[tahoe-dev] A question about the system

Jiawei Jiang jiawei at email.arizona.edu
Mon Feb 7 23:54:03 UTC 2011


When a user wants to download the file, how to make sure that pieces from
any 3 (or K) servers can be assembled into the original file.
This is what I guess: if N=10, K=3, the file will be divided into 3
pieces(note as piece 1, 2, 3), each server will has 3 pieces (piece 1,2,3)
stored. But when reconstructing, the 3 pieces must come from different
server. Pieces from same server can not be assembled due to some encryption
methods. So as long as any K servers are working, the user can get file. And
we will need 10MB totally to store a 1MB file.
Is that correct? Or is there another mechanism. Or do I need to look into
the code if it can not be easily explained?

Thanks
Jiawei

On Mon, Feb 7, 2011 at 3:23 PM, Pandelis Theodosiou <ypercube at gmail.com>wrote:

>
>
> On Tue, Feb 8, 2011 at 12:17 AM, Jiawei Jiang <jiawei at email.arizona.edu>wrote:
>
>> Hi
>>
>> I am studying your system. First, it is a very nice system.
>> I have confusion on the storage availability:
>> "Users do rely on storage servers for *availability*. The cipher text is
>> erasure-coded and distributed across *N* storage servers (the default
>> value for *N* is 10) so that it can be recovered from any *K* of these
>> servers (the default value of *K* is 3). Therefore only the simultaneous
>> failure of *N-K+1* (with the defaults, 8) servers can make the data
>> unavailable."
>> I do not know how it works, how a file is encrypted and distributed among
>> these server. Could anyone briefly explain it to me?
>>
>
> File is encrypted, then cut in small pieces and the pieces get distributed
> to the N servers.
> If any K servers are available, the file can be recontructed from the
> pieces.
> If less than K servers are available, it can't.
>
> Not a 100% accuarte explanation but brief (and hopefully clear).
>
> Pandelis
>
>
>
>> Thanks very much.
>>
>> Jiawei Jiang
>>
>>
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