System Software and Operating System - File and Input/Output System

21. In MS-DOS, relocatable object files and load modules have extensions

  • Option : A
  • Explanation : The Relocatable Object Module Format (OMF) is an object file format used primarily for software intended to run on Intel 80x86 microprocessors. It was originally developed by Intel under the name Object Module Format and is perhaps best known to DOS users as an.OBJ file.
    COM file is a type of simple executable file. On the Digital Equipment operating systems of the 1970s, .COM was used as a filename extension for text files containing commands to be issued to the operating system (similar to a batch file). With the introduction of CP/M (a microcomputer operating system), the type of files commonly associated with COM extension changed to that of executable files. This convention was later carried over to MS-DOS. Even when complemented by the more general .exe file format for executables,
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22. Primitive disk operating system (POS) manages its disk files in contiguous blocks. A file is saved to the first available space that is large enough to hold the file. Assume that a disk has only 10 contiguous blocks of available free space. Which of the following set of file operations Fa=1, Fb=3, Fc=5, Fd=6 can not be completed given the set of files and file sizes specified below

  • Option : B
  • Explanation : Given Fa=1, Fb=3, Fc=5, Fd=6

    Option (a) save (ƒa), save (ƒb), save (ƒd), delete (ƒd), save (ƒc) is possible as we can save 1,3,6 in 9 contiguous block and then delete ( fd) to get 6 contiguous space. After that we can save ( fc ) = 5 in remaining 7 blocks.. So option (a) is possible.

    Similarly we can execute Option (c) and option (d) where as option (b) is not possible
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24. A program P reads and processes 1000 consecutive records from a sequential file F stored on device D without using any file system facilities. Given the following: (i) Size of each record = 3200 bytes. (ii) Access time of D = 10 m secs. (iii) Data transfer rate of D = 800 x 103 bytes/sec (iv) CPU time to process each record = 3 m secs.

What is the elapsed time of P if F contains unblocked records and P does not use buffering?

  • Option : C
  • Explanation : In case P uses one 'Read ahead' buffer the processing and transferring of records can be overlapped.

    Elapsed time =(Access time+ Transfer time + Processing Time )x (Number of records)

    Here Access time = 10ms (given)

    Transfer time = (800 x 103 )/3200 sec = 0.004 sec = 4 ms

    Therefore Elapsed Time = (10 + 4 +3 ) * 1000 m sec = 17 sec.

    Here processing time is less than transfer time.
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