# Frequency Capping in Online Advertising

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3 first n advertisers have value 1 and demand 1, and the last advertiser has value 1 ϵ and demand n. With the same arrival sequence of users, a greedy allocation, again, gets a value of n + 1, whereas the optimal value is 2n. In fact, as we will show in 3, even when all values and demands are equal and all frequency caps are 1, no deterministic algorithm can have a competitive ratio better than 3/4. Distinction from Online Matching. Finding a matching in a bipartite graph, where one side is known and the other side is exposed one vertex at a time, is known as online matching. While the problem of online allocation with frequency capping constraints appears to be similar to online matching, they are actually quite different. In the frequency capping problem, a-priori each impression can be assigned to any of the advertisers. Now, as the impressions arrive, in the language of online matching, the existence of an edge between an advertiser and an arriving impression depends on the previous assignments made by the algorithm because of the frequency capping constraint. Specifically, if the algorithm has already assigned enough impressions from user j to advertiser i, or has exhausted i s demand, there is no edge between advertiser i and a newly arrived impression of user j; otherwise, there is an edge. This means that an adversary can no longer control the set of edges hitting each new impression; instead, the online algorithm determines the set of edges using indirect means. While we expect this property to translate into better competitive ratios for the frequency capping problem, taking advantage of it is not easy; a fact which is demonstrated by the involved analysis for the natural greedy algorithm for the problem. Results. Our online assignment problem can also be stated abstractly as follows. There are n agents, each one has a total demand d i and a value v i for items. Items of different types arrive one by one in an online fashion, and must be allocated to an agent immediately. Agent i wants no more than f i copies of any single type of item. How should an online algorithm assign each arriving item to agents to maximize value? This abstract statement suggests the following simpler algorithmic questions. Equal values, arbitrary d i, f i : Suppose agents (advertisers) have identical values for items (impressions), that is, v i = 1 for all i. Now, the goal of the online algorithm is simply to assign as many items as possible. Our main technical contribution is the analysis of a novel greedy algorithm, proving that it is 3/4-competitive; which is optimal for a deterministic algorithm. The first step towards this result is to show that we can assume without loss of generality that every advertiser has frequency cap of 1, i.e., wants no more than one impression from each user (the reduction is independent of advertisers having the same value, and also applies when advertisers have arbitrary values). This reduction is simple, yet crucial for each of the cases we study, designing algorithms directly, with arbitrary frequency caps, turns out to be rather hard. We then analyze our greedy algorithm, which assigns arriving impressions in decreasing order of total demand amongst eligible advertisers, for instances with unit frequency cap. (Assigning greedily according to maximum residual demand does not work; this algorithm cannot do better than 2/3.) The unit frequency cap means that an advertiser is eligible for an impression if she has leftover demand, and has not yet been assigned an impression of this user. We first prove that any non-lazy algorithm has a competitive ratio of 3/4 when all demands are equal (in addition to the equal values); then we build on this analysis to account for the fact that advertisers have unequal demands. 3

4 Combinatorial analysis of online algorithms is usually done via a potential function argument which shows that at each step, the change in the potential function plus the algorithm s revenue are comparable to the gain of the optimal solution. Surprisingly, our analysis considers only the final assignment, disregarding the way in which it is reached. This allows us to avoid coming up with a potential function (which in many cases seems to come out of nowhere ), and skip the tedious consideration of each possible step. Our result is especially interesting in light of the known upper bounds for unweighted online matching: 0.5 and 1 1/e 0.63 for deterministic and randomized algorithms, respectively [Karp et al., 1990]. Arbitrary values, equal integral d i /f i : The ideas used in the analysis of the equal values case can be extended to analyze the case where advertisers have different values, but the same integral ratio of demand to frequency cap. We show here that the natural greedy algorithm, which assigns in decreasing order of value, has a competitive ratio of 3/4 for this case; again, this is optimal in the sense that no deterministic algorithm can do better. Arbitrary values, d i and f i : Finally, for the general case with arbitrary values, demands and frequency caps, we design a primal-dual algorithm whose competitive ratio approaches 1 1/e 0.63 for d i /f i 1; we also show an upper bound of 1/ for this case. A competitive ratio of 1 1/e for the general case is already known from the work of [Goel and Mehta, 2007], also under an assumption similar to ours. Our online primal-dual algorithm has an interesting feature: during the execution of the algorithm, it both increases and decreases primal variables. (We refer to our online problem as the dual problem.) The same algorithm and competitive ratio also apply when advertisers have target sets, i.e., they have value v i for impressions from a set S i of users, and value of 0 for other impressions. For this case, we have a matching upper bound for deterministic online algorithms, using the upper bound on online b-matching [Kalyanasundaram and Pruhs, 2000]. (See 1.1 for a discussion regarding [Goel and Mehta, 2007] and online primal dual algorithms.) 1.1 Related Work The maximization of revenue in online ad auctions has received much attention in recent years [Blum et al., 2004, Blum and Hartline, 2005, Mehta et al., 2007, Mahdian and Saberi, 2006, Bansal et al., 2010, Buchbinder et al., 2007, Feldman et al., 2009a]. The problem of designing online algorithms to maximize advertising revenue was introduced by the adwords model [Mehta et al., 2007]: advertisers have budgets, and bids for different keywords. Keywords arrive online, and the goal is to match advertisers to keywords to maximize revenue, while respecting the advertisers budget constraints. Goel and Mehta [2007] extend the adwords model, allowing advertisers to specify bids for keywords which are decreasing functions of the number of impressions (of the keyword) already assigned to the advertiser. Our frequency capping problem is, in fact, a special case of the model of [Goel and Mehta, 2007] (but not of the adwords model of [Mehta et al., 2007]), where keywords correspond to users, and the decreasing function takes the form of a step function with a cutoff equal to the frequency cap f i of the advertiser. Hence, the (1 1/e)-competitive online algorithm of [Goel and Mehta, 2007] applies to our problem as well. On the other hand, the upper bounds in [Goel and Mehta, 2007] do not apply to our problem since the model of [Goel and Mehta, 2007] also captures online matching. Improving upon the ratio of 1 1/e in special cases is posed as an open problem in [Goel and Mehta, 2007]. 4

5 Our greedy algorithms in 3 and 4 obtain a ratio of 3/4, improving upon this ratio of 1 1/e. While the competitive ratio of our algorithm in 5 is the same as that in [Goel and Mehta, 2007], the algorithms are quite different. Moreover, our model does not inherit the upper bound of 1 1/e, and in fact, the best upper bound 4 for the case without target sets is 1/ 2. Also, while the most general problem we solve in this paper remains within the model of [Goel and Mehta, 2007], the most general and realistic version of the frequency capping problem ( 6) cannot be stated as a special case of the model of [Goel and Mehta, 2007]. For this model the question of both a competitive algorithm and an upper bound (tighter than 1 1/e) are open. The primal dual framework for online problems, first introduced by Buchbinder and Naor [2009], has been shown to be useful in many online scenarios including ad auctions, see [Bansal et al., 2012a,b, Alon et al., 2009, 2006, Buchbinder and Naor, 2006, Buchbinder et al., 2007]. Unlike these primal-dual algorithms, which update the primal variables monotonically in each round, our primaldual algorithm is novel in that it reassigns primal variables several times during the execution of the algorithm; hence, the primal variables do not necessarily increase monotonically with each round of new supply. Feldman et al. [2009b] consider frequency capping in a stochastic model, but they leave open the question of improving upon the 1 1/e ratio in this model. Finally, the work in [Abrams and Vee, 2007] also addresses user fatigue in the context of sponsored search; however, the model and algorithms substantially differ from ours. 2 Preliminaries We denote by ALG(σ) the revenue of algorithm ALG on a sequence σ of arrivals of impressions, and by OP T (σ) the revenue of the optimal offline algorithm, which knows σ in advance. Our goal is to design an online algorithm ALG that assigns each impression, immediately upon arrival, to produce a feasible allocation whose total value ALG(σ) is competitive against OP T (σ) for any arrival sequence σ of impressions. The natural greedy algorithm for the problem, denoted by GREEDY V, allocates each arriving impression to the eligible advertiser with the highest value (breaking ties arbitrarily, but consistently). The examples in the introduction show that GREEDY V is no better than 1/2-competitive. The next theorem shows that these examples are tight. The proof of the theorem is based on matching every impression assigned by GREEDY V to up to two impressions of OP T. Theorem 1. The competitive ratio of GREEDY V is 1/2. Proof. For a particular sequence of arrivals, let o(t) and g(t) be the advertisers for which an impression t is assigned by OP T and GREEDY V, respectively. Let f denote the number of impressions assigned by both OP T and GREEDY V to the same advertiser, i.e., {t o(t) = g(t)}. If v(o(t)) > v(g(t)), and the reason GREEDY V does not allocate impression t to advertiser o(t) is that GREEDY V has already allocated the maximum allowed number of impressions of the same user to o(t), then there must be t t such that t and t are impressions of the same user, g(t ) = o(t) and o(t ) o(t). Swapping the assignment of t and t in OP T (i.e., allocating t to advertiser o(t ) and t to advertiser o(t)) does not change its value, and increases the value of f by 1. Repeat this process for all such impressions. Since f is upper bounded by the total number of impressions, the process is repeated only a finite number of times. At the end of the process, 4 While 1 1/e is the best possible competitive factor for the model of [Goel and Mehta, 2007] since this model captures the adwords model of [Mehta et al., 2007], the frequency capping problem does not generalize the adwords model of [Mehta et al., 2007]. Therefore, it does not follow that 1 1/e is an upper bound for our problem. 5

7 Proof. Suppose there are two advertisers a 1 and a 2 with demand 2 each. Consider any deterministic online algorithm ALG. Assume the first three impressions belong to users u 1, u 2 and u 3. If ALG refuses to allocate any one of these impressions, then it cannot be better than 2/3-competitive, since OP T accepts them all. Otherwise, necessarily, one of the advertisers was assigned two impressions, and the other one was assigned only one impression. Assume, w.l.o.g., u 1 is the user whose impression was assigned on its own to an advertiser. Let the fourth impression belong to user u 1, then ALG cannot allocate the last impression, while OP T allocates all four impressions, yielding a ratio of 3/4. We now turn to online algorithms. A natural greedy algorithm is one that assigns an arriving impression to an eligible advertiser with the maximum residual demand. However, assigning according to residual demand, breaking ties arbitrarily, cannot have a competitive ratio better than 2/3, as the following example shows. There are two advertisers, with d 1 = 1 and d 2 = 2, with ties broken in favor of a 1. The sequence of arrivals is u 1, u 2, u 1. The residual demand algorithm allocates only two impressions: the first impression to a 2 and then the second impression to a 1. The optimal assignment, however, can assign all 3 impressions. We show that an alternative greedy algorithm, named GREEDY D, which assigns according to total demand, has a competitive ratio of 3/4. Hereby is algorithm GREEDY D : 1. Sort advertisers a 1,..., a n in a non-decreasing demand order (d 1... d n ). 2. Upon arrival of a new impression, assign it to the first eligible advertiser according to the above order. We need the following notation. Let y i denote the number of impressions assigned by GREEDY D to advertiser a i, and let y = min i y i. Let k denote the number of advertisers whose demand is exhausted by GREEDY D. In 3.1, we analyze the case of equal demands (and v i = 1), and in 3.2 we build on this analysis to deal with the case where demands are arbitrary. We include the proof of the equal demands case since it is simpler, yet gives some insight into the proof of the general case. 3.1 Equal Demand Case Algorithm GREEDY D has the property that it is non-lazy, i.e., it allocates every impression it receives, unless no advertiser is eligible for it. We show that any non-lazy algorithm, including GREEDY D, is 3/4-competitive when all advertisers have equal demand, denoted by d. Theorem 4. Let ALG be a non-lazy algorithm, and let σ be a sequence of input impressions. Then, ALG(σ) OP T (σ) 3/4. Before going into the proof of Theorem 4, consider the example depicted in Figure 1. The rectangle is divided into three areas: R 1 is the total allocation of advertisers who have exhausted their demand, R 2 is the total allocation of advertisers who have not exhausted their demand, and R 3 is unused demand. We use two upper bounds on OP T (σ) ALG(σ): and R 3 (d y ) (n k) R 2 (d y )/y, k y R 1 y /d. Note that y > 0, since an advertiser who has received no impressions can always be assigned at least one impression without violating the frequency cap constraint. The theorem now follows from these bounds together with the observation ALG(σ) = R 1 + R 2. 7

8 R 1 R 2 R 3 d-y* k n-k y* Figure 1: An assignment constructed by the online algorithm ALG. Each column is an advertiser and each row corresponds to a unit demand. Let us now begin the formal proof of Theorem 4. Let A be the set of impressions allocated by OP T, and let B A be of size OP T (σ) ALG(σ). Associate each impression of B with an advertiser, such that up to d y i impressions of B are associated with each advertiser a i. This is possible since n (d y i ) = nd ALG(σ) OP T (σ) ALG(σ) = B. i=1 Lemma 1. B = OP T (σ) ALG(σ) y k. Proof. Let a i be an advertiser for which y i = y. If y = d, then ALG(σ) = nd = OP T (σ), so we can assume y < d. Thus, each impression that ALG fails to allocate belongs to a user that already has an impression allocated to a i (otherwise, ALG could have assigned it to a i ). Hence, there are at most y users having impressions that are not allocated to all advertisers. Each such user u can have at most k more impressions allocated by OP T than by ALG (if u has an unassigned impression, all n k advertisers with non-exhausted demands must have been assigned an impression of u). Next, for the purpose of analysis, we define two types of payments received by each impression x B. Suppose impression x is associated with advertiser a i. The first payment x gets is p x = y i /(d y i ), and the second payment is p x = d/y. Lemma 2. The total payment received by all impressions of B is at most ALG(σ). Proof. Let E denote the set of advertisers whose demand is not exhausted by ALG (i.e., E = n k). Let a i E. For each impression x associated with a i, we have p x = y i /(d y i ) and the number of such impressions is at most d y i. Therefore, the first type of payment received by impressions associated with a i sums up to at most y i. Adding up over all advertisers of E, the sum of the first type payments to all impressions in B is at most a i E y i. Since payments of the second type are all equal, they add up to B d y y k d y = dk. Note that dk + a i E y i = ALG(σ), since a i E y i = d, completing the proof. Lemma 3. For each impression x B, p x + p x 3. Proof. Suppose x is associated with an advertiser a i. The total payment received by x is: y i d y i + d y y d y + d y = y 2 + d(d y ) y (d y = 3 + (2y d) 2 ) y (d y ) 3. 8

10 Proof. Let S be any optimal integral assignment. Using Observation 1, assignments S and S define flow functions f and f, respectively, in the network N. Flow function f f is a legal flow function in the residual network of f, and it can be decomposed into a collection of augmenting cycles and simple augmenting paths from s to t. Therefore, every augmenting path in f f can only increase the flow on the edges going from the advertisers to the sink t. Let f be the flow resulting from f by the addition of all augmentation paths in f f, and let S be the corresponding assignment. Inevitably, S is optimal, and has at least as many impressions assigned to each advertiser as in S. Applying Lemma 4 to the solution produced by GREEDY D, we can assume that for each advertiser a i, OP T i (σ) y i. Observation 2. If i < j, then y j y i. That is, the number of impressions allocated by GREEDY D to the advertisers is non-increasing. Proof. If y i = d i the observation holds, since y j d j d i. Else, each impression GREEDY D allocates to a j must belong to a user that already has an impression allocated to a i. We now extend Lemma 1 to the general demands setting. Lemma 5. For every advertiser a i, if k i = 0, then i [OP T j(σ) y j ] y i i k j. Proof. Every impression that GREEDY D fails to allocate to one of the advertisers a 1,..., a i must belong to a user already having an impression allocated to a i (since k i = 0). Therefore, there are at most y i users who have impressions that are not allocated to advertisers a 1,..., a i. Each such user can have at most i k j more impressions allocated to advertisers a 1,..., a i by OP T than by GREEDY D, completing the proof. We now order the impressions allocated by OP T (σ) to the advertisers. An impression x assigned to advertiser a (x) (by OP T (σ)) precedes an impression y assigned to advertiser a (y) (again, by OP T (σ)) if a (x) appears before a (y) in the order define by GREEDY D. If x and y are assigned to the same advertiser, the order between them is arbitrarily chosen. Let A be the set of impressions allocated by OP T. Define B A as the following subset of impressions. For each advertiser a i, all impressions OP T allocates to a i belong to B, except for the first y i impressions (if OP T i (σ) = y i, then none of the impressions OP T allocates to a i belongs to B.) See Figure 2 for a graphical representation of the relationship between A, B, OP T (σ) and GREEDY D (σ). For any impression x B, let B x denote the set of impressions in B which appear before x in the above order, including x. Observation 3. If impression x B is allocated to advertiser a i by OP T (σ), then k i = 0. Proof. Otherwise, y i = d i, and no impression allocated by OP T to a i belongs to B. Observation 4. OP T (σ) = GREEDY D (σ) + B. We now define, for analysis purposes, two types of payments made by the impressions allocated by GREEDY D to the impressions of B. Consider an impression x B allocated to advertiser a i by OP T. Then, the impressions allocated to a i by GREEDY D contribute p x = y i /(d i y i ) to x (recall that d i y i is not zero since x B, and therefore k i = 0). Let h be the minimal integer 10

11 B OPT(σ) A ALG(σ) Figure 2: An assignment constructed by GREEDY D. The set A contains the impressions assigned by OP T (σ). The set B contains OP T i (σ) y i impressions of A from every advertiser a i (recall that we assumed OP T i (σ) y i ). Thus, B and GREEDY V (σ) add up to OP T (σ). such that: y i h k j B x. 5 Then, the impressions allocated to a h by GREEDY D contribute p x = d h /y i to x. Lemma 6. For each impression x B allocated by OP T to a i, The payment p x is relieved from impressions that are allocated by GREEDY D to an advertiser a h with h < i and k h = 1. Proof. Since x B is assigned to a i, it must hold that: Also, by Lemma 5, B x [OP T j (σ) y j ]. [OP T j (σ) y j (σ)] y i k j. Combining both inequalities, we get B x y i i k j, and can conclude h i. Notice that k h = 1, otherwise h could be replaced by h 1 without changing y i h k j. On the other hand, k i = 0 since x B, and therefore, h i. Lemma 7. For each impression x B, p x + p x 3. Proof. Suppose OP T allocates x to advertiser a i. Let a h be the advertiser whose impressions contribute to p x. Clearly, p x + p x = y i /(d i y i ) + d h /y i. By Lemma 6, h < i, and therefore, d h d i. Hence, p x + p x = y i + d h y i + d i 3. d i y i y i d i y i y i where the last inequality follows from the same arguments used in the proof of Lemma 3. For an advertiser a h, we use the term the total contribution made by the impressions allocated to a h to denote the total payment that is associated with the impressions that were allocated to it by GREEDY D. Lemma 8. Let a h be an advertiser with k h = 0, then the total contribution made by the impressions allocated to a h (by GREEDY D ) is at most y h. 5 Following Lemma 5, every advertiser with an exhausted demand can be blamed for at most y i impressions in B. The definition of h attempts to isolate the advertiser a h which is to be blamed for the fact that x B. 11

12 Proof. By Lemma 6, the impressions allocated to a h contribute, for any impression x, only to p x (and not to p x). Each contribution is y h /(d h y h ). A contribution is made only to impressions allocated by OP T to a h which are in B, and the number of such impressions is OP T h (σ) y h d h y h. Lemma 9. Let a h be an advertiser with k h = 1, then the total contribution made by the impressions allocated to a h (by GREEDY D ) is at most y h = d h. Proof. Let x B be the last impression whose p x is contributed by the impressions allocated to a h by GREEDY D, and let a i be the advertiser to which OP T allocated x. Consider any impression z B such that B z B x y i. Necessarily, z is allocated to an advertiser a i with i i, hence, y i y i. Implying: h B z B x y i y i k j h 1 y i = y i h 1 k j y i Therefore, by definition, z gets its p z contribution from the impressions allocated to advertiser a h 1 or advertisers preceding it. By the choice of x and z this implies that there are at most y i impressions in B which get their p x contribution from the impressions allocated to a h by GREEDY D. The total contribution the impressions of a h contribute is, therefore, at most y i (d h /y i ) = d h. Corollary 2. GREEDY D (σ) 3 B. Proof. By Lemmata 8 and 9, the total contribution of all the impressions assigned to some advertiser by GREEDY D is at most GREEDY D (σ). On the other hand, by Lemma 7, the contribution for each impression of B is at least 3. The proof of Theorem 5 is now immediate: GREEDY D (σ) OP T (σ) GREEDY D (σ) GREEDY D (σ) + B 4 Equal Demands, Arbitrary Valuations k j. 3 B 3 B + B = 3 4. In this section, we assume advertisers have different values, but the same integer ratio of demand to frequency cap (this can happen, for example, when each advertiser has frequency cap f i and wants to advertise to the same number of distinct users u, so that d i = f i u). The reduction to unit frequency cap makes this equivalent to the assumption that demands are equal (d i = d) and all frequency caps are 1. The following theorem shows that the natural greedy algorithm GREEDY V, which assigns in decreasing order of value (breaking ties arbitrarily), has a competitive ratio of 3/4. Note that by Theorem 3, this ratio is optimal. Theorem 6. Let σ be a sequence of input impressions. Then, under the above assumptions, GREEDY V (σ) OP T (σ) 3/4. The rest of this section is devoted for proving Theorem 6. Algorithm GREEDY V sorts the advertisers in a non-decreasing value order, a 1, a 2,..., a n. Let y i denote the number of impressions GREEDY V allocates to advertiser a i. Let k denote the number of advertisers whose demand was exhausted (i.e., k = {i y i = d} ). Let OP T i denote the number of impressions allocated by OP T to a i and let v(z) denote the total value that a set Z of impressions contributes to OP T (σ). 12

13 Observation 5. If i < j, then y j y i. That is, the number of impressions allocated by GREEDY V to the advertisers is non-increasing. Proof. Each impression GREEDY V allocates to a j must belong to a user that already has an impression allocated to a i. Therefore, y j y i. are the first k adver- Corollary 3. The k advertisers whose demand was fulfilled by GREEDY V tisers. Lemma 10. For every advertiser a i with i > k, i OP T j y i k + i y j. Proof. Each impression that GREEDY V fails to allocate to one of the advertisers a 1,..., a i must belong to a user which already has an impression allocated to a i by GREEDY V (because i > k, and therefore, there is still demand left for a i ). Therefore, there are at most y i users who have impressions that are not allocated to the advertisers a 1,..., a i. Each such user can have at most k more impressions allocated to advertisers a 1,..., a i by OP T, than by GREEDY V. Let A be the set of impressions allocated by OP T (i.e., v(a) = OP T (σ)). We order the impressions of A according to the order of the advertisers defined by GREEDY V, breaking ties arbitrarily (i.e., the impressions of A that OP T allocates to a 1 appear first in the order, then the impressions of A that OP T allocates to a 2 and so forth.) Define B as the following set of impressions. For each advertiser a i, all impressions OP T allocates to a i belong to B, except for the first y i impressions. (If OP T i (σ) y i, then none of the impressions OP T allocates to a i belongs to B.) For every set Z A, let Z <x denote the set of impressions in Z which appear before x in the above order, excluding x, and let Z x denote the same set including x if x Z. Notice that if x Z then Z <x = Z x. Similarly, let us define Z >x and Z x. We define C B as following. An impression x B allocated by OP T to advertiser a i is in C if it satisfies the following inequality: y j < (A B) <x + B x C <x. Although this definition uses the set C, it is not cyclic because the membership of each impression x in C depends only on the membership in C of impressions that appear before it in the above order. The intuition behind the set C is that it contains the impressions that give the real difference between the solutions of OP T and GREEDY V, i.e., C is a subset of B of size equal to the difference between the number of impressions assigned by OP T and GREEDY V. The set C was not necessary in the proof of Theorem 5 because there we could assume y i OP T i, which would have implied B = C given the above definition of C. Observation 6. If impression x C is allocated by OP T to a i, then i > k (otherwise y k = d, and no impression allocated by OP T to a i could have been in B). Let x i be the last impression of A that is allocated by OP T to one of the advertisers a 1,..., a i. x i may not exist, but then no impressions arrived, and any algorithm is optimal; so it is safe to assume x i exists. Lemma 11. For 1 i n, i (OP T j y j ) C xi. 13

14 Proof. Let z i be the last impression of (B C) xi. If there is no z i then B xi = C xi, and the proof is complete since by definition i (OP T j y j ) B xi. Therefore, we can assume that z i exists. Let a h be the advertiser that z i was assigned to (by OP T ). Since z i C, and therefore, OP T j h y j (A B) <zi + B zi C <zi, = h y j + OP T j (A B) <zi B zi + C <zi h y j + A >zi A xi + C <zi. Notice that by the definition of B we have: (A B) >zi (A B) xi i j=h+1 y j. In addition, due to the way z i was chosen, each impression of A >zi A xi is either in C, or not in B, therefore: A >zi A xi = (A B) >zi (A B) xi + C >zi C xi y j + C xi C zi. j=k+1 Plugging this into the previous inequality gives: h OP T j y j + y j + C xi C zi + C <zi = j=h+1 Lemma 12. OP T (σ) GREEDY V (σ) + v(c) Proof. Let us define v n+1 = 0 for consistency, then one can easily observe that n OP T = (v i v i+1 ) OP T j. i=1 i=1 Using Lemma 11, we get: n OP T (σ) (v i v i+1 ) y j + C xi = GREEDY V (σ) + y j + C xi. n [(v i v i+1 ) C xi ] = GREEDY V (σ) + v(c), i=1 where the inequality holds since v i v i+1 0, and the last equality follows since C xi is the number of impressions of C allocated to advertisers a 1,..., a i by OP T. We now define two types of payments that the impressions allocated by GREEDY V pay to the impressions of C. Consider an impression x C allocated to advertiser a i by OP T. The impressions allocated to a i by GREEDY V pay p x = v i y i /(d y i ) to x (recall that d y i is not zero because x C, and therefore, i > k). Let h be the minimal integer such that hy i C x, then the impressions allocated to a h by GREEDY V pay p x = v h d h /y i to x. 14

15 Lemma 13. For each impression x C allocated to a i by OP T, p x is contributed by the impressions allocated (by GREEDY V ) to an advertiser a h with h k < i. Proof. Since x C: y j < (A B) <x + B x C <x. Firstly notice that C <x = C x 1, and also i OP T j A x = (A B) <x + B x. Plugging this into the previous inequality gives: y j < OP T j C x + 1 C x OP T j y j. From Lemma 10, i OP T j y i k + i y j. Combining the last two inequalities we get C x y i k, and we can immediately conclude h k. In addition, since x C, Observation 6 implies k < i. Lemma 14. For each impression x C, p x + p x 3v i. Proof. Suppose OP T allocates x to advertiser a i, and let a h be the advertiser whose impressions contribute to p x. Then p x + p x = v i y i /(d y i ) + v h d/y i. By Lemma 13, h < i, and therefore, v h v i. Hence: p x + p x = v iy i + v ( hd yi v i + d ) 3v i. d y i y i d y i y i Where the last inequality follows from the same arguments used in the proof of Lemma 3. Lemma 15. Let a h be an advertiser with h > k, then the total contribution made by the impressions allocated to a h (by GREEDY V ) is at most v h y h. Proof. By Lemma 13, the impressions allocated to a h contribute, for any impression x, only to p x (and not to p x). Each contribution is v h y h /(d h y h ). A contribution is made only to impressions allocated by OP T to a h which are in C B, and the number of such impressions cannot exceed d h y h. Lemma 16. Let a h be an advertiser with h k, then the total contribution made by the impressions allocated to a h (by GREEDY V ) is at most v h y h = v h d. Proof. Let x C be the last impression whose p x is contributed by the impressions allocated to a h, and let a i be the advertiser x is allocated to by OP T. Consider any impression z C such that C z C x y i. Necessarily, z is allocated to an advertiser a i with i i, and therefore y i y i. This implies: C z C x y i y i h y i = y i (h 1) y i (h 1). Therefore, by definition, z gets its p z payment from the impressions allocated to advertiser a h 1 or an advertiser before it. By the choice of x and z this implies that there are at most y i impressions in C whose second payment is contributed by the impressions allocated to a h by GREEDY V. The total contribution of the impressions of a h is, therefore, at most: y i (v h d/y i ) = v h d. Corollary 4. GREEDY V (σ) 3v(C) 15

16 Proof. By Lemmata 15 and 16, the total contribution of all the impressions assigned by GREEDY V is no greater than GREEDY V (σ). On the other hand, by Lemma 14, the payment received by each impression x C is at least 3v(x). The proof of Theorem 6 is now immediate: GREEDY V (σ) OP T (σ) GREEDY V (σ) GREEDY V (σ) + v(c) 3v(C) 3v(C) + v(c) = Arbitrary Valuations We now consider arbitrary valuations v i. We first prove an improved upper bound for this case. Theorem 7. No deterministic algorithm has a competitive ratio better than 1/ Proof. Consider any deterministic algorithm ALG and the following instance: There are two advertisers: a 1 with v 1 = 1, d 1 = 1 and a 2 with v 2 = 1/ 2, d 2 = 2. At step one, a user u 1 comes. If ALG assigns to a 2, the process stops. The optimal algorithm will assign to a 1. The ratio of the algorithm is 1/ 2. If ALG assigns to a 1 at step one, a new user u 2 comes with two impressions. ALG can only assign one of them to a 2, giving a total revenue of 1 + 1/ 2. However, the optimal algorithm assigns the three impressions to a 2, a 1, a 2, respectively, earning a total revenue of 1 + 2/ 2. Hence, the ratio of the algorithm is 1+1/ 2 1+2/ 2 = 1/ A Primal-Dual Algorithm In order to apply the primal-dual approach to the problem, we first formulate the offline allocation problem as a linear program as follows. We refer to the allocation problem as the dual problem, since it is a maximization problem. Let A be the set of advertisers. Let B be the set of users. Finally, for each user j B, let K(j) be the number of impressions of user j. We define variables y(i, j, k) indicating that the k-th impression of user j is assigned to advertiser a i. max s.t. v i K(j) a i A j B k=1 K(j) y(i, j, k) y(i, j, k) d i j B k=1 K(j) y(i, j, k) f i k=1 y(i, j, k) 1 a i A y(i, j, k) 0 (D) a i A a i A, j B j B, k {1, 2..., K(j)} The first set of constraints guarantees that at most d i impressions are assigned to advertiser a i. The second set of constraints guarantees the frequency cap of each advertiser. Finally, the last set of constraints guarantees that each impression is assigned only once. For consistency with previous work [Buchbinder and Naor, 2009], we refer to the maximization problem as the dual problem. We now define the primal problem. We have variable x(i) for each advertiser a i, a variable w(i, j) for each pair of advertiser a i and user j and variable z(j, k) for the k-th impression of user j. 16

17 min a i A d i x(i) + a i A,j B f i w(i, j) + j B,k z(j, k) (P ) s.t. x(i) + w(i, j) + z(j, k) v i a i A, j B, k x, w, z 0 The allocation algorithm is as follows. We assume that the reduction to the case where the frequency cap of each advertiser is 1 has already been applied. Allocation Algorithm: Upon arrival of impression k of user j: Let S(j) be the set of advertisers not yet assigned impressions of user j, and let S(j) = A \ S(j). Let m 1 S(j) be the advertiser maximizing v i x(i). Let m 2 S(j) \ m 1 be the advertiser maximizing v i x(i). a 1. Assign impression k to advertiser m For each advertiser i S(j) m 1 set: w(i, j) max{0, (v i x(i)) (v m2 x(m 2 ))}. 3. For each advertiser i S(j) \ m 1 set: w(i, j) For each impression l k of user j set: z(j, l) v m2 x(m 2 ). ) 5. For advertiser m 1 : x(m 1 ) x(m 1 ) (1 + 1 di (where c is a constant to be determined later). + vm 1 c d i a If max S(j) (v i x(i)) 0, or S(j) =, no assignment is made and no variables are updated. If there is no m 2, we view v m2 x(m 2) as equal to 0. Notice that this algorithm differs from the standard online primal-dual approach because it both increases and decreases primal variables. Theorem 8. The algorithm is (1 (c + 1) 1 )-competitive, for c = (1 + 1 d min ) d min 1, where d min is the minimum demand of any advertiser. Proof. The assignment of impressions to advertisers defines a solution to the dual program. The algorithm generates a primal solution online. We start with a few useful observations. Note first that the variables in the primal solution are always non-negative. Next, observe that for each advertiser i, x(i) is monotonically increasing during the execution of the algorithm, and v i x(i) is monotonically decreasing. Also, observe that if at some point of time in which an impression of user j arrives, v m2 x(m 2 ) = v, then in the next time in which an impression of user j arrives, v m2 x(m 2 ) v, i.e., v m2 x(m 2 ) is also monotonically decreasing. This follows since m 2 maximizes v i x(i) among advertisers i that have not yet been assigned an impression of user j (i.e., advertisers in S(j)), v i x(i) is monotonically decreasing and the set S(j) can only shrink over time. We now prove the following claims from which the theorem readily follows. 1. The primal solution produced by the algorithm is feasible. 2. In each round (arrival of an impression) the change in the dual profit is at least 1 (c + 1) 1 times the change in the primal cost. 17

18 3. The dual solution produced is feasible. Proof of (1). We prove feasibility by induction on the steps of the algorithm. Initially, there are no primal constraints and the primal solution is trivially feasible. Consider the step in which impression k of user j arrives. The algorithm does not change the value of variables w(i, j ) and z(j, k) corresponding to any user j j. Thus, since x(i) only increases, all the constraints corresponding to users j j and impression l remain feasible, i.e., x(i) + w(i, j ) + z(j, l) v i,. Next, consider the primal constraint corresponding to user j, advertiser i, and impression l k: x(i)+w(i, j)+z(j, l) v i. If advertiser i S(j) m 1, i.e., i has already been assigned an impression of user j, then: w(i, j) + z(j, l) = max{0, (v i x(i)) (v m2 x(m 2 ))} + v m2 x(m 2 ) v i x(i). and the constraint is satisfied. If advertiser i S(j)\m 1, i.e., i has not been assigned an impression of user j, then: w(i, j) + z(j, l) = 0 + v m2 x(m 2 ) v i x(i). The last inequality follows since m 2 is the advertiser that maximizes the quantity v i x(i) in the set S(j) \ m 1, yielding that the constraint is satisfied. Proof of (2). We analyze the change in the primal cost and the dual profit in the current round. Consider first the contribution to the primal cost due to the change in the w-variables and z-variables, excluding advertiser m 1 and impression (j, k): z(j, l). i S(j) w(i, j) + l<k The first sum has k 1 terms corresponding to advertisers that were already assigned impressions belonging to user j, and the second sum has also k 1 terms corresponding to the first k 1 impressions of user j. We now match each advertiser i S(j) with the variable z(j, l) corresponding to the impression l assigned to i. The new values of the variables satisfy: w(i, j) + z(j, l) = max{v i x(i), v m2 x(m 2 )}. Since both terms above can only decrease during the execution of the algorithm, the contribution to the primal cost of advertiser i S(j) and impression l < k in the current round is not higher than the contribution to the primal cost of those advertisers and impressions in the previous round. The only increase in the primal cost is, thus, due to the increase in the variables w(m 1, j), z(j, k) and x(m 1 ). Their contribution is: w(m 1, j) + z(j, k) + d m1 x(m 1 ) = v m1 x(m 1 ) + x(m 1 ) + v m 1 = v m1 (1 + 1/c). c The dual profit in this round is v m1, so the ratio between the dual profit and the primal cost is (1 + 1/c) 1 = c/(c + 1) = 1 (c + 1) 1. Proof of (3). The algorithm never assigns more than one impression of user j to advertiser i. We only need to make sure that the algorithm does not assign a total of more than d i impressions to advertiser i. It is easy to verify that after assigning l impressions to advertiser i: ( ( x(i) = v i c ) l 1). Thus, after d i rounds x(i) v i if c = (1 + 1 d min ) d min 1. At this point all primal constraints corresponding to advertiser i are satisfied and the algorithm is guaranteed not to assign any further impressions to advertiser i. d i 18

19 Targeting constraints. We assumed thus far that advertisers valued all users equally. In practice, however, when buying display ad space, advertisers can provide targeting information, specifying which subset of impressions is acceptable. That is, advertisers have value v i for acceptable impressions that meet the targeting constraints and value of 0 for others (contracts for display ads typically specify a single price-per-impression that does not vary across the set of acceptable impressions, i.e., v i does not take on different non-zero values). Suppose targeting information is user-dependent only, i.e., an advertiser may value only a subset of users with certain characteristics (age, gender, location, etc.), but does not distinguish between different impressions (e.g., when visiting different webpages) from the same user. In this case, advertiser values have the following form: v(i, j) is either v i or 0 (i.e., a i finds users with v(i, j) = v i acceptable, and the rest unacceptable). We observe that the above algorithm also works for this more general setting. The only change is that the sets S(j) and S(j) include only advertisers that accept user j. This implies the following. Theorem 9. For c = (1 + 1 d min ) d min 1, the algorithm is (1 (c + 1) 1 )-competitive also when v(i, j) {0, v i } for all i, j. Proof Sketch. In the dual objective function, advertisers do not sums up their contribution over all users, but rather only over their acceptable users. This corresponds in the primal program to constraints that only contain an advertiser i and an acceptable user j. The same analysis shows now that all primal constraints remain feasible during the execution of the algorithm. The relation between the change in primal cost and dual profit also follows by noting that if impression k of user j is assigned, then every previous impression l k is also assigned. So we can find an advertiser i to match every l k as before. Theorem 10. In the presence of targeting constraints, no deterministic algorithm has a competitive ratio higher than 1 1/e, even when demands are large. Proof. Online b-matching, which generalizes online matching, is defined as follows: requests arrive online; a request can be matched to any server to which it has an edge, but no server can be used more than b times. An instance of the b-matching problem can be mapped to an instance of the frequency capping problem with targeting constraints: advertisers correspond to servers, and each request corresponds to a new user j; j S i for the set of servers to which the request has edges. All advertisers have demand d = b, f i = 1 and v i = 1. This reduction implies an upper bound of 1 1, which follows from the upper bound on the online b-matching problem given by (1+ 1 )d d [Kalyanasundaram and Pruhs, 2000]. For d f i = 1, this upper bound tends to 1 1/e. 6 Further Directions The frequency capping problem is an important practical problem which imposes interesting algorithmic challenges. Here are two main directions for further work. Improving 1 1/e for arbitrary valuations: There is a gap between the best upper bound of 1/ 2 and the best algorithm (1 1/e) for the case of arbitrary valuations without targeting constraints, discussed in 5. The targeting constraints are to be blamed for the matching aspects, leading to the upper bound of 1 1/e in Theorem 10. By removing these constraints, the difference between our problem and online matching resurfaces, and the upper bound of 1 1/e does not hold anymore. We believe that our primal-dual algorithm is an excellent starting point for a future online algorithm for frequency capping with arbitrary values that will go beyond 1 1/e. 19

20 Content-based targeting specifications: Targeting specifications may be not only user-based, but also depend on the webpage s content. For instance, an advertiser might want to display her ads only to males (user targeting) when they browse a sports related webpage (content targeting); targeting constraints are often of this form. So, advertisers now have valuations of the form v(i, j, k) {0, v i }, i.e., the value of the k-th impression of the j-th user to advertiser i is either v i or 0 depending on what page the user was surfing on his k-th impression. Note that the model of [Goel and Mehta, 2007] does not capture this problem, which entangles a matching aspect with frequency capping. The questions of designing a good online algorithm and finding the smallest upper bound (of course, 1 1/e is a trivial upper bound since this problem generalizes arbitrary valuations with targeting) are both open. Acknowledgements. We are extremely grateful to Ning Chen for several helpful discussions, and for first suggesting the total demand algorithm. References Zoë Abrams and Erik Vee. Personalized ad delivery when ads fatigue: An approximation algorithm. In WINE 07: Workshop on Internet and Network Economics, pages , N. Alon, B. Awerbuch, Y. Azar, N. Buchbinder, and J. Naor. A general approach to online network optimization problems. ACM Transactions on Algorithms, 2(4): , N. Alon, B. Awerbuch, Y. Azar, N. Buchbinder, and J. Naor. The online set cover problem. SIAM Journal on Computing, 39(2): , Nikhil Bansal, Ning Chen, Neva Cherniavsky, Atri Rudra, Baruch Schieber, and Maxim Sviridenko. Dynamic pricing for impatient bidders. ACM Transactions on Algorithms, 6(2), Nikhil Bansal, Niv Buchbinder, and Joseph Naor. A primal-dual randomized algorithm for weighted paging. J. ACM, 59(4):19, 2012a. Nikhil Bansal, Niv Buchbinder, and Joseph Naor. Randomized competitive algorithms for generalized caching. SIAM J. Comput., 41(2): , 2012b. A. Blum and J. Hartline. Near-optimal online auctions. In SODA 05: Proceedings of the eighteenth annual ACM-SIAM symposium on Discrete algorithms, pages , Avrim Blum, Vijay Kumar, Atri Rudra, and Felix Wu. Online learning in online auctions. In Theoretical Computer Sciences, pages 324(2 3): , N. Buchbinder and J. Naor. Improved bounds for online routing and packing via a primal-dual approach. In FOCS 06: Proceedings of the 47th annual ieee Symposium on Foundations of Computer Science, pages , N. Buchbinder and J. Naor. Online primal-dual algorithms for covering and packing problems. Mathematics of Operations Research, 34(2): , May Niv Buchbinder, Kamal Jain, and Joseph (Seffi) Naor. Online primal-dual algorithms for maximizing ad-auctions revenue. In ESA 07: Proceedings of the 15th Annual European Symposium, pages ,

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