OS

Question

Consider the following characteristics of a disk system.
• 10 GB disk rotates at 10,000 rpm
• Data transfer rate of 10⁷ bytes/sec
• Average seek time is 8 ms
• Block size is 32 KB
The average service time to retrieve a single disk block from a random location on the disk is _______ (in ms).

Answer 1

We know 

total disk access time = seek time + rotational latency + transfer time

Given , seek time = 8ms

Now time taken for 10000 rotations = 60s = 60000 ms 

1 rotation time = 6 ms

Avg rotaional time = 1/2 * rotation time = 1/2* 6 = 3 ms

For transfer time , we need to find time taken to transfer one block , as it is mentioned in the question , which is given to be 32 KB.Given , data transferred in 1 s = 10⁷ bytes 

So in 1ms data transferred = 10⁴ bytes = 10 KB

So transfer time for 32 KB = 32/10 * 1ms = 3.2 ms

Hence , total time = 8 + 3 + 3.2 ms = 14.2 ms

Comments

Source: Made Easy Test Series

I believe solution given by them (Made Easy Test Series) is correct, we write 30KB= 30*1000B at many places both in CN and COA ( To find Transmission time in CN and to find I/O time in COA ) but  when we solve questions related to Disks and Data Transfer related to Disk we Consider 30KB= 30*1024B, btw my answer was also 14.2 msec.