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mainframe Online Quiz - 8

Description: mainframe Online Quiz - 8
Number of Questions: 20
Created by:
Tags: mainframe
Attempted 0/20 Correct 0 Score 0

A BUFFER LENGTH OF 0 WAS SPECIFIED FOR A BDAM DATASET FOR WHICH DYNAMIC BUFFERING WAS REQUESTED.

  1. S013 - 10

  2. S013 - 08

  3. S013 - 0C

  4. None of these


Correct Option: C

A SEQUENTIAL OR DIRECT DATASET WAS OPENED FOR INPUT,BUT THE BUFFER CONTROL BLOCK ADDRESS WAS 0.

  1. S013 - 34

  2. S013 - 38

  3. S013 - 3C

  4. S013 - 40


Correct Option: D

AN INVALID RECORD FORMAT OF FBS OR FS WAS SPECIFIED FOR A PARTITIONED DATA SET.

  1. S013 - C4

  2. S013 - C8

  3. S013 - CC

  4. S013 - D0


Correct Option: D
  1. Sequential read is possible

  2. Free space in a VSAM File is reclaimed automatically

  3. VSAM Files are device and O/S independent

  4. None of the above


Correct Option: B,C

KSDS file is loaded

  1. Sequentially with descending order of the key

  2. Sequentially with ascending order of the key

  3. Randomly with descending order of the key

  4. Randomly with ascending order of the key


Correct Option: B

Default size of Control Interval (CI) for KSDS file

  1. 4094 Bytes

  2. 4096 Bytes

  3. 4095 Bytes

  4. 4092 Bytes


Correct Option: B

what is the meaning of KEYS(20 10)?

  1. The key is 10 bytes long and starts in the 21st byte of the record

  2. The key is 10 bytes long and starts in the 20th byte of the record.

  3. The key is 20 bytes long and starts in the 10th byte of the record.

  4. The key is 20 bytes long and starts in the 11th byte of the record


Correct Option: D
  1. As a part of Listcat, used to calculate the recommended number of Index or Data buffers for a given operation, i.e., sequential, random, etc

  2. To tell VSAM the parameters for Batch Update Function - Indexed (KSDS) or Batch Update Function - Data (ESDS) processing

  3. To tell VSAM to use the default number of index and data component buffers

  4. To allocate the number of index and data component buffers used


Correct Option: D

The range of value for KEYS

  1. 1 to 255

  2. 1 to 4096

  3. 1 to 510

  4. 1 to 4095


Correct Option: A

The record keys in a CI are 50, 200, and 400. The CI is full, and there is NO free space allocated. What will happen when key 300 is added to the data set shown above?

  1. VSAM will return an error message to the program, because the file is out of space

  2. It will be placed in a new CI by itself. This is a CI split.

  3. Record 300 will be placed in the place where 400 was & Record 400 will be moved to a new CI

  4. Either 2 or 3 depending on some contraints


Correct Option: C

How are EXPORT and REPRO files different

  1. EXPORT stores data in a compressed format. REPRO stores data uncompressed.

  2. EXPORT and IMPORT are terms that are part of the REPRO command so there is actually no difference

  3. EXPORT files contain the data and index portion of the data set. REPRO files contain only the data portion

  4. EXPORT files contain all of the data set components, including attributes. REPRO files contain only the data


Correct Option: C

What is the expansion of JCL ?

  1. Job Cal Language

  2. Job Control Language

  3. Jobcal Control Language

  4. None of the above


Correct Option: B

How many fields does JCL has?

  1. 3

  2. 7

  3. 5

  4. 4


Correct Option: C

'EXEC' in JCL specifies ?

  1. None.

  2. Program

  3. Identify cobal Program

  4. Program to be executed


Correct Option: D

A JCL program help in ?

  1. To copy COBAL , ASM code on MVS operation system.

  2. To move COBAL, ASM on MVS operation system.

  3. Running a job on MVS operation system, which is a communicator btw COBAL, ASM or PL/I and OS.

  4. None of the above


Correct Option: C

JES

  1. JOB Entry Sub-System

  2. JOB Entry Syntax

  3. JOB Entry System

  4. None of the above


Correct Option: C

types of JES are:

  1. JES1

  2. JES2

  3. JES3

  4. JES4


Correct Option: B,C

AI Explanation

To answer this question, you need to understand the different types of Job Entry Subsystems (JES) in mainframe systems. Let's go through each option to understand why it is correct or incorrect:

Option A) JES1 - This option is incorrect because JES1 is not a valid type of JES. JES1 does not exist as a separate subsystem in mainframe systems.

Option B) JES2 - This option is correct. JES2 is one of the types of Job Entry Subsystems (JES) in mainframe systems. It is responsible for managing the submission and execution of batch jobs.

Option C) JES3 - This option is correct. JES3 is another type of Job Entry Subsystem (JES) in mainframe systems. It provides advanced features for job management, such as job routing and workload balancing.

Option D) JES4 - This option is incorrect because JES4 is not a valid type of JES. JES4 does not exist as a separate subsystem in mainframe systems.

The correct answer is B, C. These options are correct because JES2 and JES3 are valid types of Job Entry Subsystems (JES) in mainframe systems.

which of the below will result in the maximum level-of-detail information to be spitted out in the Log-Report

  1. MSGLEVEL=(0,1)

  2. MSGLEVEL=(1,0)

  3. MSGLEVEL=(0,0)

  4. MSGLEVEL=(1,1)


Correct Option: D
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