On Mon, 05 Oct 2009 10:26:11 -0500, Othila (JBinUp@JBinUp.local) wrote:
:Hello,Isis
:
:in this case,I chose 3% redudancy and then I moved the "Source block count" slider in order to reach the maximum "Efficiency" with the minimum number of blocks.I obtained 99.8%
:with 17 blocks.Not the right method ?
Hi Othila,
Well, efficiency can mean several things:
1) - One PAR2 block repairs one article, so if you need to repair five MBs worth
of articles, five MBs of PAR2 blocks will do it. (100% repair capability)
2) - Maximum number of PAR2 blocks per file ('Powers of Two Sizing Scheme' where
each file contains twice as many PAR2 blocks as the previous file.) (100%
capacity)
3) - CPU efficiency (Do you really care how hard your machine works? Shaving a
few seconds off your CPU time seems hardly worth it.)
Your post contained 20892 yEnc articles (85 x 245 + 67).
There were 18 Universal repair blocks (.vol11+7 = 18). A Universal block can go
anywhere in the part to repair a missing yEnc article.
In addition, each Universal block had six Contingent blocks attached to it.
These Contingent blocks are not moveable and travel along with the Universal
block they are attached to. They are only useful if they happen to land on a
missing yEnc article.
So what are the odds? We already know that you will always be able to repair up
to 18 articles wherever they might happen to be. Let's assume that there is also
a 19th article missing. Subtract the 18 articles from the total number of
articles (20892 - 18) and you are left with 20874 articles. You have 108
Contingent blocks (6 x 18), so your odds of any of those landing on that missing
19th article are 108 in 20874, or about one chance in 200. (If you were also
missing a 20th article, the odds of repairing_both_of them would be about one in
200^2, or one in 4000.)
All of the above assumes a random distribution of missing articles, which can be
calculated using Poisson's formula, q.v.,
<http://en.wikipedia.org/wiki/Poisson_distribution
Some have disagreed with me in the past, stating that missing articles are not
randomly distributed, and so a Poisson distribution is irrelevant.
I guess one person's QuickPar is another person's Poisson.
Still, even if the missing articles were_not_completely random, it is unlikely
to find very many of them residing in the same seven article-wide (your case)
block. If you were missing articles 7 & 8 you would need two of your Universal
blocks, each of which is seven times larger than the article it will repair.
In general, I find Contingent blocks about as useful as teats on a boar hog. The
chances of more than a few of them being useful in a single repair are akin to
winning the lottery and getting struck and killed by lightening on your way to
the bank to cash in your winnings.
I prefer to use QuickPar's 'Powers of Two' sizing scheme, restrict block size to
multiples of article size, set the 'Source Block Count' to maximum, and choose
the number of blocks to create from the 'Powers of Two minus One' series, i.e.,
1, 3, 7, 15, 31, 63, 127, 255, 511, 1023, 2047...
But that's just me.
Isis
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