Which of the following is the correct block size in database disk storage?

Which of the following is the correct block size in database disk storage?

Question:

2) Consider a hard disk with a sector size of 512 bytes, and 50 sectors per track. A block may be of several KBs in size. Which of the following is the valid block size?

(a) 256 bytes

(b) 786 bytes

(c) 4096 bytes

(d) 26112 bytes

Answer:

(C) 4096 bytes/4 KB

Discussion:

• A block (cluster) is a unit of memory used to specify the cluster of one or more sectors (physical disk area). Block is the term used by operating systems to specify the minimal size of memory that can be accessible from hard disk.

• Sector is the minimal size of physical storage (hard disk) and is the subdivision of a track of the hard disk.

The block size should be a multiple of sector size.

(a) 256 bytes – it is half of a sector. We need at least a sector. WRONG

(b) 786 bytes – it is not a multiple of sector size. WRONG

(c) 4096 bytes – it is a multiple of given sector size. 512 * 8 = 4096. RIGHT

(d) 26112 bytes – it is indeed a multiple of the sector size. 512 * 51 = 26112. But, it exceeds the size of a track (Given, 512 bytes * 50 tracks = 25600 bytes). The size of block (or cluster) cannot exceed the size of a track. Hence, WRONG.

Option (c), 4096 bytes is the valid block size.

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Find the capacity and the rotational latency of the hard drive

Find the capacity and the rotational latency of the hard drive / How to calculate the hard disk capacity and its rotational latency?

Question:

1) Suppose that we have a hard drive with the following characteristics. It has 10,000 cylinders, 10 tracks per cylinder, 60 sectors per track, and 500 bytes per sector. The hard drive spins at 7200 rpm (rotations per minute) and has an average seek time of 10 ms. Which of the following combination is the right i) capacity and ii) rotational latency of the hard drive?

(a) 2 GB and 3.17 ms

(b) 3 GB and 4.17 ms

(c) 4 GB and 5.17 ms

(d) 1 GB and 2.17 ms

Answer:

b) 3 GB and 4.17 ms

Discussion:

Given,

Number of cylinders                      -        10,000

Number of tracks per cylinder      -        10

Number of sectors per track          -        60

Number of bytes per sector            -        500

Hard drive spinning speed           -        7200 rpm

i) Capacity of the hard drive

Capacity = cylinders * tracks per cylinder * sectors per track * bytes per sector

                   = 10000 * 10 * 60 * 500 = 3,000,000,000 bytes

                   = 3 GB

ii) Rotational latency of the hard drive

Time taken for one rotation = 1/RPM of the hard drive

                                                = 1/7200 per minute

                                                = (1/7200) * 60 = 0.0083 seconds

                                                = 8.33 ms

Rotational latency = Half of the time consumed for one full rotation

                                      = ½ * 8.3 ms = 4.17 ms

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