Suppose we have a HDD with the following specifications:An average seek time of 10 msA rotational speed of 6000 rotations per minute.A data transfer rate of 40 MB/s.What is the expected throughput of the HDD (in KB/s) when reading a 4 KB sector from a random location on disk? You may ignore all other delay

To calculate the expected throughput of the HDD, we need to consider three factors: the seek time, the rotational delay, and the data transfer time.

1. Seek Time: This is the time it takes for the read/write head to move to the correct track on the disk. Given as 10 ms.

2. Rotational Delay: This is the time it takes for the correct sector on the disk to rotate under the read/write head. Since the disk is rotating at 6000 rotations per minute, we can calculate the average rotational delay as follows:

   Rotational speed = 6000 rotations/minute = 100 rotations/second Therefore, the time for one rotation = 1/100 = 0.01 seconds = 10 ms The average rotational delay is half the time for one full rotation, so 10 ms / 2 = 5 ms

3. Data Transfer Time: This is the time it takes to actually read the data once the read/write head is in the correct position. The data transfer rate is given as 40 MB/s. To read a 4 KB sector, the time would be:

   Data transfer time = 4 KB / 40 MB/s = 0.1 ms

Adding these three times together gives the total time to read a 4 KB sector from a random location on the disk:

Total time = Seek time + Rotational delay + Data transfer time = 10 ms + 5 ms + 0.1 ms = 15.1 ms

The throughput is the amount of data transferred per unit of time. In this case, we are transferring 4 KB in 15.1 ms. To convert this to KB/s, we do:

Throughput = 4 KB / 15.1 ms = 265 KB/s (approximately)

So, the expected throughput of the HDD when reading a 4 KB sector from a random location on the disk is approximately 265 KB/s.