Posted by
Hemos
on from the and-irradiated-testes dept.
solferino writes: "Assume engineering genius continues to allow Moore's law to hold. What are the absolute computing limits of a 1kg laptop computer as defined by the physical laws of the universe? A New Scientist article has some interesting answers."
Back of the envelope nanocomputer numbers.
by
Christopher+Thomas
·
· Score: 5
While waiting for the article to load, I did back-of-the-envelope calculations for the performance of the best possible 1 kg computer with atom-sized features.
In case anyone else is as bored as I was, here are the calculations and the numbers:
- Assume, arbitrarily, that your device is made of carbon and has one computing element (gate or memory element) per 10 atoms, average. This gives a total of about (1000 / (12 * 10)) * 6e23 = 5e24 computing elements.
- Assume that we're going for floating-point performance, and are using most of our elements for multiplication units. Assume we're cheating and using single-precision ints (32 bits). If we're allowed to pipeline arbitrarily deeply (we're runnign a toy benchmark program), then it would take somewhere in the realm of 4000 computing elements to build an IEEE-compliant floating point multiplier. This gives us 5e24/4000 = 1.25e21 multiplication units operating in parallel.
- Assume that we're signalling using light and that the light has to travel 1 nm per clock (we're very good at routing traces). This gives a clock frequency of 3e8/1e-9 = 3e17 Hz.
- This gives us a total of 1.25e21 x 3e17 = 3.75e38 FLOPS. Less than the best, but still not too shabby.
For kicks, let's compute the power requirements of this device.
- Assume that on every clock, half of the computation units change state (we're managing to use all of the computation units all of the time, with random data). This gives 5e24 / 2 = 2.5e24 transitions per clock.
- This gives 2.5e24 * 3e17 = 7.5e41 transitions per second.
- Assume that each transiton costs about 5 eV in total (split this however you like). This gives 5 eV * 1.6e-19 J/eV = 8e-19 Joules per transition.
- This gives us a power dissipation of 6e23 watts. A bit power-hungry.
For kicks, let's compute the surface temperature of this computer assuming radiative cooling:
- Assume that our computer is a 10-cm cube, with a density comparable to that of water (this is strangely-structured carbon). This gives us a surface area of 6e-2 square metres.
- Radiative energy emission from the object will therefore be equal to 6e-2 * 5.67e-8 * T^4 = 3.46e-9 * T^4 watts, where T is the object's surface temperature.
- For a power dissipation of 6e23 watts, the object's surface temperture would be (6e23 / 3.46e-9) ^ (1/4) = 1.15e8 degrees Kelvin. A bit warm.
Summary of data for the best possible nanotech computer:
5e24 computing elements.
3.75e38 FLOPS (single-precision multiplies).
3e17 Hz.
6e23 watts.
Surface temperature of 1.15e8 degrees Kelvin.
Looks like we'd have to underclock this baby.
Derivation of computing power for a comparably-sized quantum computer is left as an exercise for the reader.
In practice, we'd probably wind up building our nanocomputers as thin films with a lot less computing power but far lower power dissipation. Possibly as nano-grains, also, depending on application.
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In case anyone else is as bored as I was, here are the calculations and the numbers:
- Assume, arbitrarily, that your device is made of carbon and has one computing element (gate or memory element) per 10 atoms, average. This gives a total of about (1000 / (12 * 10)) * 6e23 = 5e24 computing elements.
- Assume that we're going for floating-point performance, and are using most of our elements for multiplication units. Assume we're cheating and using single-precision ints (32 bits). If we're allowed to pipeline arbitrarily deeply (we're runnign a toy benchmark program), then it would take somewhere in the realm of 4000 computing elements to build an IEEE-compliant floating point multiplier. This gives us 5e24/4000 = 1.25e21 multiplication units operating in parallel.
- Assume that we're signalling using light and that the light has to travel 1 nm per clock (we're very good at routing traces). This gives a clock frequency of 3e8/1e-9 = 3e17 Hz.
- This gives us a total of 1.25e21 x 3e17 = 3.75e38 FLOPS. Less than the best, but still not too shabby.
For kicks, let's compute the power requirements of this device.
- Assume that on every clock, half of the computation units change state (we're managing to use all of the computation units all of the time, with random data). This gives 5e24 / 2 = 2.5e24 transitions per clock.
- This gives 2.5e24 * 3e17 = 7.5e41 transitions per second.
- Assume that each transiton costs about 5 eV in total (split this however you like). This gives 5 eV * 1.6e-19 J/eV = 8e-19 Joules per transition.
- This gives us a power dissipation of 6e23 watts. A bit power-hungry.
For kicks, let's compute the surface temperature of this computer assuming radiative cooling:
- Assume that our computer is a 10-cm cube, with a density comparable to that of water (this is strangely-structured carbon). This gives us a surface area of 6e-2 square metres.
- Radiative energy emission from the object will therefore be equal to 6e-2 * 5.67e-8 * T^4 = 3.46e-9 * T^4 watts, where T is the object's surface temperature.
- For a power dissipation of 6e23 watts, the object's surface temperture would be (6e23 / 3.46e-9) ^ (1/4) = 1.15e8 degrees Kelvin. A bit warm.
Summary of data for the best possible nanotech computer:
Looks like we'd have to underclock this baby.
Derivation of computing power for a comparably-sized quantum computer is left as an exercise for the reader.
In practice, we'd probably wind up building our nanocomputers as thin films with a lot less computing power but far lower power dissipation. Possibly as nano-grains, also, depending on application.