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Harmony search optimization algorithmfor a novel transportation problem in aconsolidation networkSeyed Davod Hosseinia, Mohsen Akbarpour Shirazia & SeyedMohammad Taghi Fatemi Ghomiaa Department of Industrial Engineering, Amirkabir University ofTechnology, Tehran, IranPublished online: 27 Nov 2013.

To cite this article: Seyed Davod Hosseini, Mohsen Akbarpour Shirazi & Seyed MohammadTaghi Fatemi Ghomi (2014) Harmony search optimization algorithm for a novel transportationproblem in a consolidation network, Engineering Optimization, 46:11, 1538-1552, DOI:10.1080/0305215X.2013.854350

To link to this article: http://dx.doi.org/10.1080/0305215X.2013.854350

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Engineering Optimization, 2014Vol. 46, No. 11, 1538–1552, http://dx.doi.org/10.1080/0305215X.2013.854350

Harmony search optimization algorithm for a noveltransportation problem in a consolidation network

Seyed Davod Hosseini, Mohsen Akbarpour Shirazi and Seyed Mohammad Taghi Fatemi Ghomi∗

Department of Industrial Engineering, Amirkabir University of Technology, Tehran, Iran

(Received 4 April 2013; accepted 16 September 2013)

This article presents a new harmony search optimization algorithm to solve a novel integer programmingmodel developed for a consolidation network. In this network, a set of vehicles is used to transport goodsfrom suppliers to their corresponding customers via two transportation systems: direct shipment and milkrun logistics. The objective of this problem is to minimize the total shipping cost in the network, so ittries to reduce the number of required vehicles using an efficient vehicle routing strategy in the solutionapproach. Solving several numerical examples confirms that the proposed solution approach based on theharmony search algorithm performs much better than CPLEX in reducing both the shipping cost in thenetwork and computational time requirement, especially for realistic size problem instances.

Keywords: transportation; consolidation network, milk run logistics; integer programming; harmonysearch optimization algorithm

1. Introduction

Logistics is one of the most important activities in modern societies. It has been estimated that thetotal logistics cost incurred by US organizations in 1997 was 862 billion dollars, correspondingto approximately 11% of the US gross domestic product (GDP). This cost is higher than thecombined annual US government expenditure on social security, health services and defence.These figures are similar to those observed for the other North America Free Trade Agreement(NAFTA) countries and for the European Union (EU) countries (Ghiani, Laporte, and Musmanno2004). A logistics system is made up of a set of facilities linked by transportation services. Thedesign of such a system comprises three major problems: location-allocation, vehicle routingand inventory control. A lot of research in the literature has been carried out on each of theseproblems and several studies have focused on the integration of these problems (Ahmadi-Javidand Seddighi 2012). Considering all of these problems, transportation optimization is a frequentlystudied problem in the context of logistics. It arises not only in the collection and distribution ofgoods but also in service applications such as taxi driving, school bus routing, and special casessuch as transporting personnel from platform to platform using helicopters in the oil industry(Velasco et al. 2012).

There are several different kinds of road transportation network: direct shipping, milk runs,cross-docking and tailored networks (Chopra and Meindl 2001). The direct shipping network

∗Corresponding author. Email: fatemi@aut.ac.ir

© 2013 Taylor & Francis

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Engineering Optimization 1539

delivers products from suppliers to their customers without passing through any other facilities.Clearly, if a shipment consists of a full truckload (TL), it is economical to ship directly fromthe seller to the buyer(s). However, when orders are less than a truckload (LTL), the other threeapproaches can be considered (Dua, Wang, and Lu 2007). This is similar to what happens incontainer transportation: most businesses transport their containerized products by rail if they aretransporting for long distances. However, this is not the best choice when there are not enoughcontainers to be transported, and other modes of transportation (e.g. sea, river, air and road) canbe selected (Cao, Gao, and Li 2012).

Milk run logistics is a transportation system in which a truck visits various suppliers following apredefined route to collect their parts and deliver them to the customer. This type of transportationsystem is designed for suppliers located near each other. It prevents them from accumulating theirshipments to reach a full truckload or sending their shipments in volumes of less than a truckload,which would lead to an increase in transportation costs. The name of this system comes from thetraditional system for selling milk in the West, in which the milkman would walk to the doors of thecustomers’ houses with his dray on a specified route, deliver the milk in bottles to his customersand finally take back the empty bottles. This system has been used in various industries, andautomanufacturing companies around the world have been (and are) the most important users ofthis system (Sadjadi, Jafari, and Amini 2009).

The third consolidation network, the tailored network, combines both TL and LTL by allowinghigh-volume orders to be shipped from suppliers to customers directly, and low-volume ordersto be consolidated (Dua, Wang, and Lu 2007). In this article, the focus is on a tailored networkwhere two types of transportation, direct shipment and milk runs, are used simultaneously to shipthe orders from suppliers to customers directly and/or through consolidation. A novel integerprogramming model is proposed to state the transportation problem in this network where theobjective is to minimize the total transportation cost using a tailored vehicle routing strategy,but its complexity and NP-complete nature require a metaheuristic optimization algorithm forsolving.

In the past few years, there has been explosive growth in the application of harmony search(HS) metaheuristics (Geem, Kim, and Loganathan 2001) in solving NP-complete problems. In2009 the number of publications regarding HS was nine times greater than in 2005 and 2006, andit is safe to say that the growth in HS literature has been significant (Kougias and Theodossiou2012). It has been applied to various science and engineering optimization problems, includingtransportation, water distribution networks, vehicle routing and location problems (Lee and Geem2004; Geem, Lee, and Park 2005; Geem 2006, 2010; Ayvaz 2007; Ceylan et al. 2008; Nahas andThien-My 2010; Jaberipour and Khorram 2011; Gholizadeh and Barzegar 2012; Landa-Torreset al. 2012; Del Ser et al. 2012; Salcedo-Sanz et al. 2013). The success of the HS algorithmin finding good solutions for these problems distinguishes it as a strong alternative to othermetaheuristic approaches and its successful application in solving the problem proposed in thisarticle is encouraging.

The rest of article is structured as follows. In the next section, the problem is defined andthe mathematical model is stated. Section 3 introduces the solution approach based on the HSoptimization algorithm. Section 4 deals with the computational study and solves several numericalexamples. Finally, conclusions are drawn in Section 5.

2. Problem definition

The network comprises |V1| part suppliers (PS) and |V2| assembly plants (AP) (Figure 1). Theloads are shipped from the suppliers to the plants via two transportation systems.

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Figure 1. Transportation problem in the consolidation network.

The mathematical model has the following assumptions, adopted from Musa, Arnaout, andJung (2010):

• The loads to be sent from a supplier location i to a final customer j (Sij) are known and assumedto be less than the truck capacity (A), i.e. Sij ≤ A∀i ∈ V1, ∀j ∈ V2. Otherwise, the solutionis trivial since the truck for that flow will need to go directly from i to j.

• All the trucks have the same capacity (A).

The parameters and the definitions used in the model are as follows:

V1 i|i = 1, 2, . . . , |V1| is the set of part suppliers (PS)

V2 j|j = 1, 2, . . . , |V2| is the set of assembly plants (AP)

V3 r|r = 1, 2, . . . , |V3| is the set of trucks

W g|g ∈ V1 U V2Sij is flow from PSi to APj

A is truck capacity

Cqp is the cost of a truck from node q to node p

The loads from suppliers to plants can be transported directly and/or through milk runs. A milkrun trip comprises two or more suppliers and one assembly plant placed at the end of the trip(Figure 2). In a milk run trip, a supplier can take one of the three positions d, f or h. If the supplieris in position d, it will be the starting point of that trip. By contrast, if this supplier takes position

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Figure 2. Components of a milk run trip.

h, it will be the last supplier that the truck passes before moving towards the assembly plant. Thesupplier in position f will be a stopover node (Yaman, Kara, and Tansel 2007), i.e. this supplieris an intermediate node and the truck entering it has already exited from another supplier. Afterloading the stopover supplier’s shipment, the truck will move to the next supplier. Thus, the aboveconcepts can be defined as binary integer variables as follows:

uij =

1, if load from PSi to APj is sent directly

0, otherwise

zijr =

1, if load from PSi to APj is sent through a milk run trip by truck r

0, otherwise

dijr =

1, if there is a milk run trip by truck r starting from PSi and terminating at APj

0, otherwise

fijr =

1, if there is a milk run trip by truck r passing PSi and terminating at APj

0, otherwise

hijr =

1, if there is a milk run trip by truck r ending in PSi before terminating at APj

0, otherwise

xjigr =

⎧⎪⎨⎪⎩

1, if there is a milk run trip in which truck r moves from PSi to node gand terminates at APj; i = g

0, otherwise

Using the above notations, the problem can be formulated as the following pure-integerprogramming model:

Min∑i∈V1

∑j∈V2

uijCij +∑i∈V1

∑j∈V2

∑g∈W

∑r∈V3

xjigrCig (1)

uij +∑r∈V3

zijr = Min1, Sij ∀i ∈ V1, ∀j ∈ V2 (2)

dijr + fijr + hijr = zijr ∀i ∈ V1, ∀j ∈ V2, ∀r ∈ V3 (3)

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∑i∈V1

dijr −∑i∈V1

hijr = 0 ∀j ∈ V2, ∀r ∈ V3 (4)

fijr ≤∑i∈V1

dijr ∀i ∈ V1, ∀j ∈ V2, ∀r ∈ V3 (5)

fijr ≤∑i∈V1

hijr ∀i ∈ V1, ∀j ∈ V2, ∀r ∈ V3 (6)

dijr +∑g∈V1

xjgir ≤ 1 ∀i ∈ V1, ∀j ∈ V2, ∀r ∈ V3, i = g (7)

dijr + xjijr ≤ 1 ∀i ∈ V1, ∀j ∈ V2, ∀r ∈ V3 (8)

fijr + xjijr ≤ 1 ∀i ∈ V1, ∀j ∈ V2, ∀r ∈ V3 (9)

hijr − xjijr = 0 ∀i ∈ V1, ∀j ∈ V2, ∀r ∈ V3 (10)

hijr +∑g∈V1

xjigr ≤ 1 ∀i ∈ V1, ∀j ∈ V2, ∀r ∈ V3, i = g (11)

∑g∈V1

xjigr +

∑g∈V1

xjgir + xj

ijr = 2fijr + 2hijr + dijr ∀i ∈ V1, ∀j ∈ V2, ∀r ∈ V3, i = g (12)

∑i∈B

∑g∈B

xjigr ≤ |B| − 1 ∀B ⊆ Mjr , ∀j ∈ V2, ∀r ∈ V3, i = g (13)

∑i∈V1

∑j∈V2

hijr ≤ 1 ∀r ∈ V3 (14)

∑i∈V1

∑g∈W

xjigr .Sij ≤ A ∀j ∈ V2, ∀r ∈ V3, i = g (15)

uij, zijr , dijr , fijr , hijr , xjigr ∈ 0, 1 ∀i ∈ V1, ∀j ∈ V2, ∀g ∈ W , ∀r ∈ V3, i = g

The objective function (1) is the sum of direct and milk run transportation costs. Constraints (2)ensure that sending Sij has to be performed via one of the two available transportation systems.Constraints (3) show that PSi which participates in a milk run trip towards APj by truck r can takeone of the three positions dijr , fijror hijr . Constraints (4) guarantee that a milk run trip consists ofat least two suppliers (d and h). Constraints (5) and (6) show that one or more suppliers can takeposition f in a milk run trip only when two other suppliers have already taken positions d and hin the same trip. If a supplier takes position d in a milk run trip, constraints (7) and (8) guarantee,first, that none of the trucks is permitted to enter this supplier and, secondly, that the exiting truckfrom this supplier is allowed to move to a supplier, not to an assembly plant. Similarly, if a suppliertakes position f , constraints (9) ensure that the exiting truck is not allowed to enter an assemblyplant node. The only time that a truck leaving a supplier is permitted to move directly towardsan assembly plant is when the supplier takes position h. Constraints (10) and (11) reflect this factand also guarantee that the exiting truck from the supplier with position h does not have any otherchoice except moving towards the assembly plant. The routing of a truck among suppliers thattransfer their loads through a common milk run trip towards the same assembly plant is determinedby constraints (12). In constraints (13), set Mjr shows the suppliers that are intermediate nodes ina milk run trip towards APj by truck r (Mjr = i|fijr = 1; ∀j ∈ V2, ∀r ∈ V3). These constraintsstate that for every subset of set M, the number of transportation arcs among the suppliers (withposition f ) should be smaller than the size of that subset. In fact, constraints (13) along withconstraints (5) and (6) are sub-tour (Tuzun and Burke 1999) elimination constraints. Constraints

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Engineering Optimization 1543

(14) ensure that trucks being used in a milk run trip cannot be utilized for another milk run trip.Constraints (15) are capacity constraints associated with the trucks used in milk run trips.

3. Solution approach

It is well known that a binary integer programming model is NP hard (Musa, Arnaout, and Jung2010). Owing to the problem complexity and exponential growth in its size, exact solution tech-niques are limited to small size problems. In this article, a heuristics based on the HS optimizationalgorithm is developed in order to solve this NP-hard problem.

HS is a music-based metaheuristic optimization algorithm. It is inspired by the observation thatthe aim of music is to search for a perfect state of harmony. The effort to find the harmony in musicis analogous to finding the optimality in an optimization process. When a musician is improvising,he or she has three possible choices: (1) playing any famous tune exactly from his or her memory;(2) playing something similar to the aforementioned tune (e.g. adjusting the pitch slightly); or(3) composing new or random notes. Geem, Kim, and Loganathan (2001) formalized these threeoptions into three quantitative optimization processes, and the three corresponding componentsbecame use of harmony memory, pitch adjusting and randomization. The usage of harmonymemory is important because it ensures that good harmonies are considered as elements of newsolution vectors. In order to use this memory effectively, the HS algorithm adopts a parameterHMCR ∈ [0, 1], the harmony memory considering rate. If this rate is too low, only a few eliteharmonies are selected and it may converge too slowly. If this rate is extremely high (near 1), thepitches in the harmony memory are mostly used and other ones are not explored well, thus notleading to good solutions. Therefore, typically, HMCR = 0.7 ∼ 0.95 is used (Geem 2009). Thesecond component is the pitch adjustment, which has parameters such as pitch bandwidth (brange)

and pitch-adjusting rate (PAR). As the pitch adjustment in music means changing the frequency,this entails generating a slightly different value in the HS algorithm. Xnew = Xold + brange × ε

where Xold is the existing pitch stored in the harmony memory, and Xnew is the new pitch after thepitch adjusting action. Here, ε is a random number from a uniform distribution with the range of[–1, 1]. Pitch adjustment is similar to the mutation operator in genetic algorithms. A PAR can beassigned to control the degree of adjustment. A low PAR with a narrow bandwidth can slow downthe convergence of HS because of the limitation in the exploration of only a small subspace of thewhole search space. On the other hand, a very high PAR with a wide bandwidth may cause thesolution to scatter around some potential optima as in a random search. Thus, PAR = 0.1 ∼ 0.5is usually used in most applications (Geem 2009). The third component is the randomization,which increases the diversity of the solutions. Although the pitch adjustment has a similar role, itis limited to a certain area and thus corresponds to a local search. The use of randomization candrive the system further to explore various diverse solutions so as to attain the global optimality.The probability of randomization is Prandom = 1 − HMCR, and the actual probability of the pitchadjustment is Ppitch = HMCR × PAR (Geem 2009).

In the proposed HS algorithm, two matrices, named T and E, are used together to make theharmony memory. The steps of implementing the proposed solution approach are illustrated inFigure 3. These steps are explained comprehensively in the following subsections.

3.1. Matrix T

Matrix T has |V2| columns (Figure 4). Each column of this matrix represents an assembly plantand indicates the whole pairs of matching suppliers of that assembly plant. A ‘pair of matchingsuppliers’ (PMS) is defined as two suppliers whose sum of loads towards a specific assembly plant

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Figure 3. Pseudo-code of the solution approach.

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Figure 4. Matrices T , N and E (solution representation of harmony search algorithm).

is lower than the truck capacity and therefore these suppliers are allowed to form a milk run triptowards that assembly plant together. This means that PSα − PSβ is a PMS for APj if and only ifSαj + Sβj ≤ A. Constant nj represents the number of PMS for APj and nx denotes the maximumnumber of PMS(nx = maxnj | j ∈ 1, 2, . . . , |V2|). These amounts are gathered in matrix N(see Figure 4).

In Figure 4, it is assumed that the maximum number of PMS belongs to APk (nx = nk).Hence, the number of matrix rows is equal to nk . As other assembly plants have fewer PMS, 0 isplaced in their extra rows.

The important point about matrix T is that in every column of this matrix, all PMS of the relatedassembly plant are arranged in ascending order according to the transportation costs between thematching suppliers. For example, for APk in Figure 4, the transportation cost between PS1,a,k andPS1,b,k , placed on the first row, is less than that between PS2,a,k and PS2,b,k , placed on the secondrow. In fact, < PS1,a,k − PS1,b,k > is the PMS of APk that has the lowest transportation cost inbetween. To make a milk run trip for every assembly plant using matrix T , priority is given tothe PMS placed in the upper rows. This means that the PMS with lower transportation cost inbetween has more chance of attending a milk run trip towards the related assembly plant. Usingsuch a strategy in the approach is exactly in accordance with the nature of milk run logistics.

A significant question about using matrix T is raised here: how many of the existing PMS ofevery assembly plant should participate in making a milk run trip towards the assembly plant sothat the objective of minimizing the total transportation cost is achieved? Notice that suppliersthat do not participate in this trip will be allocated to the direct shipment in order to send theirloads through this system. Matrix E is utilized by the heuristics to answer this question.

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3.2. Matrix E

Matrix E has |V2| + 1 columns and 20 rows (see Figure 4). This matrix shows 20 feasible solutionsof the problem. The fitness function (FF: total transportation cost) of each solution is given in thelast column. Each element of this matrix contains one number, e. This number limits the numberof PMS of an assembly plant that should participate in a milk run trip towards that assemblyplant in each solution. To put it simply, see row t of matrix E in Figure 4. In the lth column ofthis row, related to the lth assembly plant (APl), number et,l is specified. As explained before,in the lth column of matrix T , the whole PMS of APl are listed in ascending order according totransportation cost. The algorithm applies this list to construct the tth solution; only the PMS fromrow 1 to row et,l of the list should participate in making the milk run trip to APl. Hence, Equation(16) is obtained as follows:

1 ≤ et,j ≤ nj, et,j : integer; ∀j ∈ 1, 2, . . . , |V2|&∀ t ∈ 1, 2, . . . , 20 (16)

3.3. Initial population (initial harmony memory)

In order to determine the values of e for each assembly plant in the preliminary matrix E (initialpopulation), the PMS of that assembly plant is divided equally into 20 areas. The first member ofthe initial population is made from the first part of the PMS of each assembly plant, the secondmember of the initial population from the second part, and so on. This means that to make thelast, i.e. 20th member of the initial population, the entire PMS of assembly plants are allowedto participate in making the milk run trips. Therefore, in the preliminary matrix E the followingequations are satisfied:

et,j =[

t × nj

20

]; ∀j ∈ 1, 2, . . . , |V2|&∀t ∈ 1, 2, . . . , 20

3.4. Making milk run trips

After determining which part of the PMS should participate in making milk run trips towards therelevant assembly plants, putting the selected suppliers beside each other begins in order to findthe best arrangement.

Suppose that to construct the tth solution of the memory, PMS PSα − PSβ is located at thebeginning (the first row) of the selected list of PMS for APj. First, a trip is made in the form of(PSα − PSβ). Then, the algorithm searches from the second row to the last row (row et,j) of the listto find the PMS that are in the form of PSγ − PSα or PSβ − PSδ . Wherever such pairs are found,the milk run trip takes longer and transforms into (PSγ − PSα − PSβ) or (PSα − PSβ − PSδ),provided that the sum of loads from suppliers PSα , PSβ and PSγ or that from suppliers PSα , PSβ

and PSδ towards APj does not exceed the truck capacity. Similarly, in the next step, the sameprocess continues to make the current milk run trip longer (as described in Figure 5).

Adding suppliers to a trip will continue as far as the sum of loads of suppliers participating inthe trip towards the specific assembly plant approaches the full truckload in such a way that noother supplier can be added owing to the truck capacity. Up to this stage, a milk run trip towardsan assembly plant has been made, which started with the PMS located first in the selected list.In the next stage, the same process should be repeated for the next PMS on the list not involvedin the previous trips. Until such matching pairs are left in the list, this process will be iterated inorder to make as many milk run trips towards the assembly plant as possible.

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Engineering Optimization 1547

Figure 5. Making milk run trips.

3.5. Allocating direct shipments

After all the possible milk run trips towards an assembly plant have been made using the selectedlist of PMS and all the suppliers making the trips have been determined, it is time to transport theloads of suppliers that have not participated in the trips. These suppliers have to ship their loadstowards the assembly plant by direct shipment.

3.6. Implementing the harmony search algorithm

When the preliminary matrix E has been made, the HS algorithm commences the consecutiveiterations and generates one new solution in each iteration. At the beginning of each iteration, onemember of the population (one row of matrix E) is selected randomly. HS parameters, i.e. HMCRand PAR, are utilized to determine whether the current values e should remain unchanged orshould change to one of the neighbouring values, or whether the new values should be determinedrandomly.

After the new values of e have been determined using HS parameters, one of three moves—insertion move, swap move or 2-opt move—is executed randomly in each column (AP) of matrixT from row 1 to e. These moves are commonly embedded in simulated annealing heuristics andother metaheuristics (Vincent et al.2010). Two random numbers, i and j, are required to be selectedbetween numbers 1 and e. The aim of this stage is to disturb the arrangement and priority of PMSof assembly plants (in matrix T ) in the hope of achieving better solutions. The insertion move iscarried out by inserting the ith row into the position immediately before the jth row. The swappingmove is performed by exchanging the positions of the ith and jth rows. The 2-opt move, commonlyused in solving vehicle routing-related problems, is implemented by reversing the order of rowsbetween row iand row j (Figure 6). In this stage, the order of PMS of each assembly plant altersin a specific area of each column of matrix T and therefore a new matrix, Tnew, is generated.

In the next stage, milk run trips towards each assembly plant are made based on the newmatrix Tnew and the new values of e. These new values specify how many numbers of PMS ofeach assembly plant, listed in matrix Tnew, should be used in making the milk run trips towardsthat assembly plant. In this way, new milk run trips can be made for assembly plants, and thetransportation pattern of some suppliers, which was direct shipment in the previous iteration, maychange. Then, a new solution with a new fitness function (FFnew) is gained. If this solution isbetter than the worst solution in the memory of the algorithm, matrix E will be modified. Thismeans that the old values of points e, which are in the worst row of matrix E, will be replaced withthe new ones. Likewise, matrix Tnew will be chosen as the next matrix T . Otherwise, matrices Tand E will not change for the next iteration.

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Figure 6. Insertion, swap and 2_opt move.

4. Computational study

In this section, the proposed mathematical model is solved on 40 problem instances using CPLEX10.0.0 from GAMS 22.1 System and the results are compared with the solutions of the HS-basedheuristics. These were run on a PC with an Intel Core 2 Duo CPU (2.33 GHz) and 2 GB memory.

GAMS/CPLEX is a general algebraic modelling system (GAMS) solver that allows users tocombine the high-level modelling capabilities of GAMS with the power of CPLEX optimizers.IBM ILOG CPLEX optimizers are designed to solve large, difficult problems quickly and withminimal user intervention. While numerous solving options are available, GAMS/CPLEX auto-matically calculates and sets most options at the best values for specific problems (GAMS/CPLEXManual; GAMS—A User’s Guide; IBM ILOG CPLEX V12.1—User’s Manual for CPLEX).

Figure 7 gives a small sample instance of the transportation problem of a tailored network with16 part suppliers and two assembly plants. The numbers written under each supplier indicate theloads that should be sent from that supplier to assembly plants 1 and 2, respectively. The truckcapacity is 100 and the transportation cost between each couple of nodes is supposed to be aconstant coefficient of the direct distance between them. In order to solve this instance, the HSalgorithm and CPLEX are run and their results are given in Table 1. CPLEX could obtain anobjective value of 1941 after 60 minutes, while the HS algorithm could achieve the best value of1484 for the objective function after 10,000 iterations, which took only 8 minutes. That is, HSled to 23.54% and 86.67% improvements in objective value and CPU time, respectively. A visualillustration of the solution attained by implementing HS in this instance is given in Figure 7.

4.1. Instances

In order to verify the proposed HS algorithm, 40 instances are generated as test problems. They arecategorized in four sets: set 1 (10 1) (i.e. 10 part suppliers and one assembly plant), set 2 (17 2), set3 (24 3) and set 4 (30 4). In each instance, the truck capacity is 100 and the load from each supplierto each assembly plant is generated by a uniformly distributed random number between 20 and 90.Similarly, the transportation cost between each couple of nodes in each instance network followsa uniform distribution in (0,1). For each instance, both HS and CPLEX have been run.

CPLEX is not able to solve instances like (35 4) and (50 3) on the PC used in this article. Infact, the fourth set of instances is the biggest possible problem that CPLEX can deal with on thisPC. It cannot solve these instances even on a more sophisticated PC with an Intel Core 2 DuoCPU (2.53 GHz) and 4 GB memory. The methods used to solve pure integer and mixed integerprogramming problems require dramatically more mathematical

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Figure 7. A sample transportation problem and its visual solution attained by harmony search.

Table 1. Computational results for the sample instance.

CPLEX HS Improvement by HS (%)

Problem instance PS AP Objective value CPU (min) Objective value CPU (min) Objective value CPU

Sample 16 2 1941 60 1484 8 23.54 86.67

computation than those for similarly sized pure linear programs. Many relatively small inte-ger programming models take enormous amounts of time to solve. For problems with integervariables, CPLEX uses a branch-and-cut algorithm which solves a series of linear programmingsubproblems. Because a single mixed integer problem generates many subproblems, even smallmixed integer problems can be very computationally intensive and require significant amountsof physical memory (GAMS/CPLEX Manual). Confronting these instances, CPLEX terminateswith a failure error: ‘out of memory’.

4.2. Computational results

Tables 2–5 show the results of running HS and CPLEX on the four problem sets. HS was executedfor 300, 6000, 10,000 and 20,000 iterations on problem sets 1, 2, 3 and 4, respectively, whileHMCR and PAR were set to 0.8 and 0.5, respectively, for all problem instances. The figures 300,6000, 10,000 and 20,000 show maximum improvisation in the HS algorithm, and are used as thestopping criterion of this algorithm. Optimal solutions for problem set 1 were reached immediatelyusing CPLEX, whereas it spent too much time solving problem sets 2, 3 and 4, so 30, 60 and 120minute runs were used as limits for these problem sets, respectively.

As seen in Table 2, the proposed approach was able to find the optimal solution of nine instancesin problem set 1 in a short time (on average 13.07 seconds). In addition, for the rest of the problem

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Table 2. Computational results for the problem set 1 [300 iterations for harmony search (HS)].

CPLEX HS

Problem instance PS AP Optimal objective value CPU (s) Objective value CPU (s)

1–1 10 1 3.1733 0 3.1733 14.011–2 10 1 2.0650 0 2.0760 12.931–3 10 1 3.4131 0 3.4131 13.231–4 10 1 2.9583 0 2.9583 12.961–5 10 1 4.1586 0 4.1586 12.971–6 10 1 4.5007 0 4.5007 13.061–7 10 1 5.2435 0 5.2435 12.701–8 10 1 3.3431 0 3.3431 12.751–9 10 1 4.0540 0 4.0540 13.031–10 10 1 4.3941 0 4.3941 13.08Avg. 3.7304 0 3.7315 13.07

Note: PS = part supplier; AP = assembly plant; CPU = central processing unit.

Table 3. Computational results for the problem set 2 [6000 iterations for harmony search (HS)].

CPLEX HS Improvement by HS (%)

Problem instance PS AP Objective value CPU (min) Objective value CPU (min) Objective value CPU

2–1 17 2 10.7621 30 10.7257 4.69 0.34 84.372–2 17 2 12.1387 30 12.1387 4.59 0.00 84.372–3 17 2 12.2945 30 10.5240 4.41 14.40 84.702–4 17 2 7.6792 30 7.6792 4.78 0.00 85.302–5 17 2 10.4519 30 10.4519 4.38 0.00 84.072–6 17 2 11.9337 30 12.0902 4.55 −1.31 85.402–7 17 2 10.7336 30 9.6814 4.79 9.80 84.832–8 17 2 10.7865 30 10.5455 4.52 2.23 84.032–9 17 2 11.4975 30 10.8766 4.72 5.40 84.932–10 17 2 12.1450 30 12.1450 4.47 0.00 84.27Avg. 11.0423 30 10.6858 4.59 3.09 84.63

Note: PS = part supplier; AP = assembly plant; CPU = central processing unit.

Table 4. Computational results for the problem set 3 [10,000 iterations for harmony search (HS)].

CPLEX HS Improvement by HS (%)

Problem instance PS AP Objective value CPU (min) Objective value CPU (min) Objective value CPU

3–1 24 3 31.2815 60 27.1890 8.52 13.08 85.803–2 24 3 30.1525 60 21.5221 9.32 28.62 85.803–3 24 3 36.8432 60 24.3480 8.67 33.91 84.473–4 24 3 30.5543 60 23.0819 8.89 24.46 85.553–5 24 3 26.7537 60 23.4518 9.33 12.34 85.183–6 24 3 29.3951 60 20.9725 9.29 28.65 84.453–7 24 3 19.9480 60 15.5686 9.73 21.95 84.523–8 24 3 30.7584 60 23.1520 9.01 24.73 83.783–9 24 3 23.9060 60 20.2755 9.28 15.19 84.983–10 24 3 22.5176 60 19.5945 8.94 12.98 84.53Avg. 28.2110 60 21.9156 9.10 21.59 84.91

Note: PS = part supplier; AP = assembly plant; CPU = central processing unit.

sets (i.e. 2, 3 and 4), the approach could considerably enhance the objective value obtained fromCPLEX, except for five cases in problem set 2, and CPLEX could not find any integer solutionsfor two cases in problem set 4 (4-1 and 4-3) during the 120 minute runtime. More specifically,HS was able to improve, on average, the total transportation cost to 3.09%, 21.59% and 34.97%

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Table 5. Computational results for the problem set 4 [20,000 iterations for harmony search (HS)].

CPLEX HS Improvement by HS (%)

Problem instance PS AP Objective value CPU (min) Objective value CPU (min) Objective value CPU

4–1 30 4 – 120 35.3777 22.74 – –4–2 30 4 60.1506 120 40.9334 21.15 31.95 81.054–3 30 4 – 120 38.4748 21.78 – –4–4 30 4 60.7154 120 40.8553 22.37 32.71 81.854–5 30 4 63.2004 120 41.1576 21.85 34.88 81.364–6 30 4 62.9033 120 36.4660 24.65 42.03 81.794–7 30 4 62.5417 120 37.2131 24.52 40.50 79.464–8 30 4 57.8159 120 39.2713 23.21 32.08 79.574–9 30 4 67.4558 120 43.4888 23.61 35.53 80.664–10 30 4 62.1997 120 43.4553 20.59 30.14 80.33Avg. 62.1228 120 39.6693 22.65 34.97 80.75

Note: PS = part supplier; AP = assembly plant; CPU = central processing unit.

for problem sets 2, 3 and 4, respectively. It is worth mentioning that these enhancements wereattained in about one-sixth of the CPU time spent by CPLEX.

5. Conclusions

This article introduced a novel model for the transportation problem of a consolidation networkwhere direct shipment and shipment through milk run trips are used simultaneously to send loadsfrom suppliers to customers. This is such a complex model that CPLEX is not able to solvesmall instances, e.g. 35 suppliers and four customers, owing to a shortage of memory. A heuristicapproach based on HS was proposed to solve the problem, which uses a special solution structurescheme that reduces the need for physical memory and can therefore deal with larger probleminstances. The proposed HS algorithm and CPLEX were tested on four problem sets (a total of 40instances) for verification. The computational results and comparisons of their outputs were alsopresented. It was verified that the proposed solution approach based on HS not only outperformsCPLEX solutions in reduction of shipping costs in the network, but also has a noticeably shortercomputational time requirement than CPLEX, especially for realistically sized problem instances.

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