# Catch Them if You Can: How to Serve Impatient Users

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1 Catch Them if You Can: How to Serve Impatient Users Mare Cygan Institute of Informatics, University of Warsaw, Banacha 2, Warszawa, Marcin Mucha Institute of Informatics, University of Warsaw, Banacha 2, Warszawa, Matthias Englert Department of Computer Science and DIMAP, University of Warwic, Coventry CV4 7AL, Piotr Sanowsi Institute of Informatics, University of Warsaw, Banacha 2, Warszawa, Anupam Gupta Department of Computer Science, Carnegie Mellon University, Pittsburgh, PA ABSTRACT Consider the following problem of serving impatient users: we are given a set of customers we would lie to serve. We can serve at most one customer in each time step getting value v i for serving customer i. At the end of each time step, each as-yet-unserved customer i leaves the system independently with probability q i, never to return. What strategy should we use to serve customers to maximize the expected value collected? The standard model of competitive analysis can be applied to this problem: picing the customer with maximum value gives us half the value obtained by the optimal algorithm, and using a vertex weighted online matching algorithm gives us 1 1/e fraction of the optimum. As is usual in competitive analysis, these approximations compare to the best value achievable by an clairvoyant adversary that nows all the coin tosses of the customers. Can we do better? We show an upper bound of if we compare our performance to such an clairvoyant algorithm, suggesting we cannot improve our performance substantially. However, these are pessimistic comparisons to a much stronger adversary: what if we compare ourselves to the optimal strategy for this problem, which does not have an unfair advantage? In this case, we can do much better: in particular, we give an algorithm whose expected value is at least 0.7 of that achievable by the optimal algorithm. This improvement is achieved via a novel rounding algorithm, and a non-local analysis. Categories and Subject Descriptors F.2.2 [Theory of Computation]: Nonnumerical Algorithms and Problems; G.2.2 [Discrete Mathematics]: Graph theory Graph algorithms Permission to mae digital or hard copies of all or part of this wor for personal or classroom use is granted without fee provided that copies are not made or distributed for profit or commercial advantage and that copies bear this notice and the full citation on the first page. To copy otherwise, to republish, to post on servers or to redistribute to lists, requires prior specific permission and/or a fee. ITCS 13, January 9 12, 2012, Bereley, California, USA. Copyright 2013 ACM /13/01...\$ General Terms Algorithms, Theory Keywords online algorithms, stochastic models, bipartite matching, impatience 1. INTRODUCTION The presence of impatient users is a common phenomenon in many contexts, and in this paper we initiate an algorithmic investigation of impatience. In particular, we consider the following problem of serving impatient users: we are given a set of n customers we would lie to serve. At each time step, we loo at the customers still in the system and choose one of them to serve: if we serve customer i, we get a value of v i. This customer, once served, departs from the system we serve each customer only once. Moreover, the customers are impatient: at the end of each time step, each customer still in the system may time out. We model this timingout process in a stochastic way. After every timestep each customer i decides to stay in the system independently with probability p i, and leaves, never to return, with the remaining probability 1 p i. Hence, customer i remains in the system T i Geomp i time steps, or fewer, if she was served earlier. Our goal: to devise a strategy for serving customers, so that we approximately maximize the expected total value obtained. Not only is this problem interesting in its own right, it also focuses attention on a new set of unexplored problems lying in the intersection of queueing theory/stochastic processes and approximation algorithms. It combines elements from both disciplines: the population evolves according to a natural stochastic process, but the values v i and probabilities p i are chosen in a worst-case fashion and not drawn from some underlying distribution themselves. To the best of our nowledge, this problem has not been studied in either literature. However, we believe that the problem is worth studying as it tries to answer a very basic question: what shall we do next when options can disappear? How shall we trade-off the probability of timing-out of the client with its value? Should clients with high value and low impatience be served before clients with low value and high impatience?

5 4.1 LP formulations and basic properties Consider the following linear program LP xi for a given instance I = v, p. max x itv i i,t subject to x it/p t i 1 t x it 1 i x it 0 i t LP xi To begin, we claim this linear program is a feasible relaxation of the problem. Lemma 4.1 For any instance I of the problem, we have opti opt LP I, where opti is the expected value of the optimal non-clairvoyant algorithm, and opt LP I is the optimal value of LP xi. Proof. Given any non-clairvoyant algorithm A with expected total value of V on I, we can obtain a feasible solution to LP xi with the same objective value. To do this, set x it to the probability that the algorithm A pics item i in round t. Since x it are probabilities, i xit 1. Also, define y it to be the probability of picing item i in round t conditioned on item i being alive in round t: hence x it = y it p t i. Since the algorithm can pic i in at most one round, we get t yit 1. Finally, the expected value we get is precisely it xitvi = V. As in the above proof, it will often be convenient to thin of the linear program in terms of the y it variables, so let us record this equivalent LP called LP yi here: subject to max i,t y itp t iv i y it 1 t y itp t i 1 i y it 0 i t LP yi Lemma 4.2 The integrality gap of this LP is at least 1 1/2e: i.e., for every constant ε > 0, there are instances where opti < 1 1 2e + εopt LP I. Proof. Consider the instance where we have n items, each with v i = 1 and p i = 1. By setting n xi1 = 1 n 1, xi2 = and n n all other x 2 it = 0, we obtain a feasible solution to LP xi with value 2 1. However, any non-clairvoyant algorithm can get n unit value in the first round, at most 1 1 1/n n expected value in the second round, and at most 1/n expected value in all subsequent rounds. For large enough n, this is at most 2 1/e + ε, which gives us the claimed integrality gap. This integrality gap example is interesting because it shows that no local reasoning based on the LP relaxation i.e., one that considers a single round at a time can beat the ratio of 1 1 with respect to non-clairvoyant algorithms. e Later on we show how to sidestep this problem. We show how one can introduce interactions between rounds, and impose additional structural constraints on the solutions to LP xi, which are not satisfied by the solution considered in the above proof. Finally, we note that LP xi does not capture clairvoyant algorithms, which is a positive thing, otherwise due to the results of Section 3 we would not be able to obtain competitive ratios better than using this LP. Theorem 4.3 There exists an instance I, such that the expected total value of an optimal clairvoyant algorithm is strictly larger than opt LP I. Proof. Consider an instance with two items: one with v 1 = 1, p 1 = 0, and the other with v 2 = 2, p 2 = 1/2. Here, the optimal clairvoyant algorithm pics item 2 in the first round if and only if it dies before the second round, yielding a value of 2.5. On the other hand, any solution to LP xi has an objective value of at most 2. This is because we have x 21 = 0, x 11 + x 21 1 and x x 22 1 and the objective function is the sum of the left hand sides of these two inequalities. 4.2 Randomized Rounding: the First Try The basic approach to rounding LP xi could be to assign each item i to round t using y it := x it/p t i s as probabilities. When the algorithm is run, in round t we choose the highest value item that was assigned to this round and is still alive. The expected total value of such items is equal to i yitpt iv i, exactly the contribution of round t to the objective function of LP xi. Of course, we may not be able to obtain this much value we can only pic one item each round, and more than one item assigned to round t might still be alive in round t. The following lemma gives some insight into the performance of this strategy. Lemma 4.4 Let n items have values v 1 v 2... v n 0 and let the i-th item be available with probability π i, independently of other items, where πi 1. Let X be the value of the highest valued available item, and 0 if there are no items available. Then E [X] Pr [at least one item is available] πivi n. πi We will prove this lemma in fact, a generalization of it shortly, but let us first point out its implications. Adding in a zero value item v n+1 with a probability of being available of πi, and using the fact that the probability of at least one item being available is 1 n+1 1 πi 1 exp +1 πi = 1 1/e, Lemma 4.4 gives us the following corollary. π n+1 := 1 Corollary 4.5 With v i, π i and X as defined in Lemma 4.4 we have E [X] 1 1 e πivi. Theorem 4.6 The basic rounding scheme gives us expected value 1 1 e optlp I, and hence 1 1/e times the expected value of the best non-clairvoyant algorithm. Moreover, this ratio is tight. Proof. To bound the competitive ratio, simply use Corollary 4.5 with availability probabilities set to y itp t i. To show that it is tight, consider an instance I with n identical items and all v i = 1 and all p i = 1. One optimal solution to LP yi

6 is y it = 1 for all i = 1,..., n and t = 1,..., n. Rounding n this solution using the basic rounding scheme puts us in a classic balls-and-bins scenario, where in expectation only a 1 1 e fraction of the bins are non-empty which proves the claim. This means we need to wor harder to get an improvement over 1 1/e, which we will do in the next section. But first, let us give the proof of Lemma Proof of Lemma 4.4 For the improved analysis in Section 4.3, it will be useful to prove a slightly more general result. Let n items have values v 1 v 2 v n 0. Let S j = {1, 2,..., j}. Fix a downwards closed family of subsets I 2 [n]. Consider the following process, which depends on the v i, π i values, and the family I. Let the i-th item be available with probability π i, independently of other items, and let A be the random set of available items. Pic a subset of the available items as follows: consider items in increasing order of indices, and pic item i if a it is available and b if S i 1 A, the subset of available elements previously seen, lies in I. For instance, if I consists of all -element subsets, then this process pics the top + 1 available elements with largest value this is the context in which we will use the following lemma. Lemma 4.7 Let V be the value and N be the cardinality of the set piced by the above process. Then E [V ] E [N] πivi n. πi Proof. Define B i as the event that S i 1 A I. Note that i is piced with probability π i Pr [B i]. Moreover, we have the monotonicity property that Pr [B i] Pr [B j] i j, since S j 1 A I S i 1 A I. Note that E [V ] E [N] = viπipr [Bi] πipr [Bi]. 4.3 We first consider the case where, for some, v 1 = = v = 1 and v +1 = = v n = 0. We call these sequences of values basic. In this case, E [V ] E [N] = = = πipr [Bi] πipr [Bi] πipr [Bi] πipr [Bi] + i =+1 π i Pr [B i ] πipr [Bi] πipr [Bi] + Pr [B ] πipr [B ] πipr [B ] + Pr [B ] πi πi + i =+1 π i, i =+1 π i i =+1 π i where the first inequality uses that Pr [B ] Pr [B i ] for all i >, and the second inequality uses the two facts that a Pr [B i] Pr [B ] for all i, and b reducing the numerator and denominator by the same value causes the fraction to decrease. Let us loo again at the inequality we proved for all basic sequences v i: n E [N] E [V ] π iv i, πi Note that if we fix the π i, then the last term is a constant and the other two: E [V ] and πivi are linear in vi and non-negative. Therefore, the inequality extends to all nonnegative linear combinations of basic sequences, and these are exactly the non-increasing sequences of values. Proof of Lemma 4.4: Use Lemma 4.7 with I = { }. Hence the process pics the most valuable item, and the expected size E [N] is equal the probability that at least one item appears. We note that alternatively, Lemma 4.4 can be proven via the concept of fair contention resolution due to Feige and Vondra [9]. 4.3 Randomized Rounding: Second Time s the Charm To improve on the previous analysis, the main idea is the following. We want a rounding procedure that taes advantage of a situation where none of the items assigned to some round are piced because none survived. In this case we allow picing, say, two items during the next round. We will actually pic one of these items in this round, borrowing from the future. We show that if we are allowed to tae two items in a round with constant probability, then we can obtain more than a 1 1/e fraction of the LP value in expectation. Note that conditioning on the event that in round t 1 no item was piced changes the probabilities for items being assigned to round t. However, we show that these events are positively correlated. In other words, nowing that no item was piced in round t 1 can only increase the probability that an item is assigned to round t and survives. Another technical problem we face is that for some rounds it may be the case that we always pic some item e.g., this happens in round 0. We artificially decrease the probabilities of assigning items to such rounds to ensure that even in these rounds, with constant probability, no item is piced The Algorithm We first change the LP formulation slightly as follows: we add the constraint that for a special item i, x i 0 = 1. By iterating over all items i, each time adding a constraint x i0 = 1 to the linear program LP xi, we select the best LP solution among all the n solutions considered. This does not alter the fact that the new LP optimum value is still an upper bound for opti. Before we present the rounding, here is a thought experiment. The natural rounding algorithm would be to assign item i to round t with probability y it, independently of the other items. Let Y i,t be an indicator variable corresponding to an item i being available in step t, i.e., i being assigned to round t and not dying before round t. Note that the underlying sample space is given by the random allocations of items to rounds, and by the randomness of the items coin flips. Also, let A t = { i Yi,t = 0} be the event that there are no items available in round t they all either die before this round, or are assigned elsewhere. Now for the actual algorithm: we preprocess all the probabilities so that for each round t we have Pr [A t] Q, where

7 1 2e 1 we set Q = More precisely, for each round t we compute the greatest value of α t 0, 1] such that Pr [A t] = i 1 αt yit pt i Q. Next, we randomly assign items to rounds so that item i is assigned to round t with probability α ty it. With probability 1 t αtyit, the item i remains unassigned. When the algorithm is run, in round t we consider the available items in this round: those which were assigned to this round by the random process and are still alive. Among these, we pic the most valuable item. Moreover, for t > 0, if more than one item is available, and if no item was piced in round t 1, we also pic the second most valuable available item. Note that this algorithm allows us to pic two items in some rounds, which is not allowed by our model. However, we can implement this as follows: if none of the items assigned to round t survived, we pic the most valuable item that was assigned to round t + 1 and is still alive. Then, when round t + 1 comes, we can pic an item again if any survived, effectively picing two items in this round. Note that in this way we cannot do worse than actually picing two items in round t + 1. We might do better though, if the item piced in round t were to die after that round. This brings us to a subtle technical issue: To now whether we can pic two items in round t + 2, we need to now if any of the items assigned to round t + 1 survived. However, we cannot now this if we piced an item assigned to round t + 1 in round t, and then no other item survived until round t+1. When that happens, we flip a coin for this item and declare it alive/dead in round t + 1 accordingly. We need to eep these technical issues in mind when analyzing the rounding algorithm Analysis To analyze the algorithm, we need the following improvement over Lemma 4.4, where instead of picing the most valuable item in the round, we pic the two most valuable items if possible. This is where we mae use of the more general Lemma 4.7. Lemma 4.8 Let n items have values v 1 v 2... v n 0 and let the i-th item be available with probability π i, independently of other items, where πi 1. Let X be the sum of the values of the two highest valued available item or less if there are fewer items available. Then E [X] 2 3 π iv i. e Proof. To prove this, we use Lemma 4.7 with the family I = {, {1}, {2},..., {n}}, which ensures that we pic the two most valuable items. In this case N = miny, 2, where Y is the number of items available. It is easy to see that E [N] is maximized when all the π i s are equal, in which case E [N] = Pr [Y 1] + Pr [Y 2] = 2 2Pr [Y = 0] Pr [Y = 1] = n 1 1 n 1 2 3/e, n n where the inequality holds for all n 1, which proves the lemma. The last step follows from Lemma B.1 see Appendix B. Recall that Y i,t is the probability that item i is available in round t i.e., it is assigned to round t by the algorithm, and is actually alive in that round; A t is the event that no items are available in round t. Lemma 4.9 The following facts hold true for the variables Y i,t and events A t: a Y i,t A t 1 = Y i,t Y i,t 1 = 0, y b Pr [Y i,t = 1 A t 1] = i,t p t i Pr [Y 1 y i,t 1 p t 1 i,t = 1], i c Y i,t for different items i are conditionally independent on A t 1. Proof. Statement a is obvious, b follows from a, and c is easy to verify using b and its joint version saying that the joint distribution of Y i,t and Y j,t conditioned on A t 1 is the same as their distribution conditioned on Y i,t 1 + Y j,t 1 = 0. Theorem 4.10 The rounding algorithm of Section achieves a value of at least 1 1 LPxI, i.e., at least 2e % of the optimum. Proof. By linearity of expectations we can analyze the contribution of each round separately and it is enough to prove that in each round we gain at least a 1 Q fraction of its contribution to the objective function of the LP. Let LP t = i xitvi. We consider two cases, αt < 1 and αt = 1. Case I: α t < 1. In this case we now that Pr [A t] = Q, and by Lemma 4.4 in round t we gain at least 1 Pr [A t] i viαt yit pt i αt LPt i αt yit pt i 1 Q = 1 Q LP t. α t Case II: α t = 1. We infer that t > 0, since we now that there exists an item i, such that x i,0 = 1, and consequently α 0 = 1 Q. Therefore with probability Pr [A t 1] we are eligible to pic two items in round t since we piced no items in the previous timestep, and with probability 1 Pr [A t 1] we can pic only one item in round t. Let us denote by X 1 and X 2 the value of the most valuable and second most valuable item available in round t respectively. Note that the expected gain of the algorithm in round t equals E [X 1] + E [X 2 A t 1] Pr [A t 1]. As in our first approach, where we can always only pic one item, Corollary 4.5 implies that E [X 1] 1 1/eLP t. Moreover, by Lemma 4.8 we have E [X 1] + E [X 2] = E [X 1 + X 2] 2 3/eLP t. Observe, that it suffices to show E [X 2 A t 1] E [X 2], to prove that the expected gain generated in round t is at least 1 QLP t, since then E [X 1] + E [X 2 A t 1] Pr [A t 1] E [X 1] + E [X 2] Pr [A t 1] = Pr [A t 1] E [X 1 + X 2] + 1 Pr [A t 1]E [X 1] Pr [A t 1] Pr [A t 1] e = Pr [A t 1] e e 1 1 e LP t LP t Q LP t = 1 QLP t. e e Observe that by Lemma 4.9 when conditioning on A t 1 the probability that a given item is available in round t is increasing property b and the events of corresponding to different items being available in round t are still independent

8 property c. Therefore in order to prove E [X 2 A t 1] E [X 2] we can use the coupling technique. By standard arguments we can define an auxiliary probability space, such that when a set of items S is available in round t and A t 1 holds, then in the second probability space the set of available items S is a superset of S. Therefore the second most valuable item in S is at least as valuable as the second most valuable item in S and consequently E [X 2 A t 1] E [X 2]. Since both cases give us at least 1 Q LP t, the theorem follows. 5. CONCLUSIONS The presence of impatient users is an important aspect of many real-world processes, and we initiate an algorithmic investigation in this paper. In particular, we model the problem of serving impatient users to maximize the expected value accrued, where the impatience of the users is modeled by each user leaving the system based on an independent random Geometric random variable. The first approaches to solving this problem give only a 1 1/e 0.63 approximation, so our approach is to give a more careful rounding process that gets within a factor of the non-clairvoyant optimum. We also show that any online algorithm can get only 0.64 factor of the clairvoyant optimum, so our algorithm s comparison to the non-clairvoyant optimum is not only fair but necessary. Several questions remain: can we close the gap between our approximation algorithms and the upper bounds? Can we give algorithms for the case of user arrivals? For this last problem, it is not difficult to adapt the algorithm from [12] to give an improvement over 1 1/e when the survival probabilities are bounded away from ACKNOWLEDGMENTS We than Marco Molinaro and Krzysztof Ona for many interesting discussions in the initial stages of this research. Moreover, the first, fourth and fifth author were partly supported by NCN grant N N The fourth author was also supported by ERC StG project PAAl no The second author was partly supported by EPSRC award EP/D063191/1 and EPSRC grant EP/F043333/1, while the third author was supported in part by NSF awards CCF and CCF , and by a grant from the CMU- Microsoft Center for Computational Thining. 7. REFERENCES [1] G. Aggarwal, G. Goel, C. Karande, and A. Mehta. Online vertex-weighted bipartite matching and single-bid budgeted allocations. In SODA, pages , [2] F. Baccelli and G. Hébuterne. On Queues with Impatient Customers. Rapports de recherche. Institut national de recherche en informatique et en automatique, [3] J. R. Birge and F. Louveaux. Introduction to stochastic programming. Springer Series in Operations Research. Springer-Verlag, New Yor, [4] N. Chen, N. Immorlica, A. R. Karlin, M. Mahdian, and A. Rudra. Approximating matches made in heaven. In Proceedings of the 36th International Colloquium on Automata, Languages and Programming: Part I, ICALP 09, pages , Berlin, Heidelberg, Springer-Verlag. [5] F. Y. L. Chin, M. Chroba, S. P. Y. Fung, W. Jawor, J. Sgall, and T. Tichý. Online competitive algorithms for maximizing weighted throughput of unit jobs. Journal of Discrete Algorithms, 42: , [6] F. Y. L. Chin and S. P. Y. Fung. Online scheduling with partial job values: Does timesharing or randomization help? Algorithmica, 373: , [7] B. C. Dean, M. X. Goemans, and J. Vondrá. Approximating the stochastic napsac problem: The benefit of adaptivity. Math. Oper. Res., 334: , [8] S. Dye, L. Stougie, and A. Tomasgard. The stochastic single resource service-provision problem. Nav. Res. Logist., 508: , [9] U. Feige and J. Vondrá. The submodular welfare problem with demand queries. Theory of Computing, 61: , [10] J. Feldman, A. Mehta, V. S. Mirroni, and S. Muthurishnan. Online stochastic matching: Beating 1-1/e. In FOCS, pages , [11] N. Immorlica, D. Karger, M. Minoff, and V. Mirroni. On the costs and benefits of procrastination: Approximation algorithms for stochastic combinatorial optimization problems. In 15th SODA, pages , [12] L. Jeż. One to rule them all: A general randomized algorithm for buffer management with bounded delay. In ESA, pages , [13] C. Karande, A. Mehta, and P. Tripathi. Online bipartite matching with unnown distributions. In STOC, pages , [14] R. M. Karp, U. V. Vazirani, and V. V. Vazirani. An optimal algorithm for on-line bipartite matching. In STOC, pages , [15] M. Mahdian and Q. Yan. Online bipartite matching with random arrivals: an approach based on strongly factor-revealing lps. In STOC, pages , [16] R. Ravi and A. Sinha. Hedging uncertainty: Approximation algorithms for stochastic optimization problems. In 10th IPCO, pages , [17] D. B. Shmoys and C. Swamy. An approximation scheme for stochastic linear programming and its application to stochastic integer programs. J. ACM, 536: , [18] S. Zeltyn and A. Mandelbaum. Call centers with impatient customers: Many-server asymptotics of the m/m/i+ g queue. Queueing Systems, 51: , /s APPENDIX A. PROOF OF THEOREM 2.3 Proof of Theorem 2.3: Fix a large constant c. We will define an instance, for which the algorithm that always pics an item with the highest value of q iv i, obtains in expectation at most 2 of the value obtained by the optimum online c algorithm. Tae two sets of items: S 1 contains n 1 = n items, each with value v 1 = 1 and survival probability p 1 = where 1 = n 1/c2. S 2 contains n 2 = n 1/2c items, each with

9 value v 2 = c and survival probability p 2 = where 2 = c 1 = cn 1/c2. Consider first the algorithm which always pics an item with the highest value of q iv i. Note that this value is the same for both sets, so by perturbing it slightly, we can mae the algorithm pic elements from set S 1 until none are left. How many elements will be piced from S 1. By Theorem 3.3 w.h.p. they will end after about we can upper-bound this expression by x2 2 + x x = x2 x 2 + 5x 5, 101 x which is negative in the interval 0, , 1 ] and hence 2 2 f x is negative as well, which ends the proof. n 1 log p1 n 1/c 2 ln n 1 steps. Note that by this time, the expected number of elements of S 2 that are still alive is 1, so this algorithm does not profit from these elements. Consider now the algorithm that pics elements from S 2 as long as possible. Using Theorem 3.3 again we get that it pics about n 2 1 log p2 n 1/c2 ln n c elements. For large c this is almost half of the elements piced by the first algorithm, but the values of elements in S 2 are c, which gives the claim. B. EXPLANATION FOR THE PROOF OF LEMMA 4.8 Lemma B.1 For all positive integers n n n 1 3 n n e. Proof. The claim can easily be verified for n = 1, 2. To prove it for larger values of n we show that the left-hand side of the claim is an increasing function of n for n 2. The claim follows, since the right-hand side is the limit of the left-hand side as n. Let us define x = 1 and let fx be the value of the n left-hand side of the claim. Then x 0, 1] and fx = 21 x x x 1 x = 1 x x x. We need to show that f x < 0 for x 0, 1 ]. By differentiating we 2 obtain f x = 1 x 1 x By using the Taylor expansion 3x 3 ln1 x + 2x ln1 x. x 2 1 x ln1 x = n=1 we see that the numerator in the expression for f x is equal to x2 2 + x n 3 n 2. n 1 Since for n 4 we have n=4 x n n 0 < 3 n 2 n 1 0.1

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