Question

A magnetic disk has an average seek time of 8 ms. The transferrate is 50 MB/sec. The

disk rotates at 10,000 rpm and the controller overhead is 0.3msec. Find the average time to read or write 1024 bytes.

Answer 1

11.32milisec.

Step-by-step explanation:

Average seek time = 8ms

transfer rate = 50MB/sec

disk rotate = 10000 rpm

controller overhead = 0.3 msec

Average time = ?

Average time is the addition of the seek time, transfer time, rotation time and ontroller overhead

`Average\,time = Seek\,time+Rotation\,time+Transfer\,time+Controller\,overhead`

now, find the rotation time,

Rotation time is the time at which half rotation is made.

`Rotation\,time=(0.5)/(rotation\,per\,second)`

convert rotation per minute to rotation per second.

`rps=(10000)/(60)`

substitute in the above formula.

`Rotation\,time=(0.5)/((10000)/(60) )=0.003sec`

or 3milisec

now, find the transfer time by formula:

`Transfer\,time=(Number\,of \,read\,or\,write\,bytes)/(transfer\,rate)`

so, read or write bytes is 1024

`[tex]Transfer\,time=(1024)/(50*10^(6) )=0.02milisec`[/tex]

The average time is:

`[tex]Average\,time= 8+3+0.02+0.3=11.32milisec`[/tex]