Fair (and not so fair) division

Abstract

Drawbacks of existing procedures are illustrated and a method of efficient fair division is proposed that avoids them. Given additive participants’ utilities, each item is priced at the geometric mean (or some other function) of its two highest valuations. The utilities are scaled so that the market clears with the participants’ purchases proportional to their entitlements. The method is generalized to arbitrary bargaining sets and existence is proved. For two or three participants, the expected utilities are unique. For more, under additivity, the geometric mean separates the prices where uniqueness holds and where it fails; it holds for the geometric mean except in one case where refinement is needed.

Appendix

The derivations of the procedures based on axioms are pleasingly simple in essence and illuminate the meaning, strength, possible weakening, and appropriateness of the axioms used, so we sketch them here.

Nash

The product of the agents’ utilities has a unique maximum, since the feasible set is convex and the product function has strictly convex contours in the positive orthant. Rescale the utility functions to have value 1 at the maximum. Then the hyperplane tangent to the contour of the product function at the maximum is ∑ _j u _j  = n. Therefore the feasible region is contained in the region ∑ _j u _j  ≤ n. If the latter were the feasible region, the solution would be u _j  = 1 for all j, by symmetry. Hence this is also the solution in the original problem.

Equipartite

The Kalai and Smorodinsky (1975) argument generalizes. Rescale the utility functions to have maximum value 1. Let v be the proportion gained at an efficient equipartite point. If the feasible region were the convex hull of the origin, the unit vectors, and (v,..., v), then symmetry would require the choice (v,..., v). The actual region contains this region. Hence monotonicity implies that each agent must get at least v. No one can get more, since (v,..., v) is efficient. Therefore the conditions imply this choice.

Perles–Maschler

For n = 2, every frontier can be approximated by an allocation problem with additive utilities. Suppose items 1 and 2 have the highest and lowest ratios a _i /b _i . Take part of one so a ₁ b ₁ = a ₂ b ₂. For this problem separately, rescale u ₁ so a ₁ = b ₂. Then b ₁ = a ₂ also and the problem is symmetric, so each agent gets his preferred item. The remaining items must be divided as in a separate problem, or else someone would lose. Geometric-mean prices do this, and unlike areas, they add when an item is partitioned. The integral equality results. The first step is in the Kohlberg direction since A gains a ₁ and gives up a ₂ while B gains b ₂ and gives up b ₁, and a ₁/a ₂ = b ₂/b ₁. Similar reasoning applies to later steps.

KMP

No proof of the uniqueness of the Kohlberg direction has been published. Heuristic argument: at any feasible point, the direction which maximizes min _j dw _j  / du _j equalizes the dw _j  / du _j . (Note that dw _j  ≤ 0.) If there is another equalizing direction, let Δu and Δ′u be small moves in the former and latter directions, respectively. Scale them so that Δ′u _j  ≤ Δu _j for all j and Δ′u _k  = Δu _k for some k. Then agent k loses less potential under Δ′ than Δ, because the other agents move less. In symbols, the corresponding Δ′w _k  > Δw _k . Therefore Δ′w _k  / Δ′u _k  > Δw _k  / Δu _k . This contradicts the maximin property of Δu. (Δu is the positive eigenvector of the matrix with nonnegative elements −∂w _i  / ∂u _j .)

Calculations for Example 5

To give more detail for the KMP procedure, for small u _B and u _C, agent A’s gain is maximized by giving B a fraction u _B/3 and C a fraction u _C/4 of item 3. The loss to A is u _B/3 + u _C/4. Thus the first row of the matrix Δ = [−∂w _j  / ∂u _k ] is (0, $\raise0.5ex\hbox{$\scriptstyle 1$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle 3$}$, $\raise0.5ex\hbox{$\scriptstyle 1$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle 4$}$) as long as u _B/3 + u _C/4 ≤ 1. Also, for u _A ≤ 4 and u _C ≤ 4, B’s gain is maximized by giving A a fraction u _A/4 of item 1 and C a fraction u _C/4 of item 3 at a cost to B of 3u _A/4 + 3u _C/4, and the second row of Δ is ($\raise0.5ex\hbox{$\scriptstyle 3$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle 4$}$, 0, $\raise0.5ex\hbox{$\scriptstyle 3$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle 4$}$). Similarly, the third row of Δ is ($\raise0.5ex\hbox{$\scriptstyle {\text{3}}$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle {\text{5}}$}$, $\raise0.5ex\hbox{$\scriptstyle {\text{3}}$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle {\text{4}}$}$, 0) for u _A/5 + u _B/4 ≤ 1. The positive eigenvector of Δ is proportional to the transpose of (1, 1.932, 1.851), so the Kohlberg path moves from the origin in this direction until u _B/3 + u _C/4 = 1, this being the first constraint to become binding. The result is move 1 in Table 14. A’s maximization then changes, giving C a fraction u _C/4 of item 3 and B the rest of item 3 and a fraction of item 2 as well. The first row of Δ becomes (0, $\raise0.5ex\hbox{$\scriptstyle 5$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle 4$}$, $\raise0.5ex\hbox{$\scriptstyle {15}$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle {16}$}$) and the direction (1, 0.7811, 0.7186) until u _A/5 + u _B/4 = 1, move 2. C’s maximization then changes and the third row of Δ becomes ($\raise0.5ex\hbox{$\scriptstyle 3$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle 4$}$, $\raise0.5ex\hbox{$\scriptstyle {15}$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle {16}$}$, 0) and the direction (1, 0.7819, 0.8402) until u _B = 4, move 3. Then the third row becomes ($\raise0.5ex\hbox{$\scriptstyle 3$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle 4$}$, 1, 0) and the direction (1, 0.7820, 0.8592) until u _B + 3u _C/4 = 7, move 4. Then the first row becomes (0, $\raise0.5ex\hbox{$\scriptstyle 5$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle 4$}$, $\raise0.5ex\hbox{$\scriptstyle 4$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle 3$}$) and the direction (1, 0.6990, 0.7651) until u _C = 4, move 5. Then the first row becomes (0, $\raise0.5ex\hbox{$\scriptstyle 4$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle 3$}$, $\raise0.5ex\hbox{$\scriptstyle 4$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle 3$}$) and the second row ($\raise.5ex\hbox{$\scriptstyle 3$}\kern-.1em/ \kern-.15em\lower.25ex\hbox{$\scriptstyle 3$}$, 0, 1) and the direction (1, $\raise0.5ex\hbox{$\scriptstyle 3$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle 4$}$, $\raise0.5ex\hbox{$\scriptstyle 3$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle 4$}$) until the frontier is reached, because the points (w _A, u _B, u _C), (u _A, w _B, u _C), (u _A, u _B, w _C) lie on the same the same face thereof.

Table 14 Calculations for Example 5

The discrete Raiffa procedure starts at u = ($3\raise0.5ex\hbox{$\scriptstyle 1$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle 3$}$, $3\raise0.5ex\hbox{$\scriptstyle 1$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle 3$}$, $3\raise0.5ex\hbox{$\scriptstyle 1$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle 3$}$). Here w = ($5\raise0.5ex\hbox{$\scriptstyle {11}$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle {24}$}$, 5, $5\raise0.5ex\hbox{$\scriptstyle 1$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle 8$}$). The first step moves one third of the way to w, that is, to (2u + w) / 3 = ($4\raise0.5ex\hbox{$\scriptstyle 1$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle {24}$}$, $3\raise0.5ex\hbox{$\scriptstyle 8$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle 9$}$, $3\raise0.5ex\hbox{$\scriptstyle {67}$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle {72}$}$) = (4.042, 3.889, 3.931). Here w = (4.204, 4.019, 4.073). The second step takes each u _j within 0.004 of the frontier and the third step lands on the frontier, terminating the procedure.

• Example 9 Taking turns is inefficient.

  With the values in Table 15, if the agents take turns and A goes first, A will pick item 1, then B item 2, then A item 3, and B is left with item 4. Then A’s items are worth 11 to him and B’s are worth 4 to him. If B goes first, the allocation will be reversed and B will get 16 while A gets 9. Each of these results is lopsided and creates considerable envy. It helps little to give one agent the first and fourth choices, the other the second and third. In either case, picking the agent who goes first at random gives each agent expected utility 10. This is no better than assigning every object at random. It is possible for both agents to do considerably better, even without randomization, as the Nash solution happens to be in this case. The equipartite procedure gives each agent 12.72.

  Since n = 2, the geometric-mean, superadditive, and KMP procedures agree and the geometric-mean prices can be taken as the geometric means of the two agents’ values for each item without scaling the utilities. The total of the prices is 19.02, so each agent gets a budget 19.02 / 2 = 9.51. A then buys items 2–4 and a 1.07 / 10.58 = 0.10 chance of item 1. B buys a 9.51 / 10.58 = 0.90 chance of item 1. Their total values are then 12.81 and 12.58, as shown in the table. The PZ prices are also shown. They are essentially unique in this situation. They total 20, each agent spends 10, A buys items 2–4 and a $\raise0.5ex\hbox{$\scriptstyle 4$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle {14}$}$ chance of 1, B buys a $\raise0.5ex\hbox{$\scriptstyle {10}$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle {14}$}$ chance of 1, and their values are $14\raise0.5ex\hbox{$\scriptstyle 2$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle 7$}$ and 10. Note that the utilities are normalized as they stand, and comparing the agents’ values to each other is only as valid or invalid as comparing utilities by normalizing. However, the inefficiency of taking turns, the closeness of the efficient procedures shown, and their relative advantage to A or B do not depend on the scaling of the utilities.

• Example 10 Taking turns and Nash are unfair and spiteful.

  In Table 16, where the utilities are again normalized as they stand, if the agents take turns, A will get items 3 and 4 and B will get items 1 and 2, whoever goes first. This remains true even if the order of choice is reversed on the second round, that is, the order is A B B A or B A A B. The Nash procedure is the same in this case. Then A’s utility is 16 and B’s is 27. This seems unfair, because B gets almost everything he wants while A gets little over half of what he wants, gaining nothing from the fact that B has very little value for items 3 and 4. The other procedures shown also give A items 3 and 4 and B item 1. The equipartite procedure gives A a 50% chance of item 2, the geometric-mean procedure a 60% chance, and the PZ procedure a 77% chance. If B had no value for items 3 and 4, the no-spite condition would give them to A with no penalty, leaving items 1 and 2 to be divided, and all reasonable procedures would then give A item 2 and B item 1 except for a bit of randomization of item 2. The question then is how rapidly the allocation should change as B’s values for items 3 and 4 increase from 0.

• Example 11 Divide and choose is inefficient.

  With the values in Table 17, if A divides selfishly and intelligently knowing B’s values, he will divide the items into the sets {1, 2, 3} and {4}. Then B’s best choice is {4}, worth 10 to him, leaving A with {1, 2, 3}, worth 18 to him. Similarly, if B divides, he can grant A only 10 and get 18. Each of these results is lopsided, though envy-free, and gives one item to an agent who values it little relative to the other agent by any ordinary scaling of the utilities. If the divider is chosen by a fair coin-flip, both agents have expected utility 14, considerably less than the 17 both get by any efficient, symmetric procedure, including those mentioned in Example 1 and the KMP and Raiffa continuous bargaining procedures.

• Example 12 A superadditive procedure does not exist for n > 2.

  Perles (1982) gives a very subtle example and argument showing that no efficient, superadditive procedure exists for n > 2. He expresses everything in terms of feasible sets in utility space, but as it happens, his example could arise in an additive allocation problem. Hence there is no efficient, superadditive procedure even for this more limited class of problems. Furthermore, his argument can be interpreted in terms of additive allocation, and it may shed some light on it to do so.

  Suppose n = 3 and the procedure being used is superadditive. We will construct a collection of items (vectors of agent values) which can be partitioned into two subsets in two different ways such that each subset alone would be shared in a specified way but applying superadditivity to the two partitions produces contradictory implications for the whole collection.

  First we show the existence for any procedure of two items that would be shared in a particular, simple way. If just items 1 and 2 in Table 18 are to be shared, symmetry requires that B and C receive equal shares, and efficiency requires that A receive all of item 1 before receiving any of item 2. By continuity, therefore, there is a value of a such that A receives item 1 and B and C each receive ${\text{\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/ \kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} }}$ of item 2. Then the payoffs u _j are as shown in the column after item 2.

  We now modify this pair of items in two ways to form one partition of the collection we seek. The irrelevance of B’s utility scaling implies that items 3 and 4 would be shared in the same way as items 1 and 2, with the payoffs u _j shown in the column after item 4. The irrelevance of agent ordering then implies, upon placing A last and dividing all values by a, that if only items 5 and 6 were available, C would receive item 5 and A and B would each receive ${\text{\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/ \kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} }}$ of item 6, with the payoffs u _j shown in the column after item 6.

  In the collection {3, 4, 5, 6}, items 3 and 6 are the same except for scale. Assume that combining such items does not affect the payoffs. (This is true of any procedure whose payoffs depend only on the feasible set, and it is reasonable, since the items can be reproduced by randomizing.) Then sharing items 3–6 is equivalent to sharing items 7, 4, and 5 in Table 19, where item 7 is the combination of items 3 and 6 and b = a + 1 / a. Superadditivity therefore implies that the payoffs from sharing items 7, 4, and 5 must be at least the sum of those in the second and third panels of Table 18. These sums are shown in the first u _j column of Table 19. These sums are in fact the exact payoffs, not just lower bounds, because if any payoff were larger, we would have u _A + 2u _B + 4u _C > 4a + 2 / a + 2(2 + 1 / a)+4 × 5 = 4b + 24, which is impossible because maximizing u _A + 2u _B + 4u _C item by item gives 4b + 8 + 16.

  Let c = b − 1. (Note c > 0 since b = a + 1 / a > 1.) Then item 7 is also the combination of items 8 and 9 in Table 20, and sharing items 7, 4, and 5 is equivalent to sharing items 8, 9, 4, and 5. If only item 8 were available, it would be shared equally, leading to the payoffs shown beside it. If items 9, 4 and 5 were available, each agent’s payoff would be 4, by symmetry and efficiency. Superadditivity therefore implies that the payoffs for sharing items 7, 4, and 5 must be at least the sum shown in the last column of Table 19. None can be greater, by the same argument as before. It follows that the two u _j columns in Table 19 are equal. Equating the values of u _C gives c = 3. Equating the values of u _B then gives a = $\raise0.5ex\hbox{$\scriptstyle 1$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle 4$}$. Equating the values of u _A then gives 9 = 8, a contradiction. Therefore no superadditive procedure exists.

• Example 13 Different geometric-mean allocations give the same expected utilities, as do various randomizations.

  In Table 21, for equal agent weights, the geometric-mean prices are equal and every agent’s expected utility is 4, so the expected utilities are determined, although the market can be cleared in various ways, three of which are shown.

• Example 14 Geometric-mean solutions exist with different weights and prices but the same allocations.

  In the situation of Table 22, if ${\text{\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/ \kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} }}$ < λ < 1, then the prices are as shown, and the market clears with A buying item 1, B item 2, and C item 3 and each agent spending 2√λ, so the geometric-mean allocation is the same although the relative utility scalings and prices vary with λ.

• Example 15 Equal budgets do not imply maximum welfare or the geometric-mean expected values.

  In Table 23, consider the first weights and prices shown. Let b = 2 + √2 = 3.414. Then the total of the prices is 3b, and if each agent has budget b, agent A will buy item 1, agent B will spend his budget on item 2, and agent C will buy item 3 and spend the rest of his budget on item 2. Every item is bought once and once only, but the market is not cleared in the sense of the geometric-mean procedure, and the total welfare ∑ _j λ _j ∑ _i p _ji u _ji is not maximized, because B cannot buy all of item 2 even though his scaled value for it exceeds its price, and C buys part of item 2 even though his scaled value for it is less than its price. For the weights (1, ${\text{\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/ \kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} }}$, 2), by symmetry, B would get all of item 2 and part of item 3, so equalizing the budgets does not determine the agents’ expected utilities. The geometric-mean procedure has the weights and prices shown in the last four columns. A buys item 1 and a share p of each of the others, B buys share 1 − p of item 2, and C buys share 1 − p of item 3, where p = (2 − √2) / 12 = 0.0488 to equalize the agents’ expenditures. Notice that the agents sharing items have the same scaled values for them under the geometric-mean procedure. This is essential for the uniqueness of the agents’ expected utilities and welfare maximization.

Table 15 Taking turns is inefficient

Table 16 Taking turns and Nash are unfair and spiteful

Table 17 Divide and choose is inefficient

Table 18 Three equivalent pairs of items

Table 19 A collection equivalent to items 3–6

Table 20 Another equivalent collection

Table 21 Different geometric-mean allocations with the same expected utilities

Table 22 Different geometric-mean weights and prices with the same allocations

Table 23 Equal budgets do not imply the geometric-mean procedure

Theorem 6

In additive allocation problems, the Nash and highest-scaled-value price procedures are equivalent.

Proof

At the Nash solution with the utility functions scaled to have value 1, as in the second paragraph of the Appendix, each agent must receive shares whose total value to him is 1 and which no other agent values more highly (else ∑ _i u _i  > n would be feasible). At the highest-scaled-value prices, therefore, each agent pays his scaled utility for what he receives, and his total expenditure is 1. Thus the market clears with equal expenditures and this is the solution for highest-scaled-value prices, q.e.d.