Calculate how much disk space (in sectors, tracks, and surfaces) will be require

Question

Calculate how much disk space (in sectors, tracks, and surfaces) will be required
to store 300,000 120-byte logical records if the disk is fixed-sector with 512
bytes/sector, with 96 sectors/track, 110 tracks per surface, and 8 usable surfaces.
Ignore any file header record(s) and track indexes, and assume that records cannot
span two sectors.
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P4
Consider the disk system described in Problem 3, and assume that the disk rotates
at 360 rpm. A processor reads one sector from the disk using interrupt-driven I/O,
with one interrupt per byte. If it takes 2.5 µs to process each interrupt, what
percentage of the time will the processor spend handling I/O (disregard seek
time)?


I know the answer to p3 >> i want the answer to p4 , thanks

Explanation / Answer

Each sector can hold 4 logical records. The required number of sectors is 303,150/4 = 75,788 sectors. This requires 75,788/96 = 790 tracks, which in turn requires 790/110 = 8 surfaces. There are 512 bytes/sector. Since each byte generates an interrupt, there are 512 interrupts. Total interrupt processing time = 2.5×512 = 1280µs. The time to read one sector is: (60sec/min)/(360rev/min) = 0.001736sec = 1736µs Percentage of time processor spends handling I/O: (100) × (1280/1736) = 74%

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