Revenue maximization in Stackelberg Pricing Games: beyond the combinatorial setting

Abstract

AbstractIn a Stackelberg Pricing Game a distinguished player, the leader, chooses prices for a set of items, and the other players, the followers, each seek to buy a minimum cost feasible subset of the items. The goal of the leader is to maximize her revenue, which is determined by the sold items and their prices. Most previously studied cases of such games can be captured by a combinatorial model where we have a base set of items, some with fixed prices, some priceable, and constraints on the subsets that are feasible for each follower. In this combinatorial setting, Briest et al. and Balcan et al. independently showed that the maximum revenue can be approximated to a factor of $$H_k\sim \log k$$ H k ∼ log k , where k is the number of priceable items. Our results are twofold. First, we strongly generalize the model by letting the follower minimize any continuous function plus a linear term over any compact subset of $${\mathbb {R}}^n_{\ge 0}$$ R ≥ 0 n ; the coefficients (or prices) in the linear term are chosen by the leader and determine her revenue. In particular, this includes the fundamental case of linear programs. We give a tight lower bound on the revenue of the leader, generalizing the results of Briest et al. and Balcan et al. Besides, we prove that it is strongly NP-hard to decide whether the optimum revenue exceeds the lower bound by an arbitrarily small factor. Second, we study the parameterized complexity of computing the optimal revenue with respect to the number k of priceable items. In the combinatorial setting, given an efficient algorithm for optimal follower solutions, the maximum revenue can be found by enumerating the $$2^k$$ 2 k subsets of priceable items and computing optimal prices via a result of Briest et al., giving time $$O(2^k|I|^c)$$ O ( 2 k | I | c ) where |I| is the input size. Our main result here is a W[1]-hardness proof for the case where the followers minimize a linear program, ruling out running time $$f(k)|I|^c$$ f ( k ) | I | c unless $$\mathsf {FPT} =\mathsf {W[1]} $$ FPT = W [ 1 ] and ruling out time $$|I|^{o(k)}$$ | I | o ( k ) under the Exponential-Time Hypothesis.