câu 35

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Question:

How long does it take to load a 64-KB program from a disk whose average seek time is 5 msec, whose rotation time is 5 msec, and whose tracks hold 1 MB (a) for a 2-KB page size? (b) for a 4-KB page size? The pages are spread randomly around the disk and the number of cylinders is so large that the chance of two pages being on the same cylinder is negligible.

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Answer:

Total time in Seek and rotational latency = 5 msec + 5 msec = 10 msec;
For 2 KB pages, transfer time is 1.25 msec. so total time in transfering 2 KB page is => 11.25 msec;

so to load 1024 of these pages (1024 KB = 1 MB), it requires 1024*11.25 => 11.250 sec

 

For 4 KB pages, transfer time is 2.5 msec. so total time in transfering 4 KB page is => 12.5 sec;

 

so to load 512 of these pages , it requires 512*12.5 => 6.4 sec