Consider a monopolist selling n items to an additive buyer whose item values are drawn from independent distributions $F_1,F_2,\dots ,F_n$ possibly having unbounded support. Unlike in the single-item case, it is well known that the revenue-optimal selling mechanism (a pricing scheme) may be complex, sometimes requiring a continuum of menu entries. Also known is that simple mechanisms with a bounded number of menu entries can extract a constant fraction of the optimal revenue. Nonetheless, whether an arbitrarily high fraction of the optimal revenue can be extracted via a bounded menu size remained open.

We give an affirmative answer: for every n and $\epsilon >0$, there exists $C=C(n,\epsilon )$ s.t. mechanisms of menu size at most C suffice for obtaining $(1-\epsilon )$ of the optimal revenue from any $F_1,\dots ,F_n$. We prove upper and lower bounds on the revenue-approximation complexity $C(n,\epsilon )$ and on the deterministic communication complexity required to run a mechanism achieving such an approximation.

考虑一个垄断者，其将n个商品出售给一个增值买家，而该买家的商品价值是从独立分布中得出的$F_{\mathrm{1个}}，F_{\mathrm{2个}}，\dots ，F_{ñ}$可能有无限的支持。与单项情况不同，众所周知，最佳收益销售机制（定价方案）可能很复杂，有时需要连续输入菜单。还众所周知的是，具有有限数量的菜单项的简单机制可以提取最佳收益的恒定部分。但是，是否可以通过有界菜单大小提取最佳收益的任意高的比例仍然是未知的。

我们给出肯定的答案：每n和$\epsilon >0$， 那里存在 $C=C（ñ，\epsilon ）$菜单大小最大为C的基本机制足以获得$（\mathrm{1个}-\epsilon ）$最佳收益中的任何一项 $F_{\mathrm{1个}}，\dots ，F_{ñ}$。我们证明了收入近似复杂度的上限和下限$C（ñ，\epsilon ）$ 以及运行实现这种近似的机制所需的确定性通信复杂性。