CS501 - Assignment 5 Solution Spring 2011 - June 2011

[Vuhelper]

ust follow this example and you will get through Question no: 01 !!!
Just don't forget to change your format otherwise, your assignment will be caught as a copied one !!!

A hard drive has 8 surface, with 521 tracks per surface and a constant 64 sectors per track.
Sector size is 1KB. The average seek time is 8 m sec, the track to track access time
is 1.5 m sec and the drive runs at 3600 rpm. Successive tracks in a cylinder can
be read without head movement.
a) What is the drive capacity?
b) What is the average access time for the drive?
c) Estimate the time required to transfer a 5MB file?
d) What is the burst transfer rate?


Solution:-
a) Capacity = Surfaces x tracks x sectors x sector size
= 8 x 521 x 64 x 1k = 266.752MB
b) Rotational latency : Rotation time/2 = 60/3600x2 = 8.3 m sec
Average Access time = Seek time + Rotational latency
= 8 m sec + 8.3 m sec = 16.3 m sec
c) Assume the file is stored in successive sectors and tracks starting at sector #0,
track#0 of cylinder #i. A 5MB file will need 1000 blocks and occupy from cylinder #1,
track #0, sector #0 to cylinder #(i+9), track #6, sector #7, we also assume the size of disk buffer is unlimited.

The disk needs 8ms, which is the seek time, to find the cylinder #i, 8.3 ms
to find sector #0 and 8 x (60/3600) seconds to read all 8 tracks data of this cylinder.
Then the time needed for the read to move to next adjoining track will be only 1.5 m sec.

Which is the track to track access time. Assume a rotational latency before each new track.

Access time = 8+9(8.3¸8x16.6+1.5)+8.3+6x16.6+8/64x16.6
= 1406.6 m sec
d) Burst rate = rows/sec x sectors/row x bytes/sector
= 3600/60 x 64 x 1k = 3.84 MB/sec