On Revenue-Maximizing Walrasian Equilibria for Size-Interchangeable Bidders

Abstract

We study a market setting in which bidders are single-valued but size-interchangeable, and there exist multiple copies of heterogeneous goods. Our contributions are as follows: (1) providing polynomial-time algorithms for finding a restricted envy-free equilibrium with reserve prices (EFEr); (2) posing the problem of finding a revenue-maximizing EFEr, and running experiments to show that our algorithms perform well on the metrics of revenue, efficiency, and time, without incurring too many violations of the stronger Walrasian equilibrium with reserve (envy-free plus market clearance) conditions.

Appendix

Mixed ILP to Find Optimal Allocations

Given a market $(U,C,E,{\varvec{N}},{\varvec{I}},{\varvec{R}}),$ Algorithm 5 is a mixed ILP that can be used to find an optimal allocation.

Constraints (1), (2) and (5) imply that a solution to the Mixed-ILP is a feasible allocation. Constraints (3) and (4) imply that a bidder attains reward $R_j$ if and only if it is completely fulfilled, and together with constraint (5), imply that if $y_j= 0$ then $x_{ij} = 0$ for all $i$. The objective of the mixed ILP implies that the solution maximizes bidders’ rewards over all feasible allocations and thus, it is an optimal allocation. To obtain an allocation that respects reserve price $r$, change the objective of the mixed ILP to $\sum _{j=1}^m(R_j-rI_j) y_j$, where $r\in \mathbb {R}^+$ is the reserve price parameter.

Proof of Theorem 2

Finding an optimal allocation is NP-hard. To prove this, we reduce from the following version of set packing: Given a universe $\mathcal {U} = \{u_1,u_2,\ldots ,u_n\}$ and a family of subsets $\mathcal {S} = S_1,S_2,\ldots ,S_k \subseteq \mathcal {U}$, find the maximum number of pairwise disjoint sets in $\mathcal {S}$.

Consider an input $(\mathcal {U},\mathcal {S})$ to the set packing problem as described above. Let us construct a market $(U,C,E,{\varvec{N}},{\varvec{I}},{\varvec{R}})$ from $(\mathcal {U},\mathcal {S})$ as an input to the optimal allocation problem. At a high level, the input market consists of n goods each offered in exactly 1 copy and k bidders where each bidder corresponds to a member $S_j\in \mathcal {S}$ that demand as many goods as elements in $S_j$ and attains a reward of exactly 1. A goods is connected to a bidder only if the index of the good is contained in the set $S_j$ associated with the bidder.

Formally, given $(\mathcal {U},\mathcal {S})$ where $\mathcal {U} = \{1,2,\ldots ,n\}$ and $\mathcal {S} = \{S_1,S_2,\cdots ,S_k\}$, construct $f(\mathcal {U},\mathcal {S}) = (U,C,E,{\varvec{N}},{\varvec{I}},{\varvec{R}})$ as follow: (1) let $U= \mathcal {U}$ and $N_i= 1$ for all $i= 1,2,\ldots ,n$. (2) let $C= \{1,2,\ldots , k\}$, and associate each bidder $j\in C$ to $S_j\in \mathcal {S}$ so that $I_j= |S_j|$. Also, $R_j= 1$ for all $j= 1,2,\ldots ,k$. (3) add edge $(i,j)$ to $E$ only if $i\in S_j$. Clearly the transformation f is polynomial on the size of the input $(\mathcal {U},\mathcal {S})$.

We now show that a set packing for $(\mathcal {U},\mathcal {S})$ corresponds to an optimal allocation for $f(\mathcal {U},\mathcal {S})$ and vice versa. Suppose that l is the maximum number of pairwise disjoint sets in $\mathcal {S}$ and that $S_1,S_2,\ldots ,S_l \in \mathcal {S}$ are these sets. By our transformation f we know that each bidder $j$ associated with a set $S_j$ from the previous list is connected to as many goods as $|S_j|$. Since all these sets are pairwise disjoint, all bidders are connected to different goods. Therefore, each of these bidders can be fulfilled which means that the value of the optimal allocation is at least l. Moreover, we know that is not possible to fulfill more than l bidders since l is the maximum number of pairwise disjoint sets, and selecting more than l bidders would imply, by our transformation f, that at least one good has a supply greater than 1. Therefore, l is the value of the optimal allocation of $f(\mathcal {U},\mathcal {S})$.

Suppose that l is the value of the optimal allocation of $f(\mathcal {U},\mathcal {S})$. This means that l is the maximum number of bidders that can be fulfilled. Bidder $j$ is fulfilled only if its allocation is at least $|S_j|$. By construction we know that a bidder is connected to exactly $|S_j|$ many goods. Therefore, each allocated bidder $j$ is fulfilled by exactly $|S_j|$ goods. Moreover, none of these goods are allocated to different bidders since there is exactly 1 copy of each good. Therefore, the sets associated with the selected bidders must not overlap in any element, i.e., they must be pairwise disjoint. This shows that there are at least l pairwise disjoint sets in $(\mathcal {U},\mathcal {S})$. We also know that there must be at most l pairwise disjoint sets or otherwise the value of the optimal allocation would have been more than l. Therefore, l is the maximum number of pairwise disjoint sets in $\mathcal {S}$.    $\square$

Crawford and Knoer Ascending Auction

In the unit-demand setting, it is well known that Walrasian Equilibria exist [6]. Furthermore, Crawford and Knoer [5] proposed an ascending auction mechanism which, for price increment $\epsilon$, yields an $\epsilon$WE.¹ We describe the workings of their mechanism in a unit-demand CCMM in Algorithm 6.²

This algorithm, as stated, generalizes to size-interchangeable CCMMs, except that at each step of the algorithm we must query bidders for their favorite bundles at the current prices plus $\epsilon$, rather than their favorite individual goods.