Optimization with Demand Oracles

Abstract

We study maximization subject to a budget constraint, where we are given a valuation function v, budget B and a cost $c_i$ for each item i. The goal is to find a set S that maximizes v(S) subject to $\Sigma _{i\in S}c_i\le B$. Special cases of this problem are well-studied problems from submodular optimization. In particular, when the costs are all equal (cardinality constraint), a classic result by Nemhauser et al. shows that the greedy algorithm provides an $\frac{e}{e-1}$ approximation. Motivated by a large body of literature that utilizes demand queries to elicit the preferences of agents in economic settings, we develop algorithms that guarantee improved approximation ratios in the presence of demand oracles. We are able to break the $\frac{e}{e-1}$ barrier: we present algorithms that use only polynomially many demand queries and have approximation ratios of $\frac{9}{8}+\epsilon$ for the general problem and $\frac{9}{8}$ for maximization subject to a cardinality constraint. We also consider the more general class of subadditive valuations. Here, if the valuations can only be accessed by value queries, only trivial approximation ratios can be guaranteed. In contrast, we present algorithms that use demand queries and obtain an approximation ratio of $2+\epsilon$ for the general problem and 2 for maximization subject to a cardinality constraint. We guarantee these approximation ratios even when the valuations are non-monotone. We show that these ratios are essentially optimal, in the sense that for any constant $\epsilon >0$, obtaining an approximation ratio of $2-\epsilon$ requires exponentially many demand queries.

Appendix for Section 4

1.1 Proof of Claim 4.5

The proof is basically an algebraic manipulation:

$$\begin{aligned} \gamma= & {} \underset{k_1\le k\le k_2}{\max } \alpha \cdot \frac{k_2}{k_1+k_2-k}+ \left( (1-\alpha )-\alpha \cdot \frac{k-k_1}{k_1+k_2-k}\right) \frac{k_2}{k} \nonumber \\= & {} \underset{k_1\le k\le k_2}{\max } \frac{k_2-k}{k_2-k_1}\cdot \frac{k_2}{k_1+k_2-k} +\left( \frac{k-k_1}{k_2-k_1}-\frac{k_2-k}{k_2-k_1}\cdot \frac{k-k_1}{k_1+k_2-k}\right) \frac{k_2}{k} \nonumber \\= & {} \underset{k_1\le k\le k_2}{\max } \frac{k_2-k}{k_2-k_1}\cdot \frac{k_2}{k_1+k_2-k} \cdot \left( 1-\frac{k-k_1}{k}\right) +\frac{k-k_1}{k_2-k_1}\cdot \frac{k_2}{k} \nonumber \\= & {} \underset{k_1\le k\le k_2}{\max } \frac{k_2-k}{k_2-k_1}\cdot \frac{k_2}{k_1+k_2-k} \cdot \frac{k_1}{k}+\frac{k-k_1}{k_2-k_1}\cdot \frac{k_2}{k}\nonumber \\= & {} \underset{k_1\le k\le k_2}{\max } \frac{k_2}{k(k_2-k_1)}\cdot \frac{k(k_1+k_2)-k^2-k_1^2}{k_1+k_2-k} \nonumber \\= & {} \underset{k_1\le k\le k_2}{\max } \frac{k_2}{(k_2-k_1)}\cdot \left( 1-\frac{k_1^2}{k(k_1+k_2-k)}\right) \nonumber \\= & {} \underset{k_1\le k_2}{\max } \frac{k_2}{(k_2-k_1)}\cdot \left( 1-\frac{4k_1^2}{(k_1+k_2)^2}\right) \nonumber \\= & {} \underset{k_1\le k_2}{\max } \frac{k_2}{(k_2-k_1)}\cdot \frac{k_2^2+2k_1k_2-3k_1^2}{(k_1+k_2)^2} \nonumber \\= & {} \underset{k_1\le k_2}{\max } \frac{k_2}{(k_2-k_1)}\cdot \frac{(k_2+3k_1)\cdot (k_2-k_1)}{(k_1+k_2)^2} \nonumber \\= & {} \underset{k_1\le k_2}{\max } \frac{(k_2+3k_1)\cdot k_2}{(k_1+k_2)^2} \nonumber \\= & {} \underset{x\ge 1}{\max } \frac{(x+3)\cdot x}{(x+1)^2} \nonumber \\= & {} \frac{9}{8} \end{aligned}$$

(5)

where in (5) we use the fact that the expression is maximized at $k=\frac{k_1+k_2}{2}$. In the second to last equation we set $x=\frac{k_2}{k_1}$.