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We rst consider the case for block striping. Let there be L = (Y + Z ) buffers (each Q bytes) at the client, organized as a circular buffer. Video playback starts once the rst Y buffers are completely lled with video data. The client pre lls the rst Y buffers to prevent buffer under ow, and reserves the last Z buffers for incoming data to prevent buffer over ow. Since all N S servers transmit data to a client concurrently, the client will be receiving N S video blocks simultaneously, of which (N S K ) blocks contain video data and the rest contain redundant data. This suggests that Y must be multiples of N S . Therefore, we consider groups of N S buffers (i.e., group zero consists of blocks 0 to N S 1, group one consists of blocks N S to 2N S 1, and so on.) and let y = Y /N S be the number of buffer groups pre lled.



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Remember, all these guys including me are used to being the best at what we do We made good grades in school We played first string We are meant to be right, to be making money It s human nature to not want to admit that you re wrong I know exactly how that works, I told him That pretty much sums up my disaster trade..





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The OWL-S Pro le emphasizes two aspects of the functionality of the service: The Information Transformation: Represented by inputs and outputs of the service, and The State Change produced by the Execution of the Service: Represented by the preconditions and effects of the service No schema to describe inputs/outputs/preconditions/effects (IOPE) instances is provided by the OWL-S Pro le However, such a schema exists in the Process ontology It is expected that the IOPE s published by the Pro le are a subset of those published by the Process, thus it is expected that the Process part of a description will create all the IOPE instances and the Pro le instance can simply point to these instances.

In this section, we derive the optimal rebuild rate for a system and show that the mixed distributed-baseline rebuild scheme can achieve this optimal rate. We assume that (1) unavailable data are rebuilt automatically without human intervention using erasure-correcting code computation; (2) all processing is performed by the remaining active servers and the spare

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server; and (3) the transfer capacity of the spare server is all used in rebuild. The rst assumption excludes manual and hardware replacement schemes. The second assumption excludes the use of back-up devices to reload data into the spare server. The third assumption guarantees the transfer capacity available at the spare server. Note that the rebuild of unavailable units is performed either at the spare server, the remaining (N S 1) servers, or partially done at both. Therefore, we have the next lemma. Lemma 14.1. For any unavailable data unit rebuilt and stored into the spare server, we need at least (N S 1) transmissions from the remaining active servers. Proof. If the unavailable data unit is rebuilt by the spare server, then the (N S 1) data/redundant units of the same parity group will need to be sent to the spare server, resulting in (N S 1) transmissions. If the unavailable data unit is rebuilt by one of the remaining servers, then (N S 2) data/redundant units from servers other than the rebuild server and the spare server will need to be sent to the rebuild server, resulting in (N S 2) transmissions. After that, the rebuild server will send the rebuilt data unit to the spare server for storage, incurring one more transmission. Together the whole process thus generates (N S 1) transmissions. Each server contributes its idle transfer capacity to the rebuild process either for transmission, reception, or both. Let {0 1} be the proportion of transfer capacity each server has used for transmission in the rebuild process. The sum of the transmission capacities of the remaining servers is then equal to SS (1 )(N S 1) (14.12)

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