders in the current batch 119861119897is less than c
order 119869119895can be assigned into 119861
119897The corresponding flow time
increases 119878119897
+ 119901 Based on Lemma 3 we have 119878119897minus1
lt 119879119906
le
119878119897le 1198791015840
119906+ 119882 This shows that Algorithm D1 can find optimal
solution for the problemThe time complexity of the algorithm can be established
as follows We observe that 119894 119895 119897 le 119899 119879119906
le 119879 and 119898 minus 1 of
the value 1198791 1198792 119879
119898are independent Thus the number
of different states of the recursive relations is at most n3Tmminus1For each state each use requires 119874(119898119882) Therefore theoverall time complexity of Algorithm D1 is O(mn3Tmminus1W)
The existence of such a pseudopolynomial time algorithmfor a NP-hard problemmeans that the problem is NP-hard inthe ordinary sense We have the following theorem
Theorem 10 The problem Vmrarr batch|119901119895
= 119901 119908119895| sum 119862119895
+
119883(119909) is ordinarily NP-hard
5 The Problem 119881119898 rarr batch|119901119895
= 119901 119905119895
= 119905
119908119895| sum 119862119895
+ 119883(119909)
In this section we consider a special case with identicalprocessing times and identical transportation times andderive a polynomial-time algorithm to solve it It is evidentthat the problem reduces to an optimal batching problem onthe machine for the special case Note that 119909 is a decisionvariable denoting the number of batches on the machineNote that 119899
0= lceil119899119898rceil and 119892 = 119899
0119898 minus 119899 let 119899
119906be the number
of orders transported on vehicle 119906 under a specific scheduleNote that 119899
1015840
0= lceil119899119888rceil and 119899
1015840
0le 119909 le 119899
0 if 119888 ge 119898 Hence
the number of batches 119909 on the machine can be numberedby 1198991015840
0 1198991015840
0+ 1 119899
0 Otherwise 119899
0le 119909 le 119899
1015840
0if 119888 lt 119898
Without loss of generality assume that 119888 ge 119898 in the followingdiscussion Let 120587
lowast= (1198611 1198612 119861119871) be an optimal schedule
with given 119909 = 119871 batches for the problem Vmrarr batch|119901119895
=
119901 119905119895
= 119905 119908119895| sum 119862119895
+ 119883(119909) Then we have the followinglemma
Lemma 11 For the problem Vmrarr batch|119901119895
= 119901 119905119895
=
119905 119908119895| sum 119862119895
+ 119883(119909) there exists an schedule with 119871 batches inwhich (1) 119899
0minus 1 le 119899
119906le 1198990 119906 = 1 2 119898 (2) 119887
119897= 119887119897119898 119887119897
is integral and 119897 = 1 2 119871 minus 1 and (3) 119878119897+1
= max119903119895 119878119897
+
119901 where 119903119895denotes the arrival time of some 119898 jobs on the
machine 119897 = 1 2 119871 minus 1
Proof (1) Based on Lemma 2 each vehicle has no insertedidle time during transporting Assume that there exists anoptimal schedule 120587
lowast in which the condition is not satisfiedThen there must be a pair of vehicles 119906 and V such that119899119906
ge 119899V + 2 and the last order on vehicle 119906 is processed inthe last batch If the last order on vehicle 119906 is moved to thelast position on vehicle v then the objective value will notincrease By repeating this process we can obtain a desiredoptimal schedule
(2) According to Lemma 7 in the optimal schedule 119861119897
contains all orders which finish transportation in the timeinterval (119878
119897minus1 119878119897] if the number of orders in a batch is nomore
than 119888 Since 119905119895
= 119905 for each order 119869119895 m orders arrive at the
machine together A batch contains 119898 orders or km (119896119898 le 119888)
orders if their waiting times are not greater than the limitedvalue 119882 Hence 119887
119897is integral for 119897 = 1 2 119871 minus 1
(3) It is trivial according to Lemma 3
The Scientific World Journal 7
Based on these results we can easily construct an optimalschedule for the problem
Algorithm D 2 Calculate 1198990
= lceil119899119898rceil 11989910158400
= lceil119899119888rceilFor each 119871 = 119899
1015840
0 1198991015840
0+ 1 119899
0 compute 119887
119897= 1198870
minus 1 119897 =
1 sdot sdot sdot ℎ 119887119897= 1198870 and 119897 = ℎ + 1 119871 where 119887
0= lceil1198990119871rceil and
ℎ = 1198870119871 minus 1198990
Consider the following two cases
Case 1 Consider ℎ = 0If (1198870minus1)119905 gt 119882 or119898119887
0gt 119888 then the constructed schedule
is not feasible and denote 119865(120587lowast
119871) = infin If (119887
0minus 1)119905 le 119882 and
1198981198870
le 119888 then 1198781
= 1198870119905 According to Lemma 6 there must
be 1198870119905119897 + 119905 + 119882 ge 119878
119897+ 119901 Otherwise it is not feasible Based
on Lemma 7 we can obtain
119878119897+1
= max 1198870119905 (119897 + 1) 119878
119897+ 119901 119897 = 1 2 119871 minus 1 (18)
The associated objective value can be derived as
1198651
(120587lowast
119871) = min
119871
sum
119897=1
1198981198870
(119878119897+ 119901)
minus 119892 (119878119871
+ 119901) + 119883 (119871)
(19)
Case 2 Consider ℎ gt 0If (1198870
minus 2)119905 gt 119882 or 1198981198870
gt 119888 then it is not feasible anddenote 119865(120587
lowast
119871) = infin If (119887
0minus 2)119905 le 119882 and 119898119887
0le 119888 then we
can show 1198781
= (1198870
minus 1)119905 According to Lemma 6 we have
(1198870
minus 1) 119905119897 + 119905 + 119882 ge 119878119897+ 119901 if 119897 le ℎ
(1198870
minus 1) 119905ℎ + 1198870119905 (119897 minus ℎ) + 119905 + 119882 ge 119878
119897+ 119901 if 119897 gt ℎ
(20)
Otherwise the constructed schedule is not feasible anddenote 119865(120587
lowast
119871) = infin
According to Lemma 6 we have
119878119897+1
= max (1198870
minus 1) 119905 (119897 + 1) 119878119897+ 119901 if 119897 le ℎ minus 1
119878119897+1
=max (1198870
minus 1) 119905ℎ + 1198870119905 (119897 + 1 minus ℎ) 119878
119897+ 119901 if 119897 ge ℎ
119897 = 1 2 119871 minus 1
(21)
The associated objective value can be derived as
1198652
(120587lowast
119871)
= min
ℎ
sum
119897=1
119898 (1198870
minus 1) (119878119897+ 119901) +
119871
sum
119897=ℎ+1
1198981198870
(119878119897+ 119901)
minus 119892 (119878119871
+ 119901) + 119883 (119871)
(22)
Thus the objective value is
119865 (120587lowast
119871) = min 119865
1(120587lowast
119871) 1198652
(120587lowast
119871) (23)
Theorem 12 The problem Vmrarr batch|119901119895
= 119901 119905119895
=
119905 119908119895| sum 119862119895
+ 119883(119909) can be solved by Algorithm D2 in 119874(119899119898)
time
Proof Equations (18) and (21) present that the machinecan start to process 119861
119897+1as soon as the orders assigned
into 119861119897+1
have arrived at the machine and the machine hasfinished processing119861
119897The local optimal solution in two cases
can be obtained by (19) and (22) The optimal solution isderived as (23) Hence the optimal schedule to this specialcase can be obtained by repeating the above procedure for1198991015840
0le 119871 le 119899
0and selecting the best candidate among the
solutions generated This shows that Algorithm D2 can findoptimal solution It is clear that the overall time complexityof Algorithm D2 is 119874(119899119898)
We now demonstrate the above algorithm with thefollowing numerical example
Example 13 Consider the instance with 11 orders 119898 = 2 119905 =
1 119888 = 4 119882 = 2 119901 = 2 and 119883(119909) = 5119909 Based on the abovemethod we know 119899
0= 6 119892 = 1 and 119899
1015840
0= 2 Then 119871 =
2 3 4 5 6 We have the following results
when 119871 = 2 we have 1198870
= 3 ℎ = 0 1198781
= 3 and119865(120587lowast) = 80
when 119871 = 3 we have 1198870
= 2 ℎ = 0 1198781
= 2 and119865(120587lowast) = 79
when 119871 = 4 we have 1198870
= 2 ℎ = 2 1198781
= 1 and119865(120587lowast
119871) = 101
when 119871 = 5 we have 1198870
= 2 ℎ = 4 1198870119905 sdot 3 + 119905 + 119882 lt
1198783
+ 119901 and 119865(120587lowast
119871) = infin
when 119871 = 6 we have 1198870
= 1 ℎ = 0 1198870119905 sdot 3 + 119905 + 119882 lt
1198783
+ 119901 and 119865(120587lowast
119871) = infin
Therefore we can obtain an optimal schedule120587lowast for 119871 = 3
with 1198611
= 1198691 1198692 1198693 1198694 1198612
= 1198695 1198696 1198697 1198698 and 119861
3=
1198699 11986910
11986911
6 Concluding Remarks
In this paper we address a coordinated scheduling problemwith order transportation before processing and parallel-batch production under the limited waiting time considera-tion Our goal is to optimize the sum of the total flow timeand the total processing cost First we prove that the generalproblem is NP-hard in the strong sense We also demonstratethat the problem with equal processing times on the machineisNP-hard Furthermore a dynamic programming algorithmin pseudopolynomial time is provided to prove its ordinarilyNP-hardness The problem with equal processing times andequal transportation times for each order can be solved inpolynomial time through taking into account the propertiesof the special case Our work has practical implications forthe coordination of batch-production and transportation toimprove the overall system performance in logistics andoperations management Simultaneously another importantimplication of our paper may provide a vital approach to deal
8 The Scientific World Journal
with the waiting time constraint applied to reduce the rawmaterial consumption and improve the machine utilization
There are several possible extensions to this researchFirst it is interesting to investigate the problems with otherobjective functions such as minimizing the makespan orminimizing maximum order tardinessearliness Anotherinteresting issue is to develop effective heuristics to solve thegeneral problem and investigate polynomial time algorithmsfor other special cases
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This research is partly supported by the National NaturalScience Foundation of China (Grant no 71101097) and theProgram for Liaoning Excellent Talents in University (Grantno LJQ2012017)
References
[1] P Brucker A Gladky H Hoogeveen et al ldquoScheduling abatching machinerdquo Journal of Scheduling vol 1 no 1 pp 31ndash541998
[2] C N Potts and M Y Kovalyov ldquoScheduling with batching areviewrdquo European Journal of Operational Research vol 120 no2 pp 228ndash249 2000
[3] C-Y Lee R Uzsoy and L AMartin-Vega ldquoEfficient algorithmsfor scheduling semiconductor burn-in operationsrdquo OperationsResearch vol 40 no 4 pp 764ndash775 1992
[4] L Tang J Guan and G Hu ldquoSteelmaking and refining coordi-nated scheduling problemwith waiting time and transportationconsiderationrdquo Computers and Industrial Engineering vol 58no 2 pp 239ndash248 2010
[5] C-Y Lee and V A Strusevich ldquoTwo-machine shop schedulingwith an uncapacitated interstage transporterrdquo IIE Transactionsvol 37 no 8 pp 725ndash736 2005
[6] L Tang and P Liu ldquoFlowshop scheduling problems withtransportation or deterioration between the batching and singlemachinesrdquo Computers and Industrial Engineering vol 56 no 4pp 1289ndash1295 2009
[7] H Gong and L Tang ldquoTwo-machine flowshop schedulingwith intermediate transportation under job physical spaceconsiderationrdquo Computers and Operations Research vol 38 no9 pp 1267ndash1274 2011
[8] L Tang and H Gong ldquoA hybrid two-stage transportation andbatch scheduling problemrdquo Applied Mathematical Modellingvol 32 no 12 pp 2467ndash2479 2008
[9] C-L Li and J Ou ldquoMachine scheduling with pickup anddeliveryrdquo Naval Research Logistics vol 52 no 7 pp 617ndash6302005
[10] Z-L Chen ldquoIntegrated production and outbound distributionscheduling Review and extensionsrdquo Operations Research vol58 no 1 pp 130ndash148 2010
[11] S Li J Yuan andB Fan ldquoUnbounded parallel-batch schedulingwith family jobs and delivery coordinationrdquo Information Pro-cessing Letters vol 111 no 12 pp 575ndash582 2011
[12] L Lu and J Yuan ldquoUnbounded parallel batch schedulingwith job delivery to minimize makespanrdquo Operations ResearchLetters vol 36 no 4 pp 477ndash480 2008
[13] M R Garey and D S Johnson Computers and IntractabilityA Guide To the Theory of NP-Completeness W H Freeman andCompany New York NY USA 1979
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
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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
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
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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
6 The Scientific World Journal
If 1198791
= sum119895isin1198671
119905119895
lt 119886 then it implies 1198792
= sum119895isin1198672
119905119895
gt 119886Hence the starting time of batch 119861
2is 1198782
= 1198792 It implies that
the total flow time is sum 119862119895
= 119886ℎ + (1198782
+ 119886)ℎ gt 3119886ℎ We have119865 gt 119910 which is a contradiction Thus sum
119895isin1198671119886119895
= sum119895isin1198672
119886119895
=
119886
Let 119879 = sum119899
119895=1119905119895 In the following we derive a dynamic
programming algorithm in pseudopolynomial time to solvethe problem Vmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909)
Algorithm D 1 Index the orders in the nondecreasing trans-portation time sequence that is 119905
1le 1199052
le sdot sdot sdot le 119905119899
Define 119867119877(119895 119894 119897 119879
1 1198792 119879
119898) and 119891
119877(119895 119894 119897 119879
1 1198792
119879119898
) as the minimum objective value and the minimum totalflow time of a partial schedule of orders 119869
1 119869
119895 respec-
tively where orders 1198691 119869
119895have finished transporting and
processing by using 119897 batches on themachine and the currentbatch 119861
119897contains orders 119869
119894 119869
119895
Initial Conditions Consider
119867119909
(119895 119894 119897 1198791 1198792 119879
119898)
=
0 119895 = 0 119894 = 0 119897 = 0 1198791
= 1198792
= sdot sdot sdot 119879119898
= 0
infin otherwise
(15)
for 1198781
ge 1199051 119897 = 1 2 119909 119909 isin lceil119899119888rceil 119899 and 119879
119906=
0 1 119879
Recursive Relations Consider
119891119909
(119895 119894 119897 1198791 1198792 119879
119898)
=
min1le119906le119898
119891119909
(119895 minus 1 119894 119897 1198791 119879
119906minus 119905119895 119879
119898) + 119862
119895 if 119887
119897lt 119888
infin if 119887119897ge 119888
(16)
where 119862119895
= 119878119897+ 119901|119878119897minus1
lt 119879119906
le 119878119897
le 1198791015840
119906+ 119882 119897 = 1 2 119909
for 119878119897+ 119901 le 119878
119897+1le 1198791015840
119906+ 119882 119897 = 1 2 119909 119909 isin lceil119899119888rceil 119899
and 1198791015840
119906= min
119878119897minus1lt119879119906le119878119897119879119906
Optimal Solution Consider
119865lowast
= min 119867119909
(119899 119899 + 1 119897 1198791 1198792 119879
119898) + 119883 (119909) (17)
Theorem 9 Algorithm D1 can find an optimal schedulefor the problem Vmrarr batch|119901
119895= 119901 119908
119895| sum 119862119895
+ 119883(119909) inO(mn3Tmminus1W) time
Proof Based on Lemma 5 there exists an optimal schedulewith orders assigned to each vehicle in the nondecreasingtransportation time sequence If order 119869
119895is assigned to vehicle
u then its starting time on the machine satisfies 119903119895+ 119882 le 119879
119906
If the number of orders in the current batch 119861119897is less than c
order 119869119895can be assigned into 119861
119897The corresponding flow time
increases 119878119897
+ 119901 Based on Lemma 3 we have 119878119897minus1
lt 119879119906
le
119878119897le 1198791015840
119906+ 119882 This shows that Algorithm D1 can find optimal
solution for the problemThe time complexity of the algorithm can be established
as follows We observe that 119894 119895 119897 le 119899 119879119906
le 119879 and 119898 minus 1 of
the value 1198791 1198792 119879
119898are independent Thus the number
of different states of the recursive relations is at most n3Tmminus1For each state each use requires 119874(119898119882) Therefore theoverall time complexity of Algorithm D1 is O(mn3Tmminus1W)
The existence of such a pseudopolynomial time algorithmfor a NP-hard problemmeans that the problem is NP-hard inthe ordinary sense We have the following theorem
Theorem 10 The problem Vmrarr batch|119901119895
= 119901 119908119895| sum 119862119895
+
119883(119909) is ordinarily NP-hard
5 The Problem 119881119898 rarr batch|119901119895
= 119901 119905119895
= 119905
119908119895| sum 119862119895
+ 119883(119909)
In this section we consider a special case with identicalprocessing times and identical transportation times andderive a polynomial-time algorithm to solve it It is evidentthat the problem reduces to an optimal batching problem onthe machine for the special case Note that 119909 is a decisionvariable denoting the number of batches on the machineNote that 119899
0= lceil119899119898rceil and 119892 = 119899
0119898 minus 119899 let 119899
119906be the number
of orders transported on vehicle 119906 under a specific scheduleNote that 119899
1015840
0= lceil119899119888rceil and 119899
1015840
0le 119909 le 119899
0 if 119888 ge 119898 Hence
the number of batches 119909 on the machine can be numberedby 1198991015840
0 1198991015840
0+ 1 119899
0 Otherwise 119899
0le 119909 le 119899
1015840
0if 119888 lt 119898
Without loss of generality assume that 119888 ge 119898 in the followingdiscussion Let 120587
lowast= (1198611 1198612 119861119871) be an optimal schedule
with given 119909 = 119871 batches for the problem Vmrarr batch|119901119895
=
119901 119905119895
= 119905 119908119895| sum 119862119895
+ 119883(119909) Then we have the followinglemma
Lemma 11 For the problem Vmrarr batch|119901119895
= 119901 119905119895
=
119905 119908119895| sum 119862119895
+ 119883(119909) there exists an schedule with 119871 batches inwhich (1) 119899
0minus 1 le 119899
119906le 1198990 119906 = 1 2 119898 (2) 119887
119897= 119887119897119898 119887119897
is integral and 119897 = 1 2 119871 minus 1 and (3) 119878119897+1
= max119903119895 119878119897
+
119901 where 119903119895denotes the arrival time of some 119898 jobs on the
machine 119897 = 1 2 119871 minus 1
Proof (1) Based on Lemma 2 each vehicle has no insertedidle time during transporting Assume that there exists anoptimal schedule 120587
lowast in which the condition is not satisfiedThen there must be a pair of vehicles 119906 and V such that119899119906
ge 119899V + 2 and the last order on vehicle 119906 is processed inthe last batch If the last order on vehicle 119906 is moved to thelast position on vehicle v then the objective value will notincrease By repeating this process we can obtain a desiredoptimal schedule
(2) According to Lemma 7 in the optimal schedule 119861119897
contains all orders which finish transportation in the timeinterval (119878
119897minus1 119878119897] if the number of orders in a batch is nomore
than 119888 Since 119905119895
= 119905 for each order 119869119895 m orders arrive at the
machine together A batch contains 119898 orders or km (119896119898 le 119888)
orders if their waiting times are not greater than the limitedvalue 119882 Hence 119887
119897is integral for 119897 = 1 2 119871 minus 1
(3) It is trivial according to Lemma 3
The Scientific World Journal 7
Based on these results we can easily construct an optimalschedule for the problem
Algorithm D 2 Calculate 1198990
= lceil119899119898rceil 11989910158400
= lceil119899119888rceilFor each 119871 = 119899
1015840
0 1198991015840
0+ 1 119899
0 compute 119887
119897= 1198870
minus 1 119897 =
1 sdot sdot sdot ℎ 119887119897= 1198870 and 119897 = ℎ + 1 119871 where 119887
0= lceil1198990119871rceil and
ℎ = 1198870119871 minus 1198990
Consider the following two cases
Case 1 Consider ℎ = 0If (1198870minus1)119905 gt 119882 or119898119887
0gt 119888 then the constructed schedule
is not feasible and denote 119865(120587lowast
119871) = infin If (119887
0minus 1)119905 le 119882 and
1198981198870
le 119888 then 1198781
= 1198870119905 According to Lemma 6 there must
be 1198870119905119897 + 119905 + 119882 ge 119878
119897+ 119901 Otherwise it is not feasible Based
on Lemma 7 we can obtain
119878119897+1
= max 1198870119905 (119897 + 1) 119878
119897+ 119901 119897 = 1 2 119871 minus 1 (18)
The associated objective value can be derived as
1198651
(120587lowast
119871) = min
119871
sum
119897=1
1198981198870
(119878119897+ 119901)
minus 119892 (119878119871
+ 119901) + 119883 (119871)
(19)
Case 2 Consider ℎ gt 0If (1198870
minus 2)119905 gt 119882 or 1198981198870
gt 119888 then it is not feasible anddenote 119865(120587
lowast
119871) = infin If (119887
0minus 2)119905 le 119882 and 119898119887
0le 119888 then we
can show 1198781
= (1198870
minus 1)119905 According to Lemma 6 we have
(1198870
minus 1) 119905119897 + 119905 + 119882 ge 119878119897+ 119901 if 119897 le ℎ
(1198870
minus 1) 119905ℎ + 1198870119905 (119897 minus ℎ) + 119905 + 119882 ge 119878
119897+ 119901 if 119897 gt ℎ
(20)
Otherwise the constructed schedule is not feasible anddenote 119865(120587
lowast
119871) = infin
According to Lemma 6 we have
119878119897+1
= max (1198870
minus 1) 119905 (119897 + 1) 119878119897+ 119901 if 119897 le ℎ minus 1
119878119897+1
=max (1198870
minus 1) 119905ℎ + 1198870119905 (119897 + 1 minus ℎ) 119878
119897+ 119901 if 119897 ge ℎ
119897 = 1 2 119871 minus 1
(21)
The associated objective value can be derived as
1198652
(120587lowast
119871)
= min
ℎ
sum
119897=1
119898 (1198870
minus 1) (119878119897+ 119901) +
119871
sum
119897=ℎ+1
1198981198870
(119878119897+ 119901)
minus 119892 (119878119871
+ 119901) + 119883 (119871)
(22)
Thus the objective value is
119865 (120587lowast
119871) = min 119865
1(120587lowast
119871) 1198652
(120587lowast
119871) (23)
Theorem 12 The problem Vmrarr batch|119901119895
= 119901 119905119895
=
119905 119908119895| sum 119862119895
+ 119883(119909) can be solved by Algorithm D2 in 119874(119899119898)
time
Proof Equations (18) and (21) present that the machinecan start to process 119861
119897+1as soon as the orders assigned
into 119861119897+1
have arrived at the machine and the machine hasfinished processing119861
119897The local optimal solution in two cases
can be obtained by (19) and (22) The optimal solution isderived as (23) Hence the optimal schedule to this specialcase can be obtained by repeating the above procedure for1198991015840
0le 119871 le 119899
0and selecting the best candidate among the
solutions generated This shows that Algorithm D2 can findoptimal solution It is clear that the overall time complexityof Algorithm D2 is 119874(119899119898)
We now demonstrate the above algorithm with thefollowing numerical example
Example 13 Consider the instance with 11 orders 119898 = 2 119905 =
1 119888 = 4 119882 = 2 119901 = 2 and 119883(119909) = 5119909 Based on the abovemethod we know 119899
0= 6 119892 = 1 and 119899
1015840
0= 2 Then 119871 =
2 3 4 5 6 We have the following results
when 119871 = 2 we have 1198870
= 3 ℎ = 0 1198781
= 3 and119865(120587lowast) = 80
when 119871 = 3 we have 1198870
= 2 ℎ = 0 1198781
= 2 and119865(120587lowast) = 79
when 119871 = 4 we have 1198870
= 2 ℎ = 2 1198781
= 1 and119865(120587lowast
119871) = 101
when 119871 = 5 we have 1198870
= 2 ℎ = 4 1198870119905 sdot 3 + 119905 + 119882 lt
1198783
+ 119901 and 119865(120587lowast
119871) = infin
when 119871 = 6 we have 1198870
= 1 ℎ = 0 1198870119905 sdot 3 + 119905 + 119882 lt
1198783
+ 119901 and 119865(120587lowast
119871) = infin
Therefore we can obtain an optimal schedule120587lowast for 119871 = 3
with 1198611
= 1198691 1198692 1198693 1198694 1198612
= 1198695 1198696 1198697 1198698 and 119861
3=
1198699 11986910
11986911
6 Concluding Remarks
In this paper we address a coordinated scheduling problemwith order transportation before processing and parallel-batch production under the limited waiting time considera-tion Our goal is to optimize the sum of the total flow timeand the total processing cost First we prove that the generalproblem is NP-hard in the strong sense We also demonstratethat the problem with equal processing times on the machineisNP-hard Furthermore a dynamic programming algorithmin pseudopolynomial time is provided to prove its ordinarilyNP-hardness The problem with equal processing times andequal transportation times for each order can be solved inpolynomial time through taking into account the propertiesof the special case Our work has practical implications forthe coordination of batch-production and transportation toimprove the overall system performance in logistics andoperations management Simultaneously another importantimplication of our paper may provide a vital approach to deal
8 The Scientific World Journal
with the waiting time constraint applied to reduce the rawmaterial consumption and improve the machine utilization
There are several possible extensions to this researchFirst it is interesting to investigate the problems with otherobjective functions such as minimizing the makespan orminimizing maximum order tardinessearliness Anotherinteresting issue is to develop effective heuristics to solve thegeneral problem and investigate polynomial time algorithmsfor other special cases
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This research is partly supported by the National NaturalScience Foundation of China (Grant no 71101097) and theProgram for Liaoning Excellent Talents in University (Grantno LJQ2012017)
References
[1] P Brucker A Gladky H Hoogeveen et al ldquoScheduling abatching machinerdquo Journal of Scheduling vol 1 no 1 pp 31ndash541998
[2] C N Potts and M Y Kovalyov ldquoScheduling with batching areviewrdquo European Journal of Operational Research vol 120 no2 pp 228ndash249 2000
[3] C-Y Lee R Uzsoy and L AMartin-Vega ldquoEfficient algorithmsfor scheduling semiconductor burn-in operationsrdquo OperationsResearch vol 40 no 4 pp 764ndash775 1992
[4] L Tang J Guan and G Hu ldquoSteelmaking and refining coordi-nated scheduling problemwith waiting time and transportationconsiderationrdquo Computers and Industrial Engineering vol 58no 2 pp 239ndash248 2010
[5] C-Y Lee and V A Strusevich ldquoTwo-machine shop schedulingwith an uncapacitated interstage transporterrdquo IIE Transactionsvol 37 no 8 pp 725ndash736 2005
[6] L Tang and P Liu ldquoFlowshop scheduling problems withtransportation or deterioration between the batching and singlemachinesrdquo Computers and Industrial Engineering vol 56 no 4pp 1289ndash1295 2009
[7] H Gong and L Tang ldquoTwo-machine flowshop schedulingwith intermediate transportation under job physical spaceconsiderationrdquo Computers and Operations Research vol 38 no9 pp 1267ndash1274 2011
[8] L Tang and H Gong ldquoA hybrid two-stage transportation andbatch scheduling problemrdquo Applied Mathematical Modellingvol 32 no 12 pp 2467ndash2479 2008
[9] C-L Li and J Ou ldquoMachine scheduling with pickup anddeliveryrdquo Naval Research Logistics vol 52 no 7 pp 617ndash6302005
[10] Z-L Chen ldquoIntegrated production and outbound distributionscheduling Review and extensionsrdquo Operations Research vol58 no 1 pp 130ndash148 2010
[11] S Li J Yuan andB Fan ldquoUnbounded parallel-batch schedulingwith family jobs and delivery coordinationrdquo Information Pro-cessing Letters vol 111 no 12 pp 575ndash582 2011
[12] L Lu and J Yuan ldquoUnbounded parallel batch schedulingwith job delivery to minimize makespanrdquo Operations ResearchLetters vol 36 no 4 pp 477ndash480 2008
[13] M R Garey and D S Johnson Computers and IntractabilityA Guide To the Theory of NP-Completeness W H Freeman andCompany New York NY USA 1979
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
The Scientific World Journal 7
Based on these results we can easily construct an optimalschedule for the problem
Algorithm D 2 Calculate 1198990
= lceil119899119898rceil 11989910158400
= lceil119899119888rceilFor each 119871 = 119899
1015840
0 1198991015840
0+ 1 119899
0 compute 119887
119897= 1198870
minus 1 119897 =
1 sdot sdot sdot ℎ 119887119897= 1198870 and 119897 = ℎ + 1 119871 where 119887
0= lceil1198990119871rceil and
ℎ = 1198870119871 minus 1198990
Consider the following two cases
Case 1 Consider ℎ = 0If (1198870minus1)119905 gt 119882 or119898119887
0gt 119888 then the constructed schedule
is not feasible and denote 119865(120587lowast
119871) = infin If (119887
0minus 1)119905 le 119882 and
1198981198870
le 119888 then 1198781
= 1198870119905 According to Lemma 6 there must
be 1198870119905119897 + 119905 + 119882 ge 119878
119897+ 119901 Otherwise it is not feasible Based
on Lemma 7 we can obtain
119878119897+1
= max 1198870119905 (119897 + 1) 119878
119897+ 119901 119897 = 1 2 119871 minus 1 (18)
The associated objective value can be derived as
1198651
(120587lowast
119871) = min
119871
sum
119897=1
1198981198870
(119878119897+ 119901)
minus 119892 (119878119871
+ 119901) + 119883 (119871)
(19)
Case 2 Consider ℎ gt 0If (1198870
minus 2)119905 gt 119882 or 1198981198870
gt 119888 then it is not feasible anddenote 119865(120587
lowast
119871) = infin If (119887
0minus 2)119905 le 119882 and 119898119887
0le 119888 then we
can show 1198781
= (1198870
minus 1)119905 According to Lemma 6 we have
(1198870
minus 1) 119905119897 + 119905 + 119882 ge 119878119897+ 119901 if 119897 le ℎ
(1198870
minus 1) 119905ℎ + 1198870119905 (119897 minus ℎ) + 119905 + 119882 ge 119878
119897+ 119901 if 119897 gt ℎ
(20)
Otherwise the constructed schedule is not feasible anddenote 119865(120587
lowast
119871) = infin
According to Lemma 6 we have
119878119897+1
= max (1198870
minus 1) 119905 (119897 + 1) 119878119897+ 119901 if 119897 le ℎ minus 1
119878119897+1
=max (1198870
minus 1) 119905ℎ + 1198870119905 (119897 + 1 minus ℎ) 119878
119897+ 119901 if 119897 ge ℎ
119897 = 1 2 119871 minus 1
(21)
The associated objective value can be derived as
1198652
(120587lowast
119871)
= min
ℎ
sum
119897=1
119898 (1198870
minus 1) (119878119897+ 119901) +
119871
sum
119897=ℎ+1
1198981198870
(119878119897+ 119901)
minus 119892 (119878119871
+ 119901) + 119883 (119871)
(22)
Thus the objective value is
119865 (120587lowast
119871) = min 119865
1(120587lowast
119871) 1198652
(120587lowast
119871) (23)
Theorem 12 The problem Vmrarr batch|119901119895
= 119901 119905119895
=
119905 119908119895| sum 119862119895
+ 119883(119909) can be solved by Algorithm D2 in 119874(119899119898)
time
Proof Equations (18) and (21) present that the machinecan start to process 119861
119897+1as soon as the orders assigned
into 119861119897+1
have arrived at the machine and the machine hasfinished processing119861
119897The local optimal solution in two cases
can be obtained by (19) and (22) The optimal solution isderived as (23) Hence the optimal schedule to this specialcase can be obtained by repeating the above procedure for1198991015840
0le 119871 le 119899
0and selecting the best candidate among the
solutions generated This shows that Algorithm D2 can findoptimal solution It is clear that the overall time complexityof Algorithm D2 is 119874(119899119898)
We now demonstrate the above algorithm with thefollowing numerical example
Example 13 Consider the instance with 11 orders 119898 = 2 119905 =
1 119888 = 4 119882 = 2 119901 = 2 and 119883(119909) = 5119909 Based on the abovemethod we know 119899
0= 6 119892 = 1 and 119899
1015840
0= 2 Then 119871 =
2 3 4 5 6 We have the following results
when 119871 = 2 we have 1198870
= 3 ℎ = 0 1198781
= 3 and119865(120587lowast) = 80
when 119871 = 3 we have 1198870
= 2 ℎ = 0 1198781
= 2 and119865(120587lowast) = 79
when 119871 = 4 we have 1198870
= 2 ℎ = 2 1198781
= 1 and119865(120587lowast
119871) = 101
when 119871 = 5 we have 1198870
= 2 ℎ = 4 1198870119905 sdot 3 + 119905 + 119882 lt
1198783
+ 119901 and 119865(120587lowast
119871) = infin
when 119871 = 6 we have 1198870
= 1 ℎ = 0 1198870119905 sdot 3 + 119905 + 119882 lt
1198783
+ 119901 and 119865(120587lowast
119871) = infin
Therefore we can obtain an optimal schedule120587lowast for 119871 = 3
with 1198611
= 1198691 1198692 1198693 1198694 1198612
= 1198695 1198696 1198697 1198698 and 119861
3=
1198699 11986910
11986911
6 Concluding Remarks
In this paper we address a coordinated scheduling problemwith order transportation before processing and parallel-batch production under the limited waiting time considera-tion Our goal is to optimize the sum of the total flow timeand the total processing cost First we prove that the generalproblem is NP-hard in the strong sense We also demonstratethat the problem with equal processing times on the machineisNP-hard Furthermore a dynamic programming algorithmin pseudopolynomial time is provided to prove its ordinarilyNP-hardness The problem with equal processing times andequal transportation times for each order can be solved inpolynomial time through taking into account the propertiesof the special case Our work has practical implications forthe coordination of batch-production and transportation toimprove the overall system performance in logistics andoperations management Simultaneously another importantimplication of our paper may provide a vital approach to deal
8 The Scientific World Journal
with the waiting time constraint applied to reduce the rawmaterial consumption and improve the machine utilization
There are several possible extensions to this researchFirst it is interesting to investigate the problems with otherobjective functions such as minimizing the makespan orminimizing maximum order tardinessearliness Anotherinteresting issue is to develop effective heuristics to solve thegeneral problem and investigate polynomial time algorithmsfor other special cases
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This research is partly supported by the National NaturalScience Foundation of China (Grant no 71101097) and theProgram for Liaoning Excellent Talents in University (Grantno LJQ2012017)
References
[1] P Brucker A Gladky H Hoogeveen et al ldquoScheduling abatching machinerdquo Journal of Scheduling vol 1 no 1 pp 31ndash541998
[2] C N Potts and M Y Kovalyov ldquoScheduling with batching areviewrdquo European Journal of Operational Research vol 120 no2 pp 228ndash249 2000
[3] C-Y Lee R Uzsoy and L AMartin-Vega ldquoEfficient algorithmsfor scheduling semiconductor burn-in operationsrdquo OperationsResearch vol 40 no 4 pp 764ndash775 1992
[4] L Tang J Guan and G Hu ldquoSteelmaking and refining coordi-nated scheduling problemwith waiting time and transportationconsiderationrdquo Computers and Industrial Engineering vol 58no 2 pp 239ndash248 2010
[5] C-Y Lee and V A Strusevich ldquoTwo-machine shop schedulingwith an uncapacitated interstage transporterrdquo IIE Transactionsvol 37 no 8 pp 725ndash736 2005
[6] L Tang and P Liu ldquoFlowshop scheduling problems withtransportation or deterioration between the batching and singlemachinesrdquo Computers and Industrial Engineering vol 56 no 4pp 1289ndash1295 2009
[7] H Gong and L Tang ldquoTwo-machine flowshop schedulingwith intermediate transportation under job physical spaceconsiderationrdquo Computers and Operations Research vol 38 no9 pp 1267ndash1274 2011
[8] L Tang and H Gong ldquoA hybrid two-stage transportation andbatch scheduling problemrdquo Applied Mathematical Modellingvol 32 no 12 pp 2467ndash2479 2008
[9] C-L Li and J Ou ldquoMachine scheduling with pickup anddeliveryrdquo Naval Research Logistics vol 52 no 7 pp 617ndash6302005
[10] Z-L Chen ldquoIntegrated production and outbound distributionscheduling Review and extensionsrdquo Operations Research vol58 no 1 pp 130ndash148 2010
[11] S Li J Yuan andB Fan ldquoUnbounded parallel-batch schedulingwith family jobs and delivery coordinationrdquo Information Pro-cessing Letters vol 111 no 12 pp 575ndash582 2011
[12] L Lu and J Yuan ldquoUnbounded parallel batch schedulingwith job delivery to minimize makespanrdquo Operations ResearchLetters vol 36 no 4 pp 477ndash480 2008
[13] M R Garey and D S Johnson Computers and IntractabilityA Guide To the Theory of NP-Completeness W H Freeman andCompany New York NY USA 1979
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
8 The Scientific World Journal
with the waiting time constraint applied to reduce the rawmaterial consumption and improve the machine utilization
There are several possible extensions to this researchFirst it is interesting to investigate the problems with otherobjective functions such as minimizing the makespan orminimizing maximum order tardinessearliness Anotherinteresting issue is to develop effective heuristics to solve thegeneral problem and investigate polynomial time algorithmsfor other special cases
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This research is partly supported by the National NaturalScience Foundation of China (Grant no 71101097) and theProgram for Liaoning Excellent Talents in University (Grantno LJQ2012017)
References
[1] P Brucker A Gladky H Hoogeveen et al ldquoScheduling abatching machinerdquo Journal of Scheduling vol 1 no 1 pp 31ndash541998
[2] C N Potts and M Y Kovalyov ldquoScheduling with batching areviewrdquo European Journal of Operational Research vol 120 no2 pp 228ndash249 2000
[3] C-Y Lee R Uzsoy and L AMartin-Vega ldquoEfficient algorithmsfor scheduling semiconductor burn-in operationsrdquo OperationsResearch vol 40 no 4 pp 764ndash775 1992
[4] L Tang J Guan and G Hu ldquoSteelmaking and refining coordi-nated scheduling problemwith waiting time and transportationconsiderationrdquo Computers and Industrial Engineering vol 58no 2 pp 239ndash248 2010
[5] C-Y Lee and V A Strusevich ldquoTwo-machine shop schedulingwith an uncapacitated interstage transporterrdquo IIE Transactionsvol 37 no 8 pp 725ndash736 2005
[6] L Tang and P Liu ldquoFlowshop scheduling problems withtransportation or deterioration between the batching and singlemachinesrdquo Computers and Industrial Engineering vol 56 no 4pp 1289ndash1295 2009
[7] H Gong and L Tang ldquoTwo-machine flowshop schedulingwith intermediate transportation under job physical spaceconsiderationrdquo Computers and Operations Research vol 38 no9 pp 1267ndash1274 2011
[8] L Tang and H Gong ldquoA hybrid two-stage transportation andbatch scheduling problemrdquo Applied Mathematical Modellingvol 32 no 12 pp 2467ndash2479 2008
[9] C-L Li and J Ou ldquoMachine scheduling with pickup anddeliveryrdquo Naval Research Logistics vol 52 no 7 pp 617ndash6302005
[10] Z-L Chen ldquoIntegrated production and outbound distributionscheduling Review and extensionsrdquo Operations Research vol58 no 1 pp 130ndash148 2010
[11] S Li J Yuan andB Fan ldquoUnbounded parallel-batch schedulingwith family jobs and delivery coordinationrdquo Information Pro-cessing Letters vol 111 no 12 pp 575ndash582 2011
[12] L Lu and J Yuan ldquoUnbounded parallel batch schedulingwith job delivery to minimize makespanrdquo Operations ResearchLetters vol 36 no 4 pp 477ndash480 2008
[13] M R Garey and D S Johnson Computers and IntractabilityA Guide To the Theory of NP-Completeness W H Freeman andCompany New York NY USA 1979
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