Thanks! That clears things up.
On 11/23/2022 12:31 PM, Pommier, Rex wrote:
Hi Tom, yes and no. :-)
No, it isn't true on the physical back end because all the disk is emulated on
top of FBA architectures and especially with thin provisioning, only actually
used tracks are really used. However, from a z/OS perspective yes the space is
still wasted because it is still allocated in cylinders and tracks, and from a
z/OS point of view if I have BLKSIZE=32760 RECFM=FB, z/OS can't do anything
with the remaining space.
Rex
-----Original Message-----
From: IBM Mainframe Discussion List <[email protected]> On Behalf Of Tom
Brennan
Sent: Wednesday, November 23, 2022 2:17 PM
To: [email protected]
Subject: [EXTERNAL] 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
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