2)Usually 1 track consists of 56,664 bytes on a 3390 disk.Exactly how many tracks are actually allocated by the system depends on the values specified int he LRECL and BLKSIZE parameters of the dataset na dhence the amount of storage space u have requested will vary from device to device.[/url]
SORT FIELDS = (1,5,CH,A) EQUALS
SUM FIELDS = (10,5,PD)
can any one clarify what is the use of the above statement.can i hav an example please.what is the use of equals.what do u mean by PD.
A PD field looks like this in hex: 'dd...ds' where d is 0-9 and s is the sign, usually C or F for positive and D for negative.
The DFSORT statements above sum the PD values in positions 10-14 whenever two records are found with the same CH (character) sort key in positions 1-5. Only one record of the pair is kept. EQUALS tells DFSORT to keep the first record of the pair.
If you're not familiar with DFSORT and DFSORT's ICETOOL, I'd suggest reading through "z/OS DFSORT: Getting Started". It's an excellent tutorial, with lots of examples, that will show you how to use DFSORT, DFSORT's ICETOOL and DFSORT Symbols. You can access it online, along with all of the other DFSORT books, from:
2. space=(TRKS,(20,20,10)).can any one explain how many number of tracks are created totally in thebest case and worst case.plase expalin how to calculate the number of tracks.
Each allocation (primary or 2ndary) can take 1 to 5 extents to satisfy it.
(20 *1) + (20 * 15) = 320 trks w/10 directory blks for the PDS (taken from the primary).
In this scenerio you got each alloc in 1 extent. Lucky guy.
20 * 1 (but you needed all 5 extents). That leaves 11 extents for 2ndary.
Now it takes 5 extents to satisfy the 1st 2ndary alloc and another 5 for the 2nd alloc. If you're still unlucky you can only get 1 trk for the 3rd.
Now you wind up with 20 primary and 41 2ndary or 61 total.
If the allocation cannot be filled with the remaining extent I beleive an abend (?37) will occur; you CANNOT allocate anything less than the requested 2ndary alloc as I mentioned above. If it is fulfilled w/1 extent the 2ndary tracks allocated are 60 (40 + 20) giving a total of 80 combined with the primary.
If you're REALLY unlucky you JCL out because you can't find enough DA space for the primary in 5 extents.