Posted: Wed Oct 29, 2025 9:20 am Post subject: Help needed for space allocation
All,
I have checked numerous existing threads that explained about space calculation methods based on BLKSize, Lrecl, avgrec, and the Prim and Sec spaces. However, those calculations have not provided much assistance in my current allocation, where I am struggling to interpret the JCL space parameters vs Actuals.
Data Set Name . . . : ABC
General Data Current Allocation
Management class . . : XXXXXXXX Allocated kilobytes : 4,711,572
Storage class . . . : XXXXXXXX Allocated extents . : 4
Volume serial . . . : XXXXXXXX
Device type . . . . : 3390
Data class . . . . . :
Organization . . . : PS Current Utilization
Record format . . . : FB Used kilobytes . . : 4,711,572
Record length . . . : 840 Used extents . . . : 4
Block size . . . . : 32760
1st extent kilobytes: 1216094
Secondary kilobytes : 504064
Data set name type : EXTENDED
Data set encryption : NO
From 3.4 Option with Space attribute
Command - Enter "/" to select action Tracks %Used XT
-----------------------------------------------------------------
ABC 85193 100 4
Joined: 26 Nov 2002 Posts: 12397 Topics: 75 Location: San Jose
Posted: Mon Nov 03, 2025 2:58 pm Post subject:
nbdtrjk_1 wrote:
All,
I have checked numerous existing threads that explained about space calculation methods based on BLKSize, Lrecl, avgrec, and the Prim and Sec spaces. However, those calculations have not provided much assistance in my current allocation, where I am struggling to interpret the JCL space parameters vs Actuals.
My calculation is
6000*15 = 9000 Tracks. so 90000-85193 = 4807 tracks still available to accommodate the additional volumes. Is it correct?
Thanks
nbdtrjk_1,
Well Looks like you haven't read the documentation correctly. When you specify AVGREC=K, then K indicates that PRImary and the SECondary parameters specify the number of records in thousands (so it would be the number specified multiplied by 1024). In your case , you specified a primary value of 6000 which means that the space allocation would be done for 6000 K records = 6,144,000
Looks like you did not show all of the information. Since the dataset is created with BLKSIZE of 32K, I am assuming it is a compressed dataset. A listcat of the dataset would show you that information.
nbdtrjk_1 wrote:
volume of file is 93 million records
A 3390-n device has 56,664 bytes out of which 32 bytes are reserved. So in your case (56664-32)/840 = 67
For an uncompressed dataset of 93 million records, it would take 93M/67 = 1,388,060 tracks
Check if STORCLAS that the dataset is allocated guaranteed space.
Ideally you would do this.
1. Run an IEFBR14 step and allocate the dataset.
2. Do a 3.4 listing or LISTCAT on the dataset and then you will see the real allocation values.
3. Write 93M records of data to the dataset
4. Do another 3.4 listing or LISTCAT and then see the allocation values.
nbdtrjk_1 wrote:
Any input would be appreciated which sheds light on my further analysis
People respond here voluntarily , respond when they have time and are NOT paid, So please stop bumping the thread looking for responses.
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