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A B+ tree index is built on the key ‘CS’ attribute of the relation ME16. The attribute ‘CS’ is of length 32 bytes, disk blocks are of size 2048 bytes and index pointers are of size 8 bytes. The number of pointers per node is _______.

asked in Databases by Loyal (8k points) | 112 views
0
The question is incomplete as we know leaf node and internal node structure is different in case of B+ tree..So complete information should be given.
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Key and index pointer are given. I think it's correct.
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But the question is more meaningful if it is said record and block pointers as they have different meanings in B+ tree terminology..
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A index contains two fields <key,index pointer> Means size of an index entry = 40B and block size = 2048B, so within a block there can be 2048/40 index entries or pointers.

2048/40 = 51.something

I think ans= 51, but given = 52
+1
Yes u r right..To get block factor we take floor value not ceil..Hence 51 is correct..
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they have done :

which one to consider true ?

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51 is correct !
+1

let #keys per block = x

$\therefore$ 32(x) + 8(x+1) <= 2048

$\therefore$ 40x + 8 <= 2048

$\therefore$ 40x <= 2040

$\therefore$ x = 51  (without any doubt)

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Eg taken from - https://courses.cs.washington.edu/courses/cse326/08sp/lectures/11-b-trees.pdf

So, node structure is like : <index-pointer, key, index-pointer, key,............................, index-pointer>

and not like : <key,index pointer>

Thus 52 is correct, and the solution given is also correct

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@Sushant, you have thought of 'x' as no of keys, thus there are 51 keys, but there are 52 pointers(per node)

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