IBM C9050-549

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IBM C9050-549 Exam Syllabus Topics:

IBM Certified System Programmer - IBM IMS Sample Questions:

1. A system is running low on active logs and you need to add active logs to provide some relief.
What would be the appropriate action to achieve this?

A) Allocate the new active log data sets using IDCAMS, change the ACTLGNUM DSNZPARM to a higher
value, and issue the -SET SYSPARM LOAD (new-zparm-mod) command.
B) Allocate the new active log data sets using IDCAMS, change the ACTLGNUM DSNZPARM to a higher
value, and issue the -SET SYSPARM RELOAD command.
C) Allocate the new active log data sets using IDCAMS, change the ACTLGNUM DSNZPARM to a higher
value, stop DB2, and add the active log names then start DB2.
D) Allocate the new active log data sets using IDCAMS, stop DB2, add the active log names then start
DB2.

2. Which z/OS component trace is always recommended to be active when using that componentwith IMS?

A) RRS
B) SYSAPPC
C) XES
D) SUBSYS

3. What happens when the maximum number of locks requested by an application is exceeded?

A) The application pauses until acknowledged by an Operator.
B) IMS issues a pseudo-abend of the application.
C) A message is displayed, and the application program pauses until the number of locksisreduced.
D) IMS pseudo-abends.

4. The total number of threads accessing data that can be allocated concurrently is the sum of which
combination of DB2 parameters?

A) IDBACK and IDFORE
B) MAXKEEPD and MAXOFILR
C) MAXDBAT and CONDBAT
D) MAXDBAT and CTHREAD

5. Click the Exhibit button.
Referring to the exhibit, which statement related to DB2 storage usage is correct?

A) The DBM1 address space does not seem to do any paging, but the MSTR address space seems to
suffer from paging.
B) DB2 statistics do not provide accurate information about the real storage usage. The only reliable
source of information are RMF statistics.
C) The system is over committed on virtual storage as it has more than 2 GB of real storage allocated in
the DBM1 address space.
D) The system is doing ok on virtual storage but is clearly short on real storage because the total amount
of real storage is more than 2 GB.

Solutions: