a. Total capacity 2 GB
b. The average access time is 16.33 ms
c. The estimated time to transfer a 5-MB file is approximately 79.26 milliseconds.
Disk Drive Calculations:
a. Disk Capacity:
Total sectors = 8 surfaces * 512 tracks/surface * 64 sectors/track = 2,097,152 sectors
Sector size = 1 kB = 1024 bytes
Total capacity = Number of sectors * Sector size = 2,097,152 * 1024 bytes = 2,147,483,648 bytes = 2 GB
b. Average Access Time (Seek + Rotational Delay):
Rotational speed (RPM) = 3600
Rotational period (in milliseconds) = 60,000 ms / 3600 RPM = 16.67 ms
Average rotational delay (half a rotation) = Rotational period / 2 = 16.67 ms / 2 = 8.33 ms
Assuming successive sectors are on the same track, no seek time is required.
Therefore, average access time = Seek time + Rotational delay = 8 ms + 8.33 ms = 16.33 ms
c. Time to Transfer 5-MB File:
Total bytes to transfer = 5 MB = 5,000,000 bytes
Transfer rate = Sector size / Access time = 1024 bytes / 16.33 ms = 62.89 kB/ms
Time to transfer file = Total bytes / Transfer rate = 5,000,000 bytes / 62.89 kB/ms ≈ 79.26 ms
Therefore, the estimated time to transfer a 5-MB file is approximately 79.26 milliseconds.