Distributed Storage Allocations for Optimal Service Rates
IIEEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 1
Distributed Storage Allocations for Optimal Service Rates
Pei Peng, Moslem Noori, and Emina Soljanin,
Fellow, IEEE
Abstract
Redundant storage maintains the performance of distributed systems under various forms of uncertainty. Thispaper considers the uncertainty in node access and download service. We consider two access models under twodownload service models. In one access model, a user can access each node with a fixed probability, and in theother, a user can access a random fixed-size subset of nodes. We consider two download service models. In thefirst (small file) model, the randomness associated with the file size is negligible. In the second (large file) model,randomness is associated with both the file size and the system’s operations. We focus on the service rate of thesystem. For a fixed redundancy level, the systems’ service rate is determined by the allocation of coded chunksover the storage nodes. We consider quasi-uniform allocations, where coded content is uniformly spread among asubset of nodes. The question we address asks what the size of this subset (spreading) should be. We show thatin the small file model, concentrating the coded content to a minimum-size subset is universally optimal. For thelarge file model, the optimal spreading depends on the system parameters. These conclusions hold for both accessmodels.
Index Terms
Distributed storage systems, service rate, optimal allocations, erasure coding, redundancy.
I. I
NTRODUCTION
Distributed storage systems (DSSs) are a vital part of computing and content providing environments,such as cloud data centers, caching edge networks, and fog systems. Their purpose is to ensure reliablestorage and quick access of data by end-users or computing processes. Today, both goals are commonlyaddressed by storing data redundantly, either by replication or erasure coding. The DSS performance mustbe robust to various forms of uncertainty that are either inherent or external to the system (see, e.g., [3]–[5] a r X i v : . [ c s . I T ] F e b EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 2 and references therein). This paper considers the uncertainty in network access and download services,which are common in edge computing applications [6].We address the following network access uncertainties, as considered in [7] and the follow-up work. 1)Users may only be able to access a random subset of nodes. Such users can retrieve the file if the storagecontent of the accessed nodes suffices for file decoding. 2) Even when users have access to all nodes,a node may not respond. Such users can retrieve the file if the storage content of the responding nodessuffices for file decoding. We consider two download service models. In the first download model, therandomness associated with the file size is negligible. We refer to this model as the small file model. Inthe second download model, there is randomness associated with both the file size and inherent system’soperations. Service time is distributed according to some commonly adopted distributions. We refer tothis model as the large file model.This paper adopts a redundant storage model originally proposed in [7]. In this model, a file is splitinto multiple chunks, and redundancy is introduced at some fixed level, determined by the storage budgetthat the DSS has for the file. The total storage is the only constraint. There is no limit on how manychunks a particular node can store.Two important performance measures have been considered in the literature [1], [2], [7]–[11]. One is the probability of successful data recovery and the other is the average service rate . Finding these quantitieshas been quite challenging, and the optimal allocations are known only in some special cases. Someversions of this problem are related to a long-standing conjecture by Erd ˝os on the maximum number ofedges in a uniform hypergraph [12]. In general, both measures are of interest and should be simultaneouslytaken into account. Often increasing the chance of successfully downloading a file, while desirable, shouldnot come at the cost of intolerable delivery delay. Moreover, in practice, we may often want to partiallysacrifice a successful but tardy data delivery to some users to ensure that other users, that can receive thedata, are served fast. This paper is focused on the service rate of a DSS. Service rate is an increasinglyimportant performance measure, which addresses the question of stability in distributed systems withredundancy [13]–[19].For a fixed redundancy level, the system’s service rate (as other performance measures) is determinedby the allocation of coded chunks over the storage nodes. We consider quasi-uniform storage allocations,where coded content is uniformly spread among a subset of storage nodes. The question we address askswhat the size of this subset (spreading) should be. In this paper, we consider a quasi-uniform storage
EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 3 allocation, where coded chunks are uniformly spread among a subset of storage nodes. The question weaddress asks what the size of this subset should be. Note that, depending on the allocation, some subsetsof nodes may not contain enough file chunks between them to ensure data recovery, and accessing themwill result in a zero system’s service rate. On the other hand, again depending on the allocation, somesubsets of nodes will contain redundant file chunks, and that redundancy (superfluous for file recovery)can be exploited to increase the service rate.These issues were first addressed in [1] where a nonintuitive conclusion was reached for the small filedownload model that the allocation that maximizes the probability of successful data recovery is often notthe one that maximizes the DSS service rate. Depending on the number of storage nodes and the allocatedredundancy budget, it may be beneficial for recovery to maximally spread the redundant file chunks overthe nodes, whereas concentrating the redundant chunks may increase the expected service rate.The same conclusion does not hold in the large file case, where the node service time changes with thechunk size. We consider two service time models: scaled exponential service time and shifted exponentialservice time. These and other models are considered in the context of the server-dependent scaling anddata-dependent scaling in [20], [21]. We conclude that concentrating the redundant chunks may not alwaysmaximize the expected service rate. The optimal allocation varies depending on the system parameters.The paper is organized as follows: In Sec. II, we present the system architecture and the models for theservice time and the DSS service rate. In Sec. III, we state the problem and summarize the contributionsof this paper. In Sec. IV, V, and VI, we characterize the DSS service rate and determine the optimalallocation for three common service time distributions and two different access models. Conclusions aregiven in Sec. VII. II. S
YSTEM M ODEL AND P ROBLEM F ORMULATION
A. Storage Model
A file consisting of k blocks is to be redundantly stored over a DSS with N storage nodes. To protectthe data against nodes access failure or unavailability, the file is encoded by a maximum distance separable(MDS) code into mk ( m ∈ N ) encoded blocks so that any k of them are sufficient to recover the originalfile. The mk encoded blocks are partitioned into N subsets S i ’s for i ∈ { , . . . , N } where |S i | = s i , andthus (cid:80) Ni =1 s i = mk . We refer to such partitioning as allocation . The s i blocks in S i are stored at thestorage node i . Note that ≤ s i ≤ k since storing more than k blocks on a node is unnecessary. EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 4 k/α encoded blocks are allocatedto each of the mα storege nodesWhat should α > be? A = {
1, 2, N } ← -node access set1 2 mα . . . mα + . . . mα + Nmk encoded blocks
Fig. 1. A DSS of N nodes with quasi-uniform allocation. Each node stores either k/α or data blocks of interest to some users, and thusonly ϕ = mα nodes contain data blocks. The WiFi sign indicates that the node is able to serve the user. Note that that is independent ofwhether or not the node has been accessed or has the data. Here, three nodes are successfully accessed, but only two of them have (coded)data blocks. One of the accessed nodes has data blocks but is not able to serve the user. Dealing with a general storage allocation optimization problem is computationally difficult for a generalsetup, see [8]. In this paper, we are concerned with quasi-uniform allocations [9], where a node can eitherstore a constant number of blocks k/α ( α ∈ N ) or no blocks at all. We will refer to such allocation as α quasi-uniform allocation. Fig. 1 depicts an example of α quasi-uniform allocation on N nodes.We refer to a quasi-uniform allocation where α = 1 as the minimal spreading allocation [8]. Note thatfor the minimal spreading allocation, the k file blocks are simply replicated over some m storage nodes.Similarly, an allocation with α = N/m is referred to as a the maximal spreading allocation since the filechunks are spread over all N nodes in the system. B. Data Access and Delivery Models
Fixed-size Access: In this model, the download request is forwarded to a random r -node subset of the N storage nodes. Therefore, due to an MDS code being used to store the data, the access to a given r -subset A results in the successful recovery of the data iff the nodes in A jointly contain at least k coded blocks: (cid:88) i ∈A s i ≥ k. (1)Since for α > r , it is impossible to recover the data, we only consider the ≤ α ≤ r case.Probabilistic Access: In this model, the download request is forwarded to all N nodes. However, therequest to a node fails with probability p . Assuming that A represents the set of nodes that are successfullyaccessed, the condition for data recovery is also (1). In this case, ≤ α ≤ Nm . In this access model, |A| is a Binomial random number between 1 and N .Regardless of the access model, for an accessed subset of nodes A , we denote the number of nodescontaining data by ϕ ( A ) . For instance, in Fig. 1, three nodes ( |A| =
3) are accessed while only ϕ ( A ) = 2 EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 5 of them have data. For an α quasi-uniform allocation, data recovery from this subset is successful iff ϕ ( A ) ≥ α . The probability of successful file recovery under an α quasi-uniform allocation is given by P s ( α ) = (cid:88) A : (cid:80) i ∈A s i ≥ k P ( A ) (2)where P ( A ) is the probability of accessing A . Note that the sum goes over all sets A that satisfythe condition (1). It is easy to show that for the fixed-size access, P ( A ) = ( mαϕ ( A ) )( N − mαr − ϕ ( A ) )( Nr ) , and for theprobabilistic access, P ( A ) = (cid:0) mαϕ ( A ) (cid:1) (1 − p ) ϕ ( A ) p mα − ϕ ( A ) . C. Service Rate at a Node
We assume a request is simultaneously served by all nodes in the accessed set A that contain data,where each node takes some i.i.d. random time to deliver its data blocks. In the fixed-size access model, |A| = r , while in the probabilistic access model, |A| is a Binomial random variable between 1 and N .Note that the file can be reconstructed when the accessed nodes jointly deliver k encoded blocks. We herelimit our study to the case where a node has to deliver all its blocks for the download to count.For an α quasi-uniform allocation, the download request can be served iff ϕ ( A ) ≥ α , and as soon asall blocks are downloaded from any α out of ϕ ( A ) nodes. Therefore, the average service time for the file, T s ( α | ϕ ( A )) , is the expected value of the α -th order statistics of ϕ ( A ) waiting times at the storage nodes.Different service time distributions should be considered for different file sizes. For a small file, when atask is assigned to a storage node, there is a waiting time that is needed for the content inside the node tobe available to download. We assume the waiting time follows an exponential distribution. Since the file issmall, it is reasonable to assume the time needed to download the data is negligible compared to the waitingtime. As a result, the service time also follows an exponential distribution. For a large file, except for thewaiting time, the download time also needs to be considered. It is natural to assume that the downloadtime changes with the size of the file, i.e. a larger file needs a longer time to download. Therefore, weprovide two suitable distributions for the service time: scaled exponential and shifted exponential.Exponential Service Time: We assume the service time for each node is the waiting time. The waitingtimes at all nodes are independent and identically distributed (i.i.d) random variables following an ex-ponential distribution with mean µ , denoted by Exp( µ ) . For this model, we have the average servicetime T s ( α | ϕ ( A )) = µ ( H ϕ ( A ) − H ϕ ( A ) − α ) , where H α = (cid:80) αi =1 /i denotes the α -th harmonic number. The EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 6 corresponding service rate achieved by the nodes in A (with ϕ ( A ) > α nodes containing data) is µ s ( α | ϕ ( A )) = 1 T s ( α | ϕ ( A )) = µH ϕ ( A ) − H ϕ ( A ) − α . (3)It is not hard to see that µ s ( α | ϕ ( A )) ≤ µϕ ( A ) . (4)Scaled Exponential Service Time: We assume that a node storing the whole file delivers all of its blocksin a random time, exponentially distributed with the mean /µ . It is easy to see that, equivalently, a nodestoring /α fraction of the file delivers all of its blocks in the random time exponentially distributed withthe mean / ( αµ ) . For this model, we have T s ( α | ϕ ( A )) = αµ ( H ϕ ( A ) − H ϕ ( A ) − α ) , where / ( αµ ) comesfrom the service rate scaling discussed above. The corresponding service rate from set A is µ s ( α | ϕ ( A )) = αµH ϕ ( A ) − H ϕ ( A ) − α . (5)It is not hard to see that µϕ ( A ) ≥ µ s ( α | ϕ ( A )) ≥ µ ( ϕ ( A ) − α + 1) . (6)Shifted Exponential Service Time: In this model, the data delivery consists of two steps: first, the nodetakes an exponential random time to process the request; second, it takes a constant time, proportionalto its number of the node’s stored data blocks, to deliver them to the user. Therefore, the two-stepdelivery time for a node storing /α fraction of the file can be modeled by the shifted exponentialdistribution with rate µ and the shift parameter ∆ /α , denoted by S-Exp(∆ /α, µ ) . For this model, wehave T s ( α | ϕ ( A )) = ∆ α + µ ( H ϕ ( A ) − H ϕ ( A ) − α ) . The corresponding service rate from set A is µ s ( α | ϕ ( A )) = αµ ∆ µ + α ( H ϕ ( A ) − H ϕ ( A ) − α ) . (7)As ϕ ( A ) ∈ [ α, mα ] , it is not hard to see that µϕ ( A )∆ µ + α ≥ µ s ( α | ϕ ( A )) ≥ αµ ( ϕ ( A ) − α + 1)∆ µ ( mα − α + 1) + α . (8) D. DSS Service Rate
Under an α quasi-uniform allocation, the DSS service rate, µ s ( α ) , is found by averaging over theconditional service rates, given by: µ s ( α ) = (cid:88) A : (cid:80) i ∈A s i ≥ k P ( A ) µ s ( α | ϕ ( A )) (9) EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 7 where µ s ( α | ϕ ( A )) is the service rate when the set of accessed nodes is A , given by (3), (5) or (7), and P ( A ) is the probability of accessing set A , given by (2). Some parameters we used in the rest of paperare listed below. N – number of nodes in the system r – number of accessed nodes k – number of blocks in a file p – probability of failed access m – mk is the number of encoded blocks, α – mα is the number of nodes with blockswe refer to it as the redundancy level . we refer it as allocation parameter. III. P
ROBLEM S TATEMENT AND S UMMARY OF THE C ONTRIBUTIONS
Our goal is to characterize the DSS service rate µ s ( α ) for the access and service time models definedabove. We are in particular interested in finding which α maximizes µ s ( α ) . Recall that when α = 1 , wehave the minimal spreading allocation, and when α = N/m , we have the maximal spreading allocation.When < α < N/m , we have an α quasi-uniform allocation.In this work, we conclude that the optimal allocation to maximize the service rate µ s ( α ) depends onthe size of the stored file. For the small file model, the minimal spreading allocation, i.e. α = 1 , is alwaysoptimal, while for the large file model, this is not the case and it is difficult to determine the optimalallocation. We summarize the regimes where the minimal spreading allocation is optimal and non-optimalfor large files in Tables I and II, respectively. TABLE I C ONDITIONS FOR THE MINIMAL SPREADING ALLOCATION BEING OPTIMAL FOR LARGE FILESSERVICE TIME
Scaled Exponential Shifted Exponential A CC E SS Fixed-size r ≤ min ≤ α ≤ r { N − α − (cid:113) α ( mα − α − ) } r ≤ min ≤ α ≤ r { α − (cid:114) ∆ µ + αα (∆ µm +1) ( mα − α − ) ( N − } Probabilistic p ≥ max α ≥ { − α − (cid:113) α ( mα − α − ) } p ≥ max α ≥ { − α − (cid:114) ∆ µ + αα (∆ µm +1) ( mα − α − ) } The probability of successful recovery, denoted by P s ( α ) , is another important performance metric in aDSS [8], [9]. Since P s ( α ) and µ s ( α ) may exhibit different trends by changing the allocation, we also makecomparisons between these two metrics in our numerical analysis to find an overall optimal allocation.IV. S TORAGE A LLOCATION FOR SMALL FILES
For the small file model, we assume the service time at each node follows an exponential distributionwith the mean µ . The service rate for an accessed set of nodes A is µ s ( α |A ) , where µ s ( α |A ) > if EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 8
TABLE II C ONDITIONS FOR THE MINIMAL SPREADING ALLOCATION BEING NON - OPTIMAL FOR LARGE FILESSERVICE TIME
Scaled Exponential Shifted Exponential A CC E SS Fixed-size r ≥ min ≤ α ≤ r { α − (cid:114) mmα − α + 1 · ( N − α + 1) + α − } r ≥ min ≤ α ≤ r { α − (cid:115) ∆ µm ( mα − α + 1) + mα α (∆ µ + 1)( mα − α + 1) · ( N − α + 1) + α − } Probabilistic p ≤ max α ≥ { − α − (cid:113) mmα − α +1 } p ≤ max α ≥ { − α − (cid:113) mmα − α +1 } ϕ ( A ) ≥ α , and otherwise. Thus, the DSS service rate µ s ( α ) depends on all possible sets A , whichsatisfy (cid:80) i ∈A s i ≥ k . In the following two subsections, we determine the µ s ( α ) for the two consideredaccess models. Some of results in this section were published in [1]. A. Fixed-Size Access Model
For the fixed-size access model and an exponential service time, the DSS service rate in (9) becomes µ s ( α ) = µ (cid:0) Nr (cid:1) min( r, mα ) (cid:88) ϕ = α H ϕ − H ϕ − α (cid:18) mαϕ (cid:19)(cid:18) N − mαr − ϕ (cid:19) . (10)Notice that we will use ϕ instead of ϕ ( A ) in the rest of the paper. For the service rate of the minimalspreading allocation, we have the following lemma. Lemma 1.
Under the fixed-size access model with an exponential service time, the service rate of minimalspreading allocation, i.e. α = 1 , is µ s (1) = µmrN . (11) Proof:
From (10), we get µ s (1) = µ (cid:0) Nr (cid:1) min( r,m ) (cid:88) ϕ =1 H ϕ − H ϕ − (cid:18) mϕ (cid:19)(cid:18) N − mr − ϕ (cid:19) . Notice that when ϕ > mα , (cid:0) mαϕ (cid:1) = 0 . Thus, µ s (1) = µ (cid:0) Nr (cid:1) r (cid:88) ϕ =1 ϕ (cid:18) mϕ (cid:19)(cid:18) N − mr − ϕ (cid:19) = µm (cid:0) Nr (cid:1) r (cid:88) ϕ =1 (cid:18) m − ϕ − (cid:19)(cid:18) N − mr − ϕ (cid:19) . Using Vandermonde’s convolution, one can show that (cid:80) rϕ =1 (cid:0) m − ϕ − (cid:1)(cid:0) N − mr − ϕ (cid:1) = (cid:0) N − r − (cid:1) . Therefore, µ s (1) = µm ( N − r − )( Nr ) = µmrN . EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 9
Now that we have µ s (1) , the next step is to find an upper bound on µ s ( α ) for any ≤ α ≤ r andcompare this bound with µ s (1) . Lemma 2.
Under the fixed-size access model and an exponential service time, µ s ( α ) < µmrN for ≤ α ≤ r .Proof: By applying (4) to (10), we arrive at µ s ( α ) < µα (cid:0) Nr (cid:1) min( r,mα ) (cid:88) ϕ = α ϕ (cid:18) mαϕ (cid:19)(cid:18) N − mαr − ϕ (cid:19) = µm (cid:0) Nr (cid:1) r (cid:88) ϕ = α (cid:18) mα − ϕ − (cid:19)(cid:18) N − mαr − ϕ (cid:19) < µm (cid:0) Nr (cid:1) r − (cid:88) ϕ =0 (cid:18) mα − ϕ (cid:19)(cid:18) N − mαr − − ϕ (cid:19) = µm (cid:0) Nr (cid:1) (cid:18) N − r − (cid:19) = µmrN (by Vandermonde’s convolution) . Now, using Lemma 1 and 2, we have the following theorem on the optimal storage allocation.
Theorem 1.
Under the fixed-size access model and an exponential service time, the minimal spreadingallocation ( α = 1 ) maximizes the DSS service rate.Numerical Analysis: In Fig. 2, we respectively evaluate (2) and (10) to see how the DSS service rate µ s ( α ) (left) and the successful recovery probability P s ( α ) (right) changes with the allocation parameter α .We consider a system with N = 40 storage nodes and r = 10 accessed nodes for four different levels ofredundancy m ∈ { , , , } . In the left subfigure, µ s ( α ) reaches its maximum at α = 1 , i.e. the minimalspreading allocation is optimal. When m ≤ , µ s ( α ) decreases with increasing α and approaches . When m = 4 , µ s ( α ) reaches its minimum at α = 9 . In the right subfigure, when m ≤ , P s ( α ) reaches itsmaximum at α = 1 , thus the minimal spreading allocation is optimal. When m = 4 , P s ( α ) reaches at α = 10 , i.e. the maximal spreading allocation ( α = N/m ) is optimal. From the observations, we concludethat the minimal spreading allocation is always optimal under the DSS service rate, which is consistentwith the result in Theorem 1. The optimal allocation under successful recovery probability is determinedby the level of introduced redundancy.In Fig. 3, we analyze µ s ( α ) vs. α (left) and P s ( α ) (right) vs. α for different numbers of accessednode r ∈ { , , , } . We consider a system with N = 40 storage nodes and a redundancy level of m = 3 . In the left subfigure, µ s ( α ) reaches its maximum at α = 1 , i.e. the minimal spreading allocationis optimal. µ s ( α ) decreases with increasing α except for the scenario r = 14 . In the right subfigrue, when r ≤ , P s ( α ) reaches its maximum at α = 1 , i.e. the minimal spreading allocation is optimal. When r = 14 , P s ( α ) reaches its maximum at α = 12 . From the observations, we conclude that the minimalspreading allocation is always optimal under the DSS service rate. However, it is no longer optimal under EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 10
Fig. 2. Comparisons between the DSS service rate µ s ( α ) (left) and the successful recovery probability P s ( α ) (right) as a function of theallocation parameter α . The number of storage nodes is N = 40 , and the number of accessed nodes is r = 10 . The service time follows Exp(1) . When m ≤ , the minimal spreading allocation is optimal in both figures. When m = 4 , the minimal spreading allocation and themaximal spreading allocation are optimal for the service rate and probability of success recovery, respectively. the successful recovery probability as r increases. Fig. 3. Comparisons between the DSS service rate µ s ( α ) (left) and the successful recovery probability P s ( α ) (right) as a function of theallocation parameter α . The number of storage nodes is N = 40 , and the redundancy level is m = 3 . The service time follows Exp(1) .When r ≤ , the minimal spreading allocation is optimal in both figures. When r = 14 , an allocation with α = 12 is optimal under thesuccessful recovery probability. B. Probabilistic Access Model
For probabilistic access model under exponential service time, the DSS service rate (9) becomes µ s ( α ) = mα (cid:88) ϕ = α µH ϕ − H ϕ − α (cid:18) mαϕ (cid:19) (1 − p ) ϕ p mα − ϕ . (12)We have the following result about the minimal spreading allocation. EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 11
Lemma 3.
Under the probabilistic access model with exponential service time, the service rate of theminimal spreading allocation, i.e. α = 1 , is µ s (1) = µm (1 − p ) .Proof: From (12), we get µ s (1) = µ m (cid:88) ϕ =1 ϕ (cid:18) mϕ (cid:19) (1 − p ) ϕ p m − ϕ = µm m (cid:88) ϕ =1 (cid:18) m − ϕ − (cid:19) (1 − p ) ϕ p m − ϕ = µm (1 − p ) m − (cid:88) ϕ =0 (cid:18) m − ϕ (cid:19) (1 − p ) ϕ p m − ϕ − = µm (1 − p ) (by binomial expansion) . Similar to the fixed-size access model, we find an upper bound on the DSS service rate when α ≥ . Lemma 4.
Under the probabilistic access model with an exponential service time, for any α quasi-uniformallocation, its service rate satisfies µ s ( α ) < µm (1 − p ) .Proof: By applying (4) to (12), we arrive at µ s ( α ) < µα mα (cid:88) ϕ = α ϕ (cid:18) mαϕ (cid:19) (1 − p ) ϕ p mα − ϕ = µm mα (cid:88) ϕ = α (cid:18) mα − ϕ − (cid:19) (1 − p ) ϕ p mα − ϕ = µm (1 − p ) mα − (cid:88) ϕ = α − (cid:18) mα − ϕ (cid:19) (1 − p ) ϕ p mα − ϕ − < µm (1 − p ) mα − (cid:88) ϕ =0 (cid:18) mα − ϕ (cid:19) (1 − p ) ϕ p mα − ϕ − = µm (1 − p ) Using Lemmas 3 and 4, we have the following result on the optimal storage allocation for the proba-bilistic access model.
Theorem 2.
Under the probabilistic access model with an exponential service time, the minimal spreadingallocation maximizes the DSS service rate.Numerical Analysis:
In Fig. 4, we respectively evaluate (2) and (12) to see how the DSS service rate µ s ( α ) (left) and the successful recovery probability P s ( α ) (right) changes with the allocation parameter α . We consider a system with N = 40 storage nodes and a failed access probability p = 0 . for fourdifferent levels of redundancy m ∈ { , , , } . In the left subfigure, µ s ( α ) reaches its maximum at α = 1 ,i.e. the minimal spreading allocation is optimal, and decreases with increasing α . In the right subfigure,when m = 1 , P s ( α ) reaches its maximum at α = 1 , i.e. the minimal spreading allocation is optimal. When m ≥ , P s ( α ) reaches its maximum at α = 10 , approaching for m = 3 and when α ≥ . From theobservations, we conclude that the minimal spreading allocation is always optimal under the DSS service EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 12 rate, which is consistent with the result in Theorem 2. The optimal allocation under successful recoveryprobability is determined by the level of introduced redundancy.
Fig. 4. Comparisons between the DSS service rate µ s ( α ) (left) and the successful recovery probability P s ( α ) (right) as a function of theallocation parameter α . The number of storage nodes is N = 40 , and the probability of failed access is p = 0 . .The service time follows Exp(1) . When m = 1 , the minimal spreading allocation is optimal in both figure. When m ≥ , the minimal spreading allocation is optimalconsidering µ s ( α ) , and performs the worst considering P s ( α ) . In Fig. 4, we analyze µ s ( α ) vs. α (left) and P s ( α ) (right) vs. α for p ∈ { . , . , . , . } . We considera system with N = 40 storage nodes and a redundancy level of m = 3 . In the left subfigure, µ s ( α ) reachesits maximum at α = 1 , i.e. the minimal spreading allocation is optimal. µ s ( α ) decreases with increasing α . In the right subfigrue, when p ≤ . , P s ( α ) increases with α , and reaches at about α ≥ . When p = 0 . , P s ( α ) takes values around . . When p = 0 . , P s ( α ) reaches its maximum at α = 1 . From theobservations in Fig. 4, we conclude that the minimal spreading allocation is always optimal under theDSS service rate. However, the optimal allocation under the successful recovery probability depends onthe probability of failed access.V. S TORAGE A LLOCATION FOR LARGE FILES WITH S CALED E XPONENTIAL S ERVICE T IME
For the large file model, we assume that the service time at each node follows a scaled exponentialdistribution with the mean αµ , i.e. the allocation parameter α changes the scale of an exponentialdistribution. In the following two subsections, we determine the µ s ( α ) for the two considered accessmodels. Some of the results in this section were published in [2]. EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 13
Fig. 5. Comparisons between the DSS service rate µ s ( α ) (left) and the successful recovery probability P s ( α ) (right) as a function of theallocation parameter α . The number of storage nodes is N = 40 , and the redundancy level is m = 3 . The service time follows Exp(1) .minimal spreading allocation is optimal with regards to µ s ( α ) . For p = 0 . , the minimal spreading allocation is also optimal for P s ( α ) while it performs the worst when p ≤ . . A. Fixed-Size Access Model
For the fixed-size access model under a scaled exponential service time, the DSS service rate (9)becomes µ s ( α ) = µα (cid:0) Nr (cid:1) min( r,mα ) (cid:88) ϕ = α H ϕ − H ϕ − α (cid:18) mαϕ (cid:19)(cid:18) N − mαr − ϕ (cid:19) . (13)By comparing (10) and (13), we expect that the minimal spreading allocation may not be always optimal.Indeed, we prove below that the maximal spreading allocation performs better than the minimal spreadingallocation under some system parameters’ values. Lemma 5.
Under fixed-size access model with a scaled exponential service time, the service rate of themaximal spreading allocation, i.e. α = r , is µ s ( r ) = µr ( rmr ) H r ( Nr ) . The service rate of the minimal spreadingallocation, i.e. α = 1 , is µ s (1) = µm ( N − r − )( Nr ) = µmrN .Proof: From (13), for the maximal spreading allocation, we get µ s ( r ) = µr ( Nr ) (cid:80) min( r,rm ) ϕ = r H ϕ − H ϕ − r (cid:0) rmϕ (cid:1) · (cid:0) N − rmr − ϕ (cid:1) . Since m ≥ and H = 0 , µ s ( r ) = µr ( Nr ) H r (cid:0) rmr (cid:1)(cid:0) N − rmr − r (cid:1) = µr ( rmr ) H r ( Nr ) .For the minimal spreading allocation, we get µ s (1 | ϕ ) = µϕ . Thus, by applying the same approach asin the proof of Lemma 1, we get µ s (1) = µmrN = µm ( N − r − )( Nr ) .Having Lemma 5, we arrive at the following result on the comparison between the maximal spreadingallocation and the minimal spreading allocation. Theorem 3.
Under the fixed-size access model with a scaled exponential service time, when rm ≥ N , EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 14 the maximal spreading allocation outperforms the minimal spreading allocation, i.e. µ s ( r ) ≥ µ s (1) .Proof: According to Lemma 5, we get µ s ( r ) = µr ( rmr ) H r ( Nr ) = µrm ( rm − r − ) H r ( Nr ) . Since rm ≥ N and r ≥ H r , µ s ( r ) ≥ µmH r ( N − r − ) H r ( Nr ) = µmrN = µ s (1) .Now, using Theorem 3, we have the following Corollary on the minimal spreading allocation. Corollary 1.
Under the fixed-size access model with a scaled exponential service time, the minimalspreading allocation does not always maximize the DSS service rate.1) Optimal and Non-optimality Conditions for the Minimal Spreading Allocation:
From Corollary 1,we know that the minimal spreading allocation is not always optimal. Considering the complexity of (13),finding the α that maximizes µ s ( α ) is hard. Instead, we find the optimality and non-optimality conditionsfor the minimal spreading allocation in the following. These conditions are insightful as they can help usto determine if coding is necessary in the system. Firstly, we find the optimality condition in Theorem 4. Theorem 4.
Under the fixed-size access model with a scaled exponential service time, the minimalspreading allocation maximizes the DSS service rate µ s ( α ) when r ≤ min ≤ α ≤ r,α ∈ Z { N − α − (cid:113) α ( mα − α − ) } .Proof: To prove the theorem statement, we need to find the condition ensuring µ s (1) ≥ µ s ( α ) forall ≤ α ≤ r . The expression of µ s ( α ) is given in (13). According to (6), we have µ s ( α | ϕ ) < µϕ when α ≥ , thus µ s ( α ) < µ (cid:0) Nr (cid:1) min( r,mα ) (cid:88) ϕ = α ϕ (cid:18) mαϕ (cid:19)(cid:18) N − mαr − ϕ (cid:19) = µmα (cid:0) Nr (cid:1) min( r,mα ) (cid:88) ϕ = α (cid:18) mα − ϕ − (cid:19)(cid:18) N − mαr − ϕ (cid:19) = µmα (cid:0) Nr (cid:1) min( r,mα ) (cid:88) ϕ = α α − (cid:89) i =0 mα − − iϕ − − i (cid:18) mα − αϕ − α (cid:19)(cid:18) N − mαr − ϕ (cid:19) . Since ϕ goes from α to mα , we further have µ s ( α ) < µmα (cid:0) Nr (cid:1) min( r,mα ) (cid:88) ϕ = α ( α − (cid:89) i =0 mα − − iα − − i ) (cid:18) mα − αϕ − α (cid:19)(cid:18) N − mαr − ϕ (cid:19) = µmα (cid:0) mα − α − (cid:1)(cid:0) Nr (cid:1) min( r,mα ) (cid:88) ϕ = α (cid:18) mα − αϕ − α (cid:19)(cid:18) N − mαr − ϕ (cid:19) = µmα (cid:0) mα − α − (cid:1)(cid:0) N − αr − α (cid:1)(cid:0) Nr (cid:1) (by Vandermonde’s convolution) . According to Lemma 3, we have µ s (1) = µm ( N − r − )( Nr ) . To satisfy µ s ( α ) ≤ µ s (1) , we have µmα (cid:0) mα − α − (cid:1)(cid:0) N − αr − α (cid:1)(cid:0) Nr (cid:1) ≤ µm (cid:0) N − r − (cid:1)(cid:0) Nr (cid:1) ⇔ α (cid:18) mα − α − (cid:19) ≤ α − (cid:89) i =0 N − − ir − − i . (14) As N − − ir − − i < N − − ir − − i for N > r , it can be shown that (cid:81) α − i =0 N − − ir − − i > ( N − r − ) α − , and as a result, inequality EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 15 (14) holds when α (cid:18) mα − α − (cid:19) ≤ ( N − r − α − ⇔ r ≤ N − α − (cid:113) α (cid:0) mα − α − (cid:1) . (15) If the inequality (15) holds for all ≤ α ≤ r , µ s (1) is optimal.Now that we have the optimality condition for the minimal spreading allocation, we present the non-optimality condition in Theorem 5. Theorem 5.
Under the fixed-size access model with a scaled exponential service time, the minimal spread-ing allocation does not maximize the DSS service rate µ s ( α ) when r ≥ min ≤ α ≤ r,α ∈ Z { α − (cid:112) mmα − α +1 ( N − α + 1) + α − } .Proof: To prove the theorem statement, we need to find the condition ensuring µ s (1) ≤ µ s ( α ) forat least one α ∈ [2 , r ] . The expression of µ s ( α ) is given in (13). According to (6), we have µ s ( α | ϕ ) >µ ( ϕ − α + 1) when α ≥ . Thus, µ s ( α ) > µ (cid:0) Nr (cid:1) min( r,mα ) (cid:88) ϕ = α ( ϕ − α + 1) (cid:18) mαϕ (cid:19)(cid:18) N − mαr − ϕ (cid:19) = µ (cid:0) Nr (cid:1) min( r,mα ) (cid:88) ϕ = α ( ϕ − α + 1) α − (cid:89) i =0 mα − iϕ − i (cid:18) mα − α + 1 ϕ − α + 1 (cid:19)(cid:18) N − mαr − ϕ (cid:19) . Since ϕ goes from α to mα , we further have µ s ( α ) > µ (cid:0) Nr (cid:1) min( r,mα ) (cid:88) ϕ = α ( ϕ − α + 1) (cid:18) mα − α + 1 ϕ − α + 1 (cid:19)(cid:18) N − mαr − ϕ (cid:19) = µ ( mα − α + 1) (cid:0) N − αr − α (cid:1)(cid:0) Nr (cid:1) (Vandermonde’s convolution) . According to Lemma 3, we have µ s (1) = µm ( N − r − )( Nr ) , hence, to satisfy µ s ( α ) ≥ µ s (1) , we need to have µ ( mα − α + 1) (cid:0) N − αr − α (cid:1)(cid:0) Nr (cid:1) ≥ µm (cid:0) N − r − (cid:1)(cid:0) Nr (cid:1) ⇔ mα − α + 1 m ≥ α − (cid:89) i =0 N − − ir − − i . (16) Since N − − ir − − i < N − − ir − − i for N > r , we have (cid:81) α − i =0 N − − ir − − i < ( N − α +1 r − α +1 ) α − . Hence, if mα − α + 1 m ≥ ( N − α + 1 r − α + 1 ) α − ⇔ r ≥ α − (cid:114) mmα − α + 1 ( N − α + 1) + α − , (17) inequality (16) holds, meaning that µ s ( α ) ≥ µ s (1) and µ s (1) is not optimal.From Theorems 4 and 5, we see that both conditions depend on the number of accessed nodes r . Usingthese theorems, we provide the following conjecture on the optimal value of α which maximizes µ s ( α ) . Conjecture 1.
The optimal α for maximizing the DSS service rate µ s ( α ) increases with r . That is, when r is small, the minimal spreading allocation is optimal, and when r is large, the maximal spreadingallocation becomes optimal. EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 16
2) Numerical Analysis:
In Fig. 6, we evaluate the expression of µ s ( α ) given in (13) to see how theDSS service rate changes with the allocation parameter α . We consider a system with N = 40 storagenodes. Using Theorems 4 and 5, we can easily calculate the optimality and non-optimality conditions.For example, when m = 2 , the minimal spreading allocation is optimal when r ≤ . and is non-optimalwhen r > . These conditions provide some knowledge of the optimal allocation and further insight onthe optimal allocation can be found from Fig. 6. The graph on the left shows µ s ( α ) vs. α for four differentlevel of redundancy m ∈ { , , , } , and the number of accessed nodes r = 10 . The graph on the rightshows µ s ( α ) vs. α for different numbers of accessed nodes r ∈ { , , , } , and the redundancy level m = 3 . In the left subfigure, when m ≤ , µ s ( α ) reaches its maximum at α = 1 and decreases withincreasing α , i.e. the minimal spreading allocation is optimal. When m = 3 , µ s ( α ) reaches its maximumat α = 3 . When m = 4 , µ s ( α ) increases with α and reaches its maximum at α = 10 , i.e. the maximalspreading allocation ( α = N/m ) is optimal. In the right subfigure, when r = 8 , the minimal spreadingallocation is optimal, while this is not the case for r ≥ . Since the redundancy level is m = 3 , whichmeans N/m is not an integer, we cannot apply the maximal spreading allocation. When r = 13 , theoptimal allocation is at α = 13 , where we allocate the file into of nodes. From the observationsin Fig. 6, we conclude that the minimal spreading allocation is optimal only when m or r is sufficientlysmall. By increasing either of the parameters, an α ≥ quasi-uniform allocation becomes optimal, andwhen m or r is sufficiently large, the maximal spreading allocation is optimal. Fig. 6. The DSS service rate µ s ( α ) for the fixed-size access model with scaled exponential service time as a function of the allocationparameter α . The number of storage nodes is N = 40 , and the service time follows Exp(1 /α ) . (left) µ s ( α ) vs. α with r = 10 accessednodes for four values of m . (right) µ s ( α ) vs. α with m = 3 redundancy for four values of r . Given r (or m ), the optimal allocation changesfrom the minimal spreading allocation to the maximal spreading allocation as m (or r ) increases. In Fig. 7, we analyze P s ( α ) vs. µ s ( α ) as α increases from to r . We consider a system with N = 40 EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 17 storage nodes and two values for each parameter, i.e. m ∈ { , } and r ∈ { , } . Some observationscan be made from the figure: when both m and r are sufficiently small, the minimal spreading allocationis optimal for both P s ( α ) and µ s ( α ) . When both m and r are sufficiently large, the maximal spreadingallocation is optimal for both performance metrics. Otherwise, we cannot find an optimal allocation tosimultaneously optimize both performance metrics. For example, when m = 4 and r = 8 , P s ( α ) reachesits maximum at α = 1 , while µ s ( α ) reaches its maximum at α = 4 . A performance tradeoff can beachieved by choosing α = 2 to obtain acceptable P s ( α ) and µ s ( α ) . Fig. 7. Successful recovery probability P s ( α ) vs. the DSS service rate µ s ( α ) for the fixed-size access model as a function of α for differentvalues of m and r . The number of storage nodes is N = 40 , and theservice time follows Exp(1 /α ) . The minimal (or maximal) spreadingallocation is optimal when m and r are sufficiently small (or large).Otherwise, the optimal allocation is different for different performancemetrics. B. Probabilistic Access Model
For probabilistic access model under scaled exponential service time, the DSS service rate (9) becomes µ s ( α ) = mα (cid:88) ϕ = α µαH ϕ − H ϕ − α (cid:18) mαϕ (cid:19) (1 − p ) ϕ p mα − ϕ . (18)By comparing (12) and (18), we expect that the minimal spreading allocation may not be always optimal.To this end, we first present the following result on the optimal α to maximize the service rate when noredundancy is used to store the data. Theorem 6.
Under the probabilistic access model with a scaled exponential service time and consideringa no redundancy scenario, i.e. m = 1 , the optimal α which maximizes the DSS service rate µ s ( α ) islocated in the range [(1 / − p ) /p, (1 − p ) /p ] .Proof: Since the value of α is an integer, µ s ( α ) as a function of α is discrete. To prove the optimal α islocated in the range [(1 / − p ) /p, (1 − p ) /p ] , we need to show µ s ( α ) increases with α when α ≤ (1 / − p ) /p (i.e. µ s ( α ) ≤ µ s ( α + 1) ) and decreases with increasing α when α ≥ (1 − p ) /p (i.e. µ s ( α ) ≥ µ s ( α + 1) ). EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 18
For α ≥ , the DSS service rate is µ s ( α ) = µαH α (1 − p ) α , resulting in µ s ( α ) /µ s ( α + 1) = ( αH α +1 ) / (( α +1) H α (1 − p )) . Thus, µ s ( α ) µ s ( α + 1) ≥ ⇔ αH α +1 ( α + 1) H α (1 − p ) ≥ ⇔ αα + 1 ≥ (1 − αp − p ) H α . (19) Since αα +1 > , (19) is satisfied when (1 − αp − p ) H α ≤ , resulting in α ≥ (1 − p ) /p . Similarly, µ s ( α ) µ s ( α + 1) ≤ ⇔ αα + 1 ≤ (1 − αp − p ) H α . (20) On the other hand, αα +1 ≤ / H α because H α − αα + 1 = 12 ( H α +1 + 1 α + 1 ) − α +1 (cid:88) i =2 i + 1 α + 1 ) − ≥
12 ( αα + 1 + 1 α + 1 ) −
12 = 0 . Thus, if (1 − αp − p ) ≥ / , or equivalently α ≤ (1 / − p ) /p , (20) holds.From Theorem 6, it is easy to see that p is small, the optimal α is greater than . Then we have thefollowing corollary on the minimal spreading allocation. Corollary 2.
Under the probabilistic access model with a scaled exponential service time, the minimalspreading allocation ( α = 1 ) does not always maximize the service rate.1) Optimality and Non-optimality Conditions for Minimal Spreading Allocation: From Corollary 2,we know that the minimal spreading allocation is not always optimal. Considering the complexity of(18), finding the α that maximizes µ s ( α ) is hard. Similar to the fixed-size access model, we find theoptimality and non-optimality conditions for the minimal spreading allocation. These conditions are usefulto determine if coding is beneficial in the system or not. Theorem 7.
Under the probabilistic access model with a scaled exponential service time, the minimalspreading allocation maximizes the DSS service rate µ s ( α ) when p ≥ max α ≥ ,α ∈ Z { − α − (cid:113) α ( mα − α − ) } .Proof: To prove the theorem statement, we need to find the condition ensuring µ s (1) ≥ µ s ( α ) for all α ≥ . Using (18) and considering that µ s ( α | ϕ ) < µϕ for α ≥ according to (6), we have µ s ( α ) < µ mα (cid:88) ϕ = α ϕ (cid:18) mαϕ (cid:19) (1 − p ) ϕ p mα − ϕ = µmα mα (cid:88) ϕ = α (cid:18) mα − ϕ − (cid:19) (1 − p ) ϕ p mα − ϕ = µmα mα (cid:88) ϕ = α ( α − (cid:89) i =0 mα − − iϕ − − i ) (cid:18) mα − αϕ − α (cid:19) (1 − p ) ϕ p mα − ϕ . Since ϕ goes from α to mα , we have µ s ( α ) < µmα (cid:18) mα − α − (cid:19) (1 − p ) α mα − α (cid:88) ϕ =0 (cid:18) mα − αϕ (cid:19) (1 − p ) ϕ p mα − α − ϕ = µmα (cid:18) mα − α − (cid:19) (1 − p ) α (by binomial expansion) . EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 19
From (18), µ s (1) = m (cid:88) ϕ =1 µ s (1 | ϕ ) (cid:18) mϕ (cid:19) (1 − p ) ϕ p m − ϕ = µ m (cid:88) ϕ =1 ϕ (cid:18) mϕ (cid:19) (1 − p ) ϕ p m − ϕ = µm (1 − p ) . Now, to satisfy µ s ( α ) ≤ µ s (1) , we have µmα (cid:18) mα − α − (cid:19) (1 − p ) α ≤ µm (1 − p ) ⇔ (1 − p ) α − ≤ α (cid:0) mα − α − (cid:1) ⇔ p ≥ − α − (cid:113) α (cid:0) mα − α − (cid:1) . (21) If the inequality (21) holds for all α ≥ , µ s (1) is optimal.Now, we find the non-optimality condition for the minimal spreading allocation in Theorem 8. Theorem 8.
Under the probabilistic access model with a scaled exponential service time, the mini-mal spreading allocation does not maximize the DSS service rate µ s ( α ) when p ≤ max α ≥ ,α ∈ Z { − α − (cid:112) mmα − α +1 } .Proof: We are interested in finding a condition that guarantees the existence of an α ≥ , such that µ s (1) ≤ µ s ( α ) . The expression for µ s ( α ) is given in (18). According to (6), we have µ s ( α | ϕ ) > µ ( ϕ − α +1) when α ≥ , then µ s ( α ) > µ mα (cid:88) ϕ = α ( ϕ − α + 1) (cid:18) mαϕ (cid:19) (1 − p ) ϕ p mα − ϕ = µ mα (cid:88) ϕ = α ( ϕ − α + 1)( α − (cid:89) i =0 mα − iϕ − i ) (cid:18) mα − α + 1 ϕ − α + 1 (cid:19) (1 − p ) ϕ p mα − ϕ . ϕ goes from α to mα , thus µ s ( α ) > µ mα (cid:88) ϕ = α ( ϕ − α + 1) (cid:18) mα − α + 1 ϕ − α + 1 (cid:19) (1 − p ) ϕ p mα − ϕ = µ ( mα − α + 1)(1 − p ) α mα − α (cid:88) ϕ =0 (cid:18) mα − αϕ (cid:19) (1 − p ) ϕ p mα − α − ϕ = µ ( mα − α + 1)(1 − p ) α . Since µ s (1) = µm (1 − p ) , to satisfy µ s ( α ) ≥ µ s (1) , it is sufficient to have µ ( mα − α + 1)(1 − p ) α ≥ µm (1 − p ) ⇔ (1 − p ) α − ≥ mmα − α + 1 ⇔ p ≤ − α − (cid:114) mmα − α + 1 . (22) Therefore, if the inequality (22) holds for an α ≥ , µ s (1) is not optimal.Using Theorems 7 and 8, the following result provides guidelines on the optimal α which maximizes µ s ( α ) . Conjecture 2.
The optimal α which maximizes the DSS service rate µ s ( α ) decreases with increasing p .When p is close to , the minimal spreading allocation is optimal. When p is close to , the maximalspreading allocation becomes optimal.2) Numerical Analysis: In Fig. 8, we evaluate the expression of µ s ( α ) given in (18) to investigate howthe DSS service rate changes with the allocation parameter α . We consider a system with N ≥ mα storage EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 20 nodes. Using Theorems 7 and 8, we can easily calculate the optimality and non-optimality conditions. Forexample, when m = 2 , the minimal spreading allocation is optimal when p ≥ . and is non-optimalwhen p ≤ . . These conditions provide some knowledge of the optimal allocation and further insighton the optimal allocation can be found from Fig. 8. The graph on the left shows µ s ( α ) vs. α for fourdifferent levels of redundancy m ∈ { , , , } , and the failed access probability p = 0 . . The graph on theright shows µ s ( α ) vs. α for four different values of p ∈ { . , . , . , . } , and the redundancy level m = 2 . In the left subfigure, when m = 1 , µ s ( α ) decreases with increasing α and reaches its maximumat α = 1 , i.e. the minimal spreading allocation is optimal. When m ≥ , µ s ( α ) increases with α andreaches its maximum at α = 10 , i.e. the maximal spreading allocation is optimal. In the right subfigure,when p ≤ . , the maximal spreading allocation is optimal. When p = 0 . , α = 2 allocation is optimal.When p = 0 . , the minimal spreading allocation is optimal. Therefore, the optimal allocation changeswith p . Fig. 8. The DSS service rate µ s ( α ) for the probabilistic access model with scaled exponential service time as a function of the allocationparameter α . The number of storage nodes is N ≥ mα , and the service time follows Exp(1 /α ) . (left) µ s ( α ) vs. α with the failed accessprobability p = 0 . for four values of m . (right) µ s ( α ) vs. α with redundancy of m = 2 for four values of p . Given p (or m ), the optimalallocation changes from the minimal spreading allocation to the maximal spreading allocation as m increases or p decreases. In Fig. 9, we analyze P s ( α ) vs. µ s ( α ) as α increases from to . We consider a system with N ≥ mα storage nodes and two values for each parameter m ∈ { , } and p ∈ { . , . } . When m is sufficientlysmall and p is sufficiently large, the minimal spreading allocation is optimal for both P s ( α ) and µ s ( α ) ,while for sufficiently large m and sufficiently small p , the maximal spreading allocation is optimal forboth performance metrics. For a general scenario, the optimal α for maximizing the service rate may bedifferent from that for maximizing the probability of recovery. For example, when m = 3 and p = 0 . , P s ( α ) reaches its maximum at α = 1 , and µ s ( α ) reaches its maximum at α = 10 . EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 21
Fig. 9. The successful recovery probability P s ( α ) vs. the DSS servicerate µ s ( α ) for the probabilistic access model as a function of α fordifferent values of m and r . The number of storage nodes is N ≥ mα , and the service time follows Exp(1 /α ) . The optimal allocationis affected by both m and r values. VI. S
TORAGE A LLOCATION FOR LARGE FILES WITH S HIFTED EXPONENTIAL SERVICE T IME
For the large file model, we consider the service time at each node follows a shifted exponentialdistribution with the shift ∆ /α and the rate µ , i.e. S-Exp(∆ /α, µ ) . In the following two subsections,we determine the µ s ( α ) for the two considered access models. Some of the results in this section werepublished in [2]. A. Fixed-Size Access Model
For fixed-size access model under shifted exponential service time, the DSS service rate (9) becomes µ s ( α ) = µα (cid:0) Nr (cid:1) min( r,mα ) (cid:88) ϕ = α µ + α ( H ϕ − H ϕ − α ) (cid:18) mαϕ (cid:19)(cid:18) N − mαr − ϕ (cid:19) . (23)As one may expect from (23), the minimal spreading allocation may not be always optimal. In fact, wefind a special scenario where there exists an optimal α ≥ that maximizes the DSS service rate.Let us start by assuming a constant service time of ∆ for k block. In this case, µ s ( α | ϕ ( A )) = α/ ∆ for a set A . Therefore, for the maximal spreading allocation, we have µ s ( r ) = r ( rmr ) ∆ ( Nr ) . For the minimalspreading allocation, i.e. α = 1 , using Vandermonde’s convolution, we arrive at µ s (1) = (1 − ( N − mr )( Nr ) ) . Theorem 9.
Under the fixed-size access model and a constant service time ∆ , the maximal spreadingallocation outperforms the minimal spreading allocation when rm ≥ N .Proof: Since rm ≥ N and (cid:0) N − mr (cid:1) ≥ , µ s ( r ) = r ( rmr ) ∆ ( Nr ) ≥ ( rmr ) − ( N − mr ) ∆ ( Nr ) ≥ ( Nr ) − ( N − mr ) ∆ ( Nr ) = µ s (1) .Using Theorem 9, we have the following corollary on the minimal spreading allocation. Corollary 3.
Under the fixed-size access model with a constant service time ∆ , the minimal spreadingallocation ( α = 1 ) does not always maximize the DSS service rate. EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 22
The constant service time is a special case of the shifted exponential service time. In the following, wewill find the optimality and non-optimality conditions for the minimal spreading allocation under shiftedexponential service time. Our findings show that Corollary 3 is also true for the shifted exponential servicetime.
1) Optimality and Non-optimality Conditions for the Minimal Spreading Allocation:
Considering thecomplexity of (23), finding the optimal α that maximizes µ s ( α ) in a general scenario is difficult. That beingsaid, we start by finding the optimality condition for the minimal spreading allocation in Theorem 10. Theorem 10.
Under the fixed-size access model a shifted exponential service time, the minimal spreadingallocation maximizes the DSS service rate µ s ( α ) when r ≤ min ≤ α ≤ r,α ∈ Z { α − (cid:113) ∆ µ + αα (∆ µm +1) ( mα − α − ) ( N − } .Proof: The proof approach is similar to the one for Theorem 4. Using the expression for µ s ( α ) in(23), as well as (8), we have µ s ( α ) < µ (∆ µ + α ) (cid:0) Nr (cid:1) min( r,mα ) (cid:88) ϕ = α ϕ (cid:18) mαϕ (cid:19)(cid:18) N − mαr − ϕ (cid:19) = µmα (∆ µ + α ) (cid:0) Nr (cid:1) min( r,mα ) (cid:88) ϕ = α ( α − (cid:89) i =0 mα − − iϕ − − i ) (cid:18) mα − αϕ − α (cid:19)(cid:18) N − mαr − ϕ (cid:19) < µmα (∆ µ + α ) (cid:0) Nr (cid:1) min( r,mα ) (cid:88) ϕ = α ( α − (cid:89) i =0 mα − − iα − − i ) (cid:18) mα − αϕ − α (cid:19)(cid:18) N − mαr − ϕ (cid:19) = µmα (cid:0) mα − α − (cid:1)(cid:0) N − αr − α (cid:1) (∆ µ + α ) (cid:0) Nr (cid:1) Similarly, according to (8), we have µ s (1 | ϕ ) ≥ µϕ ∆ µm +1 , and consequently µ s (1) ≥ µ (∆ µm +1) ( Nr ) (cid:80) mϕ =1 ϕ (cid:0) mϕ (cid:1) · (cid:0) N − mr − ϕ (cid:1) = µm ( N − r − ) (∆ µm +1) ( Nr ) . Now, to satisfy µ s ( α ) ≤ µ s (1) , we have µmα (cid:0) mα − α − (cid:1)(cid:0) N − αr − α (cid:1) (∆ µ + α ) (cid:0) Nr (cid:1) ≤ µm (cid:0) N − r − (cid:1) (∆ µm + 1) (cid:0) Nr (cid:1) ⇔ α (∆ µm + 1) (cid:0) mα − α − (cid:1) ∆ µ + α ≤ α − (cid:89) i =0 N − − ir − − i . (24) As N − − ir − − i < N − − ir − − i for N > r , it can be shown that (cid:81) α − i =0 N − − ir − − i > ( N − r − ) α − , and as a result, inequality(24) holds when α (∆ µm + 1) (cid:0) mα − α − (cid:1) ∆ µ + α ≤ ( N − r − α − ⇔ r ≤ α − (cid:115) ∆ µ + αα (∆ µm + 1) (cid:0) mα − α − (cid:1) ( N − . (25) If the inequality (25) holds for all ≤ α ≤ r , µ s (1) is optimal.Now that we have the optimality condition for the minimal spreading allocation, we present the conditionwhere it is not optimal in Theorem 11. Theorem 11.
Under the fixed-size access model and a shifted exponential service time, the minimal spread-
EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 23 ing allocation does not maximize the DSS service rate µ s ( α ) when r ≥ min ≤ α ≤ r,α ∈ Z { α − (cid:113) ∆ µm ( mα − α +1)+ mα α (∆ µ +1)( mα − α +1) · ( N − α + 1) + α − } .Proof: The proof approach is similar to the one for Theorem 5. Using (8), we have µ s ( α | ϕ ) > µα ( ϕ − α +1)∆ µ ( mα − α +1)+ α when α ≥ . Define A = α ∆ µ ( mα − α +1)+ α , then µ s ( α ) > µA (cid:0) Nr (cid:1) min( r,mα ) (cid:88) ϕ = α ( ϕ − α + 1) (cid:18) mαϕ (cid:19)(cid:18) N − mαr − ϕ (cid:19) > µA (cid:0) Nr (cid:1) min( r,mα ) (cid:88) ϕ = α ( ϕ − α + 1) (cid:18) mα − α + 1 ϕ − α + 1 (cid:19)(cid:18) N − mαr − ϕ (cid:19) = µA ( mα − α + 1) (cid:0) N − αr − α (cid:1)(cid:0) Nr (cid:1) . Similarly, according to (8), we have µ s (1 | ϕ ) ≤ µϕ ∆ µ +1 , and consequently µ s (1) ≤ µ (∆ µ +1) ( Nr ) (cid:80) mϕ =1 ϕ (cid:0) mϕ (cid:1)(cid:0) N − mr − ϕ (cid:1) = µm ( N − r − ) (∆ µ +1) ( Nr ) . Now, to satisfy µ s ( α ) ≥ µ s (1) , we need µA ( mα − α + 1) (cid:0) N − αr − α (cid:1)(cid:0) Nr (cid:1) ≥ µm (cid:0) N − r − (cid:1) (∆ µ + 1) (cid:0) Nr (cid:1) ⇔ A (∆ µ + 1)( mα − α + 1) m ≥ α − (cid:89) i =0 N − − ir − − i . (26) As N − − ir − − i < N − − ir − − i for N > r , we have (cid:81) α − i =0 N − − ir − − i < ( N − α +1 r − α +1 ) α − . Inequality (26) is true when A (∆ µ + 1)( mα − α + 1) m ≥ ( N − α + 1 r − α + 1 ) α − ⇔ r ≥ α − (cid:115) ∆ µm ( mα − α + 1) + mα α (∆ µ + 1)( mα − α + 1) ( N − α + 1) + α − . (27) Therefore, if the inequality (27) holds for any one of α ∈ [2 , r ] , µ s (1) is not optimal.From Theorems 10 and 11, we see that the conditions depend on the number of accessed nodes r .For the shifted exponential service time, we believe that the statement in Conjecture 1 holds, providingguidelines on finding the optimal α for maximizing µ s ( α ) .
2) Numerical Analysis:
In Fig. 10, we evaluate the expression (23) for µ s ( α ) to see how the DSSservice rate changes with α . We consider a system with N = 40 storage nodes. Using Theorems 10 and11, we can easily calculate the optimality and non-optimality conditions. For example, when m = 2 , theminimal spreading allocation is optimal when r < and is non-optimal when r > . These conditionsonly provide limited knowledge of the optimal allocation and further insight on the optimal allocationcan be found from Fig. 10. The graph on the left shows µ s ( α ) vs. α for four different redundancy levels m ∈ { , , , } where the number of accessed nodes is r = 10 . The graph on the right shows µ s ( α ) vs. α for r ∈ { , , , } , and the redundancy level m = 2 . In the left subfigure, when m ≤ , the minimalspreading allocation is optimal. When m = 3 , µ s ( α ) reaches its maximum at α = 3 . When m = 4 ,the maximal spreading allocation is optimal. In the right subfigure, when r ≤ , the minimal spreadingallocation is optimal, while for r = 17 , the allocation with α = 2 is optimal. When r = 20 , although the EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 24 allocation with α = 4 is optimal, the maximal spreading allocation provides a local maximum value whichis close to the global maximum value. From the observations, we conclude that the minimal spreadingallocation is optimal only when m or r is sufficiently small. Otherwise, an allocation with α ≥ isoptimal. Fig. 10. The DSS service rate µ s ( α ) for the fixed-size access model with shifted exponential service time as a function of the allocationparameter α . The number of storage nodes is N = 40 , and the service time follows S-Exp(3 , . (left) µ s ( α ) vs. α with r = 10 accessednodes for four values of m . (right) µ s ( α ) vs. α with m = 2 redundancy for four values of r . Given r (or m ), the optimal allocation changesfrom the minimal spreading allocation to the maximal spreading allocation as m (or r ) increases. In Fig. 11, we analyze P s ( α ) vs. µ s ( α ) as α increases from to r . We consider a system with N = 40 storage nodes, m ∈ { , } , and r ∈ { , } . Some observations can be made from the figure: when both m and r are sufficiently small, e.g. m = 3 and r = 8 , the minimal spreading allocation is optimal for both P s ( α ) and µ s ( α ) . When both m and r are sufficiently large, e.g. m = 4 and r = 10 , the maximal spreadingallocation is optimal. Otherwise, e.g. for m = 3 and r = 10 , there is no optimal α that maximizes both P s ( α ) and µ s ( α ) at the same time. Fig. 11. Successful recovery probability P s ( α ) vs. the DSS service rate µ s ( α ) for the fixed-size access model as a function of α for differentvalues of m and r . The number of storage nodes is N = 40 , and theservice time follows S-Exp(3 , . The minimal (or maximal) spreadingallocation is optimal when m and r are sufficiently small (or large).Otherwise, the optimal allocation is different for different performancemetrics. EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 25
B. Probabilistic Access Model
For the probabilistic access model and a shifted exponential service time, the DSS service rate (9)becomes µ s ( α ) = mα (cid:88) ϕ = α µα ∆ µ + α ( H ϕ − H ϕ − α ) (cid:18) mαϕ (cid:19) (1 − p ) ϕ p mα − ϕ . (28)From (28), one can expect that the minimal spreading allocation may not be always optimal. To this end,we present the following result on the optimal α to maximize the service rate when no redundancy isused to store the data.Let us start by assuming a constant service time of ∆ for k block. In this case, µ s ( α | ϕ ( A )) = α/ ∆ for a set A . Therefore, for no redundancy scenario, i.e. m = 1 , we have µ s ( α ) = α ∆ (1 − p ) α . Now, wedetermine the optimal α which maximizes the DSS service rate in Theorem 12. Theorem 12.
Under the probabilistic access model, a constant service time ∆ , and no redundancy casein data storage ( m = 1 ), the DSS service rate µ s ( α ) reaches its maximum when α = (cid:100) p/ (1 − p ) (cid:101) or α = (cid:98) p/ (1 − p ) (cid:99) .Proof: Given an integer α ≥ , the DSS service rate µ s ( α ) = α ∆ (1 − p ) α is discrete. Then we findthe optimal α by comparing the ratio µ s ( α ) /µ s ( α + 1) = α/ ( α + 1(1 − p )) with . Thus, µ s ( α ) µ s ( α +1) ≥ ⇔ αα +1(1 − p ) ≥ ⇔ α ≤ p − p Similarly, µ s ( α ) µ s ( α +1) ≤ ⇔ α ≥ p − p . Since α is an integer, µ s ( α ) reaches themaximum at (cid:100) p − p (cid:101) or (cid:98) p − p (cid:99) .Using Theorem 12, we have the following corollary on the minimal spreading allocation. Corollary 4.
Under the probabilistic access model with a constant service time ∆ , the minimal spreadingallocation ( α = 1 ) does not always maximize the DSS service rate. Now, we go one step further and find the optimality and non-optimality conditions for the minimalspreading allocation when service time follows a shifted exponential distribution.
1) Optimality and Non-optimality Conditions for the Minimal Spreading Allocation:
Considering thecomplexity of (28), finding the optimal α that maximizes µ s ( α ) in a general scenario is difficult.That beingsaid, we start by finding the optimality condition for the minimal spreading allocation in Theorem 13. Theorem 13.
Under the probabilistic access model with a shifted exponential service time, the minimalspreading allocation maximizes the DSS service rate µ s ( α ) when p ≥ max α ≥ ,α ∈ Z { − α − (cid:114) ∆ µ + αα (∆ µm +1) ( mα − α − ) } . EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 26
Proof:
The proof approach is similar to the one for Theorem 7. Using the expression for µ s ( α ) in(28), as well as (8), we have µ s ( α ) < µ ∆ µ + α mα (cid:88) ϕ = α ϕ (cid:18) mαϕ (cid:19) (1 − p ) ϕ p mα − ϕ < µmα (cid:0) mα − α − (cid:1) (1 − p ) α ∆ µ + α mα − α (cid:88) ϕ =0 (cid:18) mα − αϕ (cid:19) (1 − p ) ϕ p mα − α − ϕ = µmα (cid:0) mα − α − (cid:1) (1 − p ) α ∆ µ + α . Similarly, according to (8), we have µ s (1 | ϕ ) ≥ µϕ ∆ µm +1 , and consequently µ s (1) > µ ∆ µm +1 (cid:80) mϕ =1 ϕ (cid:0) mϕ (cid:1) (1 − p ) ϕ p m − ϕ = µm (1 − p )∆ µm +1 . Now, to satisfy µ s ( α ) ≤ µ s (1) , we need µmα (cid:0) mα − α − (cid:1) (1 − p ) α ∆ µ + α ≤ µm (1 − p )∆ µm + 1 ⇔ p ≥ − α − (cid:115) ∆ µ + αα (∆ µm + 1) (cid:0) mα − α − (cid:1) . (29) Therefore, if the inequality (29) holds for all α ≥ , µ s (1) is optimal.Now, we find the non-optimality condition for the minimal spreading allocation in Theorem 14. Theorem 14.
Under the probabilistic access model with a scaled exponential service time, the mini-mal spreading allocation does not maximize the DSS service rate µ s ( α ) when p ≤ max α ≥ ,α ∈ Z { − α − (cid:113) m (∆ µ ( mα − α +1)+ α ) α (∆ µ +1)( mα − α +1) } .Proof: The proof approach is similar to the one for Theorem 8. Using (8), we have µ s ( α | ϕ ) > µα ( ϕ − α +1)∆ µ ( mα − α +1)+ α when α ≥ . Define A = α ∆ µ ( mα − α +1)+ α , then µ s ( α ) > µA mα (cid:88) ϕ = α ( ϕ − α + 1) (cid:18) mαϕ (cid:19) (1 − p ) ϕ p mα − ϕ > µA mα (cid:88) ϕ = α ( ϕ − α + 1) (cid:18) mα − α + 1 ϕ − α + 1 (cid:19) (1 − p ) ϕ p mα − ϕ = µA ( mα − α + 1)(1 − p ) α . Similarly, according to (8), we have µ s (1 | ϕ ) ≤ µϕ ∆ µ +1 , and consequently µ s (1) < µ ∆ µ +1 (cid:80) mϕ =1 ϕ (cid:0) mϕ (cid:1) (1 − p ) ϕ p m − ϕ = µm (1 − p )∆ µ +1 . Now, to satisfy µ s ( α ) ≥ µ s (1) , we need µA ( mα − α + 1)(1 − p ) α ≥ µm (1 − p )∆ µ + 1 ⇔ p ≤ − α − (cid:115) m (∆ µ ( mα − α + 1) + α ) α (∆ µ + 1)( mα − α + 1) . (30) Therefore, if the inequality (30) holds for α ≥ , µ s (1) is not optimal.From Theorems 13 and 14, we see that the conditions depend on the probability of failed access p ,and Conjecture 2 provides an insight to the optimal α that maximizes µ s ( α ) .
2) Numerical Analysis:
In Fig. 12, we evaluate the expression for µ s ( α ) given in (28) to see howthe DSS service rate changes with the allocation parameter α . We consider a system with N ≥ mα storage nodes and the service time follows a shifted exponential distribution with ∆ = 3 and µ = 1 .Using Theorems 13 and 14, we can easily calculate the optimality and non-optimality conditions. Forexample, when m = 2 , the minimal spreading allocation is optimal when p > . and is non-optimal EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 27 when p < . . These conditions provide only limited knowledge of the optimal allocation and furtherinsight on the optimal allocation can be found from Fig. 12. The graph on the left shows µ s ( α ) vs. α forfour different redundancy levels m ∈ { , , , } , and the failed access probability p = 0 . . The graph onthe right shows µ s ( α ) vs. α for four different failed access probabilities p ∈ { . , . , . , . } , and theredundancy level m = 2 . In the left subfigure, when m = 1 , the minimal spreading allocation is optimal.When m ≥ , the maximal spreading allocation is optimal. In the right subfigure, when p = 0 . , themaximal spreading allocation is optimal, while for p = 0 . , the minimal spreading allocation is optimal.For the other two cases, an allocation with α ≥ is optimal. As can be seen, the optimal allocationchanges with the redundancy level m and the access failure probability p . Fig. 12. The DSS service rate µ s ( α ) for the probabilistic access model with a scaled exponential service time as a function of the allocationparameter α . The number of storage nodes is N ≥ mα , and the service time follows S-Exp(3 , . (left) µ s ( α ) vs. α for the access failureprobability p = 0 . and four values of m . (right) µ s ( α ) vs. α for m = 2 and four values of p . Given p (or m ), the optimal allocationchanges from the minimal spreading allocation to the maximal spreading allocation as m increases or p decreases.Fig. 13. Successful recovery probability P s ( α ) vs. the DSS servicerate µ s ( α ) for the probabilistic access model as a function of α fordifferent values of m and r . The number of storage nodes is N ≥ mα ,and the service time follows S-Exp(3 , . The optimal allocation isdetermined by both m and r . In Fig. 13, we analyze P s ( α ) vs. µ s ( α ) as α increases from to . We consider a system with N ≥ mα EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 28 storage nodes and two values for each parameter m ∈ { , } and p ∈ { . , . } . As seen in this figure,when m is sufficiently small and p is sufficiently large, the minimal spreading allocation is optimal forboth P s ( α ) and µ s ( α ) . On the other hand, for sufficiently large m and sufficiently small p , the maximalspreading allocation is optimal for both performance metrics. For other cases, there is no optimal α thatmaximizes both P s ( α ) and µ s ( α ) at the same time. For example, when m = 3 and p = 0 . , P s ( α ) reachesits maximum at α = 1 , and µ s ( α ) reaches its maximum at α = 10 .VII. C ONCLUSIONS
We considered service rates in distributed storage systems that use redundancy to improve robustnessto uncertainty in network access and download time. We focused on two access models (fixed-size andprobabilistic access) and two download service models (small and large file). Under the fixed-size accessmodel, a user can access a random fixed-size subset of nodes; under probabilistic access, a user can accesseach node with a fixed probability. In the small file download model, the randomness associated with thefile is negligible; in the large file download model, the randomness is associated with both the file sizeand inherent system’s operations. The primary performance metric of interest is the service rate of thesystem. Since redundancy for each file is fixed, the allocation of redundancy is essential for improving thesystem’s performance. The general allocation problem is hard to solve. We adopted the common modelof quasi-uniform allocation, where coded content is uniformly spread among a subset of storage nodes.Thus the subset size completely specifies the allocation.Minimal spreading concentrates coded chunks to a minimum-size subset. Maximal spreading allocatescoded chunks to each node. For the small file download model, the minimal spreading allocation is alwaysoptimal. For the large file download model, that is not the case. It is not easy to find the optimal allocation.We found the conditions under which the minimal spreading allocation is optimal. We considered scaledexponential and shifted exponential service times. Our numerical results showed that the optimal allocationunder these two service time models depends on the redundancy level, the number of accessed nodes, andthe probability of failed access. As a general rule, one should spread the data blocks to more nodes whenthe redundancy level is high, the number of accessed nodes is large, or the probability of failed access issmall. Further problems of interest include 1) finding optimal allocations for different performance metrics(e.g., service time) and 2) analyzing service rate under other service time distributions (e.g., heavy tailPareto and Weibull).
EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 29 R EFERENCES [1] M. Noori, E. Soljanin, and M. Ardakani, “On storage allocation for maximum service rate in distributed storage systems,” in , 2016, pp. 240–244.[2] P. Peng and E. Soljanin, “On distributed storage allocations of large files for maximum service rate,” in , 2018, pp. 784–791.[3] G. Joshi, Y. Liu, and E. Soljanin, “Coding for fast content download,” in . IEEE, 2012, pp. 326–333.[4] S. Kadhe, E. Soljanin, and A. Sprintson, “Analyzing the download time of availability codes,” in . IEEE, 2015, pp. 1467–1471.[5] M. Aktas and E. Soljanin, “Straggler mitigation at scale,”
IEEE/ACM Transactions on Networking , vol. 27, no. 06, pp. 2266–2279,Nov. 2019.[6] G. Yadgar, O. Kolosov, M. F. Aktas, and E. Soljanin, “Modeling the edge: Peer-to-peer reincarnated,” in , I. Ahmad and S. Sundararaman, Eds. USENIXAssociation, 2019.[7] D. Leong, A. G. Dimakis, and T. Ho, “Distributed storage allocations for optimal delay,” in . IEEE, 2011, pp. 1447–1451.[8] ——, “Distributed storage allocations,”
IEEE Transactions on Information Theory , vol. 58, no. 7, pp. 4733–4752, 2012.[9] M. Sardari, R. Restrepo, F. Fekri, and E. Soljanin, “Memory allocation in distributed storage networks,” in . IEEE, 2010, pp. 1958–1962.[10] B. Hong and W. Choi, “Asymptotic analysis of failed recovery probability in a distributed wireless storage system with limited sumstorage capacity,” in , 2014, pp. 6459–6463.[11] M. Noori and M. Ardakani, “Allocation for heterogeneous storage nodes,”
IEEE Communications Letters , vol. 19, no. 12, pp. 2102–2105,2015.[12] N. Alon, P. Frankl, H. Huang, V. Rödl, A. Ruci´nski, and B. Sudakov, “Large matchings in uniform hypergraphs and the conjecturesof erd˝os and samuels,”
Journal of Combinatorial Theory, Series A , vol. 119, no. 6, pp. 1200–1215, 2012.[13] M. Aktas, G. Joshi, S. Kadhe, F. Kazemi, and E. Soljanin, “Service rate region: A new aspect of coded distributed system design,” arXiv preprint arXiv:2009.01598 , 2020.[14] F. Kazemi, S. Kurz, and E. Soljanin, “Efficient storage schemes for desired service rate regions,” in , Apr. 2021.[15] ——, “A geometric view of the service rates of codes problem and its application to the service rate of the first order reed-mullercode,” in , June 2020.[16] F. Kazemi, E. Karimi, E. Soljanin, and A. Sprintson, “A combinatorial view of the service rates of codes problem, its equivalence tofractional matching and its connection with batch codes,” in , June2020.[17] S. E. Anderson, A. Johnston, G. Joshi, G. L. Matthews, C. Mayer, and E. Soljanin, “Service capacity region of content access fromerasure coded storage,” in
IEEE Information Theory Workshop (ITW) , Nov. 2018.[18] M. Akta¸s, S. E. Anderson, A. Johnston, G. Joshi, S. Kadhe, G. L. Matthews, C. Mayer, and E. Soljanin, “On the service capacityregion of accessing erasure coded content,” in , 2017, pp. 17–24.
EEE TRANSACTIONS ON COMMUNICATIONS, SUBMITTED JANUARY OR FEBRUARY 2021 30 [19] Y. Raaijmakers and S. C. Borst, “Achievable stability in redundancy systems,”
Proc. ACM Meas. Anal. Comput. Syst. , vol. 4, no. 3,pp. 46:1–46:21, 2020.[20] P. Peng, E. Soljanin, and P. Whiting, “Diversity vs. parallelism in distributed computing with redundancy,” in . IEEE, 2020, pp. 257–262.[21] ——, “Diversity/parallelism trade-off in distributed systems with redundancy,” arXiv preprint arXiv:2010.02147arXiv preprint arXiv:2010.02147