IBM Mainframe Forum

Hi Friends

Please help me to know how to find out expected space that will be taken by a variable length sequential file, when we know the average length and approx. number of records.

I know how to calculate space for fixed length sequential file, but never did for variable length

Thanks
Nikesh Rai

And what is it that you do not understand? This is so elementary that it should put grave doubts as to your suitability of working in the field you are working in!

How the flipping 'ell would you calculate the space for an FB file with an average length of x bytes???

SIASD!

I am using this formula till now for FB files

391,227 records of FB LRECL=90

INTEGER(27998/90) = 311. Therefor 622 records will fit one track.
391227 / 622 = 628.9823 = 629 tracks required.
629 / 15 = 41.93333 = 42 cylinders required to store your data.

but not sure if anything extra is needed for variable length file

for my case.. my file length is 25000 byte but the program will write at max 10000 byte, I am expecting average file length to be around 6000 byte

You can easily find the maximum space required -- add 4 to the record length (for the RDW) and follow EXACTLY the same calculations as for a fixed length data set (the only files in z/OS are on tape or in Unix System Services). If you want to calculate the expected space, add 4 to the average record length and follow EXACTLY the same calculations as for a fixed length data set.

Thank you Robert.

I have one more doubt. While calculating required space. which length I should consider, 1) approx average length of records i.e. around 6000 bytes or 2) Length of longest record which will have 10000 bytes.

I google it but still doubtful on this because of mixed answers.

You did not pay attention

If you want to calculate the expected space, add 4 to the average record length and follow EXACTLY the same calculations as for a fixed length data set.

Yes I did.. but was bit confused with answers on google. so wanted to confirm..

so wanted to confirm..

confirm what ???
if You do not trust our replies You have no business hanging around here

If you would rather believe some unknown source from Google over us, WHY DID YOU ASK THE QUESTION IN THE FIRST PLACE?

Since I have over three thousand posts on this forum, why would you think I would lie to you?

nikesh_rai wrote:Thank you Robert.

I have one more doubt. While calculating required space. which length I should consider, 1) approx average length of records i.e. around 6000 bytes or 2) Length of longest record which will have 10000 bytes.

I google it but still doubtful on this because of mixed answers.

Actually, it's more complex than Prino and others assume.

Knowing the longest record - e.g., the LRECL of the data set, is just about useless.

The fallacy of using the average record length is it is all to easy to figure all the records are the average length. Oops. You have 2 records, one is 10000, and the other is 2000, the average is your 6000, but the physical make up of the data set will not be what you expect.

With variable length records it is critically important to know the BLKSIZE. With BLKSIZE = 10000 and an average record length of 6000 bytes, it is all too likely you'll have blocks with one 6000 byte record. This is not so bad on tape, but it could be critical on disk.

One way to avoid space calculation problems is to use spanned records. Then the BLKSIZE, relative to the record size is less of a factor. The problem with RECFM=VBS is people tend to be afraid of it, especially Assembler programs using BSAM. Then there is the additional CPU cost to assemble the segments. Finally there are other issues with VBS. For example, you can't use DISP=MOD in situations where DISP=MOD would be natural in simple.