I read the story when it first came out and it listed the power consumption. However, when I googled to look at the “Roadrunner” articles I could not find one with the power consumption. I want to say between 10 and 20 megawatts but do not hold me to it.

Bill, please check your numbers. You’re off by 3 orders of magnitude.
Kevin had the correct number, but the wrong units.

1 petaflop = 10^15 floating operations per second.
Realizing that the journalist bungled the units, the electric power consumption is more aptly 376×10^6 flops per W.

Thus, total power = 1e15/ 376e6 = 2.6596e6 W

About 2.7 MW is the power consumption of the machine.

This makes a lot more sense that saying that a server farm-sized computer requires only a 2.7 kW service….
(For reference, most residences are in the 10-25 kW range.)

Energy consumption, is of course, a question of how much time it spends operating at that power consumption level.

kevin, there is no such thing as a watt/sec.

In fact, the article is in error, what happens if you let the publicity department play with numbers.

“performing 376 million calculations for every watt of electricity use” is meaningless, as energy is not measured in watts. Joules are the unit of energy, watts are a unit of power.

Probably the original statement said something like “376 million calculations for every Joule” but the publicity dept. changed it to watts for the reason “no one knows what a Joule is”.

1J per 376Mcalc
1 peta = 1000000 M

so it’s 1J/376Mcalc x 1000000M/p = 2700J/pcalc

So the answer is 2700J. This is the energy the computer used for every peta of calculations.

The term flop means, roughly, calculations per second, so it is a rate.

So at a RATE of 1 petaFlop of calculations per second, the computer uses 2700 joules per second, which is 2700 watts.

Hope that is clear.

calculations corresponds to joules.

flops, calc per second, corresponds to watts, joules/second.

by my calculations, 2.66 MegaWatts per second