OS Q26

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OS Q26

26A 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) The 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)?

Option : B
Explanation : n 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.

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