consider a disk with an advertised average seek time of 2 ms

consider a disk with an advertised average seek time of 2 ms, rotation speed of 45000 rpm (revolutions per minute), and 512 byte sectors with 500 sectors per track. suppose that we wish to read a file consisting of 2.56 Megabytes. Estimate the total time using 1. sequential access 2. random access

Solution

avge seek time = 2ms

no of rotations =4500rpm

sector size =512byte

file size =2.56 Megabyte (assume this is single strip)

no of sectors per track =500

file unit consists 2.54megabyte /512 bytes no of sectors

hence

file unit consists 5000 sectors

now transfer time taken as

tranfer time = (1/ sectors per track )*rotation time

=0.133ms

=0.133ms

=0.133ms

transfer time=0.133ms

rotational latency=6.5ms

seek time =2ms

now access time =seek time + rotational latency + tranfer time

=2ms+6.5ms+0.133ms

=8.633ms

now calculate the sequential access time for read the file

and also calculate the random access time for read the file

for random access the file of size 2.56 MB

the time take for access the file for reading is

first claculate trasfer rate for a file

transfer rate=no of bytes perr track/(rotationtime)

= 500*512/13ms

= 1.98MB/s

random access time = seek time+ rotational latency + tranfer time

=2ms +6.5ms + (size of the sector/transfer rate)*total no of sectors

=8.5ms+ 512/(500*512/13)*5000

=8.5ms+(13/500)*5000

=8.5ms +130ms

=138.5ms

average random access time for acccessing the 2.56Mega byte file is 138.5ms

now coming to the sequential access time calicualtion for accessing the 2.56 megabyte file for reading is

sequentail access time= seek time +rotational delay + tranfer time

here tranfer time taken for the entire file as

transfer time= (file size /no of bytes per track)* rotational time

=(2.56 Megabytes/512 bytes)* 13ms

=(5000/ 500)*13ms

=130ms

tranfer time =130ms

hence sequentail access time = 2ms+6.5ms+130ms

= 138.5ms

here sequential acess time and random access time are same for the given problem