How many FLOPS is planet Earth?

June 21, 2009 2:05 PM

How many (available, i.e., plugged in) spare GHz are there in the world?

In other words, if every computer in the world were connected via some Folding@home-like network, and were using every clock cycle not dedicated to some other "official" purpose (be it work, gaming, whatever), how many Peta/Exa/GoogleFlops would it be?

Thanks!

(and how did you come up with your answer?)

posted by mpls2 to computers & internet (6 comments total)

...including mobile devices with internet access.
posted by mpls2 at 2:06 PM on June 21

There is approximately 1 computer per person, at an average of 1.5 GHz, and 25% utilization. Their combined unutilized clockspeed is therefore ~10 exahertz (10*10^18)
posted by gensubuser at 2:25 PM on June 21

Wow - To use the slashdot poll disclaimer, 'If you're using these numbers to do anything important, you're insane.'

As a lower bound, boinc lists 22,220,102 hosts working currently, though I would have no idea how to assign an average ghz/flop number to that, or what percentage of all processors are currently taking part in the various @home projects.

And I don't know how/if you should include mobile devices: Running work units would reduce the standby time of a celphone to minutes. They rely very heavily on processor idling.
posted by Orb2069 at 4:42 PM on June 21

@gensubuser - source?

@Orb2060 - chill out, it's just a thought experiment :) i'm just wondering how much excess computing capacity is out there. perhaps for mobile devices, we could define available Hz as those that would still leave enough power for the device to be used until it could be recharged.
posted by mpls2 at 4:46 PM on June 21

...and we should definitely include GPUs
posted by mpls2 at 4:48 PM on June 21

There are certainly at least 100 million computers out there, and I would sort of be suspicious if there were 10 for every person. Therefore 1. The last time I bought a computer, it was 1.8 GHz so I rounded down for all the grandma's. The 25% number came up as the first link in my google search. I expect I underestimate as much as I overestimate my exponents, so in the end it averages out to about the right order of magnitude. This is the Fermi Problem approach, which is what the snarky removed reply about piano tuners was talking about. The idea is that it's only important to put as much thought into the calculation as is necessary to get to the next part of the problem.

Go forth and multiply.
posted by gensubuser at 9:46 PM on June 21