#Disk

Question

In addition to this Avg rotational delay is also given as 8.3 ms.

I am confused when we access 8 consecutive track then how many time we should take rotational delay and seek time.

Answer 1

First calculate time to read the file of 128 KB $\rightarrow$ $\frac{2^{17}}{2^{20}}$ $\rightarrow$ $0.125 $ sec $\rightarrow$ $125$ ms.

A) As file is contiguous in first part, so in order to read 256 sectors, overhead will come only 1 time

Total time = T avg seek + T rotn delay + T transfer time $\rightarrow$ 8.3+20+125 ms $\rightarrow$ $153.3 ms$

B) When sectors are random, overhead will come for each sector, so

Total time = 256* (T avg seek + T rotn delay + T transfer time for 1 sector)

               $\rightarrow$ 256 * (T avg seek + T rotn delay) + T total transfer time

               $\rightarrow$ 256*(28.3) + 125 $\rightarrow$ $7369.8ms$

Comments

The transfer time you have taken is to read a file from a disk.

What about time to transfer data from the disk as transfer rate  is given as 1MB/S.

---

@Jatin saini By reading a file is same as transfer file from disk. which I have done in the first line ($2^{20}$) there is for 1 MBps transfer rate only.