Final answer:
The average disk access time for a sector can be calculated using the disk parameters provided. It involves the seek time, rotational latency, and transfer time.
Step-by-step explanation:
The average disk access time for a sector can be calculated using the disk parameters provided. The average disk access time is the sum of the seek time, rotational latency, and transfer time.
1. Seek Time: The average seek time is given as 10 ms. Seek time is the time taken by the disk head to move to the desired track. Since there is no queuing delay, the seek time can be used directly in the calculation.
2. Rotational Latency: The rotational latency is determined by the disk's rotational speed. The disk spins at 9600 RPM, which means it completes one revolution every 1/9600 minutes. The rotational latency can be calculated by dividing this time by 2, as the average time for the desired sector to rotate under the disk head is half of one revolution.
3. Transfer Time: The transfer size is given as 6K bytes, and the transfer rate is 4 MB/sec. To calculate the transfer time, we need to convert the transfer size and rate to the same unit. 6K bytes is equivalent to 6 * 1024 bytes. The transfer time can be calculated by dividing the transfer size by the transfer rate.
Once we have the seek time, rotational latency, and transfer time, we can add them together to find the average disk access time for a sector.
The Average Disk Access Time for a sector, based on the given parameters, is calculated by adding the seek time (10 ms), the average rotational latency (3.125 ms), the transfer time (1.5 ms), and the controller overhead (1 ms), which equals to 15.625 ms.
The question asks for the calculation of the Average Disk Access Time for a sector, considering various parameters of a disk's performance. To calculate this, we need to consider three main components: seek time, rotational latency, and transfer time, as well as controller overhead.
Seek time is the time taken for the disk's head to move to the correct track, which is given as 10 ms. Rotational latency depends on the rotation speed of the disk, which is 9600 RPM in this case. Since there are 60,000 ms in a minute and the disk does 9600 rotations per minute, the average rotational latency is half the rotational period:
Rotational period = 60,000 ms / 9600 RPM = 6.25 ms,
Average rotational latency = 6.25 ms / 2 = 3.125 ms.
Transfer time is the time required to read or write the data. Since the transfer size is 6 KB and the transfer rate is 4 MB/sec (or 4000 KB/sec), the transfer time is:
Transfer time = Transfer size / Transfer rate = 6 KB / 4000 KB/sec = 0.0015 sec = 1.5 ms,
Finally, there's a controller overhead of 1 ms. Adding all these together, we get:
Average Disk Access Time = Seek time + Average rotational latency + Transfer time + Controller overhead,
Average Disk Access Time = 10 ms + 3.125 ms + 1.5 ms + 1 ms = 15.625 ms.