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Database file consist 1250 records. Block can hold either 3 records or (10 keys, 11 pointers). The maximum number of levels of index required for sparse B+ tree index for database file are ______.

NOTE – For maximum number of level or height there will be minimum number of keys in each node which in this case 5 so this makes number of pointers 6

Blocks required = 1250/3 = 417

Level 1 = floor(417/5) = 83

Level 2 = floor(83/6) = 13

Level 3 = floor(13/6) = 2

Level 4 = floor(2/6) = 1

So, total no. of levels = 4

Blocks required = 1250/3 = 417

Level 1 = ceil(417/11) = 38

Level 2 = ceil(38/11) = 4

Level 3 = ceil(4/11) = 1

So, total no. of levels = 3

Minimum index blocks = 38 + 4 + 1 = 43
by

so you mean instead of 11 we will use 6 .
yeah there are many question like that you can look at them under the tag indexing

I think in minimum level 1 = ceil(417/10)  not this ceil(417/11)

Blocks required = 1250/3 = 417

level 1 = ceil(417/1)

Level 2 = ceil(38/11) = 4

Level 3 = ceil(4/11) = 1

So, total no. of levels = 3

Minimum index blocks = 38 + 4 + 1 = 43