Sparse index and dense index

Question

Comments

1)157

2)16

3)3

4)2

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@arvin...post the solution of 3 and 4.....how the answer is 4 and 2?

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Post the answer

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see the solution now i have posted :p cross check for any errors :p thank  you :p

Answer 1

#records= 10000   block size=1024 B record size=100 B search key=9B pointer=7B.

index blocking factor : block size/(key size+pointer)= 1024/16 =64

table blocking factor : floor(block size/record size)= 10

1) &3) for dense index we will have one entry in index table for each record so #block entries = 10000/64 = 157 (level 1)

at level 2 #block entries = (ceil(157/64))= 3

at level 3 #block entries = ceil(3/64) =1

so we will have 157 block in level 1 and total 3 levels.

2) & 4) for sparse index , number of blocks in main table to be matched with index table = 10000/table block factor =10000/10 =1000

number of blocks needed when these are matched with index entries =(ceil)(1000/64) = 16 at level 1.

number of #block entries at level 2 = ceil(16/64) = 1

so we wil have 16 block in level 1 and total 2 levels.

 

Comments

Correct!

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ok bro :) thanks