In a message dated 11/8/2007 12:55:16 P.M. Central Standard Time, [EMAIL PROTECTED] writes:
tried to put the time for computer events into perspective. A 100-MIPS processor can execute 100 million "average" instructions per secon d, so one "average" instruction takes one hundred-millionth of a second (ten to the minus 8 power seconds). A Direct Access Storage Device read of a 4K block, if the data is not in the DASD Subsystem's cache, would take at least one millisecond, which is ten to the minus three power seconds. The difference is a factor of ten to the fifth power. If you equate one instruction with one second, then one I/O is 100000 seconds, or a little more than one day. Bill Fairchild Franklin, TN ************************************** See what's new at http://www.aol.com ---------------------------------------------------------------------- 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
