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 (revolution per minute), and 512-byte sectors with 500 sectors per track. Suppose that we wish to read a file consisting of 2.56 Mbytes (Megabytes). Estimate the total time using 1. Sequential access. 2. Random access. Submit your answer in the dropbox.

Solution

seek time = 2 ms

rotation speed = 45000 rpm

number of byte sectors =512

number of sectors per track = 500

data file size = 2.56 MB

access time = seek time + average rotational delay + transfer time + controller overhead + queuing delay

-.> average rotational delay = ( 1/2 ) * ( 1/ 45000) * 60 sec = 60 / 90000 = 1 / 1500 sec = 0.66 ms

-> transfer time = number of bytes per track / time to rotate once

-> number of bytes per track = 512 * 500 = 31000 bytes

transfer time = 31000 bytes / 2 ms = 15.5 MBps

1. sequencial access

-> here total data file consists as single .

So , access time = seek time + average rotational delay + transfer time

transfertime for 2.56 MB = (2.56 / 15.5 )sec = 0.16516 sec = 16.5 ms

total access time = 2 ms + 0.66 ms + 16 .5 ms = 19.16 ms

2. Random access

here total file divided into 512 Kbyte size each.

2.56 MB / 512 KB = 5 random files

transfertime for 2.56 MB = ( 512 KB / 15.5 MB ps ) * 5 =( 5 / 31 ) ms= `16.129 ms

total access time = 2 ms + 0.66 ms + 16 .1 ms = 18.76 ms