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1 votes
1 votes

Here answer is 42 I think because we take floor of the value? Can some one confirm it??

2 Answers

Best answer
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2 votes

In B+ tree the order of root and internal nodes and order of leaves is different.

If we take case of leaves

N * (key + block pointer) + 1 block pointer linking to next leaf $\leq$ Disk block size      (N = order of leaves)

N * (8+4) + 4 $\leq$ 512

12N $\leq$ 508

N = floor value (508/12)

N = 42

If we take case of root and internal nodes

N* size of block pointer + (N-1)* key $\leq$ Disk block size         (N = order)

4N + (N-1) 8 $\leq$ 512

N= floor value (520/12)

N=43

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1 votes
1 votes

Key size = 8 byte

Disk block size = 512 byte

Block Pointer = 4 Byte

Internal node structure of B+ tree

P* size of block pointer + (P-1)[key] <= Disk block size         {P=order i.e #f block pointer}

4P + (P-1) [8] <= 512

P= Floor value (520/12)

P=43

edited by

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