Mechanisms for combinatorial auctions with budget constraints

Abstract

This paper studies combinatorial auctions with budget-constrained bidders from a mechanism design perspective. I search for mechanisms that are incentive compatible, individually rational, symmetric, non-wasteful and non-bossy. First focusing on the greedy domain, in which any increase in a bidder’s valuation always exceeds his budget, I derive the unique mechanism, called the Iterative Second Price Auction. For the general domain, however, no such mechanism exists.

Appendix

1.1 Proof of Lemma 7

Suppose in negation that $b_i>b_j$ and $\phi ^a_j(u,b)= x_j \in V_i$ but that $\phi ^a_i(u,b)= \emptyset$. Suppose i shrinks his valued set to $\hat{V}_i = \{x_j\}$. By Lemma 3, i is still losing. By non-bossiness, bidder i’s change in report cannot change bidder j’s allocation, so bidder j still wins $x_j$. Suppose now j also shrinks his value set to $\hat{V}_j = \{x_j\}$. By Corollary 2, j must still win $x_j$. So at the new modified profile where $\hat{V}_i = \hat{V}_j =\{x_j\}$, j wins $x_j$ and i is losing. This holds true too even when $u_i(x_j) = u_j(x_j)$, i.e., when $x_j$ yields the same valuation to both bidders.

Denote the profile obtained after the above changes by $(\hat{u}, b)$. Individual rationality implies that $p_j \le b_j$. Suppose now that bidder j increases his budget to be $\hat{b}_j = b_i$ (note that $u_i(x_j)$ and $u_j(x_j)$ can be chosen to be large enough to ensure that with this new budget bidder j is still greedy). By Lemma 6, bidder j must still win $x_j$. Suppose now that bidder i lowers his budget to be $\hat{b}_i = b_j$. Lemma 6 implies he is still losing.

Effectively, bidders i and j have “swapped” their budgets and valuations while preserving their allocation bundles, contradicting symmetry which requires that their allocation bundles must be swapped also. $\square$

1.2 Proof of Theorem 1

In this proof, I restrict attention to isolated groups. I use the term value set basis to refer to the set of goods whose non-empty subsets are elements of the value set. A value set is covered by a value set basis if it contains exactly all non-empty subsets of the value set basis.

Definition 12

A value set $V_i$ is covered by a value set basis $V^b_i$ if $V^b_i \subseteq G$ and $V_i= 2^{V^b_i} - \emptyset$.

A rectangular isolated group can be thought of as a profile where each bidder’s value set basis is simply G. It will be useful to have a way of referring to groups of certain shapes like triangles or squares. I use a notion that encompasses all these shapes.

Definition 13

Let n, g, s, k be non-negative natural numbers. A group $(I,G,(V_i)_{i\in I})$ is called (n, g, s, k)-shape if

• There are n bidders in I, labeled such that $b_1>b_2>\cdots >b_n$.

• There are g goods in G.

• The value set $V_i$ of each bidder i is covered by value set basis $V^b_i$.

• $V^b_1\supseteq V^b_2 \supseteq \cdots \supseteq V^b_n$.

• $1\le s < n.$

• For $i = 1,\ldots , s$, $|V^b_i| = |G|$.

• For $s\le i<n, i\ne k$, $|V^b_i - V^b_{i+1}| = 1$.

• If $k\ne 0$, then $s\le k <n$, and $|V^b_k - V^b_{k+1}| = 2$.

In a (n, g, s, k)-shape group, value sets bases of the first s bidders (indexed by budget size, in decreasing order) contain all individual goods, and but they start to “cascade” from bidder s: each bidder’s value set basis contains all but one good from the previous bidder’s value set basis. There is an exception at bidder k, where the bidder $k+1$’s value set basis contains all but two goods from bidder k’s value set basis. Table 11 show profiles where indicates that a good is in a bidder’s value set basis. For example, a (4, 3, 2, 3)-shape group is shown in profile 1. When $k=0$, the decrease in size of value set basis is at most one. The profile 2 illustrates a “square” (4, 4, 4, 0)-shape group, and profile 3 represents a “triangular” (4, 4, 1, 0)-shape group.

Table 11 Illustration of the “shape” of a profile

I first show that at any “triangular” profile of shape (n, n, 1, 0), $\phi ^a_1=V_1(G)$. This is done through induction from the smallest triangular profile of shape (2, 2, 1, 0). Once this induction is done, I proceed to “fill out” the triangular profile to obtain the shape of a square. The base case of the shape (2, 2, 1, 0) has been done in the main text. The inductive hypothesis is that the highest-budget bidder in the group wins all the goods. I shall refer to the highest bidder as bidder 1, and his allocation simply as $\phi ^a_1$. Note that for a bidder i whose value set $V_i$ is covered by $V^b_i$, $V_i(G) = V^b_i$.

The inductive hypothesis Let n be given. At any (n, n, 1, 0)-shape group (I, G, V), $\phi ^a_1 = G$.

I now proceed to show that at any $(n\,+\,1,n\,+\,1,1,0)$-shape group (I, G, V), $\phi ^a_1 = G$. Unless stated otherwise, the inductive hypothesis is always assumed in this section. Much of the proof is based on induction on the third and forth shape argument, i.e., s and k. The essence of the argument is very much the idea of superimposing shapes to obtain new shapes while preserving the allocation rule.

I first define the notion of reducibility: a shape is reducible to another shape if two groups of the latter shape can be superimposed to form the former shape.

Definition 14

$(n,g,s',k')$ is reducible to (n, g, s, k) if for any group $(I,G,V')$ of shape $(n,g,s',k')$ there is a bidder $i\ne 1$ and there are two goods $g_1, g_2\in V^b_i$ such that

• The group $(I,G,\hat{V})$ with value set bases $\hat{V}^b_i = V^b_i - g_1$ and $\hat{V}^b_j = V^b_j$ for all $j\in I, j\ne i$, is (n, g, s, k)-shape, and

• The group $(I,G,\tilde{V})$ with value set bases $\tilde{V}^b_i = V^b_i - g_2$ and $\tilde{V}^b_j = V^b_j$ for all $j\in I, j\ne i$, is (n, g, s, k)-shape.

Basically, $(n,g,s',k')$ is reducible to (n, g, s, k) if for any $(n,g,s',k')$-shape group there are at least two ways of removing a good from the value set basis of a bidder $i\ne 1$ to attain the shape (n, g, s, k). For example, in Table 12, the shape in profile 1 is reducible to the shape in profile 2 by removing either of the boxes with the minus sign, . The shape in profile 2 is in turn reducible to the shape in profile 3 by removing either of the boxes with the minus sign.

Table 12 Illustration of reducing shapes

If $(n,g,s',k')$ is reducible to (n, g, s, k) then (n, g, s, k) is said to be expandable to $(n,g,s',k')$. $(n,g,s_1,k_1)$ is said to be chain-expandable to $(n,g,s_m,k_m)$ if there is a sequence of shapes $(n,g,s_t,k_t)_{t=1,\ldots , m}$ such that for all $t<m$, $(n,g,s_t,k_t)$ is expandable to $(n,g,s_{t+1},k_{t+1})$. The idea of superimposition is that it is possible to track allocation as one shape is expanded into another.

Remark 1

Suppose for any (n, g, s, k)-shape group (I, G, V), $\phi ^a_1 = G$. Then for any group $(I,G,V')$ with shape $(n,g,s',k')$ such that (n, g, s, k) is chain-expandable to $(n,g,s',k')$, $\phi ^a_1 = G$.

Proof

Suppose (n, g, s, k) is chain-expandable to $(n,g,s',k')$. Consider the first two shapes in the chain, (n, g, s, k) and $(n,g,\ddot{s},\ddot{k})$. By definition of reducibility, for any $(n,g,\ddot{s},\ddot{k})$-shape group $(I,G,\ddot{V})$ there is a bidder $i\ne 1$ and two groups $(I,G, \hat{V})$ and $(I,G,\tilde{V})$ of shape (n, g, s, k) such that $\hat{V}^b_j = \tilde{V}^b_j = \ddot{V}^b_j$ for all $j\in I, j\ne i$ and that $\hat{V}^b_i = \ddot{V}^b_i- g_1$ and $\tilde{V}^b_i = \ddot{V}^b_i - g_2$ for some $g_1,g_2\in \ddot{V}^b_i$.

By assumption, because $(I,G,\hat{V})$ is (n, g, s, k)-shape, $\phi ^a_1 =G$, so $\phi ^a_i = \emptyset$. At $(I,G,\tilde{V})$ also, $\phi ^a_i = \emptyset$. Since $\ddot{V}_i$ is a “combination” of $\hat{V}_i$ and $\tilde{V}_i$, by lemma 4, it must be that $\phi ^a_i = \emptyset$ for $(I,G,\ddot{V})$ as well. Because bidder i stays losing when his value set changes from $\tilde{V}_i$ to $\ddot{V}_i$, by non-bossiness, bidder 1’s allocation is unchanged as well. Therefore, for $(I,G,\ddot{V})$, $\phi ^a_1 = G$.

So if $\phi ^a_1 = G$ for a group (I, G, V) of a shape in the chain, then the same is true for a group $(I,G,\tilde{V})$ of the next shape. Hence $\phi ^a_1 =G$ for every shape in the chain. Therefore, for any $(n,g,s',k')$-shape group, $\phi ^a_1 = G$. $\square$

Now I state a series of remarks that shows which shapes can be expanded into which.

Remark 2

Let $s<n-1$. (n, n, s, 0) is expandable to $(n,n,s+1,s+1)$.

Proof

Consider any $(n,n,s+1,s+1)$-shape group (I, G, V), and consider the set $V^b_{s+1} - V^b_{s+2}$. By definition of the shape, there are 2 goods in this set, called, say, $g_1$ and $g_2$. Removing either $g_1$ or $g_2$ from $V^b_{s+1}$ gives a (n, n, s, 0)-shape. $\square$

Remark 3

$(n,n,n-1,0)$ is expandable to (n, n, n, 0).

Proof

Removing any one of the n goods in $V^b_n$ from a (n, n, n, 0)-shape group gives a $(n,n,n-1,0)$-shape group. $\square$

Remark 4

Let $s\le k<n-1$. (n, n, s, k) is expandable to $(n,n,s,k+1)$.

Proof

Consider a $(n,n,s,k+1)$-shape group and the set $V^b_{k+1} - V^b_{k+2}$. By definition, there are two goods in this set. Removing either good from $V^b_{k+1}$ gives a (n, n, s, k)-shape. $\square$

Remark 5

Let $s\ge 2$. $(n,n,s,n-1)$ is expandable to (n, n, s, 0).

Proof

Consider a (n, n, s, 0)-shape group and the set $V^b_n$. By the shape definition and the condition $s\ge 2$, there are at least 2 goods in $V^b_n$. Removing either good from $V^b_n$ gives a $(n,n,s,n-1)$-shape. $\square$

I am now ready to show that for any rectangular isolated group of size n the highest-budget bidder wins all the goods.

Remark 6

For any (n, n, n, 0)-shape group (I, G, V), $\phi ^a_1 = G$.

Proof

By Remark 2, (n, n, 1, 0) is expandable to (n, n, 2, 2). By Remark 4, (n, n, 2, 2) is expandable to (n, n, 2, 3) which is in turn expandable to (n, n, 2, 4) and so on until $(n,n,2,n-1)$. By Remark 5, $(n,n,2,n-1)$ is expandable to (n, n, 2, 0). In other words, (n, n, 1, 0) is chain-expandable to (n, n, 2, 0).

This line of reasoning can be repeated to show that (n, n, 1, 0) is chain-expandable to $(n,n,n-1,0)$. By Remark 3, $(n,n,n-1,0)$ is expandable to (n, n, n, 0). Therefore (n, n, 1, 0) is chain-expandable to (n, n, n, 0).

By Remark 1 and the inductive hypothesis that for any (n, n, 1, 0)-shape group (I, G, V), $\phi ^a_1 =G$, it must be true that for any (n, n, n, 0)-shape group (I, G, V) $\phi ^a_1 = G$. $\square$

Remark 6 shows that a triangle shape can be expanded to a square shape while preserving the allocation rule that all goods are given to the bidder with the highest budget. I now show that a triangle shape can also be expanded to a larger triangle.

Remark 7

For any $(n+1,n+1,1,0)$-shape group (I, G, V), $\phi ^a_1 =G$.

Proof

Because (n, n, 1, 0) is expandable to (n, n, 2, 0), for any (n, n, 2, 0)-shape group $(I^1,G^1,V^1)$, $\phi ^a_1 =G^1$.

Consider a $(n+1,n,2,0)$-shape group $(I^2,G^2,V^2)$. Suppose bidder $n+1$ is winning some bundle $x_{n+1}\in V^2_{n+1}$. By construction, $V^2_{n+1} \subseteq V^2_n \subseteq \ldots \subseteq V^2_1$, so by lemma 7, all $n+1$ bidders must be winning. There are however only n goods, so this cannot hold.

Therefore bidder $n+1$ is losing. If this bidder reports valuation of zero for all bundles, then non-wastefulness he is still losing and by non-bossiness he does not change the allocation of other bidders. In particular, allocation for bidder 1 must not be changed. However, with such a report the remaining bidders and the goods constitute a (n, n, 2, 0)-shape group for which bidder 1 wins all the goods. Therefore for the original $(n+1,n,2,0)$-shape group $(I^2,G^2,V^2)$, bidder 1 must still win all the goods, i.e., $\phi ^a_1 = G^2$.

Now consider a group (I, G, V) of $(n+1,n+1,1,0)$-shape. Let $g_{n+1} = V^b_1 - V^b_2$ be the good valued only by bidder 1. If bidder 1 reports $\hat{V}_1 = V_2$ then the bidders and the goods constitute a $(n+1, n, 2,0)$-shape group $(I, G-g_{n+1},\hat{V})$, at which, by the above, bidder 1 wins $G-g_{n+1}$. The good $g_{n+1}$ is now worthless to all bidders, so when bidder 1 reports $V_1$, by Lemma 5, bidder 1 must win $G-g_{n+1} + g_{n+1}$. So bidder 1 wins all the goods in G. $\square$

I have shown that, given the inductive hypothesis that the highest-budget bidder wins all goods for any (n, n, 1, 0)-shape group, the same allocation rule holds for any $(n+1,n+1,1,0)$-shape group and any (n, n, n, 0)-shape group. By induction and the fact that the allocation rule holds for the base case of shape (2, 2, 1, 0), the allocation rule holds true for any (n, n, 1, 0)-shape and any (n, n, n, 0)-shape for any n. This is summarized by the following remark.

Remark 8

Let n be any natural number. For any (n, n, 1, 0)-shape group (I, G, V), $\phi ^a_1 = G$. For any (n, n, n, 0)-shape group (I, G, V), $\phi ^a_1 =G$.

It is straightforward to extend the allocation rule to rectangular shape groups by adding or removing losing bidders.

Remark 9

For any n, g, for any (n, g, n, 0)-shape group (I, G, V), $\phi ^a_1 = G$.

Proof

There are 3 cases: $n=g, n>g$ and $n<g$. For the case $n = g$, one can apply Claim 8.

Consider the case $n>g$. Consider bidder k where $g+1\le k \le n$. If bidder k is winning, then by the structure of value sets and Lemma 7, for any $i<k$, bidder i is winning. This means there are at least $g+1$ winning bidders but there are only g goods, an impossibility. Hence bidder k must be losing.

Consider bidder n. By the above, bidder n is losing, so changing bidder n’s valuation for all bundles to zero keeps bidder n losing and by non-bossiness does not change the allocation of other bidders. Repeat the above for bidder $n-1$, and then bidder $n-2$ and so on until all bidders labeled from $g+1$ to n have zero valuation for all bundles. The remaining bidders and the goods now constitute a (g, g, g, 0)-shape group for which, by Remark 8, bidder 1 wins all the goods. Non-bossiness implies that before the changes, $\phi ^a_1 = G$.

For the case where $n<g$, first consider a (g, g, g, 0)-shape group. For this group, by Claim 8, $\phi ^a_1 =G$, so for all $i\ne 1$, $\phi ^a_i = \emptyset$. For $i\ge n$, changing bidder i’s valuation for all bundles to zero keeps bidder i losing and by non-bossiness preserves the allocation $\phi ^a_1 = G$. Doing this iteratively for all such bidders lead to a (n, g, n, 0)-shape group. Since any (n, g, n, 0)-shape group can be arrived at in this manner, it must be that for any (n, g, n, 0)-shape group, $\phi ^a_1 =G$. $\square$

Noting that any isolated rectangular group must take the shape of (n, g, n, 0) for some n and g establishes the result that for any isolated rectangular group, the highest bidder wins all the goods in the group. $\square$