On the irreducible core and the equal remaining obligations rule of

minimum cost spanning extension problems

by

Vincent Feltkamp∗

CentER and Econometrics Department, Tilburg University,

Stef TijsCentER and Econometrics Department, Tilburg University,

P.O. Box 90153, 5000 LE Tilburg, The Netherlands,

and Shigeo Muto†

Dept of Value and Decision Science,

Graduate School of Decision Science and Technology,

Tokyo Institute of Technology,

2-12-1 Oh Okayama, Meguro-ku, Tokyo 152-8552, Japan.

1st of December 1994

Abstract

Minimum cost spanning extension problems are generalizations of minimum cost spanning tree problemsin which an existing network has to be extended to connect users to a source. This paper generalizes thedefinition of irreducible core to minimum cost spanning extension problems and introduces an algorithmgenerating all elements of the irreducible core. Moreover, the equal remaining obligations rule, a one-pointrefinement of the irreducible core is presented. Finally, the paper characterizes these solutions axiomatically.The classical Bird tree allocation of minimum cost spanning tree problems is obtained as a particular casein our algorithm for the irreducible core.

1 Introduction

Consider a group of villages, each of which needs te be connected directly or via other villages to a source.Such a connection needs costly links. Each village could connect itself directly to the source, but by cooper-ating costs might be reduced. This cost minimization problem is an old problem in Operations Research, andBor

◦uvka (1926) came up with algorithms to construct a tree connecting everybody to the source with minimal

total cost. Later, Kruskal (1956), Prim (1957) and Dijkstra (1959) found similar algorithms. A historic overviewof this minimum cost spanning tree (henceforth mcst) problem can be found in Graham and Hell (1985).

In this paper, we analize a more general problem, in which a partial network of links exists already and hasto be extended to a network connecting every player to the source.

However, finding a minimal cost spanning extension is only part of the problem: if the cost of this extensionhas to be borne by the villages, then a cost allocation problem has to be addressed as well. Claus and Kleitman(1973) introduced this cost allocation problem for the original minimum cost spanning tree setting, whereuponBird (1976) treated this problem with game-theoretic methods and proposed a cost allocation closely relatedto the Prim-Dijkstra algorithm. Granot and Huberman (1981) proved that this allocation is an extremal pointof the core of the associated minimum cost spanning tree game. This game is defined as follows: the playersare the villages and the worth of a coalition is the minimal cost of connecting this coalition to the source vialinks between members of this coalition. Aarts (1992) found other extreme points of the core in case the mcstproblem has an mcst that is a chain, i.e. a tree with only two leaves. Kuipers (1993) investigated the core ofinformation games. These are games arising from mcst problems in which the costs of a connection are eitherone or zero.

∗Email: vincent.feltkamp@bigfoot.com. This author was sponsored by the Foundation for the Promotion of Research in EconomicSciences, which is part of the Dutch Organization for Scientific Research (NWO).†This author wishes to acknowledge the Canon Foundation in Europe Visiting Research Fellowship which made his visit to

CentER possible.

2

The outline of this paper is as follows.Section 2 presents a formal model of the minimum cost spanning extension (mcse) problem in which an

existing network has to be extended to a spanning network, i.e. a network connecting every village to thesource. An algorithm to find a minimum cost spanning extension and an associated set of cost allocations ispresented. In contrast to earlier work (see Feltkamp, Tijs and Muto (1994a)), this extension and set of allocationsare not generated by an algorithm a la Prim-Dijkstra, but by an algorithm similar to Kruskal’s (1956) algorithm.It is proved that the set of allocations generated is a subset of the core of the associated mcse game and thatis is independent of the extension that is constructed.

Section 3 generalizes the definition of the irreducible core, proposed in Bird (1976) for minimum cost spanningtree problems, to minimum cost spanning extension problems and proves that the set of allocations generatedby the algorithm in section 2 coincides with the irreducible core. A corollary is that Bird’s tree allocations (seeBird (1976)) for minimum cost spanning tree problems are also generated by our algorithm.

Section 4 introduces the equal remaining obligations (ERO) rule, a one-point refinement of the irreduciblecore. It is obtained as a special case of the algorithm for the irreducible core presented in section 2. Like theirreducible core, the ERO rule is independent of the extension constructed. This in contrast with Bird’s treeallocation rule, which is dependent on the tree constructed.

Section 5 axiomatically characterizes the irreducible core and the equal remaining obligations value. Amongothers, axioms we use are efficiency, consistency and converse consistency.

Finally, section 6 concludes with some remarks and suggestions for further research.The proofs of the main theorems of section 2 are provided in an appendix.

Preliminaries and notationsWe recall some standard definitions from graph theory which can be found in any elementary textbook on

graph theory to show the notational conventions. A graph G =< V,E > consists of a set V of vertices and aset E of edges. An edge e incident with two vertices i and j is identified with {i, j}1. For a graph G =< V,E >and a set W ⊆ V ,

E(W ) := {e ∈ E | e ⊆W}is the set of edges linking two vertices in W and for a subset E′ of E,

V (E′) := {v ∈ V | there exists an edge e ∈ E′ with v ∈ e}

is the set of vertices incident with E′.The complete graph on a vertex set V is the graph KV =< V,EV >, where

EV := {{v, w} | v, w ∈ V and v 6= w}.

A path from a vertex i to a vertex j in a graph < V,E > is a sequence (i = i0, i1, . . . , ik = j) of verticessuch that for all l ≤ k, the edge {il−1, il} lies in E. A cycle is an edge-disjoint path whose begin and endpoints coincide. Two vertices i, j ∈ V are connected in a graph < V,E > if there is a path from i to j in< V,E >. A subset W of V is connected in < V,E > if every two vertices i, j ∈ W are connected in thesubgraph < W,E(W ) >. A connected set W is a connected component of the graph < V,E > if no superset ofW is connected. We will usually say component when we mean connected component. If W ⊆ V , the set ofconnected components of the graph < W,E(W ) > is denoted W/E. A connected graph is a graph < V,E >with V connected in < V,E >. A tree is a connected graph that contains no cycles. A connected component ofa graph will be denoted by the letter C, and for a vertex v of the graph, the connected component containingv is denoted Cv.

The economic situations in the sequel involve a set N of users of a source ∗. For a coalition S ⊆ N , wedenote S ∪ {∗} by S∗. For two vectors x ∈ IRS and y ∈ IRT , where S and T are two disjoint coalitions, wedenote (x, y) the vector with coordinates

(x, y)k =

{xk if k ∈ Syk if k ∈ T.

Furthermore, for two coalitions S ⊆ T and a vector x ∈ IRT , we denote xS the restriction of x to S. The symbol1S is used to denote the vector in IRN with coordinates

1S,k =

{1 if k ∈ S0 if k ∈ N \ S.

For any coalition S, the simplex ∆S is defined by

∆S := {x ∈ IRS+ |∑i∈S

xi = 1}.

1This can be done because we do not consider multigraphs: two vertices are connected by at most one edge.

3

With many economic situations in which costs have to be divided one can associate a cost game (N, c)consisting of a finite set N = {1, . . . , n} of players, and a characteristic function c : 2N → IR, with c(∅) = 0.Here c(S) represents the maximal cost for coalition S if it secedes, i.e. if people of S cooperate and do not countupon help from people outside S.

The core Core(c) of a cost game (N, c), is defined by

Core(c) = {x ∈ IRN |∑i∈N

xi = c(N) and∑i∈S

xi ≤ c(S) for all S ⊆ N}.

The cardinality of a set S will be denoted by |S|.

2 Mcse problems: a solution

In this section we formally present minimum cost spanning extension problems, mcse games and an algorithmwhich for any mcse problem computes a minimum cost spanning extension and an associated set of allocations,which appears to be contained in the core of the mcse game.

A minimum cost spanning extension problem consists of a set N of users who have to extend an existingnetwork in order to be connected to a source, denoted ∗. The links are costly, and the users have to pay for theextension. Such a problem is represented by a complete graph < N∗, EN∗ > on the set N∗ containing all usersand the source, together with a set E ⊆ EN∗ of already constructed links and a weight function w : EN∗ → IR+.The cost of constructing an edge e is given by the positive weight w(e) > 0 of this edge. Because the graph ofpossible edges is always the complete graph, we denote a mcse problem with set of users N , source ∗, weightfunction w and existing edge set E by < N, ∗, w,E >. If the set of existing edges E is empty, the mcse problembecomes the classical minimum cost spanning tree problem, and instead of writing < N, ∗, w, ∅ >, we will write< N, ∗, w >.

Mcse problems can be split up into two subproblems, an Operations Research problem of connecting allusers to the source in a extended graph < N∗, E ∪ E′ > such that the cost of the extension E′ is minimal anda cost allocation problem of allocating this cost to the users in a reasonable way.

In the special case of mcst problems, an mcst is constructed by Kruskal’s algorithm, which is defined asfollows:

Algorithm 2.1 (Kruskal 1956)input : an mcst construction problem < N, ∗, w >output : an edge set E′ of a minimum cost spanning tree

1. start with the empty set E′ = ∅.

2. Find an edge e of minimum cost such that the graph < N∗, E′ ∪ {e} > does not contain a cycle.

3. Join this edge to the set E′ (E′ := E′ ∪ {e}).

4. If the graph < N∗, E′ > is not connected, go back to step 2.

5. E′ is the edge set we were seeking.

For mcse problems, a generalization of Kruskal’s algorithm is demonstrated in example 2.2.

Example 2.2 Let N = {1, 2, 3, 4, 5}, and let the weights of the edges and the graph which is already constructedbe as in figure 1. The costs of edges that are not indicated are $200. First construct the edge {1, 2}, it is thecheapest one which does not introduce a cycle. The same reasoning picks {2, 3} as second edge, as third edge{3, 4} and finally as last edge, {2, ∗} is constructed.

The algorithm demonstrated is the following:

Algorithm 2.3 (Kruskal generalized to mcse problems)input : an mcse construction problem < N, ∗, w,E >output : an mcse

1. Given M≡ < N, ∗, w,E >, define

t = 0 the initial stage,τ = |N∗/E| − 1 the number of stages,E0 = ∅ the initial extension.

4

u∗$100

u2AAAAAA

$10

$20

u1

u3������

$40

u4HHH

HHH $70

u5

The weights of links.

u∗

u2

u1

u3

u4HHH

HHH

u5

{4, 5} is already constructed.

Figure 1: A simple mcse problem

2. t := t+ 1.

3. At stage t, given Et−1, choose a cheapest edge et 6∈ E∪Et−1 such that the graph < N∗, E∪Et−1∪{et} >does not contain more cycles than the graph < N∗, E ∪ Et−1 >.

4. Define Et := Et−1 ∪ {et}.

5. If t < τ , go to step 2.

6. Eτ is the extension we were seeking. Denote the sequence of edges constructed by E = (e1, . . . , eτ ).

In lemma 7.1 it is proved that the extension constructed is indeed a mcse. Note this algorithm constructsthe edges of a minimum cost spanning extension in the order of non-decreasing costs. Because the costs of theedges in any minimum cost spanning extension are the same, it follows that any sequence E = (e1, . . . , eτ ) whichconsists of the edges of a minimum cost spanning extension ordered by non-decreasing costs can be constructedwith algorithm 2.3 by a suitable choice of the edges chosen in step 3.

Bird (1976) associates a tree allocation with every minimum cost spanning tree in an mcst problem. InFeltkamp, Tijs and Muto (1994a), we prove that this tree allocation can be associated with the sequence ofedges which Prim and Dijkstra’s algorithm generates when generating this mcst. This suggests looking forallocations associated with the sequences generated by Kruskal’s algorithm.

For an mcse problem M ≡ < N, ∗, w,E >, the minimal cost spanning extensions E′ have one edge lessthan the number of components |N∗/E| (if more edges were built, a cycle would be introduced, which cannotbe minimal in cost, as the weights of the edges are positive). Hence, defining τ := |N∗/E| − 1, we associatean allocation with every sequence E = (e1, . . . , eτ ) of edges which does not introduce new cycles in the mcseproblem M. Note that any such sequence connects all players to the source. The idea behind the allocationis that at each successive stage t, the cost of the edge et which is constructed at stage t is shared among theplayers in N according to a fraction vector f t ∈ ∆N . Three rules will be observed when allocating the cost ofet:

• At stage t, the edge et connects two components of the graph < N∗, E ∪ {e1, . . . , et−1} > creating acomponent Ct of the graph < N∗, E ∪ {e1, . . . , et} >. Only players in Ct contribute to the cost of et.

• The component Ct−1∗ of the source in the graph < N∗, E ∪ {e1, . . . , et−1} > constructed before stage tdoes not contribute to the cost of et.

• Furthermore, summing over all edges in the sequence, every component of the original graph < N∗, E >which does not contain the source pays fractions of edges to a total of one.

Hence, define the set V E (M) of sequences of fraction vectors valid for the sequence E in M by

5

V E (M) :=

(f1, . . . , fτ ) ∈ [0, 1]τ |N |

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

∑k∈Ct

f tk = 1 for all t

and∑k∈C

τ∑s=1

fsk = 1 ∀C ∈ N∗/E with ∗ 6∈ C

andf tk = 0 for all t, k s.t. k ∈ Ct−1∗

.

For a sequence F ≡ (f1, . . . , fτ ) valid for E in M, define the allocation

xE ,F (M) :=

τ∑t=1

f tw(et) ∈ IRN (2.1)

and define the set DE (M) by

DE (M) := {xE ,F | F ∈ V E}. (2.2)

If no confusion can occur, we drop the argument M.The intuition behind the first rule is that at the moment et is constructed, we do not know yet who is going

to be connected through et to the source. It can be either one of the components being connected, dependingon how the later edges are chosen. We also don’t know yet which player outside Ct will need et. Because wewant to allocate its cost immediately, it is allocated to the players in Ct. The intuition behind the two otherrules is obvious.

Lemma 2.4 For all mcse problems M, for all sequences E = (e1, . . . , eτ ) such that Eτ := {e1, . . . , eτ} is an

mcse of M and all F valid for E , the allocation x := xE ,F (M) is efficient:∑i∈N xi =

∑e∈Eτ w(e).

Proof : Validity of F implies every edge et ∈ Eτ is paid for by the component Ct it constructs. 2

Note that because the set of valid sequences of fraction vectors V E is convex and the map

xE : F 7→ xE ,F

is linear, the set DE is also convex, for any sequence E .Instead of first constructing the edges and later allocating their cost, one could allocate the cost of the

edge et immediately, because the validity of a sequence of fraction vectors can be checked stage by stage: asequence f1, . . . , fτ is valid for e1, . . . , eτ in M if and only if at every stage t it satisfies

• the component Ct constructed at stage t pays the cost of the edge et,

• for every component in the original graph < N∗, E >, the total of the fractions paid up to stage t doesnot exceed 1,

• the people outside Ct pay nothing,

• the people in the component Ct−1∗ of the source pay nothing.

In formula, this gives

∑k∈Ct

f tk = 1,

∑k∈C

t∑s=1

fsk ≤ 1 for all C ∈ N∗/E,

f tk = 0 if k 6∈ Ct,f tk = 0 if k ∈ Ct−1∗

(2.3)

for every stage t.

Example 2.5 For an mcst problem T ≡< N, ∗, w >, Prim and Dijkstra’s algorithm (cf Prim (1957) or Di-jkstra (1959)) constructs a sequence E = (e1, . . . , e|N |) of edges leading to an mcst < N∗, T > as follows:at every stage t, et is an edge which connects a player with the component of the source in the graph< N∗, {e1, . . . , et−1} > and which has minimal cost among all such edges. Without loss of generality, wenumber the players in N such that for every t, the edge et connects player t with the component of the source,

6

{∗, 1, . . . , t − 1}. Hence, the edge e1 connects player 1 to the source and using the system 2.3, we see player 1has to pay the cost of e1 and the other players do not contribute. In the second stage, player 2 is connectedto the component of the source, which now equals {∗, 1}. The first equation in system 2.3 implies players 1and 2 pay the cost of edge e2. The fourth implies that player 1, who is in the component of the source, doesnot contribute, hence, player 2 is assigned the cost of edge e2. The third equation implies the other playersdo not contribute. The inequality is satisfied, because up to now, every component in the original graph (i.e.every player) paid either one edge or no edges. By induction, we see that at every stage, the component Ct

consists of the component Ct−1∗ and the newly connected player t. Because the component of the source doesnot contribute to the cost of et, the unique valid allocation of the cost of this edge is to allocate it completely

to player t. Hence, DE (T ) consists of one allocation, in which each player i is allocated the cost of the edgeincident to i on the unique path in the tree from i to the source. This allocation is precisely Bird’s tree allocationassociated with the mcst < N∗, T >, denoted βT (cf Bird (1976)).

Example 2.6 Computing the extreme points of the set of valid fraction vectors for the sequence E constructed

in example 2.2 shows that in this case DE (M) is the convex hull of the vectors

(10, 20, 40, 100, 0), (10, 20, 40, 0, 100), (10, 20, 100, 40, 0), (10, 20, 100, 0, 40),(10, 40, 20, 100, 0), (10, 40, 20, 0, 100), (10, 100, 20, 40, 0), (10, 100, 20, 0, 40),(20, 10, 40, 100, 0), (20, 10, 40, 0, 100), (20, 10, 100, 40, 0), (20, 10, 100, 0, 40),(40, 10, 20, 100, 0), (40, 10, 20, 0, 100), (100, 10, 20, 40, 0), (100, 10, 20, 0, 40).

Inspired by Bird (1976), we associate a minimum cost spanning extension game (N, cM) with an mcseproblem M ≡ < N, ∗, w,E > as follows. Each coalition S ⊆ N , if it cannot count on the players in itscomplement, has to solve a problem similar to the problem of the grand coalition, namely, extending the

existing graph to a graph connecting all users in S to the source. The cost of this extension is the worth cM(S)of coalition S in the mcse game.

When computing the cost of a coalition S, several questions arise. Can the coalition use all or some of theedges which are already present? Is it allowed to use vertices outside S? We opt for the following answers: acoalition S is allowed to use all edges which are initially present, but can only use those vertices which lie ina component of < N∗, E > which contains members of S or the source. Let us consider an example to clarifywhat we mean.

u∗

u1u2HHH

HHHu3$4

JJ

JJ

JJ

JJ

JJ

JJ

$3

$1

$3

������

$1

Figure 2: {1, 2} is allowed to use the edge {2, 3}, but {1} is not.

Example 2.7 In the problem depicted in figure 2, edge {2, 3} is already constructed. Coalition {1, 2} is allowedto use the edge {2, 3} and can connect itself by building the edges {1, 2} and {3, ∗}, so c({1, 2}) = 1 + 1 = 2.Coalition {1} is not allowed to use the edge {2, 3} because the component {2, 3} does not have any vertices incommon with {1} or the source, hence c({1}) = 3. The other worths are c({2}) = c({3}) = 1 (connect player 2via player 3), and finally c({1, 3}) = c(N) = 2.

In general, the formula becomes

cM(S) := min

{∑e∈E′

w(e) | S ⊂ CE′

∗ and E′ contains only edges betweencomponents containing members of S∗

}

7

for all S ⊆ N , where CE′

∗ is the component of the source ∗ in the graph < N∗, E ∪ E′ >.The next theorem states an allocation associated with a sequence of edges generated by algorithm 2.3 is a

core element of the mcse game.

Theorem 2.8 For any mcse problem M, for any sequence of choices E = (e1, . . . , eτ ) in the algorithm 2.3

applied to M and any sequence of fractions F valid for E the allocation xE ,F , as defined in equation 2.1, is a

core-allocation of the mcse game (N, cM) associated with M.

The proof of this theorem is lengthy and technical, and can be found in the appendix. An immediateconsequence is

Corollary 2.9 For any sequence E of edges ordered according to non-decreasing costs leading to a minimum

cost spanning extension for an mcse problem M with associated mcse game (N, cM),

DE (M) ⊆ Core(N, cM).

Proof : For any x ∈ DE (M), there is a sequence F which is valid for E with x = xE ,F (M) ∈ Core(N, c). 2

A question which arises is, how does the set DE depend on the sequence E? It is answered in the nextproposition.

Proposition 2.10 For any mcse problemM, for any E and E constructed by the algorithm 2.3 applied toM,

DE (M) = DE (M).

A proof can be found in the appendix.

Because DE is independent of the sequence E of edges, as long as this sequence is constructed by thealgorithm 2.3, we define for an mcse problem M:

DGK(M) := DE (M)

for any sequence E obtained by the algorithm 2.3 applied to M. (The superscript GK stands for generalizedKruskal).

3 The irreducible core of mcse problems

In this section we generalize the concept of irreducible core, known from mcst problems, to mcse problems andprove that our set of allocations DGK coincides with this irreducible core.

Definition 3.1 Given an mcse problem < N, ∗, w,E >, we define the associated mcst problem < NE , ∗E , wE >as follows: NE consists of the components of < N∗, E > which do not contain the source ∗, the new source ∗Eis the component of < N∗, E > which contains the original source ∗ and wE is defined by

wE(C,D) := min{w(i, j) | i ∈ C, j ∈ D}

for all components C and D of the graph < N∗, E >.Furthermore, for an edge e = {i, j} ∈ EN∗ , define

eE := {Ci, Cj} (3.1)

and for a set of edges F ⊆ EN∗ , define

FE := {{Ci, Cj} | {i, j} ∈ F}, (3.2)

where Ci and Cj are the components of < N∗, E > containing players i and j, respectively.

8

The intuitive idea is to shrink each component not containing the source into a single player, and to shrinkthe component of the source into a new source.

It is easy to see that if F is an mcse of the mcse problem < N, ∗, w,E >, then the tree < N∗E , FE > is anmcst of the associated mcst problem < NE , ∗E , wE >. Conversely, if < N∗E , T > is an mcst of the associatedmcst problem, then there exists an mcse F with FE = T . This correspondence, though possibly not one to one,transfers the well-known structure of the collection of mcs trees of an mcst problem onto the set of mcs extensionsof an mcse problem. More about this structure will be said in the appendix, in the proof of proposition 2.10.

We now define the irreducible core of an mcse problem. It is a straightforward generalization of the definitionof irreducible core of an mcst game provided in Bird (1976). Apparently, it depends on an mcse. However, thisis not the case, as we will prove later.

Definition 3.2 Given an mcse problem M = < N, ∗, w,E > and an mcse E′, define the irreducible coreIC(M, E′) ofM with respect to E′ as follows: consider the set Var(M, E′) of all mcse problems obtained fromM by varying the weight w(e) of edges e 6∈ E∪E′, that still have E′ as mcse. Now IC(M, E′) is the intersectionof the cores of all mcse games associated with an mcse problem in Var(M, E′), i.e.

IC(M, E′) :=⋂{

Core(N, cM′) | M′ ∈ Var(M, E′)

}.

If the set E of initially present edges is empty, the present definition coincides with the definition of irreduciblecore of an mcst problem in Bird (1976). For mcst problems, it is already known that the irreducible core isindependent of the mcst used to define it.

Equivalently, one could define the irreducible core as follows: given an mcse problem M = < N, ∗, w,E >and an mcse E′, for any two players i, j ∈ N , let Pij be a path in the graph < N∗, E ∪ E′ > from i to j. It ispossible that this path is not unique, but the part of the path in E′ is. Define a new weight function

w(i, j) :=

{max{w(e) | e ∈ Pij ∩ E′} if Ci 6= Cjw({i, j}) otherwise

(3.3)

where Ci and Cj are the components of i and j, respectively, in the graph < N∗, E >. Then the following holds:

Lemma 3.3 The irreducible core IC(M, E′) of a mcse problemM = < N, ∗, w,E > coincides with the core ofthe game (N, c) = (N, c<N,∗,w,E>).

Proof : First, E′ is an mcse of the problem < N, ∗, w,E >. This can be seen as follows. Let E be aspanning extension, which is cheaper than E′. Order the edges in E by increasing weight and take the firstedge {i, j} = e ∈ E \E′. Adding e to < N∗, E ∪E′ > creates a cycle formed by e and the edges in Pij . Hence,

there exists an edge e′ ∈ Pij ∩ E′ \ E, such that substituting e with e′ in E yields another spanning extensionof the problem < N, ∗, w,E >. Now e connects two components of < N∗, E >; hence w(e) ≥ w(e′) for all edgese′ ∈ Pij ∩ E′. This spanning extension is thus an mcse, which, moreover, has one more edge in common withE′. Repeat this argument enough times to see that E′ is indeed an mcse of the problem < N, ∗, w,E >, whichimplies that < N, ∗, w,E >∈ Var(M, E′). Hence, the irreducible core of M is included in the core of the game(N, c).

On the other hand, note that for an edge e = {i, j} 6∈ E′ connecting two components Ci and Cj , max{w(e′) |e′ ∈ Pij ∩ E′} is a lower bound on the weight that can be assigned to this edge e in an mcse problem M′ ∈Var(M, E′): if a lower weight is assigned to e, then there is an edge e′ ∈ Pe ∩ E′ with higher weight than eand replacing e′ by e would yield a spanning extension E′ \ {e′} ∪ {e} which has lower cost than E′. Hence,w(e) ≤ w′(e) for all edges e ∈ EN∗ , for every problem < N, ∗, w′, E >∈ Var(M, E′). This implies that

c(S) ≤ cM′(S) for all coalitions S and for all mcse problems M′ in Var(M, E′). Because E′ is an mcse of

< N, ∗, w,E >, it also follows that c(N) =∑e∈E′ w(e) = cM

′(N). Hence, Core(c) ⊆ Core(cM

′) for all

M′ ∈ Var(M, E), which implies that the core of c is included in the irreducible core of M. Together with thefirst part of the proof, this proves the theorem. 2

In order to analyze the structure of the irreducible core of an mcse problem, we relate its structure tothe structure of the irreducible core of the associated mcst problem of definition 3.1, using the concept ofmarionettes. Zumsteg (1992) defined two players i, j in a game (N, c) to be marionettes if

c(S ∪ {i}) = c(S ∪ {j}) = c(S ∪ {i, j})

for all S ⊆ N . Considering players to be marionettes of themselves turns being marionettes into an equivalencerelation, we denote it by ∼. For any player i, the set of marionettes of i is denoted by Si.

9

Definition 3.4 For a game (N, c), the marionette-reduced game (N ′, c′) is the game in whichN ′ = {Si | i ∈ N},and which satisfies c′(C) = c(

⋃S∈C S) for all C ∈ N ′. Hence a player in the marionette-reduced game consists

of all marionettes of one player in the original game.

Equivalently, one could obtain the marionette-reduced game as a subgame of the original game: for each playerU ∈ N ′, take one representative player jU ∈ U and define T = {jU | U ∈ N ′}. Define the subgame (T, cT ) by

cT (U) = c(U)

for all U ⊆ T . For every player i in T , there is a unique player Si in N ′ satisfying jSi = i and for every playerU in N ′, there is a unique player jU in T satisfying SjU = U . Furthermore this bijection between the players ofT and N ′ turns out to be an isomorphism between the games (N ′, c′) and (T, cT ):

c′(C) = c(⋃S∈C

S) = c({jS | S ∈ C}) = cT ({jS | S ∈ C})

cT (U) = c(U) = c(⋃i∈U

Si) = c′({Si | i ∈ U})

for all coalitions C ⊆ N ′ and U ⊆ T .

Lemma 3.5 If (N, c) is a game, and (N ′, c′) is its marionette-reduced game, their cores are related as follows:

1. if x ∈ Core(N, c), then y ∈ Core(N ′, c′), where y ∈ IRN ′ is defined by yS =∑i∈S xi for all S ∈ N ′.

2. if y ∈ Core(N ′, c′) ∩ IRN ′

+ , then x ∈ Core(N, c), for all x ∈ IRN+ satisfying yS =

∑i∈S xi for all S ∈ N ′.

Moreover, such an x exists.

Proof : The proof of part 1 is trivial. To prove part 2, take y ∈ Core(N ′, c′). We first prove an x that satisfiesthe requirements exists. For all S ∈ N ′, choose a representative player iS ∈ S and assign

xi :=

{yS if i = iS for an S ∈ N ′,0 otherwise.

Because y is non-negative, so is x. Now x(N) = x({iS | S ∈ N ′}) = y(N ′) = c′(N ′) = c(N) and for anycoalition T ⊆ N , there exists a subset T ′ of T , such that

⋃i∈T Si =

⋃i∈T ′ Si, where the right-hand side is a

disjoint union. Hence,

c(T ) = c(⋃i∈T

Si) = c(⋃i∈T ′

Si) = c′({Si | i ∈ T ′})

≥∑i∈T ′

ySi =∑i∈T ′

x(Si) = x(⋃i∈T ′

Si)

≥ x(T ),

which implies x satisfies the requirements.Now take any x ∈ IRN

+ satisfying yS =∑i∈S xi for all S ∈ N ′. Then

x(N) =∑S∈N ′

x(S) =∑S∈N ′

yS = y(N ′) = c′(N ′) = c(N)

and for any coalition T , take again a subset T ′ such that⋃i∈T Si =

⋃i∈T ′ Si, where the right-hand side is a

disjoint union. Then because x is non-negative and y ∈ Core(N ′, c′),

x(T ) ≤∑i∈T ′

∑j∈Si

xj =∑i∈T ′

x(Si) =∑i∈T ′

ySi

≤ c′({Si | i ∈ T ′}) = c(T ′) = c(T )

Hence, x ∈ Core(N, c). 2

Lemma 3.6 For a given mcse problem M = < N, ∗, w,E > with associated minimum-cost spanning tree

problem T = < NE , ∗E , wE >, the mcst game (NE , cT ) associated with T coincides with the marionette-

reduction (N ′, c′) of the mcse game (N, cM).

10

Proof : We first prove that the sets NE and N ′ coincide. Two players that are in the same component of< N∗, E > are marionettes in the mcse game: if either one is connected to the source, so is the other, so that thecost of connecting one is the cost of connecting both. Hence, a player in NE , being a component of < N∗, E >,is a coalition of marionettes of the mcse game.

On the other hand, if two players are marionettes in the mcse game, it means that connecting one of themto the source is as costly as connecting the other player or connecting both players. Because the cost of alledges is positive, it follows that both players must lie in the same component.

Hence the set of players NE in the mcst game coincides with the set of players N ′ in the marionette-reductionof the mcse game.

Consider a coalition C ⊆ N ′. By definition of the associated mcst problem,

cT (C) = cM(⋃S∈C

S).

As for all S ∈ C, all players in S are marionettes in the mcse game, it follows that

cM(⋃S∈C

S) = c′(C).

Hence cT (C) = c′(C), which concludes the proof. 2

The next proposition states the relation between the irreducible core of an mcse problem and the associatedmcst problem.

Proposition 3.7 For an mcse problem M = < N, ∗, w,E > and an mcse E′, the irreducible core IC(M, E′)satisfies

IC(M, E′) = {x ∈ IRN+ | y ∈ IC(< NE , ∗E , wE >) where yS :=

∑i∈S

xi ∀S ∈ NE}

Proof : This follows easily from lemmata 3.5 and 3.6 and the fact that an mcse is transformed into an mcstby the transition from mcse problem to associated mcst problem. 2

Corollary 3.8 The irreducible core of an mcse problem is independent of the mcse used to define it.

Accordingly, we will denote the irreducible core of an mcse problem M by IC(M).

Having given some properties of the irreducible core, we now proceed to prove that it coincides with the setof allocations generated by algorithm 2.3.

Lemma 3.9 Let M≡ < N, ∗, w,E > be an mcse problem. Then DGK(M) ⊆ IC(M).

Proof : In section 2 we stated that for an mcse problem M≡ < N, ∗, w,E >, the set DGK(M) of allocations

generated by the algorithm 2.3 applied to M are core allocations of the associated game (N, cM). Since toprove this in section 7, we only use the weights of edges in the mcse E′, which is an mcse in all mcse problemsM′ ∈ Var(M, E′), it holds that DGK(M) is a subset of the core of any mcse game associated with an mcseproblem in the set Var(M, E′). Hence,

DGK(M) ⊆ IC(M). (3.4)

2

In order to prove the reverse inclusion, we need the following lemma.

Lemma 3.10 Let T = < N, ∗, w > be an mcst problem and let < N∗, T > be an mcst for T . Then Bird’s treeallocation βT lies in the set DGK(T ).

Proof : The number of edges in T equals n := |N |. Consider any sequence E = (e1, . . . , en) of edges obtainedby ordering the edges of T by non-decreasing cost. Define F = (f1, . . . , fn) by

f ti :=

{1 if et is the first edge on the path in < N∗, T > from i to the source,0 otherwise.

11

It follows that F ∈ V E (T ) and that

βT = xE ,F ∈ DE = DGK(T ).

2

Theorem 3.11 Let T be an mcst problem. Then DGK(T ) = IC(T ).

Proof : It follows from lemma 3.9 that we only have to prove IC(T ) ⊆ DGK(T ). Bird (1976) proved IC(T ) is

the convex hull of the set of all Bird allocations of the mcst problem T , with reduced weight function definedby equation 3.3. By proposition 3.10, these Bird allocations lie in DGK(T ). Now DGK(T ) equals DGK(T ),because the set DGK is obtained by only considering the weights of edges in an mcst, and these edges have thesame weight in T as in T (cf. Aarts and Driessen (1993)). Moreover DGK(T ) is convex, hence

IC(T ) = conv hull{βT | T is an mcst of T } ⊆ DGK(T ) = DGK(T ).

2

Corollary 3.12 Let M be an mcse problem. Then DGK(M) = IC(M).

Proof : Let x ∈ IC(M), and let T =< NE , ∗E , wE > be the mcst problem associated with M. We know byproposition 3.7 that the vector y ∈ IRNE , defined by yS =

∑i∈S xi for all S ∈ NE , lies in IC(T )(= DGK(T )).

Hence, there exists a sequence E = (e1, . . . , eτ ) of edges leading to an mcst of T and a sequence F = (f1, . . . , fτ )

of fraction vectors valid for E , such that y = xE ,F . Now for each edge et = {Ci, Cj}, there exists an edge et

with same weight in the weighted graph < N∗, EN∗ , w >, which connects the components Ci and Cj . Hence,

E = (e1, . . . , eτ ) is a sequence leading to an mcse of M. Define F = (f1, . . . , fτ ) by

f ti =

{f tCi

xiyCi

if yCi > 0

0 if yCi = 0.

for all t and all i ∈ N , where Ci is the component containing i. Then F is valid for E . Moreover, for any

player i, xE ,Fi = 0 if yCi = 0, but then also xi = 0, and if yCi 6= 0, then

xE ,Fi =

τ∑t=1

f tCixiyCi

=xiyCi

τ∑t=1

f tCi = xiyCiyCi

= xi.

Hence, x = xE ,F ∈ DGK(M), which completes the proof. 2

4 The equal remaining obligations value

In most cases, the irreducible core of an mcse problem M contains a continuum of allocations. If the objectiveis to choose a division of the cost, one might be better off with a one-point solution.

The equal remaining obligations value (henceforth ERO value), suggested by Jos Potters, is a one-pointrefinement of the irreducible core and is constructed by the following extension of algorithm 2.3.

Algorithm 4.1 (Equal remaining obligations solution)input : an mcse problem < N, ∗, w,E >output : an mcse and the ERO value

1. Given M≡ < N, ∗, w,E >, define

t = 0 the initial stage,τ = |N∗/E| − 1 the number of stages,E0 = ∅ the initial extension.

2. t := t+ 1.

12

3. At stage t, given Et−1, choose a cheapest edge et 6∈ E∪Et−1 such that the graph < N∗, E∪Et−1∪{et} >does not contain more cycles than the graph < N∗, E ∪ Et−1 >.

4. If Ct = Ct−1i ∪ Ct−1j is the connected component just formed by connecting the components Ct−1i and

Ct−1j of the graph < N∗, Et−1 > with the edge et = {i, j}, define the vector f t = (f tk)k∈N of fractions by

f tk =

1|Ct−1i| −

1|Ct| if k ∈ Ct−1i and ∗ 6∈ Ct,

1|Ct−1j| −

1|Ct| if k ∈ Ct−1j and ∗ 6∈ Ct,

1|Ct−1i| if k ∈ Ct−1i and ∗ ∈ Ct−1j ,

1|Ct−1j| if k ∈ Ct−1j and ∗ ∈ Ct−1i ,

0 otherwise.

5. Define Et := Et−1 ∪ {et}.

6. If t < τ , go to step 2.

7. Eτ is the mcse we sought. As before, denote E = (e1, . . . , eτ ) the sequence of edges constructed.

8. Define EROE (M) :=

τ∑t=1

f tw(et).

Applied to the mcse problem of example 2.2, this algorithm generates successively edge {1, 2}, of which players 1and 2 each pay 1− 1

2 = 12 , edge {2, 3}, of which player 3 pays 1− 1

3 = 23 and players 1 and 2 each pay 1

2 −13 = 1

6 ,edge {3, 4}, of which players 1, 2 and 3 each pay 1

3 −15 = 2

15 while players 4 and 5 pay 12 −

15 = 3

10 and finallyedge {2, ∗}, to which each player contributes 1

5 . This yields the allocation

ERO(M) =1

3(101, 101, 116, 96, 96).

Generically, the choice of edge in step 3 is unique but even in the case that the sequence E is not uniquelydefined, this algorithm yields only one allocation, independently of the choice of edges made. This in contrastwith Bird’s tree allocation rule, that may associate a different allocation with each mcst of an mcst problem.

Proposition 4.2 For any two sequences of edges E and E chosen by the algorithm 4.1 applied to an mcseproblem M,

EROE (M) = EROE (M).

The proof is similar to the proof of proposition 2.10, so we do not give it. This proposition allows us to define

ERO(M) := EROE (M)

for any sequence E constructed by algorithm 4.1. Clearly, the fractions constructed are valid for the edgesconstructed; the ERO solution is thus a refinement of the irreducible core.

To see that the ERO value deserves its name, define for an mcse problem M the initial obligation oi of aplayer i by

oi :=

{1|Ci| if ∗ 6∈ Ci0 if ∗ ∈ Ci

(4.1)

where Ci is the component of < N∗, E > containing player i.For a sequence F = (f1, . . . , fτ ) of fraction vectors, after a stage t ≤ τ a player i ∈ N has paid

∑s≤t f

si ,

while the initial obligation was oi. Hence player i’s remaining obligation oti satisfies

oti = oi −∑s≤t

fsi = ot−1i − f ti . (4.2)

13

Theorem 4.3 The algorithm 4.1 has the property that after each stage t, in each component C of the graph< N∗, E ∪ Et >, every player k in the component C has the same remaining obligation

otk =

{ 1|C| if ∗ 6∈ C,0 if ∗ ∈ C. (4.3)

Proof : The proof goes by induction on the stage t.

1. After stage zero, for k in a component C of < N∗, E >,

otk = ok − 0 =

{ 1|C| if ∗ 6∈ C,0 if ∗ ∈ C.

2. Suppose equation 4.3 holds after stage t− 1. Let Ct = Ct−1i ∪ Ct−1j be the connected component formed

at stage t by connecting the components Ct−1i and Ct−1j of the graph < N∗, E ∪ Et−1 > with the edge

et = {i, j}. Let Ctk and Ct−1k be the components of player k in < N∗, E ∪ Et > and < N∗, E ∪ Et−1 >.Then

otk = ot−1k − f tk

=

{1

|Ct−1k| − f

tk if k 6∈ Ct−1∗

0− 0 if k ∈ Ct−1∗

=

1|Ct−1i| − ( 1

|Ct−1i| −

1|Ct| ) if k ∈ Ct−1i and ∗ 6∈ Ct

1|Ct−1j| − ( 1

|Ct−1j| −

1|Ct| ) if k ∈ Ct−1j and ∗ 6∈ Ct

1|Ct−1i| −

1|Ct−1i| if k ∈ Ct−1i and ∗ ∈ Ct−1j

1|Ct−1j| −

1|Ct−1j| if k ∈ Ct−1j and ∗ ∈ Ct−1i

1|Ct−1k| if k 6∈ Ct ∪ Ct−1∗

0 if k ∈ Ct−1∗

=

1|Ct| if k ∈ Ct 63 ∗

0 if k ∈ Ct−1i and ∗ ∈ Ct−1j

0 if k ∈ Ct−1j and ∗ ∈ Ct−1i

1|Ct−1k| if k 6∈ Ct ∪ Ct−1∗

0 if ∗ ∈ Ct−1k

=

1|Ctk| if ∗ 6∈ Ctk

0 if ∗ ∈ Ctk.

Hence equation 4.3 holds after stage t as well. This completes the proof. 2

5 Axiomatic characterizations

In sections 2 and 4 we introduced the irreducible core and the equal remaining obligations value for mcseproblems. We axiomatically characterize these rules in this section.

In contrast with the characterizations in Feltkamp, Tijs and Muto (1994a) and Feltkamp, Tijs and Muto (1994b),here, a solution consists only of a set of allocations. This is possible because both the irreducible core and theERO rule are independent of the set of edges constructed.

An allocation of an mcse problem < N, ∗, w,E > is a vector x ∈ IRN+ which satisfies

∑i∈N xi ≥ cM(N). In

effect, an allocation is a vector that allocates at least the cost of a minimum cost spanning extension to theplayers.

Properties that an allocation x of an mcse problem < N, ∗, w,E > can satisfy are

14

Definition 5.1

Eff x is efficient if ∑i∈N

xi = cM(N).

MC x has the minimal contribution property if every component that does not contain the source contributesat least the cost of a minimum cost edge that connects two components. In formula: for each componentC ∈ N∗/E that does not contain the source,∑

i∈Cxi ≥ min{w(e) | e connects two components of < N∗, E >}.

FSC x has the free for source component property if xi = 0 for all i in the component of the source in thegraph < N∗, E >.

The minimal contribution property and the free for source component are motivated as follows: every componentthat has to be connected to the source has to contribute at least the cost of an edge, and the component of thesource should not contribute, because it is already connected.

A solution of mcse problems is a map ψ assigning to every mcse problem a set of allocations.

Definition 5.2

NE A solution ψ is said to be non-empty if

ψ(M) 6= ∅ for all M.

We will say a solution ψ is efficient, satisfies the minimal contribution property or the free for source componentproperty if for all M all elements of the solution ψ(M) satisfy the corresponding property.

Definition 5.3 Given an mcse problem M ≡ < N, ∗, w,E > and an edge e = {i, j} 6∈ E that connects twocomponents of < N∗, E >, define the edge-reduced mcse problem

Me =< N, ∗, w,E ∪ {e} > .

Note that the edge-reduced problem is indeed a simpler problem than the original problem: less edges have tobe constructed. However, it has the same number of players as the original problem.

The next three properties use edge-reduced mcse problems, with as extra edge an edge which constructs anew component if it is adjoined to the graph < N∗, E >, and which has minimum cost among all edges withthis property.

Definition 5.4

Loc ψ is local if for all M, for all x ∈ ψ(M), for all minimum cost edge e that when added to < N∗, E >constructs a new component C, there exists an x ∈ IRC such that

(x, xN\C) ∈ ψ(Me).

In effect, this axiom requires that when a minimum cost edge is added, this has no influence on the allocationto players that are not in the component constructed by adding this edge.

ECons ψ is minimum cost edge consistent if for all M≡ < N, ∗, w,E >, for all x ∈ ψ(M), for each minimumcost edge e that when added to < N∗, E > constructs a new component C, for each α ∈ ∆C satisfyingαw(e) ≤ xC , it holds that

x− xe,α ∈ ψ(Me),

where xe,α := (αw(e), 0N\C).

This axiom means that when a minimum cost edge is added, the savings obtained by not having to constructthis edge can be allocated arbitrarily over the players that are in the component constructed by adding thisedge. Obviously, edge consistency implies locality.

15

CECons ψ is converse minimum cost edge consistent if for all M ≡ < N, ∗, w,E >, for every minimum costedge e that when added to < N∗, E > constructs a new component C, for every x ∈ IRN that satisfies

a MC,

b FSC,

c αw(e) ≤ xC implies x− xe,α ∈ ψ(Me) for all α ∈ ∆C ,

it holds thatx ∈ ψ(M).

This axiom requires that if adding an allocation to the solution does not destroy the MC, FSC and EConsproperties, then it should be part of the solution. In effect, it requires the solution to be the largest solutionthat satisfies the other properties.

Lemma 5.5 The irreducible core satisfies the properties NE, MC, Eff, FSC, ECons and CECons.

Proof : Because of the coincidence of the irreducible core with the set DGK , the irreducible core is non-emptyfor any mcse problem: there are always valid sequences of fraction vectors for any sequence of edges constructedby the algorithm 2.3. It satisfies the minimum contribution property because every component that does notcontain the source has to pay for fractions of edges that total one, so it contributes at least the minimum costof an edge that connects two components. That it is efficient is proved in lemma 2.4. By construction, it isclear that DGK satisfies FSC.

To prove edge consistency, take an mcse problem M and suppose x ∈ IC(M) = DGK(M). For eachminimum cost edge e that when added to < N∗, E > constructs a new component C, there exists a sequenceE = (e1, e2, . . . , eτ ) starting with the edge e1 = e, that is constructed by the algorithm 2.3. Because the set

DGK(M) is independent of the sequence of edges constructed, there exists a sequence F = (f1, . . . , fτ ) ∈ V E

such that x = xE ,F . For any α ∈ ∆C satisfying αw(e) ≤ xC , define α1 ∈ ∆C by α1i := f1i for all i ∈ C. Then

∑i∈C

αi − α1i = 0,

(α− α1)w(e1) ≤τ∑t=2

f tw(et),

f tw(et) ≥ 0 for t ≥ 2.

Hence there exist vectors α2, . . . , ατ ∈ IRC satisfying

αti ≤ f ti for t ≥ 2 and for all i ∈ C,∑i∈C

αti = 0 for t ≥ 2,

(α− α1)w(e1) =

τ∑t=2

αtw(et).

(5.1)

E.g. take αt the projection of f t on the hyperplane with coordinates zero, along the line through the points∑τs=2 f

sw(es) and (α− α1)w(e1), i.e.

αt = f t −∑i∈C f

ti∑

i∈C∑τs=2 f

si w(es)

(

τ∑s=2

fsw(es)− (α− α1)w(e1))

for all t ≥ 2. Rewriting the last equation of system (5.1), we obtain

αw(e) =

τ∑t=1

αtw(et).

It follows that x− xe,α =∑τt=2(f t − αt)w(et). The first two equations of system (5.1) imply that the sequence

(f2 − α2, . . . , fτ − ατ ) is valid for (e2, . . . , eτ ). Hence, x− xe,α ∈Me.Because edge consistency implies locality, the irreducible core satisfies locality as well.To prove that the irreducible core satisfies CECons, take an mcse problem M, take a minimum cost edge e

that when added to < N∗, E > constructs a new component C, take an x ∈ IRN that satisfies

16

a MC,

b FSC,

c αw(e) ≤ xC implies x− xe,α ∈ IC(Me) for all α ∈ ∆C .

We have to prove that x ∈ IC(M).Denote by C1 and C2 the two components that are joined by e. The allocation x satisfies FSC; hence if one

of these components (say C1) contains the source, xi = 0 for i ∈ C1 and an α ∈ ∆C with α ≤ xC satisfies αi = 0for i ∈ C1. For such an α (which exists), there exists a sequence (e2, . . . , eτ ) constructed by the algorithm 2.3

applied to the problem Me and a sequence (f2, . . . , fτ ) ∈ V (e2,...,eτ ) such that x − xe,α = x(e2,...,eτ ),(f2,...,fτ ).

So with f defined by

fk :=

{αk if k ∈ C0 otherwise

for all k ∈ N , it holds that (f, f2, . . . , fτ ) is valid for the sequence (e, e2, . . . , eτ ) and

x = x(e,e2,...,eτ ),(f,f2,...,fτ ) ∈ ψ(M).

If neither of the two components contain the source, then by the minimal contribution property, bothcomponents contribute at least w(e) in the allocation x. On the other hand, both together contribute x(C), sothere exists an a1 ∈ [0, 1], such that{

x(C1) = a1w(e) + (1− a1)(x(C)− w(e))x(C2) = (1− a1)w(e) + a1(x(C)− w(e))

Define α ∈ ∆C , by

αi =

{a1xi/x(C1) if i ∈ C1,(1− a1)xi/x(C2) if i ∈ C2.

Then, αw(e) ≤ xC . Hence, there exists a mcse {e2, . . . , eτ} and a sequence (f2, . . . , fτ ) ∈ V (e2,...,eτ )(M) suchthat x − xe,α =

∑τt=2 f

tw(et). Define a21, . . . , aτ1 by at1 = (1 − a1)

∑i∈C f

ti and a22, . . . , a

τ2 by at2 = a1

∑i∈C f

ti .

Then

at1 + at2 =∑i∈C

f ti for all t ≥ 2

a1 +

τ∑t=2

at1 = a1 + (1− a1)

τ∑t=2

∑i∈C

f ti = 1

1− a1 +

τ∑t=2

at2 = 1− a1 + a1τ∑t=2

∑i∈C

f ti = 1

Defining g1 = (α, 0N\C) and

gti =

{at1xi/x(C1) if i ∈ C1

at2xi/x(C2) if i ∈ C2

we see that (g1, . . . , gτ ) ∈ V (e,e2,...,eτ )(M). Furthermore,

x(C1)− a1w(e) = (1− a1)(x(C)− w(e))

= (1− a1)

τ∑t=2

∑i∈C

f tiw(et)

=

τ∑t=2

at1w(et).

This implies that for i ∈ C1,

xi =xi

x(C1)x(C1)

=xi

x(C1)(a1w(e) +

τ∑t=2

at1w(et))

= g1iw(e) +

τ∑t=2

gtiw(et).

Similarly for i ∈ C2. Hence, x = x(e,e2,...,eτ ),(g1,...,gτ ) is in the irreducible core of M. This implies that the

irreducible core satisfies converse edge consistency. 2

17

Lemma 5.6 If a solution of mcse problems φ satisfies MC, FSC and ECons, and a solution of mcse problemsψ satisfies NE, FSC and CECons, then φ(M) ⊆ ψ(M) for each mcse problem M.

Proof : We prove the lemma for any mcse problem M by induction on the number of components of thegraph < N∗, E >. First, consider an mcse problem M ≡ < N, ∗, w,E >, where the graph < N∗, E > is con-nected. Then by FSC, x = 0 for any x ∈ φ(M) and for any x ∈ ψ(M). By NE, there has to be an x ∈ ψ(M),hence

φ(M) ⊆ {0} = ψ(M).

Now suppose the lemma holds for every mcse problem M with p− 1 components in the graph < N∗, E >.Take an mcse problemM such that < N∗, E > has p components and take x ∈ φ(M). By ECons of φ, for anyminimum cost edge that constructs a new component C when added to < N∗, E >, for each α ∈ ∆C satisfyingαw(e) ≤ xC , it holds that

x− xe,α ∈ φ(Me) ⊆ ψ(Me),

where the inclusion holds by the induction hypothesis. Because x satisfies MC, it holds that x ∈ ψ(M) byCECons of ψ. Hence φ(M) ⊆ ψ(M). 2

Theorem 5.7 The unique solution of mcse problems that satisfies NE, MC, FSC, ECons and CECons is theirreducible core.

Proof : By proposition 5.5, the irreducible core has these properties. By lemma 5.6, if two solutions have theseproperties, they contain each other and hence they coincide. 2

To characterise the ERO value, we need some other properties that a solution can have.

Definition 5.8

ET a solution ψ satisfies equal treatment if for every mcse problem M, for all x ∈ ψ(M), for each componentC of the original graph < N∗, E >, and for all pairs of players i and j ∈ C,

xi = xj .

IPCons A solution ψ is inversely proportional consistent if for every mcse problem M ≡ < N, ∗, w,E >, forevery minimum cost edge e that when added to < N∗, E > connects two components C1 and C2, neitherof which contains the source, for all x ∈ ψ(M), there exists an x ∈ ψ(Me) such that

|C1|∑i∈C1

(xi − xi) = |C2|∑i∈C2

(xi − xi).

Equal treatment is motivated by the idea that players in the same component need the same connections,hence should be allocated the same amount. Inversely proportional consistency means that if two componentsare connected by a minimum-cost edge, they contribute amounts to the cost of this edge which are inverselyproportional to their number of elements. This is motivated by the idea that a bigger component has apparentlyalready constructed more edges in E, so should pay less for the edge under consideration.

Proposition 5.9 The unique solution of mcse problems that satisfies NE, FSC, Loc, Eff, ET and IPCons isthe ERO value. Here, the ERO value is identified with the solution that assigns the singleton {ERO(M)} toevery M.

Proof : First we prove that the ERO value satisfies the required properties. That the ERO value satisfiesthe properties NE, MC, FSC, Loc and Eff is a consequence of its being a refinement of the irreducible core.That is satisfies equal treatment is also easy to see. To prove it satisfies IPCons, take an mcse problemM, and a minimum cost edge connecting the components C1 and C2, neither of which contains the source,into a component C. Then there exists a sequence of edges E = (e = e1, . . . , eτ ) a sequence of fractions

F = (f1, . . . , fτ ) constructed by the algorithm 4.1 such that ERO(M) = xE ,F . Moreover, by definition of the

algorithm, x(e2,...,eτ ),(f2,...,fτ ) = ERO(Me). Because e connects two components that do not contain the source,

xE ,Fk − x(e2,...,eτ ),(f2,...,fτ )

k =

w(e)( 1

|C1| −1|C| ) if k ∈ C1,

w(e)( 1|C2| −

1|C| ) if k ∈ C2,

0 if k 6∈ C.

18

Then ∑k∈C1

(ERO(M)− ERO(Me)) = 1− |C1||C|

=|C2||C|

and similarly, ∑k∈C2

(ERO(M)− ERO(Me)) =|C1||C|

.

Hence

|C1|∑k∈C1

(EROk(M)− EROk(Me)) =|C1||C2||C|

= |C2|∑k∈C2

(EROk(M)− EROk(Me)).

To prove uniqueness, suppose a solution ψ satisfies these six properties. We prove ψ(M) = {ERO(M)} byinduction on the number of components of the graph < N∗, E >.

Let < N∗, E > have one component. By FSC, xi(M) = 0 = EROi(M) for all i ∈ N and all x ∈ ψ(M).Suppose ψ(M) = {ERO(M)} for all mcse problems M such that < N∗, E > has less than p components.

Consider an mcse problem M such that < N∗, E > has p components. Take a minimum cost edge e thatconnects two components C1 and C2 into a new component C in < N∗, E >. By ET of ψ applied to M andMe, for all x ∈ ψ(M), for all x ∈ ψ(Me), for all i, j ∈ C1 we have

xj − xj = xi − xi =: δ1(x, x)

and for all i, j ∈ C2 we havexj − xj = xi − xi =: δ2(x, x).

Moreover, by FSC of ψ, if C1 contains the source, then δ1(x, x) = 0 and by locality and efficiency, there exists

an x ∈ ψ(Me) such that δ2(x, x) = w(e)|C2| . If C2 contains the source, then similarly one proves that δ1(x, x) and

δ2(x, x) are uniquely determined. If neither of C1 and C2 contain the source, then by IPCons, there exists anx ∈ ψ(Me) such that

|C1|∑k∈C1

δ1(x, x) = |C2|∑k∈C2

δ2(x, x),

and by locality and efficiency, ∑i∈C1

δ1(x, x) +∑i∈C2

δ2(x, x) =∑i∈C

(xi − xi) = w(e).

Hence,

δ1(x, x) =w(e)

|C1|2and δ2(x, x) =

w(e)

|C2|2.

So whether C1 or C2 contain the source or not, the numbers δ1(x, x) and δ2(x, x) are uniquely determined andindependent of x and x. Now the ERO value also satisfies the six properties and so has these same numbersδ1(x, x), δ2(x, x) characterizing the difference between ERO(M) and ERO(Me). The induction hypothesis thenimplies

x− δ1(x, x)1C1 − δ2(x, x)1C2 = x = ERO(Me) = ERO(M)− δ1(x, x)1C1 − δ2(x, x)1C2

and so x = ERO(M). 2

6 Concluding remarks and suggestions for further research

New in this paper is the integrated approach : we solve the problem of constructing an optimal network andat the same time allocate the costs. Because of this integrated view of the problem, the different algorithmsto construct a minimum cost spanning tree suggested several closely related algorithms for solutions to thecost allocation problem. For instance, Prim and Dijkstra’s algorithm appeared to be closely linked to Bird’stree allocations. This suggested to look for allocation rules that are related to Kruskal’s (1956) algorithm forconstructing an mcst. In the present paper we relate the irreducible core to Kruskal’s algorithm, and moreoverprovide a one-point refinement, the equal remaining obligations solution. A third algorithm for constructing anmcst is known, and in Feltkamp, Tijs and Muto (1994b) we associate an allocation rule with this algorithm.

Second, instead of looking only at the extreme case where no edges are present at the beginning and aspanning network has to be constructed, we consider problems where some network can be present already,construct a minimum cost spanning extension and prove that the cost of this extension can be allocated in aa stable way. This has two advantages. The mathematical advantage is that a half-solved problem is again

19

in the same class of problems, which allows for a recursive solution, the advantage from an applied point ofview is that not only problems in which all edges have yet to be constructed are treated, but also extensions ofnetworks can be solved. If the original problem was suggested by among others electrification of Moravia at thebeginning of the century, by now the problem is rather how to extend an already present network and allocatethe cost of the extension.

Third, we provide axiomatic characterizations of the irreducible core and the equal remaining obligationssolution. These axiomatic characterizations enable one to select an allocation rule, based on the properties therule should have.

Another way of approaching mcse problems is to define the cost of a coalition S as the minimal costof an extension connecting S to the source, without any restriction on the vertices of the extension. Thisapproach yields a monotonic game, with the same cost for the grand coalition, but smaller costs for the othercoalitions, because these now have more opportunities to save costs. Hence, in general the core of this variant iscontained in the core of the mcse game we defined. However, such a monotonic game associated with an mcseproblem < N, ∗, w,E > can be considered as an mcse game according to our definition associated to the mcseproblem < N, ∗, w′, E > where the weights of links have been reduced to satisfy the triangle inequality

w′({i, j}) ≤ w′({i, k}) + w′({k, j}) for all i, j, k ∈ N∗.

So this does not introduce new games. Moreover, core elements of the monotonic game can be computed byapplying algorithm 2.3 to the problem with reduced weights.

A second possible variation is not to allow a coalition to use any players in its complement when connectingto the source. This would yield a game with the same cost for the grand coalition, but larger costs for the othercoalitions. The core of this variant contains the core of our mcse game and so all algorithms presented in thispaper yield core elements of the variant.

Because the set DGK and the ERO value are not defined in function of a game but rather from the mcseproblem itself, they are independent of the game which one chooses. Moreover, for both variants, the irreduciblecore coincides with the set DGK .

All allocations introduced here lie in the irreducible core of the mcse problems. In order to get the wholecore of the corresponding games, one needs to use weights of edges that are not used in any minimum costspanning extension. In general, it is still an open problem to compute the whole core of an mcse game directlyfrom the weights of the edges, even if the attention is restricted to mcst games. Faigle, Kern, Fekete, andHochstattler (1996) proved that it is an NP -hard problem to decide whether a given vector is a member of thecore. Because of the polynomial equivalence of membership testing and optimizing linear functions relative toa polyhedron (see Grotsel, Lovasz, and Schrijver (1988)), it is therefore doubtful whether there exists efficientways to obtain the core of a mcst game.

Finally, it would be interesting to find non-cooperative games of which equilibria sustain the cooperativesolutions presented here.

Acknowledgements : Inspiring discussions with Harry Aarts, Peter Borm, Ruud Jeurissen, MichaelMaschler, Gert-Jan Otten, Jos Potters and Hans Reynierse about the subject of the paper are gratefully ac-knowledged.

7 Appendix

In this appendix, we prove theorem 2.8 and proposition 2.10. First, we need a few lemmas.

Lemma 7.1 For all sequences E chosen by algorithm 2.3 the constructed extension Eτ is an mcse for theproblem M≡ < N, ∗, w,E >.

Proof : There are τ + 1 = |N∗/E| components in < N∗, E >, at each stage two components are connected,so after stage τ , the resulting graph is connected and no new cycles have been introduced. Assume that theextension Eτ constructed is not minimal in cost, i.e. there exists a set of edges E with |E| = τ , such that< N∗, E ∪ E > is a connected graph, and ∑

e∈E

w(e) <∑e∈Eτ

w(e). (7.1)

Let the sequence E = (e1, . . . , eτ ) consist of the edges in E ordered by non-decreasing weight. Equation 7.1implies there exists a smallest t ≤ τ such that et

′= et

′for 1 ≤ t′ < t and et 6= et. Because et is a minimum

weight edge that does not introduce a cycle in < N∗, E ∪ Et−1 > =< N∗, E ∪ {e1, . . . , et−1} > and et does not

20

introduce a cycle in < N∗, E ∪ Et−1 >, it follows that w(et) ≤ w(et). Consider the end points i and j of et.They have to be connected in < N∗, E ∪ E >, hence there exists a path from i to j in < N∗, E ∪ E >. Butnot all edges in this path can be present in the graph < N∗, E ∪ Et−1 >, otherwise et would introduce a cycle.Hence there is an edge e ∈ E \Et−1 in this path that comes later in E than et, so e costs at least w(et), whichis at least w(et). Now E′ := (E \ {e}) ∪ {et} is a spanning extension of < N∗, E > such that E′ does not costmore than E, and E′ has one edge more in common with Eτ . Repeating this process enough times shows thatEτ does not cost more than E. This is a contradiction, hence the assumption that the algorithm 2.3 does notlead to an mcse is wrong. 2

In order to prove that DE is a subset of the core of the associated mcse game if E is constructed byalgorithm 2.3, we need to compare the outcome of the algorithm 2.3 applied to related mcse problems.

Suppose we have an mcse problemM≡ < N, ∗, w,E > and an edge e = {i, j} connecting the component C∗of the source with the component Cj of some player j in the graph < N∗, E >. Define E := E ∪ {e}. Consider

the mcse problem M = < N, ∗, w, E >. Distinguish the graphs, components, edges and allocations used inalgorithm 2.3 applied to the problemsM and M by giving those in the latter problem a tilde. With this setup,we prove two lemmata and a corollary, which we need to prove theorem 2.8.

Lemma 7.2 For every sequence of choices E = (e1, . . . , eτ ) in the algorithm 2.3 applied toM, one can find an

s ≤ τ such that the sequence E ≡ (e1, . . . , eτ−1) := (e1, . . . , es−1, es+1, . . . , eτ ), obtained by deleting the edge es

from E , is a sequence of edges that can be obtained by algorithm 2.3 applied to M and that satisfies

1. (N∗/(E ∪ Et)) \ {Cti , Ctj} = (N∗/(E ∪ Et)) \ {Cti} and Cti ∪ Ctj = Cti for all t < s,

that is, as long as t < s, the graphs < N∗, E ∪ Et > and < N∗, E ∪ Et > have the same components, ex-cept for the components of i and j in< N∗, E ∪ Et >, which are connected to each other in< N∗, E ∪ Et >.

2. N∗/(E ∪ Et) = N∗/(E ∪ Et−1) for all t ∈ {s, . . . , τ},that is, after stage s, the components of < N∗, E ∪ Et > coincide with the components of < N∗, E∪Et−1 >at the previous stage.

Proof : We prove the statements by induction on t. For t = 0: E ∪ E0 = E = E ∪ {e} = E ∪ E0 ∪ {e}, henceCti = Cti ∪ Ctj and (N∗/(E ∪ Et)) \ {Cti , Ctj} = (N∗/(E ∪ Et)) \ {Cti}. Hence case 1 holds.

If case 1 holds at stage t− 1, look at the effect of adding et to Et−1. Two cases can occur.

a et 6= e and adding the edge et does not introduce a cycle in the graph < N∗, E ∪ Et−1 >. Now et

is a cheapest edge which does not introduce a cycle in < N∗, E ∪ Et−1 > and any edge which does notintroduce a cycle in < N∗, E ∪ Et−1 >, does not introduce a cycle in < N∗, E ∪ Et−1 >. Hence et is alsoa cheapest edge that does not introduce a cycle in < N∗, E ∪ Et−1 >. This means et is a legitimate choicefor et. Consequently, case 1 still holds at stage t.

b et = e or adding the edge et does introduce a cycle in the graph < N∗, E ∪ Et−1 >. This meanset connects the components Ct−1i and Ct−1i of < N∗, E ∪ Et−1 >. Then Cti = Ct−1i ∪ Ct−1j = Ct−1i and

the other components are unchanged, so (N∗/(E ∪ Et)) \ {Cti} = (N∗/(E ∪ Et−1)) \ {Ct−1i , Ct−1j } =

(N∗/(E ∪ Et−1)) \ {Ct−1i }. Hence N∗/(E ∪ Et) = N∗/(E ∪ Et−1), and case 2 holds for stage t.

Suppose case 2 holds for stage t − 1. Then the edge et = {k, l} is a legitimate choice for et−1 (it doesnot introduce a cycle, and has minimal cost among the edges satisfying this). Hence, Ctk = Ct−1k ∪ Ct−1l =

Ct−2k ∪ Ct−2l = Ct−1k , which implies N∗/(E ∪ Et) = N∗/(E ∪ Et−1), and case 2 holds for stage t as well.Hence, if the first stage at which case 2 holds is called s, we see that case 1 holds for t < s and case 2 holds

for t ≥ s.Note that N∗/(E ∪ Eτ ) = {N∗} = N∗/(E ∪ Et−1), hence case 2 holds at stage τ , so s ≤ τ . 2

Lemma 7.3 Let M and M be as above, let E be a sequence of choices made by algorithm 2.3 applied to M,and let F be valid for E . Let the sequence E and the stage s be as defined in lemma 7.2. Then there exist(tk)k∈N , such that the sequence F = (f1, . . . , fτ−1) defined by

f tk :=

f tk if t < min{tk, s}τ∑

t′=tk

f t′

k if t = tk < s

f t+1k if tk ≥ t ≥ s

0 if t > tk

for all t ∈ {1, . . . , τ − 1} and all k ∈ N

21

is valid for E . In formula: F ∈ V E (M).

Proof : For all k ∈ N , define

tk := min{t | there exists a path from k to ∗ in < N∗, E ∪ Et >},

where Et is the edge set resulting in stage t in the algorithm 2.3 applied to M. Note that tk = 0 for all k ∈ Cj ,the component of < N∗, E > connected to C∗ by the edge e. Hence f tk = 0 for all stages t if k ∈ Cj . This meansthe players in Cj do not contribute to any edge. We now prove the lemma in three steps.

1. For t ≤ τ − 1, let et = {k, l} and let Ct = Ct−1k ∪ Ct−1l be the component in the graph < N∗, E ∪ Et >formed by the addition of et. Then

∑m∈Ct f

tm = 1. To prove this, we distinguish several cases:

• Suppose et is not incident to Ct−1∗ , the component of ∗ in < N∗, E ∪ Et >. Then tm = tk > t for allm ∈ Ct, hence

– if t < s then f tm = f tm form ∈ Ct. Moreover, Ct = Ct, the component in the graph< N∗, E ∪ Et >formed by the addition of et = et. Hence,∑

m∈Ctf tm =

∑m∈Ct

f tm = 1

by the assumption on F .

– if t > s then f tm = f t+1k for m ∈ Ct. Moreover, et = et+1 and Ct = Ct+1, the component in the

graph < N∗, E ∪ Et+1 > formed by the addition of et+1. Hence,∑m∈Ct

f tm =∑

m∈Ct+1

f t+1m = 1

by the assumption on F .

• Suppose that et is incident to Ct−1∗ . Then one of k and l, say k, lies in Ct−1∗ . This means tm = tl = tfor all m ∈ Ct−1l and tm < t for all m ∈ Ct−1k . Hence,

– if t < s, then f tm =∑τt′=t f

t′

m for m ∈ Ct−1l and f tm = 0 for m ∈ Ct−1k . This implies∑m∈Ct

f tm =∑

m∈Ct−1l

f tm

=∑

m∈Ct−1l

τ∑t′=t

f t′

m

=∑

m∈Ct−1l

(

τ∑t′=1

f t′

m −t−1∑t′=1

f t′

m).

Now Ct−1l = Ct−1l is a union of a number, say p, of components C1, . . . , Cp of the graph< N∗, E >.Remember that for all q ∈ {1, . . . , p}: ∑

m∈Cq

τ∑t′=1

f t′

m = 1,

hence ∑m∈Ct−1

l

τ∑t′=1

f t′

m =

p∑q=1

∑m∈Cq

τ∑t′=1

f t′

m = p (7.2)

and as Ct−1l =⋃pq=1 Cq contains exactly those players that contributed to the p − 1 edges in

{e1, . . . , et−1} that connect the components (Cq)pq=1 into Ct−1l ,

∑m∈Ct−1

l

t−1∑t′=1

f t′

m = p− 1. (7.3)

Equations 7.2 and 7.3 imply ∑m∈Ct

f tm = p− (p− 1) = 1.

22

– if t ≥ s, then f tm = f t+1m , et = et+1 and Ct = Ct+1, the component in the graph < N∗, Et+1 >

formed by the addition of et+1. Hence,∑m∈Ct

f tm =∑

m∈Ct+1

f t+1m = 1

by the assumption on F .

2. For each component C ∈ (N∗/E) that does not contain ∗: C is also a component of the graph < N∗, E >(because < N∗, E > and < N∗, E > differ only in the component of the source). Moreover, tk = tl for allk, l ∈ C and

• if tk < s for all k ∈ C, then

∑k∈C

τ−1∑t=1

f tk =∑k∈C

tk∑t=1

f tk

=∑k∈C

(

tk−1∑t=1

f tk + f tkk )

=∑k∈C

(

tk−1∑t=1

f tk +

τ∑t=tk

f tk)

=∑k∈C

τ∑t=1

f tk

= 1.

The first equality follows because f tk = 0 for t > tk, the third equality because of the definition of

f tkk , and the fifth by the assumptions on F .

• if tk ≥ s for k ∈ C, then C is not connected to ∗ in < N∗, E ∪ Es−1 > = < N∗, E ∪ Es >. Hence,according to F , nobody in C contributes to es, which is an edge incident to Cs∗ . This implies fsk = 0for all k ∈ C. But then ∑

k∈C

τ∑t=1

f tk =∑k∈C

tk∑t=1

f tk

=∑k∈C

(

s−1∑t=1

f tk +

tk∑t=s

f t+1k )

=∑k∈C

tk+1∑t=1

f tk

= 1.

3. Furthermore, tk ≤ t if t ≤ τ − 1 and k ∈ Ct−1∗ , the component of ∗ in < N∗, E ∪ Et−1 >. Hence, f tk = 0by definition.

Steps 1, 2 and 3 imply F ∈ V E (M). 2

Corollary 7.4 Let M and M be as above, let E be a sequence of choices made by algorithm 2.3 applied toM, and let F be valid for E . Then

xE ,Fk (M) ≥ xE ,Fk (M) for all k ∈ N,

where E is as defined in lemma 7.2 and F in lemma 7.3.

Proof : xE ,Fk (M) =∑tkt=1 f

tkw(et).

• If tk = 0 thentk∑t=1

f tkw(et) = 0 ≤ xE ,Fk (M)

23

• If 0 < tk < s (with s as defined in lemma 7.2) then

tk∑t=1

f tkw(et) =

tk−1∑t=1

f tkw(et) + f tkk w(etk)

=

tk−1∑t=1

f tkw(et) + (

τ∑t=tk

f tk)w(etk)

≤tk−1∑t=1

f tkw(et) +

τ∑t=tk

f tkw(et)

=

τ∑t=1

f tkw(et)

= xE ,Fk (M).

.

The second equation follows from the definition of F and the inequality holds because E is ordered bynon-decreasing weights.

• If tk ≥ s then k is not contained in Cs−1∗ = Cs∗ , so k is not allowed to contribute to es, i.e. fsk = 0. Hence,

tk∑t=1

f tkw(et) =

s−1∑t=1

f tkw(et) +

tk∑t=s

f t+1k w(et+1)

=

s−1∑t=1

f tkw(et) +

τ∑t=s+1

f tkw(et)

=

τ∑t=1

f tkw(et)

= xE ,Fk (M).

2

This now enables us to prove theorem 2.8.Theorem 2.8 For any mcse problemM, for any sequence of choices E = (e1, . . . , eτ ) in the algorithm 2.3 applied

to M and any sequence of fractions F that is valid for E the allocation xE ,F , as defined in equation 2.1, is a

core-allocation of the mcse game (N, cM) associated with M.Proof : Take any coalition S ⊆ N . We have to prove

∑i∈S xi ≤ c(S). Construct for coalition S a minimum

cost extension E′ containing only edges between components of < N∗, E > containing members of S∗, such thatS is connected to the source in < N∗, E∪E′ >. Let p be one less than the number of components of < N∗, E >containing members of S∗. Then |E′| = p and the only difference between < N∗, E > and < N∗, E ∪ E′ > isthat the component C ′∗ of ∗ in < N∗, E ∪ E′ > is a union of the component C∗ of ∗ and other componentsof < N∗, E >. Construct the nested sequence E = E0 ⊂ · · · ⊂ Ep = E ∪ E′ such that Eq \ Eq−1 consistsof exactly one edge which connects the component of the source ∗ in < N∗, Eq−1 > to another component of< N∗, Eq−1 >. Consider the mcse problems Mq =< N∗, ∗, w,Eq >, where q varies from 0 to p, and note that

for any q > 0, lemmata 7.2 and 7.3 and corollary 7.4 are applicable to the pairMq−1 andMq = Mq−1. Define

E0 = E , F0 = F and for 1 ≤ q ≤ p, define Eq = Eq−1 and Fq = Fq−1 recursively given Eq−1 and Fq−1, as inlemmata 7.2 and 7.3. Then by corollary 7.4,

xEq,F qk (Mq) ≤ x

Eq−1,F q−1

k (Mq−1)

for all k ∈ N and all 1 ≤ q ≤ p. Summing over q ∈ {1, . . . , p}, we obtain

xEp,Fpk (M′) = x

Ep,Fpk (Mp) ≤ xE0,F0

k (M0) = xE ,Fk (M)

for all k ∈ N . Summing over k ∈ N \ S yields∑k∈N\S

xEp,Fpk (M′) ≤

∑k∈N\S

xE ,Fk (M). (7.4)

Denoting (e1p, . . . , eτ−pp ) := Ep, we see that the graph < N∗, E ∪ E′ ∪ {e1p, . . . , eτ−pp } > is a spanning extension

for M. On the other hand, < N∗, E ∪ {e1, . . . , eτ} > is a minimum cost spanning extension for M, hence∑e∈E′∪{e1p,...,e

τ−pp }

w(e) ≥∑

e∈{e1,...,eτ}

w(e) = c(N) =∑k∈N

xE ,Fk (M). (7.5)

24

Now by definition of E′, ∑e∈E′

w(e) = c(S) (7.6)

and by efficiency of xEp,Fp(M′) (cf. lemma 2.4),∑e∈{e1p,...,e

τ−pp }

w(e) =∑

k∈N\S

xEp,Fpk (M′). (7.7)

Plugging equations 7.6 and 7.7 into inequality 7.5 and using inequality 7.4, we obtain

c(S) +∑

k∈N\S

xEp,Fpk (M′) ≥

∑k∈N

xE ,Fk (M)

=∑k∈S

xE ,Fk (M) +∑

k∈N\S

xE ,Fk (M).

≥∑k∈S

xE ,Fk (M) +∑

k∈N\S

xEp,Fpk (M′),

(7.8)

which is equivalent to

c(S) ≥∑k∈S

xE ,Fk (M).

As we proved in lemma 2.4 that xE ,F (M) is efficient, it is a core element of (N, cM). 2

We now give a proof of proposition 2.10.Proposition 2.10 For any mcse problem M, for any E and E constructed by the algorithm 2.3 applied to M,

DE (M) = DE (M).

Proof : First we prove that DE is independent of the order of E . Suppose that E and E are two sequencesconstructed in algorithm 2.3 applied toM, both leading to the same mcs extension E′, i.e. E and E differ onlyin their order. Because in algorithm 2.3, the edges in E and E are ordered by non-decreasing cost, this meansthat E equals E except for edges of the same cost that are swapped. If more than two edges are swapped, it ispossible to construct a series E = E0, . . . , Ep = E of sequences all leading to the same edge set E′, with for anyq ≤ p, Eq equal to Eq−1 except for exactly two subsequent edges with the same cost that are swapped.

So it suffices to prove DE = DE for E = (e1, . . . , et, et+1, . . . , eτ ) and E = (e1, . . . , et+1, et, . . . , eτ ), for somet < τ , with w(et) = w(et+1). Two cases have to be distinguished:

1. the components Ct and Ct+1 formed by adjoining et and et+1 to < N∗, E ∪ Et > and < N∗, E ∪Et+1 >,

respectively are disjoint. For F ∈ V E , define F = (f1, . . . , fτ ) by

fs = fs if s 6∈ {t, t+ 1},f t = f t+1,

f t+1 = f t.

Obviously, F ∈ V E and xE ,F (M) = xE ,F (M).

2. the components Ct and Ct+1 are not disjoint. Then we are in the situation drawn in figure 3: Ct consistsof two components C1 and C2, connected by et, and Ct+1 is formed by connecting C3 to Ct via et+1.Without loss of generality, we suppose et+1 is incident to C2.

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et et+1

Figure 3: et links C1 to C2 and et+1 links C1 ∪ C2 to C3.

25

Now let F ≡ (f1, . . . , fτ ) ∈ V E be a valid sequence of fraction vectors, and define F ≡ (fs)τs=1 by

fsk = fsk if s 6∈ {t, t+ 1} or k 6∈ C1 ∪ C2 ∪ C3,

f tk =

0f t+1k + gkf t+1

if k ∈ C1,if k ∈ C2,if k ∈ C3,

f t+1k =

f tk + f t+1k

f tk − gk0

if k ∈ C1,if k ∈ C2,if k ∈ C3,

where g ∈ IRC2 is a vector such that ∑k∈C2

gk =∑k∈C1

f t+1k

f tk ≥ gk ≥ 0 for all k ∈ C2.

(7.9)

The system of equations (7.9) are feasible, because F ∈ V E implies∑k∈C2

f tk = 1−∑k∈C1

f tk ≥∑k∈C1

f t+1k =

∑k∈C2

gk.

One easily sees that F is valid for E and that

xE ,Fk =

τ∑t=1

f tkw(et) =

τ∑t=1

f tkw(et) = xE ,Fk .

Hence DE is independent of the order of E , so we define

DE′ := DE

for any sequence E constructed by the algorithm 2.3 that leads to E′. To prove that DE′ = DE for allminimum cost spanning extensions E′ and E of M, we have to know more about the structure of the set of allmcses of M. Now constructing an mcse for the mcse problem M is equivalent to constructing an mcst on theassociated mcst problem < NE , ∗E , wE > (cf. definition 3.1). It is well known that for any two mcsts < N∗, T >and < N∗, T > of an mcst problem < N, ∗, w >, for every edge e ∈ T \ T , there exists an edge e ∈ T \ T suchthat < N∗, T ∪ {e} \ {e} > is again a minimum cost spanning tree.

Suppose that < N∗, E ∪ E′ > and < N∗, E ∪ E > are two mcses for M. Then with E′E and EE as definedin equation 3.2, note that < N∗E , E

′E > and < N∗E , EE > are mcsts of < NE , ∗E , wE >.

Now for every e′ ∈ E′\E and e′E defined as in equation 3.1, it holds that either e′E ∈ E′E\EE or e′E ∈ E′E∩EE .In the former case, there exists an edge {C,D} ∈ EE \ E′E such that

< N∗E , E′E ∪ {{C,D}} \ {e′E} >

is also a minimum cost spanning tree. By definition of EE there exists an edge e ∈ E \ E′ with eE = {C,D}and w(e) = wE(e) = wE(e′E) = w(e′). In the latter case, there exists an edge e ∈ E \ E′ with eE = e′E , so econnects the same components as e′. Because both E and E′ are mcses, w(e) = w(e′). So in both cases, weobtain that

E ∪ E′ ∪ {eE} \ {e′E}is a minimum cost spanning extension of M, which differs one edge less from E than E′ does.

Hence, to prove that DE′ is independent of the mcse E′, it suffices to prove that DE′ = DE for two mcsesE′ and E of M with |E′ \ E| = 1.

Because E′ and E are minimum cost extensions, the edge e′ ∈ E′ \ E and the edge e ∈ E \E′ have to havethe same cost. Now order E′ and E by non-decreasing cost into sequences E = (e1, . . . , es−1, e′, es+1, . . . , eτ )

and E = (e1, . . . , es−1, e, es+1, . . . , eτ ) where s equals the number of edges in E′ with cost not greater than

w(e′) = w(e). Then w(et) > w(e′) for all t > s, and moreover E and E are two sequences that can beconstructed by algorithm 2.3 applied to M.

Consider the graph < N∗, E ∪ {e1, . . . , es−1, e′} >. As the next edge es+1 has greater cost than e, it hasto be the case that adding e would introduce a cycle. But adding e to < N∗, E ∪ {e1, . . . , es−1} > does notintroduce a cycle. This means that e′ and e connect the same two components of < N∗, E ∪ {e1, . . . , es−1} >(see figure 4). Hence the components of the graphs < N∗, E ∪ Et > and < N∗, E ∪ Et > are the same for all t,

which implies that V E = V E . Together with w(e′) = w(e), this implies DE′ = DE = DE = DE . 2

26

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e′

e

Figure 4: e′ and e both link C1 to C2.

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