Project ForgeConsider a disk pack with a seek time of 4 milliseconds and rotational speed of 10000 rotations per minute (RPM). It has 600 sectors per track and each sector can store 512 bytes of data. Consider a file stored on the disk containing 2000 sectors. Assume that every sector access necessitates a seek, and the average rotational latency for accessing each sector is half of the time for one complete rotation. The total time (in milliseconds) needed to read the entire file is ____.
Explanation
One full rotation = 60000ms/10000 = 6ms. Average rotational latency = 3ms. Transfer time per sector = 6ms/600 sectors = 0.01ms. Per-sector total = seek(4) + latency(3) + transfer(0.01) = 7.01ms. For 2000 sectors: 2000 × 7.01 = 14020ms.
Source: GATE CS 2015 (Session 1 / 7th Feb Shift 1) (2015)
This question is part of a 15-day revision cycle — sign in and Project Forge will bring it back the day before you'd forget it.