Question

Consider a typical disk that rotates with 12000 rotations per minute (RPM) and has a transfer rate of 100 Mbytes/ sec. The average seek time of the disk is 5 msec. A sector capacity is 512 bytes. The average time needed to read or write a file of 5091 bytes, which is stored on sequential location, is ________ msec.

Answer 1

option (a)

$\begin{array}{cc}\text{12000 rotation time}& =\text{1 min = 60 sec = 60 * 1000 msec}\\ \text{1 rotation time}& =\frac{60\ast 1000msec}{1200}\\ & =5msec\\ \text{Average rotation delay}& =\frac{5}{2}=2.5msec\\ \text{100 MB transfer time}& =1sec\\ \text{512 B transfer time}& =\frac{1sec}{100M}\ast 512\\ & =0.00512msec\\ \text{No. of sectors required for file}& =\frac{5091B}{512B}=10\\ \text{Total file access time}& =\text{seek time + average rotational delay + 10*1}\\ \text{sector transfer time}& \\ & =5+2.5+10\ast 0.00512\\ & =7.5512msec\end{array}$