The space calculations are in terms of the virtual 3390 presented to the channel, and haven't changed since Old Man Noach cornered the market in Gopher wood. How the underlying hardware deals with it depnds on the box. In general, half track is most efficient except for load modules, where short blocks can pad out the track.
-- Shmuel (Seymour J.) Metz http://mason.gmu.edu/~smetz3 ________________________________________ From: IBM Mainframe Discussion List [[email protected]] on behalf of Tom Brennan [[email protected]] Sent: Wednesday, November 23, 2022 3:17 PM To: [email protected] Subject: Re: Bytes in a 3390 track > but it is mere waste on DASD, as 55,996 - 32760 = 23,236 bytes left > over, and because tracks can't be shared between other files Great overview, but is the note above still true with modern DS8000 boxes? It's just hard for me to imagine 3390 emulation logic holding that 23K hostage. On 11/23/2022 11:29 AM, Sri h Kolusu wrote: >>> How do I calculate the amount of space a ____ dataset needs? > > A 3390-n device has a capacity of 56,664 bytes per track, of which 55,996 > bytes are accessible by applications programmers. The largest blocksize you > can define is 32,760, which is good for tapes,but it is mere waste on DASD, > as 55,996 - 32760 = 23,236 bytes left over, and because tracks can't be > shared between other files, that leftover space would just be wasted. So, > 55,996/2 = 27,998, which is half-track blocking, the most space-efficient > blocksize to use on 3390's.If you have 3380 device types in your shop, the > maximum half-track blocksize is 23,476. > > To calculate the most efficient blocksize to use: Optimal blocksize = > INTEGER(half-track-blocksize/LRECL)*LRECL > > Assuming LRECL=500, so the optimal blocksize will be > > Integer(27,998/500) is 55 > > 55 multiplied by lrecl(500) gives you 27500 which is the optimum blksize. > > On a 3390, the best blocksize for a QSAM (Queued Sequential Access Mode) file > of record length 500 is 27,500.This will allow 55 records per block, or 110 > records per track, or 1650 records per cylinder (cylinders are 15 tracks). > > Assuming your volume of records e is 10,000, you can fit that data in less > than 7 cylinders(10,000/1650 = 6.06 rounded to 7) > > We allocate the primary space as 7 cylinders. Now the secondary space > allocation is 20% of primary space which in this case is 7 * 0.2 = 1.4 > rounded to 2 > > So, your space allocation for a dataset of lrecl 500 and a volume of 10,000 > records would be > > SPACE=(CYL,(7,2),RLSE) > > Thanks, > Kolusu > > > ---------------------------------------------------------------------- > For IBM-MAIN subscribe / signoff / archive access instructions, > send email to [email protected] with the message: INFO IBM-MAIN > ---------------------------------------------------------------------- For IBM-MAIN subscribe / signoff / archive access instructions, send email to [email protected] with the message: INFO IBM-MAIN ---------------------------------------------------------------------- For IBM-MAIN subscribe / signoff / archive access instructions, send email to [email protected] with the message: INFO IBM-MAIN
