Let us assume a disk with rotational speed of 15,000 rpm, 512 bytes per sector, 400 sectors per track and 1000 tracks on the disk, average seek time is 4ms. We want to transmit   a file of size 1 MByte, which is stored contiguously on the disk.

a. What is the transfer time for this file?

b. What is the average access time for this file?

c. What is the rotational delay in this case?

d. What is the total time to read 1 sector?

e. What is the total time to read 1 track for sequential organization?            

 

Answer:

a) 

The disk has speed of ?15000rpm? , so in one minute ?15000? times a disk can rotate. 

So, one rotation can be done in ?1/15000?? minutes ?=4ms?  

So, average rotational delay ?=4ms/2?=2ms?

The average rotational delay is half of the rotational delay. 

Now, the number of bytes stored in one track = Number of sectors in track * sector size 

?=400∗512? ?bytes?

?=204800? ?bytes?

So, we can say that ?204800? ?bytes? can be transferred in one rotation. One rotation takes 1/?15000?? minutes. Hence time taken to transfer ?204800? ?bytes? ?=1/15000?? minutes ?=4ms?

Now, the file size is ?1MByte=2^20? ?bytes.?

So, time taken to transfer ?2^20? ?bytes? ?=2^20∗4ms∗1/204800?=20.48ms?  

b) The average access time can be shown as:

Average access time = Average seek time + Average rotational delay + Transfer time 

?=4ms+2ms+20.48ms?

?=26.48ms?

c) The average rotational delay is 2 ms as calculated in (a).

d) The time taken to read one sector = Average seek time + Average rotational time + Transfer time 

Transfer time for one sector ?=512∗4?ms/=0.01ms?

So, total time to read one sector is:

?=4ms+2ms+0.01ms?

?=6.01ms?

e) The time taken to read one track = Average seek time + Average rotational time + Transfer time 

Transfer time for one track ?=400∗512∗4?ms/204800=4ms?

So, total time to read one track is:

?=4ms+2ms+4ms?

?=10ms?