On demand responsiveness in additive cost sharing

Journal of Economic Theory 125 (2005) 1–35www.elsevier.com/locate/jet

On demand responsiveness in additive cost sharing

Hervé Moulina,∗, Yves SprumontbaDepartment of Economics, Rice University, MS 22, P.O. Box 1892, Houston, TX 77251-1892, USAbDépartement de Sciences Economiques, Université de Montréal, C.P. 6128, succursale centre-ville,

Montréal H3C 3J7, Canada

Received 31 October 2003; final version received 28 May 2004Available online 11 September 2004

Abstract

Underpartial responsibility, the ranking of cost shares should never contradict that of demands.TheSolidarityaxiomsays that if agenti demandsmore,j should not paymore ifkpays less. It characterizesthequasi-proportionalmethods, sharing cost in proportion to ‘rescaled’ demands.Full responsibilityrules out cross-subsidization for additively separable costs. Restricting solidarity

to submodular cost characterizes thefixed-flowmethods, containing the Shapley–Shubik and serialmethods.The quasi-proportional methods meet—but most fixed-flow methods fail—Group Monotonicity: if

a group of agents increase their demands, not all of them pay less. Serial cost sharing is an exception.© 2004 Elsevier Inc. All rights reserved.

JEL classification:C 71; D 63

Keywords:Cost sharing; Additivity; Cross-subsidization; Group monotonicity; Solidarity; Serial method;Shapley–Shubik method

1. Introduction

The traditionalmodel of cooperativegames iswell suited todiscusscost-sharingproblemswhere the demands of the participating agents are fixed (inelastic). Shapley[14] andWeber[23] showed how the simple axioms of additivity (with respect to cost) and dummy (anagent with zero marginal cost pays nothing) determine individual cost shares as a fixedaverage of marginal costs.

∗ Corresponding author. Fax: +713-3486-329.E-mail addresses:[email protected](H. Moulin), [email protected](Y. Sprumont).

0022-0531/$ - see front matter © 2004 Elsevier Inc. All rights reserved.doi:10.1016/j.jet.2004.05.003

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2 H. Moulin, Y. Sprumont / Journal of Economic Theory 125 (2005) 1–35

Following Aumann and Shapley’s seminal contribution[1], the additive theory of costsharingwasextended to variable individual demands: agentsi = 1, . . . , nwant a quantityxiof goodi, and the problem is to split the total costC(x1, . . . , xn) using no other informationthan the cost functionC.Applications range from the pricing of utilities such aswater, phoneor electricity, where the level of individual consumption affects total cost, to the sharing ofwaiting time at a congested server[16,6], and the division of highway construction costsbetween taxpayers[8].A simple requirement of demand responsiveness has played a key role in the recent

literature on axiomatic cost sharing.Monotonicity,introduced in[10] and developed in[5], stipulates that every user’s cost share should be nondecreasing in her own demand.On the normative side, it expresses a weak form of responsibility for one’s own demand.If goods are freely disposable, monotonicity also has a positive interpretation: it preventsmanipulation by artificial (and wasteful) increases of individual demands.In this paper, we propose a handful of new demand responsiveness axioms, some of them

strengthening monotonicity, and explore their impact in the additive theory of cost sharing.Ourmodel and our threemain axioms are presented in Section 2.We have a finite number ofusers, individual demands are nonnegative integers, and the cost function is nondecreasing,zero for zero demands, and otherwise arbitrary. We always assume that individual costshares are nonnegative and depend additively upon the cost function.Group monotonicityrequires that if a group of agents simultaneously increase their de-

mands, not all of them pay less. Under the same premises,Strong groupmonotonicitystatesthat the total cost share allocated to the group should not decrease. Both properties havedual normative/positive interpretations similar to those of monotonicity. If side-paymentsare not feasible, group monotonicity is sufficient to prevent manipulations by a coordinatedartificial increase of demands; if they are feasible, we need the latter, stronger property.Our third main axiom issolidarity. It requires that a demand change by one agent affects

the share of allotheragents in the same direction: no one pays more if someone pays less.The recent literature on fair division explores the solidarity idea in a number of allocationproblems[21,20], but this is its first application to cost sharing.A natural family of methods meeting all demand responsiveness properties above are the

quasi-proportionalmethods computingi’s cost share ass(xi )∑s(xj )

C(x),wheresis an arbitrarynondecreasing “scaling function”. These simple and familiar methods only take the actualcost into account, and divide it in proportion to rescaled individual demands. Theorem 1in Section 3 characterizes this family by combining solidarity, monotonicity, and one moreaxiom calledstrong ranking. The latter requires cost shares to be ranked in the same orderas the corresponding demands, for any cost function, symmetric or not.StrongRanking is the defining axiom of the “partial responsibility” theory of cost sharing

[11]. Agents, while responsible for their demands, are not responsible for the asymmetriesof the cost function. This justifies some cross-subsidization. We submit that this viewpointinspires cost sharing rules for many publicly provided services: disabled customers are notcharged more for transportation services, the same stamp buys delivery of mail to centralor remote areas, and so on.The alternative view that agentsare responsible for cost asymmetries is the basis of

the more familiar “full responsibility” theory. This view leads to the principle ofno crosssubsidizationpervading the natural monopoly literature[2,15], and to the related dummy

H. Moulin, Y. Sprumont / Journal of Economic Theory 125 (2005) 1–35 3

axiom in the model of cooperative games. In our setting the relevant formulation is theseparabilityaxiom: if the cost function is additively separable (C(z) =∑i ci(zi) for all z),each agent should pay her stand-alone costci(xi).The impact of our demand responsiveness conditions under full responsibility is the

subject of Sections 4 and 5. Proposition 2 states that neither solidarity nor strong groupmonotonicity is compatible with separability. On the other hand, group monotonicity is avery demanding yet achievable property. The simple method charging incremental costsaccording to a fixed ordering of the users is an example. More generally, afixed-pathmethod charging incremental costs according to a fixed ordering of the different units ofdemand, is group-monotonic. However group monotonicity is typically not preserved byconvex combinations of cost sharing methods. For instance the Shapley–Shubik method[17], applying the Shapley value to the stand-alone cooperative game at the given demandprofile, is not group-monotonic. Thus this property yields a powerful critique of both theoldest cost-sharing method in the literature, and of the most popular one, the Aumann–Shapley method (the latter violates even monotonicity).Proposition 3 in Section 4 is an important positive result, allowing us to construct “rea-

sonable” group-monotonic methods in the full responsibility approach. Any convex com-bination of “nearby” fixed-path methods is group-monotonic as well. The most naturalmethod constructed in this fashion is thesubsidy-free serialmethod[10], adapting serialcost sharing to our discrete model.If solidarity is out of reach in the full responsibility theory, a weaker version of this

property is feasible when the cost function is submodular. In that case an increase in useri’s demand is a positive externality on other users, and oursubmodular solidarityaxiomin Section 5 rules out changes of opposite signs in their cost shares. We introduce the richfamily of fixed-flowmethods, corresponding to convex combinations of fixed-pathmethods,and containing for instance theShapley–Shubik and subsidy-free serialmethods.Theorem2characterizes this family by the combination of separability, monotonicity, and submodularsolidarity.Section 6 compares our results under partial and full responsibility. While the quasi-

proportional methods satisfy solidarity, the cross effects of a demand shift may not havethe expected sign. For instance, ifC is supermodular andxi increases, agentj’s cost share,s(xj )∑s(xk)

C(x),maygodown.We introduce thepositiveexternalitiesandnegativeexternalitiesaxioms: if agenti’s demand raises, all agentsj other thani benefitwhencosts are submodular,and suffer when costs are supermodular. We show in Proposition 4 that these properties areessentially incompatible with partial responsibility: no cost-sharing method satisfies strongranking and positive externalities, and the only method satisfying additivity, strong rankingand negative externalities divides costs equally no matter what. By contrast, the axiomsare easy to meet under full responsibility: the fixed-flow methods as well as the Aumann–Shapley method satisfy positive and negative externalities (Proposition 5).Section 7 takes a second look at the partial responsibility theory. Moulin and Sprumont

[11] introducedweak separability, ruling out cross-subsidization when it is not implied bystrong ranking, namely when the cost function takes the formC(z) = ∑

i c(zi). Whilethe quasi-proportional methods violate weak separability, the axiom is compatible withpartial responsibility, i.e., with strong ranking.The simplest example is thecross-subsidizingserialmethod[18]. This method is also group-monotonic (Proposition 6).Yet strong group


monotonicity is incompatible with the combination of weak separability and strong ranking(Proposition 7).

2. The model and the demand responsiveness axioms

Each agenti ∈ N = {1, . . . , n} demands an integer quantityxi ∈ N = {0,1, . . .} of apersonalized good. The cost of meeting the demand profilex ∈ NN must be split amongthe members ofN. A cost function is a mappingC : NN → R+ that is nondecreasingand satisfiesC(0) = 0; the set of such mappings is denotedC. A (cost-sharing) method�assigns to each problem(C, x) ∈ C×NN a vector of nonnegative cost shares�(C, x) ∈ RN+such that

∑�i (C, x) = C(x).We call�(., x) a (cost-sharing) method atx.

We use the following notation. Vector inequalities are written� , <,�. For anyx, x′ ∈NN, [x, x′] = {z ∈ NN | x�z�x′}, [x, x′[= [x, x′]\{x′}; ]x, x′] and]x, x′[ are definedsimilarly. We letN(x) = {i ∈ N | xi > 0} and writen(x) = |N(x)| . For anyS ⊆ N ,we denote byxS ∈ NS the restriction ofx to Sand we writex(S) = ∑

i∈S xi .We defineeS ∈ NN by eSi = 1 if i ∈ S and 0 otherwise.We sometimes writei for {i}, ij for {i, j}, and−S forN\S. If i ∈ N andC ∈ C, wedefine�iC : NN → R+ by�iC(z) = C(z+ei)−C(z).Finally, we let� = {y ∈ RN+ |

∑i∈N yi = 1}.

The requirement that cost shares be nonnegative is part of our definition of a method.Nonnegativity is a minimal form of demand responsiveness. It does rule out well-knownprocedures such asequal split beyond stand-alone costs,

�i (C, x) = C(xi,0−i )+ 1n(C(x)−

∑j∈N

C(xj ,0−j )),

and themarginal cost procedure

�i (C, x) = xi�iC(x − ei)+ 1n(C(x)−

∑j∈N

xj�jC(x − ej )),

where�iC(x − ei) is defined to be zero ifxi = 0.Throughout the paper, we restrict our attention to additive methods.

Additivity: For anyC,C′ ∈ C andx ∈ NN, �(C + C′, x) = �(C, x)+ �(C′, x).

This powerful mathematical property is devoid of ethical content. Additive cost-sharingmethods are very convenient in practice.When production can be decomposed into the sumof several independent processes (like research, production and marketing; or constructionand maintenance), applying the method to each subprocess and adding the resulting costshares is equivalent to applying the method to the consolidated cost function. The properlevel of application of the method is not a matter of dispute.We now define four demand responsiveness conditions. The first one is the familiar

monotonicity[10] requiring individual cost shares to be monotonic in the correspondingdemands.


Monotonicity: For all C ∈ C, x, x′ ∈ NN, and i ∈ N , {xi < x′i andxj = x′j for allj ∈ N\{i}} ⇒ {�i (C, x)��i (C, x

′)}.When communication between agents is easy, demand coordination is a concern. We

propose a group version of monotonicity requiring that, when several agents increase si-multaneously their demands, not all of them pay less.

Group monotonicity: For anyC ∈ C, any x, x′ ∈ NN, and any nonemptyS ⊆ N,

{xi < x′i for all i ∈ S andxi = x′i for all i ∈ N\S} ⇒ {�i (C, x)��i (C, x′) for at least

onei ∈ S}.A natural strengthening of group monotonicity evaluates the impact of demand increases

followed by side-payments. When money transfers between agents are feasible, John maybe willing to strategically raise his demand if the corresponding drop in some other agents’shares more than compensates the increase in John’s share. Strong group monotonicityprevents such manipulations.

Strong group monotonicity: For anyC ∈ C, x, x′ ∈ NN, any nonemptyS ⊆ N andi ∈ S, {xi < x′i andxj = x′j for all j ∈ N\{i}} ⇒ {∑j∈S �j (C, x)�

∑j∈S �j (C, x

′)}.Clearly stronggroupmonotonicity also rulesout theprofitability of a coordinated increase

by severalmembers ofSfollowed by side-payments withinS.Our last demand responsiveness axiom requires that all members of a group should be

treated similarly whenever a demand change occurs outside the group.

Solidarity: For anyC ∈ C, any i ∈ N, and anyx, x′ ∈ NN, {xj = x′j for all j ∈N\{i}} ⇒ {�j (C, x)��j (C, x

′) for all j ∈ N\{i}, or �j (C, x)��j (C, x′) for all j ∈

N\{i}}.We conclude this section by noting the following connection between our axioms.

Proposition 1. Every cost-sharingmethod satisfying additivity, solidarity, andmonotonic-ity satisfies strong group monotonicity.

Proof. The subsetD = {C ∈ C | C(z) ∈ {0,1} for all z ∈ NN } of cost functions takingthe values 0 or 1 plays a central role in this and several subsequent proofs, in particular thatof Theorem A.1. EveryC ∈ C is a nonnegative linear combination of cost functions inD.Thus an additive method� meets strong group monotonicity if and only if for allC ∈ Dandx ∈ NN,∑

j∈S�j (C, x)�

∑j∈S

�j (C, x + ei) for all S ⊆ N and alli ∈ S. (1)

If C(x) = 0, (1) is automatically satisfied. IfC(x) = 1, thenC(x + ei) = 1 and (1) isequivalent to

�j (C, x + ei)��j (C, x) for all distincti, j ∈ N,which is clearly implied by the combination of solidarity and monotonicity.�


3. Demand responsiveness and partial responsibility: the quasi-proportionalmethods

This section is devoted to thequasi-proportional methods discussed in theIntroduction.

Definition 1. A method� is quasi-proportionalif �i (C, x) = s(xi )∑j∈N s(xj )

C(x) for all

i, C, andx, wheres : N → R+ is nondecreasing ands(1) > 0 (with the conventionthat�(C,0) = 0). Examples include theegalitarianmethod�i (C, x) = 1

nC(x) and the

proportionalmethod�i (C, x) = xix(N)

C(x).

These methods meet additivity, monotonicity, and solidarity, hence, by Proposition 1,satisfy all the demand responsiveness axioms of Section 2. No information about “counter-factual” costs is used in computing the cost shares, which are therefore completely insensi-tive to the asymmetries in the cost structure. In particular, agents who ask more pay more,regardlessof the cost function.

Strong ranking: For allC∈ C, x ∈ NN, andi, j ∈ N, {xi�xj } ⇒ {�i (C, x)��j (C, x)}.This defines the “partial responsibility” theory of cost sharing in which agents are held

responsible for the size of their own demand but not for the idiosyncrasies of the costfunction. Note that the egalitarian method barely satisfies strong ranking: it violates thestrict version requiring that the ranking of cost shares be identical to that of demands whenthe cost function is strictly monotonic. The proportional method is more responsive todemands; its simplicity makes it very popular in practice.We now show that additivity, solidarity, monotonicity and strong ranking essentially

characterize the quasi-proportional methods. The precise statement requires a couple ofadditional definitions. Notice first that the egalitarian method possesses a property slightlystronger than solidarity,

Strict solidarity: For anyC ∈ C, any i ∈ N, and anyx, x′ ∈ NN, {xj = x′j for allj ∈ N\{i}} ⇒ {(i) �j (C, x) < �j (C, x

′) for all j ∈ N\{i} or (ii) �j (C, x) = �j (C, x′)

for all j ∈ N\{i} or (iii) �j (C, x) > �j (C, x′) for all j ∈ N\{i}},

which the proportional method fails because agents demanding zero are never affected bychanges in others’ demands. On the other hand, the proportional method meets

Zero cost for zero demand: For anyC ∈ C, any i ∈ N, and anyx ∈ NN, {xi = 0} ⇒{�i (C, x) = 0},which the egalitarian method obviously fails. This axiom is compelling if agents are notresponsible for the demands of others.1

Finally, we observe that both the egalitarian and the proportional method meet

1While automatically satisfied in themore familiar “full responsibility” theory (where it follows from additivityand separability: see Section 4), zero cost for zero demand must be explicitly required in the partial responsibilitytheory.


Positive cost for positive demand: For anyC ∈ C, anyi ∈ N, and anyx ∈ NN, {xi > 0andC(x) > 0} ⇒ {�i (C, x) > 0},which is a strong but simple interpretation of responsibility for one’s demand.2 We are nowready to state our first main result.

Theorem 1. (i) Assumen�4. A cost-sharing method� satisfies additivity, strong rank-ing, strict solidarity, and monotonicity if and only if there exists a nondecreasing functions : N → R such thats(0) > 0 and

�i (C, x) =s(xi)∑j∈N s(xj )

C(x) (2)

for all C ∈ C, i ∈ N, andx ∈ NN.

(ii) Assumen�3.A cost-sharingmethod� satisfies additivity, strong ranking, solidarity,monotonicity, zero cost for zero demand and positive cost for positive demand if and onlyif there exists a nondecreasing functions : N → R such thats(0) = 0, s(1) > 0, and(2)holds for allC ∈ C, i ∈ N, andx ∈ NN (with the convention�(C,0) = 0 ).Choosing a positive constanta and settings(t) = t + a for all t, Eq. (2) defines a hybrid

method approaching the egalitarian method whena →∞ and the proportional one whena→ 0.Any such method meets strict solidarity.Four observations are in order.(1) Theorem 1 is a corollary to Theorem A.1 and Lemma A.1, stated and proved in

the appendix, which dispense with strong ranking. Dropping strong ranking allows for afamily including more complicated methods, of which two interesting subclasses are easilydescribed.Consider first the methods�(C, x) = �(x)C(x), where�(x) is an arbitrary mapping

fromNN\{0} to �. We call such methodssimple:they completely ignore the shape of thecost function and reduce the cost-sharing problem to a “general claim arbitration” problem(where the amount to be divided may exceed or fall short of the sum of the claims) asin [7,13]. It follows from the proof of Theorem A.1 that a simple method meets additivity,solidarity andmonotonicity if andonly if it is anasymmetric quasi-proportionalmethod, i.e.,

�i (x) = si(xi)∑j∈N sj (xj )

, (3)

for all i ∈ N, wheresi : N → R+ is nondecreasing andsi(1) > 0.A second interesting subclass meeting our three axioms consists of the methods

�i (C, x) =�

n− 1 C(xi,0−i )+1

n(C(x)−

∑j∈N

�

n− 1 C(xj ,0−j )) (4)

for all i, C, andx, where� ∈ [0,1]. For� = 0, this is just the egalitarian method. Setting� = n − 1 yields “equal split beyond stand-alone costs” introduced in Section 2, which2This axiom is incompatible with the main axiom of the full responsibility theory, separability, because a

dummy agent pays nothing even if his demand is positive. By contrast, positive cost for positive demand is easilymet in the partial responsibility approach.


however does not guarantee nonnegative cost shares. The nonnegativity constraints force� between 0 and 1. These methods split equally the balance abovea fraction of stand-alone costs. The entire family of methods meeting additivity, monotonicity and solidaritycombines features of these methods and the asymmetric quasi-proportional methods: seethe appendix for details.(2)Replacing the combinationof solidarity andmonotonicity by stronggroupmonotonic-

ity (which is implied: recall Proposition 1) allows for many more methods. For instance,the simple methods�(C, x) = �(x)C(x) satisfy strong group monotonicity if and only if�i is nonincreasing inxj for any two distincti, j (which is much less restrictive than theasymmetric quasi-proportional form (3)) and strong ranking if and only if�i (x)��j (x)⇔ xi�xj for anyi, j andx. Characterizing the entire class of methods meeting additivity,strong group monotonicity and strong ranking is a challenging open problem.(3) All methods in Theorem 1 satisfy a requirement stronger than solidarity. When an

arbitrary subset of agents change their demands, the impact on the cost shares of the othersgoes in the same direction: for anyC ∈ C, anyS ⊆ N, and anyx, x′ ∈ NN, {xj = x′j forall j ∈ N\S} ⇒ {�j (C, x)��j (C, x

′) for all j ∈ N\S or �j (C, x)��j (C, x′) for all

j ∈ N\S}.(4) The egalitarian method occupies a special place within the quasi-proportional meth-

ods, because it meets a much stronger version of the solidarity axiom:

Full solidarity: For anyC ∈ C, any i ∈ N, and anyx, x′ ∈ NN, {xi < x′i andxj = x′jfor all j ∈ N\{i}} ⇒ {�(C, x)��(C, x′)}.This axiom strengthens both solidarity and monotonicity. It characterizes the fixed-

proportions methods:a cost-sharing method� satisfies additivity and full solidarity ifand only if there exists� ∈ � such that�(C, x) = C(x)� for all C ∈ C andx ∈ NN.Weomit the easy proof.3

4. Demand responsiveness and full responsibility

We turn to demand responsiveness properties within the familiar “full responsibility”approach, where agents are responsible for the idiosyncrasies of the cost function. Thedefining axiom says that everyone should pay the cost of their own demand whenever thatcost does not depend on others agents’ demands.

Separability: For all C ∈ C, x ∈ NN, {C(z) = ∑i∈N ci(zi) for all z ∈ NN, where

ci(0) = 0 for all i} ⇒ {�i (C, x) = ci(xi) for all i ∈ N}.All quasi-proportional methods violate separability. More fundamentally, the axiom is

incompatible with strong ranking even in the absence of additivity. Suppose costs are ad-ditively separable (C(z) = ∑

ci(xi)) and i’s stand-alone cost exceedsj’s at all levels(ci(t) > cj (t) for all t > 0). If xi = xj , strong ranking imposes equal cost shares fori andj while separability requires thati pays more thanj.

3 Note that ifD ∈ D andD(x) = 1, full solidarity implies�(D, x) = �(D, x + ei ) for all i. This implies firstthat�(�x , x) is independent ofx, then�(D, x) is independent of bothD andx.


Under additivity, separability is equivalent to the more familiardummyaxiom requiringthat dummy agents pay nothing: for allC ∈ C, x ∈ NN, and i ∈ N , {�iC = 0} ⇒{�i (C, x) = 0}. 4The combination of additivity and separability is well understood. Moulin andVohra[12]

offer a representation in terms of flows. Aconservative unit flow—or simply, aflow—to ademand profilex ∈ NN is a mappingf (., x) : [0, x[→ RN+ satisfying the convention thatfi(z, x) = 0wheneverzi = xi, the normalization

∑i∈N fi(0, x) = 1, and the conservation

constraints∑

i∈N fi(z, x) =∑

i∈N(z) fi(z− ei, x) for all z ∈]0, x[. Note that this implies∑i∈N(x) fi(x − ei, x) = 1.

Lemma 1. [12] A method� satisfies additivity and separability if and only if, for everyx ∈ NN , there is a(necessarily unique) flowf (., x) to x such that

�i (C, x) =∑

z∈[0,x[fi(z, x)�iC(z) (5)

for all C ∈ C and all i ∈ N ; we call f the flow representation of�.

An important consequence of this representation is the following property.

Independence of irrelevant costs: For allx ∈ NN andC1, C2 ∈ C, {C1(z) = C2(z) for allz ∈ [0, x]} ⇒ {�(C1, x) = �(C2, x)}.The simplest separable additive methods correspond to flowsf (., x)mapping[0, x[ into

{0,1}N. In this case the entire flow runs along a monotone path from 0 tox in the integergrid. Fix such a path. At each step of the path, the coordinate of exactly one agent increasesby one. Charging to this agent the corresponding cost increase and summing over the entirepath defines a cost-sharing method at the demand profilex. Formally, any sequence ofagents� : {1, . . . , x(N)} → N such that

∣∣�−1(i)∣∣ = xi for eachi ∈ N , defines a monotonepathz� to x by settingz�(0) = 0 andz�(t) = z�(t − 1)+ e�(t) for t = 1, . . . , x(N). Thispath, in turn, generates the flowfi(z, x) = 1 if for somet, z�(t − 1) = z, z�(t) = z + ei,

andfi(z, x) = 0 otherwise. The corresponding cost shares atx are

��i (C, x) =

∑t∈�−1(i)

�iC(z�(t − 1)) (6)

for all i ∈ N andC ∈ C. If, for every demand profilex, �(., x) is generated by a monotonepath tox,wesay that� ispath-generated.Wang[22] shows that amethod satisfies additivityand separability if and only if it is the pointwise limit of convex combinations of path-generatedmethods. Note that if we restrict attention to a bounded subset of demand profiles,the two axioms characterize simply the (fixed) convex combinations of path-generatedmethods.

4 See[12, Corollary 3], and notice that under additivity, their nondummy axiom is equivalent to separability.


Example 1. TheAumann–Shapley5 method�assolves every problem(C, x) by averagingthe cost-share vectors��(C, x) generated along all paths� to x. In the correspondingflow representation,f asi (z, x) equals the proportion of paths tox which go throughz and

z + ei : straightforward computations yieldf asi (z, x) = �(z)�(x−z−ei )�(x) for all i ∈ N and

z ∈ [0, x − ei], where�(z) = z(N)!/∏j∈N zj !.

Example 2. The Shapley–Shubikmethod�ss [17] averages the cost shares computedalong the paths that follow the edges of the cube[0, x[. Letting Ei(x) = {z ∈ [0, x[|∀j ∈ N\i, zj ∈ {0, xj }}, the flow associated with�ss(., x) is given byf ssi (z, x) =(n(x)−n(z,x))!(n(z,x)−1)!

n(x)! if z, z + ei ∈ Ei(x) andf ssi (z, x) = 0 otherwise, wheren(z, x) =∣∣{j ∈ N | zj < xj }∣∣ .

Example 3. The subsidy-free serialmethod�sf [10] averages the cost shares computedalong the paths that follow the “constrained diagonal” tox. Let G(x) = {z ∈ [0, x[|∀i, j ∈ N, (xi�xj ) ⇒ (|zi − zj |�1 or zi = xi < zj )} andG∗(x) = {z ∈ [0, x[|∀i, j ∈ N, (xi�xj ) ⇒ (zi − zj = 0 or zi = xi < zj )}. For eachz ∈ G(x), denote byz(x) the smallestz∗ ∈ G∗(x) such thatz∗ > z and letz(x) be the largestz∗ ∈ G∗(x) suchthatz∗�z. The flowf sf(., x) associated with the serial method�sf(., x) at x is given byf sfi (z, x) = f ssi (z− z(x), z(x)− z(x)) if z, z+ ei ∈ G(x) andf sfi (z, x) = 0 otherwise.

We are now ready to describe the consequences of our four demand responsivenessconditions under full responsibility. First, we note that the two strongest conditions areincompatible with additivity and separability.

Proposition 2. Supposen�3. Then(i) no cost-sharing method satisfies additivity, separability, and solidarity;(ii) no cost-sharing method satisfies additivity, separability, and strong group mono-

tonicity.

Proof. Statement(i): We establish the incompatibility forn = 3; the extension ton�3follows immediately by considering cost functions where all but three agents are dummies.Let � satisfy the three stated axioms and letf be its flow representation as in formula (5).Let x = e12, x′ = e123. DefineC1 ∈ C by C1(z) = 1 if z�e1 or z�e23, andC1(z) = 0otherwise. Similarly, defineC2 ∈ C by C2(z) = 1 if z�e2 or z�e13, andC2(z) = 0otherwise. Note thatC1 = D1 on [0, x], whereD1 is the additively separable functionD1(z) = 1 if z1�1,D1(z) = 0 otherwise. Thus by Independence of Irrelevant Costs andSeparability�(C1, x) = e1.On the other hand�2(C

1, x′) = f2(e3, x′).Asimilar argument

applied toC2 gives�(C2, x) = e2 and�1(C2, x′) = f1(e

3, x′).If �1(C

1, x′) < �1(C1, x) = 1, solidarity implies�2(C1, x′)� �2(C

1, x) = 0, hence�2(C

1, x′) = 0. The latter equality holds as well if�1(C1, x′) = �1(C

1, x) = 1. Thusf2(e

3, x′) = 0. A similar argument involvingC2 instead ofC1 givesf1(e3, x′) = 0. Now

5The original definition and characterizations of this method are in the model with continuous demands[3,9].We present here its counterpart in the discrete model. For two axiomatizations of this discrete version, see[4,20].


flow conservation yieldsf3(0, x′) = 0. By exchanging the roles of the agents, we getsimilarly fi(0, x′) = 0 for i = 1,2, a contradiction to the definition of a conservative unitflow.Statement(ii): The proof mimics that of statement (i): the first paragraph is unchanged;

then apply strong groupmonotonicity toC1 (with i = 3 andS = {1,3}) to get�2(C1, x′)��2(C

1, x) = 0, hencef2(e3, x′) = 0, and toC2 (with i = 3 andS = {2,3}) to getf1(e

3, x′) = 0. The rest is unchanged. �

By contrast to Proposition 2, monotonicity and group monotonicity are compatible withadditivity and separability. Both have a lot of bite. As mentioned in the Introduction, theAumann–Shapley method is not monotonic.6 The Shapley–Shubik method is monotonic,but not group-monotonic. We only prove the latter statement. LetN = {1,2,3} andC bethe cost function:

C(z)= 1 if z�(2,0,1) or z�(1,1,1) or z�(0,2,1),

0 otherwise.

Check that�ss1 (C, (1,1,1)) = �ss2 (C, (1,1,1)) = 13 >

16 = �ss1 (C, (2,2,1)) = �ss2 (C,

(2,2,1)), which contravenes group monotonicity.The characterization of all methods meeting additivity, separability and group mono-

tonicity remains an open problem. In the rest of this section, we construct a fairly richfamily of such methods.We start with thefixed-pathmethods. Any fixed sequence� : N\{0} → N generates

an infinite fixed pathz� : N → NN through the inductionz�(0) = 0, z�(t) = z�(t −1) + e�(t) for t ∈ N\{0}. If �−1(i) is infinite for eachi ∈ N, we call the sequence�unbounded; the corresponding fixed path is unbounded in each coordinate (i.e.,z�i (t) =∣∣{s ∈ �−1(i) | s� t}∣∣ goes to infinity ast grows). To any demand profilex ∈ NN , weassociate a sequence�x : {1, . . . , x(N)} → N by keeping the firstxi occurrences ofeachi : for instance, ifn = 3, x = (1,4,2) and� = 1,2,3,1,2,3,1,2,3, . . . , then�x = 1,2,3,2,3,2,2. The sequence�x in turn generates the monotone pathz�x to x andthe associated cost-sharing method��x (., x) at x via formula (6). Thefixed-path method�� based on� is defined by��(., x) = ��x (., x) for all x.To see why such methods satisfy group monotonicity, fix� andx. For eachi ∈ N let

ti (�, x) be the smallest integert such thatz�i (t) = xi. Let S ⊆ N and consider a demandprofilex′ such thatxi < x′i for all i ∈ S andxi = x′i for all i ∈ N\S. Find the (necessarilyunique) agenti0 ∈ S such thatti0(�, x)� ti (�, x) for all i ∈ S. Because� is a fixed

sequence,��i0(C, x′) is computed along a pathz�x

′to x′ which coincideswith z�x for all

t� ti0(�, x)� ti0(�, x′). Therefore��

i0(C, x)��

i0(C, x′), proving group monotonicity.

Fixed-path methods are asymmetric. To restore symmetry, we could take convex combi-nations of such methods, but the example of the Shapley–Shubik method demonstrates thatin general this operation will not preserve group monotonicity. Interestingly, taking convexcombinations of “sufficiently close” fixed-path methodsdoespreserve group monotonicity.

6A simple counterexample has two agents and the cost function�(1,1)(z) = 1 for allz� (1,1) and�(1,1)(z) = 0otherwise. Then�as(�(1,1), (1,1)) = ( 12 ,

12) and�

as(�(1,1), (2,1)) = ( 13 ,23), a contradiction to monotonicity.


Formally, let� be a fixed unbounded sequence and leti, j ∈ N. Construct�ij : N\{0} →{i, j} by deleting from� all occurrences of agents inN\{i, j}. This sequence generates theprojection ofz� onN{i,j}, denotedz�ij : N\{0} → N{i,j}, in the usual way. We say thattwo fixed unbounded sequences�, � arenearbyif for all i, j ∈ N and allt ∈ N,

{z�{™,j}(t) �= z�{i,j}(t)} ⇒ {z�{™,j}(t + 1) = z�

{i,j}(t + 1)}.

Proposition 3. If {�1, . . . , �K} is a family of unbounded sequences which are pairwisenearby, and�1, . . . , �K are nonnegative numbers summing up to1, then

∑Kk=1 �k��k sat-

isfies additivity, separability, and group monotonicity. In particular, the subsidy-free serialmethod(Example3) satisfies these three axioms.

Proof.By Lemma 1,∑K

k=1 �k��k meets additivity and separability. To check group mono-tonicity, fix C ∈ C, S ⊆ N, andx, x′ ∈ NN such thatxi < x′i for all i ∈ S andxi = x′i forall i ∈ N\S. Let i1 ∈ S be the agent inSsuch thatti1(�1, x)� ti (�1, x) for all i ∈ S. Asexplained previously,��1

i1(C, x)��1

i1(C, x′).

Next we show that��ki1(C, x)��k

i1(C, x′) for all k ∈ {2, . . . , K}. Fix k and recall that

�k is nearby�1. This implies that for alli∈N, eitherti1(�k, x)� ti (�k, x) or, if the oppositestrict inequality is true,

ti (�k, x) < s < ti1(�k, x)⇒ s /∈ �−1k (i).

If i ∈ S, the above property impliesti1(�k, x) < ti(�k, x + ei)� ti (�k, x′). It follows thatti1(�k, x) < ti(�k, x′) for all i ∈ S. Therefore��k

i1(C, x′) is computed along a path tox′

which coincides, for allt� ti1(�k, x),with the path toxalongwhich��ki1(C, x) is computed.

It follows that the subsidy-free serialmethodmeets the three axiomsbecause�sf is simplythe arithmetic average of then!methods generated by the fixed unbounded sequencesp(1),p(2), . . . , p(n), p(1), p(2), . . . , p(n), . . . corresponding to the possible permutationsponN. �

Our companion paper[11] offers a characterization of the subsidy-free serial methodbased on the property called distributivity, and expressing for the composition of costfunctions (with perfect substitute outputs) the same invariance as additivity does for theiraddition.

5. The fixed-flow methods

Within the full responsibility theory,weexplore now twoaxiomsweakening, respectively,strong ranking and solidarity. Separability is incompatible with strong ranking but is clearlycompatible with the following weaker condition:

Ranking: For allC ∈ C, x ∈ NN, andi, j ∈ N, {C is a symmetric function of all itsvariables andxi�xj } ⇒ {�i (C, x)��j (C, x)}.


The restriction to symmetric cost functions is very natural: whenC is symmetric, anydifference in cost sharesmust originate in differences in demands, hence the ranking of costshares should follow that of individual demands. Examples 1, 2, 3 all satisfy ranking.Next, separability is incompatible with solidarity (Proposition 2), but it is consistent

with a restricted version of that axiom. solidarity is very demanding when all types ofproduction externalities are allowed, thus it is natural to limit its application to those caseswhere externalities are clearly signed. For anyC ∈ C, i, j ∈ N, and z ∈ NN, define�ijC(z) = �iC(z + ej ) − �iC(z). The subsets of submodular and supermodular costfunctions, respectively, areCsub= {C ∈ C | �ijC�0 for all distincti, j ∈ N} andCsup={C ∈ C | �ijC�0 for all distincti, j ∈ N}.Submodular solidarity: For anyC ∈ Csub, anyi ∈ N, and anyx, x′ ∈ NN, {xj = x′j for

all j ∈ N\{i}} ⇒ {�j (C, x)��j (C, x′) for all j ∈ N\{i} or�j (C, x)��j (C, x

′) for allj ∈ N\{i}}. Supermodular solidarityis defined by replacingCsubwith Csup in the previousstatement.

Examples 1–3 all satisfy the restricted solidarity properties. Combining either of themwith monotonicity and separability circumscribes a very natural family of methods. Forany flowf (., x) on [0, x[ and anyx′ ∈ [0, x], theprojectionof f (., x) on [0, x′[, denotedpx′f (., x), is definedas follows: for anyi ∈ N andz ∈ [0, x′[writeK = {j ∈ N | zj = x′j }and let

px′fi(z, x)= 0 if i ∈ K,=

∑wK∈[x′K,xK ]

fi((wK, zN\K), x) otherwise,

with the convention that the sum is simplyfi(z, x) if K = ∅. Note thatpx′f (., x) is a flowto x′.For simplicity, we state our characterization result forboundedcost-sharing problems:

we fix x ∈ (N\{0})N and speak of a cost-sharing method restricted to[0, x] if x varies in[0, x]. The translation of all our axioms into this framework is straightforward: one merelyneeds to restrict their application to the demand profiles in[0, x].Definition 2. A cost-sharing method� restricted to[0, x] is afixed-flow methodif there isa flowf (., x) to x such that, for everyx ∈ [0, x], the flowpxf (., x) represents�(., x).Examples include the fixed-path, Shapley–Shubik, and subsidy-free serial methods. One

can check directlyf (., x) = pxf (., x′) for anyx, x′, x�x′, or apply our next result. On

the other hand, the Aumann–Shapley method is not a fixed-flow method. The proof of thefollowing result is in the appendix.

Theorem 2. Let x ∈ (N\{0})N . For any cost-sharing method� restricted to[0, x], thefollowing statements are equivalent:

(i) � is a fixed-flow method;(ii) � satisfies additivity, separability,monotonicity, and submodular solidarity;(iii) � satisfies additivity, separability,monotonicity, and supermodular solidarity.


We conclude this section with three comments.(1) Theorem 2 is tight. For a method satisfying all axioms but additivity, let�i (C, x) =C(xi ,0−i )+1∑

j∈N(C(xj ,0−j )+1)C(x) unlessC is additively separable, in which case we compute costshares by applying separability. A method violating only separability is the proportionalmethod�i (C, x) = xi

x(N)C(x). An example violating only monotonicity is the Aumann–

Shapley method: we prove in Section 6 that it actually satisfies properties stronger thansubmodular solidarity and supermodular solidarity. Finally, we describe a method violatingonly the latter two axioms. Letn = 3 andx = (2,2,2). Consider the two monotone pathsto x defined by the sequences

�1 = 1,2,3,1,2,3,�2 = 1,2,3,2,1,3.

Let f 1, f 2 be the flows tox corresponding to�1, �2. For eachx ∈]0, x], define the flowf (., x) to x by f (., x) = pxf

1 if x = (2,2,1), andf (., x) = pxf2 otherwise. This

generates a method restricted to[0, x] with the desired properties.(2) Fixed-flow methods satisfy a stronger property than dummy, calledstrong dummy:

for all C ∈ C, x ∈ NN, and i ∈ N, {�iC = 0} ⇒ {�i (C, x) = 0 and�j (C, x) =�j (C, (0i , x−i )) for all j ∈ N\i}. This follows by applyingf (., x′) = px′f (., x) tox′ = (0i , x−i ). However the combination of additivity, strong dummy and monotonicity isnot enough to characterize the fixed-flow methods.(3) Theorem 2 suggests interesting open questions. Many fixed-flow methods, including

the Shapley–Shubik and subsidy-free serial methods, meet ranking. Ifn = 2, all fixed-flowmethods on[0, x] wherex1 = x2 and the flowf is symmetric (f1(z, x) = f2((z2, z1), x)

for all z) meet ranking. Surprisingly, this property does not generalize ton�3. Here is anexample of a method based on a symmetric flow tox = x1e

N that violates ranking. Letn = 3, x = (3,3,3), and consider the six monotone paths tox defined by the sequence

1,2,1,2,1,3,2,3,3

and the five sequences obtained from it by permuting agents. Letf be the (fully symmetric)flow on [0, x] obtained by putting an equal weight on these six paths, and let� be theresulting fixed-flow method. Consider the cost function

C(z)= 1 if z�z∗ for somez∗ ∈ Z∗,0 otherwise,

whereZ∗ = {(1,2,2), (2,1,2), (2,2,1), (3,1,1), (1,3,1), (1,1,3)}. The functionC issymmetric in all its variables, and one computes�(C, (3,2,1)) = (12,

16,13), in contra-

diction of ranking. This raises the problem of determining which fixed-flow methods meetranking.Characterizing the class of fixed-flow methods meeting group monotonicity is another

challenging open problem.


6. Comparing the partial and full responsibility approaches

Under submodular or supermodular solidarity, the impact of an increase in agenti’sdemand onj’s cost share must go in the same direction for allj other thani but this directionis a priori unrestricted. It is natural to require more: that allj other thani benefit when costsare submodular, and suffer when costs are supermodular.

Positive externalities: For anyC ∈ Csub, any i ∈ N, and anyx, x′ ∈ NN, {xi < x′iandxj = x′j for all j ∈ N\{i}} ⇒ {�j (C, x)��j (C, x

′) for all j ∈ N\{i}}. NegativeExternalitiesis defined by replacing the subsetCsub with Csup and the inequality sign�with � in the previous statement.

All quasi-proportional methods violate positive externalities and all but one violate nega-tive externalities. The tension between these two axioms and strong ranking issystematic.

Proposition 4. (i) No cost-sharing method satisfies strong ranking and positive externali-ties.(ii) The only cost-sharing method satisfying additivity, strong ranking and negative ex-

ternalities is the egalitarian method. 7

Proof. (i) Consider the cost functionC(z) = 0 if z�2e1, = 1 otherwise. Check it issubmodular (it is a particular case of the function defined by (26) in the appendix). Bynonnegativity of cost shares and positive externalities,�(C, z) = 0 and�(C, z+ e2) = e2,

violating strong ranking. Note that additivity is not invoked.(ii) We focus on the casen = 2 and leave the extension to the reader. Letz �= 0 and

consider the supermodular cost function�z defined by (28) in Step 3 of the proof of Theorem2. Letxbe an arbitrary demand profile. If�z(x) = 0, the nonnegativity of cost shares forces�(�z, x) = (0,0). If �z(x) = 1, distinguish three cases. Ifx1 = x2, strong ranking forces�(�z, x) = (12,

12). If x1 < x2, Negative externalities and the fact that�(�z, (x2, x2)) =

(12,12) imply �2(�

z, x)� 12. Strong ranking then forces�(�

z, x) = (12,12). A symmetrical

argument handles the casex1 > x2. This shows that� is the egalitarian method on the costfunctions�z.As every cost function coincides on every[0, x] with a linear combination ofsuch functions, the claim follows by additivity. �

By contrast, the fixed-flow methods satisfy positive and negative externalities. Manymore separable methods meet these axioms: an important example is theAumann–Shapleymethod. The proof of the following result is in the appendix.

Proposition 5. The fixed-flow methods and the Aumann–Shapley method satisfy positiveexternalities and negative externalities.

To sum up, strong ranking is compatible with solidarity but the direction of the cross-effects may be counter-intuitive; separability is consistent with a limited form of solidarityonly but the cross-effects may be guaranteed to have the expected sign.

7 Note that negative externalities is implied by full solidarity and compare the current result with observation(4) following Theorem 1.


7. Weak separability

This section takes a second look at the partial responsibility theory.We argue in[11] thata sound theory of partial responsibility should perform cross-subsidization to correct forcost asymmetries, and for that purpose only. Subsidization is not justified when the costfunction is symmetric: in such cases, the separability principle still applies. Thus if the costfunction is not only symmetric but also additively separable, each agent will pay her “own”separable cost:

Weak separability: For allC ∈ C, x ∈ NN, {C(z) = ∑i∈N c(zi) for all z ∈ NN } ⇒

{�i (C, x) = c(xi) for all i ∈ N}.All quasi-proportionalmethods violateweak separability. For instance, ifn = 2, x1 < x2,

andC(z) = c(z1)+ c(z2), the proportional method cross-subsidizes agent 2 at 1’s expensewhenc is convex, and vice versa whenc is concave.Yet, weak separability is compatible with strong ranking. A complete description of the

methods meeting additivity, weak separability, and strong ranking is offered in LemmaA.2in the appendix. A key member of that class is the cross-subsidizing serial method definedin [18]. 8

Example 4. For anyx ∈ NN∗ = {x ∈ NN : x1� · · · �xn} and all i ∈ N , definexi =xie

N ∧ x. The cross-subsidizing serialmethod�cs assigns to every problem(C, x) ∈C× NN∗ the vector of cost shares�cs(C, x) = 1

nC(x1)eN + 1

n−1[C(x2) − C(x1)]eN\1+ · · ·+ [C(xn)−C(xn−1)]en. The cost shares for an arbitrary problem obtain by applyingthe formula after reordering the coordinates of the demand profile in nondecreasing order.

Proposition 6. The cross-subsidizing serial method satisfies strong ranking,weak separa-bility, and group monotonicity. It fails solidarity and strong group monotonicity.

We omit the easy proof of these two statements. The violation of strong group mono-tonicity is a consequence of a general incompatibility recorded in the next proposition; seethe appendix for a proof.

Proposition 7. If n�3, no cost-sharing method satisfies additivity, strong ranking, weakseparability, and strong group monotonicity.

We conjecture that a similar incompatibility holds when strong group monotonicity isreplaced with solidarity.

Acknowledgments

Weare grateful for their comments to seminar participants at University College London,CORE Louvain-La Neuve, Sabançi and Koç Universities in Istanbul, University of Sydney,

8 See[11] for a characterization based on the distributivity axiom mentioned at the end of Section 4.


University of Melbourne, Universitat Autònoma de Barcelona, and the 2003 Social ChoiceandWelfare Economics conference inMalaga.Moulin’s researchwas supported by theNSFunder grant SES 011 2032. Sprumont’s research was supported by grants from the SSHRCof Canada and the Spanish Ministry of Education.

Appendix A.

A.1. Shared-flow representation of additive methods

Define asharing rulefor x ∈ NN to be a mappingr(., ., x) : [0, x[×N → �. The vectorr(z, j, x) specifies how the flowfj (z, x) betweenzandz+ ej is shared among the agents.Choosing for each demand profilex a flowf (., x) and a sharing ruler(., ., x), the formula

�(C, x) =∑

z∈[0,x[

∑j∈N

�jC(z)fj (z, x)r(z, j, x) (7)

for all C ∈ C defines an additive cost-sharing method. Keep in mind thatr is vector-valuedwhile �jC andfj are scalar-valued.Conversely, Moulin and Vohra[12] show that all additive methods possess at least one

suchshared-flow representation(f, r). Observe that these methods meet independence ofirrelevant costs.

A.2. Statement and proof of Theorem A.1

Theorem A.1 and Lemma A.1 below constitute a more general characterization resultthan Theorem 1, in which we drop the strong ranking assumption. Denote byR+ andR++the sets of nonnegative and strictly positive real numbers, respectively. For eachi ∈ N,

choose a functionsi : N → R+ such thatsi(xi) > 0 wheneverxi > 0. If x > 0, writeFi(x) = si (xi )∑

N sj (xj )andF(x) = (F1(x), . . . , Fn(x)) Next, choose a functioni assigning

a real number to every pair(xi, ti) ∈ (N\{0})2 such thatti�xi. For every cost-sharingproblem(C, x) ∈ C×NN and eachi ∈ N, define

sai (C, xi) =xi∑ti=1

i (xi, ti)�iC((ti − 1)ei), (8)

which may be viewed as a generalized stand-alone cost for agenti. The formula

�s,i (C, x) = sai (C, xi)+ Fi(x)(C(x)−∑j∈N

saj (C, xj )) (9)

definesanadditive cost-sharingmethodprovidedcost sharesarealwaysnonnegative.Setting�+j (xj ) = max{0, j (xj , tj ) | tj ∈ [1, xj ]} and�−j (xj ) = min{0, j (xj , tj ) | tj ∈ [1, xj ]},


this amounts to imposing the following restrictions onfj , j for all i andx:

Fi(x)

1−∑N\i

�+j (xj )

+ (1− Fi(x))�−i (xi)�0. (10)

This claim follows easily from the inequalities�−i (xi)C(xi,0−i )�sai (C, xi)��+i (xi)C(xi,0−i ); we omit the straightforward details. By construction, we have

Lemma A.1. Everymethod�s, constructedas in(9)and(10)satisfiessolidarity. Ifsi(0) >0 for all i ∈ N , �s, also satisfies strict solidarity.

Monotonicity imposes further restrictions ons, . One of them is that eachsi is non-decreasing (to see this, consider in (9) a functionC such thatC(xi,0−i ) = 0 for all i).Necessary and sufficient conditions seem difficult to formulate in a simple way; in partic-ular, the inequalitiesi�0 are not implied. In the particular case wherei is independentof both ti andxi , however, one checks that�s, meets monotonicity if and only ifsi isnondecreasing andi�0 for all i. Then inequalities (10) reduce to

∑j∈N\i j �1 for all i.

Adding the requirement of symmetry, we obtain the family of methods

�i (C, x) =�

n− 1C(xi,0−i )+s(xi)∑N s(xj )

(C(x)−∑j∈N

�

n− 1C(xj ,0−j ))

where 0��1, ands : N → R+.Whens(xi) = 1 for all xi, this formula reduces to (4)in Section 3.

Theorem A.1. (i) Assumen�4 and let� be a cost-sharing method satisfying additivity,strict solidarity, and monotonicity. For all i ∈ N, there exists a nondecreasing functionsi : N → R++ and a real-valued functioni such that� = �s,, where�s, is definedin (9).(ii) Assumen�3 and let� be a cost-sharing method satisfying additivity, solidarity,

monotonicity, zero cost for zero demand, and positive cost for positive demand. For alli ∈ N there exists a nondecreasing functionsi : N → R+ with si(0) = 0 < si(1), and areal-valued functioni such that� = �s,.

Proof. Statement(i): We fix throughout the proof a method� meeting additivity, strictsolidarity, and monotonicity. Recall thatD is the subset of cost functionsD such thatD(z) = 0,1. Writing �D = {z ∈ D | z − ei /∈ D for all i ∈ N(z)}, independence ofirrelevant costs yields�(D, x) = �(D′, x) wheneverD,D′ ∈ D andx ∈ �D ∩ �D′.Wemay therefore define� : NN → � by

�(x) = �(D, x),

whereD is any cost function inD such thatx ∈ �D.We spend Steps 1–5 computing�(D, x) for D ∈ D. After the preliminary Steps 1–3,

we construct the desired functionssi, i = 1, ..n, and derive the quasi-proportional form


�i (x) = Fi(x) in Step 4. Step 5 computes�(D, x) for anyD, x. Using additivity, Step 6then derives�(C, x) for all C ∈ C.We use the following notation. IfD ∈ D andx ∈ NN, ti(D, x−i ) = inf {xi |(xi, x−i ) ∈

D}, with the conventionti = +∞ if this set is empty. For anyS ⊆ N, �(S) = {y ∈ RS+ |∑i∈S yi = 1}.Step1: We show that ifD ∈ D, x ∈ D and there exist two distinct agentsi, j ∈ N

such thatti (D, x−i ) > 0 andtj (D, x−j ) > 0, then�(D, x) � 0. It follows that for anyx ∈ NN , {n(x)�2} ⇒ {�(x)� 0}.To prove the first statement, assume its premises and consider the increase ini’s demand

from 0 toxi. By strict solidarity,�−i (D, x) − �−i (D, (x−i ,0i )) is either strictly positiveor zero. In the latter case,�(D, x) = ei, and in the former case�−i (D, x)� 0. Repeatingthis argument with agentj, either�(D, x) = ej or �−j (D, x) � 0. The only possibilityis therefore that�−i (D, x) � 0 and�−j (D, x) � 0, hence�(D, x) � 0. The secondstatement follows from the first becauseti (D, x−i ) = xi for all i wheneverx ∈ �D.Step2:We show that for anyD ∈ D, i ∈ N, x ∈ D such thatx−i �= 0, andx′ = (x′i , x−i )

such thatx′i > xi,

�−i (D, x′)− �−i (D, x) = −��−i (x′) for some��0. (11)

Monotonicity implies�i (D, x′)−�i (D, x)�0, whereas

∑N (�j (D, x

′)−�j (D, x)) =0 by budget balance. Thus, by solidarity,

�j (D, x′)− �j (D, x)�0 for all j �= i. (12)

ChooseD′ ∈ D such thatx′ ∈ �D′ and let� ∈ R++.Applying strict solidarity to�D+D′whenxi increases tox′i , we obtain thatsign{�(�j (D, x′) − �j (D, x)) + (�j (D

′, x′) −�j (D

′, x))} is independent ofj, j �= i, wheresign(u) is 1,0, or −1 if u is, respectively,positive, zero or negative. From Step 1,�j (D

′, x′) − �j (D′, x) = �j (x′) > 0 because

n(x′)�2.On the other hand�j (D, x′)−�j (D, x) = uj �0. By strict solidarity,u−i � 0oru−i = 0. In the latter case, (11) holdswith� = 0. In the former, the vector�u−i+�−i (x′)is either strictly positive, zero, or strictly negative. Since this holds for all� > 0, u−i and�−i (x′) are parallel and of opposite sign, and (11) holds with� > 0.Step3: We show that for any two distincti, j ∈ N and anyx−ij �= 0, the direction of

�−ij (yij , x−ij ) is independent ofyij ∈ N2\{0}. That is, there exists(x−ij ) ∈ �(N\ij)such that, for allyij ∈ N2\{0}, �−ij (yij , x−ij ) = �(x−ij ) for some� > 0.Takei = 1, j = 2, fix x−12 �= 0, and write�−12(y12, x−12) = s(y) ∈ R

N\12+ , where

y = (y1, y2) ∈ N2\{0}. From now on, we do not repeat the restrictiony �= 0. By Step 1,s(y)� 0.Pick y, y′ noncomparable (i.e., neithery�y′ nor y′�y holds) andD such that�D

containsy andy′. Applying (11) twice, respectively, to the increase fromy to y ∨ y′, andfrom y′ to y ∨ y′,

�−12(D, y ∨ y′)− s(y) = −�s(y ∨ y′),�−12(D, y ∨ y′)− s(y′) = −�′s(y ∨ y′).


•••

• •

•y y’ z’ w’

z w

Fig. 1.

•

•

• •

•y y’

t t’

z0

Fig. 2.

Taking the difference of these equalities,

s(y)− s(y′) = s(y ∨ y′) for some ∈ R. (13)

This property holds for any noncomparabley, y′.Recall that our goal is to show that the direction ofs(y) is independent ofy. To this end,

assume first that there existy, y′ with y2 = y′2 > 0, y1 < y′1 ands(y) �= s(y′). Constructz, z′ such thatz1 = z′1 > y′1 andz2 < z′2 = y′2 (see Fig. 1). Two applications of (13) give

s(z)− s(y) = �s(z′),s(z)− s(y′) = �′s(z′).

Taking differences we find thats(z′) ands(y) − s(y′) have the same direction, � 0.Let L be the straight line borne by and containings(y). It containss(z) ands(y′).Now considerw,w′ such thatw1 = w′1 > z′1 andw2 < w′2 = z′2 (see Fig. 1). By the

above argument,s(w′) is borne by, andL containss(w).Applying (13) to z′, w gives

s(z′)− s(w) = s(w′).

Thuss(z′) ∈ L, implying thatL is the line borne by through 0. Nows is the direction ofs(t) for the six pointst of our construction.We check next thats(t) is borne by for any t such thatt2�y2 (and t �= 0). We just

proved this under the assumption thatt1 > y′1. Next supposet1�y′1 and 0< t2�y2. Setz0 = (y′1 + 1,0) andt ′ = t ∨ z0 (see Fig. 2). By (13) applied tot, z0, s(t) is borne by.The remaining values oft aret = (t1,0) with 0 < t1�y′1. There we apply (13) to t andt ′ = (t1− 1,1).Summing up, we have shown that ifs is not constant on a given horizontal line{y |

y2 = ε > 0}, then the direction ofs(y) is constant on the whole band{y | y2�ε; y �= 0}.If there are such integersε as large as we want, this establishes the desired statement atonce. The remaining case is when there is a numberε such thats is constant on every line{y | y2 = ε′}, ε′�ε. For anyy such thaty1 > 0, apply (13) to y andy′ = (y1 − 1, ε′)


whereε′� max{y2+1, ε} :we deduce thats(y) is borne by the value ofson{y | y2 = ε′}.The remaining casey1 = 0 is handled similarly.Step4: We show that there existn functionssi : N → R++ such that, for allx ∈ NN,

{n(x)�2} ⇒ {�(x) = F(x)}.We begin by considering the demand profilesx such thatn(x) = n. In particular,

�(x) � 0. By Step 3, the ratio�i�j(x) is independent ofxk for all k �= i, j (recalln�4).

For i, j, k all distinct,

�i�j(x) = sij (xi, xj )⇒ sij (xi, xj )sjk(xj , xk)ski(xk, xi) = 1.

A standard argument (omitted for brevity), shows the existence ofn functionssi onN\{0}such that�i�j

(x) = si (xi )sj (xj )

for all i, j, all x. Thus the desired form�(x) = F(x) holds

whenevern(x) = n.

Next, we treat the casen(x) = n−1, definingsi(0) in the process. Consider somexwithN(x) = N\1. For all i, j�2, we have

�i�j(x) = �i

�j(1, x−1) = si(xi)

sj (xj ),

where the first equality follows from Step 3. By Step 3 again,�1�2(x) does not depend upon

xj , for anyj �= 1,2, providedxj remains positive:

�1�2(x) = �1

�2(0, x2,1, . . .1) =: 1

g2(x2).

Defininggi for i = 3, . . . , n similarly and combining the two properties above,si(xi)

sj (xj )= gi(xi)

gj (xj )for all i, j �= 1 and allx−1� 0.

Thus sigidepends neither oni nor onxi.Calling this ratios1(0)we now have

�i�j(x) = si (xi )

sj (xj )

for all i, j . A similar construction deliverssi(0) for all i, and proves�(x) = F(x) whenn(x) = n− 1.Wecomplete the argument by decreasing induction onn(x).Fix q, 2�q�n−2.Assume

�(x) = F(x)whenevern(x)�q+1 and considerxwith n(x) = q, sayN(x) = {1, . . . , q}.For any distincti, j ∈ N\{1, n}, Step 3 and the inductive assumption imply

�i�j(x) = �i

�j(x−n,1n) = si(xi)

sj (xj ).

Note thatq�2 ensuresx−i,j �= 0, as required to apply Step 3. Moreover,q�n − 2guarantees that we have at least two choices for agent 1 inN(x), ditto for agentn inN\N(x). The equality�(x) = F(x) is now clear.


Step5: We derive an explicit formula for�(D, x) for all D ∈ D andx ∈ D. Using thenotation�i (D) = ti (D,0−i ), note thatxiei ∈ D ⇔ �i (D)�xi and defineK(D, x) = {i ∈N | �i (D)�xi}. DefineH : RN+\{0} × RN → RN by

Hi(a, b) = ai∑N aj

1+∑N\i

(bi − bj )aj

. (14)

We show that there existn functions�i such that, for allD ∈ D andx ∈ D,�(D, x)=H(a, b), where for alli, (15)

ai = si(xi),bi = �i (xi, �i (D)) if i ∈ K(D, x),

0 otherwise.

In Step 5a, we prove (15) whenK(D, x) = �. In Step 5b, we construct the functions�iand prove (15) when|K(D, x)| = 1. Step 5c completes the proof by an induction argumenton |K(D, x)|.Two simple facts aboutH will be useful. First

∑N Hi(a, b) = 1. Next, for anyz ∈ RN,{∑

N

zi = 1 andzi

ai− zj

aj= bi − bj for all i, j such thatai, aj �= 0

}(16)

⇒{z = H(a, b)}.We will use the notationx−i = (0i , x−i ).Step5a: WhenK(D, x) = �, (15) reduces to�(D, x) = F(x).We prove this equality

by induction onn(x).The smallest possible size isn(x) = 2 : if x = x1e

1, say, thenx ∈ D forces�1(D)�x1.

So we fixxwith N(x) = {1,2}. Definex∗ = x + e3 and chooseD∗ such that

D∗ ∩ [0, x] = D ∩ [0, x] andx−i∗ ∈ �D∗ for i = 1,2.This is possible becausex−1∗ − e3 = x−1 = x2e

2 /∈ D and similarlyx−2∗ − e3 /∈ D.By construction ofD∗ and Step 4,�(D∗, x−1∗ ) = �(x−1∗ ) = F(x−1∗ ). Applying (11) to

D∗, x∗ andx−1∗ ,

�−1(D∗, x∗)− F−1(x−1∗ ) = −�F−1(x∗)⇒ �−1(D∗, x∗) = F−1(x∗).

This, the symmetric property upon exchanging 1 and 2, and the fact thatF(x∗) � 0, givefinally �(D∗, x∗) = F(x∗).Next we apply (11) toD∗, x∗ andx−3∗ = x :

�−3(D∗, x∗)− �−3(D∗, x) = −�F−3(x∗) = −�′F−3(x).

Independence of irrelevant costs implies�(D∗, x) = �(D, x), so we conclude�−3(D, x)= F−3(x) for some�0.Repeating this construction with coordinate 4 instead of 3 yields


2

1

3

•

••

••

••

• ••

•

••

•

•

x2e2

x3e3

x1e1

x-3

x-1

t1

x

x-2

Fig. 3.

�−4(D, x) = �F−4(x) for some��0. The conclusion�(D, x) = F(x) follows as in theprevious paragraph.Now the induction argument. FixD, x such thatK(D, x) = �. Supposeti (D, x−i ) =

0⇔ x−i ∈ D for at least two agents, sayi = 1,2. BecauseK(D, x−i ) = � for i = 1,2,the inductive assumption yields�(D, x−i ) = F(x−i ) and, by (11), �−i (D, x) = �F−i (x)for i = 1,2. This implies�(D, x) = F(x) as above.If ti (D, x−i ) = 0 for at most onei, then we can assumetj (D, x−j ) > 0 for j = 1,2.We

nowconstructx∗ = x+e3 andD∗ exactly as above.This is possible becauset1(D, x−1) > 0meansx−1 /∈ D, thereforex−1∗ −e3 /∈ D, and similarlyx−2∗ −e3 /∈ D. The same argumentends the proof.Step5b: We prove (15) when|K(D, x)| = 1.In this and the next substep, we use one more piece of notation: givenz, z′ ∈ RN,

a ∈ RN++ and two distincti, j ∈ N,

z ∼aij z′ ⇔zi

ai− zj

aj= z′iai− z′jaj.

We start by choosingxwith n(x) = 3, say,N(x) = {1,2,3} andD such thatK(D, x) ={1} andx−1 ∈ D.We have drawn in Fig. 3 the traces ofD on the three coordinate hyper-planes. We writez = �(D, x) andzi = �(D, x−i ), ai = si(xi) for all i.Applying (11) to x, x−1 givesz−1− z1−1 = −�F−1(x), hence, by Step 5a,

zi

ai= zj

ajfor all i, j �= 1. (17)


Similarly, (11) applied tox, x−2 andx, x−3 gives

z ∼aij z2 for all i, j �= 2, (18)

z ∼aij z3 for all i, j �= 3. (19)

These three properties imply

z21

a1− z23

a3= z1

a1− z3

a3= z1

a1− z2

a2= z31

a1− z32

a2. (20)

By independence of irrelevant costs,z2 = �(D, x−2) only depends uponx−2 and thetrace ofD on [0, x−2]. Similarly z3 depends only uponx−3 and the trace ofD on [0, x−3].Thus the common quantity in (20) only depends onx1 and the trace ofD on [0, x1e1],namely�1(D).We write this quantity as�1(x1, �1(D)) or �1 for simplicity in the rest ofthis step.Combining (17), (20), we get

z1

a1− zi

ai= �1 for all i �= 1,

and by (16),

z = H(a, (�1,0−1)),

which is precisely (15) for (D, x).Next we derive a similar formula forz2 and forw1 = �(D, x1e1).Applying (11) to x−2,

x1e1, we obtain

z2 ∼a2ij w1 for all i, j �= 3, with a22 = s2(0), a2i = ai, i �= 2. (21)

Combining (21) and (18), and the fact that∼aij is an equivalence relation, we get for alli�4,

w1 ∼a21i z2 ∼a1i z⇒ w1 ∼a1i z⇔w11

a1− w1i

ai= �1,

whereai = si(0). Now w1 only depends onx1 and �1(D), as does�1. If we repeatthe construction of Step 5b while exchanging the roles of 2 and 4, i.e., forx′ such thatN(x′) = {1,3,4} andD′ such thatK(D′, x′) = {1} andx′−1 ∈ D, we get

w11

a1− w12

s2(0)= �1.

Combining these properties ofw1, including a similar equality where 3 replaces 2, and (16),we conclude

w1 = H(a1, (�1,0−1)),

wherea11 = s1(x1) anda1i = si(0) for i�2 : this is precisely (15) for (D, x1e1). As �(D,x1e

1) only depends onx1 and�1(D), this establishes (15) for any(D, x) wheren(x) = 1.


Next we computez2. From (18) and (21), we obtain, respectively

z21

a1− z22

s2(0)= �1,

z21

a1− z2i

ai= �1 for all i�3.

Thus (16) impliesz2 = H(a2, (�1,0−1)). Again,�(D, x−2) only depends uponx−2 andthe restriction ofD to [0, x−2]. Our choice ofD in Step 5b places no constraint on therestriction ofD to [0, x−2], exceptK(D, x−2) = {1}. Therefore, we have proven (15) forany(D, x) wheren(x) = 2 and|K(D, x)| = 1.To complete the proof of (15) for any (D, x) such thatn(x) = 3 and|K(D, x)| = 1,

it only remains to take care of the case wherex−1 /∈ D (recall we assumedx−1 ∈ D

instead at the beginning of Step 5(b). Consider such a configuration illustrated in Fig. 3.We already know thatz2 = �(D, x−2) andz3 = �(D, x−3) are given by (15). Applying(11) to z = �(D, x), z2 and toz, z3 delivers, respectively, the direction ofz−2 and ofz−3,hence determineszby (16). We omit the details.To complete Step 5b, it remains to establish formula (15) when |K(D, x)| = 1 and

n(x) > 3. We proceed by induction onn(x). The argument is essentially the same asabove, once we observe thatK(D, x) = {1} implies that for alli �= 1, xiei /∈ D, henceti (D, x−i ) = 0.Assume the desired conclusion holds whenevern(x)� t − 1 and let(D, x) be such that

K(D, x) = {1} andn(x) = q.Without loss of generality, assumeN(x) contains 2,3. Asbefore (see (21)) we definea by ai = si(xi) for all i, andai by aii = si(0) andaij = aj for

all j �= i. ClearlyK(D, x−2) = {1}, so the inductive assumption givesz2 = �(D, x−2) = H(a2, (�1,0−1)).

Applying (11) to z = �(D, x) andz2 gives

z1

a1− zi

ai= �1 = z21

a1− z2i

aifor all i�3.

The symmetrical argument involvingzandz3 gives z1a1− z2

a2= �1 and the proof of Step 5b

is complete.Step5c: We prove (15) by induction on|K(D, x)|.Fix k�2 and assume (15) holds whenever|K(D, x)|�k − 1. Consider(D, x) with

|K(D, x)| = k, say,K(D, x) = {1, . . . , k}. Definez = �(D, x), zi = �(D, x−i ), a,andai exactly as in Step 5b. ClearlyK(D, x−i ) = {1, . . . , k}\{i} for i�k, therefore theinductive assumption implies

zi = H(ai, (b−i ,0i )) for i = 1, . . . , k,whereb is defined bybj = �j (xj , �j (D)) if j�k , andbj = 0 otherwise. Applying (11)successively toz, z2 andz, z3 gives, for alli�3,

z ∼a1i z2⇒z1

a1− zi

ai= b1− bi


and

z ∼a12 z3⇒z1

a1− z2

a2= b1− b2

and (16) then yieldsz = H(a, b), as desired.Step6: We construct the functionsi , i ∈ N, and complete the proof.Rewrite (14) as

Hi(a, b) = ai∑N aj

(1−

∑N

bjaj

)+ aibi .

Definei , i ∈ N, by

i (xi, ti) = si(xi)�i (xi, ti) if ti�xi,

0 otherwise.

Recalling our convention�i (D) = +∞ if D does not intersect thei-axis, formula (15) canbe rewritten as follows. For alli, x, D such thatx ∈ D,

�i (D, x) = i (xi, �i (D))+ Fi(x)

(1−

∑N

j (xj , �j (D))

).

This formula coincideswith (9) because (8) reduces tosai (D, xi) = i (xi, �i (D))whenD ∈ D. An additive method is entirely determined by its behavior on such pairs(D, x),

thus the proof is complete.

statement(ii): We fix throughout an additive method� meeting the four axioms in thestatement, denoted for brevity SOL,MON,ZCZDandPCPD.The structure of the proof, andthe numbering of the corresponding steps and substeps, are identical to those of the proofof statement (i): we compute�(x) first, thus constructingF, then�(D, x) by constructing. Some of the steps are considerably shorter, however.Step1: ZCZD and PCPD imply at once for allx,D,

x ∈ D ⇒ {�i (D, x) > 0⇔ i ∈ N(x)}

and in particular�i (x) > 0⇔ xi > 0.Step2:We showhere a stronger property than (11), namely for allx,D, i such thatxi > 0

andx−i �= 0,

x−i ∈ D ⇒ �−i (D, x)− �−i (D, x−i ) = −��−i (x) for some� > 0, (22)

x−i /∈ D ⇒ �−i (D, x) = ��−i (x) for some� > 0. (23)

Note that by taking the difference of (22) for xi andx′i , we get (11).


To prove (22) and (23), we note that�i (D, x) < �i (D, x−i ) (by ZCZD and PCPD),

hence∑

N\i �j (D, x) >∑

N\i �j (D, x−i ). By SOL, �j (D, x)��j (D, x−i ) for each

j ∈ N\i. Now suppose that for somej, k ∈ N(x)\i we have�j (x) < �j (x−i ) whereas

�k(x) = �k(x−i ). PickD′ containingxbut notx′ and apply SOL to�D+D′ betweenx−i

andx :{�(�j (D, x)− �j (D, x

−i ))+ �j (D′, x)}�k(D′, x)�0,

which contradicts PCPD for� large enough. Thus�j (D, x) − �j (D, x−i ) < 0 ⇔ j ∈

N(x)\i, and for all� > 0, all j, k ∈ N(x)\{i},{�(�j (D, x)− �j (D, x

−i ))+ �j (D′, x)}{�(�k(D, x)− �k(D, x

−i ))+ �k(D′, x)}�0,

implying�−i (D, x)− �−i (D, x−i ) = −��−i (D′, x) for some� > 0. If we chooseD′ sothatx ∈ �D′ we get (22), and (23) follows.Step3: The direction of the vector�−i (x) is independent ofxi ∈ N\{0}, whenx−i �= 0.Indeed, pickxi, x′i such that 0< xi < x′i , andD such that(xi, x−i ) ∈ �D. By (23),

�−i (D, (x′i , x−i )) = ��−i ((x′i , x−i )) for some� > 0 and by (11) �−i (D, (x′i , x−i )) −�−i (D, (xi, x−i )) = �−i (D, (x′i , x−i )) − �−i ((xi, x−i )) = −�−i (x) for some�0.Step4: We construct the nondecreasing functionssi such thatsi(0) = 0 < si(1) and

�(x) = F(x) for all x such thatn(x)�3.SettingN0 = N(x) and�0(x) to be the projection of�(x) onRN0, we note that�0(x)

is strictly positive and in�(N0). By Step 3,�0i�0jdepends onxi, xj only, and a standard

argument gives then the functionssi onN\{0} such that�0(x) = F(x) for all x such thatN(x) = N0. The argument is essentially the same as in Step 4 of the previous proof, exceptthat we only needN0 to contain 3 or more agents. This is why we only need to assumen�3in statement (ii) of TheoremA.1.In view of Step1, it is now routine to show that the functionssi do not depend onN0,

and to extend the equality�(x) = F(x) to all coordinates by settingsi(0) = 0.Step5: We show, as in the previous proof, the existence of functions�i allowing the

representation of�(D, x) by (15).Step5a: For any(D, x) with x ∈ D andK(D, x) = �, �(D, x) = F(x).

Prove this first forxsuch thatN(x) = {1,2}.Setx∗ = x+e3, pickD∗ such thatD∗ = D

on[0, x] andx−1∗ , x−2∗ /∈ D∗.By (23)�−i (D∗, x∗) is borne byF−i (x∗) for i = 1,2. In viewof Step 4 this implies�(D∗, x∗) = F(x∗). Invoke next (22): �−3(D∗, x∗)−�−3(D∗, x) =−�F−3(x∗). Therefores1(x1)s2(x2)

= �1�2(D∗, x) = �1

�2(D, x), where the latter equality follows

from independence of irrelevant costs.The ascending induction onn(x) is now immediate by (22) and Step 1.Step5b: We assume now|K(D, x)| = 1, and of coursex ∈ D. If n(x) = 1 there is

nothing to prove (by ZCZD), contrary to the previous proof. Ifn(x) = 2, we simply copyStep 5b to define the functions�i and computez, z2, z3. By Step 1 we need not worry aboutwhether or notx−1 ∈ D. The ascending induction onn(x) is unchanged.Step5c: The ascending induction on|K(D, x)| is unchanged.Step6: is unchanged. �


A.3. Proof of Theorem 1

The “if” part of statements (i) and (ii) is clear. To prove the “only if” part in eitherstatement, we assume that a method given by (9) satisfies strong ranking, and prove thati ,hencesai is identically zero for alli.We check firstsi = sj for all i, j. Pick x = teN , t�1, andC,C(z) = 1 if n(z)�2,

C(z) = 0 otherwise. Then�(C, x) = F(x), so strong ranking impliessi(t) = sj (t)

for t�1. Under zero cost for zero demand, this is enough. On the other hand ifn�4, wechoosex′ = teN\{1,2} and the same cost functionC to get s1(0) = s2(0), and we are done.Assume now there is an agenti andx1, t1 such that1(x1, t1) �= 0. ChooseC = �(t1,0−1)

as in (28) so thatsai (C, xi) = 0 for i�2, sa1(C, x1) = 1(x1, t1). Then fori�2,

�1(C, x1eN) = 1+

1

n(1− 1) �=

1

n(1− 1) = �i (C, x1e

N),

contradicting strong ranking.

A.4. Proof of Theorem 2

Step1: (i) implies (ii) and (iii). Let� be a fixed-flow method. It is easy to check that�satisfies additivity, separability and monotonicity; we prove here that it also meets submod-ular solidarity and supermodular solidarity. Letf (., x) be the fixed-flow associated with�and writef (., x) = pxf (., x) for all x ∈ [0, x]. Observe thatpxf (., x + ei) = f (., x) forall i, all x ∈ [x − ei].We fix i, j, i �= j, x ∈ [x − ei], andz ∈ [x − ej ].First we comparefj (z, x) andfj (z, x + ei). If zi < xi, thenK contains neitheri nor j,

hence(x+ ei)K = xK , implyingfj (z, x) = fj (z, x+ ei). If zi = xi, thenK containsi butnot j, hence[xK, (x + ei)K ] = {xK, xK + eiK} so thatfj (z, x) = fj (z, x + ei)+ fj (z+ei, x + ei).

Now we compute

�j (C, x + ei)− �j (C, x)

=∑

z∈[0,x]:zi=xi ,zj<xj(�jC(z+ ei)fj (z+ ei, x + ei)

+ �jC(z)(fj (z, x + ei)− f (z, x)))

=∑

z∈[0,x]:zi=xi ,zj<xj(�jC(z+ ei)− �jC(z))fj (z+ ei, x + ei),

implying submodular solidarity and supermodular solidarity at once.Step2: (ii) implies (i). Fix a method� satisfying additivity, separability, monotonicity,

and submodular solidarity. Letf be its flow representation as in (5). We showf (., x) =pxf (., x) for anyx ∈ [0, x].Step2a: For alli ∈ N, x ∈ [0, x − ei], andz ∈ [0, x − ei],

fi(z, x) = fi(z, x + ei). (24)

This is a well-known consequence of Monotonicity: see[10] or [19] for a proof. Eq. (24)does not imply the propertyfj (z, x) = fj (z, x+ei) for j �= i (except whenn = 2, where it


follows from flow conservation); that property is derived in the next step using submodularsolidarity.Step2b: For alli ∈ N, x ∈ [0, x − ei], andz ∈ [0, x − ei],

f (z, x) = f (z, x + ei). (25)

The proof of (25) is by induction onz(N), the sum of the coordinates ofz. Let z = 0.By Step 2a,fi(0, x) = fi(0, x + ei). If (25) fails, as total outflow from 0 is 1 underf (., x) andf (., x + ei), there existj, k ∈ N\i such thatfj (0, x) < fj (0, x + ei) andfk(0, x) > fk(0, x + ei). Consider the cost function

�0(w)= 0 if w = 0,1 otherwise.

Note that�0 ∈ Csub. But �j (�0, x) = fj (0, x) < fj (0, x + ei) = �j (�0, x + ei) and�k(�0, x) = fk(0, x) > fk(0, x + ei) = �k(�0, x + ei), contradicting submodular soli-darity.Next, fixk > 0 and assume that (25) is true for allz ∈ [0, x− ei[ such thatz(N)�k−1.

Fix z ∈ [0, x − ei[ such thatz(N) = k. By the induction hypothesisfj (z − ej , x) =fj (z− ej , x + ei) for all j ∈ N such thatzj > 0 , hence the total incoming flow atz is thesame underf (., x) andf (., x + ei):∑

j∈N(z)fj (z− ej , x) =

∑j∈N(z)

fj (z− ej , x + ei).

Conservation of flows and (24) imply now∑

j∈N\i fj (z, x) =∑

j∈N\i fj (z, x + ei). Ifzj < xj for at most onej ∈ N\i, we are done. Otherwise suppose, by contradiction, thatf (z, x) �= f (z, x + ei) : there existj, k ∈ N\i such thatfj (z, x) < fj (z, x + ei) andfk(z, x) > fk(z, x + ei). Consider the cost function

�z(w)= 0 if w�z, (26)

1 otherwise.

Again,�z ∈ Csub. Compute�j (�z, x + ei)− �j (�z, x).

=∑

wN\j∈[0,zN\j ](fj ((zj , wN\j ), x + ei)− fj ((zj , wN\j ), x))

= fj (z, x + ei)− fj (z, x) > 0

and, symmetrically,�k(�z, x+ei)−�k(�z, x) = fk(z, x+ei)−fk(z, x) < 0, contradictingsubmodular solidarity.Step2c: For alli ∈ N , x ∈ [0, x − ei], andz ∈ [0, x[,

f (z, x) = pxf (z, x + ei). (27)

Sincepx′pxf (., x) = px′f (., x) for x′�x�x, this will complete the proof of Step 2.By Step 2b, (27) is true forz ∈ [0, x − ei[ as for such az we havei /∈ K andxK =

(x+ei)K. It remains to extend it to allz ∈ [xiei, x[.Consider firstz = xiei and suppose, by


way of contradiction, thatf (xiei, x) �= pxf (xiei, x+ei) : there existj, k ∈ N\i such that

fj (xiei, x) > fj (xie

i, x+ei)+fj ((xi+1)ei, x+ei) andfk(xiei, x) < fk(xiei, x+ei)+

fk((xi + 1)ei, x + ei). Recalling Step 2b, this yields a violation of submodular solidarityfor the cost function�(xi+1)ei . Thusf (z, x) = pxf (z, x + ei) for z = xie

i . An inductionargument mimicking that in Step 2b completes Step 2c, and the proof that� is a fixed-flowmethod.Step3: (iii) implies(i). The argument in Step 2 is easily adapted. The submodular cost

functions�z are merely replaced with the supermodular functions�z defined by

�z(w)= 1 if w�z, (28)

0 otherwise

and the induction argument is carried onx(N)− z(N) rather than onz(N).

A.5. Proof of Proposition 5

Step 1 of the proof of Theorem 2 shows that the fixed-flow methods meet positive exter-nalities and negative externalities.Next, we show that the Aumann–Shapley method (Example 1) also does. To shorten

notation, we denote� the Aumann–Shapley method, andf its flow representation. Fix ademandprofilex, twoarbitraryagents1and2, andz−1 ∈ [0−1, x−1[.Forz1 = 0, . . . , x1+1,wewritef (z1, x) insteadoff ((z1, z−1), x)andf (z1, x+e1) insteadoff ((z1, z−1), x+e1).We claim that

x1∑z1=0

f2(z1, x) =x1+1∑z1=0

f2(z1, x + e1) (29)

and, fork = 0, . . . , x1− 1,k∑

z1=0f2(z1, x)�

k∑z1=0

f2(z1, x + e1). (30)

The claim implies that the sequence(f2(z1, x); z1 = 0, .., x1), augmented by 0 as the lastterm, stochastically dominates(f2(z1, x + e1); z1 = 0, .., x1 + 1). Using (5), this implies�2(C, x) � �2(C, x + e1) if C is submodular and the reverse weak inequality holds ifCis supermodular.Eq. (29) follows immediately from the fact that the Aumann–Shapley method satisfies

strong dummy (see comment 2 after Theorem 2). We omit the easy proof.To prove (30), recall the notation and formula in Example 1 and define k = �((k, z−1)),

�k = �(x − (k, z−1) − e2) for k = 0, . . . , x1 − 1, and�−1 = �(x − (0, z−1) + e1 − e2).

Note that

�(x + e1)

�(x)= x(N)+ 1

x1+ 1 ,

k =

k∏t=1(z(N\1)+ t)

k! 0,


�k =

k∏t=1(x1− t + 1)

k∏t=1(x(N)− z(N\1)− t)

�0.

Writing ak = z(N\1)+ k andbk = x(N)− z(N\1)− k, (30) reads

x(N)+ 1x1+ 1

(1+ a1x1

b1+ a1a2x1(x1− 1)

2b1b2+ · · · + a1 . . . akx1 . . . (x1− k + 1)

k!b1 . . . bk)

� b0

x1+ 1 + a1+ a1a2x1

2b1+ · · · + a1 . . . akx1 . . . (x1− k + 2)

k!b1 . . . bk−1 .

Usingx(N)+ 1 = a1 + b0, subtractb0

x1+1 from both sides of this inequality and divideeach side bya1 to obtain the equivalent inequality

1

x1+ 1 +x(N)+ 1x1+ 1

(x1

b1+ a2x1(x1− 1)

2b1b2+ . . .+ a2 . . . akx1 . . . (x1− k + 1)

k!b1 . . . bk)

�1+ a2x1

2b1+ · · · + a2 . . . akx1 . . . (x1− k + 2)

k!b1 . . . bk−1 .

Subtract 1x1+1 from both sides and divide byx1 to obtain

x(N)+ 1x1+ 1

(1

b1+ a2(x1− 1)

2b1b2+ · · · + a2 . . . ak(x1− 1) . . . (x1− k + 1)

2 . . . kb1 . . . bk

)� 1

x1+ 1 +a2

2b1+ · · · + a2 . . . ak(x1− 1) . . . (x1− k + 2)

2 . . . kb1 . . . bk−1,

in whicha1 andb0 no longer appear.Next, usingx(N)+ 1= a2+ b1, a similar two-step argument reduces that inequality to

the equivalent

x(N)+ 1x1+ 1

(1

b2+ a3(x1− 2)

3b2b3+ · · · + a3 . . . ak(x1− 2) . . . (x1− k + 1)

3 . . . kb2 . . . bk

)� 1

x1+ 1 +a3

3b2+ · · · + a3 . . . ak(x1− 2) . . . (x1− k + 2)

3 . . . kb2 . . . bk−1,

wherea2 andb1 no longer appear.Repeating this procedurek − 1 times establishes that the original inequality (30) is

equivalent to

x(N)+ 1x1+ 1

(1

bk−1+ ak(x1− k + 1)

kbk−1bk

)� 1

x1+ 1 +ak

kbk−1.

Usingx(N)+1= ak+bk−1, subtracting 1x1+1 from both sides andmultiplying by

bk−1ak,

we get 1x1+1 + (x(N)+1)(x1−k+1)

(x1+1)kbk � 1k, which reduces tox(N)+ 1�bk.


A.6. Statement and proof of Lemma A.2

Wework out the restrictions imposed by strong ranking andweak separability on formula(7). For anyx ∈ NN, let �(x) = {y ∈ � |for all i, j ∈ N, xi�xj ⇒ yi�yj }. Write{xi | i ∈ N(x)} = {�1, . . . , �K}, where 0< �1 < . . . < �K, and set�0 = 0. Fork = 1, . . . , K, letNk = {i ∈ N | xi��k}. Using this notation, the extreme points of�(x)are the vectorsbk = 1

|Nk | eNk , k = 1, . . . , K.We sometimes writeK(x) to emphasize that

K depends uponx.

Lemma A.2. An additive cost-sharing method� satisfies strong ranking and weak sep-arability if and only if it possesses a shared-flow representation(f, r) such that, for allx ∈ NN, z ∈ [0, x[, j ∈ N, andk ∈ {1, . . . , K(x)},

{�k−1�zj < �k} ⇒ {r(z, j, x) = bk}. (31)

Proof. Step1: We show that an additive cost-sharing method� satisfies strong rankingif and only if it possesses a shared-flow representation(f, r) such that, for allx ∈ NN,

z ∈ [0, x[, andj ∈ N, r(z, j, x) ∈ �(x).The “if” statement is clear from the representation in (7). Conversely, assume� meets

additivity and strong ranking, and fixx ∈ NN such thatx1� · · · �xn. Define the linearisomorphismh : RN → RN by

h(y) = (ny1, (n− 1)(y2− y1), . . . ,2(yn−1− yn−2), yn − yn−1)

andobserve that itmaps�(x) into a face�(x)of�.For instance, ifx1, . . . , xn areall distinct,h is an isomorphism from�(x) into�; if N = {1,2,3,4,5} andx1 = x2 < x3 < x4 = x5,

then�(x) is the facey2 = y5 = 0.Define�(C, x) = h(�(C, x)) for all C ∈ C. By strong ranking,�(C, x) is borne by a

vector in�(x), hence�(C, x)�0. Therefore�(., x) is abona fideadditive cost-sharingmethod atx. Thus it has a shared-flow representation

�(C, x) =∑

z∈[0,x[

∑j∈N

�jC(z)gj (z, x)r(z, j, x) (32)

for all C ∈ C.If yi = 0 for ally ∈ �(x), then�i (C, x) = 0 for allC ∈ C, implying thatri(z, j, x) = 0

for all z ∈ [0, x[ andj ∈ N. Thusr(z, j, x) ∈ �(x) for all zandj. Applying h−1 to bothsides of (32), we obtain

�(C, x) =∑

z∈[0,x[

∑j∈N

�jC(z)gj (z, x)h−1(r(z, j, x))

for all C ∈ C. Sinceh−1(r(z, j, x)) ∈ �(x), we are done.Step2: We complete the proof. Let� be a cost-sharing method satisfying additivity,

strong ranking, and weak separability, and fixx ∈ NN . For a ∈ N\{0} and i ∈ N,

consider the function�aei

defined in (28): �aei

(z) = 1 if zi�a and�aei

(z) = 0 otherwise.


By weak separability,�(∑

i∈N �aei

, x) = e{i∈N |xi �a} for all a ∈ N\{0}. Equivalently,

for all k ∈ {1, . . . , K} and�k−1 < a��k,∑i∈Nk

�(�aei

, x) = eNk . (33)

By strong ranking�(�aei

, x) ∈ �(x) for all i. By (33) and the fact thatbk is an extreme

point of�(x), �(�aei

, x) = bk whenever�k−1 < a��k andi ∈ Nk. If � is represented by

(f, r), �(�aei

, x) =∑z:zi=a−1 fi(z, x)r(z, i, x), therefore the same extremality argument

yields r(z, i, x) = bk for all z ∈ [0, x[ such thatzi = a − 1 andfi(z, x) > 0. Note thatzi varies from�k−1 to �k − 1. If fi(z, x) = 0, the choice ofr(z, i, x) is irrelevant and wemay setr(z, i, x) = bk in that case too. This proves the “only if” part of the lemma. Theproof of the converse statement is similar because every symmetric separable cost function

C(z) =∑i∈N c(zi) may be written as a linear combination of cost functions�aei

. �

It is interesting to compare the shared-flow representation in Lemma A.2 with that ofseparable methods given in formula (5) in Section 4. In both cases the share systemr in thegeneral formula (7) is entirely determined and the flowf is arbitrary. Thus the two familiesof methods—separable on one hand, weakly separable and meeting strong ranking on theother hand—are comparably rich. In the former case, however, every flowf yields a differentmethod, whereas in the latter case different flows may lead to the same method. Considerfor instance the cross-subsidizing serial method�cs (Example 4) and fixx ∈ NN∗ . Everyflow f (., x) that goes entirely throughx1, x2, .., xn, together with the sharing ruler(., ., x)defined by formula (31), determines a valid representation of�cs(., x).

A.7. Proof of Proposition 7

Assumen�3 and let� be a cost-sharing method satisfying all four stated axioms. ByLemmaA.2,� has a shared-flow representation(f, r) such that for allx ∈ NN, z ∈ [0, x[,j ∈ N, andk ∈ {1, . . . , K(x)}, equation (31) holds. In particular, this implies zero cost forzero demand becausexi = 0 impliesbki = 0 for all k = 1, . . . , K(x). As a consequence,any contradiction derived forn = 3 extends to more agents by considering demand profileswhere all agents but three demand zero.Fix thusn = 3 and letx be such that 2�x1 < x2 < x3.Write f for f (., x). As already

noted in the proof of Proposition 1, strong group monotonicity is equivalent to the propertythat for allD ∈ D, all distinct agentsi, j ∈ N, and allx ∈ NN,

D(x) = 1⇒ �j (D, x + ei)��j (D, x). (34)

In the following argument, we writeD ∨ D′ for the supremum of two cost functionsD,D′ ∈ D, that is,(D ∨D′)−1 = D−1(1) ∪D′−1(1). Choosex1 ∈ [0, x1[ and set

D = �(0,0,x3) ∨ �(x1+1,0,0),

where�z is the cost function defined in (28). By (31) and independence of irrelevant costs,�(D, (x1, x2, x3)) = e3 becauseD coincides with�(0,0,x3) on [0, (x1, x2, x3)]. By (34),


�2(D, x)��2(D, (x1, x2, x3)),hence�2(D, x) = 0.By (31) again, this givesf1(z−e1) =0 for allzsuch thatz1 = x1+1 andz3 < x3. Since this is true for allx1 ∈ [0, x1[, the entireflow must be contained inA = {z ∈ [0, x[| z1 = 0 or z3 = x3}, that is to sayfi(z) > 0requiresz3 = x3 or z3 < x3, z1 = 0, andi �= 1.Next choosex′1 ∈ [1, x1] and set

D = �(0,x2,x3−1) ∨ �(0,x′1,x3).

NowD coincides with�(0,x2,x3−1) on [0, x− e3], (31) and independence of irrelevant costsimply that�(D, x−e3) is a convex combination ofe3 and12e23, hence�1(D, x−e3) = 0.By (34), �1(D, x) = 0. This implies thatf2(z − e2) = 0 for all z such that 0�z1�x1,

1�z2�x1, andz3 = x3.Next pickx′2 ∈ [x1, x2[ and set

D = �(x1,x1,x3) ∨ �(0,x′2+1,x3).

As D coincides with�(x1,x1,x3) on [0, (x1, x′2, x3)], (31), independence of irrelevant costsand the fact thatf (., (x1, x′2, x3)) is contained inA together imply that�(D, (x1, x′2, x3))= 13e123. Now (34) yields�3(D, x)� 1

3. By strong ranking, this forces�(D, x) = 13e123.

Thusf2(z − e2) = 0 for all z such that 0�z1�x1 − 1, x1 + 1�z2�x2, andz3 = x3.

Moreover,f3(z− e3) = 0 for all zsuch thatz1 = 0, x1+ 1�z2�x2, andz3 = x3.

Gathering our results, we conclude that (i) the entire flow must go through the edgebetween(0, x1, x3 − 1) and(0, x1, x3), i.e., f3(0, x1, x3 − 1) = 1, and (ii) within {z ∈[0, x[| z3 = x3}, the flow follows the path(0, x1, x3)→ (x1, x1, x3)→ (x1, x2, x3).

Finally, consider

D = �(x1−1,0,x3) ∨ �(0,x1,x3).

By the shape of the flowf (., x − e1) and (31), �(D, x − e1) = 13e123. By the shape of

f (., x) and (31), �(D, x) = e3, a contradiction to monotonicity.

References

[1] R.J. Aumann, L. Shapley, Values of Nonatomic Games, Princeton University Press, Princeton, NJ, 1974.[2] W. Baumol Panzar, R. Willig, Contestable Markets and the Theory of Industry Structure, Harcourt BraceJanovitch, NewYork, 1982.

[3] L. Billera, D. Heath, Allocation of shared costs: a set of axioms yielding a unique procedure, Math. Oper.Res. 7 (1982) 32–39.

[4] E. Calvo, J.C. Santos, A value for multichoice games, Math. Soc. Sci. 40 (2000) 341–354.[5] E. Friedman, H. Moulin, Three methods to share joint costs or surplus, J. Econ. Theory 87 (1999) 275–312.[6] M. Haviv, The Aumann–Shapley price mechanism for allocating congestion costs, Oper. Res. Letters 29(2001) 211–215.

[7] C. Herrero, M. Maschler, A. Villar, Individual rights and collective responsibility: the rights-egalitariansolution, Math. Soc. Sci. 37 (1999) 59–77.

[8] D. Lee, Game-theoretic procedures for determining pavement thickness and traffic lane costs in highway costallocation, PhD. Thesis, Texas A&M University, 2002.

[9] L. Mirman,Y. Tauman, Demand compatible equitable cost sharing prices, Math. Oper. Res. 7 (1982) 40–56.[10] H. Moulin, On additive methods to share joint costs, Japan. Econ. Rev. 46 (1995) 303–332.


[11] H. Moulin, Y. Sprumont, Responsibility and cross-subsidization in cost sharing, Mimeo, 2002.[12] H. Moulin, R. Vohra, A representation of additive cost sharing methods, Econ. Letters 80 (2003) 399–407.[13] N. Naumova, Non symmetric equal sacrifice solutions for cost allocation problems, Math. Soc. Sci. 43 (2002)

1–18.[14] L. Shapley,A value forn-person games, H.W. Kuhn,W. Tucker (Eds.), Contributions to the Theory of Games

II, Annals of Mathematics Studies, vol. 28, Princeton University Press, Princeton, NJ, 1953.[15] W. Sharkey, The Theory of Natural Monopoly, Cambridge University Press, Cambridge, 1982.[16] S. Shenker, Making greed work in networks: a game-theoretic analysis of gateway service discipline,

IEEE/ACM Trans. Networking 3 (1995) 819–831 (circulated 1989).[17] M. Shubik, Incentives, decentralized control, the assignment of joint costs, and internal pricing, Manage. Sci.

8 (1962) 325–343.[18] Y. Sprumont, Ordinal cost sharing, J. Econ. Theory 81 (1998) 126–162.[19] Y.Sprumont,Aumann–Shapleypricing: a reconsiderationof thediscrete case,Mimeo,Université deMontréal,

2004.[20] Y. Sprumont, L. Zhou, Pazner–Schmeidler rules in large societies, J. Math. Econ. 31 (1999) 321–339.[21] W. Thomson, Welfare-domination under preference replacement: a survey and open questions, Soc. Choice

Welfare 16 (1999) 373–394.[22] Y. Wang, The additivity and dummy axioms in the discrete cost sharing model, Econ. Letters 64 (1999)

187–192.[23] R.Weber, Probabilistic values for games, in: A. Roth (Ed.), The ShapleyValue, Cambridge University Press,

Cambridge, 1988.

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On demand responsiveness in additive cost sharing