3 ms·
> The problem is that encryption with larger key sizes is computationally much more than 2x expensive. In fact, compute costs are based on a cube of key size, m
by spicyj 16y ago
> The problem is that encryption with larger key sizes is computationally much more than 2x expensive. In fact, compute costs are based on a cube of key size, meaning encryption using a 2048 bit private key can be as much as 30x more expensive than 1024 bit.
How'd you get 30? From what you said, it sounds like it should be 2^3 = 8 times as expensive.
- john_nl 16y agoI'm also curious where the 30x comes from. If cost is indeed the square of key size I would expect: Cost(1024) = (1024*x)^2 Cost(2048) = (2048*x)^2 = (2*1024*x)^2 = 4 * (1024*x)^2 Which is a 4x increase in cost, not 30x.
- illumin8 16y agoCube, not square. Also, implementation matters. 2^3, or 8 times more expensive.
- illumin8 16y ago1024^3 = 1073741824 2048^3 = 8589934592 / 1024^3 = 8. It depends on the implementation. Some implementations are more computationally expensive than others. See: http://devcentral.f5.com/weblogs/macvittie/archive/2010/09/10/f5-friday-the-2048-bit-keys-to-the-kingdom.aspx http://devcentral.f5.com/weblogs/macvittie/archive/2010/09/1... Average of 5x reduction in transactions per second when moving from 1024 bit to 2048 bit keys.