Disk access time

Question

If there is a magnetic disk as following: Average seek time: 12 ms Rotation rate: 3600 RPM Transfer rate: 3.5 MB/second Number of sectors per track: 64 Sector size: 512 bytes Controller overhead: 5.5 ms What's the average time to read a single sector?

Answer 1

SEEK TIME=12MS

AVG RL=1/2(RL)=1/2(60/3600)=1/2(16.66MS)=8.33MS

NOW THE MAJOR PART:

ONE TRACK HAS 64 SECTOR

IN 16.66 MS WE CAN READ 64 SECTOR

SO TO READ 1 SECTOR WE NEED 16.66/64 MS=0.26MS

(NOTE THAT : THE TRANSEFER TIME OF ONE SECTOR IS 0.406 MS WHICH IS NOT NEEDED HERE COZ WE ARE ASKED TO READ ONLY  IF ASKED WE ALSO INCLUDE THIS FACTOR ALSO...)

TOTAL TIME REQUIRED IS

Seek time (12 MS) + Average RL (8.3 MS) + Sector read time (0.26MS) + Overhead (5.5 MS) = 26.1 MS(ANS)

Answer 2

Avg time to read=T_seek+T_rotational+T_Transfer+T_Controller

T_rotational=1/2(Rotational Time)

Rotational time:

3600 rotaion in 1 min

1 rotation in (1000*60)/3600 msec=16.66 msec

so Rotational time=16.66 msec

T_rotational=1/2(16.66)=8.33msec

T_Transfer:

3.5MB in 1 sec

As sectro contain 512 B

so time taken to read 512B=(512*1000)/(3.5*2^20)

T_Transfer=0.139 msec

Hence

Total time to read=12+8.33+0.139+5.5=25.969 msec=26 msec

Comments

to read the sector which time we should include 0.26ms or 0139 ms and why??