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Suppose we have a block-addressable disk drive. With such block organized disk non data overhead of subblocks and, interblock gaps have to be accounted for.

There are $40,000$ bytes per track and the amount of space taken up by subblocks and interblock gaps is equivalent to  $250$ $Bytes$ per block. A file contains records and record size is $200$ $Bytes$ to be stored on the disk. If a total of $32$ blocks can be stored per track then the blocking factor is ________

The term "blocking factor" is used to indicate the number of records that are to the stored in each block in a file. A block is organized to hold an integral number of logical records.

## 3 Answers

Best answer
Record size = 200B
Let there are x records to be stored in each block.
then Each block holds data = 200xB  and nondata overhead = 250B.
The number of blocks that can fit on a track = 40,000 / (200x + 250) = 32
200x+250 =1250 => x = 5

so, There are five 200B records per block ,  that means Blocking factor is 5
by

### 4 Comments

Can you give me source of this question?
here blocks are basically sectors only, arent they?

@thor

@ritwik

Block IS NOT the sector.

A block  is a group of sectors that the operating system can point to .

A block might be one sector, or it might be several sectors (2,4,8, or even 16). The bigger the drive, the more sectors that a block will hold. [1]

http://www.alphaurax-computer.com/computer-tips/hard-drive-knowledge-blocks-vs-sectors

http://stackoverflow.com/questions/12345804/difference-between-blocks-and-sectors

so, blocking factor = number of records per block.

Answer : 5

Size of each track = 40000 B    , No. of blocks/track = 32   , Gap after each block = 250 B  ,  Record Size = 200 B

Total blocks size = 40000 - total size of Gap = 40000 - (32*250) = 32000

size of each block = Total blocks size / no. of blocks = 32000/32 = 1000

no. of records in each blocks = size of each block / Record Size = 1000/200  = 5

Track size=40,000B

Block overhead=250B

Record size=200B

No. of block=32

Block size=x+overhead=x+250   (where x is space for records)

No. of block=32=Track size/Block size

32=(40,000)/(x+250)

x=1000B

No. of record=x/Record size=1000/200=5
Answer:

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