Because a PDS/E member occupies at minimum 4096 bytes ?
Regards, Thomas Berg _________________________________________ Thomas Berg Specialist A M SWEDBANK > -----Ursprungligt meddelande----- > Från: IBM Mainframe Discussion List [mailto:ibm-m...@bama.ua.edu] För Eric > Bielefeld > Skickat: den 11 februari 2010 16:39 > Till: IBM-MAIN@bama.ua.edu > Ämne: Re: PDS vs. PDSE > > I just discovered something about PDS/Es that I don't remember being > discussed. This discussion inspired me to copy my JCL PDS to a PDS/E on > one of my accounts. Notice that the % full went from 62 to 95%. I used > the same blksize. I figured that since the PDS was 62% full, I'd make the > PDS/E 2 cylinders less. Here are the results: > > Tracks % XT Device Dsorg Recfm Lrecl Blksz > ------------------------------------------------ > $IQLG.JCL.CNTL > 120 62 1 3390 PO FB 80 7520 > ------------------------------------------------ > $IQLG.JCLN.CNTL > 120 95 2 3390 PO-E FB 80 7520 > > That's almost a third more space used. Any comments? I'm sure there are > a few who know why that is out there. This PDS has just over a thousand > members. > > Eric > > -- > Eric Bielefeld > Systems Programmer > IBM MVS Technical Services > Dubuque, Iowa > 563-845-4363 > > ---------------------------------------------------------------------- > For IBM-MAIN subscribe / signoff / archive access instructions, > send email to lists...@bama.ua.edu with the message: GET IBM-MAIN INFO > Search the archives at http://bama.ua.edu/archives/ibm-main.html ---------------------------------------------------------------------- For IBM-MAIN subscribe / signoff / archive access instructions, send email to lists...@bama.ua.edu with the message: GET IBM-MAIN INFO Search the archives at http://bama.ua.edu/archives/ibm-main.html
