# 128-bit storage: are you high?

**Note:** This was originally published to http://blogs.sun.com/bonwick/date/20040925 by Jeff Bonwick. It came up for discussion recently but since Oracle no longer has any published copies, it was retrieved from The Wayback Machine and reposted here.

## Saturday Sep 25, 2004

One gentle reader offered this feedback on our recent ZFS article:

64 bits would have been plenty … but then you can’t talk out of your ass about boiling oceans then, can you?

Well, it’s a fair question. Why did we make ZFS a 128-bit storage system? What on earth made us think it’s necessary? And how do we know it’s sufficient?

Let’s start with the easy one: how do we know it’s necessary?

Some customers already have datasets on the order of a petabyte, or 2^{50} bytes. Thus the 64-bit capacity limit of 2^{64} bytes is only 14 doublings away. Moore’s Law for storage predicts that capacity will continue to double every 9-12 months, which means we’ll start to hit the 64-bit limit in about a decade. Storage systems tend to live for several decades, so it would be foolish to create a new one without anticipating the needs that will surely arise within its projected lifetime.

If 64 bits isn’t enough, the next logical step is 128 bits. That’s enough to survive Moore’s Law until I’m dead, and after that, it’s not my problem. But it does raise the question: what are the theoretical limits to storage capacity?

Although we’d all like Moore’s Law to continue forever, quantum mechanics imposes some fundamental limits on the computation rate and information capacity of any physical device. In particular, it has been shown that 1 kilogram of matter confined to 1 liter of space can perform at most 10^{51} operations per second on at most 10^{31} bits of information [see Seth Lloyd, “Ultimate physical limits to computation.” Nature 406, 1047-1054 (2000)]. A fully-populated 128-bit storage pool would contain 2^{128} blocks = 2^{137} bytes = 2^{140} bits; therefore the minimum mass required to hold the bits would be (2^{140} bits) / (10^{31} bits/kg) = 136 billion kg.

That’s a lot of gear.

To operate at the 10^{31} bits/kg limit, however, the entire mass of the computer must be in the form of pure energy. By E=mc^{2}, the rest energy of 136 billion kg is 1.2x10^{28} J. The mass of the oceans is about 1.4x10^{21} kg. It takes about 4,000 J to raise the temperature of 1 kg of water by 1 degree Celcius, and thus about 400,000 J to heat 1 kg of water from freezing to boiling. The latent heat of vaporization adds another 2 million J/kg. Thus the energy required to boil the oceans is about 2.4x10^{6} J/kg * 1.4x10^{21} kg = 3.4x10^{27} J. Thus, fully populating a 128-bit storage pool would, literally, require more energy than boiling the oceans.

Posted at 01:03AM Sep 25, 2004 by bonwick in ZFS