Tag: mainframe

Questions Related to mainframe

  1. Actual Assembly Area

  2. Agent Arrival Area

  3. Agent Assembly Area

  4. Actual Assembly Agent


Correct Option: C
  1. Nothing, the program will behave normally.

  2. The program will go in an infinite loop

  3. The program will not get complied.

  4. Output files wont be generated.


Correct Option: B
  1. PIC 9.99 is a THREE-POSITION field where as PIC 9v99 is FOUR-POSITION numeric field.

  2. No difference.

  3. PIC 9.99 is a FOUR-POSITION field where as PIC 9v99 is THREE-POSITION numeric field

  4. PIC 9.99 takes less space then PIC 9V99.


Correct Option: C
  1. Load Module not found

  2. Output files not defined in JCL

  3. DD name is incorrect

  4. Indicates a time out abend


Correct Option: D
  1. Use IF ABEND statement in the JCL

  2. Use COND

  3. Both a and b

  4. None


Correct Option: A
  1. Use GDG name(-3)

  2. Use GDG name(-2)

  3. Use GDG name(-1)

  4. Use GDG name(0)


Correct Option: B

For how long a job can be executed continuously in a Mainframe

  1. 248 Days

  2. 250 Days

  3. 252 Days

  4. 258 Days


Correct Option: A

AI Explanation

To answer this question, we need to understand the maximum limit for job execution in a Mainframe.

In a Mainframe system, the maximum limit for a job execution is typically determined by the maximum value that can be represented by a signed 32-bit integer. This value is 2,147,483,647.

To calculate the maximum time a job can be executed continuously, we need to convert this value from seconds to days. There are 86,400 seconds in a day (24 hours * 60 minutes * 60 seconds).

Maximum time = 2,147,483,647 seconds / 86,400 seconds per day

Simplifying the calculation, we find that the maximum time a job can be executed continuously is approximately 24,855.82 days.

Among the given options: A. 248 Days - This option is incorrect because it is much lower than the maximum limit of approximately 24,855.82 days. B. 250 Days - This option is incorrect because it is much lower than the maximum limit of approximately 24,855.82 days. C. 252 Days - This option is incorrect because it is much lower than the maximum limit of approximately 24,855.82 days. D. 258 Days - This option is incorrect because it is much lower than the maximum limit of approximately 24,855.82 days.

Therefore, the correct answer is A) 248 Days.

  1. Minimum Size : 1 track Maximum size : 10 cylinder

  2. Minimum Size : 10 track Maximum size : 1 cylinder

  3. Minimum Size : 0 track Maximum size : 1 track

  4. Minimum Size : 1 track Maximum size : 1 cylinder


Correct Option: D

How many bytes does a S9(7) SIGN TRAILING SEPARATE field occupy ?

  1. 3 bytes

  2. 4 bytes

  3. 7 bytes

  4. 8 bytes


Correct Option: D
Explanation:

To solve this question, the user needs to be familiar with the format of COBOL numeric fields.

In COBOL, S9(7) SIGN TRAILING SEPARATE represents a numeric field with a maximum of 7 digits, including the sign (positive or negative) and a separator character (usually a comma or a period) that separates the integer and decimal portions of the number. The "SIGN TRAILING SEPARATE" clause specifies that the sign character appears after the digits of the number, while the separator character appears between the last two digits of the integer portion and the first decimal digit.

Now, let's go through each option and explain why it is right or wrong:

A. 3 bytes: This option is incorrect. A numeric field with the "SIGN TRAILING SEPARATE" clause always occupies an even number of bytes. Since the maximum number of digits in S9(7) is 7, and the sign and separator characters each occupy one byte, the field must occupy at least 9 bytes (7 digits + 1 sign + 1 separator).

B. 4 bytes: This option is incorrect for the same reason as option A. A S9(7) SIGN TRAILING SEPARATE field must occupy an even number of bytes, which is at least 9 bytes.

C. 7 bytes: This option is incorrect. As explained above, a S9(7) SIGN TRAILING SEPARATE field must occupy an even number of bytes. Additionally, 7 bytes is not enough to store a 7-digit number with a sign and a separator character.

D. 8 bytes: This option is correct. A S9(7) SIGN TRAILING SEPARATE field must occupy an even number of bytes, which is at least 9 bytes. However, if the number is positive, the sign character is represented by a low-value (i.e., hex 'F') in the leftmost byte of the field. Therefore, if the number is positive, the field occupies only 8 bytes, which is the correct answer.

The Answer is: D