Transcript

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

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

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

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 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

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

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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

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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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

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

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

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 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

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

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

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

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

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


Recommended