>So assuming you could use the whole drive, the number of these current >generation drives that you would need to back a single address space is > > (2**64)/(2**38) = 2**(64-38) = 2**26 = 67,108,864
Back in my "old life" I once did a presentation at an IBM customer meeting trying to illustrate how much 2**64 is. Besides a similar DASD device calculation, I also did a simpified swap time calculation. At 100MB/s sustained transfer speed, it takes 5850 years to transfer all bytes of a single 64bit AS. Make that 10GB/s and use 100 devices in parallel it still takes 210+ days. According to Word, the PoP has about 4'000'000 characters and is roughly 5cm thick. So 4 of them fit into one 16MB AS and this corresponds to a pile of paper of 20cm height. Going to 31bit multiplies these figures by 128: 512 PoPs and 25.6 meters. Going to 64bit from 31bit multiplies them by another 8 billions: 4400 billion PoPs and you can travel all the way from the Earth to the sun and half this distance further on on that "bridge of PoPs"..... have good journey :-) History has shown all these "this will sufficce forever" statements have provben that "forever" is a relatively short period of time. So, I concluded my presentation stating that 64bit will do it for a couple of years. Peter Hunkeler CREDIT SUISSE ---------------------------------------------------------------------- For IBM-MAIN subscribe / signoff / archive access instructions, send email to [EMAIL PROTECTED] with the message: GET IBM-MAIN INFO Search the archives at http://bama.ua.edu/archives/ibm-main.html
