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?
| A. | 12 sec |
| B. | 14 sec |
| C. | 17sec |
| D. | 21sec |
Answer : C Explanation :
In case P uses one 'Read ahead' buffer the processing and transferring of records can be overlapped. 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. Appasami said: (7:47pm on Saturday 2nd June 2018)
Transfer time = 3200 /(800 x 10^3 ) sec = 0.004 sec = 4 ms
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Option: A Explanation : Explanation will come here. Explanation will come here. Explanation will come here. Explanation will come here. Explanation will come here. |
