26 ans

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+20 votes

The order of an internal node in a $B+$ tree index is the maximum number of children it can have. Suppose that a child pointer takes $6$ bytes, the search field value takes $14$ bytes, and the block size is $512$ bytes. What is the order of the internal node?

- $24$
- $25$
- $26$
- $27$

+28 votes

Best answer

+16 votes

Answer: C

lets assume order of b+ tree is n.

B+ tree internal node has n-1 keys and n node (child) pointer.

Given : key size: 14 bytes

Child pointer : 6 bytes

we have ,

block size = (n-1) * key size + n * child pointer

512>=(n-1)*14+n*6

512>=14n-14+6n

n=(512+14)/20

n=526/20

n=26.3

n=26

0

block size = (n-1) * key size + n * child pointer

this formula is block size = (n-1) *( key size+data pointer) + n * child pointer

Data pointer is not considered here why?

this formula is block size = (n-1) *( key size+data pointer) + n * child pointer

Data pointer is not considered here why?

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