An Improved Upper Bound for the Ring Loading Problem
aa r X i v : . [ c s . D M ] A p r An Improved Upper Bound for the Ring Loading Problem
Karl DäubelInstitut für Mathematik, Technische Universität Berlin, Germany [email protected]
Abstract
The
Ring Loading Problem emerged in the 1990s to model an important special case oftelecommunication networks (SONET rings) which gained attention from practitioners andtheorists alike. Given an undirected cycle on n nodes together with non-negative demandsbetween any pair of nodes, the Ring Loading Problem asks for an unsplittable routing of thedemands such that the maximum cumulated demand on any edge is minimized. Let L bethe value of such a solution. In the relaxed version of the problem, each demand can besplit into two parts where the first part is routed clockwise while the second part is routedcounter-clockwise. Denote with L ∗ the maximum load of a minimum split routing solution.In a landmark paper, Schrijver, Seymour and Winkler [SSW98] showed that L ≤ L ∗ + 1 . D ,where D is the maximum demand value. They also found (implicitly) an instance of the Ring Loading Problem with L = L ∗ + 1 . D . Recently, Skutella [Sku16] improved thesebounds by showing that L ≤ L ∗ + D , and there exists an instance with L = L ∗ + 1 . D .We contribute to this line of research by showing that L ≤ L ∗ + 1 . D . We also take a firststep towards lower and upper bounds for small instances. Given an undirected cycle on n nodes together with non-negative demands between any pair ofnodes, the Ring Loading Problem asks for an unsplittable routing of the demands such that themaximum cumulated demand on any edge is minimal. Formally, we are given a graph G = ( V, E ) with nodes V = [ n ] := { , . . . , n } , edges { i, i + 1 } for each i ∈ V , where we assume throughoutthe paper that { n, n + 1 } := { n, } , and demands for each pair of nodes i < j of value d i,j ≥ .By a slight abuse of notation, we refer to both the demand from i to j and its value as d i,j . Anunsplittable solution decides for each demand whether it should be routed clockwise, sending allof its value along the path { i, i + 1 , . . . , j } , or counter-clockwise, sending all of its value along thepath { i, i − , . . . , , n, . . . , j } . The load of an edge, for a given solution, is the sum of all demandvalues that are routed on paths that use the edge. We call the maximum load on any edge ofthe ring the load of the solution. The problem is to find an unsplittable routing that minimizesthe load. We denote with L the load of such an optimal unsplittable solution. See Fig. 1 for anexample.The problem was introduced by Cosares and Saniee [CS94] to mathematically model surviv-able networks with respect to the emerging standard of synchronous optical networks (SONET).The underlying structure to this technology, the SONET ring, is a set of network nodes and linksthat are arranged in a cycle. In this way, even in the event of a link failure, most of the trafficcould be recovered. See [BC02, Gor02, VPD04] for further resources on technical details. Tothe best of our knowledge, Cosares and Saniee [CS94] also established the name Ring LoadingProblem . They further showed via a reduction of the
Partition Problem that the problem isNP-hard and provided an algorithm that returns an unsplittable solution with load at most L .Using a result from Schrijver et al. [SSW98], Khanna [Kha97] showed that there exists a PTAS,i.e. a class of poly-time algorithms that return a solution with load at most (1 + ε ) L , for each1 2 34567 8 1 2 34567 8Figure 1: An instance of the Ring Loading Problem on nodes and non-zero demands with d , = d , = d , = d , = 1 (left) together with an optimum unsplittable routing of load (right).fixed ε > . If all non-zero demands have the same value, Frank [Fra85] showed that the RingLoading Problem can be solved in polynomial time.Although a PTAS for the
Ring Loading Problem exists, there remain unsolved problems thatconnect unsplittable solutions to a relaxed version of the
Ring Loading Problem . To this end,consider the
Ring Loading Problem where demands are allowed to be routed splittably, i.e. ademand can be routed partly clockwise while the remaining part is routed counter-clockwise.The definition of the load of an edge and the load of a solution generalize naturally to therelaxed version. We denote with L ∗ the optimum load of a split solution. The relaxed version ofthe Ring Loading Problem has a linear programming formulation [CS94] and can thus be solvedin polynomial time. Further effort was put into finding more efficient algorithms (see [VSKW96,MKT97, SSW98, DLM99, MK04, Wan05]). It was also shown in [MKT97, SSW98, DLM99] that L ≤ L ∗ , and this bound is tight ([MKT97, SSW98]).In a landmark paper, Schrijver, Seymour and Winkler [SSW98] proved in this context that L ≤ L ∗ + D , where we denote with D := max i The following theorem is the main contribution of this work. Theorem 1. Any split routing solution to the Ring Loading Problem can be turned into anunsplittable routing while increasing the load on any edge by at most D . In particular, we have L ≤ L ∗ + D . In order to prove the theorem, we first define a general framework that unifies structuralresults of split routings introduced by Skutella [Sku16]. We then apply this framework in a newway to obtain better upper bounds. This result is the first progress towards closing the remainingadditive gap since Skutella [Sku16].As all previous lower bound examples are of relative small size, it is interesting to settle thesecases conclusively. We take a step into this direction by showing upper and lower bounds for2mall instances. The upper bounds are deduced from a mixed integer linear program that verifiesfor a given instance size that no worse examples can exist. Although the lower bounds also followfrom this formulation, we provide further examples to enrich the view on instances where thedifference L − L ∗ is large with respect to D . In fact, we give an infinite family of instances with L > L ∗ + D .A summary of previous results on lower and upper bounds together with new advancementsis shown in Fig. 2 on the right vertical line, while on the left results are given with respect to δ ∈ (cid:2) , (cid:3) that parametrizes instances of the Ring Loading Problem and indicates whether ademand of medium size exists.Just as Schrijver et al. [SSW98] and Skutella [Sku16] before, we mention a nice combinatorialimplication of our result. Schrijver et al. [SSW98] define β to be the infimum of all reals α suchthat the following combinatorial statement holds: For all positive integers m and nonnegativereals u , . . . , u m and v , . . . , v m with u i + v i ≤ , there exist z , . . . , z m such that for every k, z k ∈ { v k , − u k } and (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k X i =1 z i − m X i = k +1 z i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ α. Schrijver et al. [SSW98] prove that β ∈ (cid:2) , (cid:3) . Skutella [Sku16] reduces the size of theinterval to β ∈ (cid:2) , (cid:3) . As a result of our work, we obtain β ∈ (cid:2) , (cid:3) . Further Related Work In the Ring Loading Problem with integer demand splitting, eachdemand is allowed to be split into two integer parts which are routed in different directions alongthe ring. The objective is to find an integer split routing that minimizes the load. Let L ′ bethe load of an optimal integer split routing solution. Lee et al. [LC97] showed an algorithmthat returns an integer split routing solution with load at most L ′ + 1 . Schrijver et al. [SSW98]found an optimal solution in pseudo-polynomial time. Vachani et al. [VSKW96] provided an O (cid:0) n (cid:1) algorithm. In [Myu01] Myung presented an algorithm with runtime O ( nk ) where k isthe number of non-zero demands. Wang [Wan05] proved the existence of an O ( k + t S ) algorithmwhere t S is the time for sorting k nodes.More recently, the weighted Ring Loading Problem was introduced where each edge has aweight associated with it, and the weighted load of an edge is the product of its weight and thesmallest integer greater or equal than its load. In the case where demand splitting is allowed,Nong et al. [NYL09] gave an O (cid:0) n k (cid:1) algorithm. If integer demand splitting is allowed, theauthors present a pseudo-polynomial time algorithm. Later, Nong et al. [NCN10] present an O (cid:0) n k (cid:1) algorithm. If the demands have to be send unsplittably, Nong et al. [NYL09] prove theexistence of a PTAS.In a broader context, the Ring Loading Problem is a special case of unsplittable multicom-modity flows. We mention the case of single source unsplittable flows, as similarities betweentheorems and conjectures for these problems exist (see [DGG99, Sku02, GC07, MSS07]). Wealso refer to the survey of Shepherd [She09]. Outline In Section 2, we introduce some notation and provide useful results from Schrijver etal. [SSW98] and Skutella [Sku16] that we need. We then continue in Section 3 with the proofof Theorem 1. In Section 4, we turn our attention to upper and lower bounds for small instances.We wrap everything up with our conclusions in Section 5. In this section, we introduce further notation and mention results already presented in [SSW98,Sku16]. We start with a preprocessing step to reduce the size and complexity of an instance tothe Ring Loading Problem . 3 . . . . . D . D . D . D . D . D upper boundslower bounds [Tab. 2][SSW98] [SSW98] . D . D [Sku16][Sku16] [Sku16] D . D [Lem. 5] . D Figure 2: Summary of known results dependent on δ (left) and independent of δ (right). Thecurrently best bounds are due to Theorem 1 together with the lower bound in [Sku16].Two demands d i,j and d k,l are parallel if there exists a path from i to j and a path from k to l that are edge-disjoint, otherwise they are crossing . Note that the demands d i,j and d i,k areparallel.As Theorem 1 only argues about the load increase on all edges for split routing solutions, wecan ignore and delete all demands that are routed unsplittably. The following observation showsthat we can assume that there are not too many remaining demands. Observation 1 ([SSW98]) . Given a split routing of two parallel demands d and d . The routingcan be altered such that at most one demand is routed splittably, without increasing the load onany edge.Proof. By the definition of parallel demands, we know that there are paths P i , i ∈ [2] , connectingthe nodes of demand d i , such that P ∩ P = ∅ . Let Q i = E \ P i , i ∈ [2] . Note that P is completelycontained in Q and vice versa. We denote with x i the amount of flow that demand d i is routingalong Q i , i ∈ [2] . If we now decrease the demand value routed along Q and Q by min { x , x } and increase the demand value routed along P and P by the same amount, the load on theedges of P and P remain unchanged while the load on Q ∩ Q decreases. Afterwards either d or d is routed unsplittably.If we apply Observation 1 and delete afterwards all demands that are routed unsplittably, wecan concentrate on instances with pairwise crossing demands, implying in particular that everynode is end point of at most one demand. If a node is not the end point of a demand, the loadon its adjacent edges have the same value, allowing us to delete the node and merge the edges.After this process we are left with a ring on n = 2 m nodes, demands d i := d i,i + m > for i ∈ [ m ] and a split routing. We denote for all i ∈ [ m ] with u i > the amount of flow from demand d i routed clockwise and likewise with v i > the remainder of flow routed counter-clockwise. Notethat u i + v i = d i , i ∈ [ m ] . From now on we refer to an instance with this structure as split routingsolution . An example is given in Fig. 3 on the left.The following definition describes for a given δ ∈ (cid:2) , (cid:3) all split routing solutions withoutdemands of medium size (with respect to δ ) and ensures the existence of a demand on theboundary to medium demands. Formally we call these split routing solutions δ -instances: Definition 1. Let δ ∈ (cid:2) , (cid:3) . We call a split routing solution a δ -instance , if for all i ∈ arg min j ∈ [ m ] (cid:0)(cid:12)(cid:12) D − d j (cid:12)(cid:12)(cid:1) holds d i ∈ { δD, (1 − δ ) D } . 22 12 21311 1 1 2 3456789 10 + + − − + − − + + − m = 5 pairwise crossing demands with u =(2 , , , , and v = (2 , , , , together with the load on each edge (left). The correspondingunsplittable solution for z = ( v , − u , − u , v , v ) = (2 , − , − , , together with load changeson every edge (right). The additive performance of z is .A -instance for example has a demand of value D , whereas a -instance only has demandsof value D . An important property of δ -instances is that d i ∈ [0 , δD ] ∪ [(1 − δ ) D, D ] for all i ∈ [ m ] .Any unsplittable solution has to decide for each demand d i whether u i units of flow arererouted to use the counter-clockwise direction, or whether v i units of flow are rerouted to usethe clockwise direction. We encode this decision using z = ( z , . . . , z m ) , with z i ∈ { v i , − u i } forall i ∈ [ m ] , where z i = v i means that we send the demand completely in clockwise direction,whereas z i = − u i means that we completely send the demand in counter-clockwise direction.In either case the z i values model exactly the increase of load on the clockwise edges from i to i + m , and the decrease of load on the counter-clockwise edges. For k ∈ [ m ] the load on an edge { k, k + 1 } changes by k X i =1 z i − m X i = k +1 z i , while the load on the opposite edge { k + m, k + m + 1 } changes by the negative amount. Themaximum increase of load on any edge is therefore max k ∈ [ m ] (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k X i =1 z i − m X i = k +1 z i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . As described by Skutella [Sku16], we refer to this quantity as the additive performance of z .In Fig. 3 an example of the load change and the additive performance is given.Let x ∈ R be fixed, we define p z ( k ) := x + P ki =1 z i , for k ∈ [ m ] . We refer to p z as a pattern starting at x = p z (0) and ending at y = p z ( m ) . We denote with a := min k ∈ [ m ] p z ( k ) and b := max k ∈ [ m ] p z ( k ) the minimum and maximum of pattern p z , respectively. We refer to [ a, b ] asstrip and say that the pattern p z lives on the strip [ a, b ] of width b − a . As p z ( k ) − p z ( k − 1) = z i ,when we refer to a pattern p z we also refer to the corresponding unsplittable solution. As thechoice of x might vary, multiple patterns correspond to a single unsplittable solution. A patterncan be visualized as seen in Fig. 4. Observation 2 ([Sku16]) . Given an unsplittable solution z with corresponding pattern p z withstart point x , end point y living on a strip of [ a, b ] , then the additive performance of pattern p z is max k ∈ [ m ] (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k X i =1 z i − m X i = k +1 z i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = max { b − x − y, x + y − a } . (1)5 = 1 x = 2 y = 3 k b = 4 p z ( k ) Figure 4: An example of a pattern p z that corresponds to the split routing given in Fig. 3with z = (2 , − , − , , and start point x = 2 , end point y = 3 , minimum value a = 1 and maximum value b = 4 . The additive performance of the pattern due to Observation 2 is max { b − x − y, x + y − a } = 3 . Proof. By the definition of p z ( k ) we have k X i =1 z i − m X i = k +1 z i = 2 p z ( k ) − x − y. The claim then follows from the fact that the maximum in | p z ( k ) − x − y | is obtained at anindex k where p z ( k ) is either maximum or minimum. Observation 3 ([Sku16]) . Let ε > . Given an unsplittable solution z with corresponding pattern p z with start point x and end point y living on a strip of [ a, b ] , the additive performance of pattern p z is at most b − a + ε if and only if the pattern starts at x and ends at y ∈ [ y opt − ε, y opt + ε ] ∩ [ a, b ] with y opt := a + b − x .Proof. The claim follows from the definitions together with Observation 2.Given a strip of width D , say [0 , D ] . Let x ∈ [0 , D ] , we denote throughout with ¯ x := D − x the reflection of x across D . We can construct a pattern with start point x living on a strip [ a, b ] ⊆ [0 , D ] by applying iteratively the following observation. Observation 4. Let d = u + v with u, v ≥ . If I is an interval of size at least d and x ∈ I ,then x + v ∈ I or x − u ∈ I (or both). Formally, we construct the pattern p z by setting p z (0) = x for some x ∈ [0 , D ] and choose z k ∈ {− u i , v i } iteratively such that p z ( k ) = p z ( k − 1) + z i ∈ [0 , D ] for all k = 1 , . . . , m , whichalways works by Observation 4. If this decision is not unique, we set z k such that (cid:12)(cid:12) D − p z ( k ) (cid:12)(cid:12) is minimal, i.e. p z ( k ) is as close as possible to the middle D of the interval [0 , D ] . Remainingties are broken arbitrarily. A pattern that is constructed with respect to this procedure is calleda forward greedy pattern . For technical reasons, we call a forward greedy pattern p z proper if itsstart point is far enough away from the boundary, i.e. x ∈ (cid:2) δ D, (cid:0) − δ (cid:1) D (cid:3) . This requirement isused in Lemma 2.We obtain a backward greedy pattern p z by applying this procedure backwards. We define p z ( m ) = y , for some y ∈ [0 , D ] , and iteratively choose p z ( k − 1) = p z ( k ) − z k , for all k = m, . . . , ,such that (cid:12)(cid:12) p z ( k − − D (cid:12)(cid:12) is minimal. We call a backward greedy pattern proper if its end pointis far enough away from the boundary, i.e. y ∈ (cid:2) δ D, (cid:0) − δ (cid:1) D (cid:3) .A pattern is called a (proper) greedy pattern if it is either a (proper) forward greedy patternor a (proper) backward greedy pattern.Using a forward greedy pattern starting at D together with Observation 3, Schrijver etal. [SSW98] showed that any split routing solution to the Ring Loading Problem can be turnedinto an unsplittable solution while increasing the load on any edge by at most D .Although the following structural properties of (greedy) patterns are crucial for our results,we refer the reader for complete proofs to [Sku16].6 efinition 2 ([Sku16]) . Let ε ≥ . Two patterns p z and p z ′ are said to be ε -close if | p z ( k ) − p z ′ ( k ) | ≤ ε for some k ∈ { , , . . . , m } . The following lemma combines two ε -close patterns to a single pattern while preserving crucialproperties. Lemma 1 ([Sku16]) . Consider a fixed split routing solution. Let p z ′ be a pattern with start point x ′ living on strip [ a ′ , b ′ ] , and p z ′′ a pattern with end point y ′′ living on strip [ a ′′ , b ′′ ] . If the twopatterns are ε -close for some ε ≥ , then there is a pattern p z living on a sub-strip of (cid:20) min (cid:8) a ′ , a ′′ (cid:9) − ε, max (cid:8) b ′ , b ′′ (cid:9) + 12 ε (cid:21) with start point x and end point y such that x + y = x ′ + y ′′ . This lemma describes situations where δ D -close patterns exist. Lemma 2 ([Sku16]) . Consider three proper greedy patterns p z a , p z b , p z c , all three living on sub-strips of [0 , D ] . If the sorting of the patterns by their end points is not a cyclic permutation ofthe sorting of their start points, then (at least) two of the three patterns are δD -close. At the end of the section reconsider Fig. 2. Both previous and new results are shown withrespect to δ on the left and the consequences for all instances independent of δ on the right. In this section we prove Theorem 1. We start by defining a general framework that allows us touse Lemmas 1 and 2 in a very unified manner. The following definition is at the heart of thisframework (see Fig. 5). Definition 3. Given a greedy pattern p z a living on a sub-strip of [0 , D ] with start point x a andend point y a , we call a forward greedy pattern p z b induced by p z a , if it lives on a sub-strip of [0 , D ] with start point x b := ¯ y a + x a . Likewise, we call a backward greedy pattern p z c induced by p z a , if it lives on a sub-strip of [0 , D ] with end point y c := ¯ x a + y a . If a greedy pattern p z a and its induced patterns are proper, the following lemma ensuresthe existence of a pattern with an additive performance that only depends on the start and endpoints of p z a together with δ . It is therefore possible to pick a single pattern, check if its inducedpatterns are proper, and obtain a strong bound on the additive performance. Lemma 3. For a δ -instance with δ ∈ (cid:2) , (cid:3) , let p z a be a greedy pattern living on a sub-strip of [0 , D ] with start point x a and end point y a . Denote with p z b a forward greedy pattern induced by p z a and with p z c a backward greedy pattern induced by p z a . If all three greedy patterns are proper,then there exists a pattern with additive performance at most max (cid:26) D − 13 ( x a + y a ) + δ D, D + 13 ( x a + y a ) + δ D (cid:27) . Proof. We first show that the sorting of the start points is not a cyclic permutation of the sortingof the end points. This allows us to use Lemma 2 that guarantees the existence of two patternsthat are δ D -close. We then conclude the lemma by showing that if any two of the three patternsare δ D -close, that there exists a pattern with the required additive performance. In Fig. 5 is anillustration of the procedure.By the definition of p z b as forward greedy pattern induced by p z a and p z c as backward greedypattern induced by p z a , we know that x b = ¯ y a + x a and y c = ¯ x a + y a . The definitions are7 · · · m − m Dx a x b ¯ y c ¯ y a ¯ x a ¯ x b y c y a x c x c y b y b Figure 5: An illustration of a greedy pattern p z a with induced forward greedy pattern p z b andinduced backward greedy pattern p z c as defined in Definition 3.such that the interval between x a and ¯ y a is divided into three equal parts by the points x b and ¯ y c . A straightforward computation shows that ¯ y c = x a + x b , ¯ x b = y a + y c , (2)i.e. the optimal start point for pattern p z c is in the middle between x a and x b , and the optimalend point of p x b is in the middle between y a and y c . Because | x a − ¯ y a | = | ¯ x a − y a | , and thesymmetric definitions of x b and y c , it also holds that | x a − x b | = | y a − y c | .For the sake of brevity we define ε := max (cid:8) D − ( x a + y a ) + δ D, ( x a + y a ) − D + δ D (cid:9) .In fact, we want to show that there exists a pattern with additive performance at most D + ε .We now show that | x a − x b | ≤ ε , which also implies that | y a − y c | ≤ ε . By definition of x b = ¯ y a + x a = D − y a + x a , we can conclude that | x a − x b | = max { x b − x a , x a − x b } = max (cid:26) D − 23 ( x a + y a ) , 23 ( x a + y a ) − D (cid:27) ≤ (cid:26) D − 13 ( x a + y a ) + δ D, 13 ( x a + y a ) − D + δ D (cid:27) = 2 ε, (3)where the inequality follows from the fact that δ ≥ . With Eqs. (2) and (3), it follows that | ¯ y c − x b | = | ¯ y c − x a | = | ¯ x b − y a | = | ¯ x b − y c | ≤ ε. (4)If for the start point of p z c holds x c ∈ [¯ y c − ε, ¯ y c + ε ] , we know by Observation 3 that theadditive performance of p z c is at most D + ε , from which the lemma follows. We can thusassume that x c ∈ [0 , ¯ y c − ε ] ∪ [¯ y c + ε, D ] . Using Eq. (4), we can therefore conclude that either x c ≤ ¯ y c − ε ≤ min { x a , x b } or max { x a , x b } ≤ ¯ y c + ε ≤ x c .Equivalently, if for the end point of p z b holds y b ∈ [¯ x b − ε, ¯ x b + ε ] , we know by Observation 3that the additive performance of p z b is at most D + ε , from which the lemma follows. We canthus assume that y b ∈ [0 , ¯ x b − ε ] ∪ [¯ x b + ε, D ] . Using Eq. (4), we can therefore conclude thateither y b ≤ ¯ x b − ε ≤ min { y a , y c } or max { y a , y c } ≤ ¯ x b + ε ≤ y b .Assume first that ¯ y a ≤ x a (as shown in Fig. 5), then ¯ y a ≤ x b ≤ ¯ y c ≤ x a and ¯ x a ≤ y c ≤ ¯ x b ≤ y a ,which implies that either x c ≤ x b ≤ x a or x b ≤ x a ≤ x c and that either y b ≤ y c ≤ y a or y c ≤ y a ≤ y b . In either case, the sorting of the patterns by their start points is not a cyclicpermutation of the patterns by their end points.Assume now that x a ≤ ¯ y a , then x a ≤ ¯ y c ≤ x b ≤ ¯ y a and y a ≤ ¯ x b ≤ y c ≤ ¯ x a , which impliesthat either x c ≤ x a ≤ x b or x a ≤ x b ≤ x c and that either y b ≤ y a ≤ y c or y a ≤ y c ≤ y b . In eithercase, the sorting of the patterns by their start points is not a cyclic permutation of the patternsby their end points. 8s p z a , p z b and p z c are proper greedy patterns, we can apply Lemma 2, ensuring the existenceof two patterns that are δ D -close. We conclude the proof by showing that the closeness of anytwo patterns guarantees the existence of a pattern with the claimed additive performance.i) Assume that p z a and p z b are δ D -close. Then Lemma 1 assures the existence of a patternwith start point x and end point y such that x + y = x b + y a = D + ( x a + y a ) on asub-strip of (cid:2) − δ D, D + δ D (cid:3) . Using Observation 2, a straightforward calculation shows thatthis pattern has additive performance at most max (cid:26) D − 13 ( x a + y a ) + δ D, D + 13 ( x a + y a ) + δ D (cid:27) . ii) Assume that p z a and p z c are δD -close. Then Lemma 1 assures the existence of a patternwith start point x and end point y such that x + y = x a + y c = D + ( x a + y a ) on asub-strip of (cid:2) − δ D, D + δ D (cid:3) . Using Observation 2, a straightforward calculation shows thatthis pattern has additive performance at most max (cid:26) D − 13 ( x a + y a ) + δ D, D + 13 ( x a + y a ) + δ D (cid:27) . iii) Assume that p z b and p z c are δD -close. Then Lemma 1 assures that there exists a patternwith start point x and end point y such that x + y = x b + y c = D − ( x a + y a ) on asub-strip of (cid:2) − δ D, D + δ D (cid:3) . Using Observation 2, a straightforward calculation shows thatthis pattern has additive performance at most max (cid:26) D + 13 ( x a + y a ) + δ D, D − 13 ( x a + y a ) + δ D (cid:27) . In either case, the lemma follows.If both the start point and the end point of a greedy pattern p z a are far enough away fromthe boundary, the next lemma ensures that its induced patterns are proper. Note that this is astronger requirement on p z a than being a proper greedy pattern. Lemma 4. For a δ -instance with δ ∈ (cid:2) , (cid:3) , let p z a be a greedy pattern living on a sub-strip of [0 , D ] with start point x a and end point y a . Denote with p z b a forward greedy pattern induced by p z a and with p z c a backward greedy pattern induced by p z a . If x a , y a ∈ (cid:2) δ D, (cid:0) − δ (cid:1) D (cid:3) , then p z a , p z b and p z c are proper.Proof. By definition, p z a is proper. For x ∈ [0 , D ] holds that ¯ x ∈ (cid:2) δ D, (cid:0) − δ (cid:1) D (cid:3) if and onlyif x ∈ (cid:2) δ D, (cid:0) − δ (cid:1) D (cid:3) . By assumption, we therefore also know that ¯ x a and ¯ y a are far enoughaway from the boundary. By the definition of induced patterns, we know that x b = ¯ y a + x a ,which implies in particular that min { x a , ¯ y a } ≤ x b ≤ max { x a , ¯ y a } . The start point x b of theinduced forward greedy pattern is consequently far enough away from the boundary. Using thesame argumentation for the definition of y c = ¯ x a + y a , the lemma follows.A crucial part of the proof of Theorem 1, and the main contribution of this work is thefollowing auxiliary lemma. Lemma 5. For a δ -instance with δ ∈ (cid:2) , (cid:3) there exists a pattern with additive performance atmost (cid:0) + δ (cid:1) D . m − m · · · x a D D y a := y a ( D + d m ) − v m y a Figure 6: An example of the construction step in Lemma 5. If x a ≤ D , we extend the patternwith y a = ( D + d m ) . Proof. We start the proof by modifying the instance such that the special demand of value either δD or (1 − δ ) D is the last demand. These two cases will be treated separately. In either case,we then use the nice structure of the newly created instance to find a greedy pattern that canbe used with Lemma 3.Let d i be the demand that minimizes (cid:12)(cid:12) D − d i (cid:12)(cid:12) over all i ∈ [ m ] . By the definition of a δ -instance, we know that d i ∈ { δD, (1 − δ ) D } .We now rotate the instance such that the specially chosen demand d i has index m , andis thus the last demand of the instance. By a slight abuse of notation we will refer to thisnewly created instance again as instance. Recall that now the demand d m has the property that d m ∈ { δD, (1 − δ ) D } .The following procedure is similar to the one described by Skutella [Sku16] when dealing withdemands of medium size. An example is depicted in Figure 6. We first delete the last demand m to obtain a smaller instance. We define a backward greedy pattern ending at ( D + d m ) − v m and starting at some x a ∈ [0 , D ] . This backward greedy pattern can be extended in two possibleways to create a pattern that includes demand m , once with end point y a := ( D − d m ) andonce with end point y a := ( D + d m ) . A crucial observation is that both possible extensionsproduce a valid backward greedy pattern for the original instance. Depending on the particularstart point x a , we choose in which way the pattern will be extended: If x a ≤ D , we extend thepattern with end point y a , otherwise we extend the pattern with y a . For the rest of the proof,we may assume that x a is at most D and the pattern is therefore extended with end point y a := y a . This assumption can be made, as the following construction is highly symmetric withrespect to y a and y a , in fact, all arguments remain valid if we change start and end points ofsubsequent patterns by reflecting their value around D . Let p z a denote the resulting backwardgreedy pattern starting at x a and ending at y a .We consider two cases, first that d m = (1 − δ ) D and second that d m = δD . For the sake ofbrevity, we define ε := D + δ D . In fact, we want to find a pattern with additive performanceat most D + ε . Case a) If d m = (1 − δ ) D , we can rewrite y a = ( D + d m ) = D − δ D . The lemma followsimmediately from Observation 3 if x a ∈ [¯ y a − ε, ¯ y a + ε ] . We can therefore assume that x a fallseither into the interval (cid:2) , δ D − D (cid:3) or into the interval (cid:2) D + δD, D (cid:3) . Recall the assumptionthat x a is at most D . It is easy to see that δ D − D is negative for all δ ∈ (cid:2) , (cid:3) . It followsthat x a ∈ (cid:2) D + δD, D (cid:3) . Note that this interval is also empty for all δ > , and the lemma istrivially correct. In fact, this is exactly the argumentation used by Skutella [Sku16] in his proofof Lemma 6.As y a = D − δ D ∈ (cid:2) δ D, (cid:0) − δ (cid:1) D (cid:3) , the backward greedy pattern p z a is proper. Because D ≥ x a ≥ D + δD ≥ δ D , it furthermore holds that x a is far enough away from theboundary. We can thus apply Lemma 4 together with Lemma 3 and the fact that x a + y a ∈ D + δ D, D − δ D (cid:3) to obtain a pattern with additive performance at most max (cid:26) D + 718 δD, D + δ D (cid:27) = (cid:18) 76 + δ (cid:19) D. Case b) If d m = δD , we can rewrite y a = ( D + d m ) = D + δ D . The lemma followsimmediately from Observation 3 if x a ∈ [¯ y a − ε, ¯ y a + ε ] . We can therefore assume that x a fallseither into the interval (cid:2) , D − δD (cid:3) or into the interval (cid:2) D − δ D, D (cid:3) . Recall the assumptionthat x a is at most D . It is easy to see that D − δ D ≥ D for all δ ∈ (cid:2) , (cid:3) . It follows that x a ∈ (cid:2) , D − δD (cid:3) . Note that this interval is also empty for all δ > , and the lemma istrivially correct. We need this assumption, when arguing that we can apply Lemma 3.As y a = D + δ D ∈ (cid:2) δ D, (cid:0) − δ (cid:1) D (cid:3) , the backward greedy pattern p z a is proper. As x a mightbe zero, we cannot apply Lemma 4. We therefore have to argue that the induced patterns areproper. Let p z b be a forward greedy pattern induced by p z a and p z c be a backward greedy patterninduced by p z a . By definition, we have x b = ¯ y a + x a and y c = ¯ x a + y a . By substituting thedefinitions and bounds of y a and x a , we obtain x b = 13 D − δ D + 13 x a ≥ D − δ D ≥ δ D. The start point x b is therefore far enough away from the boundary and the pattern p z b is thusproper. We similarly obtain y c = 56 D + δ D − x a ≤ D + δ D ≤ D − δ D, for all δ ∈ (cid:2) , (cid:3) . As we assumed that δ ≤ , the backward greedy pattern p z c induced by p z a isproper. We can thus apply Lemma 3 together with the fact that x a + y a ∈ (cid:2) D + δ D, D − δ D (cid:3) to obtain a pattern with additive performance at most max (cid:26) D + δ D, D + 718 δD (cid:27) = (cid:18) 76 + δ (cid:19) D. In either case, the lemma follows.An easy consequence of Lemma 5 is that there exists for any split routing solution a patternwith additive performance at most D , which already improves upon the best known previousresult of D from Skutella [Sku16]. However, when combined with Skutellas [Sku16] result oninstances with medium demands (see Lemma 6), we obtain our main Theorem 1. Lemma 6 ([Sku16]) . For any δ -instance with δ ∈ (cid:2) , (cid:3) there exists a pattern with additiveperformance at most (cid:0) − δ (cid:1) D .Proof of Theorem 1. Let δ ∈ (cid:2) , (cid:3) be such that the given split routing solution is a δ -instance.If δ ≥ , the theorem follows from Lemma 6, as (cid:0) − δ (cid:1) D ≤ . D . Otherwise, the theoremfollows from Lemma 5, as (cid:0) + δ (cid:1) D ≤ . D . In this section, we show lower and upper bounds for small instances of the Ring Loading Problem .This is of interest, as strong lower bound examples seem to exist for fairly small values of m (see Section 4.2). Therefore, dealing with these cases conclusively might lead to a deeperunderstanding on the correct bounds.The main result of this section are bounds on the maximal load increase while turning a splitrouting solution into an unsplittable solution for instances of the Ring Loading Problem whereat most demands are routed splittably. 11 heorem 2. Let m ≥ be an integer. Any split routing solution to the Ring Loading Problem with m split demands can be turned into an unsplittable solution without increasing the load onany edge by more than • (1 + ε ) D , if m ≤ and • (cid:0) + ε (cid:1) D , if m = 7 ,for ε ≤ × − . Furthermore, there are instances of the Ring Loading Problem with m pairwisecrossing demands with L = L ∗ + D , for m ≤ , and L = L ∗ + D , for m = 7 . Note that the dependency on ε is unavoidable, as our technique for the upper bounds de-pends on solutions to large mixed integer linear programs that rely on floating point arithmetic.However, the theorem implies that L ≤ L ∗ + αD for instances of the Ring Loading Problem wherean optimal split routing solution exists such that at most m demands are routed splittably, with α = 1 + ε if m ≤ and α = + ε if m = 7 .The remainder of this section is structured as follows. In Section 4.1, we introduce the mixedinteger linear program that provides us with the upper bounds for m ≤ . In Section 4.2, weprovide a more detailed view on lower bounds, some of which provide the matching lower boundsin Theorem 2. On the way, we show a lemma that turns any split routing solution into aninstance of the Ring Loading Problem while retaining the load increase. We conclude the sectionby proving Theorem 2 in Section 4.3. We introduce a mixed integer linear program (MILP) that outputs for a given integer m a splitrouting solution with m demands that cannot be turned into an unsplittable routing withoutincreasing the load on some edge by at least αD , for α ≥ as large as possible. This gives usthe claimed upper bounds in Theorem 2.We first introduce our notation. Let P be the set of all patterns, i.e. z ∈ P with z =( z , . . . , z m ) corresponds to a particular choice of values z i ∈ { v i , − u i } for all i ∈ [ m ] . Notethat |P| = 2 m . Without loss of generality we assume that D = 1 , as we can otherwise scale theinstance. We furthermore assume that every pattern z ∈ P starts at x p = 0 . We use the followingvariables: For all i ∈ [ m ] , we denote with u i and v i the continuous variables that correspondto a particular split routing. For all patterns z ∈ P we refer to the maximum and minimumvalues obtained by the pattern with a z and b z , respectively. For this purpose we furthermorehave binary variables w min z,i and w max z,i that indicate at which position i ∈ { , , . . . , m } of pattern z ∈ P the minimal and maximal values are. In fact, w min z,i = 1 if i ∈ arg min j ∈{ ,...,m } P jk =1 z j (and equivalently for w max z,i ). The variables y z correspond to the end point and c z to the additiveperformance of pattern z ∈ P . In order to decide where the maximum value of the additiveperformance is obtained (see Observation 2) we use a binary variable w z .Our MILP formulation is as follows: max E s.t. E ≤ c z , ∀ z ∈ P (5a) u i + v i ≤ , ∀ i ∈ [ m ] (5b) P mi =0 w min z,i ≥ , ∀ z ∈ P (5c) P mi =0 w max z,i ≥ , ∀ z ∈ P (5d) a z ≤ P ij =1 z j , ∀ z ∈ P , ∀ i ∈ { , . . . , m } (5e) a z + W · (cid:0) − w min z,i (cid:1) ≥ P ij =1 z j , ∀ z ∈ P , ∀ i ∈ { , . . . , m } (5f) b z ≥ P ij =1 z j , ∀ z ∈ P , ∀ i ∈ { , . . . , m } (5g)12 z − W · (cid:0) − w max z,i (cid:1) ≤ P ij =1 z j , ∀ z ∈ P , ∀ i ∈ { , . . . , m } (5h) y z = P mj =1 z j , ∀ z ∈ P (5i) c z ≥ b z − y z , ∀ z ∈ P (5j) c z ≥ y z − a z , ∀ z ∈ P (5k) c z − W w z ≤ b z − y z , ∀ z ∈ P (5l) c z − W (1 − w z ) ≤ y z − a z , ∀ z ∈ P (5m) u i , v i ≥ , ∀ i ∈ { , . . . , m } (5n) w z , w min z,i , w max z,i ∈ { , } , ∀ z ∈ P , ∀ i ∈ { , . . . , m } (5o)Note that the z i values are no variables but place holders for either − u i or v i variables (dependingon the particular pattern z ∈ P ). The constants W and W should be large enough suchthat Eqs. (5f), (5h), (5l) and (5m) are trivially satisfied if the respective binary term doesn’tvanish. Because we assumed that D = 1 , we can fix W = W = m . The basic idea isthat we compute for each pattern z ∈ P the additive performance c z (Eqs. (5c) to (5m)) andthen ensure that the auxiliary variable E is upper bounded by all of these values (Eq. (5a)).The objective then maximizes E in order to find feasible u and v values that maximize theminimum additive performance. More specifically, we first ensure that the split routing is feasible(Eqs. (5b) and (5n)); secondly, we ensure that for each pattern z ∈ P the variables a z , b z , y z and c z are set to the correct values. To see this we will focus on the a z values, the othervariables follow analogously. Let z ∈ P be an arbitrary but fixed pattern. We have to show that a z = min i ∈{ ,...,m } P ij =1 z j . By Eq. (5e) we know that a z ≤ min i ∈{ ,...,m } P ij =1 z j . If we nowknow that there exists an index i ∈ { , . . . , m } such that a z ≥ P ij =1 z j , the claim follows. Thishowever is ensured by Eq. (5e) together with Eq. (5c), as there exists at least one index i suchthat w min z,i = 1 and thus a z ≥ P ij =1 z j − W (cid:16) − w min z,i (cid:17) = P ij =1 z j .This very naïve approach has many drawbacks, most prominently the massive amount ofused binary variables (overall (2 m + 1) 2 m ), the large number of constraints, many symmetrieswith respect to the u i and v i variables and the use of big-M constraints (Eqs. (5f), (5h), (5l)and (5m)). However, we only want to solve the MILP for small values of m .The first issue can be improved by the following observation: Assume we are given a fixedpattern z ∈ P with the property that there exists an index < i < m such that z i = v i and z i +1 = v i +1 . Then the index i will never contribute to a maximum or a minimum of the a z and b z variables, as p z ( i − ≤ p z ( i ) ≤ p z ( i + 1) . It is therefore not necessary to maintain thevariables w min z,i and w max z,i . In fact, the binary variable w min z,i has only to be maintained if z i = − u i and z i +1 = v i +1 , and likewise w max z,i has only to be maintained if z i = v i and z i +1 = − u i +1 , for < i < m . Similar arguments hold for the border cases i ∈ { , m } . One can think of theserestrictions as local minima and maxima for any fixed pattern. A local minima is − u i followedby v i +1 (excluding the extreme cases), and equivalently a maxima is v i followed by − u i +1 . Asall local minima and maxima are the same, independent of the specific u and v values, the globalminima and maxima (depending on the concrete realization of u and v ) are obtained at one ofthose spots. Consequently every pattern has instead of m + 1 binary variables at most m + 1 ,as each index is associated with at most one of the variables w min z,i and w max z,i .To break (at least some) symmetries, we added the constraints u ≤ u i and u ≤ v i for all i ∈ [ m ] . To see that these inequalities are valid, consider Fig. 7. By relabelling the ring such thatan arbitrary but fixed node has label , while the remaining nodes are labelled consecutive inclockwise direction, we can assume that the split demand u is the smallest among all other splitdemands. Note that the number of constraints increased only slightly, while the search spacedecreased significantly. Implementation Details and Experiments We implemented the MILP with the reducednumber of binary variables and the symmetry breaking constraints in C++ and solved it using13 23456 u u u v v v Figure 7: A different view on split routing solutions to visualize necklace symmetries. By rela-belling the nodes such that an arbitrary node has label , while remaining nodes are labelledconsecutive in clockwise direction, we obtain the same instance with different labels. m add. perf. − ε < < s < MB + ε 528 36 : h . GBTable 1: Additive performance bounds on instances of different sizes together with requiredcomputing resources. The numerical error satisfies ε ≤ × − . gurobi [GO18]. The computations were carried out on a computer cluster with two Xeon E5-2630 v4 CPU’s and GB RAM. Table 1 shows results of the experiments together with runtimes, memory consumption and the number of binary variables after gurobis preprocessing. Wecan see that the approach produced (almost) optimal solutions for all instances with m ≤ .In particular we see that all split routing solutions of instances with m ≤ can be turned intounsplittable solutions while increasing the load on any edge by no more than (1 + ε ) D . Wefurthermore see that all split routing solutions with m = 7 can be turned into unsplittablesolutions while increasing the load on any edge by at most (cid:0) + ε (cid:1) D . As m increases, theamount of used memory and run times are growing rapidly. Our approach is therefore incapableto output a solution for m ≥ . Skutella disproved in [Sku16] with a counterexample Schrijver et al.’s conjecture [SSW98] that L ≤ L ∗ + D . Particularly, Skutella found an instance of the Ring Loading Problem on nodesand demands with D = 10 and δ = , where the optimum split routing has load L ∗ = 39 whilean optimum unsplittable routing has load L = L ∗ + D + 1 = 50 , overall providing an additivegap of D (see Fig. 10). In this section we present further counterexamples that will extendthe view on known results. We like to highlight that despite our best efforts, we were unable tofind counterexamples that improve upon the additive gap of D given by Skutella [Sku16].Note that in general, a split routing solution where the best additive performance is high withrespect to D does not yield a counterexample to Schrijver et al.’s conjecture, as they are notoptimum. In fact an optimum split routing for an instance with m pairwise crossing demandssplits every demand evenly. Furthermore, the increase of load on some edge by αD , for some α ≥ , does not imply that any unsplittable routing increases the load on the edge with themaximum load by αD .However, the following lemma justifies our restriction to split routing solutions in search ofinstances of the Ring Loading Problem where the difference L − L ∗ is large with respect to D .Note that this can be used in conjunction with the split routing solution of Schijver et al. [SSW98]to provide a counterexample to their own conjecture, namely an instance of the Ring LoadingProblem with L = L ∗ + D . 14 emma 7. Let α ≥ . Any split routing solution that cannot be turned into an unsplittablerouting without increasing the load on some edge by at least αD can be turned into an instanceof the Ring Loading Problem with L − L ∗ ≥ αD . 18 14 181281810 (a) A split routing solution together with in-curred edge loads. 18 14 18128181010 0 8161614181020124 4 6 2 (b) Introduction of short demands to equalizethe loads to . 18 14 1812818101010 00 88 1616161614 1418 1810 1020 20121244 44 66 22 (c) Subdividing edges such that short demandsare routed along their edge. The highlightedarea contains demands of value > D . 18 14 18 12818102222 1010 00 88 161616 1614 1418 1810 1018 1812 1244 44 66 22 (d) Enlarged section of Fig. 8c showing the sub-division of edges to reduce the demand value ofshort demands. Figure 8: An example how to create an instance of the Ring Loading Problem from a given splitrouting, where the worst-case load increase is preserved. The split routing has m = 7 pairwisecrossing demands with D = 18 and δ = such that any unsplittable routing increases the loadon some edge by at least D + 1 = 19 . Proof. An example of the following procedure can be found in Fig. 8 with a split routing solutionon nodes and non-zero demands such that any unsplittable routing increases the load onsome edge by at least D for D = 18 .Let l : E → N be the function that maps every edge to its load value with respect to thegiven split routing solution. We define l max := max e ∈ E l ( e ) to be the maximum edge load. Forevery edge { i, i + 1 } of the ring, we introduce a new demand d i,i +1 of value l max − l ( { i, i + 1 } ) .We call the edge { i, i + 1 } the edge of demand d i,i +1 . If all new demands at distance one arerouted unsplittably along their edge, we equalize all edge loads, as d i,i +1 + l ( { i, i + 1 } ) = l max .Because the load is the same on every edge, this configuration is also an optimum split routingfor the newly created instance. Figure 8b shows the made changes to the instance.15here are two issues that can occur. First that not every unsplittable routing to this en-hanced instance increases the edge load of an optimum split routing by the same amount as anunsplittable solution for the initial split routing did. This cannot happen, if the newly introducedshort demands are routed along their short edge in an optimum unsplittable routing. And secondthat the introduced demands have a value larger than D , which would consequently reduce themaximum increase of edge load relative to D .The first issue is fixed by the following technique: We subdivide every edge, adding betweenevery neighbouring pair of nodes i and i + 1 an additional node i ′ . All pairwise crossing demandsremain in their current state, whereas the short demands d i,i +1 are divided together with theedge. This means we delete the demand d i,i +1 and introduce two new demands d i,i ′ and d i ′ ,i +1 of the same value (see Fig. 8c). Note that an optimum split routing has the same structureafter the subdivision process. We now argue that any unsplittable routing can be turned intoa different unsplittable routing where all short demands are routed along their respective edge,without increasing the load on any edge. As the node i ′ is adjacent to the demands d i,i ′ and d i ′ ,i +1 only, ignoring these demands ensures that the load on { i, i ′ } and { i ′ , i + 1 } is the same.We consider two cases, first that both d i,i ′ and d i ′ ,i +1 are routed the long way, and second thatonly one of them, say d i,i ′ , is routed the long way while the other demand d i ′ ,i +1 is routed onits edge. Assume therefore that both d i,i ′ and d i ′ ,i +1 are routed the long way. Rerouting bothof them to use their edges decreases the load on every edge different form { i, i ′ } and { i ′ , i + 1 } by d i,i +1 . The load on { i, i ′ } and { i ′ , i + 1 } remains the same, implying the claim. If now only d i,i ′ is routed the long way, rerouting the demand decreases the load on every edge different from { i, i ′ } by d i,i +1 , while the load on { i, i ′ } is now increased to match the new load of { i ′ , i + 1 } .Thus showing that the maximum load did not increase. We finally relabel all nodes from to n clockwise along the ring.We now fix the second issue (see Fig. 8d): We assume in the following, that in an optimumunsplittable routing the short demands are routed along the short edge (see the previous step).Note that only newly introduced demands may have a demand value larger than D . Let d j,j +1 >D be such a demand. We again subdivide the edge, introducing a node j ′ between j and j + 1 on the ring. For the pairwise crossing demands nothing changes. We introduce two newdemands d j,j ′ = d j ′ ,j +1 = d j,j +1 − D and update the original demand value to d j,j +1 = D .Note that all demand values strictly decreased with respect to the original demand value d j,j +1 .It furthermore holds that the load on the edges { j, j ′ } and { j ′ , j + 1 } is unchanged, if all newdemands are routed along the short paths. By the assumption, we know that this is fulfilledfor the “outer” demand d j,j +1 . For the two created small demands at distance one this followsfrom the same argumentation as described in the previous modification step. As the invariant ispreserved, we can repeat this process until no more demands of value greater than D remain.Overall we obtain an instance of the Ring Loading Problem with L − L ∗ ≥ αD , as the loadincrease of unsplittable routings on some edge by αD guarantees the increase of load by αD onan edge of maximum load.Skutella [Sku16] found the split routing solution that led to the counterexample (see Fig. 10)by brute force enumeration over instances with specific structural properties on pairwise cross-ing integer demands of value at most D = 10 . We show in the remainder of this section how tomodify this split routing solution in order to obtain further counterexamples that are δ -instancesfor < δ ≤ (see also Fig. 2).Consider the split routing solution in the first row of Table 2 (see also Fig. 10 on the left).For ε = 0 , the instance corresponds to one of the split routings given by Skutella [Sku16]. Byadding ε ≥ to certain split demands, the instance changes continuously such that D = 10 + 2 ε and δ = D while any unsplittable solution increases the load on some edge by at least D + 1 .To see that any unsplittable solution increase the load on some edge by at least 11 + 2 ε , one hasto consider all of the different patterns and observe that its additive performance is at least 11+ 2 ε . In order to obtain the additive performance for a pattern, the minimum a , the maximum16ond. D δ add. perf. Ref. u = (4 , , , , , , , 2) +( ε, ε, ε, , ε, ε, ε, ε ≥ ε D D + 1 = (cid:18) δ (cid:19) D Fig. 10 v = (6 , , , , , , , 2) +( ε, ε, ε, , ε, ε, ε, u = (4 , , , , , , , 2) +( ε, ε, ε, ε, ε, ε, ε, ε ) ε ∈ [0 , 1] 10 + 2 ε εD D + 1 = (cid:18) − δ (cid:19) D Fig. 10 v = (6 , , , , , , , 2) +( ε, ε, ε, ε, ε, ε, ε, ε ) u = (7 , , , , , , – 18 8 / D + 1 = 19 Fig. 8 v = (11 , , , , , , Table 2: Lower bound examples and their properties; addition of vectors is component-wise. b and the end point y with respect to ε has to be calculated and the maximum in Observation 2computed. One problem that occurs, is that the minimum and maximum values consideredthroughout might not be unique, as the dependence on ε might change the results. The examplein Fig. 9 illustrates this issue; the minimum a for this pattern varies for different values of ε .The minimum is therefore either − − ε or − . In this particular case however, the additiveperformance of the pattern is either way at least 11 + 2 ε , as max { b − y, y − a } = ( max { 11 + 2 ε, } , if a = − and max { 11 + 2 ε, ε } , if a = − − ε, which is exactly 11 + 2 ε . A thorough analysis of the remaining cases reveals that the additiveperformance of all patterns is at least 11 + 2 ε .For the second split routing solution given in the second row of Table 2 (see also Fig. 10 onthe right) a similar approach can be used to show that the additive performance of all patternsis at least 11 + 2 ε , where we need the fact that ε ∈ [0 , . p z ( k ) = 0 − − ε ε ε − ε − − − u + v + v − u − u + v − u + v ε = k ε = k Figure 9: A pattern for the unsplittable solution z = ( − u , v , v , − u , − u , v , − u , v ) of thesplit routing solution in the first row of Table 2 with ε = 0 (top) and ε = 2 (bottom). Theminima of the two variations (highlighted circles) are different.17 εε 44 ++ εε 64 ++ εε εε εε ε ε + ε + ε + ε ε + ε + ε 46 ++ εε 44 ++ εε 64 ++ εε 22 ++ εε 73 ++ εε εε ε ε ε ε + ε + ε + ε + ε ε + ε + ε + ε Figure 10: Two examples of split routings with pairwise crossing demands and D = 10+2 ε suchthat any unsplittable routing increases the load on some edge by at least D +1 = 11+2 ε , for ε ≥ (left) and ε ∈ [0 , (right). The instances coincide with the split routing from Skutella [Sku16]for ε = 0 . (a) m = 3 , D = 4 (b) m = 5 , D = 4 (c) m = 6 , D = 2 Figure 11: Three split routing solutions with m pairwise crossing demands of value D . With Sections 4.1 and 4.2 at hand, we can turn our attention to the proof of Theorem 2. Proof of Theorem 2. We see in Table 1 that any split routing solution with m ≤ can be turnedinto an usplittable solution while increasing the load on any edge by no more than (1 + ε ) D (after rescaling by D ). For m = 7 the worst-case increase of load is (cid:0) + ε (cid:1) D . Thus the upperbounds follow. The lower bounds can be produced with Lemma 7 together with the found splitrouting solutions given in Figs. 8 and 11a to 11c. The remaining even cases follow with the sametype of instance as Fig. 11c. We showed that any split routing solution to the Ring Loading Problem can be turned into anunsplittable solution while increasing the load on any edge by at most D . We furthermore18howed that split routing solutions with at most pairwise crossing demands cannot yield lowerbounds with additive performance worse than (cid:0) + ε (cid:1) D , for a small ε . On the way, we alsoproved that any split routing solution with large additive performance can be turned into aninstance of the Ring Loading Problem while maintaining the load increase. 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