a worst-case analysis for the split delivery vehicle routing problem with minimum delivery amounts

Optim LettDOI 10.1007/s11590-012-0554-9

ORIGINAL PAPER

A worst-case analysis for the split delivery vehiclerouting problem with minimum delivery amounts

Yupei Xiong · Damon Gulczynski · Daniel Kleitman · Bruce Golden ·Edward Wasil

Received: 14 September 2011 / Accepted: 1 September 2012© Springer-Verlag 2012

Abstract In the vehicle routing problem (VRP), a fleet of vehicles must service thedemands of customers in a least-cost way. In the split delivery vehicle routing problem(SDVRP), multiple vehicles can service the same customer by splitting the deliveries.By allowing split deliveries, savings in travel costs of up to 50 % are possible, andthis bound is tight. Recently, a variant of the SDVRP, the split delivery vehicle routingproblem with minimum delivery amounts (SDVRP-MDA), has been introduced. Inthe SDVRP-MDA, split deliveries are allowed only if at least a minimum fraction ofa customer’s demand is delivered by each visiting vehicle. We perform a worst-caseanalysis on the SDVRP-MDA to determine tight bounds on the maximum possiblesavings.

Keywords Vehicle routing · Split deliveries · Minimum delivery amounts ·Worst-case analysis

Y. XiongMercury Systems, Inc., Princeton, NJ 08540, USA

D. Gulczynski (B)RouteSmart Technologies, Inc., Columbia, MD 21045, USAe-mail: [email protected]

D. KleitmanDepartment of Mathematics, Massachusetts Institute of Technology,Cambridge, MA 02139, USA

B. GoldenRobert H. Smith School of Business, University of Maryland,College Park, MD 20742, USA

E. WasilKogod School of Business, American University,Washington, DC 20016, USA

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1 Introduction

In the vehicle routing problem (VRP), a fleet of vehicles must service the demands ofcustomers. A vehicle begins and ends its route at the same depot and the sum of thedemands of the customers on a route cannot exceed a vehicle’s capacity. A customermust have all of its demand delivered at one time by a single vehicle. The objective isto minimize the total distance traveled by the fleet.

In the split delivery vehicle routing problem (SDVRP), more than one vehicle isallowed to service a customer, so that a customer’s demand can be split among severalvehicles on different routes. The objective in the SDVRP is to minimize the totaldistance traveled by the fleet, while satisfying the demand of each customer. (Forpapers that cover the VRP and SDVRP in detail, consult the book edited by Goldenet al. [2]).

Archetti et al. [1] provide a worst-case analysis of the SDVRP. They show that,by allowing split deliveries, travel distance can be reduced by at most 50 %, and thisis a tight bound. That is, they show Z(V R P)

Z(SDV R P)≤ 2, where Z(V R P) is the distance

of an optimal VRP solution to an instance (no splits allowed), and Z(SDV R P) isthe distance of an optimal SDVRP solution to the same instance (splits allowed).Furthermore, there are instances where Z(V R P)

Z(SDV R P)is arbitrarily close to 2. Archetti

et al. [1] assume the triangle inequality holds for travel among locations, and weassume the same throughout this paper. In addition, vehicles have the same capacity(homogeneous fleet), and no customer’s demand can exceed the vehicle capacity.

Recently, Gulczynski et al. [3] considered the split delivery vehicle routing problemwith minimum delivery amounts (SDVRP-MDA). In the SDVRP-MDA, split deliver-ies are allowed only if at least a minimum fraction of a customer’s demand is deliveredby each vehicle visiting the customer. For example, if p = 1

5 is the minimum deliveryfraction, and qi = 10 is the demand of customer i , then each vehicle visiting customeri must deliver at least pqi = 2 units. The objective of the SDVRP-MDA and theSDVRP is the same, that is, minimize the total distance traveled by the fleet.

In Fig. 1, we provide an example of the SDVRP-MDA given by Gulczynski et al. [3].There are three customers (nodes 1, 2, and 3) and a depot (node 0). Edge labels aredistances and node labels in parentheses are demands. The vehicle capacity is 120.In Fig. 1a, we show that the optimal solution to the VRP has three direct round-tripsfor a total distance of 30. In Fig. 1b, we show the optimal solution to the traditionalSDVRP with no minimum delivery amounts (p = 0). The total distance traveled bythe fleet in the SDVRP (25 units) is smaller than the distance traveled in the VRP (30units). In Fig. 1c, we show the optimal solution to the SDVRP-MDA with p = 3

10 ,that is, each customer must have at least 30 % of its demand delivered by a vehicle.The solution in Fig. 1b is not feasible when p = 3

10 (the split delivery to customer 1has only 25 % of its demand (20 units) on one of the two routes).

Notice that, in general, the minimum delivery fraction p must be at least 0 and atmost 1. When p = 0, the SDVRP-MDA reduces to the SDVRP, and when p > 1

2 ,the SDVRP-MDA reduces to the VRP. In this paper, we focus on the cases when0 < p ≤ 1

2 . We extend the theoretical work of Gulczynski et al. [3]. One of theirresults gives bounds for a worst-case SDVRP-MDA scenario. Let Z(V R P) be thedistance of an optimal VRP solution to an instance, and let Z p(M D A) be the distance

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A worst-case analysis for the split delivery vehicle

v v

v

0

1 3

2

(60)

(100)

(80)

3

2

5

55

VRP

Total Distance = 30

v v

v

0

1 3

2

(60)

(100)

SDVRP

Total Distance = 25

v v

v

0

1 3

2

(20)

(100)

(80)

SDVRP-MDA

Total Distance = 27

(60)

(20)

p = 0 p = 3/10

(40)

(a) (b) (c)

Fig. 1 An SDVRP-MDA example with four nodes. The edge labels are distances and the node labels inparentheses are demands. The vehicle capacity is 120

of an optimal SDVRP-MDA solution to the same instance, with minimum deliveryfraction p. Further, let M(p) be the least upper bound of Z(V R P)

Z p(M D A). They showed that

2 − p ≤ M(p) ≤ 2, when 0 ≤ p ≤ 12 .

The result of Gulczynski et al. [3] offers the possibility that M(p) depends onp. As in Fig. 1, when we try to solve instances of the SDVRP-MDA, we observethat, in general, as p increases from 0 (within the range 0 < p ≤ 1

2 ), the totaldistance increases. Therefore, we expect the bound to be a decreasing function of p.Surprisingly, this turns out to not be the case for almost all p. In this paper, we provethat for 0 < p < 1

2 , M(p) = 2. That is, the worst-case bound for the SDVRP-MDAis independent of p, and it is the same as that for the SDVRP. The single case p = 1

2is the only minimum delivery fraction for which this result does not hold. For p = 1

2the worst-case bound is 3

2 , not 2, that is, M( 12 ) = 3

2 .The remainder of this paper is organized as follows. In Sect. 2, we state and prove

our results for the cases where 0 < p < 12 . In Sect. 3, we state and prove our result

for the single case where p = 12 . Finally, we provide some brief, concluding remarks.

2 Worst-case bound for 0 < p < 12

Before presenting the main result of this section, we state and prove the followinglemma.

Lemma 1 For the SDVRP-MDA, where p is a positive rational number less than 12 ,

M(p) ≥ 2.

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T23T

T4T1

Depot

Fig. 2 The example SDVRP-MDA used in the proof of the lemma

Proof of Lemma 1 Let rs be the rational representation of p in reduced form. That

is, rs = p, where r and s are integers whose greatest common divisor is 1. Since

0 < rs < 1

2 , s > 2r , so s − r − 1 ≥ r and s > 2. Let n be a positive integer. Consideran SDVRP-MDA with minimum delivery fraction p, and node locations as given inFig. 2. The distance between the depot and any customer is 1, and the distance betweenany pair of customers is 0. (The argument holds if distances between customers mustbe strictly positive by letting the distance between any pair of customers be a andconsidering the limiting case as a goes to 0.) The customers are located on n tracks,T1, . . . , Tn , emanating from the depot. There are 3s customers located at the end oftrack T1; s customers are denoted by a square (SC), s customers are denoted by a circle(CC), and s customers are denoted by a triangle (TC). There are 2s customers at theend of each of track T2 through track Tn−1; s are SCs, and s are CCs. There are 2scustomers at the end of track Tn ; s are SCs, and s are TCs. All customers have demand1, and the vehicle capacity is Q = 2 − 1

sn . In Fig. 2, r = 1, s = 3, p = 13 , and n = 4.

An optimal VRP solution consists of direct trips to the customers for a totaldistance of 2(3s +2s(n −2)+2s) = 2(2sn +s). Consider the SDVRP-MDA solutionconsisting of n + 1 sets of routes, R1, . . . , Rn+1. There are s routes in R1 each ofdistance 2. Each route travels up T1 and back, delivering 1 unit to an SC, (s−r)n−1

sn

units to a CC , and rs units to a TC. Since (s−r)n−1

sn ≥ (s−r)n−nsn = s−r−1

s ≥ rs , at

least the minimum amount is delivered to each customer. The load of each route is1 + (s−r)n−1

sn + rs = Q. In total, the routes in R1 completely satisfy the demand of

each SC at the end of T1; they satisfy all but rn+1sn units of demand of each CC at the

end of T1, and they satisfy all but s−rs units of demand of each TC at the end of T1.

For i = 2, . . . , n −1, there are s routes in Ri , each of distance 2. Each route travelsup Ti , delivering 1 unit to an SC, and (s−r)n−i

sn units to a CC . It travels back to the

depot along Ti−1 after delivering the remaining rn+i−1sn units to a CC . Since (s−r)n−i

sn ≥(s−r)n−n

sn = s−r−1s ≥ r

s , and rn+i−1sn = r

s + i−1sn > r

s , at least the minimum amount is

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Table 1 Route sets constructed to prove Lemma 1

Route sets(direction)

Number ofroutes

Tracktaken

Amountdeliveredon a routeto SCs

Amount deliveredon a route to CCs

Amount deliveredon a route to TCs

Totaldistance

R1 (Up) s T1 1 (s−r)n−1sn

rs 2s

R1 (Back) T1 0 0 0

Ri (Up) s Ti 1 (s−r)n−isn 0 2s

Ri (Back) Ti−1 0 rn+i−1sn 0

Rn (Up) s Tn 1 0 s−r−1s 2s

Rn (Back) Tn−1 0 rn+n−1sn 0

Rn+1 (Up) s T1 0 0 s−rs 2s

Rn+1 (Back) Tn 0 0 r+1s

Note: 2 ≤ i ≤ n − 1

delivered to each customer. The load of each route is 1 + (s−r)n−isn + rn+i−1

sn = Q. Intotal, the routes in Ri completely satisfy the demand of each SC at the end of Ti ; theysatisfy all but rn+i

sn units of demand of each CC at the end of Ti , and they satisfy theremaining rn+i−1

sn units of demand of each CC at the end of Ti−1.There are s routes in Rn each of distance 2. Each route travels up Tn , delivering

1 unit to an SC, and s−r−1s to a TC. It travels back to the depot along Tn−1 after

delivering the remaining rn+n−1sn units to a CC . From above, we see that s−r−1

s ≥ rs

and rn+n−1sn ≥ r

s , so at least the minimum amount is delivered to each customer. Theload of each route is 1 + s−r−1

s + rn+n−1sn = Q. In total, the routes in Rn completely

satisfy the demand of each SC at the end of Tn ; they satisfy all but r+1s units of demand

of each TC at the end of Tn , and they satisfy the remaining rn+n−1sn units of demand

of each CC at the end of Tn−1.There are s routes in Rn+1 each of distance 2. Each route travels up T1, delivering

the remaining s−rs units to a TC. It travels back to the depot along Tn after delivering

the remaining r+1s to a TC. Since r

s < 12 , s−r

s = 1 − rs > r

s , and so since r+1s > r

s ,at least the minimum amount is delivered to each customer. The load of each route iss−r

s + r+1s = s+1

s = 1+ 2n−nsn < 1+ sn−1

sn = Q (since s > 2). After the routes in Rn+1are carried out, the demand at each customer is completely satisfied. Thus, the union ofR1, . . . , Rn+1 constitutes a feasible SDVRP-MDA solution. Since there are s(n + 1)

routes in this union, and since each route has a distance of 2, the total distance traveledin the solution is 2s(n + 1). Thus, M(p) ≥ Z(V R P)

Z p(M D A)≥ 2(2sn+s)

2s(n+1)= 2n+1

n+1 = 2 − 1n+1 .

By letting n → ∞, we get M(p) ≥ 2, and the lemma is proved. ��In Table 1, we give a breakdown of the route sets constructed in the proof of

Lemma 1. For each set, we give the track used from the depot (up) on a route inthe set, the track used to the depot (back) on a route in the set, the number of routes inthe set, the amounts delivered to each type of customer on each track of a route in theset, and the total distance traveled among all routes in the set.

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Next, we prove the general case.

Theorem 1 For the SDVRP-MDA with minimum delivery fraction p, 0 < p < 12 ,

M(p) = 2.

Proof of the Theorem 1 Let p be any real number in the open interval (0, 12 ). Since the

rational numbers are a dense subset of the real numbers, there exists a rational numberq such that p < q < 1

2 . Since any feasible solution with minimum delivery fraction qis feasible with minimum delivery fraction p, we have Z p(M D A) ≤ Zq(M D A). ByLemma 1, for every ε > 0, there exists an SDVRP-MDA such that Z(V R P)

Zq (M D A)> 2 − ε.

Thus, Z(V R P)Z p(M D A)

> 2 − ε, as well. By letting ε → 0, we get M(p) ≥ 2, since M(p) is

an upper bound of Z(V R P)Z p(M D A)

.

We have now proved M(p) ≥ 2 when 0 < p < 12 . Since Gulczynski et al. [3]

showed M(p) ≤ 2 when 0 ≤ p ≤ 12 , we have M(p) = 2 when 0 < p < 1

2 , and thetheorem is proved. ��

The only case not addressed by Theorem 1 is when p = 12 . We consider this in the

next section.

3 A worst-case bound for p = 12

3.1 Definitions and observations

We first need to introduce notation and give definitions. Let S be a solution to the VRPor the SDVRP-MDA with route set R. We denote the distance traveled on route r ∈ Rby d(r) and the total distance traveled across all routes in R by d(R).

Definition 1 A dedicated route is a route on which a single customer is delivered itsentire demand and no other customers are visited.

Definition 2 The rank of a customer with split service on a route is its position ona list of all split customers on that route ordered from greatest demand to smallestdemand.

For example, if route r visits four customers, u1, u2, u3, and u4, with split serviceand demands 10, 20, 30, and 20, respectively, then u3 is the rank 1 customer on r , u2is the rank 2 customer, u4 is the rank 3 customer, and u1 is the rank 4 customer. Thetie between u2 and u4 is broken arbitrarily.

Observation 1 Given sets of routes R = {r1, r2, . . . , rm} and R′ = {r ′1, r ′

2, . . . , r ′m}

such that r ′i visits a subset of the customers on ri in the same order as on ri and visits

no other customers, for all i = 1, . . . , m, then d(R′) ≤ d(R).

The result in Observation 1 is a straightforward consequence of the triangleinequality.

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Observation 2 Given sets of routes R = {r1, r2, . . . , rm} and R′ = {r ′1, r ′

2, . . . , r ′m′ }

with the property that for each r ′i ∈ R′ there exists two distinct routes r1

i , r2i ∈ R, such

that d(r ′i ) ≤ min {d(r1

i ), d(r2i )}, and r1

i = r1j , r1

i = r2j , and r2

i = r2j , for all i, j =

1, . . . , m′ where i = j, then d(R′) ≤ 12 d(R).

This is true because2d(R′) = 2

∑m′i=1 d(r ′

i ) ≤ 2∑m′

i=1 min {d(r1i ), d(r2

i )} ≤ ∑m′i=1(d(r1

i ) + d(r2i )) ≤

d(R). The last inequality holds, because, by the property stated in the observation,no route appears more than once in the sum. Thus, we are summing the distances ofroutes in a subset of R and this sum is not greater than the sum of the distances overall routes in R.

Observation 3 Consider a non-increasing sequence of non-negative real numbersn1, . . . , nm, for odd m. Let

∑mi=1 ni = η. Any subsequence of n1, . . . , nm obtained by

choosing exactly one of n2 j and n2 j+1 for j = 1, . . . , �m2 � has a sum that does not

exceed η2 . This observation can be extended to the case where m is even, by considering

the sequence with nm+1 = 0 added to the end of the list.

This observation holds because, for each number in the subsequence we choose,there is a distinct larger number that we do not choose. As an example, consider thesequence 7, 6, . . . , 1 for which η = 28. If we choose one of 6 and 5, one of 4 and 3,and one of 2 and 1, then the sum of the numbers we chose will not be greater than14 = 28

2 . (In fact, the largest the sum can be is 6 + 4 + 2 = 12.)

3.2 Constructing a VRP solution from an SDVRP-MDA solution

Before we formally present and prove the main theorem of this section, we point outthat, in general, from any SDVRP-MDA solution we can construct a VRP solution.This usually requires additional routes and an increase in cost. Below, we describe aconstruction algorithm with a provable upper bound on the cost increase. Proving thisbound constitutes the major part of the proof of the theorem.

Given an SDVRP-MDA solution S with p = 12 (even splits) with route set R, we

demonstrate how to construct VRP solution S (no splits) with route set R = R1 ∪ R2,such that d(R) ≤ 3

2 d(R). R1 is constructed in such a way that for each r ∈ R1,there exists a distinct corresponding route r ∈ R, such that the customers visitedon r constitute a subsequence of the customers visited on r . By Observation 1, itfollows that d(R1) ≤ d(R). The set R2 is composed entirely of dedicated routes. Weshow how to construct the routes in R2 in such a way that, by using Observation 2,d(R2) ≤ 1

2 d(R). Thus, we prove that d(R) ≤ 32 d(R).

An overview of the construction of R1 is as follows. For each route r ∈ R, letu1, u2, . . . , um be the customers with split service on r ordered by their ranks. Thatis, uk is a split customer on r with demand qk , and qk ≥ qk+1 for all k = 1, . . . , m −1(for now, assume m is odd). Since a split delivery must satisfy exactly half of acustomer’s demand, on route r we deliver a total of exactly ϕ

2 units to the split cus-tomers, where ϕ = ∑m

i=1 qi . By Observation 3, if we choose number q2 j or number

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12

3

4

56

8

9

(20) (90)

(70)

(65)

(40)

7(50)

(50)

Depot

Route 2

Route 1

Route 4

Route 3

Route 5

(60)

(55)

Fig. 3 An SDVRP-MDA solution with a depot and nine customers. Customer demands are given inparentheses. There are five routes

q2 j+1 (but not both) for j = 1, . . . , �m2 �, then the sum of the chosen numbers is at

most ϕ2 . Thus, if we follow route r , delivering the full demand to either customer

u2 j or u2 j+1 for j = 1, . . . , �m2 �, delivering the full demands to all non-split cus-

tomers (those that already have their full demands satisfied on r ), and skipping allthe other customers on r , then we do not violate the vehicle capacity. Each route inR1 is constructed in this manner, so each route r ∈ R1 is a subsequence of a cor-responding route r ∈ R. We will show how to proceed in a way that ensures eachcustomer is serviced in full on some route in R1, except possibly the customers thathave rank 1 on a route. We visit these remaining rank 1 customers on dedicated routesin R2.

In the discussion above, we assumed that m is odd so that u2 j+1 is well-defined whenj = �m

2 �. To handle the case when m is even, we let um+1 be a dummy customerwith demand qm+1 = 0. We interpret visiting the customer um+1 as explicitly notvisiting the customer um . In this case, we still visit a subsequence of customers on r ,and Observation 3 still holds, so we do not violate the vehicle capacity constraint.

To construct the routes in R2, we construct a dedicated route for each customerwith split service in S that is not visited on a route in R1. Only customers that haverank 1 for some route in R are visited on a dedicated route in R2. We will show how toconstruct these dedicated routes so that the conditions in Observation 2 are satisfiedand d(R2) ≤ 1

2 d(R).

Before describing the details of the construction procedure of S, we give anexample. In Fig. 3, we have solution S with route set R. There is a depotand nine customers labeled 1 through 9. Customer demands are given in paren-theses. There are five routes, three depicted with solid lines (routes 1, 3, and5), two depicted with dashed lines (routes 2 and 4). Route 1 is givenby the node sequence depot–6–1–2–3–depot. Route 2 is given by depot–3–4–5–depot.Route 3 is given by depot–5–7–6–depot. Route 4 is given by depot–1–2–8–depot.Route 5 is given by depot–4–9–depot. Notice that customers 1 through 6 each have

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6

12

3

4

5

7

9

8

Depot

Route 5

Route 4 Route 2

DedicatedRoute 3

Route 3

Route 1

DedicatedRoute 5

Fig. 4 The VRP solution after removing the splits from the solution given in Fig. 3

their service split evenly between two vehicles, and customers 7 through 9 each havetheir demand delivered in full by a single vehicle. In this example, the vehicle capacityis 100 units. Travel distances on the routes are 200 units for route 1, 100 units forroute 2, 150 units for route 3, 200 units for route 4, and 50 units for route 5. Thus,d(R) = 700.

In Fig. 4, we show solution S with route set R = R1 ∪ R2. We have four routes inR1 (routes 1, 2, 3, and 4) and two routes in R2 (dedicated routes 3 and 5). Each routein R1 visits a subsequence of customers on its corresponding route in R. Therefore,d(R1) ≤ 700. Dedicated Routes 3 and 5 in R2 service the rank 1 customers on routes3 and 5 in R, respectively. By the triangle inequality, d(R2) ≤ 150 + 50 = 200, andd(R) ≤ 900 < 1050 = 3

2 d(R).

To determine which customers in S to assign to which routes in R, we con-struct an auxiliary graph G in which the vertices are the routes in S and theedges are the split customers in S. Vertices r and r ′ are connected via edge eif and only if service at customer e is split between routes r and r ′ in S (apair of vertices can be connected via multiple edges). In addition, each edgehas two labels, one for each of its vertices, giving the ranks of the customeron its respective routes. We denote the label of edge e for vertex r by l(e, r),and we let lmin(e) be the smaller of the two labels of e. Further, we defineg : Z → Z by g(x) = x + 1 if x is even, and g(x) = x − 1 if x isodd. We will use the functions l and g in our assignment procedure describedbelow.

We now show how to partition the edges of G into a set P of edge-disjoint paths thatwill allow us to assign customers to routes in R and ultimately construct S. Pseudo-code for this construction procedure is given in Algorithm 1 in the Appendix. We inputthe graph G and output the set P . We start constructing a first path ρ1 by arbitrarilyidentifying an edge with label 1 for a vertex. Additional paths are constructed one byone. Given paths ρ1, ρ2, . . . , ρh−1 already in P , we construct path ρh by identifyingedge e1, where lmin(e1) is the smallest among all edges not used in ρk , for any k < h(ties are broken arbitrarily). Then, we construct ρh = (r0, e1, r1, e2, . . . , em, rm) by

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Fig. 5 The graph G for thesolution given in Fig. 3. Verticesare split customers, edges areroutes, and labels are the ranksof edges for the respectivevertices

letting r0 be the vertex incident to e1 with the smaller label (lmin(e1) = l(e1, r0)),letting r1 be the other vertex incident to e1, and letting l(ei+1, ri ) = g(l(ei , ri )),for i = 1, . . . , m − 1. Path ρh is maximal in the sense that there exists no edge ein G incident to vertex rm not already on a path such that l(e, rm) = g(l(em, rm)).After constructing path ρh , we add it to P and proceed to path ρh+1. We continueconstructing paths until all the edges of G appear on a path.

Using the paths in P , we assign customers to routes in R. The first step inthis procedure assigns all customers whose deliveries are carried out in full ona single route r ∈ R to their corresponding route r ∈ R1. Next, given pathρ = (r0, e1, r1, e2, . . . , em, rm) in P , we distinguish two cases: 1) l(em, rm) > 1,and 2) l(em, rm) = 1. In the first case, we assign customer ei to route ri ∈ R1 fori = 1, . . . , m. In the second case, if route r0 has a smaller distance traveled than routerm , then we assign customer e1 to a dedicated route in R2, and we assign customerei to route ri−1 ∈ R1 for i = 2, . . . , m. If route rm has a smaller distance traveledthan route r0, then we assign customer em to a dedicated route in R2, and we assigncustomer ei to route ri ∈ R1 for i = 1, . . . , m − 1. Pseudo-code for this procedureis given in Algorithm 2 in the Appendix. We input the sets R and P and output theset R.

As an example, in Fig. 5, we give the auxiliary graph for the SDVRP-MDA solutiongiven in Fig. 3. In this graph, we see that vertices r1 and r2 are connected via edge e3,since routes 1 and 2 split the delivery to customer 3. The label of edge e3 for vertexr1 is 1, since customer 3 has the largest demand among all split customers on route 1.The label for vertex r2 is 2, since customer 3 has the second largest demand amongall split customers on route 2.

In this example, we start constructing the edge disjoint paths by choosing an arbi-trary edge with label 1 for some vertex, say, edge e3. We begin path ρ1 at r1 thevertex incident to e3 with the smaller label, and travel along e3 to vertex r2. Sincel(e3, r2) = 2, we next move along edge e5 since g(2) = 3 = l(e5, r2) and arriveat r3. Since l(e5, r3) = 2, we stop constructing ρ1 since there is no edge with labelg(2) = 3 incident to vertex r3. We start path ρ2 at, say, vertex r2, since l(e4, r2) = 1,and we move along edge e4 to vertex r5 before stopping since there are no otheredges incident to r5. We start path ρ3 at, say, vertex r3, since l(e6, r3) = 1. Wemove along edge e6 to vertex r1, and then we move along edge e1 to vertex r4, sinceg(l(e6, r1)) = 3 = l(e1, r1). We stop here since l(e1, r4) = 1 (no edge has a label ofg(1) = 0). We start path ρ4 at r4 since l(e2, r4) = 2 is the smallest label among edges

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not already assigned to paths (in fact e2 is the only such edge), and we move alongedge e2 to vertex r1 before stopping since all edges are now on a path. We have con-structed the paths, ρ1 = (r1, e3, r2, e5, r3), ρ2 = (r2, e4, r5), ρ3 = (r3, e6, r1, e1, r4),and ρ4 = (r4, e2, r1).

By the assignment procedure, we assign customer e3 to route r2 and customer e5 toroute r3 in R1. We assign customer e4 to a dedicated route in R2. We assign customere1 to route r1 in R1 and customer e6 to a dedicated route in R2 (note, route r3 is smallerthan route r4). We assign route e2 to r1 in R1. These are the routes in the VRP solutiongiven in Fig. 4.

3.3 Theorem for p = 12

We now present the main theorem of this section.

Theorem 2 For the SDVRP-MDA with p = 12 , M(p) = 3

2 .

Proof of the Theorem 2 Since Gulczynski et al. [3] showed that 2 − p ≤ M(p) when0 ≤ p ≤ 1

2 , we only need to demonstrate that M( 12 ) ≤ 3

2 to prove the theorem. Tothis end, we show that given SDVRP-MDA solution S with p = 1

2 with route set

R, the construction algorithm described above generates VRP solution S with routeset R, such that d(R) ≤ 3

2 d(R). Thus, the case when S is an optimal solution givesM( 1

2 ) ≤ 32 .

Our proof relies on three properties of a path ρ = (r0, e1, r1, e2, . . . , em, rm) in P .

Property 1 l(ei , ri ) > 1 and l(ei+1, ri ) > 1, for i = 1, . . . , m − 1.

This is true because all labels are positive integers, g is one-to-one, g(0) = 1,g(1) = 0, and l(ei+1, ri ) = g(l(ei , ri )), for i = 1, . . . , m − 1.

Property 2 Given consecutive edges ei , ei+1 incident to vertex ri on path ρ, one ofthe two labels, l(ei , ri ) and l(ei+1, ri ), is equal to 2 j and the other is equal to 2 j + 1,for some integer j .

This is true because, by definition, l(ei+1, ri ) = g(l(ei , ri )), g(2 j) = 2 j + 1, andg(2 j + 1) = 2 j , for any integer j .

Property 3 If l(em, rm) = 1, then l(e1, r0) = 1, as well.

This property holds because, by definition, l(e1, r0) is a minimum among all labelsof edges not already on a path. Therefore, l(e1, r0) ≤ l(em, rm).

Notice that in the assignment procedure described earlier, we do not assign customere1 to route r0 except possibly on a dedicated route. Also, we do not assign customerem to route rm except when l(em, rm) > 1 (case 1) or possibly on a dedicated route(case 2). Thus, by Property 1, we do not assign any customer to a route in R1 if it hasrank 1 on its corresponding route in R. Also, since the paths in P are edge-disjoint

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and since we assign edges to incident vertices in G, by Property 2, we might assignthe customer with rank 2 j or the customer with rank 2 j +1 to a route in R1, but neverboth. Therefore, no route in R1 violates the vehicle capacity constraint. By visitingthe customers assigned to the routes in R1 in the same order in which they are visitedon their corresponding routes in R, by Observation 1, we have d(R1) ≤ d(R).

When l(em, rm) = 1 (case 2), by Property 3, l(e1, r0) = 1, as well. Thus, in theassignment procedure above, we assign to a dedicated route in R2 a customer that hasrank 1 on a route in R. Since we always chose this customer to be on the smaller of twodistinct routes, by the triangle inequality and Observation 2, we have d(R2) ≤ 1

2 d(R).

Thus, d(R) ≤ 32 d(R), and the theorem is proved. ��

In Algorithm 3 in the Appendix we provide the pseudo-code for the entire con-struction algorithm. We input the SDVRP-MDA solution S and output the VRPsolution S.

4 Conclusions

In this paper, we presented a worst-case analysis for the SDVRP-MDA. We showedthat for the SDVRP-MDA with a minimum delivery fraction p, when 0 < p < 1

2 , wehave a savings of up to 50 % by allowing split deliveries. This bound is tight, and themaximum savings for the SDVRP-MDA is the same as for the traditional SDVRP. Inthe case where p = 1

2 , a maximum savings of 33 13 % is possible, and this bound is

tight.

Acknowledgements We thank Claudia Archetti for her input on an early version of this paper.

5 Appendix

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References

1. Archetti, C., Savelsbergh, M., Speranza, M.: Worst-case analysis for split delivery vehicle routingproblems. Transport. Sci. 40, 226–234 (2006)

2. Golden, B., Raghavan, S., Wasil, E.: The Vehicle Routing Problem: Latest Advances and NewChallenges. Springer, New York (2008)

3. Gulczynski, D., Golden, B., Wasil, E.: The split delivery vehicle routing problem with minimum deliveryamounts. Transport. Res. Part E 46, 612–626 (2010)

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a worst-case analysis for the split delivery vehicle routing problem with minimum delivery amounts

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