In a message dated 11/8/2007 1:02:40 P.M. Central Standard Time, [EMAIL PROTECTED] writes:
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. I made a mistake. A track not in the cache would take on the order of 20 milliseconds, so that would equate to 20 days instead of one day. A track already cached would result in an access time of one millisecond. If the 4K block can be found in a buffer somewhere in virtual storage inside the processor, it might take from 100 to 1000 instructions to find and access that data, which would equate to 100 to 1000 seconds, or roughly one to 17 minutes. And that assumes that the page containing the 4K block of data can be accessed without a page fault resulting in a page-in operation (another I/O), in which case we are back to several days to do the I/O. By the way, it takes at least 5000 instructions in z/OS to start and finish one I/O operation, so you can add about two hours of overhead to perform the I/O that lasts for 20 days. You really want to avoid doing an I/O if at all possible. 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
