The number 1 response to this question was great.
Now, the only problem is that I don't really understand it. Here's my example of why it's not clear to me. Please let me know what I'm doing wrong.
Applying the above analysis to the FFT:
Say I want to compute the fft of a vector of doubles length n. Then I need to do approximately nlog(n) flops. The bytes required to do the computation are 8n, assuming a 64bit system. So $\beta$ = log(n)/8.
Suppose my processor works at 2GHz=Fmax and I have 8Gbytes=Bmax of RAM. Then, I have equality when:
log(n)/8 (flops/byte) = (Bmax/Fmax) (bytes/Hz) = 4 This implies
log(n) = 32 => n = 2^32 = 4,294,967,296
So, I should be able to compute the fft of a vector of size n? I don't know. Also, why don't the units cancel?