andrew@orca.UUCP (Andrew Klossner) (12/20/84)
In discussing addresses larger than 32 bits, remember that those extra
bits pile up additional memory awfully fast.
To build a computer with 48-bit addresses and to give it 2**48 memory
cells, you would need more cells than there are atoms in the Earth.
Addresses bigger than 48 bits are probably not worthwhile, unless you
need a discontiguous (sparse) address space.
-- Andrew Klossner (decvax!tektronix!orca!andrew) [UUCP]
(orca!andrew.tektronix@csnet-relay) [ARPA]eder@ssc-vax.UUCP (Dani Eder) (12/21/84)
> In discussing addresses larger than 32 bits, remember that those extra > bits pile up additional memory awfully fast. > > To build a computer with 48-bit addresses and to give it 2**48 memory > cells, you would need more cells than there are atoms in the Earth. > Addresses bigger than 48 bits are probably not worthwhile, unless you > need a discontiguous (sparse) address space. > > -- Andrew Klossner (decvax!tektronix!orca!andrew) [UUCP] > (orca!andrew.tektronix@csnet-relay) [ARPA] 2**48 = 281.475 trillion. If you have 8-bit words, and 1 megabit RAM's, you would need 2,147,483,648 of them. Assuming you need one cubic inch per RAM, they would fit in a box 100x100x125 feet. While this is a very large amount of memory, it is a long way from even the number of grains of sand on a large beach, let alone the atoms in the Earth (about 10**50). Dani Eder / Boeing Aerospace Company / ssc-vax!eder / (206)773-4545
geoff@desint.UUCP (Geoff Kuenning) (12/22/84)
In article <1258@orca.UUCP> andrew@orca.UUCP (Andrew Klossner) writes: >To build a computer with 48-bit addresses and to give it 2**48 memory >cells, you would need more cells than there are atoms in the Earth. >Addresses bigger than 48 bits are probably not worthwhile, unless you >need a discontiguous (sparse) address space. I think you were thinking 10**48, Andy. 2**48 is only 256x10**12. This is an easily achievable number. Terabit memories have been available for over 15 years; you only need 256 of them to use up 48 bits. Another way to think of this is that 2**48 is 2**32 times 2**16. 2**32 is 4 gigabytes, or about 8 Eagles. So you need 65536*8, or 2**16x2**3, or 512K Eagles to hold 2**48 bits. That's probably more than the number of Eagles that will ever be manufactured, but clearly it is a number that will be achievable in the next decade or two. -- Geoff Kuenning ...!ihnp4!trwrb!desint!geoff