On Tue, 23 Dec 2008 03:07:44 -0600, Parin Gangar wrote: >The dataset has following details - > >Device type . . . . :,3390 >Record format . . . :,FB >Record length . . . :,1526 >Block size . . . . :,7630 >1st extent tracks . :,15000 >Secondary tracks . :,1500 > >Current Allocation, >,Allocated tracks . :,297,141 >,Allocated extents . :,190 >, , >, , >Current Utilization, >,Used tracks . . . . :,297,141 >,Used extents . . . :,190 > >I tried searching this on Google, but didn't get anything. I know that there is >some formula for this.
Why searching Google if you can get the data from your z/OS system (I have got only 6 blocks in a track): bytes/block * block/track * #tracks 7630 * 6 * 297141 = 13603114980, less then 13 GiB By the way: using the blocksize of 27468, you would need less then 248000 tracks, it is nearly 50000 tracks difference. -- Zaromil ---------------------------------------------------------------------- For IBM-MAIN subscribe / signoff / archive access instructions, send email to lists...@bama.ua.edu with the message: GET IBM-MAIN INFO Search the archives at http://bama.ua.edu/archives/ibm-main.html
