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Consider a file consist 30,000 fixed length record.Block size 1024 bytes,record size 100 bytes.Search key 9 bytes, block pointer size 6 bytes.

A) How many levels of index required if 1st level uses dense index.

B) How many levels of index required if 1st level uses sparse index.

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No.of records = 30000, Block Size = 1024 bytes, record size = 100 bytes, search key = 9 bytes, block pointer size  6 bytes

index blocking factor : $\left \lfloor (block size/(key size+pointer) \right \rfloor$= 1024/15 =68

table blocking factor : $\left \lfloor (block size/(record size) \right \rfloor $= 10

A) for dense index we will have one entry in index table for each record so no of block entries = $\left \lceil 30000/68  \right \rceil $ = 442(1st level)

2nd level no of block entries = $\left \lceil 442/68  \right \rceil $ = 7

3rd level no of block entries = $\left \lceil 7/68  \right \rceil $ = 1

3 level.

 

B) for spars index, number of entries in index file is equal to number of data blocks = 30000/table block factor =30000/10 =3000

1st level no of block entries = $\left \lceil 3000/68  \right \rceil $ = 45

2nd level no of block entries = $\left \lceil 45/68  \right \rceil $ = 1

2 level

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