On Wed, Sep 17, 2008 at 9:30 AM, (IBM Mainframe Discussion List) <[EMAIL PROTECTED]> wrote: > From the article on Stretch: > "[Stretch] ... could perform 100 billion computations a day and handle half > a million instructions per second." > > There are 86400 seconds in one day. Half a million instructions per second > for one day equals 43 billion instructions, which somehow were able to > perform 100 billion computations. I don't know of any z/OS instructions > that can > perform more than one computation per instruction, but I must confess I > haven't read about all the newest ones yet. How was Stretch able to perform > over > two computations for each instruction handled?
Well, obviously it had more than one CPU! (JOKE) You want reporters should be able to do math?? Not on this planet, I'm afraid... Actually, given CISC architecture, it seems possible that "half a million instructions per second" is an *average*, and that computations (say, BCTR -- "subtract one", hey, that's a computation!) would use less than the average, so more than 0.5MIPS was possible. Of course, like all benchmarks, that would be meaningless in the real world. Alternatively, maybe they meant "100 billion computations a day...across all 9 machines"! ---------------------------------------------------------------------- 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
