Consider the problem of finding the total time to access and read data stored as tracks on a computer disk. This can be described, at its simplest, as access time plus transfer time. Access time is comprised of the average time for the reading head to find the correct track (known as seek time) plus the rotational delay, or latency. Seek time is zero for fixed-head systems. The rotational latency is the average time taken for the disc to spin around to the correct place to start reading. This is considered to be, on average, half the time taken for one complete disk rotation. The time taken to read data, the transfer time, is given by the amount of data on the track to be read, divided by the total data on the track, all multiplied by the time taken for one disk rotation. Clearly if all of the data on a track is to be read, then this reduces to the time taken for one disk rotation. There may, in fact, be delays caused by I/O queuing (see Stallings, section Disk Performance Parameters, pages 225-227, tenth global edition), but in the following problem we will just consider access time plus transfer time. Given a moveable-head system with a constant disk rotation speed of 12,000 revolutions per minute (rpm), an average seek time of 6 milliseconds and 512 byte sectors with 500 sectors per track, answer the following questions, giving all your working. Give your final answers to (a) and (b) in milliseconds (ms).

(a) The file is sequentially organised, and is stored on 6 complete tracks followed by exactly one half of a track. [12 marks]

(b) The same file as that in part (a) is now distributed at random across the disk, i.e. each sector of the file is randomly placed on the disk. [12 marks]

(c) If the answer to part (a) is X, and the answer to part (b) is Y, express X as a percentage of Y, giving your answer to one significant figure.

its about disk rotation. yes i think its also about operating system