web.stanford.edu · On efficient and almost budget balanced allocation mechanisms Hervé Moulin Rice University May 2008 We consider two resource allocation problems where one ...

On efficient and almost budget balanced allocation mechanisms Hervé Moulin Rice University May 2008 We consider two resource allocation problems where one dimensional mechanisms achieve allocative efficiency in equilibrium under quasi-linear preferences, but are unable to balance the budget. We define the efficiency loss of a mechanism as its largest ratio of budget imbalance to efficient surplus, over all profiles of individual preferences. We look for mechanisms minimizing the efficiency loss. To assign one (desirable or undesirable) object among n agents, we find a subsidy-free VCG mechanism guaranteeing voluntary participation and an efficiency loss of 2n/2n. This mechanism is optimal in the VCG class, and among all strategy-proof mechanisms treating equals equally. The result generalizes partially to the assignment of multiple identical objects, where the optimal mechanism is no longer a VCG one, and the tradeoff between voluntary participation and minimizing the efficiency loss can be severe. To exploit a one input-one output convex technology, n users request a quantity of output and the mechanism assigns cost shares. In the canonical residual* mechanism, total surplus is maximized in any Nash equilibrium of a natural potential game. Total payment may differ from the actual cost and a user with a null demand may be subsidized. If the cost function is totally monotone (e.g., polynomial with positive coefficients, or exponential), participation is voluntary and total payment covers at least actual cost. The efficiency loss is at most min{2/log n,1}. For power cost functions, C(a)=ap, p>1, the above ratio converges to zero as 1/np-1. Participation appears to be voluntary however a small budget deficit is possible. These properties are lost if the cost function is not smooth. References H. Moulin, Auctioning or assigning an object: some remarkable VCG mechanisms, ACM SIGecom Exchanges, Vol. 7, No 1, December 2007 H. Moulin, Efficient, strategy-proof and almost budget-balanced assignment, www.ruf.rice.edu/~econ/faculty/Moulin/spassgnt.pdf, forthcoming Journal of Economic Theory H. Moulin, An efficient and almost budget-balanced cost sharing method, www.ruf.rice.edu/~econ/faculty/Moulin/residual17.pdf, forthcoming Games and Economic Behavior

Auctioning or assigning an object: some remarkableVCG mechanisms

HERVE MOULIN

Rice University

1. AUCTIONING A GOOD

A second price Vickrey auction is a simple mechanism to transfer a valuable object(a good) between a seller and n potential buyers. It treats buyers fairly, elicits theirtruthful valuations for the good, and allocates the good efficiently. But the divisionof the surplus between the seller and the buyers is hardly compelling. Writing a∗k

for the k-th highest valuation among buyers, and setting without loss of generalitythe seller’s valuation at zero, of the total surplus a∗1 the seller gets a∗2 and thebuyers a∗1 − a∗2, thus either share can be arbitrarily large or small, depending onthe distribution of valuations. Moreover at most one buyer gets any surplus at all.

The familiar Vickrey-Clarke-Groves (thereafter VCG, see [Green and Laffont1979]) mechanisms preserve the incentives and efficiency properties of the Vick-rey auction, but allow more flexibility in the division of the surplus. Suppose thatexplicit guidelines regulate the division of surplus: the seller should get λa∗1, thebuyers (1− λ)a∗1. We construct a VCG mechanism achieving such division with amargin of error as small as possible. For any numbers λ−, λ+ such that

0 ≤ λ− ≤ λ+ ≤ 1 and λ+ − λ− =n

2n−1 − 1(1)

our mechanism guarantees to the seller, at all profiles of non-negative valuations(aj , j ∈ N), a revenue between λ−a∗1 and λ+a∗1. For instance with 10 potentialbuyers, any given share λ can be approximated within 2% at all profiles.

The simplest way to describe this (or any other VCG) mechanism is by the netutility Ui(a) of agent i at the profile a = (aj , j ∈ N) ∈ RN

+ , where no object andno cash transfer yields Ui = 0. Thus if i does not get the object, e.g., if ai < a∗1,Ui(a) is a cash transfer to i, whereas if i receives the object he pays a∗1 − Ui(a)for it. We use the notation Bk

m =∑m

l=k

(ml

), which decreases as k grows. Assume

n ≥ 3 and pick any λ−, λ+ as in (1). Then we define

Ui(a) = a∗1 − n− (1− λ−)n− 1

{n−1∑k=1

(−1)k−1(n−2k−1

) Bkn−1

B1n−1

a∗k−i} for any a ∈ RN+ (2)

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ACM SIGecom Exchanges, Vol. 7, No. 1, December 2007.

2 · H. Moulin

where a∗k−i is the k-th highest valuation among buyers other than i. We have for alla

(1− λ+)a∗1 ≤∑i∈N

Ui(a) ≤ (1− λ−)a∗1

We also show that no tighter bounds on the respective shares of seller and buyerscan be implemented by a strategyproof mechanism (including a non VCG one)treating equals equally.

An interesting special case is λ+ = 1, i.e., we give most of the surplus to theseller. In this case the best lower bound on its share is λ− = 1 − n

2n−1 (slightlybetter than suggested by (1)), so the buyers get a non-negative share of surplus notlarger than n

2n−1 .Symmetrically, if we wish to give most of the surplus to the buyers, we set

λ− = 0. The corresponding mechanism (introduced in [Guo and Conitzer 2007a;Moulin 2007a])

Ui(a) = a∗1 − {n−1∑k=1

(−1)k−1(n−2k−1

) Bkn−1

B1n−1

a∗k−i} (3)

is meaningful if the object is the common property of the agents, so the seller isreplaced by a residual claimant burning the cash surplus generated by the mech-anism. We speak in this case of an assignment problem. The cash transfer to theresidual claimant never exceeds the share ρ = n−1

2n−1−1 of the efficient surplus:

0 ≤ a∗1 −∑i∈N

Ui(a) ≤ ρa∗1 for all a ∈ RN+

The left-hand inequality above states that the mechanism is self sufficient (alsoknown as feasible), it never generates a budget deficit.

Importantly, participation in the mechanism (3) is voluntary (Ui(a) ≥ 0 for alla), because the coefficients of a∗k−i in equation (3) alternate in sign and decrease inabsolute value.

By contrast, for any strictly positive share λ−, participants in the mechanism(2) may end up with a net loss: the payment by the agent who gets the object isalways non-negative, but can exceed a∗1 slightly 1; similarly an agent receiving noobject generally receives some cash, but could end up losing some2.

2. ASSIGNING A GOOD FAIRLY

In the assignment problem, the mechanism (3) is ”fair” because it treats equalsequally, and guarantees voluntary participation. The same is true of the Vickreyauction, but in the Vickrey auction the budget surplus (that we call a budget loss)may equal the entire efficient surplus a∗1, while for our mechanism (3) it neverexceeds the share ρ of a∗1.

A natural strenghtening of voluntary participation is often discussed in the liter-ature (e.g., [Moulin 1992; Cramton et al. 1987]):

1It is as high as (1 +λ−n−1

)a∗1 at the profile where exactly two agents have the same positivevaluation, and other valuations are zero.2He will pay as much as

λ−n−1

a∗1 at the profile where only one agent has a positive valuation.

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Auctioning or assigning an object: some remarkable VCG mechanisms · 3

—Unanimity lower bound : Ui(a) ≥ ai

n , everyone has a claim on a fair chance to getthe object.

Unfortunately, there exists no self sufficient strategyproof assignment mechanismmeeting the Unanimity lower bound: see Lemma 1 in [Moulin 2007b]3. A relatedbut not directly comparable lower bound on individual net gains was recently in-troduced by [Porter et al. 2004] (see also [Atlamaz and Yengin 2006]):

—q-fairness: Ui(a) ≥ a∗k

n , everyone has a claim on a fair share of the q-th highestvaluation.

The 1-fairness property forces the equal division of the surplus a∗1 (given self suf-ficiency), and is obviously out of reach for a self sufficient strategyproof mechanism(for instance, 1-fairness is stronger than the unanimity lower bound). It is easyto check that 2-fairness is equally impossible for such mechanisms. The followingVCG mechanism (introduced by [Bailey 1997], see also [Cavallo 2006]) is 3-fair:

U3i (a) = a∗1 − a∗1−i +

a∗2−i

n

It is self sufficient and its relative budget loss is at most 2n :∑

i∈N

U3i (a) = a∗1 − 2

n(a∗2 − a∗3) ⇒ 0 ≤ a∗1 −

∑i∈N

U3i (a) ≤ 2

na∗1

Letting q vary between 3 and n, we examine the tradeoff between the less andless generous individual guarantees under q-fairness, and our ability to minimizethe relative budget loss. For each q = 3, · · · , n, we seek the smallest number ρ(q)for which we can find a q-fair strategyproof mechanism treating equals equally andsuch that

0 ≤ a∗1 −∑i∈N

Ui(a) ≤ ρ(q)a∗1 for all a ∈ RN+

With the notation Bk,k′

m =∑k′

l=k

(ml

), and q∗ = q− 1 if q is odd,= q− 2 if q is even,

we find

ρ(q) =n− 1B1,q∗

n−1

for all q = 3, · · · , n (4)

and the correponding optimal VCG mechanism is

Uqi (a) = a∗1 − {

q∗∑k=1

(−1)k−1(n−2k−1

) Bk,q∗

n−1

B1,q∗

n−1

a∗k−i} for all a ∈ RN+ (5)

Equations (4) and (5) imply that for any odd q, ρ(q) = ρ(q + 1) and Uq = Uq+1.Therefore q-fairness with q odd does not cost more in terms of the index ρ thanthe weaker (q + 1)-fairness. If we fix an odd q and let n grow, we see from (4) that

3This is true among deterministic mechanisms. If lotteries are allowed, random assignment withuniform probability on all participants is vacuously strategyproof and achieves the Unanimitylower bound.

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4 · H. Moulin

ρ(q) goes to zero as 1nq−2 ; for instance

ρ(3) = ρ(4) =2n

; ρ(5) = ρ(6) =24

n(n2 − 5n + 18)

Interestingly n-fairness is free if n is odd, as in this case Un is precisely the mech-anism (3) and ρ(n) = ρ. And if n is even, ρ(n) = n−1

2n−1−2 is hardly larger thanρ = n−1

2n−1−1 so the price of n-fairness is very small.Writing {n

2 } for the integer n2 or n+1

2 , then {n2 }-fairness guarantees to everyone

a fair share of the median valuation. Although this is a considerably strongerrequirement than n-fairness, we note that it only requires to double the cap ρ(q)on the relative budget loss. Indeed it is easy to check, with the help of Stirlingformula, that B1,n∗

n−1 ' 2B1,{n

2 }∗

n−1 when n grows.

3. ASSIGNING A BAD FAIRLY

We now assume that one of the n agents must perform a costly task, for which theyare equally responsible; individual costs ci of doing the job are private information.This indivisible task is a common property ”bad”, and effciency requires to assignit to one of the agents with lowest cost. Cash transfers may be used to compensatethis agent. For examples of this problem see [Porter et al. 2004] and the classicNIMBY problem ([Kunreuther 1996]).

We write c∗k for the k-th lowest cost, and Vi for agent i’s net disutility, whereperforming no task and getting no cash yields Vi = 0. Thus in a budget balancedand efficient allocation we have

∑i∈N Vi = c∗1, whereas if the task is assigned

efficiently but transfers leave a surplus,∑

i∈N Vi ≥ c∗1.We wish to compare the budget loss generated by a self sufficient and strate-

gyproof mechanism to a meaningful notion of the ”efficient surplus”. The intuitivechoice of the efficient cost c∗1 proves misguided. Indeed the constant ρ caps theratio of the budget loss to c∗1 if and only if

0 ≤∑i∈N

Vi(c)− c∗1 ≤ ρc∗1 for all c ∈ RN+ (6)

There is in fact only one strategyproof mechanism treating equals equally for whichthe ratio

PVi−c∗1

c∗1 remains non-negative and bounded above. This is the familiarpivotal mechanism ([Green and Laffont 1979; Moulin 1986]) defined by

Vi(c) = c∗1 for all c ∈ RN+

Here the agent who is assigned the task is not compensated, while every other agentpays the minimal cost to the residual claimant. Thus ρ = n−1 and this huge wasteof money occurs for all c, disqualifying the pivotal mechanism.

A better estimate of the surplus from assigning the task efficiently is c∗n − c∗1,namely the difference between the worst and the best possible assignment of thetask4. Now the number ρ caps the relative budget loss of a given mechanism if and

4Another natural choice is c − c∗1, where c = 1n

PN ci is the average cost of the task. Here

the benchmark is the random assignment of the taks, a perfectly incentive compatible and fairmechanism. The corresponding formulas are different, but the gist of the results is preserved.

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Auctioning or assigning an object: some remarkable VCG mechanisms · 5

only if

0 ≤∑i∈N

Vi(c)− c∗1 ≤ ρ(c∗n − c∗1) for all c ∈ RN+ (7)

The pivotal mechanism does not satisfy (7) for any bounded number ρ. On theother hand the tightest cap ρ that a strategyproof and self sufficient mechanismtreating equals equally can achieve is

ρ =n− 1

2n−1 − 1if n is odd; =

n− 12n−1 − 2

if n is even

and a mechanism implementing ρ is

Vi(c) = c∗1 −{n∗∑

k=1

(−1)k−1(n−2k−1

) Bkn−1

B1n−1

c∗k−i} where n∗ = n− 1 if n odd, = n− 2 if n even

Note the close analogy with the tightest cap ρ and the optimal mechanism (3)in the good assignment problem: the formulas are in fact identical for n odd. Theparallel between the two models extends to the discussion of individual guarantees.

The natural Stand Alone upper bound Vi(c) ≤ ci cannot be met by a self sufficientstrategyproof mechanism unless its relative budget loss in (7) is unbounded; andthe much stronger Unanimity upper bound Vi(c) ≤ ci

n cannot be true for any selfsufficient strategyproof mechanism.

Now q-fairness places the upper bound Vi(c) ≤ c∗k

n on individual disutility. It isnot comparable to the Stand Alone upper bound, yet it is typically much tighter.We find that equation (5) where c replaces a defines a q-fair mechanism achievingthe cap ρ(q) = ρ(q) ((4)). Moreover this is the smallest feasible cap on the relativebudget loss of any q-fair strategyproof mechanism treating equals equally.

Therefore the tradeoffs between q-fairness and the worst relative budget loss areidentical in both models.

4. CONCLUDING COMMENTS

1. For proofs and more detailed discussion of the results described here, the readeris referred to [Moulin 2007a; 2007b].

2. Our results in section 1 beg for a generalization to the auction or assignmentof multiple, possibly heterogenous, objects.

For the case of multiple identical objects, the generalization of the VCG mech-anisms discussed here is straightforward: see [Guo and Conitzer 2007a; 2007b;Moulin 2007a]. However it can be shown that VCG mechanisms no longer mini-mize the (worst case) relative efficiency loss, therefore there is no reason to believethey achieve the optimal tradeoff between efficiency and q-fairness.

Finally the case of multiple heterogenous objects remains entirely open: a plau-sible conjecture is that we can improve somewhat the efficiency performance of thecanonical pivotal mechanism.

REFERENCES

M. Atlamaz and D. Yengin, Fair Groves mechanisms, mimeo, Rochester University, 2006.

M.J. Bailey, The demand revealing process: to distribute the surplus, Public Choice, 91:107-126,1997.

ACM SIGecom Exchanges, Vol. 7, No. 1, December 2007.

6 · H. Moulin

R. Cavallo, Optimal decision-making with minimal waste: strategyproof redistribution of VCGpayments, International Conference on Autonomous Agents and Multi-agents Systems, (AA-MAS) Hakodate, Japan, 2006.

P. Cramton, R. Gibbons and P. Klemperer, Dissolving a partnership efficiently, Econometrica,55, 3, 615-632, 1987.

M. Guo and V. Conitzer, Worst case optimal redistribution of VCG payments, Proceedings ofthe 8th ACM Conference on Electronic Commerce (EC-07), pp. 30-39, San Diego, CA, USA.

M. Guo and V. Conitzer, Worst case optimal redistribution of VCG payments (extended version)

J. Green, and J.J. Laffont , Incentives in public decision making, Amsterdam: North-Holland,1979.

H. Kunreuther, The Role of Compensation in Siting Hazardous Facilities, Journal of PolicyAnalysis and Management, September 1996

H. Moulin, Characterizations of the Pivotal Mechanism, Journal of Public Economics, 31, 53–78,1986.

H. Moulin, An Application of the Shapley Value to Fair Division with Money, Econometrica, 60,6, 1331–1349, 1992.

H. Moulin, Efficient, strategy-proof and almost budget-balanced assignment,www.ruf.rice.edu/˜econ/faculty/Moulin/spassgnt.pdf

H. Moulin, Auctioning or assigning an object: some remarkable VCG mechanisms,www.ruf.rice.edu/˜econ/faculty/Moulin/auctass1obj.pdf

R. Porter, Y. Shoham and M. Tennenholtz, Fair imposition, Journal of Economic Theory, 118,209-228, 2004.

ACM SIGecom Exchanges, Vol. 7, No. 1, December 2007.

An e¢ cient and almost budget balanced costsharing method

Hervé Moulin�

Rice University, Houston,Texas, USA

Revised, April 2008

Abstract

For a one input, one output convex technology C we character-ize n-person cost sharing games where the Nash equilibrium demandsmaximize total surplus. Cost shares can cover the exact cost if and onlyif C is a polynomial of degree n�1 or less. For general C, the residual�cost shares cover costs exactly only if at least one agent demands nooutput.If the cost function is totally monotone (e.g., polynomial with pos-

itive coe¢ cients, or exponential), a null demand agent receives a cashsubsidy and total payments always cover the actual cost. The ratio ofexcess payment to the e¢ cient surplus is no larger than minf 2

logn ; 1g.For power cost functions, C(a) = ap, p > 1, payments may yield a

surplus or a de�cit, and the ratio of budget imbalance to the e¢ cientsurplus converges to zero as 1

np�1 .For analytic cost functions, the ratio converges to zero exponen-

tially along a given sequence of users.All properties above are lost if the cost function is not smooth.

1 Introduction

The exploitation of a one input-one output technology (a cost function C)with increasing marginal costs occupies a special place in the mechanism

�I am greatly indebted to Fedor Nazarov at Michigan State University, for the crucialinsight into the proof of Theorem 1, and to Doug Hensley at Texas A&M, who providednumerical simulations in support of a key conjecture. I am also grateful to participants inthe 2006 AGATE workshop in Bertinoro, and in the meetings of the Society for EconomicDesign in Bodrum. This work is supported by the NSF under grant SES-0414543.

1

design literature, because in this simple model the complex tradeo¤s betweene¢ ciency and incentive compatibility can be studied in great detail. Wefocus here on cost sharing mechanisms, where users of the technology submitdemands of output which the mechanism must meet, and are charged apersonalized price (an amount of input)1.

The appeal of these mechanisms is the simplicity of their messages, aquantity of output, as opposed to the report of an entire preference or utilityfunction. It is well known that for most2 cost functions, we cannot guaran-tee e¢ ciency of all Nash equilibrium outcomes at all pro�les of individualpreferences (even restricted by quasi-linearity in either input or output).We take here the second best route, comparing cost sharing mechanisms interms of the relative ine¢ ciency of their Nash equilibrium outcomes.

We assume that preferences are quasi-linear in the input, so we can speakof the e¢ cient surplus with respect to the "no production" outcome, andcompare it to the e¢ ciency loss in a given equilibrium outcome (the di¤er-ence between e¢ cient surplus and equilibrium surplus). The performanceof our mechanism is measured by the worst (largest) ratio of these twonumbers, over all preference pro�les and equilibrium outcomes. This crude,prior-free index eschews the equilibrium selection problem, and provides acomplete ordering of cost sharing mechanisms.

An earlier paper [28] computed this index for three familiar cost sharingmethods, average, serial and incremental cost sharing. Average and serial3

cost sharing are budget balanced mechanisms: cost shares cover exactly thecost of production. The Nash equilibrium is unique4 and involves ine¢ cientoverproduction of output. The incremental method yields a di¤erent kindof ine¢ ciency. It achieves the surplus maximizing demand in any Nashequilibrium (as explained below) but collects more money than necessary:user i is charged yinci = C(xN ) � C(xN�i), where xN is total demand andxN�i the demand of other users, and convexity of C implies

Pi(C(xN ) �

C(xN�i)) � C(xN ). In practice the mechanism relies on a residual claimantwho absorbs the budget surplus.

1Examples include telephone billing ([4]), access to a network ([31, 24, 17]), queuinggames ([8, 30]).

2We give the precise meaning of most in Proposition 1.3Let C be the cost function and xi user i�s demand. Average cost shares are yi =

xixNC(xN ); and serial cost shares yi =

C(xi)n�i+1 �

Pi�1j=1

C(xj)(n�j+1)(n�j) , where x

i = (n � i +1)xi +

Pi�1j=1 xj when x1 � x2 � �� � xn.

4For the average cost games, this requires quasi-linearity of preferences (or at leastbinormality: [39]); not so for the serial games, that are in addition strictly dominancesolvable ([30]).

2

The main message of [28] is negative: when the number n of agentsgrows, the relative e¢ ciency loss (for the worst choice of individual pref-erences) converges to 1 for all three methods, though slower for serial (as1 � 1

logn) than for average or incremental (as 1 �1n). Whether the waste

comes from overproduction or from burning cash, it may exhaust the entirebene�t o¤ered by the technology.

Here we reach instead a positive conclusion: we propose a canonicalmodi�cation of the incremental mechanism, the residual� mechanism, withidentical non cooperative properties, in particular surplus maximizing Nashequilibrium demands. The payments it collects may generate a cash de�citor a surplus. For a large class of smooth (analytic) cost functions, whenn is large the residual� mechanism burns a vanishingly small amount ofcash, or requires a vanishingly small in�ow of cash, relative to the e¢ cientsurplus. Therefore the choice of a residual claimant is neither di¢ cult norcontroversial, though it is always easier to �nd one ready to eat some surplusthan to �nance a de�cit.

We provide an explicit upper bound of the relative e¢ ciency loss whenall derivatives of C are non negative, and an asymptotic result when C isanalytic. However we do not claim any optimality property for this up-per bound, be it among general cost sharing mechanisms, or among thoseguaranteeing surplus maximization in equilibrium.

The latter class of mechanisms admits a simple characterization, namelyuser i is charged the total cost C(xN ), minus an arbitrary charge indepen-dent of his own demand xi. Equivalently if vi(xi) is i�s willingness to payfor xi units of output, the surplus function P (x) =

Pi vi(xi) � C(xN ) is a

cardinal potential ([26]) of the corresponding demand game. Thus any sur-plus maximizing demand pro�le x is a Nash equilibrium at which the cashsurplus or de�cit is the only source of welfare loss. Although multiple Nashequilibria are possible, most familiar learning algorithms such as �ctitiousplay and best reply dynamics converge to one of them because the potentialP increases along any path ([26, 27, 20]). Therefore simple patterns of de-centralized behavior by the participants achieve one of the Nash outcomes,even when information on preferences is entirely private. On the other handcoordination on a particular Nash equilibrium in a one-shot version of thegame requires complete information or preplay communication.

An important feature of the residual� mechanism distinguishes it fromthe incremental one, and from most cost sharing mechanisms in the litera-ture (e.g., [31, 24]): there a null demand is neither charged nor subsidized(xi = 0) yi = 0), whereas in our residual� mechanism a user with a null orsmall demand typically receives a cash subsidy. This makes sense because

3

marginal costs increase: an inactive user bene�ts all active users by refrain-ing from using the technology, so she has a legitimate claim on the availablesurplus. However the manager of the mechanism must now monitor entry,lest "spam" users show up to demand no output and receive some cash.

Relation to the literatureClassical results on implementation in Nash equilibrium ([23]) do not

apply to our model because we work in the quasi-linear domain. The resid-ually e¢ cient mechanisms that we characterize are closer in spirit and inform to the Vickrey-Clarke-Groves mechanism ([12]), implementing in thatsame domain the surplus maximizing outcome, at the cost of some budgetimbalance. However the restriction to a one dimensional message rules outimplementation in dominant strategy.

The worst case methodology is standard fare in the optimization liter-ature (e.g., [37]). In the mechanism design literature, its �rst success isfor congestion and routing games ([19, 35, 33, 2]), where the correspondingperformance index was dubbed the price of anarchy. An upper bound onthe price of anarchy can be computed for very general con�gurations of therouting network and congestion costs5, which in turn gives some insightsinto the network design for which the index is minimal ([33, 2]).

For cost sharing problems where user gets at most one unit of indivisible"service", [31, 16] used the worst absolute e¢ ciency loss to compare variouscost sharing methods, respectively for the case of increasing and decreasingmarginal costs. For sharing the cost of a divisible commodity, this paperand its predecessor [28] were inspired by the substantial recent literature onthe capacity sharing problem.

A �xed amount of a divisible good (e.g., bandwidth) must be dividedbetween a given set of users. After eliciting a "bid" from each user, amechanism divides the good among and assign monetary charges ("costshares") to the bidders. The goal is to ensure that in equilibrium the pro�leof shares xi maximizes the surplus

Pi vi(xi): [13, 5, 22, 36, 40] . Just like

here, the restriction to a one dimensional scalar bid is regarded as compellingfor practical implementation among many users (e.g., [14]). However thedesign problem is more complicated for capacity sharing than for our costsharing model. Since the sum of the actual shares xi is �xed, the bid cannotbe a simple demand of output, and the surplus

Pi vi(xi) is not a potential

for any of the games discussed in the capacity sharing literature6. The

5 see Chapter 3 in [34], in particular Theorem 3.2.6.6Hence there is no analog to our characterization of surplus maximizing mechanisms

in Proposition 1.

4

restriction that cost shares should be proportional to capacity shares, as inKelly�s seminal paper ([18]), simpli�es the design problem, but allows onlyto recover 75% of the e¢ cient surplus (in the worst case): ([13, 22, 40]). Ifthis restriction is lifted, full surplus maximization is possible ([14, 40]). Seealso [15] who adapt Kelly�s mechanism to the case of an elastic supply andshow that it recovers 66% of the e¢ cient surplus. Finally [9] discusses thecase of multiple goods, each with a given capacity, when each user wantsthe same quantity of all goods; similar results obtain with a 2-dimensionalmessage.

Another key feature of our analysis is the parsimonious accounting ofthe charges collected from the users. We strive to recover almost exactly theactual cost of total demand, because an excess payment will be lost to theusers, and a de�cit will require help from an outside banker. By contrast thecapacity model is cost-free, but the charges collected are not a small fractionof the e¢ cient surplus. Some authors interpret the cash payments as therevenue that a seller wishes to maximize ([40]), others simply view them asa neutral redistribution that does not a¤ect social welfare ([13, 22, 36, 14]).

The parsimonious accounting of excess payments is also the viewpoint in[10, 11, 29], proposing a modi�ed Vickrey auction to assign (identical) indi-visible objects, in such a way that the worst cash surplus is an exponentiallysmall fraction of the e¢ cient surplus as the number of participants grows.

There is a symmetric class of output sharing mechanisms (often calledcooperative production games) where agents contribute input shares and themechanism divides the resulting output: e.g., [6, 21]. Close to home, Beviaand Corchon ([3]) discover independently the output sharing version of theresidual� mechanism (which they call "incremental") for polynomial costfunctions, and prove the counterpart of our Propositions 2. We comment inthe concluding section 9 on the possibility of adapting our other results tothat context.

2 Overview of the results

Section 3 introduces the model. Proposition 1 characterizes the cost sharingmechanisms guaranteeing the surplus maximizing output production at all(or at least one) Nash equilibrium outcomes for all preference pro�les. Wecall such mechanism residually e¢ cient. The general form of agent i�s costshare takes the simple form

yi = C(xN )� hi(x�i; C) (1)

5

for an arbitrary function hi (note the analogy with VCG mechanisms).Equivalently the surplus P (x) is a cardinal potential of the demand game,for all preference pro�les. If C is di¤erentiable all Nash equilibria of alldemand games maximize the surplus; if it is not, at least one equilibriummaximizes P , but one can choose preferences so that some equilibria do not.

The de�nition of the residual� mechanism �� in section 4 stems from theanswer to a natural question: for which cost function C can we �nd mech-anisms guaranteeing �rst best e¢ ciency of all Nash equilibrium outcomes,i.e., surplus maximizing demand and budget balance? By Propositions 2and 3 the only cost functions for which this is possible are the polynomialsof degree at most n� 17. And in this case the residual� mechanism is char-acterized by the additional property of anonymity (symmetric treatment ofthe agents).

The combinatorial expression (11) of residual� cost shares is not intuitivefor large n. In this overview we illustrate it only for n = 3; the cost shareof agent 1 is

y�1 = C(x123)� 2C(x23) + C(x2) + C(x3) (2)

Check that a null demand results in a negative cost share: x1 = 0 ) y1 =C(x2) +C(x3)�C(x23) � 0, by convexity of C and C(0) = 08. The budgetimbalance at the demand pro�le x is

�� =Xi

yi � C(x123) = 2fC(x123)�Xij

C(xij) +Xi

C(xi)g (3)

If C is not a quadratic polynomial, �� is typically non zero; however itvanishes whenever at least one agent is inactive: xi = 0 for some i) �� = 0.This property is apparent in (3) above; it holds for all n and all C, and infact characterizes �� among residually e¢ cient mechanisms: Proposition 3.

It is often easy to predict if the residual� mechanism will generate a cashsurplus (�� � 0), or a de�cit (�� � 0). Lemma 2 shows that the former(resp. the latter) always happens if the n-th derivative C(n) is non negative(resp. non positive) everywhere.

Section 5 de�nes the "worst case" performance index of a residuallye¢ cient mechanism � ((1)). The budget imbalance at the demand pro�le xis

�(x;C; �) =Xi

yi � C(xN ) = (n� 1)C(xN )�Xi

hi(x�i; C)

7This result is proven in [3] in the equivalent context of an output sharing game.8By contrast the incremental cost shares yinci = C(xN )�C(xN�fig) are characterized

by (1) and the requirement that an inactive user be neither charged nor subsidized: xi =0) yi = 0.

6

It can be positive or negative. The residual cost of the residually e¢ cientmechanism � is then

r(C; �) = supv

maxx2E(C;v)

j�(x;C; �)jmaxx0 P (x0; v; C)

where we maximize over all quasi-linear and convex preferences and all corre-sponding Nash equilibria. This seemingly untractable computation is greatlysimpli�ed by Lemma 3, adapting to our context a result of [13]. To computer(C; �) it is enough to maximize over pro�les v made of linear utilities, hencethe simple form

r(C; �) = supx2RN+

j�(x;C; �)jxNC 0(xN )� C(xN )

(4)

Theorems 1 and 2 in section 6 make a strong case for the residual�

mechanism if the cost function is totally monotone, namely in�nitely di¤er-entiable with all its derivatives non-negative. The mechanism never gener-ates a cash de�cit, no one su¤ers a net welfare loss from participating in themechanism (Voluntary Participation) and cost shares are co-monotonic todemands (Ranking). Most importantly, the residual cost is at most 2

logn forall n.

Theorems 3 and 4 in section 7 consider other classes of smooth costfunctions. For power functions C(a) = ap; p > 1, and their positive linearcombinations, the asymptotic e¢ ciency property of the residual� mechanismis strong: the residual cost vanishes as 1

np�1 when n grows (Theorem 3).Numerical evidence strongly suggests that both Voluntary Participation andRanking still hold for these functions, but I have been unable to prove ordisprove this conjecture.

A weaker result holds for analytic and convex cost functions: the ratio ofthe budget imbalance to the e¢ cient surplus converges to zero exponentiallywhen the set of users increases (Theorem 4).

We do not make any claim of optimality for our residual� mechanism.Fixing a cost function and a number of users, one would like to �nd the lowestfeasible residual cost among all residually e¢ cient mechanisms; the samequestion applies to the class of budget balanced cost sharing mechanisms,and to the class of mechanisms restricted by neither budget balance nor thee¢ ciency of the equilibrium output level. All three questions are widelyopen.

In section 8 we note that the regularity of the cost function is critical tothe good behavior of the residual� mechanism. For a piecewise linear cost

7

function, and more generally for a non di¤erentiable cost function, surplusesand de�cits will typically occur, and the residual cost grows exponentiallywith n. Voluntary Participation and Ranking no longer hold in general.

Some concluding comments are gathered in section 9, and many proofsare in the appendix, section 10.

3 Residually e¢ cient demand mechanisms

A �nite set N of agents share a technology C producing a units of outputat cost C(a). We assume that C is non decreasing and convex on R+, withC(a) = 0 , a = 0, and lim1

C(a)a = 1. We write C for the set of such

functions.Each user i 2 N requests the quantity xi; xi � 0, of output and is

charged $yi: this charge can be negative, namely user i can be subsidizedby the mechanism (for instance as a reward for a null demand). User i�sallocation is zi = (xi; yi) 2 R+ � R. A pro�le of individual allocations isdenoted z 2 (R+ � R)N and is called an outcome.

If S is a subset of N , we use the notation xS =Pi2S xi and x; = 0. The

budget imbalance of outcome z given C is

�(z; C) = yN � C(xN )

which we call a surplus if �(z; C) > 0, and a de�cit if �(z; C) < 0. If�(z; C) = 0 we call outcome z budget-balanced.

We write D for the set of functions u, concave and nondecreasing on R+,and normalized by u(0) = 0. The preferences of user i are represented by aquasi-linear utility function vi(xi) � yi, where vi 2 D. We abuse languageby speaking of D as the preference domain of our agents.

We writeP (x;C; v) =

XN

vi(xi)� C(xN ) (5)

for the social surplus at demand x, and

S(C; v) = maxx2RN+

P (x;C; v)

for the e¢ cient surplus of the economy (N;C; v). The normalizations C(0) =0 and vi(0) = 0 ensure S(C; v) � 0. We call a demand pro�le x e¢ cientif P (x;C; v) = S(C; v). Our assumptions guarantee that the set of e¢ cientdemand vectors x is non empty, compact and convex.

8

De�nition 1The outcome z is residually e¢ cient in (N;C; v) if the correspondingdemand pro�le x is e¢ cient.

In the quasi-linear context, outcome z is e¢ cient (Pareto optimal) if andonly if it is budget balanced and residually e¢ cient.

De�nition 2Given N and C 2 C, a demand mechanism is a game form associatingto each pro�le of demands x 2 RN+ a pro�le of charges y = �(x) 2 RN .A preference pro�le v 2 DN generates a non cooperative demand game(N; v; �), where xi is user i�s strategy and vi(xi)� �i(x) is her payo¤.

Our �rst result characterizes the mechanisms for which at any pro�lev 2 DN , the Nash equilibria of the demand game (N; v; �) are all e¢ cient,or at least one of these equilibria is e¢ cient. Recall from ([26]) that thefunction Q(x) de�ned on RN+ , is a cardinal potential for the normal formgame (N ;ui; i 2 N) if we have for all i:

ui(x0i; x�i)� ui(x) = Q(x0i; x�i)�Q(x) for all xi; x0i and all x�i

Lemma 1Fix N;C 2 C, and a mechanism �. The three following statements areequivalent:

i) for some v 2 DN the social surplus P ((5)) is a cardinal potential ofthe demand game (N; v; �);

ii) for all v 2 DN the social surplus P ((5)) is a cardinal potential ofthe demand game (N; v; �);

iii) there exist for all i a function hi is arbitrary over RN�i+ such that

yi = �i(x) = C(xN )� hi(x�i; C) for all i 2 N and x 2 RN+ (6)

ProofCheck i)) iii). Note that P is a cardinal potential for the game (N ;ui; i 2N) if and only if for all i, ui(x)�P (x) is independent of the variable xi. Ap-plying this to ui(x) = vi(xi)��i(x) we see that C(xN )��i(x)�

PN�fig vj(xj)

is independent of xi, therefore C(xN ) � �i(x) = hi(x) is independent of xias well.Statements iii)) ii) is just as easy and ii)) i) requires no proof.

Proposition 1For any N ,C 2 C, and mechanism �, the three following statements areequivalent:

9

i) for any utility pro�le v 2 DN , every e¢ cient demand x is a Nash equilib-rium of the demand game (N; v; �) (the corresponding outcome is residuallye¢ cient);ii) for any utility pro�le v 2 DN , the game (N; v; �) has at least one e¢ cientNash equilibrium;iii) the demand mechanism takes the form (6).

We call such a mechanism residually e¢ cient. Moreoveriv) if C is di¤erentiable, for all v 2 DN the only Nash equilibria of thegame (N; v; �) are the e¢ cient demands;v) if C is not di¤erentiable at a, there exist v 2 DN and a Nash equilibriumx of (N; v; �) such that xN = a and x is not e¢ cient.

ProofStep 1: iii)) i)) ii): Fix a pro�le of utilities v 2 DN . That a maximizerof P is a Nash equilibrium of (N; v; �) is clear because P is a potential forthis game (Lemma 1).

Step 2: ii)) iii)We �x a mechanism � meeting ii), an arbitrary x 2 RN+ , and we choose � inthe subgradient ([32]) @C(xN ) of C at xN . This means that the line withslope � through (xN ; C(xN )) supports the graph of C

C(a)� C(xN ) � �(a� xN ) for all a

Assume �rst x 6= 0 so that � > 0 by de�nition of C. Choose "; 0 < " < �and construct a utility pro�le v as follows

vi(a) = (�+ ")a for 0 � a � xi; vi(a) = (�� ")a+ 2"xi for a � xi

We claim that x is the only e¢ cient demand pro�le at v. Consider anotherdemand pro�le y such that yN > xN , and pick i such that yi > xi. Choosey0i, y

0i < yi, such that y0i � xi and y0i + yN�fig > xN . We check that

P (y) < P (y0i; y�i), implying that y is not e¢ cient. Pick � in @C(y0i+yN�fig),

so � � � by convexity of C. Then compute

vi(yi)� vi(y0i) = (�� ")(yi � y0i) < �(yi � y0i) � C(yN )� C(y0i + yN�fig)

completing the check.Consider next y such that yN < xN and pick similarly an agent i such

that yi < xi; a symmetrical argument, omitted for brevity, shows that asmall increase of yi increases P (y), so y is not optimal either. Finally if y issuch that yN = xN but y 6= x, we choose i such that yi > xi and compute

vi(yi)� vi(xi) = (�� ")(yi � xi) < �(yi � xi) � C(yi + xN�fig)� C(xN )

10

so that, again, decreasing yi to xi improves P . The claim is proven.By assumption ii); x is a Nash equilibrium of the game (N; v; �). Write

the equilibrium property for agent i:

vi(xi)� �i(x) � vi(ai)� �i(ai; x�i) for all ai � 0

, �i(ai; x�i)��i(x) � (�+")(ai�xi) if ai � xi; � (��")(ai�xi) if ai � xiLetting " go to zero we conclude

�i(ai; x�i)� �i(x) � �(ai � xi) for all ai � 0 (7)

Earlier we assumed x 6= 0. We check now that (7) holds for x = 0 as well.If � > 0 we choose " < �, de�ne the pro�le vi(a) = (� � ")a for all i anda; and repeat the argument above: 0 is the only e¢ cient demand, hence aNash equilibrium, and the equilibrium property boils down to (7). If � = 0we simply use the pro�le vi(a) = 0 for all i and a. The null demand is thenuniquely e¢ cient by our assumption a > 0) C(a) > 0.

Fix now an agent i and the demands x�i. Write for all a � 0 : f(a) =�i(a; x�i) and (a) = @C(a+ xN�i). Inequality (7) holds for all xi; ai, andfor all � 2 @C(xN ), so it can be written as

f(b)� f(a) � �(b� a) for all a; b � 0 and all � 2 (a)

) f(b) = supa�0ff(a) + (a)(b� a)g

Hence f is convex, as the supremum of the a¢ ne functions b! f(a)+�(b�a)over all a and � 2 (a). Moreover its subgradient @f(a) contains (a), itselfa subgradient. Therefore (Theorem 24.3 in [32]) @f(a) = (a) and f aprimitive of : f(a) = C(a+ xN�i) + hi, where hi depends only upon x�i.This is the desired property (6).

Step 3 statement iv)Assume C is di¤erentiable, �x a residually e¢ cient mechanism (6) and v 2DN , and let x be a Nash equilibrium. We prove that x is e¢ cient. Forall i the equilibrium property says that the convex function a ! g(a) =�i(a; x�i) � vi(a) reaches its minimum at xi, or equivalently 0 2 @g(xi).Now the subgradient correspondence is additive (Theorem 23.8 in [32]) andfor a di¤erentiable function, it is just the derivative:

@g(xi) =@

@xi�i(x) + @f�vig(xi) = C 0(xN ) + @f�vig(xi)

11

therefore 0 2 @g(xi) , �C 0(xN ) 2 @f�vig(xi), namely the line with slopeC 0(xN ) supports the function vi from above at xi:

vi(bxi)� vi(xi) � C 0(xN )(bxi � xi) for all bxi � 0 (8)

Summing up these inequalities for all i, and taking into account C 0(xN )(bxN�xN ) � C(bxN )�C(xN ), gives P (bx) � P (x) for all bx and completes the proof.Step 4 statement v)Given a residually e¢ cient mechanism � and a preference pro�le v, thesurplus P (x) =

PN vi(xi) � C(xN ) is a potential for the demand game

(N; v; �), therefore x is a Nash equilibrium of that game if and only if it is acoordinatewise maximum of P . Suppose C is convex but not di¤erentiableat a, where his subgradient is @C(a) = [�; �] with � < �. Assume N =f1; 2g for simplicity (the general case is just as easy) and consider the twolinear utilities v1(a) = �a; v2(a) = �a with � � � < � � �. The demandx� = (a2 ;

a2 ) is a Nash equilibrium because vi(xi) � C(xi + a

2 ) reaches itsmaximum at a2 for i = 1; 2. However P (x

�) < P ((0; a)). �Statement v) uncovers a limitation of residually e¢ cient mechanisms

when the cost function is not di¤erentiable: decentralized convergence toan e¢ cient demand is not guaranteed. Our results below on the asymp-totic budget balance of one canonical such mechanism, require much moreregularity from the function C.

For any residually e¢ cient mechanism, user i�s cost share yi is non de-creasing in her demand xi, however the sign of this cost share is not re-stricted: a null demand xi = 0 may result in a payment to i, yi < 0. Asdiscussed in the introduction, subsidizing an inactive user is justi�able underincreasing marginal costs. What is not desirable is the con�guration xi = 0and yi > 0, in which user i would rather not participate in the mechanism.

De�nition 3A residually e¢ cient mechanism � satis�esVoluntary Participation (VP)if for any v 2 DN and any Nash equilibrium outcome z, we have vi(xi) ��i(x) � 0 for all i 2 N (recall vi(0) = 0).It satis�es Ranking (RKG) if for all demand pro�le x and all i; j 2 N :

xi � xj ) �i(x) � �j(x)

Ranking rules out a very coarse form of inequity, whereby a user ischarged more for demanding less output9.

9A stronger property is No Envy : no user prefers another user�s equilibrium outcometo his own. In fact No Envy is out of reach for residually e¢ cient mechanisms, with theexception of the two person problem with quadratic costs. We omit the straightforwardproof.

12

Note that a mechanism � satis�es VP if and only if a null demand isnot charged: for all x and all i 2 N : xi = 0 ) �i(x) � 0. AssumeVP and pick x such that xi = 0. At the utility pro�le vi � 0; vj(bxj) =minfC 0(xN )bxj ; C 0(xN )xjg for all j 6= i, x is a Nash equilibrium and VPapplied to i gives �i(x) � 0. Conversely if xi = 0 ) �i(x) � 0 agent iguarantees a non negative utility by a null demand, and VP follows.

The most natural residually e¢ cient mechanism is the incremental one,de�ned by the property that a null demand is neither charged nor subsidized:xi = 0) �i(x) = 0. This gives :

yinci = C(xN )� C(xN�i) for all i 2 N and x 2 RN+ (9)

Its budget imbalance at x is yincN �C(xN ) = (n� 1)C(xN )�PN C(xN�i).

Convexity of C and C(0) = 0 imply yincN � C(xN ), by summing up theinequalities C(xN�i) � xi

xNC(0) +

xN�ixN

C(xN ).Consider next the incremental� mechanism, also residually e¢ cient:

y�i = C(xN )�n� 1n

C(n

n� 1xN�i) for all i 2 N and x 2 RN+

Inequality y�N � C(xN ) follows directly from the convexity of C. We letthe reader check that both mechanisms meet Ranking and VP, and thaty�i � yinci for all x and i.

4 The residual� mechanism

Can a residually e¢ cient mechanism be Pareto optimal, i.e., budget balancedfor every pro�le of demands? Given a cost function C can we choose thefunctions hi so that

(n� 1)C(xN ) =XN

hi(x�i; C) for all x 2 RN+ (10)

(where n is the cardinality of N)? The answer to this question singles outthe polynomial cost functions, and leads to the canonical mechanism whichis the object of this paper.

Say the mechanism � is anonymous if it treats symmetrically the n de-mands, namely x! �(x) is symmetric in all variables. A residually e¢ cientmechanism (6) is anonymous if and only if hi(x�i; C) = h(x�i; C), wherethe function h is symmetric in its n� 1 variables.

Proposition 2

13

If C 2 C is not a polynomial of degree at most n�1, no n-person residuallye¢ cient mechanism achieves budget balance at all demand pro�les.

The proof is in the Appendix. It is technically similar to Walker�s CubicalArray Lemma (p.1529 in [38]), stating a necessary and su¢ cient conditionfor the existence of a VCG mechanism ([12]) balancing the budget at allutility pro�les. See also [25] discussing a class of budget-balanced VCGmechanisms.

De�nition 4For any C in C, de�ne the n-residual� mechanism:

y�i = ��i (x) = C(xN )� (n� 1)

n�1Xk=0

(�1)kk + 1

XT�N�i;jT j=k

C(xN�fi[Tg) (11)

(recall the convention x; = 0). It is residually e¢ cient and anonymous.

With two agents, equation (11) gives h�(x�1) = C(x2) hence the 2-residual� is just the incremental one:

n = 2 : y�1 = yinc1 = C(x1 + x2)� C(x2)

For n = 3, they are already di¤erent:

n = 3 : y�1 = C(x123)� 2C(x23) + C(x2) + C(x3) (12)

The budget imbalance is

��3(x;C) = 2fC(x123)�X

ij�(1;2;3gC(xij) +

Xi2f1;2;3g

C(xi)g (13)

Unlike the incremental one, for a judicious choice of C the 3-residual� mech-anism may generate a surplus or a de�cit, depending upon the demandpro�le. An example is the cost function C(a) = ea � a3

3 � a � 1, in C, forwhich the function

��3((a

3;a

3;a

3); C) = ea � 3e

2a3 + 3e

a3 � 2

27x3 � 1

changes sign in a: see Figure 1. On the other hand Lemma 2 below showsthat if the 3rd derivative of C is of constant sign on R+, ��3(x;C) is of thesame sign for all x.

Check now that 3-residual� meets Voluntary Participation. If agent 1 isinactive, x1 = 0, he receives a cash subsidy, as y�1 = C(x2)+C(x3)�C(x23) �

14

0. Ranking holds true as well, because y�1 � y�2 , C(x2)�C(x1) � C(x23)�C(x13) and C is convex.

However, Voluntary participation does not hold anymore when 4 or moreagents are present. Agent 1�s 4-residual� cost share is

n = 4 : y�1 = C(x1234)� 3C(x234) +3

2

Xij�(2;3;4g

C(xij)�X

i2f2;3;4gC(xi)

For x1 = 0 and xi = 1 for i = 2; 3; 4, we get y�1 = �2C(3)+ 92C(2)�3C(1), so

y�1 = +72 for C(a) = maxfa; 8a�7g. For brevity we let the reader check that

Ranking holds for the 4-residual�, but not for the 5-residual� mechanism andbeyond.

We turn to the two characteristic properties of the residual� mechanism:for any C it is budget balanced whenever at least one demand is zero; if Cis a polynomial of degree at most n � 1, it is budget balanced everywhere.The former statement is clear with 3 agents using (12). That the budgetimbalance (13) cancels when C is a quadratic polynomial is also easy tocheck. It generalizes.

Proposition 3i) The budget imbalance of the residual� mechanism at the demand x andcost C 2 C is

��n(x;C) = (n� 1)fnXk=0

(�1)kX

T :jT j=kC(xN�T )g (14)

ii) Given C, the residual� mechanism is characterized by the combinationof three properties: anonymity, residual e¢ ciency, and guaranteed budgetbalance whenever at least one demand is zero:

fxi = 0 for at least one i 2 Ng ) ��n(x;C) = 0 (15)

iii) Assume C is a is a polynomial of degree at most n � 1. Then then-residual� mechanism is budget balanced for all x 2 Rn+. Moreover it isthe only anonymous demand mechanism such that for any v 2 DN all itsNash equilibrium demands are e¢ cient, and the corresponding cost sharesare budget balanced.ProofStep 1: equation (14)Fix T; T � N , and check that the term C(xN�T ) appears in the sum (11)only if i 2 T and then with coe¢ cient (�1)jT j n�1jT j .

15

Step 2: auxiliary resultFix N and a set function V from subsets of N , including N and the emptyset, into R. De�ne B(N;V ) =

Pn0 (�1)k

PT :jT j=k V (N�T ). Call agent

i 2 N a dummy in (N;V ) if V (T [fig) = V (T ) for all T � N�fig. Then ifsome agent i is a dummy, B(N;V ) = 0. Indeed T ! T [fig is a one-to-onemapping from 2N�fig to 2N�2N�fig, and in the sum de�ning B, the termsV (T ) and V (T [ fig) cancel each other.

Now let G be an arbitrary function from x 2 RN+ to G(x) 2 R. For anyx, de�ne BN (x;G) =

Pn0 (�1)k

PT :jT j=kG(xN�T ; 0T ). Applying the above

remark to the set function V (T ) = G(xN�T ; 0T ), we see that BN (x;G) = 0whenever at least one coordinate xi is null.

Step 3: statement ii)Equation (14) reads ��n(x;C) = (n � 1)BN (x;G), where G is the functionG : G(x) = C(xN ), so (15) follows at once from step 2.

Next we claim that h� is the only function h symmetric in its n � 1variables, such that for all x, (n � 1)C(xN ) �

PN h(x�i; C) = 0 whenever

at least one coordinate of x is null.Pick such a function h(x�i) (for simplicity we omit C in h(x�i; C)).

We prove the claim by induction on the number of non-zero coordinates ofex 2 Rn�1+ . For x = 0 2 Rn+, budget balancePN h(x�i) = (n � 1)C(xN )

gives h(0) = C(0). Assume h(ex) is determined for each ex with at mostk � 1 non zero coordinates, and consider ex = (x1; � � � ; xk; 0; � � � ; 0) 2 Rn�1+ .Set x = (ex; 0) 2 Rn+; budget balance at x gives Pk

1 h(x�i) + (n � k)h(ex) =(n� 1)C(xf1;��� ;kg), which determines h(ex).Step 4: statement iii)First we claim that the budget imbalance (14) is identically zero if C isa polynomial of degree at most n � 1. We prove this whether or not thepolynomial C is a convex function, or satis�es the other assumptions in thede�nition of C. The following notation will be useful; for all k; 0 � k � n

Ank(x;C) =X

T�N;jT j=kC(xN�T ) ; Bn(x;C) =

nX0

(�1)kAnk(x;C) (16)

De�ne for any a � 0, Ca(b) = C(a+b). We check �rst the following identity,for all i, all k = 0; � � � ; n� 1, all C and x:

Ank(x;C)�Ank((0; x�i); C) = An�1k (x�i; Cxi � C) (17)

For any T containing i, the two corresponding terms in the LHS cancel; forT � N�fig, they are C(xN�T ) � C(xN�fi[Tg) = (Cxi � C)(x(N�fig)�T ),

16

establishing (17). As Bn(x;C) is a linear combination of Ank(x;C), andAnn(x;C) = A

nn((0; x�i); C) = C(0), (17) gives Bn(x;C)�Bn((0; x�i); C) =

Bn�1(x�i; Cxi � C). Step 2 gives Bn((0; x�i); C) = 0, hence we have for alli; C;and x:

Bn(x;C) = Bn�1(x�i; Cxi � C) (18)

The claim follows by an easy induction on n. Start with B1(x;C) = C(x)�C(0), which is null if C is a constant. Next if C is a polynomial of degreeat most n� 1, then Cxi � C is a polynomial of degree at most n� 2.

The second statement in iii) follows at once from Proposition 1 andstatement ii) above.�

If C is su¢ ciently di¤erentiable, our next result helps to predict whetherthe residual� mechanism will produce a budget surplus or a de�cit.

Lemma 2Suppose the function C 2 C is n times di¤erentiable on R+�f0g.

i) If C(n) � 0 (resp. > 0) on R+�f0g, then ��n(x;C) is weakly in-creasing (resp. strictly increasing) in x and ��n(x;C) � 0 on Rn+ (resp.��n(x;C) > 0 on Rn+�f0g).

ii) If C(n) � 0 (resp. < 0) on R+�f0g, then ��n(x;C) is weakly de-creasing in x (resp. strictly decreasing), and ��n(x;C) � 0 on Rn+ (resp.��n(x;C) < 0 on Rn+�f0g).

ProofBy linearity of B in C, both statements are equivalent. We prove �rst byinduction on n the following property P(n)

fC(n) � 0 on R+�f0gg ) Bn(x;C) weakly increasing in x

Property P(1) is clear from B1(x;C) = C(x) � C(0). If C is n times dif-ferentiable, Cx1 � C is n� 1 times, and C(n) � 0 implies (Cx1 � C)(n�1) �0. The inductive argument follows at once from (18). Next P(n) andBn((0; x�1); C) = 0 ((15)), imply Bn(x;C) � 0. A similar argument shows

fC(n) > 0 on R+�f0gg ) Bn(x;C) strictly increasing in x

�Consider a power function Cp(a) = ap; p > 1. If p is an integer, all its

derivatives are non negative so the residual� mechanism generates a surplusfor any n. The same is true of any polynomial C(a) =

PKk=0 �ka

k withpositive coe¢ cients �k.

If p is not an integer, the sign of the transfer to the residual claimantdepends on the parity of n. Write dpe for the smallest integer larger than

17

p: the �rst dpe derivatives of Cp are positive, then they are alternativelynegative (for dpe+ 1; dpe+ 3; : : :) and positive (for dpe+ 2; dpe+ 4; : : :).

5 The residual cost of a residually e¢ cient mech-anism

In this and the next two sections we assume that the cost function C isdi¤erentiable. By Proposition 1 for all v 2 DN the set E(C; v) of e¢ cientdemand pro�les x equals then the set of Nash equilibrium demands in thegame (N; v; �), for any residually e¢ cient �.

We measure the e¢ ciency loss of a residually e¢ cient mechanism by theworst ratio of the budget imbalance of a Nash equilibrium to the e¢ cientsurplus. Fix N;C and a preference pro�le v. We write the budget imbalanceof � at x as follows

�(x;C; �) =XN

�i(x)� C(xN ) = (n� 1)C(xN )�XN

hi(x�i; C)

In De�nition 5 and Lemma 3 we use the following convention for a ratio f(t)g(t)

where g(t) = 0 is possible:

fg(t) = 0 and f(t) = 0g ) f(t)

g(t)= 0; fg(t) = 0 and f(t) 6= 0g ) f(t)

g(t)= +1

De�nition 5Fix N and a di¤erentiable cost function C 2 C.i) The residual cost of the residually e¢ cient outcome z = (x; y) at pro�lev 2 DN is

r(z; C; v) =j�(z; C)jS(C; v)

=jyN � C(xN )jPN vi(xi)� C(xN )

ii) The residual cost of the residually e¢ cient mechanism � is

r(C; �) = supv2DN ;x2E(C;v)

j�(x;C; �)jS(C; v)

Note that we count surplus and de�cit alike. A surplus is inherently eas-ier to implement than a de�cit, yet we submit that a large budget imbalance,irrespective of its sign, makes the residual claimant scenario implausible.

Observe that if the residually e¢ cient mechanism � never generates abudget de�cit, and satis�es Voluntary Participation, then r(C; �) � 1 for

18

all C. Pick any preference pro�le v, e¢ cient demand x, and cost sharesy = �(x). VP implies vi(xi) � yi for all i. Summing upXN

vi(xi) � yN = C(xN )+�(x;C; �) =)j�(x;C; �)jS(C; v)

=�(x;C; �)P

N vi(xi)� C(xN )� 1

(where ifPN vi(xi) = C(xN ) we have �(x;C; �) = 0 and the ratio is zero

by convention).The following key result is a variant of the "linearity lemma" (Lemma 4

in [13]). It reduces the computation of the residual cost of a mechanism to atractable maximization problem. We speak of unanimous utilities when allusers share the same utility function. For any demand pro�le x, the worstresidual cost over all utility pro�les for which x is a Nash equilibrium obtainsfor the unanimous linear utilities of slope C 0(xN ). If C 0 increases strictly atxN , bx is then a Nash equilibrium if and only if bxN = xN . In general any bxsuch that C 0(bxN ) = C 0(xN ) is an equilibrium.

Notation. We de�ne

�(a;C) = aC 0(a)� C(a) for all a � 0

namely the e¢ cient surplus at the unanimous linear utilities with slopeC 0(a).

Lemma 3Fix N , a di¤erentiable C 2 C, and a residually e¢ cient mechanism �; wehave

r(C; �) = supRN+

j�(x;C; �)j�(xN ; C)

(19)

ProofPick an arbitrary utility pro�le v in D and a Nash equilibrium x in thecorresponding demand game. As in Step 3 of the proof of Proposition 1,inequality (8) holds true. Applying it at bxi = 0 gives vi(xi) � C 0(xN )xi.Next consider the unanimous utilities v�i (a) = C

0(xN )a for all i and a � 0.Clearly x is still a Nash equilibrium in (N; v�; �), and we have

S(C; v) =XN

vi(xi)� C(xN ) �XN

v�i (xi)� C(xN ) = �(xN ; C)

If �(xN ; C) > 0, we get

j�(x;C; �)jS(C; v)

� j�(x;C; �)j�(xN ; C)

=j�(x;C; �)jS(C; v�)

(20)

19

If �(xN ; C) = 0, our convention implies that the RHS of (20) is zero if�(x;C; �) = 0 and 1 else. Hence (20) still holds. �

We illustrate Lemma 3 with the two incremental mechanisms (section 3).The budget surplus of �inc is �(x;C; �inc) = (n�1)C(xN )�

PN C(xN�i) �

0. Fixing a = xN , the maximum over x obtains for xi = xj all i; j. Therefore(19) reads

rn(C; �inc) = (n� 1) sup

a>0

C(a)� nn�1C(

n�1n a)

aC 0(a)� C(a) (21)

We claim limn!1 rn(C; �inc) = 1: when n grows large, the transfer to the

RC may exhaust almost the entire surplus. By VP and �inc � 0, we havern(C; �

inc) � 1 (see the argument just after De�nition 5). On the other hand

rn(C; �inc) � (n� 1)

C(1)� nn�1C(

n�1n )

C 0(1)� C(1)

and by di¤erentiability of C the right-hand side converges to 1 as n grows.The performance of the incremental� mechanism is worse. We have

j�(x;C; ��)j = (n�1)f 1nPN C(

nn�1xN�i)�C(xN )g, and the maximum for

a = xN �xed obtains when all but one demands are null, hence

rn(C; ��) = (n� 1) sup

a>0

n�1n C( n

n�1a)� C(a)aC 0(a)� C(a) (22)

Using inequality C( nn�1a) � C(a) +

1n�1aC

0(a) in the RHS of (22), we get

rn(C; ��) � (n�1)

n , so for large n, the worst case de�cit is at least the entiresurplus. For a power function Cp(a) = ap; p > 1, check limn rn(Cp; ��) = 1;on the other hand for E(a) = ea � 1, pick a = n(n� 1) in (22) to concludelimn rn(E; �

�) =1.Contrast the two incremental mechanisms: if the burden of budget im-

balance is borne by the users, that extra cost will at worse eat up all thesurplus; if we ask the mechanism to cover a de�cit, the latter may growarbitrarily larger than the e¢ cient surplus.

6 Totally monotone cost functions

Our �rst two theorems identify a class of cost functions for which theresidual� mechanism, despite the intricate computation of its cost shares,is an appealing device to exploit the technology C: no agent is coerced intousing the mechanism (VP); cost shares are comonotonic to demands (RKG);and the relative e¢ ciency losses vanishes no slower than 1

logn .

20

De�nition 6The function C is totally monotone if it is inde�nitely di¤erentiable on R+,and all its derivatives are non negative. Such a function is in C if C(0) = 0and C is not linear.

Totally monotone functions are increasing and convex, as well as ana-lytic10, stable by positive linear combinations, multiplication, and composi-tion. Examples in C include polynomials with positive coe¢ cients and noconstant term, the functions ea�1, and, up to a constant ensuring C(0) = 0,the functions eP (a), eP (a)+ e�P (a), eP (a)� eQ(a) if P;Q; and P �Q are poly-nomials with positive coe¢ cients, etc..

Theorem 1i) If C 2 C is n� 1 times di¤erentiable and C(k) � 0 for k = 1; � � � ; n� 1,the residual� mechanism satis�es RKG, and VP; furthermore

y�i � yinc = C(xN )� C(xN�i) for all x 2 Rn+; all i 2 N (23)

ii) If C 2 C is n times di¤erentiable and C(k) � 0 for k = 1; � � � ; n

y�i � y�i = C(xN )�n� 1n

C(n

n� 1xN�i) for all x 2 Rn+; all i 2 N (24)

Corollary to Lemma 2 and Theorem 1If C 2 C and C(k) � 0 for k = 1; � � � ; n, the residual� mechanism neverruns a de�cit, and guarantees Voluntary Participation and Ranking.

The following notation is used in our next result, and in subsequentsections: an = O(bn) means that an

bnis bounded away from in�nity, and

an = �(bn) means that it is bounded away from zero and in�nity.

Theorem 2i) If C is totally monotone, the residual cost r�n(C) of the n-residual

� mech-anism is bounded as follows:

r�n(C) � minf2

log n; 1g for all n (25)

ii) If E(a) = ea � 1, then r�n(E) = �( 1logn)

The proof of both Theorems is in the Appendix.Recall that ��n(x;C) � 0 and VP together guarantee r�n(C) � 1. Thus

the bound r�n(C) � 2logn starts to bite for n � 8. For instance

r�10(C) � 0:87; r�25(C) � 0:62; r�50(C) � 0:51; r�100(C) � 0:4310because their Taylor development at 0 implies C(a) �

PK0 C

(k)(0)ak

k!for all K and

all a � 0.

21

The upper bound (25) does not depend on the particular cost in the totallymonotone class. For speci�c cost functions it can typically be improved. Anexample is the exponential function E

n 2 5 10 25 50r�n(E) 0:5 0:22 0:20 0:17 0:15

These numerical values, as well as those in the next section, follow fromLemma 6 in the Appendix providing a simple expression for r�n(C) wheneverC is n-times di¤erentiable and its derivatives have constant sign.

7 Other smooth cost functions

7.1 Power functions

Power functions results from a Cobb Douglas technology under �xed factorprices, and are ubiquitous in empirical models. We saw at the end of section4 that it is easy to sign the budget imbalance of the residual� mechanism.We prove now that the residual cost r�n(C) converges to zero in n at apolynomial rate.

For any p > 1 recall the notation Cp(a) = ap. If p is not an integer, Cpis not totally monotone. As discussed after Lemma 3, the sign of ��n(x;Cp)depends on the parity of n once n � dpe.

Theorem 3i)For any p, p > 1, we have r�n(Cp) = O(

1np�1 );

ii) Fix a positive generalized polynomial C: C(a) =PK0 �ka

pk with pk > 1and �k > 0 for all k. Then r�n(C) = O(

1np

��1 ), where p� = mink pk.

Proof in the Appendix. Some numerical examples for p = 1:5 and p = 2:5suggest that the convergence is actually faster than 1

np�1 :n 2 5 10 25 50

r�n(C1:5) 0:58 0:1 0:04 0:017 0:009r�n(C2:5) 0:43 0:01 0:0012 0:00012 0:000025

Note that in Theorem 3, the convergence of r�n(Cp) to zero is not uniformin p: there exists a constant Kp depending upon p such that r�n(Cp) �

Kp

np�1 .Although one checks easily that �n = supp>1 r

�n(Cp) is �nite for all n, it is

unclear whether �n converges to zero as n grows. Numerical computationsfor small values of n reveal �n = limp!1 r

�n(Cp) and give

n 2 5 10 25 50�n 0:69 0:47 0:36 0:28 0:17

Do the Voluntary Participation and Ranking properties hold for powerfunctions ? By Theorem 1 they do if p � n � 2, in particular for any p if

22

n = 3. For larger n, I conjecture that VP and RKG hold for all powerfunctions, and 1140 numerical tests in the case n = 7 did not disprove theconjecture11.

On the other hand the inequality y�i � y�i certainly does not hold forn � 7. A counterexample12 is p = 1:2; x = (0; 3; 6; 10; 12; 14; 20), for whichy�1 = �38:7 < �4:7 = y�1 .

7.2 Analytic functions and generalized polynomials

Totally monotone functions are simply analytic functions C(a) =P10 �ka

k

where the coe¢ cients �k are non negative. When we remove the sign re-striction on the �k, the budget imbalance ��n(x;Cp) cannot be signed, andneither is VP or RKG guaranteed. However Theorem 4 establishes a weakerform of asymptotic e¢ ciency. For a given C and an arbitrary in�nite se-quence of users (utility pro�les), we show that the residual cost among the�rst n users in the sequence vanishes in n13. In contrast to Theorems 2and 3, Theorem 4 below o¤ers no insight about the asymptotic behavior ofr�n(C). In particular I do not know if r

�n(C) converges to zero for all analytic

functions C.

De�nition 7Fix C 2 C and for all n a n-person residually e¢ cient mechanism �n.For any utility pro�le vn 2 Dn, de�ne the corresponding residual costbr(C; vn; �n)

br(C; vn; �n) = supx2E(vn;�n)

r(z; C; vn) where z = (x; �n(x))

We call the sequence of mechanisms f�ng asymptotically e¢ cient at C,if for any sequence fvn; n = 1; 2; � � � g in D, we have

limn!1

br(C; (v1; v2; � � � ; vn); �n) = 0 (26)

We check that the sequence of incremental+ mechanisms is not asymp-totically e¢ cient. Fix a > 0, and consider the sequence of identical linearpreferences with slope C 0(a). With n users, the unanimous demand pro�le

11 I am grateful to Doug Hensley for running these random tests.12again provided by Doug Hensley.13Reference [7] develops a related analysis for two simple budget balanced demand

games, but for a given distribution of preferences.

23

xi =an is the worst Nash equilibrium, hence we get as in (21)

brn(C; (v1; v2; � � � ; vn); �+) = (n� 1)C(a)� nC(n�1n a)

aC 0(a)� C(a)

and this ratio goes to 1 as n goes to in�nity. The same conclusion holds forthe incremental� mechanism.

Turning to the residual� mechanism, we rewrite (19) as

r�n(C) = supa>0

supx:xN=a j��n(x;C)j

�(a;C)

The convergence of r�(C) to zero implies thatsupx:xN=a j��n(x;C)j

�(a;C) goes to zerouniformly in a. We show next that asymptotic e¢ ciency of �� only requiresthis ratio to converge pointwise in a.

Lemma 4The residual� mechanism is asymptotically e¢ cient at C 2 C if

for all a > 0: limn!1

supx:xN�a

j��n(x;C)j = 0 (27)

ProofFix C satisfying (27), and the sequence fvng. Pick for each n an arbitraryNash equilibrium xn of the demand game for vn = (v1; v2; � � � ; vn). We mustshow limn r�n(z

n; C; vn) = 0. In the proof below, we assume for simplicitythat C has no �at part, namely C 0 is strictly increasing on R+. The detailsof the proof for the general case are omitted for brevity.

If for some n at least one coordinate of xn is null, then zn = (x; ��n(x)) isbudget balanced (Proposition 3) so r�n(z

n; C; vn) = 0. Thus we focus on thesubsequence of fvng such that xni > 0 for all i, and we need only consider thecase where an in�nite subsequence exists. We denote it fvng for simplicity,keeping in mind that from vn to vn+1 we may add more than one new user.Below we write v0+i and v0�i for the right and left derivatives of the concavefunction vi.

Let an = xnf1;��� ;ng be the total demand at xn. We check �rst that the

sequence fang is bounded above. Consider user 1: xn1 > 0 implies v0�1 (xn1 ) �C 0(an), therefore C 0(an) � v0+1 (0), implying a

n � ba, where ba is de�ned byC 0(ba) = v0+1 (0).

Next we show that the sequence fang is weakly increasing. Supposean > an+1. Then for each i = 1; � � � ; n; the equilibrium property gives

v0+i (xn+1i ) � C 0(an+1) < C 0(an) � v0�i (x

ni )) xn+1i � xni

24

a contradiction.Finally the inequalities vi(xni ) � v0�i (xni )xni � C 0(an)xni imply

P (zn; C; vn) =XN

vi(xni )� C(an) � C 0(an)an � C(an) = �(an; C)

Now f�(an; C)g is strictly positive and weakly increasing in n because fangis:

r�n(zn; C; vn) � j��n(xn; C)j

�(an; C)�supx:xN�ba j��n(x;C)j

�(an; C)�supx:xN�ba j��n(x;C)j

�(a1; C)

and assumption (27) implies limn r�n(zn; C; vn) = 0.�

Theorem 4The residual� mechanism is asymptotically e¢ cient at C 2 C if C is analyticon R+, or is a generalized polynomial, C(a) =

PK0 �ka

pk , with �k 2 R; pk >1 for all k.i) If C is analytic the convergence is exponential:

r�n(C; (v1; v2; � � � ; vn)) = O( 12n )ii) If C is a generalized polynomial the convergence is polynomial:

r�n(C; (v1; v2; � � � ; vn)) = O( 1np

��1 ), where p� = mink pk

The proof is in the Appendix.

8 Non smooth cost functions

When we make no di¤erentiability assumption on C, the residual� mecha-nism is still characterized by Proposition 3, but non e¢ cient Nash equilib-rium demands may occur. The properties of VP and RKG are entirely lost,and so is the convergence to zero of the residual cost of ��. We show theseclaims for the two-piece linear cost C(a) = a+maxfa� 1; 0g.Write eBn(a;C) = Bn(( an ; an ; � � � ; an); C) so that (n�1) eBn(a;C) is the budgetimbalance when all demands are equal. Check that

eBn(a;C) = nX0

(�1)k�n

k

�maxfn� k

na� 1; 0g

is linear for nn�k � a �

nn�k�1 and compute

eBn( n

n� k ;C) =1

n� k

kXj=0

(�1)j�n

j

�(n� j)�

kXj=0

(�1)j�n

j

�

25

=1

n� kfn(�1)k

�n� 2k

�g � f(�1)k

�n� 1k

�g = (�1)k�1

n� 1

�n� 1k � 1

�Figure 2 shows the function a ! eBn(a;C) for n = 10; 20. As a variesbetween 1 and n, the budget imbalance ��n((

an ;

an ; � � � ;

an); C) varies between

a large surplus and a large de�cit.We deduce that r�n(C) grows exponentially with n. Let bn2 c be the largest

integer bounded above by n2 , and choose a =

nbn2c . We have

(n� 1)j eBn( nbn2 c ; C)j =�n� 1bn2 c � 1

�� 2n

�(n

bn2 c; C) � �(2; C) = 1

where the asymptotic behavior of� n�1[n2 ]�1

�follows from Stirling formula.

We let the reader check that VP and RKG fail, e.g., for n = 4 andx = (0; 1; 1; 1).

9 Concluding comments

1. All results are preserved if the (non decreasing, convex, di¤erentiable)technology C has a �nite capacity A, and lima!AC(a) = 1. Among npotential users, each individual demand is capped at A

n , and the strategicanalysis of the demand game is otherwise unchanged. For example the costC(a) = B

A�a is totally monotone, and Theorems 1,2 apply.

2. The output sharing model. We interpret C as a production function, x asthe pro�le of input contributions, and y as that of output shares. Individualpreferences are quasi-linear in output, so utilities take the form yi � vi(xi)where vi is convex and increasing. If C is increasing and concave, equation(6) still captures all mechanisms for which all Nash equilibria of the out-put sharing game maximize total surplus. Proposition 3 characterizing theresidual� formula is clearly preserved, and so is Lemma 2. Feasibility nowrequires that the residual claimant burns some output, i.e.,

Pi yi � C(xN );

for the residual� mechanism, this means ��n(x;C) � 0. The Voluntary Par-ticipation constraint is similarly reversed to xi = 0) �i(x) � 0, namely aninactive agent may get a piece of output. For what class of concave pro-duction functions does the residual� output sharing mechanism shares theproperties of the residual� cost sharing mechanism in our four Theorems?This question is worthy of future research.

26

3. The residual� formula can be used to generate an almost budget balancedVCG mechanism ([12]) in those models where individual types are one-dimensional and total surplus depends only upon the aggregate type. Anexample is the familiar provision of a public good model: the convex costof producing z units of public good is (z), and user i�s utility is ui =xiz� ti when z units are produced and he is charged $ti. In a general VCGmechanism each user i reports xi, then the public good is produced at thelevel z� maximizing (

Pi xi)z � (z) over R+, and user i is charged

ti = (z�)� xN�iz� + hi(x�i)

where the function hi(x�i) is arbitrary. To see the connection with our costsharing problem, write the e¢ cient surplus at x as

C(xN ) = maxz�0

fxNz � (z)g

and check that the net utility Ui of user i is Ui = C(xN )�hi(x�i). First beste¢ ciency is

Pi Ui = C(xN ), or equivalently tN = (z

�): users pay exactlythe cost of the e¢ cient level of output. By Proposition 2 we can achieve thisfor all x if and only if C is a polynomial of the right degree. If (z) = zq; q >1, then up to a constant C(z) = z

qq�1 . Setting hi(x�i) = h�(x�i) as in (11)

for this C gives a VCG mechanism with the same e¢ ciency properties as inTheorem 3), generating alternatively a budget surplus or a de�cit dependingon the parity of n (Lemma 2).

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[23] E. Maskin, The theory of implementation in Nash equilibrium: a survey,in Social Goals and Social organization, Hurwicz, Schmeidler and Son-nenschein Eds, Cambridge University Press, Cambridge Mass., 1985.

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[28] H. Moulin, The price of anarchy of serial, average, and incrementalcost sharing, mimeo 2006, forthcoming, Economic Theory.

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[30] H.Moulin and S. Shenker, Serial cost sharing, Econometrica, 60, 1009-1037, 1992.

29

[31] H.Moulin and S. Shenker, Strategyproof Sharing of Submodular Costs:Budget Balance versus E¢ ciency, Economic Theory, 18, 3, 511-533,2001.

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10 Appendix: remaining proofs

10.1 Proposition 2

We prove it for a more general class of functions, namely any function Cbounded on every interval (e.g., continuous). We show that equation (10)has a solution only if C is a polynomial of degree at most n� 1. Fix n, C,and some functions hi satisfying(10). Using the same notations as in theproof of Proposition 3, in particular (16) and Ca(�) = C(a + �), we haveBn(x; hi) = 0 for all x. By linearity of Bn(x;C) with respect to C; this givesBn(x;C) = 0 for all x.

30

We write POL(m) for the set of polynomials of degree at most m, andconsider the two properties A(n) and B(n) of a function C bounded on everyinterval of R+:

A(n) : fBn(x;C) = 0 for all x 2 Rn+g ) C 2 POL(n� 1) (28)

B(n) : fCa � C 2 POL(n� 1) for all a > 0g ) C 2 POL(n) (29)

Assume B(n) is true for all n, and deduce A(n) by induction on n and (18).Property A(1) is clear; next Bn(x;C) = 0 for all x implies Bn�1(x�i; Cxi �C) = 0 for all x, and Cxi � C is bounded on every interval because C is.By the inductive assumption A(n� 1), we get Ca �C 2 POL(n� 2) henceC 2 POL(n� 1) by B(n� 1).

We prove B(n) by induction on n. Start with B(0): if Ca(b) � C(b)is constant in b, we have C(a + b) = C(a) + f(b) for some function f ,and the conclusion C 2 POL(1) follows from Cauchy equation, given ourassumption that C is bounded on every interval ([1]). Assume now B(m) upto m = n� 1 and consider C such that Ca � C 2 POL(n� 1). There existfunctions fk; k = 0; � � � ; n � 1; de�ned on R+ such that for all a; x � 0, wehave

C(a+ x)� C(x) =n�1Xk=0

fk(a)xk (30)

Observe that fk(0) = 0, and fk is bounded on every interval. To check thelatter claim, �x n distinct values xk, and use the corresponding equations(30), a non singular linear system in (f0(a); � � � ; fn�1(a)), to compute fk asa linear combination of functions Cxk .

Apply (30) successively to C(a + b + x) � C(a + x) then to C(a + b +x)� C(x):

n�1Xk=0

fk(a+ b)xk =n�1Xk=0

fk(a)xk +n�1Xk=0

fk(b)(a+ x)k

Identify the coee�cients of xk on both sides of this equation: we see thatfkb � fk 2 POL(n� 1� k) for k = 0; � � � ; n� 1. The inductive assumptiongives fk 2 POL(n�k) for k = 1; � � � ; n�1, then for k = 0 the identi�cationgives

f0(a+ b) = f0(a) + f0(b) +R(a; b) for all a; b � 0 (31)

where R is a symmetric polynomial in a; b, a linear combination of termsapbq where p+ q � n and p; q � 1 (the latter because fk(0) = 0). If f and g

31

are two solutions of (31), both bounded on every interval, f � g satis�es theCauchy equation '(a + b) = '(a) + '(b) and is therefore in POL(1). Nowfor any polynomial R as above, there is a solution f of (31) in POL(n) (weomit the straightforward details). This establishes that f0 2 POL(n), andby (30), so does C. The proof of B(n) is complete.

10.2 Proof of Theorem 1

For any function f on R+, we use the notation

Rn(x; f) =nXk=0

(�1)kk + 1

Ank(x; f)

where Ank is given by (16). Now the residual� cost shares write

��i (x) = C(xN )� (n� 1)Rn�1(x�1; C)

Step 1. Proof of VP and RKGRecall from the discussion after De�nition 3 that VP amounts to xi = 0)�i(x) � 0. The latter is clearly a consequence of (23), which is proven in thesteps 2 to 4 below.As for Ranking, it is equivalent to the property that Rn�1(x�1; C) is weaklyincreasing in all variables xi. Changing n� 1 into n for simplicity, we showthat Rn(x; f) is weakly increasing if f is n-di¤erentiable. Taking derivativesin (17) we have

@1Ank(x; f) = A

n�1k�1(x�1; f

0x1) (32)

implying @1Rn(x; f) = Rn�1(x�1; f 0x1). Hence an induction argument on n:if the �rst n derivatives of f are non negative, the �rst n� 1 derivatives off 0x1 are non negative as well.

Step 2. The two inequalities (23),(24) reduce to, respectively, the left andright part of

C(xN�i) � h�(x�1; C) �n� 1n

C(n

n� 1xN�i) for all x�i 2 Rn�1+ (33)

Changing n into n+ 1, this is

1

nC(xN ) � Rn(x;C) �

1

n+ 1C(n+ 1

nxN ) for all x 2 Rn+

For any function f on R+, not necessarily such that f(0) = 0, we considertwo slightly more general inequalities, denoted (L34) and (R34) for left andright

1

nf(xN )�

1

n(n+ 1)f(0) � Rn(x; f) �

1

n+ 1f(n+ 1

nxN ) (34)

32

We show in step 3 that (L34) holds if f is n times di¤erentiable, and f (k) � 0on R+ for k = 1; � � � ; n; then in step 4 we show (R34) if f is n + 1 timesdi¤erentiable, and f (k) � 0 on R+ for k = 1; � � � ; n+1. This establishes (33)under the premises of statements i) and ii) respectively.

Step 3. Proof of (L34).Write (L34) as P(n; q), where q is the number of non-zero coordinates

in x1; � � � ; xn. We prove P(n; q) by a double induction on n and q. BothP(1; q) and P(n; 0) are equalities, the latter because of the identity

nXk=0

(�1)kk + 1

�n

n� k

�=

1

n+ 1

Assume P(n; q) holds whenever n+ q � r and prove it for n; q s.t. n+ q =r + 1. Identity (17) implies

Rn(x; f) = Rn((0; x�1); f) +Rn�1(x�1; fx1 � f)

By induction (L34) holds for Rn((0; x�1); f) and Rn�1(x�1; fx1 � f), thelatter because (fx1 � f)(k) � 0 for k = 1; � � � ; n� 1. Therefore:

Rn((0; x�1); f) � 1

nff(xN�1)�

1

n+ 1f(0)g

Rn�1(x�1; fx1 � f) � 1

n� 1ff(xN )� f(xN�1)�1

n(f(x1)� f(0))g

) Rn(x; f) �1

nff(xN )�

1

n+ 1f(0)g+ 1

n(n� 1)ff(xN )�f(xN�1)�f(x1)+f(0)g

Convexity of f implies f(xN ) � f(xN�1) � f(x1) + f(0) � 0, and step 3 iscomplete.

Step 4. Proof of (R34)Substep 4.1

We show that for any b > 0, the maximum of Rn(x; f) over all x such thatxN = b is achieved for b� = ( bn ; � � � ;

bn). This is clear for n = 1 and for n = 2

by convexity of f , so we assume n � 3. As Rn(x; f) is symmetric in thevariables xi it is enough to show for all x

x1 < x2 =) @1Rn(x; f) � @2Rn(x; f) (35)

where we use the notation @i = @@xi. For all k = 1; � � � ; n � 1, check the

identity

@1Ank(x; f)� @2Ank(x; f) = An�1k (x�2; f

0)�An�1k (x�1; f0)

33

whereas @1Ann(x; f)� @2Ann(x; f) = @1An0 (x; f)� @2An0 (x; f) = 0. Therefore

@1Rn(x; f)� @2Rn(x; f) =

n�1Xk=1

(�1)kk + 1

(An�1n�k(x�2; f0)�An�1n�k(x�1; f

0))

@2Rn(x; f)� @1Rn(x; f) =

n�2Xk=0

(�1)kk + 2

(An�1n�k�1(x�2; f0)�An�1n�k�1(x�1; f

0))

We claim now thatPn�2k=0

(�1)kk+2 A

n�1n�k�1(x�i; f

0) is weakly increasing in x�i,which completes the proof of (35) and of substep 4.1. The claim is obviousfor n = 1, and for n � 2 we use (32) to reduce it to the following inequality

n�2Xk=0

(�1)kk + 2

An�2n�2�k(ex; f (2)b ) � 0 (36)

where ex 2 Rn�2+ and b � 0. We set m = n � 2, write ex simply as x, andrewrite (36) as the inequality

Tm(x; g) =

mXk=0

(�1)kk + 2

Amm�k(x; g) � 0

whenever g is m times di¤erentiable, g and g(k) � 0 on R+ for k = 1; � � � ;m.We prove a stronger statement namely, under the same premises for g

Tm(x; g) �1

m(m+ 1)(g(xM )�

2

m+ 2g(0)) (37)

where M = f1; � � � ;mg. The proof of (37) parallels that of (L34) in step 3.Refer to (37) as Q(m; q), where q is the number of non-zero coordinates

in x1; � � � ; xm and use a double induction onm+q. BothQ(1; q) andQ(m; 0)are actually equalities, the latter because

mXk=0

(�1)kk + 2

�m

m� k

�=

1

(m+ 1)(m+ 2)

For m � 2, (17) implies as above

Tm(x; g) = Tm((0; x�1); g) + Tm�1(x�1; gx1 � g)

We can apply the inductive assumption because gx1�g � 0 and (gx1�g)(k) �0 for k = 1; � � � ;m� 1, hence Tm(x; g) is bounded below by

1

m(m+ 1)(g(xM�1)�

2

m+ 2g(0))+

1

(m� 1)m(g(xM )�g(xM�1)�2

m+ 1(g(x1)�g(0)))

34

=1

m(m+ 1)(g(xM )�

2

m+ 2g(0))+

2

(m� 1)m(m+ 1)(g(xM )�g(xM�1)�g(x1)+g(0))

The desired conclusion follows by convexity of g.

Substep 4.2By substep 4.1 it remains to prove Rn(b�; f) � 1

n+1f(n+1n b) for any b > 0

and b� = ( bn ; � � � ;bn), namely

Rn(b�; f) =

nXk=0

(�1)kk + 1

�n

n� k

�f(n� kn

b) � 1

n+ 1f(n+ 1

nb)

,n+1Xk=0

(�1)k�n+ 1

k

�f(n+ 1� k

nb) � 0 (38)

De�ne

Bn(b; f) =nXk=0

(�1)k�n

k

�f(n� kn

b)

Our last claim, denoted S(n), is that for any n, ff (k) � 0 for k = 1; � � � ; ngimplies Bn(b; f) � 0. Then (38) is simply Bn+1(n+1n b; f) � 0 and we aredone. De�ne g = f b

n� f and compute

Bn�1(n� 1n

b; g) =n�1Xk=0

(�1)k�n� 1k

�(f(n� kn

b)�f(n� k � 1n

b)) = Bn(b; f)

Property S(1) is clear, and S(n) follows from the above equality by inductionon n.

10.3 Proof of Theorems 2, 3 and 4

10.3.1 Preliminary result

The following property simpli�es the computation of the residual cost r�n(C).If the n-th derivative of C is of constant sign, the largest budget transfer (inabsolute value) is achieved for identical demands. For any a � 0, we writeea= ( an ; an ; � � � ; an), and eBn(a;C) = Bn(ea;C) =Pn

k=0(�1)k�nk

�C(n�kn a).

Lemma 6Suppose the function C is continuous on R+ and n times di¤erentiable onR+�f0g. If the sign of C(n) is constant on R+�f0g, then for all a > 0

maxx:xN=a

jBn(x;C)j = j eBn(a;C)j (39)

35

r�n(C) = (n� 1) supa>0

j eBn(a;C)j�(a;C)

(40)

ProofBy Lemma 2 (39) holds if we show that for all a > 0, ea is a solution ofmaxx:xN=aBn(x;C) if C

(n) � 0, or of minx:xN=aBn(x;C) if C(n) � 0. Bothproofs are identical so we only discuss the case C(n) � 0. Then (40) followsfrom (39) and Lemma 3.

The proof parallels that of substep 4.1, where we established thatRn(x;C),a di¤erent linear combination of the Ank(x;C) than Bn(x;C), reaches itsmaximum over fx : xN = ag at ea. We want to prove (35) for Bn(x;C)instead of Rn(x; f), and by exactly the same argument, this reduces to theanalog of (36) namely

n�2Xk=0

(�1)kAn�2n�2�k(ex;C(2)b ) � 0, Bn�2(ex;C(2)b ) � 0 for all ex 2 Rn�2+ and b � 0

The desired conclusion follows now from Lemma 2. Incidentally if C(n) isstrictly positive everywhere, the proof is easily adapted to show that ea isthe unique maximizer of Bn(x;C) over the relevant simplex.

10.3.2 Theorem 2 statement ii)

We consider the exponential function E(a) = ea � 1. We apply (40) with�(a;E) = ea(a � (1 � e�a)) and eBn(a;E) = Pn

0 (�1)k�nk

�en�kna = ea(1 �

e�an )n:

r�n(E) = (n� 1) supa�0

(1� e� an )n

a� (1� e�a) � supa�0

(1� e�a)n

a� 1n(1� e�na)

(41)

For n � 3 the RHS function to maximize is zero at a = 0 and a =1, witha single critical point an solving

ean = nan + e�(n�1)an

It follows easily that an !1 and eannan

! 1, then that anlogn ! 1. Plugging

this back into (41) gives r�n(E) � e�1

logn as desired.

10.3.3 Theorem 2, statement i)

For any integers n;m de�ne

�n;m =n

meBn(1; Cm) = n

m

nX0

(�1)k�n

k

�(n� kn

)m

36

Recall �n;m = 0 if m < n (Proposition 2). We claim

�n;m �2

log nfor all m (42)

Compute for all a � 01Xm=1

m

n�n;m

am

m!=

1Xm=1

1

m!eBn(a;Cm) = eBn(a;E) = (e an � 1)n (43)

In the development of ean � 1 as a series, the m-th term is no larger than

the m-th term in the development of ean . Because all such terms are non

negative, the m-th term in the development of (ean � 1)n is no larger than

the m-th term in that of (ean )n = ea. Thus mn �n;m � 1 for all m. This gives

(42) for m � n logn2 . It remains to deal with those m such that

1 � m

n� log n

2(44)

Equation (43) implies

m

n�n;m

mm

m!� (e

mn � 1)n = em(1� e�

mn )n

Combine this with the upper bound for m! given by Stirling�s formula

m! � 2p2�m(

m

e)me

112m � 3 2

pm(m

e)m

where e1

12m � 1:1 for m � 2 gives the last inequality. Now we have

�n;m � 3n2pm(1� e�

mn )n � 3 2

pn(1� 1

2pn)n

where the last inequality uses the two bounds in (44). It is now a simplematter to check (e.g., by plotting y = 3 2

px(log x)(1 � 1

2px)x with Map-

ple/Mathematica)

3 2pn(1� 1

2pn)n � 0:6

log nfor all n

completing the proof of (42).

37

We �x next a totally monotone function C(a) =P1k=0 �ka

k, with �k � 0.Inequality (42) caps j

eBn(a;C)�(a;C) independently of a or C:

eBn(a;C)�(a;C)

=

P1k=n �k

eBn(a;Ck)P1k=0 �k�(a;Ck)

=

P1k=n �k

kn�n;ka

kP1k=0(k � 1)�kak

� 2

n log n

P1k=n k�ka

kP1k=n(k � 1)�kak

� 2

(n� 1) log n

where kk�1 �

nn�1 gives the last inequality, and completes the proof of (25).

10.3.4 Theorem 3

Statement i)Fix p, p > 1 and not an integer, and n, n � dpe + 1 = k�. Apply (18) atx = ea and use the linearity of Bn�1 in C:

eBn(a;Cp) = eBn�1(n� 1n

a; (Cp) an)� eBn�1(n� 1

na;Cp)

Recall from the discusion immediately after Lemma 2 that for n � k�,the sign of Bn alternates in n. Thus for all a; a0; a00, eBn�1(a0; Cp) andeBn�1(a0; (Cp)a00) have the same sign and the opposite sign of eBn(a;Cp).Therefore j eBn(a;Cp)j � j eBn�1(n�1n a;Cp)j. Repeating this argument forn� 1; � � � ; k� + 1 gives

j eBn(a;Cp)j � j eBk�(k�na;Cp)j = (

a

n)pj eBk�(k�; Cp)j (45)

Thus j eBn(a;Cp)j is a constant multiple of ( an)p. The desired estimationof r�n(Cp) follows at once from �(a;Cp) = (p� 1)ap and (40).Statement ii)Fix n and two n�times di¤erentiable functions Ci; i = 1; 2. We claim thatif the sign of C(n)1 and that of C(n)2 are constant, though not necessarilyidentical,

r�n(�1C1 + �2C2) � maxfr�n(C1); r�n(C2)g for all �1; �2 � 0 (46)

Lemma 3, and the linearity of Bn and � in C imply

j�1Bn(x;C1) + �2Bn(x;C2)j�1�(x;C1) + �2�(x;C2)

� �1jBn(x;C1)j+ �2jBn(x;C2)j�1�(x;C1) + �2�(x;C2)

� �1 eBn(xN ; C1) + �2 eBn(xN ; C2)�1�(x;C1) + �2�(x;C2)

38

hence the claim. If now C(a) =PK0 �ka

pk with pk > 1 and �k > 0, weset C = C1 + C2, where C1 (resp. C2) collects all terms of which the n�thderivative is positive (resp. negative). We can apply (46) repeatedly tothe C1-sum because any partial sum has a positive n�th derivative. Inview of statement i), this gives r�n(C1) = O(

1

np�1�1), where p�1 is the smallest

exponent in the C1-sum. Similarly r�n(C2) = O(1

np�2�1), and we use (46) one

last time between C1 and C2.

10.3.5 Theorem 4

ProofStatement i)Fix an analytic function C(a) =

P10 �ka

k. From Bn(x;Ck) = 0 if k < n,we get Bn(x;C) =

P1n �kBn(x;Ck). By (39) and the proof of Statement

i) in Theorem 2

Bn(x;Ck) � eBn(xN ; Ck) = xkN eBn(1; Ck) � xkN(the last inequality is just kn�n;k � 1 and it holds for all n; k). Therefore

xN � a) jBn(x;C)j �1Xn

j�kjBn(x;Ck) �1Xn

j�kjak �1

3n

1Xn

j�kj(3a)k

Because C is analytic on R+,P1n j�kj(3a)k converges to zero when n is

large. Thus n supx:xN�a jBn(x;C)j = O(12n ). From the the proof of Lemma

4, we conclude as desired that the sequence r�n(C; (v1; v2; � � � ; vn)) is alsoO( 12n ).Statement ii)The proof is essentially the same as for Theorem 3. Inequality (45) gives forall p > 1

n supx:xN�a

jBn(x;Cp)j = O(1

np�1)

We conclude by linearity of Bn in C. The sign of the coe¢ cients �k doesnot matter any more because we are not dividing Bn(x;C) by the e¢ cientsurplus.

39

0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0­0.02

0.00

0.02

0.04

0.06

0.08

0.10

0.12

0.14

0.16

0.18

0.20

x

y

FIGURE 1

An example where the 3-residual� mechanism generates a surplus or a de�cit,depending upon the demand pro�le. the smooth and convex cost function is

C(a) = ea � a3

3� a� 1

and the budget imbalance for identical demands

��3((a

3;a

3;a

3); C) = ea � 3e 2a3 + 3e a3 � 2

27x3 � 1

changes sign in a

1

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5

­14

­12

­10

­8

­6

­4

­2

0

2

4

6

8

10

12

14

x

y

FIGURE 2.a: For the two-piece linear cost function C(z) = z + (z � 1)+,and 10 agents demanding x units each, the graph shows Bn((x; � � � ; x); C) =19�

�((x; � � � ; x); C) when x varies on [0; 1]. The budget is balanced for 0 � x �110 .

1

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

­5000

­4000

­3000

­2000

­1000

0

1000

2000

3000

4000

5000

x

y

FIGURE 2.b: The same computation with 20 agents. Budget balance is for0 � x � 1

20 .

2