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Author Topic: Some Statistics  (Read 6855 times)
lfm
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August 18, 2010, 07:53:51 PM
 #21

I'll offer a BTC 100.00 bounty for an addition to
http://nullvoid.org/bitcoin/statistix.php
that shows the inverse (1/x) data: Computing power.

Specifically, total khash/sec for the entire network, based on either current difficulty or (preferably) observed block intervals / moving-average.

How "strong" is the bitcoin community now, and how is it trending?  That's what I'm looking for.  I'm not picky.



I think this shows what you want. This is the total difficulty per day. to get total hashes /day multiply by 2^32.

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Ground Loop
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August 18, 2010, 08:02:28 PM
 #22

The simplest way is to derive it from the hash target (difficulty).

If you take the present target value
0000000000800e00000000000000000000000000000000000000000000000000
and divide it by the total search space (all 0xF), you get the odds.

In decimal: 5.26786429665E+64 divided by the total search size (2^256) shows you have a one in 2.19e12 chance of finding a new block with every hash try.

We know that the network, as a whole is guided toward 6/hr, and the difficulty reflects that.  If the average generation rate is one every 10 minutes (600 seconds) then we know that the network as a whole is trying
3,663,473,554
hash attempts per second.

So the aggregate network power is around 3,663,473 khash/second.  Call it 3.7 gigahash/sec. Smiley   Further, if my machine is plugging along at 4,000 khash/sec, then I know I represent about 0.1% of the active network, and have a 1/1000 chance of finding "the next block".

If you use the *actual* recent block find rates (instead of the 10 minute average), which you currently track and graph, then you can see the interesting graph over time of how many total attempts likely produced that result.

I believe the 1/x characteristic would be more interesting to watch for trends than the seconds/block you currently graph, since the number of seconds is forced into a normalized 600 over time.  Total network strength should show an always-upward trend, with some notable spikes, dips, and steps.

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August 18, 2010, 09:55:20 PM
 #23

I think this shows what you want. This is the total difficulty per day. to get total hashes /day multiply by 2^32.


How does the Difficulty number (511) relate to the hash target?

511.77353426 * 2^32 = 2.2e12 hash/day, or 25 Mhash/sec.  I'm not sure how to interpret that, but it doesn't seem to reflect the 3.7 Ghash/sec total.

Also, the difficulty web page here:
http://www.alloscomp.com/bitcoin/calculator.php

seems to think that 3,600,000 khash/sec is a 10minute average rate.

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August 18, 2010, 10:49:28 PM
 #24

How does the Difficulty number (511) relate to the hash target?

Difficulty = 0xFFFF0000000000000000000000000000000000000000000000000000 / HashTarget

For easier calculation you can divide numerator and denominator by 2^193 (i.e. shift right 193 bits).
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August 18, 2010, 10:58:24 PM
 #25

Ah, I see!  Simpler than I thought.
(difficulty * 2^32) is the odds of a hit on EACH try.
That's the 2.20e12 value again.

So total network power, in hash/sec, is then
(difficulty * 2^32) / 600

Where 600 is a smoothed approximation of last-block-interval-in-seconds.

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August 18, 2010, 11:11:49 PM
 #26

So there are about 2000 people generating coins. (1,5Mhash/s average and a 1Ghash/s cluster)
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September 27, 2010, 03:07:31 PM
 #27

I will be moving over next few days and nullvoid.org and bitbot may be unavailable during this period.  I will try to prepare availability as much as possible.
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December 06, 2010, 06:52:17 PM
 #28

So there are about 2000 people generating coins. (1,5Mhash/s average and a 1Ghash/s cluster)

How did you compute that 2,000 number?  I'ld like to recalc it using the current difficulty.

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January 13, 2011, 08:24:32 PM
 #29

I will be moving over next few (105?) days and nullvoid.org and bitbot may be unavailable during this period.  I will try to prepare availability as much as possible.

I moved and have reliable/stable Internet access again, thus http://nullvoid.org/bitcoin is available to those that asked about it (and to everyone else).
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