Fortunately your explanation was clear enough to find out each
"mini-record" has at least 10 cells (9 control plus a t least one for
data).
However you used word "SECTORS". I used "34-byte DATA CELL", following
the 3390 documentation (other names can be found elsewhere).
It seems the documentation use SECTOR for something else. The
SA22-1025-00 says the 3390 device has 224 sectors per track and 222 for
3380. So, the SECTOR cannot be the same as DATA CELL.
I have no idea what is the SECTOR.
BTW: I also do not understand the formula. For me it should be just
Numberofcells=9+RoundUP(KL/34)
of course you provided right formula as it is documented, but I simply
don't understand the purpose of KN and 6 and 232. i'm pretty sure it is
not just a magic, but what is the rationale behind?
Last but not least: THANK YOU VERY MUCH for the explanations you gave
me, I appreciate it.
Regards
--
Radoslaw Skorupka
Lodz, Poland
W dniu 2018-01-23 o 15:48, Christopher Y. Blaicher pisze:
With all I put in the last post, I forgot to answer WHY 20 sectors at a minimum
for a record. It is because at a minimum a record consists of a COUNT and a
DATA section. Each one takes a minimum of 10 sectors, so with even a 1 byte
record you need 20 sectors. If you had a 1 byte key and 1 byte of data you
would need 30 sectors, 10 for each part.
Chris Blaicher
Technical Architect
Mainframe Development
P: 201-930-8234 | M: 512-627-3803
E: [email protected]
Syncsort Incorporated
2 Blue Hill Plaza #1563
Pearl River, NY 10965
www.syncsort.com
Data quality leader Trillium Software is now a part of Syncsort.
-----Original Message-----
From: IBM Mainframe Discussion List [mailto:[email protected]] On Behalf
Of Christopher Y. Blaicher
Sent: Tuesday, January 23, 2018 8:30 AM
To: [email protected]
Subject: Re: CKD details
See the following for how to calculate a sector
https://www.ibm.com/support/knowledgecenter/en/SSB27H_6.2.0/fa2mr_sectval.html
Question 2: why 20 sectors at a minimum. It's a long answer.
We tend to think of a record as three unified parts, COUNT, KEY and DATA, but
really they are three mini-records (for lack of a better description). Each
part needs enough cells for the data, plus 9 cells of CRC and other control
information that we can never see.
So, for a COUNT field it takes 10 cells; for a KEY field, if present, and a
DATA field the calculation is:
SECTORS = 9 + (KL + (6 * KN) + 6) / 34)
Where KN = (KL + 6) / 232
KL = Key length
Change KL to DL and do the same calculation for the data area.
Question 3: Are you running under VM and using a mini-disk? VM formats their
R0 differently, I believe.
Hope this helps.
Chris Blaicher
Technical Architect
Mainframe Development
P: 201-930-8234 | M: 512-627-3803
E: [email protected]
Syncsort Incorporated
2 Blue Hill Plaza #1563
Pearl River, NY 10965
www.syncsort.com
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