irreduciblecore and equal remaining obligations rule of minimum cost spanning tree games-2014

8/13/2019 IrreducibleCore and Equal Remaining Obligations rule of Minimum cost spanning tree games-2014

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On the irreducible core and the equal remaining obligations rule of

minimum cost spanning extension problems

by

Vincent Feltkampi

Maastricht School of Management, POBox 1203, 6201BE Maastricht, The Netherlands

Stef Tijs

CentER and Econometrics Department, Tilburg University,

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

and Shigeo Mutoii

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.

95-96?

Abstract

Minimum cost spanning extension problems are generalizations of minimum cost spanning tree problems

in which an existing network has to be extended to connect users to a source. This paper generalizes the

definition of irreducible core to minimum cost spanning extension problems and introduces an algorithm

generating all elements of the irreducible core. Moreover, the equal remaining obligations rule, a one-point

refinement 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 case

in 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 minimaltotal 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 either

one or zero.iEmail: [email protected]. This author was sponsored during the major part of this research by the Foundation for

the Promotion of Research in Economic Sciences, which is part of the Nehterlands Organization for Scientific Research (NWO).iiThis author wishes to acknowledge the Canon Foundation in Europe Visiting Research Fellowship which made his visit to

CentER possible.


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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 Kruskals (1956) algorithm.It is proved that the set of allocations generated is a subset of the core of the associated mcse game and that

is is independent of the extension that is constructed.Section 3 generalizes the definition of the irreducible core, proposed in Bird (1976) for minimum cost spanning

tree 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 Birds 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 Birds 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 asetEof edges. An edgee incident with two verticesi and j is identified with{i, j}1. For a graphG =< V, E >and a set WV,

E(W) :={e E| e W}

is the set of edges linking two vertices in Wand for a subset E ofE,

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 andv =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 {il1, 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 setW is a connected componentof the graph < V, E > if no superset ofW is connected. We will usually say component when we mean connected component. IfW V, the set ofconnected components of the graph < W,E(W) > is denoted W/E. A connected graph is a graph < V, E >withV connected in< V, E >. Atreeis 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 Tare two disjoint coalitions, wedenote (x, y) the vector with coordinates

(x, y)k =

xk ifk Syk ifk T .

Furthermore, for two coalitions S Tand a vectorx IRT, we denotexS the restriction ofx to S. The symbol1S is used to denote the vector in IR

N with coordinates

1S,k=

1 ifk S0 ifk N\ S.

For any coalition S, the simplex S is defined by

S :={xIRS+|iS

xi= 1}.

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


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$100

2

$10

$20

1

3

$40

4 $705

The weights of links.

2

1

3

45

{4, 5} is already constructed.

Figure 1: A simple mcse problem

2. t:= t + 1.

3. At staget, givenEt1, choose a cheapest edge etE Et1 such that the graph< N, E Et1 {et}>does not contain more cycles than the graph < N, E Et1 >.

4. DefineEt :=Et1 {et}.

5. Ift < , go to step 2.

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

In lemma 35 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, . . . , e) whichconsists of the edges of a minimum cost spanning extension ordered by non-decreasing costs can be constructedwith algorithm 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 Dijkstras algorithm generates when generating this mcst. This suggests looking forallocations associated with the sequences generated by Kruskals 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 := |N/E| 1, we associatean allocation with every sequence E = (e1, . . . , e) 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 Naccording to a fraction vectorft 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, . . . , et1} > creating acomponentCt of the graph < N, E {e1, . . . , et}>. Only players in Ct contribute to the cost ofet.

The component Ct1 of the source in the graph < N, E {e1, . . . , et1} > constructed before stage t

does not contribute to the cost ofet.

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 VE(M) of sequences of fraction vectors validfor the sequence E in M by


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VE(M) :=

(f1, . . . , f )[0, 1]|N|

kCt

ftk = 1 for all t

andkC

s=1

fsk = 1 C N/Ewith C

and

ftk = 0 for all t, k s.t. k Ct1

.

For a sequence F (f1, . . . , f ) valid forE inM, define the allocation

xE,F(M) :=

t=1

ftw(et) IRN (2.1)

and define the set DE(M) by

DE(M) :={xE,F | F VE}. (2.2)

If no confusion can occur, we drop the argument M.The intuition behind the first rule is that at the momentet 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 dont know yet which player outsideCt 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 4 For all mcse problemsM, for all sequencesE= (e1, . . . , e) such that E :={e1, . . . , e}is an mcse

ofM and all Fvalid for E, the allocation x := xE,F(M) is efficient:

iNxi=

eEw(e).

Proof : Validity ofF implies every edge et E is paid for by the component Ct it constructs.

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

xE

:F 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: asequencef1, . . . , f is valid for e1, . . . , e in M if and only if at every stage t it satisfies

the componentCt constructed at staget 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 Ct1 of the source pay nothing.

In formula, this gives

kCt

ftk = 1,

kC

ts=1

fsk 1 for all C N/E,

ftk = 0 if k Ct,

ftk = 0 if k Ct1

(2.3)

for every staget.

Example 5 For an mcst problem T < N, , w >, Prim and Dijkstras 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, . . . , et1} > 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,


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{, 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 ofe1 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 componentCt

consists of the component Ct1 and the newly connected playert. Because the component of the source does

not contribute to the cost ofet, 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 toion the unique path in the tree from i to the source. This allocation is precisely Birds tree allocationassociated with the mcst < N, T >, denoted T (cf Bird (1976)).

Example 6 Computing the extreme points of the set of valid fraction vectors for the sequence E constructed

in example 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 inSto the source. The cost of this extension is the worthcM(S)of coalition Sin 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 ofSor the source. Let us consider an example to clarifywhat we mean.

123

$4

$3

$1

$3

$1

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

Example 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

eE

w(e)| S CE

and E contains only edges between

components containing members ofS


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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 3 is acore element of the mcse game.

Theorem 8 For any mcse problem M, for any sequence of choices E= (e1, . . . , e) in the algorithm 3 applied to

M and any sequence of fractions Fvalid for E the allocationxE,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 9 For any sequenceEof 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 Fwhich is valid for E with x = xE,F(M) Core(N, c).

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

Proposition 10 For any mcse problem M, for any E andEconstructed by the algorithm 3 applied to M,DE(M) =D

E(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 3, we define for an mcse problem M:

DGK

(M) :=DE(M)

for any sequence E obtained by the algorithm 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 11 Given an mcse problem < N, , w , E >, we define the associated mcst problem < NE, E, wE>as follows: NEconsists of the components of< N

, E > which do not contain the source , the new sourceE

is the component of< N, E > which contains the original source and wEis defined by

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

for all components Cand 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)

whereCi andCj are the components of< N

, E >containing players i and j, respectively.


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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 ifF is an mcse of the mcse problem < N, , w , E >, then the tree < NE, FE > is anmcst of the associated mcst problem < NE, E, wE >. Conversely, if< N

E, T > is an mcst of the associated

mcst problem, then there exists an mcse F withFE=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 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 12 Given an mcse problem M = < N, , w , E > and an mcse E, define the irreducible coreIC(M, E) of M with respect to E as follows: consider the set Var(M, E) of all mcse problems obtainedfrom M by varying the weight w(e) of edges e E E, that still have E as mcse. Now IC(M, E) is theintersection of 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 setEof 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

} ifCi =Cjw({i, j}) otherwise

(3.3)

whereCiand Cj are the components ofi and j , respectively, in the graph< N, E >. Then the following holds:

Lemma 13 The irreducible core IC(M, E) of a mcse problem M = < N, , w , E >coincides with the core of

the game (N, c) = (N, c

).

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 extension

of the problem< N, , w , E >. Nowe connects two components of< N, E >; hencew(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 ofM is included in the core of the game(N, c).

On the other hand, note that for an edge e = {i, j} E connecting two componentsCiand 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 e

and replacing e bye 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) =

eEw(e) = cM(N). Hence, Core(c) Core(cM

) for allM Var(M, E), which implies that the core ofc is included in the irreducible core ofM. Together with thefirst part of the proof, this proves the theorem.

In order to analyze the structure of the irreducible core of an mcse problem, we relate its structure to thestructure of the irreducible core of the associated mcst problem of definition 11, using the concept of marionettes.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 allS N. Considering players to be marionettes of themselves turns being marionettes into an equivalencerelation, we denote it by . For any playeri, the set of marionettes ofi is denoted by Si.


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Definition 14 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(

SCS) for all C N

. Hence a player in the marionette-reduced game consistsof 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 playerUN, take one representative player jUUand defineT ={jU |UN

}. Define the subgame (T, cT) by

cT(U) =c(U)

for all UT. For every playeri in T, there is a unique player Si in N satisfyingjSi =i and for every player

U inN, there is a unique player jU inT satisfyingSjU =U. Furthermore this bijection between the players ofT andN turns out to be an isomorphism between the games (N, c) and (T, cT):

c(C) = c(SC

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

cT(U) = c(U) = c(iU

Si) = c({Si| i U})

for all coalitions C N and UT.

Lemma 15 If (N, c) is a game, and (N, c) is its marionette-reduced game, their cores are related as follows:

1. ifx Core(N, c), then y Core(N, c), where y IRN

is defined by yS=

iSxi for all S N.

2. ify Core(N, c) IRN

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

iSxi 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 Sand assign

xi:= yS ifi = 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 any

coalition T N, there exists a subset T ofT, such that

iTSi =

iTSi, where the right-hand side is adisjoint union. Hence,

c(T) = c(iT

Si) = c(iT

Si) = c({Si| i T

})

iT

ySi =iT

x(Si) = x(iT

Si)

x(T),

which implies x satisfies the requirements.Now take any x IRN+ satisfying yS= iSxi for all S N. Then

x(N) =SN

x(S) =SN

yS=y(N) =c(N) =c(N)

and for any coalition T, take again a subset T such that

iTSi =

iTSi, where the right-hand side is adisjoint union. Then becausex is non-negative and y Core(N, c),

x(T) iT

jSi

xj =iT

x(Si) =iT

ySi

c({Si| i T}) = c(T) = c(T)

Hence,x Core(N, c).

Lemma 16 For a given mcse problemM = < N, , w , E >with associated minimum-cost spanning tree prob-

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

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


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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 playersNEin 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(SC

S).

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

cM(SC

S) =c(C).

HencecT(C) =c(C), which concludes the proof.

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

Proposition 17 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>) whereyS :=iS

xiS NE}

Proof : This follows easily from lemmata 15 and 16 and the fact that an mcse is transformed into an mcst bythe transition from mcse problem to associated mcst problem.

Corollary 18 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 3.

Lemma 19 LetM < 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 3 applied to M are core allocations of the associated game (N, cM). Since to

prove 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)

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

Lemma 20 Let T =< N, , w > be an mcst problem and let < N, T > be an mcst for T. Then Birds treeallocation T lies in the setDGK(T).

Proof : The number of edges in T equals n:= |N|. Consider any sequence E= (e1, . . . , en) of edges obtained

by ordering the edges ofTby non-decreasing cost. Define F= (f1, . . . , f n) by

fti :=

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


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It follows thatF VE(T) and that

T =xE,F DE =DGK(T).

Theorem 21 LetT be an mcst problem. Then DGK

(T) = IC(T).

Proof : It follows from lemma 19 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 problemT, with reduced weight function defined byequation 3.3. By proposition 20, these Bird allocations lie in DGK(T). NowDGK(T) equalsDGK(T), becausethe set DGK is obtained by only considering the weights of edges in an mcst, and these edges have the sameweight inT as in T(cf. Aarts and Driessen (1993)). Moreover DGK(T) is convex, hence

IC(T) = conv hull{T |Tis an mcst ofT } DGK(T) =DGK(T).

Corollary 22 LetM 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 17 that the vector y IRNE , defined by yS =

iSxi for all SNE, lies in IC(T)(= D

GK(T)).Hence, there exists a sequenceE= (e1, . . . , e) of edges leading to an mcst ofTand a sequenceF= (f1, . . . , f )

of fraction vectors valid for E, such that y = xE,F. Now for each edgeet = {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, . . . , e) is a sequence leading to an mcse ofM. DefineF= (f1, . . . , f) byfti =

ftCi

xiyCi

ifyCi >0

0 ifyCi = 0.

for all t and all i N, where Ci is the component containing i. ThenF is valid forE. Moreover, for anyplayeri, x

E,Fi = 0 ifyCi = 0, but then also xi= 0, and ifyCi = 0, then

xE,Fi =

t=1

ftCixi

yCi=

xiyCi

t=1

ftCi =xiyCiyCi

=xi.

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

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 3.

Algorithm 23 (Equal remaining obligations solution)input : an mcse problem < N, , w , E >output : an mcse and the ERO value

1. GivenM < N, , w , E >, define

t = 0 the initial stage,

= |N/E| 1 the number of stages,E0 = the initial extension.

2. t:= t + 1.


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3. At staget, givenEt1, choose a cheapest edge etE Et1 such that the graph< N, E Et1 {et}>does not contain more cycles than the graph < N, E Et1 >.

4. If Ct = Ct1i Ct1j is the connected component just formed by connecting the components C

t1i and

Ct1j of the graph< N, Et1 > with the edge et ={i, j}, define the vector ft = (ftk)kNof fractions by

ftk =

1|Ct1

i |

1|Ct| ifk Ct1i and C

t,

1|Ct1

j |

1|Ct| ifk Ct1j and C

t,

1|Ct1

i |

ifk Ct1i and Ct1j ,

1|Ct1

j |

ifk Ct1j and Ct1i ,

0 otherwise.

5. DefineEt :=Et1 {et}.

6. Ift < , go to step 2.

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

8. Define EROE(M) :=

t=1

ftw(et).

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

2 = 1

2, edge{2, 3}, of which player 3 pays 1 1

3= 2

3and players 1 and 2 each pay 1

2 1

3= 1

6,

edge{3, 4}, of which players 1, 2 and 3 each pay 13 15

= 215

while players 4 and 5 pay 12

15

= 310

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 Birds tree allocation rule, that may associate a different allocation with each mcst of an mcst problem.

Proposition 24 For any two sequences of edges E andE chosen by the algorithm 23 applied to an mcseproblemM,

EROE(M) = EROE(M).

The proof is similar to the proof of proposition 10, so we do not give it. This proposition allows us to define

ERO(M) := EROE(M)

for any sequence E constructed by algorithm 23. 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 obligationoi of aplayeri by

oi:=

1|Ci|

if Ci

0 if Ci(4.1)

whereCi is the component of< N, E > containing player i.

For a sequence F = (f1, . . . , f ) of fraction vectors, after a stage t a player i N has paid

st fsi,

while the initial obligation was oi. Hence player isremaining obligationoti satisfies

oti = oi

stfsi =o

t1i f

ti . (4.2)


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Theorem 25 The algorithm 23 has the property that after each stage t, in each component C of the graph< N, E Et >, every player k in the component Chas the same remaining obligation

otk =

1|C| if C,

0 if C. (4.3)

Proof : The proof goes by induction on the staget.

1. After stage zero, fork in a component Cof< N, E >,

otk =ok 0 =

1|C|

if C,

0 if C.

2. Suppose equation 4.3 holds after staget 1. LetCt =Ct1i Ct1j be the connected component formed

at stage t by connecting the components Ct1i and Ct1j of the graph < N

, E Et1 > with the edge

et ={i, j}. Let Ctk and Ct1k be the components of playerk in < N

, E Et > and < N, E Et1 >.Then

otk = ot1k f

tk

= 1|Ct1k | ftk ifk Ct10 0 ifk Ct1

=

1|Ct1

i |

( 1|Ct1

i |

1|Ct| ) ifk Ct1i and C

t

1|Ct1

j |

( 1|Ct1

j |

1|Ct| ) ifk Ct1j and C

t

1|Ct1

i |

1|Ct1

i |

ifk Ct1i and Ct1j

1|Ct1

j |

1|Ct1

j |

ifk Ct1j and Ct1i

1|Ct1

k |

ifk Ct Ct1

0 ifk Ct1

=

1|Ct| ifk C

t

0 ifk Ct1i and Ct1j

0 ifk Ct1j and Ct1i

1|Ct1

k |

ifk Ct Ct1

0 if Ct1k

= 1

|Ctk| if C

tk

0 if Ctk.

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

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 allocationof an mcse problem < N, , w , E > is a vector x IRN+ which satisfies iNxi cM(N). Ineffect, 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


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Definition 26

Eff x is efficientif iN

xi= cM(N).

MC x has the minimal contribution propertyif 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 component

C N/Ethat does not contain the source,iC

xi min{w(e)| e connects two components of< N, E >}.

FSC x has the free for source componentproperty 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 solutionof mcse problems is a map assigning to every mcse problem a set of allocations.

Definition 27

NE A solution is said to benon-empty if

(M) = for all M.

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

Definition 28 Given an mcse problem M < N, , w , E > and an edge e = {i, j} E that connects two

components of< N, E >, define theedge-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 29

Loc is local if for all M, for all x (M), for all minimum cost edge e that when added to < N, E >constructs a new componentC, there exists an xIRC such that

(x, xN\C)(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 is minimum cost edge consistentif for all M < N, , w , E >, for all x (M), for each minimumcost edge e that when added to < N, E > constructs a new component C, for each C satisfyingw(e) xC, it holds that

x xe, (Me),

wherex

e,

:= (w(e), 0

N\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.


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

a MC,

b FSC,

c w(e) xC impliesx xe, (Me) for all C,

it holds thatx(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 30 The irreducible core satisfies the properties NE, MC, Eff, FSC, ECons and CECons.

Proof : Because of the coincidence of the irreducible core with the setDGK, the irreducible core is non-empty

for any mcse problem: there are always valid sequences of fraction vectors for any sequence of edges constructedby the algorithm 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 4. By construction, it is clearthatDGK 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, . . . , e) starting with the edge e1 = e, that is constructed by the algorithm 3. Because the set

DGK(M) is independent of the sequence of edges constructed, there exists a sequence F = (f1, . . . , f ) VE

such that x = xE,F. For any C satisfyingw(e) xC, define1 C by1i :=f1i for all i C. Then

iC

i 1i = 0,

( 1)w(e1)

t=2

ftw(et),

ftw(et) 0 for t 2.

Hence there exist vectors2, . . . , IRC satisfying

ti fti for t 2 and for all i C,

iC

ti = 0 for t 2,

( 1)w(e1) =

t=2

tw(et).

(5.1)

E.g. take t

the projection offt

on the hyperplane with coordinates zero, along the line through the pointss=2 f

sw(es) and ( 1)w(e1), i.e.

t =ft

iCf

ti

iC

s=2 f

siw(e

s)(

s=2

fsw(es) ( 1)w(e1))

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

w(e) =

t=1

tw(et).

It follows that x xe, =

t=2(f

t t)w(et). The first two equations of system (5.1) imply that the sequence

(f2

2

, . . . , f

) is valid for (e2

, . . . , e

). Hence, x xe,

M

e

.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


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a MC,

b FSC,

c w(e)xC impliesx xe, IC(Me) for all C.

We have to prove that x IC(M).Denote byC1 andC2 the two components that are joined by e. The allocationx satisfies FSC; hence if one

of these components (say C1

) contains the source, xi= 0 fori C1

and an C with xC satisfiesi = 0for i C1. For such an (which exists), there exists a sequence (e2, . . . , e

) constructed by the algorithm 3

applied to the problem Me and a sequence (f2, . . . , f ) V(e2,...,e) such that x xe, = x(e

2,...,e),(f2,...,f).So with f defined by

fk :=

k ifk C0 otherwise

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

x= x(e,e2,...,e),(f,f2,...,f) (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 contributex(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 C, by

i=

a1xi/x(C1) ifi C1,(1 a1)xi/x(C2) ifi C2.

Then, w(e) xC. Hence, there exists a mcse{e2, . . . , e} and a sequence (f2, . . . , f ) V(e2,...,e)(M) such

that x xe, =

t=2 ftw(et). Definea21, . . . , a

1 bya

t1 = (1 a

1)

iCfti and a

22, . . . , a

2 bya

t2 =a

1

iCfti .

Then

at1+ at2 =

iC

fti for allt2

a1

+

t=2 a

t

1 = a1

+ (1 a1

)

t=2iCfti = 1

1 a1 +

t=2

at2 = 1 a1 + a1

t=2

iC

fti = 1

Definingg1 = (, 0N\C) and

gti =

at1xi/x(C1) ifi C1at2xi/x(C2) ifi C2

we see that (g1, . . . , g)V(e,e2,...,e)(M). Furthermore,

x(C1) a1w(e) = (1 a1)(x(C) w(e))

= (1 a1)

t=2iCfti w(e

t)

=

t=2

at1w(et).

This implies that fori C1,

xi = xix(C1)

x(C1)

= xix(C1)

(a1w(e) +

t=2

at1w(et))

= g1i w(e) +

t=2 gtiw(e

t).

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

irreducible core satisfies converse edge consistency.


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Lemma 31 If a solution of mcse problems satisfies MC, FSC and ECons, and a solution of mcse problems satisfies NE, FSC and CECons, then (M) (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 problemM < N, , w , E >, where the graph < N, E > is con-nected. Then by FSC, x= 0 for any x (M) and for any x (M). By NE, there has to be an x (M),hence

(M) {0}= (M).

Now suppose the lemma holds for every mcse problem M with p 1 components in the graph < N, E >.Take an mcse problem M such that< N, E > has p components and takex (M). By ECons of, for anyminimum cost edge that constructs a new componentCwhen added to < N, E >, for each C satisfyingw(e) xC, it holds that

x xe, (Me) (Me),

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

Theorem 32 The unique solution of mcse problems that satisfies NE, MC, FSC, ECons and CECons is theirreducible core.

Proof : By proposition 30, the irreducible core has these properties. By lemma 31, if two solutions have theseproperties, they contain each other and hence they coincide.

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

Definition 33

ET a solution satisfies equal treatmentif for every mcse problem M, for all x (M), for each componentCof the original graph < N, E >, and for all pairs of players i and j C,

xi= xj .

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

|C1|iC1

(xi xi) =|C2|iC2

(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 34 The unique solution of mcse problems that satisfies NE, FSC, Loc, Eff, ET and IPCons is theERO value. Here, the ERO value is identified with the solution that assigns the singleton{ERO(M)} to everyM.

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, . . . , e) a sequence of fractions

F = (f1, . . . , f ) constructed by the algorithm 23 such that ERO(M) =xE,F. Moreover, by definition of the

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

xE,Fk x(e2,...,e),(f2,...,f)k =

w(e)(

1

|C1| 1

|C| ) ifk C1,

w(e)( 1|C2| 1|C| ) ifk C2,

0 ifk C.


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Then kC1

(ERO(M) ERO(Me)) = 1 |C1|

|C| =

|C2|

|C|

and similarly, kC2

(ERO(M) ERO(Me)) = |C1|

|C|.

Hence

|C1|kC1

(EROk(M) EROk(Me)) =

|C1||C2|

|C| =|C2|

kC2

(EROk(M) EROk(Me)).

To prove uniqueness, suppose a solution satisfies these six properties. We prove (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 Nand all x (M).Suppose (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 applied to M andMe, for all x (M), for all x (Me), for all i, j C1 we have

xj xj =xi xi =: 1(x,x)

and for all i, j C2 we havexj xj =xi xi=: 2(x,x).

Moreover, by FSC of , ifC1 contains the source, then 1(x,x) = 0 and by locality and efficiency, there exists

an x (Me) such that2(x,x) = w(e)|C2|

. IfC2 contains the source, then similarly one proves that 1(x,x) and

2(x,x) are uniquely determined. If neither ofC1 and C2 contain the source, then by IPCons, there exists anx(Me) such that

|C1|kC1

1(x,x) =|C2|kC2

2(x,x),

and by locality and efficiency,

iC1

1(x,x) + iC2

2(x,x) =iC

(xi xi) =w(e).

Hence,

1(x,x) = w(e)

|C1|2 and 2(x,x) =

w(e)

|C2|2.

So whether C1 orC2 contain the source or not, the numbers 1(x,x) and 2(x,x) are uniquely determined andindependent ofx and x. Now the ERO value also satisfies the six properties and so has these same numbers1(x,x), 2(x,x) characterizing the difference between ERO(M) and ERO(M

e). The induction hypothesis thenimplies

x 1(x,x)1C1 2(x,x)1C2 = x= ERO(Me) = ERO(M) 1(x,x)1C1 2(x,x)1C2

and sox = ERO(M).

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 Dijkstras algorithm appeared to be closely linked to Birdstree allocations. This suggested to look for allocation rules that are related to Kruskals (1956) algorithm forconstructing an mcst. In the present paper we relate the irreducible core to Kruskals 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


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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 the

rule should have.Another way of approaching mcse problems is to define the cost of a coalition S as the minimal cost

of 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 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, and

Hochstattler (1996) proved that it is an N P-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 8 and proposition 10. First, we need a few lemmas.

Lemma 35 For all sequences Echosen by algorithm 3 the constructed extensionE is an mcse for the problemM < N, , w , E >.

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

eEw(e)=< N, E {e1, . . . , et1}> and et does not


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introduce a cycle in < N, E Et1 >, it follows that w(et) w(et). Consider the end points i and j ofet.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 Et1 >, otherwise et would introduce a cycle.Hence there is an edge e E\ Et1 in this path that comes later inEthan et, soe 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, andE has one edge more in common with E. Repeating this process enough times shows thatE does not cost more than E. This is a contradiction, hence the assumption that the algorithm 3 does not

lead to an mcse is wrong.

In order to prove that DE is a subset of the core of the associated mcse game if E is constructed byalgorithm 3, we need to compare the outcome of the algorithm 3 applied to related mcse problems.

Suppose we have an mcse problemM < N, , w , E >and an edgee = {i, j}connecting the componentCof the source with the component Cj of some playerj in the graph < N

, E >. Define E:= E {e}. Consider

the mcse problemM = < N , , w, E >. Distinguish the graphs, components, edges and allocations used inalgorithm 3 applied to the problemsM andM 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 8.

Lemma 36 For every sequence of choices E = (e1, . . . , e) in the algorithm 3 applied to M, one can find an

s such that the sequence

E (e1, . . . , e1) := (e1, . . . , es1, es+1, . . . , e), obtained by deleting the edge es

fromE, is a sequence of edges that can be obtained by algorithm 3 applied toMand that satisfies1. (N/(E Et)) \ {Cti , C

tj}= (N

/(E Et)) \ {Cti} and Cti C

tj =

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 ofi andjin < N, E Et >, which are connected to each other in < N, E Et >.

2. N/(E Et) =N/(E Et1) for all t {s , . . . , },that is, after stages, the components of< N, E Et >coincide with the components of< N, EEt1 >at the previous stage.

Proof : We prove the statements by induction on t. Fort= 0: E E0 = E=E {e}= E E0 {e}, henceCti =C

ti C

tj 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 Et1. Two cases can occur.

a et= e and adding the edge et does not introduce a cycle in the graph < N, E Et1 >. Now et

is a cheapest edge which does not introduce a cycle in < N, E Et1 > and any edge which does notintroduce a cycle in < N, E Et1 >, does not introduce a cycle in < N, E Et1 >. Henceet is alsoa cheapest edge that does not introduce a cycle in < N, E Et1 >. This meanset 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 Et1 >. This meanset connects the componentsCt1i and C

t1i of< N

, E Et1 >. Then Cti =Ct1i C

t1j =

Ct1i and

the other components are unchanged, so (N/(E Et))\ {Cti} = (N/(E Et1))\ {Ct1i , C

t1j } =

(N/(E Et1)) \ {Ct1i }. Hence N/(E Et) =N/(E Et1), 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 et1 (it doesnot introduce a cycle, and has minimal cost among the edges satisfying this). Hence, Ct

k = Ct1

k Ct1

l =

Ct2k Ct2l =

Ct1k , which impliesN/(E Et) =N/(E Et1), and case 2 holds for staget as well.

Hence, if the first stage at which case 2 holds is called s, we see that case 1 holds for t < sand case 2 holdsforts.

Note thatN/(E E) ={N}= N/(E Et1), hence case 2 holds at stage , so s .

Lemma 37 LetM andMbe as above, letEbe a sequence of choices made by algorithm 3 applied to M, andletFbe valid forE. Let the sequenceEand the stage s be as defined in lemma 36. Then there exist (tk)kN,such that the sequenceF= (f1, . . . , f1) defined by

ftk :=

ftk ift tk

for all t {1, . . . , 1}and all k N


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is valid forE. In formula:F VE(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 3 applied toM. Note that tk = 0 for all k Cj ,the component of< N, E >connected toC by the edgee. Hence ftk = 0 for all stagest ifk Cj . This meansthe players in Cj do not contribute to any edge. We now prove the lemma in three steps.

1. Fort 1, let et ={k, l} and let Ct = Ct1k Ct1l be the component in the graph< N

, E Et >

formed by the addition of et. Then

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

Suppose et is not incident to Ct1 , the component of in < N, E Et >. Thentm = tk > t for all

m Ct, hence

ift < s then ftm = ftmfor m C

t. Moreover, Ct =Ct, the component in the graph < N, E Et >formed by the addition of et =et. Hence,

mCtftm=

mCtftm = 1

by the assumption on F.

ift > sthen ftm = ft+1k for m

Ct. Moreover, et =et+1 and Ct =Ct+1, the component in thegraph< N, E Et+1 >formed by the addition ofet+1. Hence,

mCt

ftm =

mCt+1

ft+1m = 1


Suppose that et is incident to Ct1 . Then one ofk and l, sayk , lies in Ct1 . This meanstm= tl = t

for all m Ct1l and tm< t for all m Ct1k . Hence,

ift < s, then ftm =

t=t f

t

m for m Ct1l and

ftm = 0 for m Ct1k . This implies

mCt

ftm = mCt1

l

ftm

=

mCt1l

t=t

ft

m

=

mCt1l

(

t=1

ft

mt1t=1

ft

m).

NowCt1l =Ct1l is a union of a number, sayp, of components C1, . . . , C pof the graph < N

, E >.Remember that for all q {1, . . . , p}:

mCq

t=1

ft

m= 1,

hence mCt1

l

t=1

ft

m =

pq=1

mCq

t=1

ft

m = p (7.2)

and as Ct1l =p

q=1 Cq contains exactly those players that contributed to the p 1 edges in

{e1, . . . , et1} that connect the components (Cq)pq=1 into C

t1l ,

mCt1lt1

t=1ft

m = p 1. (7.3)

Equations 7.2 and 7.3 imply mCt

ftm = p (p 1) = 1.


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ift s, then ftm =ft+1m , e

t =et+1 and Ct =Ct+1, the component in the graph < N, Et+1 >formed by the addition ofet+1. Hence,

mCt

ftm =

mCt+1

ft+1m = 1


2. For each componentC (N/E) that does not contain: Cis 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 < sfor all k C, then

kC

1t=1

ftk =kC

tkt=1

ftk

=kC

(

tk1t=1

ftk+ ftkk )

= kC

(

tk1

t=1

ft

k+

t=tk

ft

k)

=kC

t=1

ftk

= 1.

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

ftkk , and the fifth by the assumptions on F.

if tk s for k C, then C is not connected to in < N, E Es1 > = < N, E Es >. Hence,

according toF, nobody in Ccontributes to es, which is an edge incident to Cs. This implies fsk = 0

for all k C. But then

kC

t=1

ftk = kC

tkt=1

ftk

=kC

(s1t=1

ftk+

tkt=s

ft+1k )

=kC

tk+1t=1

ftk

= 1.

3. Furthermore, tk t ift 1 and k Ct1 , the component of in < N

, E Et1 >. Hence, ftk = 0by definition.

Steps 1, 2 and 3 implyF VE(M). Corollary 38 LetM andM be as above, letEbe a sequence of choices made by algorithm 3 applied to M,and letFbe valid for E. Then

xE,Fk (M) xE,Fk (

M) for all k N,whereE is as defined in lemma 36 andF in lemma 37.Proof : x

E,

F

k (

M) =

tkt=1 f

tkw(e

t).

Iftk = 0 thentkt=1

ftkw(et) = 0 xE,Fk (M)


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If 0< tk < s(with s as defined in lemma 36) then

tkt=1

ftkw(et) =

tk1t=1

ftkw(et) + ftkk w(e

tk)

=

tk1t=1

ftkw(et) + (

t=tk

ftk)w(etk)

tk1t=1

ftkw(et) +

t=tk

ftkw(et)

=

t=1

ftkw(et)

= xE,Fk (M).

.

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

Iftk s thenk is not contained in Cs1 =C

s, sok is not allowed to contribute to e

s, i.e. fsk = 0. Hence,

tkt=1

ftkw(et) =

s1t=1

ftkw(et) +

tkt=s

ft+1k w(et+1)

=s1t=1

ftkw(et) +

t=s+1

ftkw(et)

=

t=1

ftkw(et)

= xE,Fk (M).

This now enables us to prove theorem 8.

Theorem 8For any mcse problem M, for any sequence of choices E= (e1, . . . , e) in the algorithm 3 appliedto 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

iSxi c(S). Construct for coalition S a minimum

cost extensionE containing only edges between components of< N, E >containing members ofS, such thatSis connected to the source in < N, E E >. Letp be one less than the number of components of< N, E >containing members ofS. 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 \ Eq1 consistsof exactly one edge which connects the component of the source in < N, Eq1 > to another component of< N, Eq1>. Consider the mcse problems Mq =< N

, , w , E q >, where qvaries from 0 to p, and note that

for any q >0, lemmata 36 and 37 and corollary 38 are applicable to the pair Mq1 and Mq =Mq1. DefineE0 =E, F0 =Fand for 1 q p, define Eq = Eq1 and Fq =Fq1 recursively given Eq1 and Fq1, as inlemmata 36 and 37. Then by corollary 38,

xEq,Fqk (Mq) x

Eq1,Fq1k (Mq1)

for all k Nand all 1 q p. Summing over q {1, . . . , p}, we obtain

xEp,Fpk (M

) =xEp,Fpk (Mp) x

E0,F0k (M0) =x

E,Fk (M)

for all k N. Summing over k N\ SyieldskN\S

xEp,Fpk (M

)

kN\S

xE,Fk (M). (7.4)

Denoting (e1p

, . . . , epp

) :=Ep, we see that the graph < N, E E {e1

p

, . . . , epp

}> is a spanning extensionforM. On the other hand, < N, E {e1, . . . , e}> is a minimum cost spanning extension for M, hence

eE{e1p,...,epp }

w(e)

e{e1,...,e}

w(e) =c(N) =kN

xE,Fk (M). (7.5)


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Now by definition ofE, eE

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

and by efficiency ofxEp,Fp(M) (cf. lemma 4),

e{e1p,...,epp }w(e) =

kN\Sx

Ep,Fpk (M

). (7.7)

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

c(S) +

kN\S

xEp,Fpk (M

) kN

xE,Fk (M)

=kS

xE,Fk (M) +

kN\S

xE,Fk (M).

kS

xE,Fk (M) +

kN\S

xEp,Fpk (M

),

(7.8)

which is equivalent to

c(S) kS

xE,Fk (M).

As we proved in lemma 4 that xE,F(M) is efficient, it is a core element of (N, cM).

We now give a proof of proposition 10.Proposition 10For any mcse problem M, for any E andEconstructed by the algorithm 3 applied to M,

DE(M) =DE(M).

Proof : First we prove that DE is independent of the order ofE. Suppose thatE and

E are two sequences

constructed in algorithm 3 applied to M, both leading to the same mcs extension E, i.e. E and

Ediffer only

in their order. Because in algorithm 3, the edges in E andEare ordered by non-decreasing cost, this meansthatE equals Eexcept 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 = Eof sequences all leading to the same edge set E, with for anyq p, Eq equal toEq1 except for exactly two subsequentedges with the same cost that are swapped.

So it suffices to prove DE =DE forE= (e1, . . . , et, et+1, . . . , e) andE= (e1, . . . , et+1, et, . . . , e), for some

t < , with w(et) =w(et+1). Two cases have to be distinguished:

1. the componentsCt andCt+1 formed by adjoininget andet+1 to< N, E Et >and < N, E Et+1 >,

respectively are disjoint. For F VE, defineF= (f1, . . . , f) byfs = fs ifs {t, t + 1},

ft = ft+1,

f

t+1

= f

t

.

Obviously,F VE and xE,F(M) =xE,F(M).2. the componentsCt andCt+1 are not disjoint. Then we are in the situation drawn in figure 3: Ct consists

of two components C1 and C2, connected by et, and Ct+1 is formed by connecting C3 to C

t via et+1.Without loss of generality, we suppose et+1 is incident to C2.

C1

C2

C3

et et+1

Figure 3: et linksC1 to C2 and et+1 links C1 C2 to C3.


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25

Now letF (f1, . . . , f )VE be a valid sequence of fraction vectors, and defineF (fs)s=1 byfsk = f

sk ifs {t, t + 1} or k C1 C2 C3,

ftk =

0ft+1k + gkft+1

ifk C1,ifk C2,ifk C3,

ft+1k = f

tk+ f

t+1k

ftk gk0

ifk C1,ifk C2,ifk C3,

whereg IRC2 is a vector such that kC2

gk =kC1

ft+1k

ftk gk 0 for all k C2.

(7.9)

The system of equations (7.9) are feasible, because F VE implies

kC2 ftk = 1 kC1 f

tk kC1 f

t+1k = kC2 gk.

One easily sees thatF is valid forEand thatxE,Fk =

t=1

ftkw(et) =

t=1

ftkw(et) =xE,Fk .

HenceDE is independent of the order ofE, so we define

DE

:=DE

for any sequenceEconstructed by the algorithm 3 that leads to E. To prove that DE

=DE for all minimumcost spanning extensions E and E ofM, we have to know more about the structure of the set of all mcsesof M. Now constructing an mcse for the mcse problem M is equivalent to constructing an mcst on the

associated mcst problem< NE, E, wE>(cf. definition 11). 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 EE andEEas defined

in equation 3.2, note that < NE, EE>and< N

E,

EE> are mcsts of< NE, E, wE>.Now for everye E\EandeEdefined as in equation 3.1, it holds that either e

EE

E\EEor e

EE

EEE.

In the former case, there exists an edge {C, D} EE\ EE such that

< NE, EE {{C, D} } \ {e

E}>

is also a minimum cost spanning tree. By definition ofEE there exists an edge e E\ E with eE = {C, D}

and w(e) = wE(e) = wE(eE) = w(e

). In the latter case, there exists an edge e E\ E with eE = eE, so e

connects the same components as e. Because both E and E are mcses, w(e) = w(e). So in both cases, we

obtain that E E {eE} \ {eE}

is a minimum cost spanning extension ofM, 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 EofM with |E \ E|= 1.

BecauseE and Eare minimum cost extensions, the edgee E \ Eand the edge e E\ E have to havethe same cost. Now orderE and Eby non-decreasing cost into sequences E = (e1, . . . , es1, e, es+1, . . . , e)

andE = (e1, . . . , es1, e, es+1, . . . , e) where s equals the number of edges in E with cost not greater thanw(e) = w(e). Then w(et) > w(e) for all t > s, and moreover E andE are two sequences that can beconstructed by algorithm 3 applied to M.

Consider the graph < N, E {e1, . . . , es1, 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, . . . , es1} > does not

introduce a cycle. This means that e and e connect the same two components of< N, E {e1, . . . , es1} >(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 VE =VE. Together with w(e) =w(e), this implies DE =DE =DE =DE.


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C1

C2

e

e

Figure 4: e

and e both link C1 to C2.

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