Approximation Algorithms for Data Placement on Parallel Disks

Transcription

1 Approximation Algorithms for Data Placement on Parallel Diss L. Golubchi. Khanna y. Khuller z R. Thurimella x A. Zhu { Abstract We study an optimization problem that arises in the context of data placement in multimedia storage systems. We are given a collection of M multimedia data objects that need to be assigned to a storage system consisting of N diss d d 2::: d N. We are also given sets U U 2 ::: U M such that U i is the set of clients requesting the ith data object. Each dis d j is characterized by two parameters, namely, its storage capacity C j which indicates the maximum number of data objects that may be assigned to it, and a load capacity L j which indicates the maximum number of clients that it can serve. The goal is to nd a placement of data objects on diss and an assignment of clients to diss so as to maximize the total number of clients served, subject to the capacity constraints of the storage system. We study this data placement problem for two natural classes of storage systems, namely, homogeneous and uniform ratio. Our rst main result is a tight upper and lower bound on the number of items that can always be paced for any input instance to homogeneous as well as uniform ratio storage systems. We show that an algorithm given in [] for data placement, achieves this bound. Our second main result is a polynomial time approximation scheme for the data placement problem in homogeneous and uniform ratio storage systems, answering an open question of []. Finally, we also study the problem from an empirical perspective. Introduction We study a data placement problem that arises in the context of multimedia storage systems. In this problem, we are given a collection of M multimedia data objects that need to be assigned to a storage system consisting of N diss d d 2 ::: d N. We are also given sets U U 2 ::: U M such that U i is the set of Department of Computer cience and Institute for Advanced Computer tudies, University of Maryland, College Par, MD Research supported by NF CAREER Award CCR leana@cs.umd.edu. y Department of Computer and Informationcience, University of Pennsylvania, Philadelphia, PA 904. This wor was done when the author was at Bell Laboratories, Murray Hill, NJ sanjeev@cis.upenn.edu. z Department of Computer cience and Institute for Advanced Computer tudies, University of Maryland, College Par, MD Research supported by NF Award CCR and NF CAREER Award CCR samir@cs.umd.edu. x Department of Computer cience, The University ofdenver, Denver, CO This wor was done while the author was visiting the University of Maryland. rami@cs.du.edu. { Computer cience Department, tanford University, tanford, CA, This wor was done while the author was with the Department of Computer cience at the University ofmary- land. anzhu@cs.stanford.edu. clients requesting the ith data object. Each dis d j is characterized bytwo parameters, namely, its storage capacity C j which indicates the maximum number of data objects that may be assigned to it, and its load capacity L j which indicates the maximum number of clients that it can serve simultaneously. The goal is to nd a placement of data objects on diss and an assignment of clients to diss so as to maximize the total number of clients served, subject to the capacity constraints. We can view each data object simply as a color, and each client as a unit size item with a color corresponding to the data object in which they are interested. Each dis d j can hold at most L j items, with the additional constraint that the total number of distinct colors of the items in d j does not exceed C j. The data placement problem described above arises naturally in the context of storage systems for multimedia objects where one sees to nd a placement ofthe data objects, e.g., movies, on a system of diss. These and other issues related to the design of multimedia storage systems are discussed in [8]. We study this data placement problem for the following two natural types of storage systems. Homogeneous torage ystems: In a homogeneous storage system, all diss are identical. We denote by and L the storage capacity and the load capacity, respectively, of each dis and refer to this variant as -HDP (homogeneous data placement). Uniform Ratio torage ystems: In a uniform ratio storage system, the ratio L j =C j of the load to the storage capacity isidentical for each dis. We denoteby C min and C max the minimum and the maximum storage capacity ofany dis in such a system and refer to this variant asurdp (uniform ratio data placement). Notice that homogeneous storage systems are a special case of uniform ratio storage systems. In the remainder of this paper, we assume that (i) the total number of clients does not exceed the total P P M load capacity, i.e., i= ju N ij j= L j, and (ii) the total number of data objects P does not exceed the total N storage capacity, i.e., M j= C j.. Related Wor The data placement problem described above bears some resemblance to the classical multi-dimensional

2 napsac problem [6, 0, 2]. We may view each dis as a napsac with a load as well as a storage dimension, and each client as a unit size item with a color associated with it. However, in our problem, the storage dimension of a dis behaves in a nonaggregating manner in that assigning additional items of an already present color does not increase the load along the storage dimension. It is this particular aspect of our problem that maes it dicult to adapt nown techniques for multi-dimensional pacing problems. hachnai and Tamir [] studied the above data placement problem they refer to it as the class constrained multiple napsac problem. The authors gave an elegant algorithm, called the sliding window algorithm, and P showed that this algorithm pacs all items N whenever j= C j M + N ; for URDP. An easy corollary of this result is that one can always pac a ( ; +C min )-fraction of all items for URDP. (This is seen by increasing the capacity ofeachdisby,pac- ing all the items, and then from each dis dropping the color with the fewest items.) The authors showed that the problem is NP-hard when each dis has an arbitrary load capacity, and unit storage. Recently, hachnai and Tamir [2] study a variation of the data placement problem above where in addition to having a color, each item u has a size s(u) and a prot p(u) associated with it. For the special case when s(u) = p(u) for each item, and the total number of dierent colors M is constant, the authors give adualapproximation scheme whereby for any >0, they give a polynomial time algorithm to obtain a ( ; =4)-approximate solution provided the load capacity is allowed to be exceeded by a factor of ( + ). The paper by Dawande, Kalagnanam and ethuraman [5] deals with a related bin-pacing problem, where there are n items, each with a color and size, and m distinct bin sizes. The objective is to pac all items into bins such that no bin contains items of more than p colors. The objective istominimize the total capacity ofthe bins that are used. For the case where the total number of colors is constant they give a polynomial time approximation scheme (PTA). The authors also describe some practical heuristics for the problem which give a constant approximation guarantee..2 Our Results Our rst main result is a tight upper and lower bound on the number of items that can always be paced for any input instance to homogeneous as well as uniform ratio storage systems. It is worth noting that in the case of completely arbitrary storage systems no such absolute bounds are possible. Theorem.. (ection 3) It is always possible to pac a ( ; (+ p ) 2 )-fraction of items for any instance of -HDP, ormore generally, a (; (+ p )-fraction C min) 2 of items can always be paced for any instance ofurdp. Moreover, there exists a family of instances for which it is infeasible to pac any larger fraction of items. The upper bounds above are constructive asthey are achieved by the sliding window algorithm of []. A side-result of our proof technique here is a simple alternate proof of the result that all items can be paced P N whenever j= C j M + N ;. Our second main result is a polynomial time approximation scheme for the data placement problem in uniform ratio storage systems, answering an open question of []. Theorem.2. (ection 4) For any xed > 0, one can obtain in polynomial time a (; )-approximate solution to the data placement problem for any instance of URDP. We also strengthen the NP-hardness results of [] by showing that the data placement problem is NPhard even for very special cases of homogeneous storage systems. Theorem.3. (Appendix A) The -HDP problem is NP-complete for homogeneous diss with storage capacity =2and strongly NP-hard for 3. Both reductions above are from NP-hard partitioning problems and illustrate how item colors can eectively encode large non-uniform sizes arising in the instances of these partitioning problems, even though each item in our problem is of unit size itself. We also note here that the case = is easily solvable in polynomial time. Finally, we also study the problem from an empirical perspective. We study the homogeneous case on instances generated by a Zipf distribution [9] (this corresponds to measurements performed in [3]for amovies- on-demand application) and compare the actual performance of the sliding window algorithm with the bounds obtained above aswell as the bounds in []. The results of this study are presented in ection 3.2. We also show how to implement the sliding window algorithm so that it runs in O((N + M)log(N + M)) steps, improving on the O(NM) running time of [] (details of this implementation will be provided in the full version of the paper). Note that the input size for our problem is measured by N and M and not the total number of items (which canbemuch larger). The algorithm will output the subset of data objects assigned to each dis and the number of items of each color assigned to each dis. We next describe in some detail the motivating application for our data placement problem.

3 .3 Motivational Application Recent advances in high speed networing and compression technologies have mademultimedia services, such as video-on-demand (VOD) servers, feasible. The enormous storage and bandwidth requirements of multimedia data necessitates that such systems have very large dis farms. One viable architecture is a parallel (or distributed) system with multiple processing nodes in which each nodehasitsown collection of diss and these nodes are interconnected, e.g., via a high-speed networ. We note that diss are a particularly interesting resource. Firstly, diss can be viewed as \multidimensional" resources the dimensions being storage capacity and load capacity. Based on the application one or the other resource can be the bottlenec. econdly, all dis resources are not equivalent since a dis's utility is determined by the data stored on it. It is this \partitioning" of resources (based on data placement) that contributes to some of the diculties in designing costeective parallel multimedia systems, and I/O systems in general. In a large parallel VOD system improper data distribution can lead to a situation where requests for (popular) videos cannot be served even when the overall load capacity of the system is not exhausted because these videos reside on highly loaded nodes, i.e., the available load capacity and the necessary data are not on the same node. One approach to addressing the load imbalance problem is to partition each video across all the nodes in the system and thus avoid the problem of \splitting resources", e.g., as in the staggered striping technique []. However, this approach suers from a number of implementation-related shortcomings that are detailed in [4]. An alternate system is described in [4] where the nodes are connected in a shared-nothing manner [3]. Each nodej has a nite storage capacity, C j (in units of continuous media (CM) objects), as well as a nite load capacity, L j (in units of CM access streams). These nodes are constructed by combining groups of diss into dis clusters. In fact, in this paper we will mostly view nodes as \logical diss". For instance, consider a server that supports delivery of MPEG-2 video streams where each stream has a bandwidth requirement of 4 Mbits/s and each corresponding video le is 00 mins long. If each nodeinsuchaserver has 20 MBytes/s of load capacity and 36 GB of storage capacity, then each such node can support L j = 40 simultaneous MPEG-2 video streams and store C j = 2 MPEG-2 videos. In general, dierent nodes in the system may dier in their storage and/or load capacities. In our system each CM object resides on one or more nodes of the system. The objects may be striped on an intra-node basis but not on the inter-node basis. For example, objects that require more than a single node's load capacity (to support the corresponding requests) are replicated on multiple nodes. The number of replicas needed to support requests for a CM object is a function of the demand. This should result in a scalable system which can growon a node by node basis. The diculty here is in deciding on: () how many copies of each video to eep, which can be determined by the demand for that video, e.g., as in [4], and (2) how to place the videos on the nodes so as to satisfy the total anticipated demand for each video within the constraints of the given storage system architecture. Our data placement problem tries to capture these issues..4 Organization We start with an overview of the sliding window algorithm in ection 2. In ection 3, we present a tight analysis of the sliding window algorithm to derive the upper bounds of Theorem.. We also present herea family of instances that give the matching lower bound as wellasnumerical results. Finally, in ection 4 we present our approximation schemes for homogeneous as well as uniform ratio storage systems and thus establish Theorem.2. A proof of Theorem.3 appears in Appendix A. 2 liding Window Algorithm For sae of completeness, we describe here the sliding window algorithm of hachnai and Tamir []. We maintain a list R[] ::: R[m], m M of colors where R[i] denotes the number of items of color i that remain. The list R is sorted in non-decreasing order. At step j, we assign items to dis d j. For the sae of readability, R[i] always refers to the number of currently unassigned items of a particular color (i.e., we do not explicitly indicate the current stepj of the algorithm in this notation). Assume that the diss are numbered in a non-decreasing order of their capacities, i.e., C C 2 C N.Weassign items and remove from R the colors that are paced completely, and we move the partially paced colors to their updated places (in the sorted set) according to the remaining number of unpaced items of that color. The assignment of colors to a dis d j follows the general rule that we select the rst consecutive sequence of C j or less colors, R[u] ::: R[v], whose total number of items either equals to or exceeds the load capacity L j. We then assign colors R[u] ::: R[v] to d j. In order to not exceed the load capacity, we will split the items of the last color R[v]. It could happen that no such sequence of colors is available, i.e., all colors have relatively few items. In this case, we greedily select

4 the colors with the largest number of items to ll the current dis. The selection procedure is as follows: we rst examine R[], which is the color with the smallest number of items. If these items exceed the load capacity, we will assign R[] to the rst dis and re-locate the remaining piece of R[] (which for R[] will always be the beginning of the list). If not, we then examine the total numberofitemsinr[] and R[2], and so on until either we nd a sequence of colors with a suciently large number of items ( L j ), or the rst C j colors have a total number of items <L j. In the latter case, we go on to examine the next C j colors R[2] ::: R[C j +] and so on, until either we ndc j colors with a total number of items at least L j or we are at the end of the list, in which case we simply select the last sequence of C j colors which has the greatest total number of items. We note that the liding Window algorithm can be implemented to run in O((N + M) log(n + M)) steps, where N is the number of diss and M is the number of colors. Note that this is a signicant improvement on the running time in []. The reason for the improvement is that we do not have to start the \window", of length C j, from the beginning of the list R in each iteration j. At the end of iteration i, after we have determined the colors R[m] ::: R[n] to place in dis d i,we now that R[j] < Li C i = r for all j m ;. This indicates that all \windows" ending with R[j], for all j m ; will under-ll the load of any dis. Thus the right end of the \window" will always monotonically move to the right. We need to use a 2-3 tree data structure to implement this algorithm. (ip-lists can be used as well to get the same expected worst case running time, with a simpler implementation.) Details are defered to the full version. 3 Analysis We now show that the liding Window Algorithm guarantees to pac (; (+ p ) 2 ) fraction of all items in the homogeneous case. We assume each dis has load capacity L and storage capacity. The liding Window Algorithm guarantees to pac (; (+ p ) fraction C min) 2 of items in the uniform ratio case, where the minimum capacity of a dis is denoted by C min. We also show that these bounds are tight. We rst discuss the homogeneous case. Note that if there are any unpaced items, then every dis is lled to the maximum either on the number of items it can hold or on the number of colors that can be stored. We will call the former as load saturated and the latter (the rest) as storage saturated. (Therefore, if a dis is storage saturated, then it still has some unlled load capacity.) Denote the number ofload- saturated diss and the number of storage-saturated diss by N L and N,respectively. It is easy to see that R[] ::: R[N L ] are load-saturated diss, and the rest are storage-saturated diss. Let m j denote the number of colors assigned to dis d j. Obviously for storagesaturated diss, m j =. Letc be the smallest fraction of load to which a storage-saturated dis is lled. Note that this dis must store a color with a number of items of that color being at most cl=. (Minimumis at most the average.) Now every color on the unassigned list has no more than c L= remaining items of that color. (Otherwise, the liding-window algorithm would have put this color on the lightest-loaded dis and increased greedily the total number of paced items.) Lemma 3.. Using the liding Window Algorithm, the number of unpaced items is at most clnl. Proof. The main thing we needtoprove is that there are at most N L colors left after we run the liding- Window algorithm. For each left over color we now that the number of items is at most cl, so the total number of unpaced items is at most clnl. We examine the number of colors stored in the loadsaturated diss. If there is a load-saturated dis d j with m j <colors, then there are no colors left when the algorithm terminates, i.e., all items are paced, which can be explained as follows. The reason that less than colors P are paced into d j is due to the fact that at m step j j i= R[i] L. ince we sort the colors in a nondecreasing order, at this point any consecutive sequence of ; colors in the list has the total size L. ince at step j, we \split" at most one color, which is always added to the beginning of R, atany step P t j we have a guarantee that, for the new list R, i= R[i] L, unless we have less than colors. This implies that we ll the diss to their load capacity until we run out of colors. Hence we can pac all items. We can now assume that all the load-saturated diss have colors. The storage-saturated diss have colors as well. We start with M N colors. During the process, we can split at most N L colors, i.e., we can generate at most N L new \instances" of originally existing colors. This is because only lling diss that are load-saturated can result in generating new \instances" of colors. o the number of new \instances" of colors generated is upper bounded by the number of loadsaturated diss. Thus the number of colors left is M + N L ; N N L. 2 By splitting a color we mean that, in the current iteration of the algorithm, only some of the items of that color are paced into the current dis the remaining items might be paced in future iterations of the algorithm. Hence, this color might be paced into multiple diss.

5 Corollary 3.. For \uniform ratio" diss, if P N j= C j M + N ;, then all colors can be paced using the liding Window algorithm. Proof. This is an alternate proof for the claim in []. Our analysis of the algorithm maes the proof simpler. Let r = Lj C j, denote the uniform ratio. ince the ratios' Lj C j are uniform, once any dis becomes storagesaturated, the rest of the diss will be storage-saturated as well. The main claim we need to prove is that after we ll dis d N;,wehave atmostc N colors left. We will prove this shortly. If d N; is storage-saturated, then we can safely assign the remaining C N colors to d N. If d N; is load-saturated, then all previous diss are load-saturated. ince the total number of items does not exceed the total load capacity, we will not exceed the load capacity. We argue that if there is a load-saturated dis d j with m j <C j, then all the items will be paced. At this stage, R[`] Lj m Cj j C j; r C Ct t; r for all `>m j and all t j. Recall that diss are sorted in non-decreasing order of C i.thus we have the following result: at any stept > j, P C t i= R[i] P C t i=2 R[i] C t r = L t, and so all items are paced without sliding the window. ince all load-saturated diss have C j colors, after we ll dis d N;,wehave generated at most N ; new \instances" of colors. The total number of colors left is M + N ; ; P N; i= C i C N. 2 Lemma 3.2. Using the liding Window Algorithm, the number of unpaced items is at most (;c)ln. Proof. At least LN L +cln items are paced. ubtracting this quantity from an upper bound on the total numberofitemsn L gives N L + N L L ; L N L ; c L N, which yields the claim. 2 Theorem 3.. The liding Window Algorithm guarantees to pac ( ; (+ p ) 2 ) fraction of items in the homogeneous case. Proof. The above two lemmas give ustwo upper bounds on the number ofunpaced items. Hence the number of unpaced items is at most min( clnl ( ; c) L N ). The number of paced items is at least L N L + c L N. Hence the ratio of unpaced(u) to paced() items is at most U This yields min( clnl ( ; c) L N ) L N L + c L N : U + ; ( + p ) 2 which proves the claim. (The details of this derivation are given in Appendix B.) 2 This proof can also be extended to the uniform-ratio case. The motto in the homogeneous case is that the bigger the dis the better the performance of the liding Window Algorithm. o in a uniform-ratio system one should expect the algorithm to do at least as well as the homogeneous case where all diss assume the smallest dis size in the uniform-ratio system. The following theorem formally proves this intuition. Theorem 3.2. The liding Window Algorithm guarantees to pac ( ; (+ p ) fraction of items C min) 2 in the uniform ratio case where C min denotes the minimum capacity of a dis in our system. Proof. Let r = Lj C j for j = :::N denote the uniform ratio. From the proof of Corollary 3. above, if there is a load-saturated dis d j with m j <C j,then all items will be paced. Thus we will focus on the case where for all j we have m j = C j. Let M L denote the total number of colors (we count the same colors in dierent diss as dierent multiple colors) in the load-saturated diss ::: N L,soM L = P N L j= m j = P NL j= C j. Let M denote the total number of colors in the P storage-saturated P diss N L + ::: N,so M = N N j=n L+ m j = j=n L+ C j. Again let c be the smallest fraction of load to which a storage-saturated dis is lled. Thus we have the following similar results: For each leftover color we now that the number of items is at most c r. There are at most N L colors left unassigned. We have N L ML m min = ML C min. The number of unpaced items is at most c r M L C min. At least r M L + c r M items are paced. The number of unpaced items is at most r M L + r M ; r M L ; c r M =(; c) r M. Hence the ratio of the unpaced(u) to paced() items is at most U min( crml C min ( ; c) r M ) : r M L + c r M Let y = M ML M L+M and thus M =; L+M y. implifying the upper bound for this expression, we obtain min( cy C min ( ; c)( ; y)) : y + c( ; y) Note that this is the same expression as in Theorem 3. for the homogeneous system. Optimizing this expression gives the same bound as in Equation 4. (Appendix B) with replaced by C min. This proves the claim. 2

6 3. Tight Example We nowgive an example to show that the bound of (; (+ p ) 2 ) is tight. In other words, there are instances for which no solution will pac more than ( ; (+ p ) ) 2 fraction of items. Assume that, the storage capacity of a dis is a perfect square 2 and 2. (When =, the tight example is trivial.) Let N, thenumber of diss, be + p, and let L = + p. There are p colors with a large number of items each, U ::: U p with ju i j =2+ p for i p we will refer to these as \large colors". And, there are ( ;)(+ p )+ colors with a small number of items each, U p + ::: U (+ p ) with ju i j =for p + i ( + p ) we will refer to these as \small colors". We will show that there are always at least p items that do not get paced. In this case, the fraction of items that are not paced is at least exactly (+ p ) p (+ p )(+ p ) 2. This proves the claim. which is We rst consider the p large colors. An unsplit set U i has all its items paced in a single dis. A split set U i has its items paced in several diss. For a dis that contains at least one large unsplit color, the available load capacity left is at most ; 2. (Note that after pacing one large unsplit color, the available load capacity is smaller than the storage capacity.) We can exchange any of the remaining colors with j 2 items of the same color, with any j small (distinct color) items in any other dis, while still pacing the same number of items. These diss now have one large unsplit color, and at most ; 2 small colors. The remaining diss have only large split colors. p In fact, assume that there are exactly p (0 p ) large colors that do not get split U ::: U p, with dis d i containing U i. Now consider the remaining N ; p diss we are left with at least N ; p( ; ) = (N ; p) +p colors, but we only have (N ; p) storage capacity left. ince the remaining p ; large colors are all split, this generates an additional p ; p \instances" of colors. Thus we have at least (N ; p)+p + p ; p colors. This will create an excess of p items that we cannot pac. 3.2 Numerical Results We have implemented the liding Window algorithm and compared its performance to the theoretical results developed in this section. Refer to Figs. {3 for plots of the fraction of unassigned items given by our worst case bound (i.e., (+ p ) 2 ), the bound corresponding to 2 When is not a perfect square we can get arbitrarily close to the bound by modifying this family of examples. the corollary of the result in [] (i.e., + ), as well as the liding Window algorithm. The results presented here are for the homogeneous case only. For the purposes of this comparison we generated the test cases using the Zipf distribution to determine the sewness in the number of items of each color. The Zipf distribution is dened as follows [9]: Prob[item of color i] = where c = H (;) M c i (;) and 8 i = ::: M and 0 H (;) M = MX j= j (;) where determines the degree of sewness. For instance, = :0 corresponds to the uniform distribution whereas = 0:0 corresponds to the sewness in access patterns often attributed to movies-on-demand type applications, e.g., similar to the measurements performed in [3]. We experimented with dierent values of and computed the percentage of items that can be paced by the liding Window algorithm as a function of, the load capacity of a dis. The results of these experiments are given in Figs. {3. In all cases L = 00 and N =5. % of unassigned items sliding window algorithm # of colors/dis Figure : =:0. ( + p ) 2 We can draw the following conclusions from these gures: the theoretical bound is reasonably tight when the the numberofitemsofeach color is fairly sewed (as in Figs. 2 and 3), as is the case in a VOD server, which is our motivational application furthermore, the performance of the liding Window algorithm is very close to the theoretical bound for inputs which are \similar" to the tight example given in ection 3 (as in Fig. 2)

7 % of unassigned items % of unassigned items sliding window algorithm # of colors/dis Figure 2: =0:5. ( + p ) 2 sliding window algorithm # of colors/dis Figure 3: =0:0. ( + p ) 2 the performance of the liding Window algorithm can be signicantly better than the theoretical bound when the number of items of each color is approximately the same and each dis has a relatively small storage capacity (as in Fig. ) however, the theoretical bound is reasonably tight for larger values of (which again, is reasonable for our motivational application). 4 Polynomial Time Approximation chemes We now present an approximation scheme our data placement problem. We rst present the approximation scheme for homogeneous storage systems and then briey setch the ideas used to get an approximation scheme for uniform ratio systems. 4. Homogeneous torage ystems We design an algorithm that for any xed >0, gives a ( ; )-approximation in polynomial time. If the error parameter =( + p ) 2, then we can simply use the liding Window algorithm to obtain a ( ; )- approximation. In the rest of this section, we focus on the case when <=( + p ) 2. In other words, can be assumed to be a constant when is a xed constant. Our approximation scheme involves the following steps:. First we showthatanygiven input instance can be approximated by another instance I 0 such that no color class in I 0 contains \too many" items. 2. Next we show that for any input instance there exists a near-optimal solution that satises certain structural properties concerning how items are assigned to the diss. 3. Finally, we give an algorithm that in polynomial time nds the near-optimal solution referred to in step (2) above, provided that the input instance is as determined by step () above. We now describe in detail each of these steps. In what follows, we use OPT(I) to denote an optimal solution to the instance I and to denote =. Also, for any solution, weusejj to denote the number of items paced by it. 4.. Preprocessing the Input Instance We say that an instance I is B-bounded if the size of each color class is at most B. Lemma 4.. For any instance I, we can construct in polynomial time another instance I 0 such that I 0 is (L)-bounded, any solution 0 to I 0 can be mapped to a solution to I of identical value, and jopt(i 0 )j( ; )jopt(i)j. Proof. Consider a color c j in the instance I such that ju j j >L. Replace c j with a new set of colors c j c2 j ::: cs j where s = dju jj=(l)e. Let Uj i denote the set of items with color c i j where i s. Then juj s; j = ::: = juj j = L and juj s j = ju jj;(s;)l: Repeat this procedure for any color class that has more than L items in I. Wenowhave our instance I 0. It is easy to see that any feasible solution to I 0 gives a feasible solution of same value to I simply replace each color c i j with c j. Now consider a solution for instance I. We show that it can be mapped to a solution 0 of size ( ; )jj for I 0.IfjU j jl for j M, then clearly is also a feasible solution of the same value for I 0. Otherwise, x a color class U j in I such that ju j j >L. Label the occurrences of the items of color c j as 2 ::: as we move fromd to d N in solution. Replace the ith occurrence of a color c j item with an item of color c l j where l = di=le. The resulting solution may no longer be a feasible solution for I 0. A dis may nowcontain items of two dierent colors, say c l j and cl+ j, in place

8 of a single color c j and hence the total number of colors in the dis may become +. We simply discard all the items in any diss where this event occurs. Repeat this procedure for every color class with more than L items in I. We claim that we have discarded no more than an -fraction of paced items. The reason is that we throwaway at most L items from a color class at a crossover dis but this event occurs only once in every L occurrences of items paced from a color class. Thus what we discard is at most an -fraction of what is paced tructured Approximate olutions Let us call a color class U j small if ju j jl=, and large otherwise. Also, for a given solution, we say that a dis is light if it contains less than L items, and it is called heavy otherwise. The lemma below shows that there exists a ( ; )-approximate solution where the interaction between light diss and large color classes, and between heavy diss and small color classes, obeys some nice properties. Lemma 4.2. For any instance I, there exists a solution satisfying the following properties: at most one light dis receives items from any large color class, aheavy dis is assigned either zero or all items in a small color class, and pacs at least ( ; )OPT(I) items. Proof. Let n i denote the number of items assigned to the ith dis in the solution OPT(I). Relabel the diss through N such that n n 2 ::: n N.Assume w.l.o.g. that OPT(I) is a lexicographically maximal solution in the sense that among all optimal P solutions, i OPT(I) is one that maximizes the sums j= n j for each i 2 [::N]. It is easy to see that the rst property follows from the maximal property ofopt(i). To establish that a heavy dis in OPT(I) receives either zero or all items from a small color class in the solution, wemay need to discard some items from the heavy diss in OPT(I). Let X be the set of heavy diss that contain at most ( ; )L items from large color classes. Consider any dis d i 2 X that receives some but not all items from a small color class U j. imply move all items of U j to d i. Repeat this process till no dis in X violates this property. ince a small color class has at most L= items, clearly the capacity of no dis is violated in this process. Finally, for the remaining diss, simply discard any items from small color classes. Clearly, the resulting solution is ( ; )-approximate. 2 For a given solution, a dis is said to be -integral w.r.t. a color class U j if it is assigned dle items from U j, where 0 < and is a non-negative integer. Lemma 4.3. Any solution can be transformed into a solution 0 such that each heavy dis in is ( 2 =)-integral in 0 w.r.t. each large color class, and 0 pacs at least ( ; )jj items. Proof. To obtain the solution 0 from, ineach heavy dis, round down the number of items assigned from any heavy color class to the nearest integral multiple of d( 2 =)Le. Then the total number of items discarded from any heavy dis in this process is at most ( 2 )L ; ( 2 )L (L): ince each heavy dis contains at least L items, the total number of items discarded in this process can be bounded by jj. Thus 0 satises both properties above The Approximation cheme We start by preprocessing the given input instance I so as to create an (L)-bounded instance I 0 as described in Lemma 4.. We now give an algorithm to nd a solution to I 0 such that satises the properties described in Lemmas 4.2 and 4.3 and pacs the largest number of items. Clearly, jj ( ; ) 2 jopt(i 0 )j( ; ) 3 jopt(i)j: Let O be an optimal solution to the instance I 0 that is lexicographically maximal. Assume w.l.o.g. that we now thenumber of heavy diss in O, say N 0. Let H be the set of diss d through d N 0 and let L be the remaining diss, d N 0 + through d N. The algorithm consists of two steps, corresponding to the pacing of items in H and L respectively. Pacing items in H: We rst guess a vector hl l 2 ::: l N 0i such that l i denotes the number of small color classes to be assigned (completely) to a dis d i 2 H. ince all diss are identical, we can guess such avector in O(N + )timeby guessing a compact representation of the following P form. We guess a vector hq 0 q q i such that i=0 q i = N 0 and q i denotes the number of diss in H that are assigned i small color classes (completely). It is easily seen that any such vector can be mapped to a vector of the form hl l 2 ::: l N 0i and vice versa. Now proceeding from through N 0,we assign to a dis d i the largest size l i small color classes that remain. Next we use a dynamic program by moving across the diss from through N 0 so as to nd an optimal ( 2 =)-integral solution for pacing the largest number of items from the large color classes. For the purpose of

9 this pacing, the capacity ofeachheavy dis is restricted to be ( ; )L and the number of color classes allowed in dis d i is given by ; l i. Let = = 3 and q = d( 2 L)=e. The dynamic programming solution is based on maintaining a -tuple ~v = hv v 2 ::: v i where v i denotes the number of large color classes that have i q elements available in them. Proceeding from i = through N 0,we compute a table entry T [~v i] for each possible state vector ~v. The entry indicates the largest number of items that can be paced in the diss d through d i subject to the constraint that the resulting state vector is ~v. ince there are at most N color classes, the total number ofstatevectors is bounded by (N) =3, which is polynomial for any xed. Pacing items in L: We now that our solution need not assign items from a large color class to more than one dis in L. Moreover, at most L items from any large color class are paced in a dis in L. o at this stage we can truncate down the size of each large color class to blc. Wenow construct an instance of the b-matching problem on a weighted bipartite graph G =(X [ Y E) where X has a vertex for each dis in L and Y hasavertex for each remaining color class. For each x 2 X and y 2 Y, there is an edge (x y) 2 E whose weight is equal to the number of items remaining in the color class corresponding to y. Nowwe nd a maximum weight matching such that the degree of each vertex in X is restricted to be while the degree of each vertex in Y is restricted to be. Clearly, such a matching gives a feasible solution of maximum weight. This completes our approximation scheme. 4.2 Uniform Ratio torage ystems We now briey setch how the PTA result above can also be extended to the case of uniform ratio storage systems. If =( + p C min ) 2 then we can use the liding Window algorithm to obtain a ( ; )- approximation. On the other hand, if C min as well as C max are bounded by a constant (parameterized by ), the approach of the preceding subsection easily extends to give apta. The diculty thus lies in the case when C min is small but C max is relatively large. In other words, our system contains diss of widely varying storage capacities. We handle this case by showing that every \large" dis can be approximately represented by a collection of diss with bounded storage capacity such thatwe loose at most an -fraction of items due to this approximate representation. Once this transformation is made, we can once again use the approach of the preceding subsection to obtain a PTA. We defer the details to the full version. References []. Berson,. Ghandeharizadeh, R. R. Muntz, and X. Ju. taggered triping in Multimedia Information ystems. IGMOD, pages 79{90, 994. [2] C. Cheuri and. Khanna. On multidimensional pacing problems. In ACM ymp. on Discrete Algorithms, pages 85{94, 999. [3] A. L. Chervena. Tertiary torage: An Evaluation of New Applications. Ph.D. Thesis, UC Bereley, 994. [4] C. F. Chou, L. Golubchi, and J. C.. Lui. A performance study of dynamic replication techniques in continuous media servers. Technical Report C-TR- 3948, University of Maryland, October 998. [5] M. Dawande, J. Kalagnanam, and J. ethuraman. Variable ized Bin Pacing With Color Constraints. Technical report, IBM Research Division, T.J. Watson Research Center, 999. [6] A. M. Frieze and M. R. B. Clare. Approximation algorithms for the m-dimensional 0- napsac problem: worst-case and probabilistic analyses. European Journal of Operational Research, pages 00{09, 984. [7] M. R. Garey and D.. Johnson. Computers and intractability: A guide to the theory of NP-completeness. Freeman, an Francisco, 979. [8]. Ghandeharizadeh and R. R. Muntz. Design and Implementation of calable Continuous Media ervers. Parallel Computing Journal, 24():9{22, January 998. [9] D. E. Knuth. The Art of Computer Programming, Volume 3. Addison-Wesley, 973. [0] P. Raghavan. Probabilistic construction of deterministic algorithms: approximating pacing integer programs. Journal of Computer and ystem ciences, pages 30{43, 988. [] H. hachnai and T. Tamir. On two class-constrained versions of the multiple napsac problem. To appear in Algorithmica. [2] H. hachnai and T. Tamir. Polynomial time approximation schemes for class-constrained pacing problems. Manuscript, 999. [3] M. tonebraer. A Case for hared Nothing. Database Engineering, 9():4{9, 986. [4] J. Wolf, H. hachnai, and P. Yu. DAD Dancing: A Dis Load Balancing Optimization cheme for Videoon-Demand Computer ystems. In ACM IGMET- RIC/Performance Conf., pages 57{66, 995. Appendix A: NP-hardness proof In this appendix we give our proof of Theorem.3. We rst dene the PARTITION problem, and then describe a polynomial time reduction from it to the homogeneous data placement problem. PARTITION: Given a nite set A, with a size s(a) for each element a 2 A, is there a subset A 0 A such that X a2a 0 s(a) = X a2a;a 0 s(a)?

10 This remains NP-complete even if we require ja 0 j = jaj=2 [7]. Let n = jaj. Proof. The problem is easily seen to be in NP, since a proposed solution can be trivially checed in polynomial time. We nowshow a polynomial time reduction from PARTITION. Let a max be the maximum size item. For the reduction, dene K = n C s(a max ) where C is a large constant. Let L = 3 K and N = n. o there are n 2 diss with P = 2 (storage capacity) and P a large L value. n Let D = i= (K + s(a n i)) = n K + i= s(a i). Let M = n + 2, with x i = K 2 ; s(a i) for i n and x n+ = D and x 2 P n+2 = D. 2 P n+2 Note that i= x n i = i= x i + D = 3 n 2 K which is exactly NL, the storage capacity. We claim that if there is a solution to the partition problem with ja 0 j = jaj=2 and with s(a 0 )=s(a ; A 0 ) then there is a way topac all items into the N diss. The items are paced as follows. Put x i items of color i in dis i. Dis i now has space for one new color, and exactly 3K 2 ; ( K 2 ; s(a i)) items. This is exactly K + s(a i ). If item a i 2 A 0 then add items of color n + to dis i, otherwise add items of color n +2 to dis i. ince each dis can hold two colors, this does not violate the color. Note that the P number of items of color n+ that we pac is exactly a i2a 0(K +s(a i )) = n 2 K + s(a0 )= D. The calculation is identical for 2 items of color n + 2, and this concludes the proof that all items are paced. We now argue that if there is a solution to 2-HDP where all items get paced, then there is a solution to the PARTITION problem. We rst claim that if all items are paced, then items of colors i and j, with i j n cannot be paced into the same dis. This is the case since: (a) only two colors can be paced in a dis and hence no other color can go into that dis, and (b) the total capacity usedupby items of color i and j, i j n, would not exceed K, which ismuch less than the capacity of the dis. If we cannot saturate the dis to full load capacity, thenwe cannot pac all items (since the number of items exactly equals the total load capacity). If each items of color i is in a distinct dis, then without loss of generality we pac items of color i in dis i and now we are left with items of only two colors that we need to split equally between the diss, and each dis can only tae an item of one color, with K + s(a i ) items of that color. ince K >> s(a i )we must pac items of color n + in exactly n=2 diss, and the items of color n + 2 in the remaining n=2 diss. Let A 0 = fa i jitems of color n +arepaced in dis ig. As said earlier ja 0 j = n,ands(a0 )=s(a ; A 0 ). This completes the proof. 2 When 3, the problem can be seen to be strongly NP-hard for even the homogeneous case by a simple reduction from 3-PARTITION [7]. Appendix B: Details of Proof of Theorem 3. Given that the ratio of unpaced(u) to paced() items is at most U min( clnl ( ; c) L N ) L N L + c L N let y = NL N N and thus N =; y. implifying the upper bound for the numberofunpaced to paced items, we obtain (4.) This is the same as min( cy ( ; c)( ; y)) : y + c( ; y) cy min( y + c( ; y) ( ; c)( ; y) y + c( ; y) ): We can simplify the two functions to the following min( ( c + ;y y ) ( ; c)( ; y) ; ( ; c)( ; y) ): The rst term is strictly increasing as c or y increases, while the second term is strictly decreasing as c or y increases. o in order to maximize the expression, we need to set the two terms equal, which means cy =(; c)( ; y): This gives y = ; c ; c + c : ubstituting for y gives us that the upper bound for this ratio is at most c ; c 2 ; c + c : 2 This achieves its maxima when c =(; + p ). The fraction of all items that are paced is U + = + U : Replacing the bound that we derived for c we get that This yields U +2 p : U + ; ( + p ) 2 :