🚀 go-pugleaf

Anders Sundqvist (Anders.Sundqvist@soo.mts.dec.com) wrote:
: While we are at it, how much *IS* a 128 bit wide byte address?
: - 2^128 is approximately 10^38
(snip)
: Anders
: PS. I hope I got my calculations right, if not PLEASE correct me!

Your calculation looks ok to me (2^128=3.4x10^38 on my calculator).
But while looking at your calculation, the following thought occurred
to me.  Suppose you actually had 10^38 bytes of memory.  And suppose
you had a 128-bit bus, capable of moving 16 bytes per cycle.  And
finally suppose you had a 1-Ghz clock (10^9 cycles/sec), fast by
today's standards.  Then it would take you 10^38/16/10^9 = 6x10^27secs
= 2x10^20years (more or less) just to move all that memory through
the bus.  Obviously (imho), that much memory is totally useless
on any machine available today (and probably for several tomorrows).

By comparison, a 32-bit machine and a 32-bit bus, at the same 1-Ghz
clock rate, takes 2^32/4/10^9=1.07sec to move all its memory through
the bus.  So the move from 32-bit to 128-bit corresponds to a
"time scale" (so to speak) that moves from 1sec to 10^20years.
A much faster processing architecture would have to accompany a 128-bit
memory address.  Of course, a 128-bit bus is useful all by itself,
just to push bytes through the system that much faster.  But to
actually install all that memory wouldn't be very useful.

John
forkosh@panix.com