Theoretical Paper

Journal of the Operational Research Society (2008) 59, 119–136. doi:10.1057/palgrave.jors.2602328; Published online 29 November 2006

An exact algorithm for a cross-docking supply chain network design problem

C S Sung1 and W Yang1

1Korea Advanced Institute of Science and Technology, Daejon, Korea

Correspondence: CS Sung, Department of Industrial Engineering, Korea Advanced Institute of Science and Technology, Daejon 305-701, South Korea. E-mail: cssung@kaist.ac.kr

Received November 2005; Revised August 2006; Published online 29 November 2006.

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Abstract

This paper proposes a branch-and-price algorithm as an exact algorithm for the cross-docking supply chain network design problem introduced by one of the authors of this paper. The objective is to optimally locate cross-docking (CD) centres and allocate vehicles for direct transportation services from the associated origin node to the associated CD centre or from the associated CD centre to the associated destination node so as to satisfy a given set of freight demands at minimum cost subject to the associated service (delivery) time restriction. A set-partitioning-based formulation is derived for the problem for which some solution properties are characterized. Based on the properties, a branch-and-price algorithm is derived. The properties can also be used in deriving any efficient local search heuristics with the move operation (neighbourhood search operation) of modifying assignment of some freight demands from current CD centres to other CD centres. Computational experiments show that the branch-and-price algorithm is effective and efficient and also that the solution properties contribute to improve the efficiency of the local search heuristics.

Keywords:

logistics, integer programming, networks and graphs

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Introduction

This paper proposes a branch-and-price algorithm as an exact algorithm for the cross-docking supply chain network design problem (CDSCNDP) introduced by Sung and Song (2003). The problem (CDSCNDP) is concerned with optimally locating cross-docking (CD) centres and allocating vehicles for direct transportation services from the associated origin node to the associated CD centre or from the associated CD centre to the associated destination node so as to satisfy a given set of freight demands at minimum cost subject to the associated service (delivery) time restriction. Each freight demand is assumed to be delivered through just one CD centre (where operations including sorting and consolidating are handled) located between origin and destination nodes. Thus, it can be checked at a pre-processing stage whether each demanded freight can be delivered through some CD centre within the associated delivery time limit or not. This will be discussed more in the next section. Moreover, it is assumed that two types of vehicles are available in the network and that one of the types has the smaller capacity than the other type. The vehicle cost of the smaller type is assumed to be less than that of the other type, while being larger in the vehicle cost per unit capacity than that of the other type.

As discussed in Sung and Song (2003), the problem (CDSCNDP) is different from many ordinary network design problems including hub networks design problems (Campbell, 1994; Aykin, 1995; Sung and Jin, 2001), less-than-truck-load (LTL) transportation networks design problems (Powell, 1986; Powell and Sheffi, 1989; Crainic and Jacques, 1992; Akyilmaz, 1994; Farvolden and Powell, 1994) and air networks design problems for express shipment delivery (Barnhart and Schneur, 1996; Kim et al, 1999; Armacost et al, 2002) in the sense that it considers the integrated issue incorporating discrete transportation units (vehicles) and service time restrictions explicitly as well as facility location and vehicle allocation.

The expected contributions of this paper can be specified as follows:

  • It presents an exact algorithm. An exact algorithm is necessary for the following reasons even if the problem was proven to be NP-hard in Sung and Song (2003). First, the problem (CDSCNDP) is concerned with decisions on the strategic and tactical levels of system designs including facility location and investment on vehicles so that it may be desirable to find an optimal solution. Second, an exact algorithm may provide a strong basis to evaluate the performance of any other heuristics for the problem (CDSCNDP). In other words, the proposed exact algorithm may be used as an effective strategic planning tool for small-scale problem instances. It may also be used as an evaluation basis for any existing heuristics or those to be developed for large-scale problem instances as well.
  • Some of the solution properties characterized in this paper can also be used for deriving any efficient local search heuristics with the move operation (neighbourhood search operation) of modifying assignment of some freight demands from current CD centres to other CD centres. The move operation, denoted by MOVEMD, is easy to use, referring to Sung and Song (2003). As the operation MOVEMD is performed, the delivery quantity associated with each direct service can be updated. Therewith, the contribution of the operation MOVEMD to the objective total cost can be computed by finding how many vehicles are needed for the direct services modified by the operation, which corresponds to solving vehicle allocation problems associated with the modified direct services. This implies that the efficiency of such a local search heuristic is dependent on how efficiently the vehicle allocation problems can be solved. In order to solve the vehicle allocation problems, several solution properties were characterized in Sung and Song (2003). In this connection, more efficient solution properties are characterized in this paper.

The remainder of this paper is organized as follows. The next section presents the path-based formulation derived in Sung and Song (2003), and also derives a set-partitioning-based formulation for the problem (CDSCNDP). It is shown that the proposed set-partitioning-based formulation (SPBF) can be used to derive a better lower bound than the existing path-based formulation. Next, we follow-up with a section that characterizes some solution properties that are used to derive a branch-and-price algorithm based on the SPBF. The penultimate section presents the computational results to evaluate the performance of the derived algorithm and the contribution of the solution properties in improving the efficiency of the aforementioned local search heuristics. Finally, the last section makes concluding remarks.

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Problem formulations

Path-based formulation

For the problem (CDSCNDP), a path-based formulation was derived in Sung and Song (2003) by use of the following notation.

Sets

I
set of origin nodes (terminals representing manufactures and suppliers)
J
set of destination nodes (terminals representing retailers and customers)
K
set of intermediate nodes (terminals representing CD centres)
V
set of all nodes, V=IcupJcupK
EIN
set of edges (representing potential direct services) from origin nodes to intermediate nodes, that is EIN={e=(i,k)|iset symbolI,kset symbolK}
EOUT
set of edges (representing potential direct services) from intermediate nodes to destination nodes, that is EOUT={e=(k,j)|kset symbolK,jset symbolJ}
E
set of all edges, that is E=EINcupEOUT
Q
set of freight demands each of which is represented by an ordered pair of two nodes (i,j), where i and j denote the associated origin node and the associated destination node, respectively
Qk
set of freight demands that can be transported through the intermediated node (CD centre) k

Parameters

fk
setup cost of the intermediate node (CD centre) k
hik1,hik2,hkj1,hkj2
unit vehicle costs of type-1 and type-2 vehicles for each direct service (i,k)set symbolEIN and (k,j)set symbolEOUT, respectively (hik1<hik2 and hkj1<hkj2)
lambda1,lambda2
capacities of type-1 and type-2 vehicles, respectively (lambda1<lambda2andhe1/lambda1>he2/lambda2)
tik,tkj
transportation time required for each direct service (i,k)set symbolEIN and (k,j)set symbolEOUT, respectively
pk
handling (sorting and consolidating) time at the intermediate node k
dij
quantity of freight demand (i,j)set symbolQ
TLij
delivery time limit for freight demand (i,j)set symbolQ

Decision variables

Yik,Zik,Ykj,Zkj
numbers of type-1 and type-2 vehicles allocated for each direct service (i,k)set symbolEIN and (k,j)set symbolEOUT, respectively
Xijk
1 if the freight demand (i,j)set symbolQ is transported through the intermediate node kset symbolK;
0 otherwise
Wk
1 if the CD centre k is established located);
0 otherwise

For notational convenience, notation (or subscript) e and (i,k) (or (k,j)) will be used alternately to denote the associated edge, and q and (i,j) will be used alternately to denote the associated freight demand. The two terms 'edge' and 'direct service' will also be used alternatively.

Formulation (PBF):

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The objective function (1) represents the two major cost functions including the cost of locating CD centres and the cost of allocating any vehicles for the associated direct services. Constraints (2) represent the situation where all the freight demands have to be serviced. Constraints (3) and (4) require that the total amount of demands transported through any direct service should not exceed the total capacity of any allocated vehicles. Constraints (5) represent the condition that freight demand (i,j) may be consolidated at the intermediate node k if and only if the associated CD centre is located at the node. Note that constraints (5) can be replaced with the following constraints (5') which are the disaggregated version of the constraints (5).

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The aggregated constraints (5) are weaker than the disaggregated constraints (5') so that in order to find a better lower bound, it is necessary to replace constraints (5) with constraints (5') even though it may require more time to solve the associated LP relaxations because of the larger number of constraints (Sung and Song, 2003). The service time restrictions can be satisfied with only the X-variables Xijk in the formulation (PBF) such that for each freight demand (i,j), the sum of transportation times between nodes and handling time at an intermediate node k does not exceed the predefined time limit TLij (tik+pk+tkjless than or equal toTLij) (Sung and Song, 2003). Thus, the set Qk can be identified at a pre-processing stage. Under the assumption that each freight demand is transported through a single path, the X-variables Xijk have binary values.

Set-partitioning-based formulation

In this section, a set-partitioning-based formulation (SPBF) is derived for the problem (CDSCNDP). Such SPBFs have been considered for various problems including vehicle routing problem with time windows (Desrochers et al, 1992), generalized assignment problem (Cattrysse et al, 1994; Savelsbergh, 1997) and p-median problem (Garfinkel et al, 1974).

For the formulation, the following notation will be used in addition.

Sets

Qik
set of freight demands that can be transported from node i to node k, Qik={(i',j)set symbolQk|i'=i}
Qkj
set of freight demands that can be transported from node k to node j, Qkj={(i,j')set symbolQk|j'=j}
OmegaINik
set of freight demands to be transported and the associated vehicle allocation on the edge from origin node i to intermediate node k (ie set of integral points consisting of X, Y and Z variables feasible to the associated vehicle capacity constraint (3) on the edge),
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OmegaOUTkj
set of freight demands to be transported and the associated vehicle allocation on the edge from intermediate node k to destination node j (ie set of integral points consisting of X, Y and Z variables feasible to the associated vehicle capacity constraint (4) on the edge),
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Qik(r),Qkj(s)
set of freight demand corresponding to rset symbolOmegaINik and sset symbolOmegaOUTkj, respectively

Parameters

crik
vehicle allocation cost induced by r=(X,Yik,Zik)set symbolOmegaINik, crik=hik1Yik+hik2Zik
cskj
vehicle allocation cost induced by s=(X,Ykj,Zkj)set symbolOmegaOUTkj, cskj=hkj1Ykj+hkj2Zkj

Decision variables

deltarik
1 if the associated freight demands with rset symbolOmegaINik is transported by the associated vehicles;
0 otherwise
deltaskj
1 if the associated freight demands with sset symbolOmegaOUTkj is transported by the associated vehicles;
0 otherwise
Wk
1 if the CD centre k is established;
0 otherwise

Note that Qk=Unioniset symbolI Qik=Unionjset symbolJ Qkj.

Formulation (SPBF):

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The objective function (7) is equivalent to the objective function (1) in the (PBF). Constraints (8) and constraints (11) are equivalent to constraints (2) and constraints (5) in (PBF), respectively. Constraints (9) mean that at most one set of freight demands associated with variable deltarik is delivered for direct service (i,k), and constraints (10) mean that at most one set of freight demands associated with variable deltaskj is delivered for direct service (k,j). In a similar approach to constraints (5) in the (PBF), constraints (11) can be replaced with constraints (11') which are the disaggregated version of constraints (11).

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Note that the variables deltarik and deltaskj are defined to satisfy the constraints (3) and (4), respectively.

A set of valid inequalities to the formulation (PBF) has been derived and used in Sung and Song (2003) to propose a cutting plane scheme based on the set of valid inequalities so as to obtain a stronger lower bound than that obtained from the LP relaxation of the formulation (PBF). The set of valid inequalities represents a set of Chvátal–Gomory valid inequalities for the convex hull of OmegaINik (or OmegaOUTkj). However, only for a subset of the inequalities with some coefficient fixed, separation algorithm was derived to find out any violated constraints from a current solution in the work. Therefore, the associated convex hull of OmegaINik (or OmegaOUTkj) may not be described well.

Now, denote by LB1, LB2 and LB3, the lower bounds obtained from the LP relaxation of the formulation (PBF), the cutting plane scheme proposed by Sung and Song (2003) and the LP relaxation of the formulation (SPBF) presented in this paper, respectively.

Property 1
 

The following relation holds:

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Proof
 

The relation LB1less than or equal toLB2 is obvious. The proof for the relation LB2less than or equal toLB3 can be made by showing that the formulation (SPBF) can be derived via Dantzig–Wolfe decomposition of the following formulation (VPBF). The formulation (VPBF) can, however, be derived from the formulation (PBF) by introducing a set of variables Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author equivalent to the set of variables Xijk and adding a set of constraints Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author.

Formulation (VPBF):

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The lower bound obtained from the LP relaxation of the formulation (VPBF) is denoted by LB1'. Obviously, LB1' is equal to LB1. Now, Dantzig–Wolfe decomposition of the formulation (VPBF) is made as follows. Assume that the set OmegaINik={(X,Yik,Zik)set symbolB|Qik| times Z+ times Z+|sum(i,j)set symbolQik dijXijkless than or equal tolambda1Yik+lambda2Zik} contains points (rik)'s. For notational convenience, r will be used instead of rik. Then, it follows that

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where xrik, yrik and zrik are the incidence vector of Qik(r)subset equalsQ (xrik being composed of xjrik's having 0–1 incidence values in the associated position), the number of type-1 vehicles to deliver Qik(r), and the number of type-2 vehicles to deliver Qik(r), respectively. Similarly, assume that the set

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contains points (skj)'s. For notational convenience, s will be used instead of skj. Then, it follows that

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where xskj, yskj and zskj are the incidence vector of Qkj(s)subset equalsQ (xskj being composed of (xiskj)'s having 0–1 incidence values in the associated position), the number of type-1 vehicles to deliver Qkj(s), and the number of type-2 vehicles to deliver Qkj(s), respectively. Recall the relations crik=hik1yrik+hik2zrik and cskj=hkj1yskj+hkj2zskj. Now, let us replace the variables Xijk, Yik, Zik, Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author, Ykj, Zkj with the variables xjrik, yrik, zrik, deltarik, xiskj, yskj,zskj and deltaskj. Then, the terms sumrset symbolOmegaINik xjrikdeltarik and sumsset symbolOmegaOUTkj xiskjdeltaskj can be represented by sum{rset symbolOmegaINik|(i,j)set symbolQik(r)}deltarik and sum{sset symbolOmegaOUTkj|(i,j)set symbolQkj(s)} deltaskj, respectively. However, if the sets Qik(r) and Qkj(s) are empty, then the associated variables deltarik and deltaskj, respectively, are not necessary to generate for solving the problem (CDSCNDP), so that the relations sumrset symbolOmegaINikdeltarik=1 and sumsset symbolOmegaOUTkj deltaskj=1 can be rewritten as the relations sumrset symbolOmegaINik deltarikless than or equal to1 and sumsset symbolOmegaOUTkjless than or equal to1, respectively. Therewith, the formulation (VPBF) can be transformed to the proposed formulation (SPBF) by Dantzig–Wolfe decomposition. Because the separation algorithm proposed by Sung and Song (2003) is not an exact separation algorithm for the convex hull of OmegaINik (or OmegaOUTkj), the associated lower bound, LB2, is equal to or lower than that obtained from the LP relaxation of the above Dantzig–Wolfe decomposition being equivalent to the proposed formulation (SPBF), LB3. square

Property 1 implies that a stronger lower bound for the problem (CDSCNDP) can be obtained from the LP relaxation of the proposed formulation (SPBF), which is very desirable to derive an exact algorithm for the problem. Note that a similar property to Property 1 can be derived in the situation where the aggregated constraints (5) and (11) are replaced with the disaggregated constraints (5') and (11'), respectively.

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Solution approach

The proposed formulation (SPBF) may have a huge number of variables because the variables deltarik and deltaskj are associated with various subsets of freight demand sets Qik and Qkj, respectively. In spite of that, a small subset of all the variables is required to find an optimal solution. To exploit this aspect, the column generation technique is used to generate such required variables in this paper. Specifically, a branch-and-price algorithm for the problem (CDSCNDP) is derived based on the formulation (SPBF) with the column generation technique applied throughout the associated branch-and-bound tree.

Consider the LP relaxation of the proposed formulation (SPBF), which will be called the 'master problem', denoted by (SPBFMP). Denote by (SPBFRMP) a 'restricted master problem' containing only a subset of all the variables of the master problem (SPBFMP).

Initial restricted master problem

The initial restricted master problem used in the proposed algorithm includes all the W variables, (Wk)'s, and a subset of all the delta-variables. The delta-variables, (deltarik)'s and (deltaskj)'s, to be included in the initial restricted master problem are generated as follows. First, for each freight demand, the cheapest path to make its delivery is found. Second, once all such cheapest paths are found in association with all the given freight demands, the set of freight demands to be delivered for each direct service can be identified from among all the cheapest paths associated with the direct service. Third, for each direct service associated with only a non-empty set of freight demands, the corresponding variables deltarik and deltaskj are generated and added to the initial restricted master problem. Then, the vehicle cost crik (or cskj) associated with deltarik (or deltaskj) can be computed by finding how many vehicles are needed to deliver the associated freight demands at minimum vehicle cost.

To calculate the number of vehicles at minimum vehicle cost, the next section characterizes several solution properties and proposes an associated efficient calculation procedure.

Calculating the number of vehicles for direct service

Suppose that a set of freight demands to be delivered for direct service e is determined, denoted by Q(e). Then, the problem to calculate the number of vehicles at minimum vehicle cost for the delivery can be defined as follows:

Problem CNV(e):

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Let D(e)=sum(i,j)set symbolQ(e) dij, Yemax=left ceilingD(e)/lambda1right ceiling and Zemax=left ceilingD(e)/lambda2right ceiling. Then, the problem CNV(e) needs to be solved to construct the initial restricted master problem, and to reduce the size of the pricing problems of generating delta-variables, (deltarik)'s and (deltaskj)'s, that will be fully explained in the next section.

Moreover, the problem CNV(e) shall be solved in local search heuristics using the operation MOVEMD for the problem (CDSCNDP). As mentioned earlier, the contribution of the operation MOVEMD to the objective total cost can be measured by finding how many vehicles are needed for the direct services in association with the modified delivery quantity due to the operation, which corresponds to solving the problem CNV(e)'s associated with the modified direct services. This implies that the efficiency of such a local search heuristic is dependent on how efficiently the problem CNV(e) can be solved.

In order to solve the problem CNV(e), several solution properties were characterized in Sung and Song (2003). In this connection, more efficient solution properties are characterized in this section.

Property 2
 

(Referring to Sung and Song, 2003) Given Ze=m(0less than or equal tomless than or equal toZemax), left ceiling(D(e)-mlambda2)/lambda1right ceiling and left floor(ZemaxYemax-mYemax)/Zemaxright floor are derived as lower and upper bounds of Y-variable Ye, respectively.

By use of Property 2, the number of vehicles to transport D(e) for direct service e at minimum vehicle cost could be determined by comparing the combinations of (Ye,Ze)'s, which results in the number summ=0Zemax{left floor(ZemaxYemax-mYemax)/Zemaxright floor-left ceiling(D(e)-mlambda2)/lambda1right ceiling+1}. The above combinations have been compared in Sung and Song (2003) to solve the problem CNV(e).

Property 3
 

Given Ze=m(0less than or equal tomless than or equal toZemax), the optimal number of type-1 vehicles, Ye, is left ceiling(D(e)-mlambda2)/lambda1right ceiling.

Proof
 

It is obvious, because a solution left ceiling(D(e)-mlambda2)/lambda1right ceiling is the minimum number of type-1 vehicles to satisfy (22). square

Property 3 obviously dominates over Property 2, so that it is enough to compare Zemax+1 combinations of (Ye,Ze)'s in total.

To obtain an optimal (Ye,Ze) for the problem CNV(e) more efficiently, some more properties are to be characterized. Let Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author, Re=lambda1Ye+lambda2Ze-D(e) and Ce=he1Ye+he2Ze. Note that if Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author is given, the associated Ye, Ze, Re and Ce can be obtained easily. The associated Ye, Ze, Re and Ce with Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author are denoted by Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author, Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author, Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author and Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author throughout this paper, respectively. Also, given D(e), the associated Ye, Ze, Re, Ce and Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author with an optimal solution for the problem CNV(e) are denoted by Ye*|D(e), Ze*|D(e), Re*|D(e), Ce*|D(e) and Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author, respectively.

Property 4
 

Given Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author and Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author, it holds that if r1less than or equal tor2, then xi1<xi2, where (r1,xi1) and (r2,xi2) are the values of (Re,Ce) associated with Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author and Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author, respectively.

Proof
 

Denote by (y1,z1) and (y2,z2) the values of (Ye,Ze) associated with Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author and Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author, respectively. Then,

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Corollary 5
 

If Re associated with Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author is 0, then it is unnecessary to consider the associated solutions for the problem CNV(e) with Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author.

Proof
 

It is obvious by Property 4. square

Property 4 and Corollary 5 can be used as dominance relations for solving the problem CNV(e) efficiently.

Property 6
 

If (lambda2 mod lambda1) is 0 (ie lambda2 being a multiple of lambda1), then Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author is either 0 or 1.

Proof
 

Let D(e)=lambda2k+A where kset symbolZ+ and A=(D(e) mod lambda2). For n(1less than or equal tonless than or equal toZemax), Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author, since lambda2 mod lambda1=0. By Property 4, Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author dominates over Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author. Thereupon, Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author is either 0 or 1. square

Property 6 implies that the problem CNV(e) can be solved in O(1)-bounded operations when (lambda2 mod lambda1) is 0 as a special case. A similar result was characterized in Sung and Song (2003). For general cases, the following properties are characterized.

Property 7
 

With the values D(e)=lambda2k+A>lambda1dotlambda2 where kset symbol{lambda1,lambda1+1,...} and A=(D(e) mod lambda2), Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author is periodic in Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author starting at 1 with the period lambda1.

Proof
 

This can be made by showing Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author, where 0<n<n+lambda1less than or equal toZemax. Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author since Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author. Also, Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author since Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author and lambda2lambda1 mod lambda1=0. Thus, Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author. square

According to Property 3, the problem CNV(e) can be solved by comparing Zemax+1 solutions. Note that Zemax gets larger as D(e) increases and it can be much larger than lambda1 for some D(e). Property 7 implies that the problem CNV(e) can be solved by comparing at most lambda1+1 solutions for any arbitrary case. Thereupon, the problem CNV(e) can be solved by comparing Zemax+1 solutions when D(e) is small such that Zemax<lambda1 (by Property 3), and comparing lambda1+1 solutions when D(e) is large such that Zemaxgreater than or equal tolambda1 (by Property 7).

The periodic characteristics presented in Property 7 seem to provide a motivation to characterize more solution properties to solve the problem CNV(e) effectively. Accordingly, the following solution properties are characterized such that as D(e) increases, the associated optimal solutions to the problem CNV(e) also have periodic characteristics. To implement the solution properties, a preprocessing procedure is derived to keep some information corresponding to one period at a cost of a small computation load and memory consumption. After the preprocessing procedure, the problem CNV(e) can be solved with any given D(e) in O(1)-bounded operations.

Property 8
 

With the values D(e)=klambda2 (kset symbolZ+), the following relations hold:

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Proof
 

It is easy to see that Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author. Thus, the conclusion can be made by Corollary 5. square

Property 9
 

With the values D(e)1>0 and D(e)2=D(e)1+klambda2>0 (kset symbolZ), the following relations hold:

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Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author

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Proof
 

It is obvious by the definition of Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author. square

Corollary 10
 

With the values D(e)1greater than or equal to(lambda1-1)lambda2+l (lset symbol{1,...,lambda2-1}) and D(e)2=D(e)1+klambda2 (kset symbolZ+), the following relations hold:

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Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author

Proof
 

This can be proved by showing Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author. By Property 7, (Ye*|D(e)2,Ze*|D(e)2) can be obtained by comparing lambda1+1 solutions associated with Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author. For Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author, it is easy to see that Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author is greater than Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author exactly by the amount khe2 as shown in Property 9. Thus, Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author. square

Let Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author for the given values D(e) ((lambda1-1)lambda2+1less than or equal toD(e)less than or equal tolambda1lambda2-1) and m (0less than or equal tomless than or equal tolambda1).

Corollary 11
 

With the values D(e)1greater than or equal to(lambda1-1)lambda2+l (lset symbol{1,...,lambda2-1}) and D(e)2=D(e)1-klambda2>0 (kset symbolZ+), the following relations hold:

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Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author

where Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author with m=Zemax|D(e)2=lambda1-k.

Proof
 

This can be proved by showing Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author. By Property 3, (Ye*|D(e)2,Ze*|D(e)2) can be obtained by comparing Zemax|D(e)2+1 solutions associated with Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author. For Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author, it is easy to see that Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author is less than Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author exactly by the amount khe2 as shown in Property 9, Thus, Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author. square

In order to implement the results of Corollary 10 and Corollary 11, all the data Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author, Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author, Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author, Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author, Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author for (lambda1-1)lambda2+1less than or equal toD(e)less than or equal tolambda1lambda2-1, 0less than or equal tonless than or equal tolambda1) are required. Once all the required data are found, the problem CNV(e) can be solved with any freight demand quantity for the associated direct service e in O(1)-bounded operations by use of Property 8, Corollaries 10 and 11. The procedure for solving the problem CNV(e), denoted by PROCCNV, is then derived as follows.

Procedure PROCCNV:

The demand quantity to be delivered for direct service e is given as D(e)=sum(i,j)set symbolQ(e) dij. Let k=left floorD(e)/lambda2right floor-(lambda1-1) and l=D(e) mod lambda2. Then, the optimal solution (Ye*|D(e),Ze*|D(e)) and the optimal cost Ce*|D(e) can be determined as in the following situations.

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The step of calculating all required data needs to be implemented as a preprocessing step in O((lambda2-1)dot(lambda1+1))-bounded operations and memory by the definitions of the required data, which may be time-consuming though. However, the PROCCNV may be beneficial in the aspect of efficiency for both the proposed branch-and-proce algorithm and the local search heuristics using the operation MOVEMD because they may require a lot of the problem CNV(e)'s to be solved iteratively.

In the case of the proposed branch-and-proce algorithm, to reduce the size of the pricing problems of generating delta-variables, (deltarik)'s and (deltaskj)'s, that will be presented in the next section in detail, it is necessary to know the upper bound on the number of each vehicle type when demand quantity to be delivered is an arbitrary value within some bounds, but not a given value as in the problem CNV(e). The upper bounds can be found by consecutively repeating the procedure that D(e) is fixed each time at a value within the bounds and then the associated problem CNV(e) is solved whenever it is necessary to solve the pricing problem. However, the method requires multiple number of the repetitive problems CNV(e)'s to be solved for the single pricing problem. Futhermore, it is necessary to solve a number of such pricing problems. Therefore, a lot of such repetitive problems CNV(e)'s need to be solved in the proposed branch-and-price algorithm.

Similarly, in the case of local search heuristics using the operation MOVEMD, a lot of repetitive problems CNV(e)'s need to be solved. This is because the operation MOVEMD is often required to be performed many times for solving the problem (CDSCNDP) in such a local search heuristic. By the way, it is necessary to evaluate the contribution of each move operation MOVEMD to the objective total cost, for which it needs to solve the problems CNV(e)'s associated with the modified direct services by the operation.

Pricing problem

This section considers two types of pricing problems to generate delta-variables: one for deltarikset symbolOmegaINik, denoted by (PPik) and the other for deltaskjset symbolOmegaOUTkj, denoted by (PPkj). Let PiijPAR (unrestricted), PiikIN(less than or equal to0), PikjOUT(less than or equal to0), PiijkLOC(less than or equal to0) and PiijkF (unrestricted) be the optimal values of the dual variables for constraints (8), (9), (10), (11') and (12) in the current formulation (SPBFRMP), respectively. The reduced cost Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author of variable deltarikset symbolOmegaINik and the reduced cost Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author of variable deltaskjset symbolOmegaOUTkj are given as follows:

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Recall that xjrik and xiskj are 0–1 incidence parameters indicating whether demand (i,j) is included in the sets Qik(r) and Qkj(s) or not, respectively, when the proposed formulation (SPBF) is derived by Dantzig–Wolfe decomposition (as presented in the proof of Property 1). However, in the pricing problems, xjrik and xiskj are 0–1 decision variables indicating whether demand (i,j) is included in the sets Qik(r) and Qkj(s) (xjrik or xiskj=1) or not (xjrik or xiskj=0), respectively. Then, the pricing problem (PPik) can be derived as follows:

Problem (PPik):

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Similarly, the pricing problem (PPkj) can be formulated by replacing the objective cost function Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author, and variables Yik,Zik and xjrik in the problem (PPik) with Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author, and Ykj, Zkj and xiskj, respectively. If a variable deltarikset symbolOmegaINik (or deltaskjset symbolOmegaOUTkj) with negative reduced cost is obtained from (PPik) (or (PPkj)), the variable is added to the current formulation (SPBFRMP).

Some of xjrik-variables in the problem (PPik) and some of xiskj-variables in the problem (PPkj) can be fixed as in the following property so as to reduce the size of the pricing problems.

Property 12
 
  1. If (PiijPAR+PiijkLOC+PiijkF)less than or equal to0 for some j in the problem (PPik), then there is an optimal solution with xjrik=0;
  2. If (PiijPAR+PiijkLOC+PiijkF)greater than or equal toCij*|D(e)=dij for some j in the problem (PPik), then there is an optimal solution with xjrik=1;
  3. If (-PiijkF)less than or equal to0 for some i in the problem (PPkj), then there is an optimal solution with xiskj=0;
  4. If (-PiijkF)greater than or equal toCij*|D(e)=dij for some i in the problem (PPkj), then there is an optimal solution with xiskj=1.
Proof
 

For the proof of (a), let us compare the situations of xjrik=1 and xjrik=0 for some j such that (PiijPAR+PiijkLOC+PiijkF)less than or equal to0 in the problem (PPik). The value of the objective function (23) for the former case cannot be smaller than that for the latter case, because the coefficient of the variable in (23) is non-negative. Also, the former case makes the constraint (24) tighter than the latter case. Therefore, the former case cannot give a better solution than the latter case. The proofs for (b), (c) and (d) can be made similarly by comparing the situations of xjrik (or xiskj)=1 and xjrik (or xiskj)=0 in the same manner as in the proof of (a). square

Let Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author be a subset of the set Qe (the associated Qik or Qkj) such that (PiijPAR+PiijkLOC+PiijkF)less than or equal to0 in the problem (PPik) (ie satisfying the statement (a) in Property 12) or (-PiijkF)less than or equal to0 in the problem (PPkj) (ie satisfying the statement (c) in Property 12) for Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author. Also, let Unfortunately we are unable to provide accessible alternative text for this. If you require assistance to access this image, please contact help@nature.com or the author be a subset of the set Qe (the associated Qik or Qkj) such that (PiijPAR+PiijkLOC+PiijkF)greater than or equal toCij*|D(e)=dij in the problem (PPik) (ie satisfying the statement (b) in Property 12) or (-PiijkF)greater than or equal toCij*