3 Some Generalizations of the Ski Rental Problem

Transcription

1 CS 655 Design and Analysis of Algorithms November 6, 27 If It Looks Like the Ski Rental Problem, It Probably Has an e/(e 1)-Competitive Algorithm 1 Introduction In this lecture, we will show that Certain online problems that generalie the ski-rental problem we call them repeating ski rental generaliations are no more complicated than the ski rental problem, and thus, have the same optimal deterministic and randomied competitive ratios. There is a framework that allows us to build the optimally-competitive algorithms for these repeating ski rental generaliations. To do this, we will create a randomied algorithm for one such generaliation, the dynamic TCP acknowledgment problem, and then use that algorithm as a basis for creating algorithms for other ski rental generaliations. 2 Ski Rental Problem The ski rental problem can be stated thus: suppose you are about to go skiing for the first time in your life. Naturally, you ask yourself whether to rent skis or to buy them. Renting skis costs, $r, whereas buying skis costs more, say, $b. Your goal is to minimie your total cost including all future ski trips. Unfortunately, you don t know how many such trips there will be. You must make the decision online. In a previous lecture, we saw an optimal deterministic 2-competitive algorithm for this problem. It works by renting skis until the cost of renting so far equals the cost of buying, and then buying the skis. So in our example above, we would keep renting the skis until we d spent $b so far on renting, and then buy. We also saw an optimal randomied algorithm that works by picking on which day to buy skis according to some probability distribution. Using the right probability distribution (it happened to be e e 1 ) gives the optimal competitive ratio of e/(e 1), or about The ski rental problem is very important because a number of other systems and networking problems are, at heart, simply the ski rental problem. Here are a few. 3 Some Generaliations of the Ski Rental Problem In the spin block problem, a process in an operating system has to decide the most efficient way to wait for a lock. It could spin, at a cost of r units per cycle, or it could block at a large cost, say, b units. The b units are supposed to represent the cost of context-switching or restarting the process and restoring its state. One 2-competitive deterministic algorithm (which is optimal) is to spin until the cost of spinning so far equals the context-switch cost and then to context-switch, and to repeat the process until the lock becomes available. In the dynamic TCP acknowledgment problem, a server receiving for TCP packets over the network has to send acknowledgments for each of these packets efficiently. It is possible to acknowledge multiple incoming packets with a single acknowledgment packet. We can measure the cost of acknowledging a sequence of packets as follows: say sending an acknowledgment has a high fixed cost, b, while for every time unit that a packet remains unacknowledged, it accrues r latency cost. One algorithm for minimiing the cost of acknowledging a sequence of packets (that is optimally 1

2 CS 655 Design and Analysis of Algorithms November 6, 27 2-competitive) is to wait until the total latency equals the cost of sending an acknowledgment, and then to send the acknowledgment; and of course, to repeat the process continuously as packets arrive. These problems where buying, or context-switching or acknowledging, occurs not once but repeatedly, we shall call repeating ski rental generaliations. These generaliation all have deterministic 2-competitive algorithms that follow the same basic pattern, and you might wonder if e/(e 1)- competitive randomied algorithms can be created for them by following a pattern similar to skirental s algorithm. The answer, unfortunately, is, No. Using the deterministic algorithms we gave above to form a randomied algorithm in the style of ski rental leads to a greater than e/(e 1) competitive ratio. In 21, Anna Karlin, Claire Kenyon and Dana Randall discovered the optimal e/(e 1) algorithm. Their algorithm used the same basic framework as ski-rental s randomied algorithm i.e. choose from a one parameter family of deterministic algorithms according to a probability density function, e /e 1. However, what was unique was the structure of the deterministic algorithms that they came up with. We will look at their algorithm now, but first let us define the TCP acknowledgment problem properly. 4 Dynamic TCP Acknowledgment Problem Formally, there is a sequence of n packet arrival times R = a 1,..., a n. The goal is to partition R into k subsequences σ 1,..., σ k where a subsequence end is defined by an acknowledgment. We use σ i to denote the set of arrivals in the partition and t i to denote the time when the acknowledgment for σ i is sent. To capture the fact that all arrivals must be acknowledged, we must have k 1 and t k a n, the time of the last arrival. By delaying the acknowledgments for some packets, we reduce the number of acknowledgments sent (and the associated costs) from n to k. However, delaying an acknowledgment adds to the latency costs 1, which we want to avoid since it can result in bursty TCP traffic, with potentially disastrous consequences. Thus, one wants to optimally balance the costs of an acknowledgment with the costs associated with the increased latency. We model the trade-off between the cost of an acknowledgment and the cost for the latency by using the objective function C ALG = k + k latency(σ i, t i ), i=1 where ALG is an algorithm that partitions R into σ 1,..., σ k ; latency(σ i, t i ) measures the extra latency for the arrivals in i when the acknowledgment is sent at time t i i.e. latency(σ, t) = a σ(t a). N.B. For ease of calculation, we ll assume that all packets arrive between times and 1. Figure 1 shows a pictorial description of the latency cost. 4.1 Another 2-Competitive Deterministic Algorithm Now we can describe a one-parameter family of deterministic algorithms whose analysis, while tricky, is still possible. Algorithm A. Let P (t, t ) be the set of packets that arrive between time t and time t, i.e. the set of i such that t < a i < t. Suppose that the ith acknowledgment occurred 1 We use latency to refer to the delay introduced by postponing the transmission of the acknowledgment message. There is additional latency introduced by the transmission of messages, but we do not affect this by delaying acknowledgments. 2

3 CS 655 Design and Analysis of Algorithms November 6, 27 Figure 1: A pictorial representation of the dynamic TCP acknowledgment problem. Figure 2: Algorithm A at time t i (and assume that t = ), algorithm A performs the next acknowledgment at the first time t i+1 > t i for which there is a time τ i+1, t i τ i+1 t i+1, such that P (t i, τ i+1 )(t i+1 τ i+1 ) =. Intuitively, this time is chosen so that, given that the previous acknowledgment occurred at time t i, in hindsight, units of latency cost would have been saved by performing an additional acknowledgment at time τ i+1. See Figure 2. As a warmup, we will show that A 1 is 2-competitive. analysis. The following lemma will help in this Lemma 1. Without loss of generality, we can assume that the optimal algorithm sends an acknowledgment between any pair of successive acknowledgments of algorithm A 1. Proof. Consider an arbitrary input sequence I, and suppose that A 1 acknowledges at times t i. Consider any sequence S of acknowledgments, and assume that it does not send any acknowledgment in (t i, t i+1 ). Enrich this sequence by sending an additional acknowledgment at time τ i+1. The acknowledgment cost increases by 1, and the latency cost decreases by at least 1, so this new sequence is at least as good as S. Hence there is an optimal sequence which sends at least one acknowledgment in each interval (t i, t i+1 ). 3

4 CS 655 Design and Analysis of Algorithms November 6, 27 Figure 3: Proof of A 1 s 2-competitiveness Theorem 2. A 1 is 2-competitive. Proof. Consider an arbitrary input sequence I. From Lemma 1, the cost C A1, of A 1 on input I satisfies C A1 n OP T + L(OP T ) + L(A 1 \ OP T ) where OP T is an optimal algorithm, n OP T is the number of acknowledgments performed by OP T on input I, L(OP T ) is the total latency cost incurred by A, and L(A 1 \ OP T ) is the total latency incurred by A 1 that is not incurred by OP T. You can see from Figure 3 that L(A 1 \OP T ) is precisely the area of a set of rectangles, where each rectangle has its left side at the time when OP T sends an acknowledgment, and its right side at the following time when A 1 sends an acknowledgment. By definition of algorithm A 1, all these rectangles have area at most 1. Hence, L(A 1 \OP T ) n OP T and we obtain that C A1 C OP T + n OP T 2C OP T. 4.2 An e/(e 1)-competitive Randomied Algorithm This algorithm, A happens to be more analyable than the algorithm we described earlier, when used as the basis for a randomied algorithm for the dynamic TCP acknowledgment problem. First, we need to understand how the cost of A relates to the cost of OP T on any input. Let n (I) denote the number of acknowledgments of algorithm A on input I. Looking at Figure 4, we see that the latency cost of A is bounded by the area above the OP T curve, plus the area under OPT and over A, minus the area, denoted E (I), under A and over OPT. The first term is just C OP T (I) n OP T (I), the latency cost of OP T. The second term can be analyed as in the proof of Theorem 2; it is just a set of n OP T (I) rectangles, each of which has area at most by definition of A, for a total of at most n OP T (I). Hence C A n (I) C OP T n OP T (I) + n OP T (I) E (I). 4

5 CS 655 Design and Analysis of Algorithms November 6, 27 Figure 4: Definition of E Lemma 3. Let n denote the number of acknowledgments of algorithm A and n OP T denote the number of acknowledgments of the optimal algorithm on some input I. Then the area E above the optimal curve and below the A curve on input I is at least: E n w dw (1 )n OP T. Proof. Fix an input I. Let D(n, ) be the minimum, over all acknowledgment sequences S with n acknowledgments, of the area above the S curve that is below the A curve. We will prove a lower bound on D(n u, ) for all u. We claim that for any u > v, D(n u, ) (v )(n v n u ) + D(n v, ). (1) The proof is illustrated in Figure 5. The left side shows three acknowledgments sequences for the give input: S, the acknowledgment sequence with n u acknowledgments that minimies D(n u, ), and the acknowledgment sequences of A v and A. The shaded areas in Figure 5 represent the n v area v rectangles which caused algorithm A v to send an acknowledgment. At most n u such rectangles intersect the curve S, since this is exactly the number of times the curve S meets the arrival curve. Therefore, there are at least n v n u of these area v rectangles which lie strictly above S; the upper left corners of these are circled in Figure 5. Let T be the set of times at which these n v n u rectangles begin (if there are more than n v n u of these, then we choose any n v n u of them to define the set T). We define a new acknowledgment sequence S = S T. The resulting curve is shown in the right side of Figure 5. Since S = n u and T = n v n u, the number of acknowledgments in S is precisely n v. The n v n u rectangles of T are all between S and S. Each of them has area v, of which an area of at most can lie above A (by definition of A ). Thus the area above S is at least (v )(n v n u ) plus the area above S. These facts combine to give us Equation 1. Taking u = v + dv, we obtain from Equation 1 D(n v+dv, ) (v )(n v n v+dv ) + D(n v, ). Rewriting and integrating from to t, for any < t 1, we obtain t dd(n v, ) t (v )dn v 5

6 CS 655 Design and Analysis of Algorithms November 6, 27 Figure 5: Proof of lemma 3 which implies, by integration by parts on the right-hand side D(n t, ) D(n, ) t n v dv (t )n t. Observing that D(n, ) =, and that n v n t for v > t, we have D(n t, ) n v dv (1 )n t. Taking n t = n OP T and noting that D(n OP T, ) is a lower bound on E gives the lemma. Letting tend towards ero, the A curve tends to the curve of packet arrivals, sos that lim E is equal to the latency of OP T, and lemma 3 yields C OP T 4.3 Analysis of the Randomied Algorithm n d. (2) With these results, we can now analye our randomied TCP acknowledgment algorithm. Algorithm A. This algorithm chooses between and 1 according to the probability density function p() = e /(e 1) and then runs A. Theorem 4. Let A be the randomied algorithm that picks between and 1 according to the probability density function p() and then runs the resulting A. For any input I, the ratio between the expected cost incurred by A on I and the optimal cost on I satisfies where P () = p(x)dx C A (I) C OP T (I) 1 + (p() P ()) n d Proof. Let C A denote the expected cost incurred by the algorithm A. Combining the calculation below with Equation 2 yields C A C OP T n OP T + p()(n + n OP T E )d 6

7 CS 655 Design and Analysis of Algorithms November 6, 27 C OP T n OP T + = C OP T n OP T + = C OP T n OP T + = C OP T n OP T + = C OP T + because = C OP T + p()n d p()n d = 1 p()n d p()(n + n OP T p()n + p()n OP T p() p()n + p()n OP T p() (n w dw (1 )n OP T ) d p()n d + n OP T p() n w dw p()(1 )n OP T d n w dwd p() p() n w dwd by changing the order of integration = C OP T + = C OP T + p()n d (p() P ())n d w n w P ()ddw n w P (w)dw n w dwd If we plug in the ski rental problem distribution, p() = e /(e 1), we get C A (I) C OP T (I) 1 + (p() P ())n d n d = 1 + ( e e 1 e 1 e 1 )n d n d = e e 1 5 Conclusion The proof that algorithm A is e/(e 1)-competitive suggests that the repeating ski rental generaliations are not harder than ordinary ski rental. Karlin et. al. actually show that if we restrict the inputs of the TCP acknowledgment problem so they can be defined by a single parameter (just like for ski rental), then the expression for the competitive ratio of running A on these basis inputs is exactly the same as that of ski rental. This is further proof that the repeating ski rental problems are fundamentally the same as ski rental. We also saw a framework for designing optimally competitive randomied algorithms for ski rental generaliations; define a one-parameter family of algorithms in terms of savings the algorithm would have gained had it acted earlier, then pick from this family of algorithms according to the probability density function e /(e 1). In other words, if it looks like the ski rental problem, it probably has an e/(e 1)-competitive algorithm. 7

8 CS 655 Design and Analysis of Algorithms November 6, 27 6 Acknowledgment These lecture notes borrowed heavily from Karlin et. al. s paper, Dynamic TCP acknowledgment and Other Stories about e/(e-1), Symposium on Theory of Computing, 21. 8