##### Average latency of disk drive

Plz explain it in detail  ...

Given ans are  P=12.5msec  Q = 0.3125MB/s

Shreyans Dhankhar
16 Dec 2014 11:52 am

2400 rotations in 60*1000 msec
1 rotation will be in (60*1000)/2400 =25msec
so latency will be 25/2=12.5 msec so the value of P is 12.5 msec

in 25msec u cover 62500 bits
in 1000msec(1sec) u can cover (62500*1000)/25=2500 Kbits per second  or 312.5KBps

Arvind Rawat
16 Dec 2014 12:21 pm

The average latency of a device is equivalent to the time taken by the device to complete half rotation. Given, the disk rotation speed = 2400 rpm i.e. the device completes 2400 rotations in 1min or 60 sec, then the time required for completing half rotation is given as

$P = {60*1000 \over {2*2400}} msec$= 12.5 msec

The data transfer rate is defined as the amount of data transferred per unit time. The device completes half rotation in 12.5 msec and the data transferred by the device in half rotation is given as

$D = {62500 \over 2} bits$

So, the data transfer rate can be calculated as follows:

$Q = {D \over P}$

$Q = {62500 \over 2*12.5} bits/msec$= 0.3125 MB/s

Parimal Andhalkar
16 Dec 2014 01:00 pm

is there any formula to calculte seek time using no. of tracks and other given information ??

How we calculte seek time ??