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# Operating System - File System

26:

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 uses one buffer(i.e. it always reads ahead into the buffer)?

 A. 12 sec B. 14 sec C. 17 sec D. 21 sec Answer Report Discuss Option: B Explanation : In case P uses one 'Read ahead' buffer the processing and transferring of records can be overlapped. ∴ Elapsed time =(Access time+ Transfer time)x (Number of records) = (10 + 4) 1000 m sec = 14 sec. Here processing time is less than transfer time. Click on Discuss to view users comments. Write your comments here:
27:

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 records of F are organized using-a blocking factor of 2 (i.e. each block on D contains two records of F) and P uses one buffer?

 A. 12 sec B. 14 sec C. 17 sec D. 21 sec Answer Report Discuss Option: A Explanation : In this case 500 accesses are sufficient to transfer 1000 records. ∴ Elapsed time=(Time taken to transfer Z records in one acess +Processing time of two records) x 500 = [(10 + 4 + 4) + 3 + 3] x 500m sec = 12 sec. Click on Discuss to view users comments. Write your comments here:
28:

Access time is the highest in the case of

 A. floppy disk B. cache C. swapping devices D. magnetic disks Answer Report Discuss Option: D Explanation : Click on Discuss to view users comments. Write your comments here:
29:

A file organization component of a VSAM file is

 A. relative record data set. B. keyed sequential data set. C. entry sequential data set. D. all of these Answer Report Discuss Option: D Explanation : VSAM stands for Virtual Storage Access Method. VSAM is a file storage access method used in MVS, ZOS and OS/390 operating systems. It was introduced by IBM in 1970's. It is a high performance access method used to organize data in form of files in Mainframes. VSAM is used by COBOL and CICS in Mainframes to store and retrieve data. VSAM makes it easier for application programs to execute an input-output operation.  VSAM comprises four data set organizations: Key Sequenced Data Set (KSDS), Relative Record Data Set (RRDS), Entry Sequenced Data Set (ESDS) and Linear Data Set (LDS). The KSDS, RRDS and ESDS organizations contain records, while the LDS organization simply contains a sequence of pages with no intrinsic record structure, for use as a memory-mapped file. VSAM records can be of fixed or variable length. They are organised in fixed-size blocks called Control Intervals (CIs), and then into larger divisions called Control Areas (CAs). Control Interval sizes are measured in bytes — for example 4 kilobytes — while Control Area sizes are measured in disk tracks or cylinders. Control Intervals are the units of transfer between disk and computer so a read request will read one complete Control Interval. Control Areas are the units of allocation so, when a VSAM data set is defined, an integral number of Control Areas will be allocated. Click on Discuss to view users comments. Write your comments here:
30:

The file structure that redefines its first record at a base of zero uses the term

 A. relative organization B. key fielding C. dynamic reallocation D. hashing Answer Report Discuss Option: A Explanation : Click on Discuss to view users comments. Write your comments here:

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