How does the OPSLOG archive process determine the amount of space it needs?
Every so often we have one that will abend (B37-04) space issue and the TRACK parameters passed to the archive proc are always (1313,1313).
//OPSSARCH JOB MSGLEVEL=1
//STARTING EXEC OPSSARCH,SSID=OPSS,LOGNAME='OPSLOG1',TRACK=(1313,1313)
XXOPSSARCH PROC SSID=,
XX LOGNAME=,
XX TRACK=
XXARCHPROC EXEC PGM=OI,
XX PARM='ARCHNTRK &SSID,&LOGNAME,&TRACK' IEFC653I SUBSTITUTION JCL - PGM=OI,PARM='ARCHNTRK OPSS,OPSLOG1,(1313,1313)'
Release : 14.0
Component : OPS/MVS
The calculation performed by the archive process is as follows:
- record size 676
- bytes per track 47476
total tracks = number of records x 676 + 288 divided by 47476
This number of total tracks is then divided by 16 to get the primary and secondary allocations.