Access vs. Bandwidth in Codes for Storage

Maximum distance separable (MDS) codes are widely used in storage systems to protect against disk (node) failures. A node is said to have capacity $l$ over some field $\mathbb{F}$, if it can store that amount of symbols of the field. An $(n,k,l)$ MDS code uses $n$ nodes of capacity $l$ to store $k$ information nodes. The MDS property guarantees the resiliency to any $n-k$ node failures. An \emph{optimal bandwidth} (resp. \emph{optimal access}) MDS code communicates (resp. accesses) the minimum amount of data during the repair process of a single failed node. It was shown that this amount equals a fraction of $1/(n-k)$ of data stored in each node. In previous optimal bandwidth constructions, $l$ scaled polynomially with $k$ in codes with asymptotic rate $<1$. Moreover, in constructions with a constant number of parities, i.e. rate approaches 1, $l$ is scaled exponentially w.r.t. $k$. In this paper, we focus on the later case of constant number of parities $n-k=r$, and ask the following question: Given the capacity of a node $l$ what is the largest number of information disks $k$ in an optimal bandwidth (resp. access) $(k+r,k,l)$ MDS code. We give an upper bound for the general case, and two tight bounds in the special cases of two important families of codes. Moreover, the bounds show that in some cases optimal-bandwidth code has larger $k$ than optimal-access code, and therefore these two measures are not equivalent.Comment: This paper was presented in part at the IEEE International Symposium on Information Theory (ISIT 2012). submitted to IEEE transactions on information theor

MDS Array Codes with Optimal Rebuilding

MDS array codes are widely used in storage systems to protect data against erasures. We address the rebuilding ratio problem, namely, in the case of erasures, what is the the fraction of the remaining information that needs to be accessed in order to rebuild exactly the lost information? It is clear that when the number of erasures equals the maximum number of erasures that an MDS code can correct then the rebuilding ratio is 1 (access all the remaining information). However, the interesting (and more practical) case is when the number of erasures is smaller than the erasure correcting capability of the code. For example, consider an MDS code that can correct two erasures: What is the smallest amount of information that one needs to access in order to correct a single erasure? Previous work showed that the rebuilding ratio is bounded between 1/2 and 3/4 , however, the exact value was left as an open problem. In this paper, we solve this open problem and prove that for the case of a single erasure with a 2-erasure correcting code, the rebuilding ratio is 1/2 . In general, we construct a new family of r-erasure correcting MDS array codes that has optimal rebuilding ratio of 1/r in the case of a single erasure. Our array codes have efficient encoding and decoding algorithms (for the case r = 2 they use a finite field of size 3) and an optimal update property