Mechanism Design for Perturbation Stable Combinatorial Auctions

Abstract

Motivated by recent research on combinatorial markets with endowed valuations by (Babaioff et al., EC 2018) and (Ezra et al., EC 2020), we introduce a notion of perturbation stability in Combinatorial Auctions (CAs) and study the extent to which stability helps in social welfare maximization and mechanism design. A CA is γ-stable if the optimal solution is resilient to inflation, by a factor of γ ≥ 1, of any bidder’s valuation for any single item. On the positive side, we show how to compute efficiently an optimal allocation for 2-stable subadditive valuations and that a Walrasian equilibrium exists for 2-stable submodular valuations. Moreover, we show that a Parallel 2nd Price Auction (P2A) followed by a demand query for each bidder is truthful for general subadditive valuations and results in the optimal allocation for 2-stable submodular valuations. To highlight the challenges behind optimization and mechanism design for stable CAs, we show that a Walrasian equilibrium may not exist for γ-stable XOS valuations for any γ, that a polynomial-time approximation scheme does not exist for (2 − ε)-stable submodular valuations, and that any DSIC mechanism that computes the optimal allocation for stable CAs and does not use demand queries must use exponentially many value queries. We conclude with analyzing the Price of Anarchy of P2A and Parallel 1st Price Auctions (P1A) for CAs with stable submodular and XOS valuations. Our results indicate that the quality of equilibria of simple non-truthful auctions improves only for γ-stable instances with γ ≥ 3.

Appendix: A

1.1 A.1 Technical Details Missing from the Proof of Theorem 6

In this section, we complete the proof of Theorem 6, working along the lines of the proof of [11, Lemma 3.10]. We have already showed that there exists a structured submenu of exponential size with the following properties:

• 1. For all $S\in \mathcal {S}$, bidder 2 can bid in such a way such that he is allocated S.

• 2. For each $S,T\in \mathcal {S}$: $|p_{S} - p_{T}| \leq \frac {1}{m^{5}}$.

• 3. For all $S,T \subseteq M$ such that $S\in \mathcal {S}$, S ⊂ T: $p_{T} - p_{S} \geq \frac {1}{m^{3}}$.

• 4. For all $S\in \mathcal {S}$: p_S ≤ m.

• 5. For each $S,T\in \mathcal {S}$: |S| = |T|.

With this submenu, we are ready to create a valuation where it is hard to find the demand bundle. Given a set $O\in \mathcal {S}$, we use, as in the proof of [11, Lemma 3.10], the following valuation function:

$$ v^{O}(S) = \left\{\begin{array}{ll} t\cdot|S|, &\textit{if } |S| < k \\ t\cdot k - 1/m^{4}, &\textit{if } S\in \mathcal{S} \textit{and } S\neq O\\ t\cdot k, &\textit{if } S=O \textit{or } \exists T\in \mathcal{S} \textit{s.t. } T\subset S\\ t\cdot(k-\frac{1}{2^{m-|S|}}), &\textit{otherwise} \end{array}\right. $$

(19)

where t is a large number whose value we will set later. To establish the submodularity of v^O, we show that for any S ⊂ T and j∉T, we have that 0 ≤ v(S ∪ j) − v(S) ≥ v(T ∪ j) − v(T). Since each marginal is at most t, any case where v(S ∪ j) − v(S) = t is trivial. Now we notice:

• If $S\cup j \in \mathcal {S}$, then v(S ∪ j) − v(S) ≥ t − 1/m⁴. However the value of the marginal v(T ∪ j) − v(T) is one of the following:

  • 1/m⁴, if $T\in \mathcal {S}-\{O\}$.

  • 0, if T = O.

  • $\frac {t}{2^{m-|T|}}$, if both v(T) and v(T ∪ j) are calculated by the 4th case of the valuation.

  • $\frac {t}{2^{m-|T|}}$, if v(T) is calculated by the 4th case of the valuation and v(T ∪ j) is calculated by the 3rd case of the valuation.

  Because |T| < m, in all cases t − 1/m⁴ is greater for a large enough t, e.g. t > 2^m.

• If $\exists S^{\prime }\in \mathcal {S} \textit {s.t. } S^{\prime }\subset S$, then v(S ∪ j) − v(S) = v(T ∪ j) − v(T) = 0.

• If $v(S) = t\cdot (k-\frac {1}{2^{m-|S|}})$, then $v(S\cup j) - v(S) = \frac {t}{2^{m-|S|}}$. Because $S\subseteq T$, the value of v(T ∪ j) − v(T) is either going to be 0 or $\frac {t}{2^{m-|T|}}$. In any case the marginal of S is greater.

Thus we see that in any case the marginal of S is both positive and greater than the marginal of T. This proves the submodularity and monotonicity.

Now we notice that finding bundle O by doing value queries in v^O(⋅) is hard: it requires $|\mathcal {S}|-1$ queries. This is because for any $O^{\prime }\in \mathcal {S}$ distinguishing between v^O(⋅) and $v^{O^{\prime }}(\cdot )$ requires querying either O or $O^{\prime }$, which in the worst case requires querying almost every bundle in $\mathcal {S}$. Since $\mathcal {A}$ does polynomially many value queries, bidder 2 will not be allocated bundle O, since the size of $\mathcal {S}$ is exponential.

Now all that is left to do is to prove is that if bidder 2, who can pick any set S at price p_S, has valuation v^O(⋅), then he strictly demands O. To this end, we need to prove the following inequality (see also [11, Claim 3.13]):

$$ v^{O}(O) - p_{O} > v^{O}(S) - p_{S} $$

(20)

for any S≠O. We examine the following cases:

• If |S| < k then v(S) = t ⋅|S|. Manipulating inequality (20) leads to proving t ⋅ (k −|S|) > p_O − p_S. The lhs is at least t (since |S|≤ k − 1) and the rhs is at most m (because of submenu property 4 and p_S ≥ 0). Since t is as large as we want it to be, it holds that t > m.

• If $S\in \mathcal {S}$ then inequality (20) becomes 1/m⁴ > p_O − p_S. This indeed holds because of submenu property 2.

• If O ⊂ S, then v(S) = v(O). However because of submenu property 3, the price of S is greater than O’s. This makes it have strictly less utility.

• If $\exists T\in \mathcal {S}$ s.t. T ⊂ S (T≠O), then v(S) = v(T) + 1/m⁴ and also p_S ≥ p_T + 1/m³ (submenu property 3). This makes S at less favorable than T. However T already has less utility than O because of the second bullet.

• If $v(S) = t\cdot (k-\frac {1}{2^{m-|S|}})$, then inequality (20) becomes $\frac {t}{2^{m-|S|}} > p_{O} - p_{S}$. This holds because we can take t as large as we want, making the LHS as large as we want, while at the same time the RHS is at most m (because of submenu property 4 and p_S ≥ 0).

Thus we have proven (20), for every S≠O. This makes bidder 2 strictly demand the bundle O. However the mechanism cannot allocate that bundle with polynomially many queries. This entails that the best strategy for bidder 2 is to bid untruthfully, e.g. bidding according to v(S) = |S ∩ O|. This makes $\mathcal {A}$ not DSIC which leads to a contradiction and completes the proof.