The Complexity of Optimal Multidimensional Pricing
We resolve the complexity of revenue-optimal deterministic auctions in the unit-demand single-buyer Bayesian setting, i.e., the optimal item pricing problem, when the buyer’s values for the items are independent. We show that the problem of computing a revenue-optimal pricing can be solved in polynomial time for distributions of support size , and its decision version is NP-complete for distributions of support size . We also show that the problem remains NP-complete for the case of identical distributions.
Consider the following natural pricing scenario: We have a set of items for sale and a single unit-demand buyer, i.e., a consumer interested in obtaining at most one of the items. The goal of the seller is then to set prices for the items in order to maximize her revenue by exploiting stochastic information about the buyer’s preferences. More specifically, the seller is given access to a distribution from which the buyer’s valuations for the items are drawn, i.e., , and wants to assign a price to each item in order to maximize her expected revenue. We assume, as is commonly the case, that the buyer here is quasi-linear, i.e., her utility for item is , and she will select an item with the maximum nonnegative utility or nothing if no such item exists. This is known as the Bayesian Unit-demand Item-Pricing Problem (BUPP) [CHK07], and has received considerable attention in the CS literature during the past few years [GHK05, CHK07, Bri08, CHMS10, CD11, DDT12b].
Throughout this paper we focus on the well-studied case [CHK07, CHMS10, CD11] that is a product distribution, i.e., the valuations of the buyer for the items are mutually independent random variables. We assume that the (marginal) distributions are discrete and are known to the seller (i.e., the values of the support and the corresponding probabilities are rational numbers given explicitly in the input). This seemingly simple computational problem appears to exhibit a very rich structure. Prior to our work, even the (very special) case that the distributions have support was not well understood: First note that the search space is apparently exponential, since the support size of is . What makes things trickier is that the optimal prices are not necessarily in the support of (see [CD11] for a simple example with two items with distributions of support ). So, a priori, it was not even clear whether the optimal prices can be described with polynomially many bits in the size of the input description.
Revenue-optimal pricing is well-studied by economists (see, e.g., [Wil96] for a survey and [MMW89] for a simple additive case with two items). The pricing problem studied in this work fits in the general framework of optimal multi-dimensional mechanism design, a central question in mathematical economics (see [MV07] and references therein). Finding the optimal deterministic mechanism in our setting is equivalent to finding the optimal item-pricing. A randomized mechanism, on the other hand, would allow the seller to price lotteries over items [BCKW10, CMS10], albeit this may be less natural in this context.
Optimal mechanism design is well-understood in single-parameter settings for which Myerson [Mye81] gives a closed-form characterization for the optimal mechanism. Chawla, Hartline and Kleinberg [CHK07] show that techniques from Myerson’s work can be used to obtain an analogous closed-form characterization (and also an efficient algorithm) for pricing in our setting, albeit with a constant factor loss in the revenue. In particular, they obtain a factor approximation to the optimal expected revenue (subsequently improved to in [CHMS10]). Cai and Daskalakis [CD11] obtain a polynomial-time approximation scheme for distributions with monotone hazard-rate (and a quasi-polynomial time approximation scheme for the broader class of regular distributions). That is, prior to this work, closed-form characterizations (and efficient algorithms) were known for approximately optimal pricing. The question of whether such a characterization exists for the optimal pricing has remained open and was posed as an open problem in these works [CHK07, CD11].
Our Results. In this paper, we take a principled complexity-theoretic look at the BUPP with independent (discrete) distributions. We start by showing (Theorem 1) that the general decision problem is in NP (and as a corollary, the optimal prices can be described with polynomially many bits). We note that the membership proof is non-trivial because the optimal prices may not be in the support. Our proof proceeds by partitioning the space of price-vectors into a set of (exponentially many) cells (defined by the value distributions ), so that the optimal revenue within each cell can be found efficiently by a shortest path computation. One consequence of the analysis is that the optimal pricing problem has the integrality property: if the values in the supports are integer then the optimal prices are also integer (though they may not belong to the support).
We then proceed to show (Theorem 2) that the case in which each marginal distribution has support at most can be solved in polynomial time. Indeed, by exploiting the underlying structure of the problem, we show that it suffices to consider price-vectors to compute the optimal revenue in this case.
Our main result is that the problem is NP-hard, even for distributions of support (Theorem 3) or distributions that are identical but have large support (Theorem 4). This answers an open problem first posed in [CHK07] and also asked in [CD11, DDT12b]. The main difficulty in the reductions stems from the fact that, for a general instance of the pricing problem, the expected revenue is a highly complex nonlinear function of the prices. The challenge is to construct an instance such that the revenue can be well-approximated by a simple function and is also general enough to encode an NP-hard problem.
Previous Work. We have already mentioned the main algorithmic works for the independent distributions case with approximately-optimal revenue guarantees [CHK07, CHMS10, CD11]. On the lower bound side, Guruswami et al. [GHK05] and subsequently Briest [Bri08] studied the complexity of the problem when the buyer’s values for the items are correlated, respectively obtaining APX-hardness and inapproximability, for some constant . More recently, Daskalakis, Deckelbaum and Tzamos [DDT12b] showed that the pricing problem with independent distributions is SQRT-SUM-hard when either the support values or the probabilities are irrational. We note that their reduction relies on the fact that, for certain carefully constructed instances, it is SQRT-SUM-hard to compare the revenue of two price-vectors. This has no bearing on the complexity of the problem under the standard discrete model we consider, for which the exact revenue of a price-vector can be computed efficiently.
Related Work. The optimal mechanism design problem (i.e., the problem of finding a revenue-maximizing mechanism in a Bayesian setting) has received considerable attention in the CS community during the past few years. The vast majority of the work so far is algorithmic [CHK07, CHMS10, BGGM10, Ala11, DFK11, HN12, CDW12a, CDW12b], providing approximation or exact algorithms for various versions of the problem. Regarding lower bounds, Papadimitriou and Pierrakos [PP11] show that computing the optimal deterministic single-item auction is APX-hard, even for the case of bidders. We remark that, if randomization is allowed, this problem can be solved exactly in polynomial time via linear programming [DFK11]. In very recent work, Daskalakis, Deckelbaum and Tzamos [DDT12a] show -hardness for computing the optimal randomized mechanism for the case of additive buyers. We remark that their result does not have any implication for the unit-demand case due to the very different structures of the two problems.
The rest of the paper is organized as follows. In Section 2 we first define formally the problem, state our main results, and prove some preliminary basic properties. In Section 3 we show that the decision problem is in NP. In Section 4 we give a polynomial-time algorithm for distributions with support size 2. Section 5 shows NP-hardness for the case of support size 3, and Section 6 for the case of identical distributions. We conclude in Section 7.
2.1 Problem Definition and Main Results
In our setting, there are one buyer and one seller with items, indexed by . The buyer is interested in buying at most one item (unit demand), and her valuation of the items are drawn from independent discrete distributions, one for each item. In particular, we use , , to denote the support of the value distribution of item , where . We also use , , to denote the probability of item having value , with . Let . We use to denote the probability of the valuation vector being , i.e., the product of ’s over such that and .
In the problem, all the distributions, i.e., and , are given to the seller explicitly. The seller then assigns a price to each item. Once the price vector is fixed, the buyer draws her values from the distributions independently, i.e., with probability . We assume that the buyer is quasi-linear, i.e., her utility for item equals . Let
If , the buyer selects an item that maximizes her utility , and the revenue of the seller is . If , the buyer does not select any item, and the revenue of the seller is .
Knowing the value distributions as well as the behavior of the buyer described above, the seller’s objective is to compute a price vector that maximizes the expected revenue
We use Item-Pricing to denote the following decision problem: The input consists of discrete distributions, with and all being rational and encoded in binary, and a rational number . The problem asks whether the supremum of the expected revenue over all price vectors is at least , where we use to denote the set of nonnegative real numbers.
We note that the aforementioned decision problem is not well-defined without a tie-breaking rule,
i.e., a rule that specifies which item the buyer selects when there are multiple
items with maximum nonnegative utility.
Throughout the paper, we will use the following maximum
price111It may also be called the maximum value tie-breaking rule,
since an item with the maximum price among a set of items with
the same utility must also have the maximum value. tie-breaking rule (which is convenient for our arguments): when there are multiple items with maximum nonnegative utility, the buyer selects the item with the smallest index among items with the highest price. (We note that the critical part is that an item with the highest price is selected. Selecting the item with the smallest index among them is arbitrary — and does not affect the revenue; however we need to make such a choice so that it makes sense to talk about “the” item selected by the buyer in the proofs.) We show in Section 2.2 that our choice of the tie-breaking rule does not affect the supremum of the expected revenue (hence, the complexity of the problem).
We are now ready to state our main results. First, we show in Section 3 that Item-Pricing is in NP.
Item-Pricing is in NP.
Second, we present in Section 4 a polynomial-time algorithm for Item-Pricing when all the distributions have support size at most .
Item-Pricing is in P when every distribution has support size at most .
As our main result, we resolve the computational complexity of the problem. We show that it is NP-hard even when all distributions have support size at most (Section 5), or when they are identical (Section 6).
Item-Pricing is NP-hard even when every distribution has support size at most .
Item-Pricing is NP-hard even when the distributions are identical.
2.2 Tie-Breaking Rules
In this section, we show that the supremum of the expected revenue over is invariant to tie-breaking rules. Formally, a tie-breaking rule is a mapping from the set of pairs with to an item such that .
We will need some notation. Let be the maximum price tie-breaking rule described earlier. We will denote by the expected revenue of under , and by the seller’s revenue under when the valuation vector is . Given a price vector and a valuation vector , we also denote by the set of items with maximum nonnegative utility (so iff ).
We show the following:
The supremum of the expected revenue over is invariant to tie-breaking rules.
Let and denote the numbers that specify the distributions. Let be a tie-breaking rule. We will use to denote the expected revenue of under and use to denote the seller’s revenue under when the valuation vector is .
It is clear that for any and , we have since picks an item with the highest price among those that maximize the utility. Hence, it follows that .
On the other hand, given any price vector , we consider
where and is the rank of sorted in increasing order (when there are ties, the item with the smaller index is ranked higher). We claim that
It then follows from (1) that , which gives the proof of the lemma.
To prove (1), we show that the following holds for any valuation vector :
To prove (2), we consider two cases. If , then we have when is sufficiently small, and thus, . When , we make the following three observations. First, the utility of an item under is at least as high as that under . Second, if for some items , then under the utility of item remains strictly higher than that of item , for sufficiently small. Third, if and (in particular, ) for some , then under the utility of item is strictly higher than that of item when , as . It follows from these observations that when is sufficiently small, must pick, given and , an item such that . (2) then follows from the definition of . ∎
We will henceforth always adopt the maximum price tie-breaking rule, and use to denote the revenue of the seller with respect to this rule. One of the advantages of this rule is that the supremum of the expected revenue is always achievable, so it makes sense to talk about whether is optimal or not. In the following example, we point out that this does not hold for general tie-breaking rules.
Example: Suppose item has value with probability , item has value with probability and value with probability , and in case of tie the buyer prefers item . The supremum in this example is : set for item and for item . The buyer will buy item with probability (if her value for item is ) and item with probability (if her value for item is ). However, an expected revenue of is not achievable: if we give price to item , then the buyer will always buy item and the revenue is . Note that the expected revenue for this tie-breaking rule is not a continuous function of the prices.
Before proving that the supremum is indeed always achievable under the maximum price rule, we start by showing that without loss of generality, we may focus the search for an optimal price vector in the set
denote the minimum and maximum values in the support , respectively.
For any price vector , there exists a such that
First, it is straightforward that no price should be above ; if such a price exists, we can simply replace it by and this will not decrease the expected revenue.
The non-trivial part is to argue that it is no loss of generality to assume that no price is below . Let Suppose that there exists such that , i.e., the set is nonempty; otherwise, there is nothing to prove.
Fix an arbitrarily and let We consider the price vector defined by for and otherwise. As , it follows that and therefore (in particular, now). It is also clear that It suffices to show that .
Indeed, note that so this process will terminate in at most stages. After the last stage we will obtain a vector whose expected revenue is lower bounded by all the previous ones.
To prove that , we proceed as follows. Given any valuation vector , we compare the revenue to and consider the following two cases:
Case 1: On input , the item selected by the buyer is not from . We claim that the same item is selected on input . Indeed, we did not decrease prices of items in , hence their utilities did not go up, while the utilities of the remaining items did not change. Therefore, the revenue does not change in this case, i.e., .
Case 2: On input , the item selected is from . Then by the definition of , the revenue we get is certainly less than . On input , we know that (since item must have nonnegative utility, i.e., ) and thus, . We claim that . To see this, we consider two sub-cases. If , then we must have and the claim follows from our choice of the maximum price tie-breaking rule. If , then every must satisfy ; otherwise, by definition of we have and , a contradiction. From and , we have .
The lemma follows by combining the two cases. ∎
Now we show that the supremum can always be achieved under the maximum price rule .
There exists a price vector such that .
By the compactness of , it suffices to show that if a sequence of vectors approaches , then
To this end, it suffices to show that, for any valuation vector ,
Given any valuation , it is easy to check that when is sufficiently large. (Again consider two cases: and .) (3) then follows, since is the highest price of all items in under the maximum price tie-breaking rule. ∎
3 Membership in NP
In this section we prove Theorem 1, i.e., Item-Pricing is in NP.
Proof of Theorem 1.
We start with some notation. Given a price vector and a valuation , let denote the item picked by the buyer under the maximum price tie-breaking rule, with iff . We will partition into equivalence classes so that two price vectors from the same class yield the same outcome for all valuations: for all .
Consider the partition of induced by the following set of hyperplanes. For each item and each value , we have a hyperplane . For each pair of items and pair of values and , we have a hyperplane , i.e., . These hyperplanes partition our search space into polyhedral cells, where the points in each cell lie on the same side of each hyperplane (either on the hyperplane or in one of the two open-halfspaces).
We claim that, for every valuation , all the vectors in each cell yield the same outcome. Consider any cell . It is defined by a set of equations and inequalities. Given any price vector and any value , let be the set of valuation vectors such that and the buyer ends up buying item on . We claim that does not depend on , i.e., it is the same set over all . To this end, first, if the points of satisfy then . So suppose that satisfies . Consider any valuation vector with . The valuation is in iff for all , we have , and in case of equality we have (iff due to the equality), and in case of further equality we have . Because all points of the cell lie on the same side of each hyperplane , it follows that does not depend on . As a result, for any cell and any , all the points yield the same outcome .
Next, we show that it is easy to compute the supremum of the expected revenue over , for each cell . To this end, let denote the set of valuations for which the buyer picks item if the prices lie in the cell , and let be the probability of : It turns out that can be computed efficiently, since the probability of can be computed efficiently as shown below (and is the disjoint union of , ).
Given , to compute the probability of , we note that is actually the Cartesian product of subsets of , . For each , we can determine efficiently the subset of values such that the buyer prefers item to if has value and has value from . As a result, we have
and thus, we multiply the probabilities of these subsets , for all , and the probability of . Summing up the probabilities of over gives us , the probability of .
Finally, the supremum of the expected revenue over all is the maximum of over all in the closure of . Let denote the closure of ; this is the polyhedron obtained by changing all the strict inequalities of into weak inequalities. The supremum of over all points can be computed in polynomial time by solving the linear program that maximizes subject to . In fact, as we will show below after the proof of Theorem 1, that this LP has a special form: The question of whether a set of equations and inequalities with respect to a set of hyperplanes of the form and is consistent, i.e., defines a nonempty cell, can be formulated as a negative weight cycle problem, and the optimal solution for a nonempty cell can be computed by solving a single-source shortest path problem. It follows that the specification of a cell in the partition is an appropriate yes certificate for the decision problem Item-Pricing , and the theorem is proved. ∎
Next we describe in more detail how to determine whether a set of equations and inequalities defines a nonempty cell, and how to compute the optimal solution over a nonempty cell. The description of a (candidate) cell consists of equations and inequalities specifying (1) for each item , the relation of to every value , and (2) for each pair of items and each pair of values and , the relation of to . Construct a weighted directed graph over nodes where nodes correspond to the items. For each inequality of the form or , include an edge with weight , and call the edge strict or weak accordingly as the inequality is strict or weak. In fact, there is a tightest such inequality (i.e., with the smallest value ) since the cell is in , and it suffices to include the edge for this inequality only. Similarly, for each inequality of the form or (or only for the tightest such inequality, i.e. the one with the largest value ) include an edge with weight . For each inequality of the form or (or only for the tightest such inequality) include a (strict or weak) edge with weight . Similarly, for every inequality of the form or (or only for the tightest such inequality) include a (strict or weak) edge with weight .
We prove the following connections between and the cell :
1. A set of equations and inequalities defines a nonempty cell if and only if
graph does not contain a negative weight cycle or a zero weight cycle with a strict edge.
2. The supremum of the expected revenue for a nonempty cell is achieved by the price vector that
consists of the distances from node 0 to the other nodes of the graph .
1. Considering node 0 as having an associated variable with fixed value 0, the given set of equations (i.e., pairs of weak inequalities) and (strict) inequalities can be viewed as a set of difference constraints on the variables , and it is well known that the feasibility of such a set of constraints can be formulated as a negative weight cycle problem. If there is a cycle with negative weight , then adding all the inequalities corresponding to the edges of the cycle yields the constraint (which is false); if there is a cycle with zero weight but also a strict edge, then summing the inequalities yields .
Conversely, suppose that does not contain a negative weight cycle or a zero weight cycle with a strict edge. For each strict edge , replace its weight by for a sufficiently small (we can treat symbolically), and let be the resulting weighted graph. Note that does not contain any negative weight cycle, hence all shortest paths are well-defined in . Compute the shortest (minimum weight) paths from node to all the other nodes in , and let be the vector of distances from . For each edge the distances and (where ) must satisfy hence all the (weak and strict) inequalities are satisfied.
To determine if a set of equations and inequalities defines a nonempty cell, we can form the graph and test for the existence of a negative weight cycle using for example the Bellman-Ford algorithm.
2. Suppose that cell specified by the constraints is nonempty. Then we claim that the vector of distances from node 0 to the other nodes in the graph is greater than or equal to any vector in all coordinates. We can show this by induction on the depth of a node in the shortest path tree of rooted at node . Letting , the basis is trivial. For the induction step, consider a node with parent in . By the inductive hypothesis . The edge implies that or , and the presence of the edge in the shortest path tree implies that . Therefore, .
The supremum of the expected revenue over the cell is given by the optimal value of the linear program that maximizes subject to , where is the closure of the cell . Observe that all the coefficients of the objective function are nonnegative, and clearly is in the closure . Therefore achieves the supremum of the expected revenue over . ∎
The NP characterization of Item-Pricing and the corresponding structural characterization of the optimal price vector of each cell have several easy and useful consequences.
First, we get an alternative proof of Lemma 2.3 regarding the maximum tie-breaking rule:
Second Proof of Lemma 2.3.
Suppose that the supremum of the expected revenue is achieved in cell . Let be the corresponding graph, and let be the price vector of the distances from node 0 to the other nodes. If then the conclusion is immediate, so assume . From the proof of the above lemma we have that coordinate-wise for all .
We claim that for any valuation , the revenue is at least as large as the revenue under any . Suppose that the buyer selects item under for prices . Then and thus also (since is in the closure of ) and thus is also eligible for selection under . If the buyer selects under then we know that and the conclusion follows. Suppose that the buyer selects another item under and that and hence . Then we must have due to the tie-breaking rule. The facts that is in the closure of and imply that for all , and therefore the buyer should have picked instead of under prices , a contradiction.
We conclude that for any , for any , and the lemma follows. ∎
Another consequence suggested by the structural characterization of Lemma 3.1 is that the maximum of expected revenue can always be achieved by a price vector in which all prices are sums of a value and differences between pairs of values of items. This implies for example the following useful corollary.
If all the values in , , are integers, then there must exist an optimal price vector with integer coordinates.
4 A polynomial-time algorithm for support size 2
In this section, we present a polynomial-time algorithm for the case that each distribution has support size at most . In Section 4.1, we give a polynomial-time algorithm under a certain “non-degeneracy” assumption on the values. In Section 4.2 we generalize this algorithm to handle the general case.
4.1 An Interesting Special case.
In this subsection, we assume that every item has support size , where satisfies , for all . Let denote the probability of the value of item being . For convenience, we also let . In addition, we assume in this subsection that the value-vectors and satisfy the following “non-degeneracy” assumption:
Non-degeneracy assumption: , and for all .
As we show next in Section 4.2, this special case encapsulates the essential difficulty of the problem.
Let OPT denote the set of optimal price vectors in that maximize the expected revenue . Next we prove a sequence of lemmas to show that, given and that satisfy all the conditions above one can compute efficiently a set of price vectors such that and . Hence, by computing for all , we get both the maximum of expected revenue and an optimal price vector.
We start with the following lemma:
If satisfies for all , then either or we have .
Assume for contradiction that satisfies , for all but . It then follows from the maximum price tie-breaking rule that for all . Moreover, there is at least one such that : If , then consider with and for all other . It follows that as we assumed that for all and thus, . ∎
Next we show that there can be at most one such that ; otherwise . We emphasize that all the conditions on are assumed in the lemmas below, the non-degeneracy assumption in particular.
If has more than one such that , then we have .
Assume for contradiction that has more than one such that . We prove the lemma by explicitly constructing a new price vector from such that for all and for at least one . This implies that and thus, is not optimal. We will be using this simple strategy in most of the proofs of this section.
Let denote the item with the smallest among all with . By the non-degeneracy assumption, is unique. Recall that . We let denote the set of such that , so . By the non-degeneracy assumption again, we have for all . We now construct as follows: For each , set if ; otherwise set for some sufficiently small . Next we show that for all . Fix a . We consider the following three cases:
If , then by the definition of . When is sufficiently small, we have
If and , then we have since for all other . We claim that in this case. To see this, note that there exists an item such that and by our choice of . As , we must have and thus, and is not obtained from selling item . Therefore, we have
Finally, if neither of the cases above happens, then we have (note that this includes the case when ). For this case we have and
The lemma then follows because in the second case above, we indeed showed that the following valuation vector in satisfies : and for all . ∎
Lemma 4.2 reduces our search space to such that either or for some , where we use to denote the set of price vectors such that and for all other .
The next lemma further restricts our attention to such that for all .
If but for some , then we have .
Assume for contradiction that . As , we also have . Now we use to denote the set of all such that . It is clear that . We use to denote the following new price vector: for all , and for all , where is sufficiently small. We use the same proof strategy to show that . Fix any . We have
If , then clearly as well and thus, .
If , then by the definition of . When is sufficiently small,
If but , then it is easy to see that , because and for all . It follows that and .
The lemma follows by combining all three cases. ∎
As suggested by Lemma 4.3, for each , we use to denote the set of such that and for all other . In particular, must be if (, by the non-degeneracy assumption). The next lemma shows that we only need to consider such that for all .
If satisfies for some , then we have .
We construct from as follows. Let denote the set of all such that . By our assumption, is nonempty. Then set for all and for all , where is sufficiently small. Similarly we show that by considering the following cases:
If and , we consider the following cases. If , then
Otherwise, there exists a such that . This implies that is not obtained from any item in . As a result, and .
If the case above does not happen, then we must have (this includes the case when ). As a result, we have and .
The lemma follows by combining the two cases. ∎
Finally, we use for each to denote the set of such that ; for all ; , for all such that ; and , for all other . However, may still be exponentially large in general. Let denote the set of such that . Given , our last lemma below implies that, if is the smallest index in such that , then for all larger than ; otherwise is not optimal. In other words, has to be monotone in setting , , to be ; otherwise is not optimal. As a result, there are only many price vectors that we need to check, and the best one among them is optimal. We use to denote this set of price vectors.
Given and , if there exist two indices such that , but , then we must have .
We use to denote for convenience. Also we may assume, without loss of generality, that there is no index between and in ; otherwise we can use it to replace either or , depending on its price.
We define two vectors from . First, let denote the vector obtained from by replacing by . Let denote the vector obtained from by replacing by . In other words, the th and th entries of are , respectively, while all other entries are the same. Our plan is to show that if , then . This implies that cannot be optimal and the lemma follows.
We need some notation. Let denote the projection of onto all but the th and th coordinates:
We use to index entries of vectors in . Let