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Author Topic: [MINING] Question  (Read 468 times)
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February 22, 2012, 11:48:42 AM
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How does one calculate a ratio between MHash/s and Shares? I would like to know how many MHash/s it takes me per share.
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February 22, 2012, 03:37:47 PM
 #2

You can also use CGminer. It gives you an average accepted shares per minute, per card and total. It would become more accurate the longer it runs.


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February 22, 2012, 03:49:54 PM
 #3

For "normal" pools they are difficulty 1 shares = 2^32 shares or roughly 4.295 billion shares.

So 4295 / MH per s = seconds between shares.  Remember shares are random.  You will on average find 1 share every 2^32 hashes but in the short run there will be variance.

Likely your next question is hashes per block.  
Share difficulty = 1
Current block difficulty = 1,376,302
It will on average* take 1376302 (difficulty 1) shares to find a block with difficulty 1376302.
1376302*2^32 = 5,911,172,079,419,390 (5.9 quadrillion)

Remember a share is worthless.  It has no value it is just used as a proxy.  Since a miner will find roughly 1 share every 4.3 billion hashes and can't over the long run increase share count without increasing hashing power a pool can use shares as proof of work.  When you submit a share you are proving you did ~ 2^32 hashes.

Now for p2pool it uses variable difficulty.  Currently share difficulty is ~550.  Thus it takes 550 * 2^32 hashes for one share = 2,362,232,012,800 (~2.4 trillion).


* We can only calculate the average since hashes are random.  Swings in # of hashes to solve a block of 10% to 500% of expected value are pretty common.  There have been blocks are short as <1% of difficulty and as long as 11,000% difficulty (and possibly longer or shorter).
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