OS Q25

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OS Q25

25A 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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