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Showing posts with label Disk Storage Access Exercises. Show all posts
Showing posts with label Disk Storage Access Exercises. Show all posts

## Find the transfer rate of and capacity of a hard disk in DBMS, How to find the data transfer rate and the capacity of hard disk for efficient storage access in dbms?

#### Question:

5) Consider a disk with the following specifications: 512 bytes sector size, 8 double sided platters (2 surfaces per platter), 5000 tracks per surface, 100 sectors per track, average seek time of 10 msec, and the disk rotates at 5400 revolutions per minute (rpm). Which of the following is the correct combination of number of cylinders, average rotational delay, and transfer rate (if an entire track of data can be transferred per revolution of disk)?

a) 5000 cylinders, 5.55 milliseconds, 4.5 KB
b) 5000 cylinders, 4.5 milliseconds, 12 KB
c) 4000 cylinders, 5.55 milliseconds, 10 KB
d) 4000 cylinders, 2 milliseconds, 2.5 KB

a) 5000 cylinders, 5.55 milliseconds, 4.5 KB

Discussion:

Given,
Number of platters                -        8
Tracks per disk surface         -        5000
Sectors per track                    -        100
Sector size                               -        512 bytes
Average seek time                -        10 msec
rpm                                        -        5400.

Number of cylinders is same as the number of tracks. Hence, the number of cylinders is 5000.
The average rotational delay is half of the rotation time. One complete rotation takes 1/5400 in a minute = (1/5400) * 60 seconds = 0.0111 seconds = 11.1 milliseconds. The average rotational delay = 11.1/2 = 5.55 msec (approx).
We need the capacity of the track to find the transfer rate. The capacity of the track can be found as follows;
Number of bytes per track = bytes per sector * sectors per track
= 512 * 100 = 51200 bytes = 50 KB.
We know that, from above, the time for one complete rotation is 11.1 milliseconds. Hence, the transfer rate is 50 KB/11.1 milliseconds = 4.5 KB/msec (approx.).

## Find the capacity of a hard disk in database, How many recrods can be stored in a hard disk track? Which of the following is the correct combination of capacity of track, capacity of each surface, and capacity of disk?

#### Question:

4) Consider a disk with a sector size of 512 bytes, 14409 tracks per surface, 16947 sectors per track, 8 double-sided platters, and an average seek time of 9 ms. Which of the following is the correct combination of capacity of track, capacity of each surface, and capacity of disk (approximately)?

a) 100 MB, 200 GB, and 2 TB
b) 8 MB, 116 GB and 2 TB
c) 8 MB, 200 GB and 1 TB
d) 80 MB, 116 GB and 1 TB

b) 8 MB, 116 GB and 2 TB

Discussion:

Given,
Sector size                                                  - 512 bytes
Number of sectors per track                    – 16947
Number of tracks per surface                  14409
Number of platters                                   8 double-sided

Capacity of a track = sector size * number of sectors per track
= 512 * 16947 = 8 MB approx.
Capacity of each surface = capacity of track * number of tracks per surface
= (512*16947) * 14409 = 116 GB approx.
Capacity of the disk = capacity of each surface * number of platters * number  of pages per platter
= 116 GB * 8 * 2 = 2 TB approx.

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