Question

Consider a magnetic disk drive with 8 surfaces, 512 tracks per surface, and 64 sectors per track. Sectorsize is 1 kB. The average seek time is 8 ms, the track-to-track access time is 1.5 ms, and the drive rotates at 3600 rpm. Successive tracks in a cylinder can be read without head movement.

a. What is the disk capacity?
b. What is the average access time? Assume this file is stored in successive sectors and tracks of
successive cylinders, starting at sector 0, track 0, of cylinder i. c. Estimate the time required to transfer a 5-MB file.

Answer 1

The disk capacity is 256 MB. The average access time is 9.5 ms. The time required to transfer a 5-MB file is approximately 0.39 ms.

Step-by-step explanation:

a. To calculate the disk capacity, we need to multiply the number of surfaces by the number of tracks per surface by the number of sectors per track by the sector size. So, the disk capacity is 8 surfaces * 512 tracks per surface * 64 sectors per track * 1 kB per sector. This equals 262,144 kB or 256 MB.

b. The average access time is calculated by adding the average seek time and the track-to-track access time. So, the average access time is 8 ms + 1.5 ms = 9.5 ms.

c. To estimate the time required to transfer a 5-MB file, we divide the file size by the data transfer rate. The data transfer rate is the rotational speed in revolutions per minute (RPM) divided by 60 (to convert minutes to seconds) times the number of sectors per track times the sector size. So, the data transfer rate is (3600 RPM / 60) * 64 sectors per track * 1 kB per sector. This equals 12,800 kB/s.

Now, we can calculate the transfer time. The transfer time is the file size divided by the data transfer rate. So, the transfer time is 5 MB / 12,800 kB/s. This equals 0.000390625 s or approximately 0.39 ms.

Answer 2

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.